Universe in Problems - User contributions [en]

Evolution of Universe

2012-10-04T17:58:08Z

Den: /* Problem 56 */

[[Category:Standard Cosmological Model|2]]

__NOTOC__
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=== Problem 1 ===
Rewrite the first Friedman equation in terms of redshift and analyze the contributions of separate components at different stages of Universe evolution.
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<div id="SCM11"></div>
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=== Problem 2 ===
Find the time dependence for the scale factor and analyze asymptotes of the dependence. Plot $a(t)$.
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<div id="SCM12"></div>
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=== Problem 3 ===
Determine the redshift value corresponding to equality of radiation and matter densities.
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$$
\begin{gathered}
\rho _r(t) = \rho _m(t);\\
\rho _{r0}\frac{a_0^4}{a^4} = \rho _{m0}\frac{a_0^3}{a^3};\\
\frac{\rho _{m0}}{\rho _{r0}} = \frac{\Omega _{m0}}{\Omega _{r0}} = \frac{a_0}{a} = 1 + z;\\
\Omega _{r0} = 5 \cdot {10^{ - 5}} \\
z = \frac{\Omega _{m0}}{\Omega _{r0}} - 1 \approx 5400\\
\end{gathered}
$$
</p>
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<div id="SCM_9"></div>
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=== Problem 4 ===
Construct effective one-dimensional potential (see [[Solutions_of_Friedman_equations_in_the_Big_Bang_model#dyn15|problem]] of Chapter 3)
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<div id="V(x)"></div>
$$
V(x) = - \frac{1}{2}\sum\limits_i \Omega _{i0}x^{ - (1 + 3w_i)};\quad x \equiv \frac{a}{a_0}
$$
In the SCM
$$
V(x) = - \frac{1}{2}\Omega _{m0}x^{-1}-\frac{1}{2}\Omega _{\Lambda 0}x^2 \simeq - 0.135x^{-1}-0.365x^2
$$

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=== Problem 5 ===
Show that the following holds: $\dot H= -4\pi G\rho_m$ and $\ddot H= 12\pi G\rho_m H$.
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<div id="SCM_18"></div>
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=== Problem 6 ===
Expand the scale factor in Taylor series near the time moment
$t_0$:
\[\frac{a(t)}{a(t_0)}=1+\sum\limits_{n=1}^\infty\frac{A_n(t_0)}{n!}[H_0(t-t_0)]^n;\quad A_n\equiv\frac{1}{aH^n}\frac{d^na}{dt^n}\]
and calculate values for few first coefficients $A_n$.
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The coefficients $A_1$ and $A_2$ equal to:
$$A_1 = \frac{\dot a}{aH} = 1;\quad A_2 = \frac{\ddot a}{aH^2} = - q = 1 - \frac{3}{2}{\Omega _m}$$
The parameters $A_n(n > 2)$ can be calculated by consequent differentiation of the relation
$$\ddot a = a\left( \dot H + H^2 \right)$$
The time derivatives of the scale factor can be determined making use of the fact that in SCM $\dot H = - 4\pi G\rho _m$, and $\rho _m = \rho _0a^{ - 3}$. For example,
$$A_3 = 1 + \frac{\ddot H}{H^3} + 3\frac{\dot H}{H^2}$$
Using $\ddot H = 12\pi G{\rho _m}H,\;\dot H = - 4\pi G\rho _m$, one finds that
$$A_3 = 1$$
Let us present the expressions for some consequent expansion coefficients
\[\begin{array}{l}
{A_4} = 1 - \frac{3^2}{2}\Omega _m;\\
{A_5} = 1 + 3\Omega _m + \frac{3^3}{2}\Omega _m^2;\\
{A_6} = 1 - \frac{3^3}{2}\Omega _m - 3^4\Omega _m^2 - \frac{3^4}{4}\Omega _m^3
\end{array}\]
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=== Problem 7 ===
Show that all the coefficients $A_n$ can be expressed through
elementary functions of the deceleration parameter $q$ or the
density parameter \[\Omega_m=\frac23(1+q).\]
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<div id="SCM15"></div>
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=== Problem 8 ===
Consider the case of flat Universe filled by non-relativistic matter and dark
energy with state equation $p_X=w\rho_X$, where the state parameter $w$ is parameterized as the following \[w=w_0+ w_a(1-a)=w_0 + w_a \frac{z}{1+z}.\]
Express current values of cosmographic parameters through $w_0$ and
$w_a$. % (See [http://arxiv.org/abs/0802.1583 Cosmography of f(R) gravity])
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In the considered case
\[\frac{H^2(z)}{H_0^2} = \Omega _m(1 + z)^3 + \Omega _X(1 + z)^{3\left( 1 + w_0 + w_a \right)}e^{ - \frac{3w_az}{1 + z}}\]
Calculation of the cosmographic parameters requires in general to integrate $H(z)$ in order to obtain $a(t).$ One can avoid this procedure with help of the relation
\[\frac{d}{dt} = - (1 + z)H(z)\frac{d}{dz}.\]
We can, given $H(z)$, apply that relation to calculate $\dot H,\ddot H, \dddot H, \ddddot H $ and so on. Using the expressions for time derivatives of Hubble parameter obtained in the problem (\ref{equ61_6}), Chapter 2,
\[\begin{array}{l}
\dot H = - H^2(1 + q);\\
\ddot H = H^3\left( j + 3q + 2 \right);\\
\dddot H = H^4\left[ s - 4j - 3q(q + 4) - 6 \right];\\
\ddddot H = H^5\left[ l - 5s + 10\left( q + 2 \right)j + 30(q + 2)q + 24 \right]
\end{array}\]
one can express the current values of the cosmological parameters (for $z = 0$) in terms of $w_0,w_a:$
\[\begin{array}{l}
q_0 = \frac{1}{2} + \frac{3}{2}\left( 1 - \Omega _m \right)w_0;\\
j_0 = 1 + \frac{3}{2}\left( 1 - \Omega _m \right)\left[3w_0\left( 1 + w_0 \right) + w_a \right];\\
s_0 = - \frac{7}{2} - \frac{33}{4}\left( 1 - \Omega _m \right)w_a - \frac{9}{4}\left( 1 - \Omega _m \right)\left[ 9 + \left(7 - \Omega _m \right)w_a \right]w_0 - \\
- \frac{9}{4}\left( 1 - \Omega _m \right)\left( 16 - 3\Omega _m \right)w_0^2 - \\
- \frac{27}{4}\left( 1 - \Omega _m \right)\left(3 - \Omega _m \right)w_0^3;\\
l_0 = \frac{35}{2} + \frac{1 - \Omega _m}{4}\left[ 213 + (7 - \Omega _m)w_a \right]w_a + \\
+ \frac{1 - \Omega _m}{4}\left[ 489 + 9(82 - 21\Omega _m)w_a \right]w_0 + \\
+ \frac{9}{2}\left( 1 - \Omega _m \right)\left[ 67 - 21\Omega _m + \frac{3}{2}(23 - 11\Omega _m)w_a \right]w_0^2 + \\
+ \frac{{27}}{4}\left( {1 - {\Omega _m}} \right)(47 - 24{\Omega _m})w_0^3 + \\
+ \frac{81}{2}\left( 1 - \Omega _m \right)(3 - 2\Omega _m)w_0^4
\end{array}\]
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<div id="SCM16"></div>
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=== Problem 9 ===
Show that the results of the previous problem applied to SCM coincide with
the ones obtained in the [[#SCM_18|problem]].
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In the SCM $\left( {{w_0},{w_a}} \right) = \left( { - 1,0} \right)$ and therefore
\[\begin{array}{l}
{q_0} = \frac{1}{2} - \frac{3}{2}\left( {1 - {\Omega _m}} \right);\\
{j_0} = 1;\\
{s_0} = 1 - \frac{9}{2}{\Omega _m};\\
{l_0} = 1 + 3{\Omega _m} + \frac{{27}}{2}\Omega _m^2
\end{array}\]
</p>
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=== Problem 10 ===
Photons with $z=0.1,\ 1,\ 100,\
1000$ are registered. What was the Universe age $t_U$ in the moment of
their emission? What period of time $t_{ph}$ were the photons on the
way? Plot $t_U(z)$ and $t_{ph}(z)$
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$$
\begin{gathered}
t_U=\int_z^\infty\frac{dz}{(1 + z)H(z)} ;\; t_UH_0=\int_z^\infty\frac{dz}{(1 + z)\sqrt{\Omega _{m0}(1 + z)^3 + \Omega _{\Lambda 0} } }; \\
t_{ph}H_0=\int_0^z \frac{dz}{(1 + z)\sqrt{\Omega _{m0}(1 + z)^3 +\Omega _{\Lambda 0}}}\\
\end{gathered}
$$
At $$z = 0.1\,\,\, \to \,\,\,\,t_U H_0 = 0.899\,\,\,\,\,t_{ph} H_0 = 0.093$$
$$z = 1\,\, \to \,\,\,t_U H_0 = 0.431\,\,\,\,\,t_{ph} H_0 = 0.561$$
$$z = 100\,\,\, \to \,\,t_U H_0 = 1.264 \times 10^{ - 3} \,\,\,\,\,t_{ph} H_0 = 0.991$$
$$z = 1000\,\, \to \,\,\,t_U H_0 = 3.663 \times 10^{ - 5} \,\,\,\,\,t_{ph} H_0 = 0.993$$

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=== Problem 11 ===
Determine the present physical distance to the object that emitted light with current redshift $z$
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The physical distance to the object emitted a photon is related to the redshift $z$ by
$$
R(z) = \frac{1}{1 + z}\int_0^z\frac{dz'}{H(z')}
$$
In the SCM
$$
H =H_0\sqrt {\Omega _{r0}(1 + z)^4 + \Omega _{m0}(1 + z)^3+\Omega _{\Lambda 0}}
$$
The parameters ${H_0},\;\Omega _{r0},\Omega _{m0},\Omega _{\Lambda 0}$ are fixed by the model.
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<div id="SCM_15"></div>
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=== Problem 12 ===
A photon was emitted at time $t$ and registered at time $t_0$ with red
shift $z$. Find and plot the dependence of emission time on the redshift $t(z)$.
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$$
\begin{gathered}
a = \frac{1}{1 + z}; \\
a(t) = A^{1/3}\sinh ^{2/3}\left(t/{t_\Lambda } \right);\quad A \equiv \frac{\Omega_{ m0}}{\Omega _{\Lambda 0}}; \\
\frac{t_0}{t_\Lambda }= Arth\sqrt {\Omega _{\Lambda 0}} ; \\
\frac{t}{t_0} = \frac{1}{Arth\sqrt{\Omega _{\Lambda 0}}}Arsh\left[\sqrt {\frac{\Omega _{\Lambda 0}}{\Omega _{m0}}} \frac{1}{\left(1 + z\right)^{3/2}} \right] \\
\end{gathered}
$$
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<div id="SCM18"></div>
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=== Problem 13 ===
Find the time dependence for the scale factor and analyze its asymptotes. Plot $a(t)$
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In the SCM
$$
\left(\frac{\dot a}{a}\right)^2 = H_0^2\left[ \Omega _{m0}\left(\frac{a_0}{a} \right)^3 + \Omega _{\Lambda 0} \right]
$$
Solution of the equation in the case $\Omega _{\Lambda 0} > 0$ reads
$$
\begin{gathered}
a(t) = a_0\left( \frac{\Omega _{m0}}{\Omega _{\Lambda 0}}\right)^{1/3}\left[ sh\left( \frac{3}{2}\sqrt {\Omega _{\Lambda 0}}H_0t \right) \right]^{2/3}; \\
a\left( t \ll H_0^{ - 1}\right) \propto t^{2/3};\; a\left( t \gg H_0^{ - 1} \right) \sim e^{H_0t} \\
\end{gathered}
$$
Below it is convenient to use the expression for time dependence of the scale factor in the following form
$$
a(t) = A^{1/3}\sinh ^{2/3}\left( t/{t_\Lambda }\right);\quad
A \equiv \frac{\Omega_{m0}}{\Omega _{\Lambda 0}} =
\frac{1 - \Omega _{\Lambda 0}}{\Omega _{\Lambda 0}};\quad t_\Lambda \equiv \frac{2}{3}\left(H_0 \right)^{-1}\left( \Omega _{\Lambda 0}\right)^{ - 1/2}
$$
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=== Problem 14 ===
Using the explicit solution for scale factor $a(t)$, obtained in the previous problem, find the cosmic horizon $R_h$ (see Chapter 3)
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<p style="text-align: left;">
As we have seen above
\[a(t)=A\sinh^{2/3}{(t/t_\Lambda)},\]
where \[A=\frac{\Omega_{m0}}{\Omega_{\Lambda0}},\ t_\Lambda=\frac{2}{3H_\infty}.\]
Consider \[f(t)\equiv\ln{a(t)}=\ln A + \frac23\ln\sinh\frac{t}{t_\Lambda},\] then
\[\dot f=\frac{2}{3t_\Lambda\tanh(t/t_\Lambda)}=\frac{H_\infty}{\tanh(3tH_\infty/2)},\]
and finally
\[R_h=\frac{c}{\dot f}=\frac{c}{H_\infty}\tanh\left(\frac32tH_\infty\right).\]
</p>
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=== Problem 15 ===
Show that the Universe becomes $R_h$-delimited only after the cosmological constant starts to dominate.
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<p style="text-align: left;">
According to the result of the previous problem
\[\dot R_h=\frac32\left[1-\tanh^2\left(\frac32tH_\infty\right)\right]c,\]
so for early times $\dot R_h>c$ and an observer does not experience a divergent redshift with increasing $R$. Then after some ''transition'' time $t^*$, estimated by the condition
\[ct^*=R_h(t^*),\]
he (or she) will begin to encounter an observational limit at a finite radius $R_h$. Numerical solution of the equation gives $t^*\approx0.86H_\infty^{-1}$. This time is roughly the point at which the Universe transits from matter-dominated to $\Lambda$-dominated epoch. Eventually, the Universe becomes de Sitter one and therefore $\dot R_h\rightarrow0$ with $R_h$ settling at the fixed value $c/H_\infty$.
</p>
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<div id="SCM21"></div>
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=== Problem 16 ===
Analyze the stability of Friedmann equations.
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<p style="text-align: left;">
Friedmann equations are given by
\[H^2=\frac{8\pi G}{3}\rho_m+\frac\Lambda3,\]
\[\dot H=-4\pi G\rho_m,\]
\[\dot\rho_m+3H\rho_m=0\]
Let us introduce the following dimensionless variables (recall that $\rho_m,\ \Lambda>0$):
\[x=\frac{\sqrt{8\pi G\rho_m}}{\sqrt3H},\ y=\frac{\sqrt\Lambda}{\sqrt3H}\]
Equations for $x$ and $y$ read (with $N\equiv\ln a$)
\[x'=\frac{dx}{dN}=\frac32x(x^2-1),\]
\[y'=\frac{dy}{dN}=\frac32yx^2.\]
The critical points, i.e. the solutions corresponding to $x'=0,\ y'=0$ and their eigenvalues are given by
\[x_{cr}=0,\ y_{cr}=1 (\Lambda-dominant\, case,)\ \mu_1=0, \mu_2=-\frac32;\]
\[x_{cr}=1,\ y_{cr}=0 (matter-dominant\, case,)\ \mu_1=3, \mu_2=\frac32.\]
From the eigenvalues we see that the matter dominant phase is unstable $\mu_{1,2}>0$ while the cosmological constant dominant phase has $\mu_2<0$ being stable. The existence of a zero eigenvalue $\mu_1=0$ in the first case originates from the fact that the two variables x and y are connected by the relation $x^2+y^2=1$. Therefore in this case one can reduce the dynamics to one-dimensional space.
</p>
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<div id="SCM22"></div>
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=== Problem 17 ===
Determine solutions for the perturbations $\delta\rho_m$ and $\delta H$. Make sure that the solutions are stable.
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<p style="text-align: left;">
The variation of Friedmann equations yields
\[2H\delta H=\frac{\kappa^2}{3}\delta\rho_m,\]
\[\delta\dot H=-\frac{\kappa^2}{2}\delta\rho_m,\]
\[\delta\dot\rho_m+3\rho_m\delta H+ 3H\delta\rho_m=0,\]
where $\kappa^2\equiv8\pi G$.
Solution for the above equation is
\[\delta H=\frac{t_\Lambda}{4\kappa^2}C\tanh(\tau) e^{-f(\tau)},\ \delta\rho_m= \frac{C}{4\kappa^4}e^{-f(\tau)},\]
where \[\tau\equiv\frac{t}{t_\Lambda},\]
\[f(\tau)=-2\tau +\ln\left(-1+e^{4\tau}\right)+2\ln\tanh\tau.\]
Here $C$ is an arbitrary constant. We see that $f(\tau)$ approaches $2\tau$ as $t\rightarrow\infty$ and both $\delta H$ and $\delta\rho_m$ decay which implies that the considered solution is stable.
</p>
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<div id="SCM23"></div>
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=== Problem 18 ===
Rewrite the first Friedman equation in terms of conformal time.
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<p style="text-align: left;">
\[\frac{1}{a^4}\left( \frac{da}{d\eta } \right)^2=H_0^2\left[ \Omega_m a(\eta )^{-3}+\Omega_{\Lambda} \right]\]
</p>
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<div id="SCM24"></div>
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=== Problem 19 ===
Find relation between the scale factor and conformal time
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<p style="text-align: left;">
\[\eta =\frac{1}{H_0}\int\limits_0^a\frac{dx}{\sqrt{x}{{\left( {{\Omega }_{m}}+{{\Omega }_{\Lambda }}{{x}^{3}} \right)}^{1/2}}}\]
</p>
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<div id="SCM25"></div>
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=== Problem 20 ===
Find explicit dependence of the scale factor on the conformal
time.
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<p style="text-align: left;">
The integral obtained in the previous problem can be expressed in terms of elliptic integral of the first kind $F\left( \varphi ,k \right)$. For that purpose we rewrite it in the form
$$
\Omega^{1/6}_{\Lambda} \Omega^{1/3}_m H_{0} \eta = \int\limits_0^{u} \dfrac{dx}{\sqrt{x}\sqrt{1 + x^3}}.
$$
The upper limit equals to $u=\left(\dfrac{\Omega_{\Lambda}}{\Omega_m}\right)^{1/3}a.$ The integral can be presented in the form
$$3^{1/4} \Omega^{1/6}_{\Lambda} \Omega^{1/3}_m H_{0} \eta = F\left(\arccos\dfrac{1+(1-\sqrt{3})u}{1+(1+\sqrt{3})u}, \dfrac{\sqrt{2+\sqrt{3}}}{2}\right).$$

Here $y={{3}^{1/4}}\Omega _{\Lambda }^{1/6}\Omega _{m}^{1/3}{{H}_{0}}\eta ,\quad k=\frac{\sqrt{2+\sqrt{3}}}{2}\approx 0.97.$
</p>
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<div id="SCM25"></div>
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=== Problem 21 ===
Find relative density of dark energy $10^9$ years later.
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<p style="text-align: left;">
$$
\begin{gathered}
t = t_\Lambda Arcth\sqrt {\Omega _\Lambda} ;\quad \Omega _{\Lambda } = th^2\frac{t}{t_\Lambda };\\
t_\Lambda \simeq 10.768 \cdot 10^{9}\;\mbox{years}\;\;t = 14.7\cdot 10^{9}\;\mbox{years}\\
\Omega _{\Lambda 0} \simeq 0.77\\
\end{gathered}
$$
</p>
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<div id="SCM_17"></div>
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=== Problem 22 ===
Find variation rate of relative density of dark energy. What are its asymptotic values? Plot its time dependence.
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<p style="text-align: left;">
$$
\begin{gathered}
\frac{d\Omega _\Lambda }{dt} = \frac{2}{t_\Lambda }\frac{sh(t/{t_\Lambda })}
{ch^3(t/{t_\Lambda })}; \\
x \equiv \frac{t}{t_\Lambda };\quad \frac{d\Omega _\Lambda }{dx} = 2\frac{sh(x)}{ch^3(x)} \\
\end{gathered}
$$

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<div id="SCM26"></div>
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=== Problem 23 ===
Estimate size of the cosmological horizon.
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<p style="text-align: left;">
The cosmological horizon is defined as
$$
L_{H0} = a_0\int\limits_0^{t_0}\frac{dt}{a\left( t \right)}.
$$
Using the results of the problem \ref{A_t_lambda}, one obtains
$$
L_{H0}= A^{-1/3}\int\limits_0^{t_0}\frac{dt}{sh^{2/3}\left( t/{t_\Lambda } \right)};
$$
</p>
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<div id="SCM_19"></div>
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=== Problem 24 ===
Find time dependence of Hubble parameter and plot it.
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<p style="text-align: left;">
$$
H = \frac{\dot a}{a} =\frac{2}{3}\left(t_\Lambda \right)^{ - 1}cth\left( {\frac{t}{t_\Lambda }} \right)
$$

In the SCM ${t_0} \simeq 1.2t_\Lambda .$ Therefore
$$
H{t_0} \simeq 0.8cth\left( {1.2t/{t_0}} \right)
$$

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File:12_16.JPG|Time dependence of the Hubble parameter (in units of the age of the Universe).
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</p>
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</div></div>

<div id="SCM27"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 25 ===
Find time dependence of dark matter density.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Present the first Friedmann equation in the form
\[\rho_m=\frac{1}{8\pi G}(3h^2-\Lambda)\]
and make use of results of the previous problem to obtain
\[\rho_m=\frac{\Lambda}{8\pi G}\frac{1}{\sinh^2(t/t_\Lambda)}.\]

\item {\bf Find the asymptotic (in time) value of the Hubble parameter.}

The required asymptote can be obtained in two ways: first, from the time dependence of the Hubble parameter (see the previous problem), and second, immediately from the first Friedmann equation in the SCM. Taking into account that $\rho_{m} \to 0$ at $t\to \infty$, one gets
$$
H_{\infty} = \frac{2}{3t_\Lambda} = \sqrt {\frac{\Lambda}{3}}.
$$
</p>
</div>
</div></div>

<div id="SCM28"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 26 ===
At present the age of the Universe $t_0\simeq13.7\cdot10^9$ years is close to the Hubble time $t_H=H_0^{-1}\simeq14\cdot10^9$ years.
Does the relation $t^*\simeq t_H(t^*)=H^{-1}(t^*)$ for the age $t^*$
of the Universe hold for any moment of its evolution
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
This relation cannot be satisfied in arbitrary moment of time, as the Hubble parameter has limit value in the SCM (see the previous problem):
$$
\frac{t^*}{t_H(t^*)} = H\left( t^*\right)t^* = \frac{2}{3}\frac{Arcth\sqrt {\Omega _\Lambda(t^*)}}{\sqrt{\Omega_\Lambda (t^*)}}.
$$
For $t\to \infty,\;\Omega _\Lambda\left( t^* \right) \to 1,$ and $$\frac{t^*}{t_H(t^*)} \to \infty. $$

\item {\bf Determine values of state parameter for the dark energy, that provide accelerated expansion of Universe in the present time.}

$$
\begin{gathered}
\sum\limits_i (\rho _{i0} + 3p_{i0}) < 0\; \Rightarrow \rho _{m0} + \rho _{DE0} + 3w_{DE}\rho _{DE0} < 0; \\
w_{DE} < - \frac{1}{3}\left( 1 + \frac{\Omega _{m0}}{\Omega _{DE0}}\right); \\
w_{DE} < - \frac{1}{3}\Omega _{DE0}^{-1};\; w_{DE} < - 0.46\\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM29"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 27 ===
Find current value of the deceleration parameter.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For a multi-component system the deceleration parameter equals to
$$
q = \frac{\Omega}{2} + \frac{3}{2}\sum\limits_i {w_i}\Omega _i
$$
In the SCM $\Omega = 1,w_m = 0,\;w_\Lambda= - 1,$ and
$$
\begin{gathered}
q = \frac{1 - 3\Omega_{\Lambda 0}}{2};\\
q_0 \simeq - 0.6 \\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM_24"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 28 ===
Find the redshift dependence of the deceleration parameter. Analyze the limiting cases.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
q \equiv - \frac{a\ddot a}{\dot a^2} = - \frac{\ddot a}{a}\frac{1}{H^2}; \\
\frac{\ddot a}{a} = - \frac{4\pi G}{3}\sum\limits_i \left( \rho _i + 3p_i \right) = - \frac{1}{2}H_0^2\sum\limits_i \Omega _{i0}\left(1 + 3w_i\right)(1 + z)^{3\left(1 + w_i\right)} ; \\
H^2 = H_0^2\sum\limits_i \Omega _{i0}(1 + z)^{3\left( 1 + w_i \right)}; \\
q = \frac{3}{2}\frac{\sum\limits_i \Omega _{i0}\left( 1 + w_i \right)(1 + z)^{3\left( 1 + w_i \right)}}
{\sum\limits_i \Omega _{i0}(1 + z)^{3\left( 1 + w_i\right)} } - 1 \\
\end{gathered}
$$
In the SCM
$$
q = \frac{1}{2}\frac{\Omega _{m0}(1 + z)^3- 2\Omega _{\Lambda 0} }
{\Omega _{m0}(1 + z)^3+\Omega_{\Lambda 0}}
$$

$$
q(z \to \infty ) = \frac{1}{2},\;q(z \to - 1) = - 1
$$

<gallery widths=600px heights=500px>
File:12_24.JPG|Dependence of the deceleration parameter on the redshift.
</gallery>
</p>
</div>
</div></div>

<div id="SCM_25"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 29 ===
Find and plot the time dependence of the deceleration parameter.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
q = - \frac{\ddot aa}
{\dot a^2} = - \frac{\ddot a}
{aH^2}; \\
a(t) = A^{1/3}sh^{2/3}\left(t/{t_\Lambda}\right);\quad A\equiv \frac{\Omega_ {m0}}
{\Omega _{\Lambda 0}}; \\
\frac{t_0}
{t_\Lambda } = Arth\sqrt {\Omega _{\Lambda 0}} ;\quad H = \frac{2}
{3}\frac{1}
{t_\Lambda }cth\left( \frac{t}
{t_\Lambda }\right); \\
q = \frac{1}
{2}\left[ 1 - 3th^2\left( \frac{t}
{t_\Lambda }\right)\right] \\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM29"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 30 ===
Find the moment of time when the dark energy started to dominate over dark matter. What redshift did it correspond to?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
\left( \frac{\dot a}{a} \right)^2 = \frac{8\pi G}
{3}\rho _{cr}\left( \Omega _{m0}(1 + z)^3+\Omega_{\Lambda 0}\right); \\
\Omega _{m0}(1 + z^*)^3 =\Omega _{\Lambda 0};\\
z^* = \left(\frac{\Omega _{\Lambda 0}}
{\Omega _{m0}}\right)^{1/3}-1; \\
\Omega _{\Lambda 0}\simeq 0.73;\quad \Omega _{m0}\simeq 0.27; \\
z^* \simeq 0.39; \\
\Omega _\Lambda = th^2\left(\frac{t}
{t_\Lambda }\right);\quad \Omega _m = 1 - \Omega _\Lambda = ch^{-2}\left( \frac{t}
{t_\Lambda } \right); \\
\Omega _\Lambda(\tilde t) = \Omega _m(\tilde t) \to \tilde t = \frac{Arsh(1)}{Arth(\sqrt {\Omega _{\Lambda 0} })}t_0 \simeq 9.5\cdot10^9 \,years \\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM30"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 31 ===
Determine the moment of time and redshift value corresponding to the transition from decelerated
expansion of the Universe to the accelerated one.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
\frac{\ddot a}
{a} = \frac{4\pi G}
{3}\rho_{cr}\left[2\Omega_{\Lambda 0} - \Omega _{m0}(1 + z)^3\right]; \\
2\Omega _{\Lambda 0} = \Omega _{m0}(1 + z^*)^3; \\
z^* = \left( \frac{2\Omega_{\Lambda 0}}
{\Omega _{m0}}\right)^{1/3} - 1; \\
\Omega _{\Lambda 0}\simeq 0.73;\quad \Omega _{m0}\simeq 0.27; \\
z^* \simeq 0.745 \\
\end{gathered}
$$

Let us now determine the time moment $t^*$ of the transition
$$
\begin{gathered}
a^* = \frac{1}
{1 + z^*} = \left(\frac{\Omega _{m0}}
{2\Omega _{\Lambda 0}} \right)^{1/3}; \\
t(a) = \frac{2}{3\sqrt{\Omega_{\Lambda 0}}H_0}Arsh\left(\sqrt{ \frac{\Omega_{\Lambda 0}}
{\Omega_{m0}}}a^{3/2} \right); \\
t^* = \frac{2}
{3\sqrt {\Omega _{\Lambda 0}}}H_0Arsh\left(\sqrt {1/2} \right) \\
\end{gathered}
$$
There is an alternative way to solve the problem.
In the multi-component Universe with the corresponding state equations $p_i = w_i\rho _i$ the deceleration parameter $q \equiv - \frac{\ddot aa}{\dot a^2}$ equals to
$$
q = \frac{\Omega }
{2} + \frac{3}
{2}\sum\limits_i w_i\Omega _i
$$
In the case of the SCM $\Omega =1$ and
$$
q = \frac{1}
{2} - \frac{3}
{2}\frac{\Omega _{\Lambda 0}}
{\left[\left(1 + z\right)^3\Omega _{m0}+\Omega_{\Lambda 0}\right]}
$$

From the condition $q=0$ it follows that
$$
z^* =\left(\frac{2\Omega_{\Lambda 0}}
{\Omega _{m0}} \right)^{1/3}- 1
$$
For $\Omega _{\Lambda 0} = 0.73,\;\Omega _{m0} = 0.27$ one obtains that
$$
z^* = 0.745
$$
</p>
</div>
</div></div>

<div id="SCM31"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 32 ===
Solve the previous problem using the derivative $d\eta/d\ln a$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\frac{d\eta}
{d\ln a} = a\frac{dt}{ada} = \frac{1}{aH} = \frac{1}{\dot a}
$$

If the quantity $aH = \dot a$ grows, then $\ddot a > 0,$ which corresponds to the accelerated expansion (inflation) of the Universe. In the SCM the quantity
$$
\frac{aH}{H_0} \simeq \sqrt{ \frac{0.27 + 0.73a^3}{a}}
$$
starts to grow at $a=0.573,$ which corresponds to $z=0.745$
</p>
</div>
</div></div>

<div id="SCM32"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 33 ===
Is dark energy domination necessary for transition to the accelerated expansion?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
No. The transition to the accelerated expansion takes place at $(z=0.75)$, which is considerably earlier than the transition to the dark energy domination stage at $(z=0.45).$
</p>
</div>
</div></div>

<div id="SCM33"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 34 ===
Consider flat Universe composed of matter and dark energy in form of cosmological constant. Find the redshift value corresponding to equality of densities of the both components $\rho_m(z_{eq})=\rho_\Lambda(z_{eq})$ and the one corresponding to beginning of the accelerated expansion $q\left(z_{accel} \right) = 0$. Obtain relation between $z_{eq}$ and $z_{accel}$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
z_{eq} = \left(\frac{\Omega _\Lambda}
{\Omega _{m0}} \right)^{1/3}- 1; \\
z_{accel}= \left(\frac{2\Omega_\Lambda }
{\Omega _{m0}}\right)^{1/3}- 1; \\
z_{accel} = 2^{1/3}\left(z_{eq}+ 1\right) - 1 \\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM34"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 35 ===
Show that density perturbations stop to grow after the transition from dust to $\Lambda$-dominated era.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
On the $\Lambda$-dominated stage the non-relativistic matter density perturbation size depends on conformal time as
\begin{equation}\label{1SCM30}
a(\eta)=-\frac{1}{H\eta}, \,\, \eta=-const\cdotp e^{-Ht},\,\,\eta<0
\end{equation}
where the Hubble parameter $H$ is constant, $H^2=(8\pi/3)G\rho_\Lambda$.
The linearized Einstein equations for the case of single-component medium take the following form in the momentum representation:
\begin{equation}\label{3SCM30}
k^2\Phi+3\frac{a^{'}}{a}+3\frac{a^{'2}}{a^2}\Phi=-4\pi Ga^2\delta\rho,
\end{equation}
\begin{equation}\label{2SCM30}
\Phi^{''}+3\frac{a^{'}}{a}\Phi^{'}+\left(2\frac{a^{''}}{a} - \frac{a^{'2}}{a^2}\right)\Phi=4\pi Ga^2\delta p.
\end{equation}
Assuming that the $\Lambda$-term is constant (which is exactly true for the vacuum energy density) one obtains $\delta\rho_\Lambda=\delta p_\Lambda=0$.
Using \eqref{2SCM30} and taking into account \eqref{1SCM30} we come up with the equation: $${\Phi}^{''}-\frac{3}{\eta}{\Phi}^{'}-\frac{3}{\eta^2}\Phi=0,$$
Solutions of the equation behave as
$$\Phi\propto\eta\propto\frac 1a$$ and $$\Phi\propto\eta^3\propto\frac{1}{a^3}.$$
Use now \eqref{3SCM30} for the modes beyond the horizon and take into account that for dust density $\rho_M\propto a^{-3}.$ Keeping only the first solution for the gravitational potential (the second decays too fast), one can see that the density perturbations do not grow: for the modes under horizon the relation
$$\delta_M\propto a^3\frac{k^2}{a^2}\Phi=const$$ holds.

Therefore the perturbations stop to grow at the transition from the dust-dominated to the $\Lambda$-dominated stage. If the present accelerated expansion of the Universe is actually caused by non-zero cosmological constant, then the structure formation from small primary perturbations is eventually close to end and it will never resume again: the structures larger than the presently observed ones will never appear.
</p>
</div>
</div></div>

<div id="SCM35"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 36 ===
What happens to the velocity fluctuations of non-relativistic matter and radiation with respect to Hubble flow in the epoch of cosmological constant domination?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM36"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 37 ===
Find the ratio of baryon to non-baryon components in the galactic halo.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM37"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 38 ===
Imagine that in the Universe described by SCM the dark energy was instantly switched off. Analyze further dynamics of the Universe.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="local-group"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 39 ===
Estimate density of dark energy in form of cosmological constant using the Hubble diagram for the neighborhood of the Local group.
<gallery widths=600px heights=500px>
File:im_eps.JPG|Hubble diagram for the neighborhood of the Local group.
</gallery>
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
An elegant and simple method was recently proposed (see A.D.Chernin, UFN 178, 267 (2008)), which allows not only to become convinced in the existence of the dark energy but also to estimate its density. The method is based on the assumption, that the dark energy (in form of the cosmological constant) governs the Hubble flow even on comparably small distances $\left( \sim \,Mpc \right)$. Consider the resulting force $\vec{F}$ (per unit mass), acting on a galaxy from the picture \ref{hubble_diagram}. The force is a sum of two contributions:
$$
\vec{F}=\vec{F}_N+\vec{F}_\Lambda
$$
Here $\vec{F}_N$ is the force produced by the total mass of the Local group, and $F_{\Lambda }$ is the force created by the dark energy. For a crude estimate let us neglect the non-spherical mass distribution in the Local group. Then the force acting on a satellite galaxy in a given point $R$ outside the local Group equals to
$$
F=-\frac{GM_{LG}}{R^2}+\frac{8\pi G}{3}{\rho_{\Lambda }}R,
$$
where $M_{LG}\simeq (1\div 3) \times 10^{12}M_{\odot }$ is the total mass of the Local group. As can be seen from the picture, the transition from domination of the local group gravity to the vacuum domination takes place on the distance of order $R_{0}=1.8\div 2\; \, Mpc.$ It allows to make a rough estimate of the dark energy density from the condition $F\left( R=R_0 \right)=0.$ The estimate results in the following
$$
(0.1\pm 0.03)<\rho_V<(1\pm 0.3)\times 10^{-26} \, kg/m^3.
$$
The proximity of the obtained crude estimate to the actual value of the dark energy density in the SCM is striking.
$$
\rho_{V}\simeq (0.75\pm 0.05)\times 10^{-26}\, kg/m^3.
$$
</p>
</div>
</div></div>

<div id="SCM38"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 40 ===
Estimate the Local group mass by methods used in the previous problem.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM39"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 41 ===
Find "weak" points in the argumentation of the two preceding problems.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM40"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 42 ===
The product of the age of the Universe and current Hubble parameter (the Hubble's constant) is a very important test (Sandage consistency test) of internal consistency for any model of Universe. Analyze the
parameters on which the product $H_0 t_0$ depends in the Big Bang model and in the SCM.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM41"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 43 ===
Show that for a fixed source of radiation the luminosity distance for high redshift values in flat Universe is greater for the dark energy dominated case compared to the non-relativistic matter dominated one.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the case of Universe filled by dark energy and non-relativistic matter, the photometric distance is equal to
$$
d_L = \frac{1 + z}
{H_0}\int_0^z \frac{dz'}
{\sqrt{\Omega _{m0}\left(1 + z'\right)^3+ \Omega _{DE0}\left(1 + z' \right)^{3\left( 1 + w_{DE}\right)}}}
$$

In the case $\Omega _{m0} = 1,\;\Omega _{DE0} = 0$
one obtains
$$
d_L = \frac{2}
{H_0}\left[1 + z -(1 + z)^{1/2}\right]
$$
and in the case $\Omega _{m0} = 0,\;\Omega _{DE0} = 1$
$$
d_L = \frac{1}{H_0}z(1 + z)
$$
instead.

In means that the supernova appear dimmer in the dark energy dominated Universe.
</p>
</div>
</div></div>

<div id="SCM42"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 44 ===
In the observations that discovered the accelerated expansion of the Universe the researchers in particular detected two $Ia$ type supernovae: $1992P$, $z=0.026$, $m=16.08$ and $1997ap$, $z=0.83$,
$m=24.32$. Show that these observed parameters are in accordance with the SCM.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The apparent $m$ and absolute $M$ stellar magnitudes are related with the photometric distance $d_L$ by the expression
\begin{equation}\label{p(1)}
m - M = 5\log _{10}\left(\frac{d_L}
{\mbox{Mpc}} \right) + 25
\end{equation}
The photometric distance in the SCM is equal to
\begin{equation}\label{p(2)}
d_L = \frac{1 + z}
{H_0}\int_0^z \frac{dz'}
{\sqrt {\Omega _{m0}(1 + z)^3 + \Omega _\Lambda}}
\end{equation}
For small $z$ $d_L(z) \simeq z/H_0.$ Using the supernova $1992P$ with small $z=0.026$ one can find $M = - 19.09.$ Assuming that the $1a$-type supernova are standard candles (they have equal absolute magnitudes), one obtains from the relation \ref{p(1)} supernova with $z=0.83$
$$H_0d_L\simeq 1.16$$
From the other hand it follows from the relation \ref{p(1)} that in the case
$\Omega _{m0} = 1,\;\Omega_\Lambda = 0$
$$
H_0d_L\simeq 0.95
$$
In the SCM one gets much better agreement:
$$
H_0d_L \simeq 1.2
$$
</p>
</div>
</div></div>

<div id="SCM43"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 45 ===
Find the redshift value, at which a source of linear dimension $d$ has minimum visible angular
size.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The apparent angular size equals to $$\delta \theta = {{(1 + z)^2 d } \over {d_L (z)}},$$
where $d_L (z)$ is the photometric distance from a source with the redshift $z$. In the case of cosmological constant domination the photometric distance is:
$$d_L (z) = - {z \over {H_0 \Omega _\Lambda ^{1/2} }},$$
In the SCM (see the previous problem)
$$
d_L = \frac{1 + z}
{H_0}\int_0^z \frac{dz'}
{\sqrt {\Omega _{m0}(1 + z)^3 + \Omega _\Lambda}}
$$
then the angular size is
$$\delta \theta = - {{H_0 \Omega _\Lambda ^{1/2} d} \over 2}{{(1 + z)^2 } \over z},$$
and $z = - 1$. It corresponds to infinitely far future.
</p>
</div>
</div></div>

<div id="SCM44"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 46 ===
Compare the observed value of the dark energy density with the
one expected from the dimensionality considerations (the cosmological constant problem).
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The dimensionality considerations suggest the following value for density of the dark energy in form of the cosmological constant
\[\rho_\Lambda^{(dim)}\sim M_{Pl}^4\approx10^{76}GeV^4.\]
At the same time the ''observed'' value fixed by the SCM is
\[\rho_\Lambda^{(obs)}\approx10^{-48}GeV^4.\]
This huge disagreement
\[\frac{\rho_\Lambda^{(dim)}}{\rho_\Lambda^{(obs)}}>10^{120}\]
is called the ''cosmological constant problem''. It presents a fundamental problem for elementary particles theory and physics in general, rather than for cosmology itself. (Look for details in R.Bousso, [http://arxiv.org/abs/0708.4231 TASI Lectures on the cosmological constant])
</p>
</div>
</div></div>

<div id="SCM45"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 47 ===
Determine the density of vacuum energy using the Planck scale as cutoff parameter.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For a quantum field with a given mode of frequency $\omega$, the zero-point energy is $\pm\omega/2$ with the plus sign for a bosonic field and the minus sign for a fermionic field. The total zero-point energy is then $\pm\sum\limits_i\omega_i/2$. In the continuum limit, we have for a free field (bosonic or fermionic)
\[\frac12\sum\limits_i\omega_i=\frac12\int\frac{d^3xd^3k}{(2\pi)^3}(k^2+m^2)^{1/2}=V\int\frac{k^2dk}{4\pi^2}(k^2+m^2)^{1/2}.\] This integral diverges, thus we need to introduce the physical cutoff. In cosmological context this cutoff must be the Planck mass at largest. With the cutoff $k_{max}\sim M_{Pl}$ one obtains
\[\rho_\Lambda=\int_0^{k_{max}}\frac{k^2dk}{4\pi^2}(k^2+m^2)^{1/2} \approx\frac{k_{max}^4}{16\pi^2}=\frac{M_{Pl}^4}{16\pi^2} \approx10^{74}GeV^4.\]
This result only slightly modifies the dimensional considerations given in the previous problem.
</p>
</div>
</div></div>

<div id="SCM46"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 48 ===
Identifying the vacuum fluctuations density with the observable dark energy value in SCM, find the required frequency cutoff magnitude in the fluctuation spectrum.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[\nu_{max}=10^{12}Hz.\]
</p>
</div>
</div></div>

<div id="SCM47"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 49 ===
What purely cosmological problem originates from the divergence of the zero-point energy density?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
If the zero-point energy density is infinite, then our Universe is infinitely curved and the space size is infinitely small.
</p>
</div>
</div></div>

<div id="SCM48"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 50 ===
With exact supersymmetry, the bosonic contribution to cosmological constant is canceled by its fermionic counterpart. However, we know that our world looks like not supersymmetric. Supersymmetry, if exists, has to be broken above or around $100GeV$ scale. Compare the observed value of the dark energy density with the one expected from broken supersymmetry.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
If we take the $k_{max}^{(SUSY)}\sim100GeV$ cutoff, then
\[\rho_\Lambda^{(SUSY)}=\frac{\left(k_{max}^{(SUSY)}\right)^4}{16\pi^2}\approx10^6GeV^4.\]
\[\frac{\rho_\Lambda^{(SUSY)}}{\rho_\Lambda^{(obs)}}>10^{54}.\]
Thus supersymmetry does not solve the problem of the cosmological constant.
</p>
</div>
</div></div>

<div id="SCM49"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 51 ===
Determine duration of the inflation period.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Let us see what happens with the relative density during all the history of Universe. From the first Friedmann equation it follows that in the matter-dominated epoch $\left| \Omega - 1 \right| \sim t^{2/3},$ and in the radiation-dominated epoch $\left| \Omega - 1 \right| \sim t^{2/3}.$ Then assume that inflation started at $t_i,$ and finished at $t_f.$ Then in the period of $t_f < t < t_{eq} = 50000$ years the radiation dominated and the range $t_{eq} < t < t_0 $ corresponds to the matter domination epoch. Then the presently observed difference
$$
\left|\Omega (t_0 ) - 1 \right| = \left| \Omega (t_i ) - 1 \right|e^{ - 2H(t_f - t_i )} \left( \frac{t_{eq} }
{t_f } \right)\left( \frac{t_0}
{t_{eq} }\right)^{2/3}
$$
According the SCM $\left|\Omega \left( t_0\right) - 1 \right| < 0.04.$ If one assumes that before the beginning of the inflation there was $\left|\Omega \left( t_i \right) - 1 \right| \sim 1,$ then
$$
N > \frac{1}{2}\ln \left[\frac{\Omega \left( t_i \right) - 1}{\Omega \left(t_0\right) - 1}\left( \frac{t_{eq}}{t_f } \right)\left( \frac{t_0}{t_{eq}} \right)^{2/3} \right] \approx 60
$$

Assume that radiation dominated before the inflation started. Then $H = \frac{1}{2t}.$ If at the moment of time $t_i$ the vacuum dark energy started to dominate and then the expansion continued with constant rate, then $H\approx 1/t_i$ during the inflation. And finally one obtains that
$$
N = H\left(t_f - t_i\right) \approx \frac{t_f - t_i }
{t_i }
$$
it is commonly assumed that the inflation finishing time is of order of the Grand Unification time $t_f\approx 10^{-35}.$ Therefore $t_f\approx 10^{-38}.$ Evidently the inflation could last longer as we used only lower estimate for $N.$
</p>
</div>
</div></div>

<div id="SCM50"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 52 ===
Plot the dependence of luminosity distance $d_L$ (in units of $H_0^{-1}$) on the redshift $z$ for the two-component flat Universe with non-relativistic liquid ($w=0$) and cosmological constant
($w=-1$). Consider the following cases:

a) $\Omega_\Lambda^0=0$;

b) $\Omega_\Lambda^0=0.3$;

c) $\Omega_\Lambda^0=0.7$;

d) $\Omega_\Lambda^0=1$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM51"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 53 ===
For a Universe filled by dark energy with state equation $p_{DE}=w_{DE}\rho_{DE}$ and non-relativistic matter obtain the Taylor series for $d_L$ in terms of $z$ near the observation point
$z_0=0$. Explain the obtained result.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the universe filled by dark energy with the state equation $p_{DE}=w_{DE}\rho_{DE}$ and non-relativistic dark matter the photometric distance takes on the form:
\begin{equation}
\label{d_L}
d_L = \frac{1 + z}
{H_0}\int_0^z \frac{dz'}
{\sqrt{\Omega _{m0}\left(1 + z'\right)^3+ \Omega _{DE0}\left(1 + z' \right)^{3\left( 1 + w_{DE}\right)}}}
\end{equation}
Expand the integrand in Taylor series in $z$ in vicinity of the observation point $z_0=0$ to obtain
$$
\begin{gathered}
\label{tey}
\frac{1}{\sqrt{\Omega _{m0}\left(1 + z'\right)^3+ \Omega _{DE0}\left(1 + z' \right)^{3\left( 1 + w_{DE}\right)}}}\simeq \\
\simeq 1 -\frac{3}{2}\left(1+w_{DE}\Omega _{DE0}\right)z+\frac{3}{4}\left(3w_{DE}\Omega _{DE0}-1\right)z^2.\nonumber
\end{gathered}
$$
Substitution of the decomposition (\ref{tey}) into the integral (\ref{d_L}) results in the following:
\begin{equation}
\label{d_L1}
d_L = \frac{1 + z}
{H_0}\left(z -\frac{3}{4}\left(1+w_{DE}\Omega _{DE0}\right)z^2+\frac{1}{4}\left(3w_{DE}\Omega _{DE0}-1\right)z^3\right).
\end{equation}
</p>
</div>
</div></div>

<div id="SCM_45"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 54 ===
Determine position of the first acoustic peak in the CMB power spectrum produced by baryon oscillations on the surface of the last scattering.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Early Universe was filled by photon-baryon plasma, which can be treated as a single-component liquid. The baryons were trapped by the potential walls generated by density fluctuations, and they were eventually compressed. The contraction lead to heating of the plasma and therefore to increasing radiative pressure of the photons which is directed outside. Finally the radiative pressure stop the compression and lead to expansion of the plasma. During the expansion the plasma cools down and its radiative pressure decreases. The gravity starts to dominate again leading to repeated compression. The concurrence between the gravity and the pressure lead to longitudinal (acoustic) oscillations in the photon-baryon fluid. When the matter and the radiation get decoupled in the recombination process, the picture of the acoustic oscillations remains frozen into the CMB. Today we detect the evidence of the primordial acoustic waves (dense and diluted regions) in form of the primary anisotropy in the CMB.

It is well known that any acoustic wave, whatever complicatedly shaped, can be represented in form of superposition of modes with different wave numbers $k$, $k\propto 1/\lambda $. Every mode $\lambda $ corresponds to certain angular scale $\theta $ on the sky. Therefore in order to facilitate the comparison of the theory with observations, one should use the angular (multipole) decomposition in terms of the Legendre polynomials $P_{l} (\cos \theta )$, instead of the Fourier transform in terms of sines and cosines. Order of the polynomial $l$ plays the same role as the index $k$ in the Fourier decomposition. For $l\ge 2$ the Legendre polynomials are oscillating functions in the interval $\left[1,-1\right]$. Number of the oscillations increases with growth of $l$. Therefore
\begin{equation} \label{acust1_}
l\propto \frac{1}{\theta }
\end{equation}
The temperature fluctuations can be observed and qualitatively analyzed with help of paired measurements in the directions $\stackrel{\frown}{n},\stackrel{\frown}{n}'$, so that $\stackrel{\frown}{n}\cdot \stackrel{\frown}{n}'=\cos \theta $. Taking average over all the pairs in the assumptions of Gaussian fluctuations, one obtains the two-point correlator $C(\theta )$, which can be presented in terms of the multipole decomposition
\begin{equation} \label{acust2_}
\left\langle \delta T(\stackrel{\frown}{n})\cdot \delta T(\stackrel{\frown}{n}')\right\rangle \equiv C(\theta )=\sum _{l}\frac{2l+1}{4\pi } C_{l} P_{l} (\cos \theta )
\end{equation}
where $C_{l} $ are the coefficients.

As we already mentioned, the analysis of the temperature fluctuations enables us to clarify the structure of the longitudinal oscillations. The mode with the maximum wavelength corresponds to the maximum angular size of the primary anisotropy. This fundamental mode was detected the first. There are reliable evidences of the fact that the second and the third modes were also already detected. The distance $r_{s} $, passed by the acoustic wave during the time period before the recombination, is called the sound horizon. The sound horizon is fixed by (or rather fixes) the physical scale o the last scattering surface. Size of the sound horizon depends on values of the physical parameters. Distance to the last scattering surface $d_{sls} $ depends on the cosmological parameters too. Together they determine the angular size of the sound horizon$\theta _{s} $ (see Fig.)

<gallery widths=600px heights=500px>
File:acustic_pic.JPG|Dependence of the deceleration parameter on the redshift.
</gallery>

\begin{equation} \label{acust3_}
\theta _{s} \approx \frac{r_{s} }{d_{sls} }
\end{equation}
Analysis of the temperature anisotropy allows to determine $\theta _{s} $. Varying the cosmological parameters in $r_{s} ,d_{sls} ,$ one achieves the best agreement with the observations and thus fixes the cosmological parameters.

We can estimate the sound horizon size as the length passed by sound from the moment $t=0$ to the recombination time $t_{*} $
\begin{equation} \label{acust4_}
r_{s} (z_{*} ;\Omega _{b} ,\Omega _{r} )\approx \int _{0}^{t_{*} }c_{s} dt
\end{equation}
where $z_{*} $ is the redshift of the recombination moment $\left(z_{*} \approx 1100\right)$, $c_{s} $ is the sound speed in the photon-baryon medium
\begin{equation} \label{acust5_}
c_{s} \approx c\left[3\left(1+\frac{3\Omega _{b} }{\Omega _{r} } \right)\right]^{-1/2}
\end{equation}
To determine $dt$ one should use the first Friedmann equation in the form
\begin{equation} \label{acust6_}
\left(\frac{da}{dt} \right)^{2} =H_{0}^{2} \left[\Omega _{r0} \left(\frac{a_{0} }{a} \right)^{2} +\Omega _{m0} \frac{a_{0} }{a} +\Omega _{k0} +\Omega _{\Lambda 0} \left(\frac{a}{a_{0} } \right)^{2} \right]
\end{equation}
Making use of the relations $\frac{a_{0} }{a} =1+z$ and $\Omega _{r} +\Omega _{m} +\Omega _{\Lambda } +\Omega _{k} =1$, one obtains
\begin{equation} \label{acust7_}
dt=H_{0}^{-1} (1+z)^{-1} \left\{\left(1+z\right)^{2} \left(1+\Omega _{m,0} z\right)+z(z+2)\left[\left(1+z)^{2} \Omega _{r.0} -\Omega _{\Lambda ,0} \right)\right]\right\}^{-1/2} dz
\end{equation}
Distance to the last scattering surface, which corresponds to its angular size, is determined by the quantity called the angular diameter distance. It is connected by the photometric distance $d$ by the simple relation
\begin{equation} \label{acust8_}
d_{sls} =\frac{d}{(1+z_{*} )^{2} }
\end{equation}
The position of the first acoustic peak is determined by
\begin{equation} \label{acust9_}
l\approx \frac{d_{sls} }{r_{s} }.
\end{equation}
Let us now predict the position of the first peak for the case of flat Universe. To the leading order the sound speed is
\[c_{s} =\frac{c}{\sqrt{3} }. \]
Let us assume also that in the recombination epoch the Universe was matter-dominated. Under such assumptions one gets
\begin{equation} \label{acust10_}
r_{s} =\frac{c_{s} }{H_{0} \sqrt{\Omega _{m} } } \int _{z_{*} }^{\infty }(1+z)^{-5/2} dz =r_{s} =\frac{2c_{s} }{3H_{0} \sqrt{\Omega _{m} } } \left(1+z_{*} \right)^{-3/2}
\end{equation}
Using the standard procedure (see section 4), we obtain for the comoving radial coordinate $r_{sls} $ of the last scattering surface the following
\begin{equation} \label{acust11_}
r_{lsr} =\frac{c}{H_{0} } \int _{0}^{z_{*} }\left[\Omega _{m} (1+z)^{^{3} } +\Omega _{\Lambda } \right] ^{-1/2} dz
\end{equation}
Replacing the integrand by its binomial decomposition
\[\Omega _{m}^{-1/2} (1+z)^{-3/2} -\left(\Omega _{\Lambda } /2\Omega _{m}^{3/2} \right)\left(1+z\right)^{-9/2} \]
and taking into account that
\[d_{sls} =r_{sls} /(1+z_{*} ),\]
we obtain
\begin{equation} \label{acust12_}
d_{sls} =\frac{2c}{7H_{0} (1+z_{*} )} \left\{7\Omega _{m}^{-1/2} -2\Omega _{\Lambda } \Omega _{m}^{-3/2} +{\rm O} \left[\left(1+z_{*} \right)^{-1/2} \right]\right\}.
\end{equation}
Using the relation $\Omega _{\Lambda } =1-\Omega _{m} $ and neglecting the higher order terms, one obtains
\begin{equation} \label{acust13_}
d_{sls} \approx \frac{2c\Omega _{m}^{-1/2} }{7H_{0} (1+z_{*} )} \left(9-2\Omega _{m}^{3} \right)
\end{equation}
Combination of the relations \eqref{acust10_} and \eqref{acust13_} gives for the first acoustic peak the following
\begin{equation} \label{acust14_}
l\approx \frac{d_{sls} }{r_{s} } \approx 0.74\sqrt{1+z_{*} } \left(9-2\Omega _{m}^{3} \right)\simeq 220.
\end{equation}
This result agrees very well with numerous observations.
</p>
</div>
</div></div>

<div id="SCM52"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 55 ===
\bf Compare the asymptotes of time dependence of the scale factor
$a(t)$ for the SCM and de Sitter models. Explain physical reasons of
their distinction.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM53"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 56 ===
Redshift for any object slowly changes due to the acceleration (or deceleration) of the Universe expansion. Estimate change of velocity in one year.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

$$
\Delta z = \dot z \Delta t = H_0 \left(1 + z - {H(z)\over H_0} \right)\Delta t
$$

$$H(z)=H_0 \left[\Omega_{m0}(1+z)^3+\Omega_{\Lambda 0}\right]^{1/2}$$
$$\Omega_{m0}+\Omega_{\Lambda 0}=1,~\Omega_{m0}\approx 0.3$$
$$\Delta z \approx 2\cdot 10^{-11}$$
$$\Delta v = c{\Delta z\over 1+z}\approx 0.25~\mbox{cm/s}$$

</p>
</div>
</div></div>

<div id="SCM54"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 57 ===
Determine the lower limit of ratio of the total volume of the Universe to the observed one?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
If the Universe is the 3-sphere, then its radius is $R_0=a_0$
$$
R_0 = \frac{1}{H_0 \sqrt {\left| \Omega _{curv} \right|} }
$$
For the SCM parameters the cosmological horizon equals to
$$
L_{p,0} = a_0 \int_0^{t_0 } \frac{dt}{a(t)} = \frac{3.6}{H_0}
$$
We assumed that $a(t)\sim t^{2/3}.$ Therefore
$$
\frac{R_0}
{L_{p,0} } = \frac{1}
{3.6\sqrt {\left| \Omega _{curv}\right|} }
$$
Taking into account that $\left|\Omega _{curv} \right| < 0.02,$ one obtains $\frac{a_0}{L_{p0} } > 2$.
Number of the regions similar to the one inside our horizon equals to the ratio of the 3-sphere volume $2\pi ^2 a_0^3 $ to the volume of the observed Universe
$$
N = \frac{2\pi ^2 a_0^3}
{(4\pi /3)L_{p0}^3 } \simeq 4.7\left(\frac{a_0 }
{L_{p0} } \right)^3 > 38
$$
Thus we can see not more than $3 \%$ of the Universe volume.
</p>
</div>
</div></div>

<div id="SCM55"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 58 ===
What is the difference between the inflationary expansion in the early Universe and the present accelerated expansion?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The difference is in the fact that in the present time there is about $27 \%$ of non-relativistic matter which was absent in the inflation period. The numerical values of Hubble parameter in the beginning of the inflation was very different from the one at the beginning of the current period of accelerated expansion of Universe.
</p>
</div>
</div></div>

<div id="SCM56"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 59 ===
Compare the values of Hubble parameter at the beginning of the
inflation period and at the beginning of the present accelerated
expansion of Universe.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
h^2 (x) = \Omega _{m0} x^3 + (1 - \Omega _{m0} )x^\alpha ;\quad \alpha = 3(1 + w)
$$
Assume for simplicity that the Universe is spatially flat. From the definition $O(x)$
it follows that
$$
O(x) = \Omega _{m0} + \left( 1 - \Omega _{m0}\right)\frac{x^\alpha - 1}
{x^3 - 1}
$$
The required result follows from the fact that $\alpha =0$ for cosmological constant, $\alpha >0$ for quintessence and $\alpha <0$ for phantom energy.
</p>
</div>
</div></div>

<div id="SCM57"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 60 ===
Plot the dependencies $h(x)=H(x)/H_0(x)$, where $x=1+z$, in the SCM, for quintessence and for phantom energy cases.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Plots of the dependencies $
h(x) = H(x)/H_0 (x);\quad x = 1 + z
$ in the SCM are presented on Fig. \ref{fig:52} for phantom energy and quintessence.
<gallery widths=600px heights=500px>
File:12_52.JPG|Dependence of the deceleration parameter on the redshift.
</gallery>
</p>
</div>
</div></div>

<div id="SCM58"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 61 ===
Find the constraints imposed by the Weak Energy Condition (WEC) for the dark energy on the redshift-dependent Hubble parameter. ([http://arxiv.org/abs/astro-ph/0703416 Inspired by A.Sen, R.Scherrer])
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The redshift-dependent Hubble parameter $H(z)$ is defined by
\[H^2(z)=\frac{8\pi G}{3}(\rho_m+\rho_{DE})\]
This equation can be rewritten as
\begin{equation}\label{SCM_eq_76_1}
E^2(z)\equiv\frac{H^2(z)}{H^2_0(z)}=\Omega_{m0}(1+z)^3+(1-\Omega_{m0})\frac{\rho_{DE}}{\rho_{DE0}}
\end{equation}
The WEC forces \[\frac{\rho_{DE}}{\rho_{DE0}}\ge1 \ \left(\frac{d\rho}{dz}\ge0\right)\] for $z>1$, so equation (\ref{SCM_eq_76_1}) becomes
\begin{equation}\label{SCM_eq_76_2}
E^2(z)\ge\Omega_{m0}(1+z)^3+(1-\Omega_{m0})
\end{equation}
Going back to equation (\ref{SCM_eq_76_2}) and taking the derivative with respect to $z$ (which we denote throughout with a prime) gives:
\[
2E(z)\frac{dE}{dz}=3\Omega_{m0}(1+z)^2+\frac{1-\Omega_{m0}}{\rho_{DE0}}\frac{d\rho_{DE}}{dz}.
\]
Now the WEC implies that \[\frac{d\rho_{DE}}{dz}\ge0,\] so we get
\begin{equation}\label{SCM_eq_76_3}
E(z)\frac{dE}{dz}\ge\frac32\Omega_{m0}(1+z)^2.
\end{equation}
Equations (\ref{SCM_eq_76_2}) and (\ref{SCM_eq_76_3}) give the constraints that the WEC for the dark energy places on the redshift-dependent Hubble parameter.
</p>
</div>
</div></div>

<div id="SCM59"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 62 ===
Using the statefinders, show that the power law cosmology mimics SCM model at (see Chapters 3 and 9)
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Recall that the statefinders are defined as
\[r\equiv\frac{\dddot a}{aH^3},\ s\equiv\frac{r-1}{3(q-1/2)}, \ q\ne\frac12.\]
The SCM model corresponds to the point $r=1,\, s=0$ in the $(s,r)$ plane. In power-law cosmology (see Chapter 3) the statefinders are given by
\[r=2q^2+q,\, s=\frac23(q+1).\]
We note that $r=1,\, s=0$ for $q=-1$. Thus the power law cosmology mimics SCM model with $q=-1$.
</p>
</div>
</div></div>

Evolution of Universe

2012-10-04T17:57:28Z

Den: /* Problem 53 */

[[Category:Standard Cosmological Model|2]]

__NOTOC__
<div id="SCM-10"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 1 ===
Rewrite the first Friedman equation in terms of redshift and analyze the contributions of separate components at different stages of Universe evolution.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM11"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2 ===
Find the time dependence for the scale factor and analyze asymptotes of the dependence. Plot $a(t)$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM12"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 3 ===
Determine the redshift value corresponding to equality of radiation and matter densities.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
\rho _r(t) = \rho _m(t);\\
\rho _{r0}\frac{a_0^4}{a^4} = \rho _{m0}\frac{a_0^3}{a^3};\\
\frac{\rho _{m0}}{\rho _{r0}} = \frac{\Omega _{m0}}{\Omega _{r0}} = \frac{a_0}{a} = 1 + z;\\
\Omega _{r0} = 5 \cdot {10^{ - 5}} \\
z = \frac{\Omega _{m0}}{\Omega _{r0}} - 1 \approx 5400\\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM_9"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 4 ===
Construct effective one-dimensional potential (see [[Solutions_of_Friedman_equations_in_the_Big_Bang_model#dyn15|problem]] of Chapter 3)
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
<div id="V(x)"></div>
$$
V(x) = - \frac{1}{2}\sum\limits_i \Omega _{i0}x^{ - (1 + 3w_i)};\quad x \equiv \frac{a}{a_0}
$$
In the SCM
$$
V(x) = - \frac{1}{2}\Omega _{m0}x^{-1}-\frac{1}{2}\Omega _{\Lambda 0}x^2 \simeq - 0.135x^{-1}-0.365x^2
$$

<gallery widths=600px heights=500px>
File:12_9.JPG|
</gallery>
</p>
</div>
</div></div>

<div id="SCM13"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 5 ===
Show that the following holds: $\dot H= -4\pi G\rho_m$ and $\ddot H= 12\pi G\rho_m H$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM_18"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 6 ===
Expand the scale factor in Taylor series near the time moment
$t_0$:
\[\frac{a(t)}{a(t_0)}=1+\sum\limits_{n=1}^\infty\frac{A_n(t_0)}{n!}[H_0(t-t_0)]^n;\quad A_n\equiv\frac{1}{aH^n}\frac{d^na}{dt^n}\]
and calculate values for few first coefficients $A_n$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The coefficients $A_1$ and $A_2$ equal to:
$$A_1 = \frac{\dot a}{aH} = 1;\quad A_2 = \frac{\ddot a}{aH^2} = - q = 1 - \frac{3}{2}{\Omega _m}$$
The parameters $A_n(n > 2)$ can be calculated by consequent differentiation of the relation
$$\ddot a = a\left( \dot H + H^2 \right)$$
The time derivatives of the scale factor can be determined making use of the fact that in SCM $\dot H = - 4\pi G\rho _m$, and $\rho _m = \rho _0a^{ - 3}$. For example,
$$A_3 = 1 + \frac{\ddot H}{H^3} + 3\frac{\dot H}{H^2}$$
Using $\ddot H = 12\pi G{\rho _m}H,\;\dot H = - 4\pi G\rho _m$, one finds that
$$A_3 = 1$$
Let us present the expressions for some consequent expansion coefficients
\[\begin{array}{l}
{A_4} = 1 - \frac{3^2}{2}\Omega _m;\\
{A_5} = 1 + 3\Omega _m + \frac{3^3}{2}\Omega _m^2;\\
{A_6} = 1 - \frac{3^3}{2}\Omega _m - 3^4\Omega _m^2 - \frac{3^4}{4}\Omega _m^3
\end{array}\]
</p>
</div>
</div></div>

<div id="SCM14"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 7 ===
Show that all the coefficients $A_n$ can be expressed through
elementary functions of the deceleration parameter $q$ or the
density parameter \[\Omega_m=\frac23(1+q).\]
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM15"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 8 ===
Consider the case of flat Universe filled by non-relativistic matter and dark
energy with state equation $p_X=w\rho_X$, where the state parameter $w$ is parameterized as the following \[w=w_0+ w_a(1-a)=w_0 + w_a \frac{z}{1+z}.\]
Express current values of cosmographic parameters through $w_0$ and
$w_a$. % (See [http://arxiv.org/abs/0802.1583 Cosmography of f(R) gravity])
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the considered case
\[\frac{H^2(z)}{H_0^2} = \Omega _m(1 + z)^3 + \Omega _X(1 + z)^{3\left( 1 + w_0 + w_a \right)}e^{ - \frac{3w_az}{1 + z}}\]
Calculation of the cosmographic parameters requires in general to integrate $H(z)$ in order to obtain $a(t).$ One can avoid this procedure with help of the relation
\[\frac{d}{dt} = - (1 + z)H(z)\frac{d}{dz}.\]
We can, given $H(z)$, apply that relation to calculate $\dot H,\ddot H, \dddot H, \ddddot H $ and so on. Using the expressions for time derivatives of Hubble parameter obtained in the problem (\ref{equ61_6}), Chapter 2,
\[\begin{array}{l}
\dot H = - H^2(1 + q);\\
\ddot H = H^3\left( j + 3q + 2 \right);\\
\dddot H = H^4\left[ s - 4j - 3q(q + 4) - 6 \right];\\
\ddddot H = H^5\left[ l - 5s + 10\left( q + 2 \right)j + 30(q + 2)q + 24 \right]
\end{array}\]
one can express the current values of the cosmological parameters (for $z = 0$) in terms of $w_0,w_a:$
\[\begin{array}{l}
q_0 = \frac{1}{2} + \frac{3}{2}\left( 1 - \Omega _m \right)w_0;\\
j_0 = 1 + \frac{3}{2}\left( 1 - \Omega _m \right)\left[3w_0\left( 1 + w_0 \right) + w_a \right];\\
s_0 = - \frac{7}{2} - \frac{33}{4}\left( 1 - \Omega _m \right)w_a - \frac{9}{4}\left( 1 - \Omega _m \right)\left[ 9 + \left(7 - \Omega _m \right)w_a \right]w_0 - \\
- \frac{9}{4}\left( 1 - \Omega _m \right)\left( 16 - 3\Omega _m \right)w_0^2 - \\
- \frac{27}{4}\left( 1 - \Omega _m \right)\left(3 - \Omega _m \right)w_0^3;\\
l_0 = \frac{35}{2} + \frac{1 - \Omega _m}{4}\left[ 213 + (7 - \Omega _m)w_a \right]w_a + \\
+ \frac{1 - \Omega _m}{4}\left[ 489 + 9(82 - 21\Omega _m)w_a \right]w_0 + \\
+ \frac{9}{2}\left( 1 - \Omega _m \right)\left[ 67 - 21\Omega _m + \frac{3}{2}(23 - 11\Omega _m)w_a \right]w_0^2 + \\
+ \frac{{27}}{4}\left( {1 - {\Omega _m}} \right)(47 - 24{\Omega _m})w_0^3 + \\
+ \frac{81}{2}\left( 1 - \Omega _m \right)(3 - 2\Omega _m)w_0^4
\end{array}\]
</p>
</div>
</div></div>

<div id="SCM16"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 9 ===
Show that the results of the previous problem applied to SCM coincide with
the ones obtained in the [[#SCM_18|problem]].
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the SCM $\left( {{w_0},{w_a}} \right) = \left( { - 1,0} \right)$ and therefore
\[\begin{array}{l}
{q_0} = \frac{1}{2} - \frac{3}{2}\left( {1 - {\Omega _m}} \right);\\
{j_0} = 1;\\
{s_0} = 1 - \frac{9}{2}{\Omega _m};\\
{l_0} = 1 + 3{\Omega _m} + \frac{{27}}{2}\Omega _m^2
\end{array}\]
</p>
</div>
</div></div>

<div id="SCM10"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 10 ===
Photons with $z=0.1,\ 1,\ 100,\
1000$ are registered. What was the Universe age $t_U$ in the moment of
their emission? What period of time $t_{ph}$ were the photons on the
way? Plot $t_U(z)$ and $t_{ph}(z)$
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
t_U=\int_z^\infty\frac{dz}{(1 + z)H(z)} ;\; t_UH_0=\int_z^\infty\frac{dz}{(1 + z)\sqrt{\Omega _{m0}(1 + z)^3 + \Omega _{\Lambda 0} } }; \\
t_{ph}H_0=\int_0^z \frac{dz}{(1 + z)\sqrt{\Omega _{m0}(1 + z)^3 +\Omega _{\Lambda 0}}}\\
\end{gathered}
$$
At $$z = 0.1\,\,\, \to \,\,\,\,t_U H_0 = 0.899\,\,\,\,\,t_{ph} H_0 = 0.093$$
$$z = 1\,\, \to \,\,\,t_U H_0 = 0.431\,\,\,\,\,t_{ph} H_0 = 0.561$$
$$z = 100\,\,\, \to \,\,t_U H_0 = 1.264 \times 10^{ - 3} \,\,\,\,\,t_{ph} H_0 = 0.991$$
$$z = 1000\,\, \to \,\,\,t_U H_0 = 3.663 \times 10^{ - 5} \,\,\,\,\,t_{ph} H_0 = 0.993$$

<gallery widths=600px heights=500px>
File:12_10(1).JPG|
</gallery>

<gallery widths=600px heights=500px>
File:12_10(2).JPG|
</gallery>
</p>
</div>
</div></div>

<div id="SCM17"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 11 ===
Determine the present physical distance to the object that emitted light with current redshift $z$
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The physical distance to the object emitted a photon is related to the redshift $z$ by
$$
R(z) = \frac{1}{1 + z}\int_0^z\frac{dz'}{H(z')}
$$
In the SCM
$$
H =H_0\sqrt {\Omega _{r0}(1 + z)^4 + \Omega _{m0}(1 + z)^3+\Omega _{\Lambda 0}}
$$
The parameters ${H_0},\;\Omega _{r0},\Omega _{m0},\Omega _{\Lambda 0}$ are fixed by the model.
</p>
</div>
</div></div>

<div id="SCM_15"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 12 ===
A photon was emitted at time $t$ and registered at time $t_0$ with red
shift $z$. Find and plot the dependence of emission time on the redshift $t(z)$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
a = \frac{1}{1 + z}; \\
a(t) = A^{1/3}\sinh ^{2/3}\left(t/{t_\Lambda } \right);\quad A \equiv \frac{\Omega_{ m0}}{\Omega _{\Lambda 0}}; \\
\frac{t_0}{t_\Lambda }= Arth\sqrt {\Omega _{\Lambda 0}} ; \\
\frac{t}{t_0} = \frac{1}{Arth\sqrt{\Omega _{\Lambda 0}}}Arsh\left[\sqrt {\frac{\Omega _{\Lambda 0}}{\Omega _{m0}}} \frac{1}{\left(1 + z\right)^{3/2}} \right] \\
\end{gathered}
$$
<gallery widths=600px heights=500px>
File:12_13.JPG|
</gallery>
</p>
</div>
</div></div>

<div id="SCM18"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 13 ===
Find the time dependence for the scale factor and analyze its asymptotes. Plot $a(t)$
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the SCM
$$
\left(\frac{\dot a}{a}\right)^2 = H_0^2\left[ \Omega _{m0}\left(\frac{a_0}{a} \right)^3 + \Omega _{\Lambda 0} \right]
$$
Solution of the equation in the case $\Omega _{\Lambda 0} > 0$ reads
$$
\begin{gathered}
a(t) = a_0\left( \frac{\Omega _{m0}}{\Omega _{\Lambda 0}}\right)^{1/3}\left[ sh\left( \frac{3}{2}\sqrt {\Omega _{\Lambda 0}}H_0t \right) \right]^{2/3}; \\
a\left( t \ll H_0^{ - 1}\right) \propto t^{2/3};\; a\left( t \gg H_0^{ - 1} \right) \sim e^{H_0t} \\
\end{gathered}
$$
Below it is convenient to use the expression for time dependence of the scale factor in the following form
$$
a(t) = A^{1/3}\sinh ^{2/3}\left( t/{t_\Lambda }\right);\quad
A \equiv \frac{\Omega_{m0}}{\Omega _{\Lambda 0}} =
\frac{1 - \Omega _{\Lambda 0}}{\Omega _{\Lambda 0}};\quad t_\Lambda \equiv \frac{2}{3}\left(H_0 \right)^{-1}\left( \Omega _{\Lambda 0}\right)^{ - 1/2}
$$
</p>
</div>
</div></div>

<div id="SCM19"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 14 ===
Using the explicit solution for scale factor $a(t)$, obtained in the previous problem, find the cosmic horizon $R_h$ (see Chapter 3)
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
As we have seen above
\[a(t)=A\sinh^{2/3}{(t/t_\Lambda)},\]
where \[A=\frac{\Omega_{m0}}{\Omega_{\Lambda0}},\ t_\Lambda=\frac{2}{3H_\infty}.\]
Consider \[f(t)\equiv\ln{a(t)}=\ln A + \frac23\ln\sinh\frac{t}{t_\Lambda},\] then
\[\dot f=\frac{2}{3t_\Lambda\tanh(t/t_\Lambda)}=\frac{H_\infty}{\tanh(3tH_\infty/2)},\]
and finally
\[R_h=\frac{c}{\dot f}=\frac{c}{H_\infty}\tanh\left(\frac32tH_\infty\right).\]
</p>
</div>
</div></div>

<div id="SCM20"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 15 ===
Show that the Universe becomes $R_h$-delimited only after the cosmological constant starts to dominate.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
According to the result of the previous problem
\[\dot R_h=\frac32\left[1-\tanh^2\left(\frac32tH_\infty\right)\right]c,\]
so for early times $\dot R_h>c$ and an observer does not experience a divergent redshift with increasing $R$. Then after some ''transition'' time $t^*$, estimated by the condition
\[ct^*=R_h(t^*),\]
he (or she) will begin to encounter an observational limit at a finite radius $R_h$. Numerical solution of the equation gives $t^*\approx0.86H_\infty^{-1}$. This time is roughly the point at which the Universe transits from matter-dominated to $\Lambda$-dominated epoch. Eventually, the Universe becomes de Sitter one and therefore $\dot R_h\rightarrow0$ with $R_h$ settling at the fixed value $c/H_\infty$.
</p>
</div>
</div></div>

<div id="SCM21"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 16 ===
Analyze the stability of Friedmann equations.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Friedmann equations are given by
\[H^2=\frac{8\pi G}{3}\rho_m+\frac\Lambda3,\]
\[\dot H=-4\pi G\rho_m,\]
\[\dot\rho_m+3H\rho_m=0\]
Let us introduce the following dimensionless variables (recall that $\rho_m,\ \Lambda>0$):
\[x=\frac{\sqrt{8\pi G\rho_m}}{\sqrt3H},\ y=\frac{\sqrt\Lambda}{\sqrt3H}\]
Equations for $x$ and $y$ read (with $N\equiv\ln a$)
\[x'=\frac{dx}{dN}=\frac32x(x^2-1),\]
\[y'=\frac{dy}{dN}=\frac32yx^2.\]
The critical points, i.e. the solutions corresponding to $x'=0,\ y'=0$ and their eigenvalues are given by
\[x_{cr}=0,\ y_{cr}=1 (\Lambda-dominant\, case,)\ \mu_1=0, \mu_2=-\frac32;\]
\[x_{cr}=1,\ y_{cr}=0 (matter-dominant\, case,)\ \mu_1=3, \mu_2=\frac32.\]
From the eigenvalues we see that the matter dominant phase is unstable $\mu_{1,2}>0$ while the cosmological constant dominant phase has $\mu_2<0$ being stable. The existence of a zero eigenvalue $\mu_1=0$ in the first case originates from the fact that the two variables x and y are connected by the relation $x^2+y^2=1$. Therefore in this case one can reduce the dynamics to one-dimensional space.
</p>
</div>
</div></div>

<div id="SCM22"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 17 ===
Determine solutions for the perturbations $\delta\rho_m$ and $\delta H$. Make sure that the solutions are stable.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The variation of Friedmann equations yields
\[2H\delta H=\frac{\kappa^2}{3}\delta\rho_m,\]
\[\delta\dot H=-\frac{\kappa^2}{2}\delta\rho_m,\]
\[\delta\dot\rho_m+3\rho_m\delta H+ 3H\delta\rho_m=0,\]
where $\kappa^2\equiv8\pi G$.
Solution for the above equation is
\[\delta H=\frac{t_\Lambda}{4\kappa^2}C\tanh(\tau) e^{-f(\tau)},\ \delta\rho_m= \frac{C}{4\kappa^4}e^{-f(\tau)},\]
where \[\tau\equiv\frac{t}{t_\Lambda},\]
\[f(\tau)=-2\tau +\ln\left(-1+e^{4\tau}\right)+2\ln\tanh\tau.\]
Here $C$ is an arbitrary constant. We see that $f(\tau)$ approaches $2\tau$ as $t\rightarrow\infty$ and both $\delta H$ and $\delta\rho_m$ decay which implies that the considered solution is stable.
</p>
</div>
</div></div>

<div id="SCM23"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 18 ===
Rewrite the first Friedman equation in terms of conformal time.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[\frac{1}{a^4}\left( \frac{da}{d\eta } \right)^2=H_0^2\left[ \Omega_m a(\eta )^{-3}+\Omega_{\Lambda} \right]\]
</p>
</div>
</div></div>

<div id="SCM24"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 19 ===
Find relation between the scale factor and conformal time
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[\eta =\frac{1}{H_0}\int\limits_0^a\frac{dx}{\sqrt{x}{{\left( {{\Omega }_{m}}+{{\Omega }_{\Lambda }}{{x}^{3}} \right)}^{1/2}}}\]
</p>
</div>
</div></div>

<div id="SCM25"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 20 ===
Find explicit dependence of the scale factor on the conformal
time.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The integral obtained in the previous problem can be expressed in terms of elliptic integral of the first kind $F\left( \varphi ,k \right)$. For that purpose we rewrite it in the form
$$
\Omega^{1/6}_{\Lambda} \Omega^{1/3}_m H_{0} \eta = \int\limits_0^{u} \dfrac{dx}{\sqrt{x}\sqrt{1 + x^3}}.
$$
The upper limit equals to $u=\left(\dfrac{\Omega_{\Lambda}}{\Omega_m}\right)^{1/3}a.$ The integral can be presented in the form
$$3^{1/4} \Omega^{1/6}_{\Lambda} \Omega^{1/3}_m H_{0} \eta = F\left(\arccos\dfrac{1+(1-\sqrt{3})u}{1+(1+\sqrt{3})u}, \dfrac{\sqrt{2+\sqrt{3}}}{2}\right).$$

Here $y={{3}^{1/4}}\Omega _{\Lambda }^{1/6}\Omega _{m}^{1/3}{{H}_{0}}\eta ,\quad k=\frac{\sqrt{2+\sqrt{3}}}{2}\approx 0.97.$
</p>
</div>
</div></div>

<div id="SCM25"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 21 ===
Find relative density of dark energy $10^9$ years later.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
t = t_\Lambda Arcth\sqrt {\Omega _\Lambda} ;\quad \Omega _{\Lambda } = th^2\frac{t}{t_\Lambda };\\
t_\Lambda \simeq 10.768 \cdot 10^{9}\;\mbox{years}\;\;t = 14.7\cdot 10^{9}\;\mbox{years}\\
\Omega _{\Lambda 0} \simeq 0.77\\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM_17"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 22 ===
Find variation rate of relative density of dark energy. What are its asymptotic values? Plot its time dependence.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
\frac{d\Omega _\Lambda }{dt} = \frac{2}{t_\Lambda }\frac{sh(t/{t_\Lambda })}
{ch^3(t/{t_\Lambda })}; \\
x \equiv \frac{t}{t_\Lambda };\quad \frac{d\Omega _\Lambda }{dx} = 2\frac{sh(x)}{ch^3(x)} \\
\end{gathered}
$$

<gallery widths=600px heights=500px>
File:12_15.JPG|
</gallery>
</p>
</div>
</div></div>

<div id="SCM26"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 23 ===
Estimate size of the cosmological horizon.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The cosmological horizon is defined as
$$
L_{H0} = a_0\int\limits_0^{t_0}\frac{dt}{a\left( t \right)}.
$$
Using the results of the problem \ref{A_t_lambda}, one obtains
$$
L_{H0}= A^{-1/3}\int\limits_0^{t_0}\frac{dt}{sh^{2/3}\left( t/{t_\Lambda } \right)};
$$
</p>
</div>
</div></div>

<div id="SCM_19"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 24 ===
Find time dependence of Hubble parameter and plot it.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
H = \frac{\dot a}{a} =\frac{2}{3}\left(t_\Lambda \right)^{ - 1}cth\left( {\frac{t}{t_\Lambda }} \right)
$$

In the SCM ${t_0} \simeq 1.2t_\Lambda .$ Therefore
$$
H{t_0} \simeq 0.8cth\left( {1.2t/{t_0}} \right)
$$

<gallery widths=600px heights=500px>
File:12_16.JPG|Time dependence of the Hubble parameter (in units of the age of the Universe).
</gallery>

</p>
</div>
</div></div>

<div id="SCM27"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 25 ===
Find time dependence of dark matter density.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Present the first Friedmann equation in the form
\[\rho_m=\frac{1}{8\pi G}(3h^2-\Lambda)\]
and make use of results of the previous problem to obtain
\[\rho_m=\frac{\Lambda}{8\pi G}\frac{1}{\sinh^2(t/t_\Lambda)}.\]

\item {\bf Find the asymptotic (in time) value of the Hubble parameter.}

The required asymptote can be obtained in two ways: first, from the time dependence of the Hubble parameter (see the previous problem), and second, immediately from the first Friedmann equation in the SCM. Taking into account that $\rho_{m} \to 0$ at $t\to \infty$, one gets
$$
H_{\infty} = \frac{2}{3t_\Lambda} = \sqrt {\frac{\Lambda}{3}}.
$$
</p>
</div>
</div></div>

<div id="SCM28"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 26 ===
At present the age of the Universe $t_0\simeq13.7\cdot10^9$ years is close to the Hubble time $t_H=H_0^{-1}\simeq14\cdot10^9$ years.
Does the relation $t^*\simeq t_H(t^*)=H^{-1}(t^*)$ for the age $t^*$
of the Universe hold for any moment of its evolution
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
This relation cannot be satisfied in arbitrary moment of time, as the Hubble parameter has limit value in the SCM (see the previous problem):
$$
\frac{t^*}{t_H(t^*)} = H\left( t^*\right)t^* = \frac{2}{3}\frac{Arcth\sqrt {\Omega _\Lambda(t^*)}}{\sqrt{\Omega_\Lambda (t^*)}}.
$$
For $t\to \infty,\;\Omega _\Lambda\left( t^* \right) \to 1,$ and $$\frac{t^*}{t_H(t^*)} \to \infty. $$

\item {\bf Determine values of state parameter for the dark energy, that provide accelerated expansion of Universe in the present time.}

$$
\begin{gathered}
\sum\limits_i (\rho _{i0} + 3p_{i0}) < 0\; \Rightarrow \rho _{m0} + \rho _{DE0} + 3w_{DE}\rho _{DE0} < 0; \\
w_{DE} < - \frac{1}{3}\left( 1 + \frac{\Omega _{m0}}{\Omega _{DE0}}\right); \\
w_{DE} < - \frac{1}{3}\Omega _{DE0}^{-1};\; w_{DE} < - 0.46\\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM29"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 27 ===
Find current value of the deceleration parameter.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For a multi-component system the deceleration parameter equals to
$$
q = \frac{\Omega}{2} + \frac{3}{2}\sum\limits_i {w_i}\Omega _i
$$
In the SCM $\Omega = 1,w_m = 0,\;w_\Lambda= - 1,$ and
$$
\begin{gathered}
q = \frac{1 - 3\Omega_{\Lambda 0}}{2};\\
q_0 \simeq - 0.6 \\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM_24"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 28 ===
Find the redshift dependence of the deceleration parameter. Analyze the limiting cases.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
q \equiv - \frac{a\ddot a}{\dot a^2} = - \frac{\ddot a}{a}\frac{1}{H^2}; \\
\frac{\ddot a}{a} = - \frac{4\pi G}{3}\sum\limits_i \left( \rho _i + 3p_i \right) = - \frac{1}{2}H_0^2\sum\limits_i \Omega _{i0}\left(1 + 3w_i\right)(1 + z)^{3\left(1 + w_i\right)} ; \\
H^2 = H_0^2\sum\limits_i \Omega _{i0}(1 + z)^{3\left( 1 + w_i \right)}; \\
q = \frac{3}{2}\frac{\sum\limits_i \Omega _{i0}\left( 1 + w_i \right)(1 + z)^{3\left( 1 + w_i \right)}}
{\sum\limits_i \Omega _{i0}(1 + z)^{3\left( 1 + w_i\right)} } - 1 \\
\end{gathered}
$$
In the SCM
$$
q = \frac{1}{2}\frac{\Omega _{m0}(1 + z)^3- 2\Omega _{\Lambda 0} }
{\Omega _{m0}(1 + z)^3+\Omega_{\Lambda 0}}
$$

$$
q(z \to \infty ) = \frac{1}{2},\;q(z \to - 1) = - 1
$$

<gallery widths=600px heights=500px>
File:12_24.JPG|Dependence of the deceleration parameter on the redshift.
</gallery>
</p>
</div>
</div></div>

<div id="SCM_25"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 29 ===
Find and plot the time dependence of the deceleration parameter.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
q = - \frac{\ddot aa}
{\dot a^2} = - \frac{\ddot a}
{aH^2}; \\
a(t) = A^{1/3}sh^{2/3}\left(t/{t_\Lambda}\right);\quad A\equiv \frac{\Omega_ {m0}}
{\Omega _{\Lambda 0}}; \\
\frac{t_0}
{t_\Lambda } = Arth\sqrt {\Omega _{\Lambda 0}} ;\quad H = \frac{2}
{3}\frac{1}
{t_\Lambda }cth\left( \frac{t}
{t_\Lambda }\right); \\
q = \frac{1}
{2}\left[ 1 - 3th^2\left( \frac{t}
{t_\Lambda }\right)\right] \\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM29"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 30 ===
Find the moment of time when the dark energy started to dominate over dark matter. What redshift did it correspond to?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
\left( \frac{\dot a}{a} \right)^2 = \frac{8\pi G}
{3}\rho _{cr}\left( \Omega _{m0}(1 + z)^3+\Omega_{\Lambda 0}\right); \\
\Omega _{m0}(1 + z^*)^3 =\Omega _{\Lambda 0};\\
z^* = \left(\frac{\Omega _{\Lambda 0}}
{\Omega _{m0}}\right)^{1/3}-1; \\
\Omega _{\Lambda 0}\simeq 0.73;\quad \Omega _{m0}\simeq 0.27; \\
z^* \simeq 0.39; \\
\Omega _\Lambda = th^2\left(\frac{t}
{t_\Lambda }\right);\quad \Omega _m = 1 - \Omega _\Lambda = ch^{-2}\left( \frac{t}
{t_\Lambda } \right); \\
\Omega _\Lambda(\tilde t) = \Omega _m(\tilde t) \to \tilde t = \frac{Arsh(1)}{Arth(\sqrt {\Omega _{\Lambda 0} })}t_0 \simeq 9.5\cdot10^9 \,years \\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM30"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 31 ===
Determine the moment of time and redshift value corresponding to the transition from decelerated
expansion of the Universe to the accelerated one.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
\frac{\ddot a}
{a} = \frac{4\pi G}
{3}\rho_{cr}\left[2\Omega_{\Lambda 0} - \Omega _{m0}(1 + z)^3\right]; \\
2\Omega _{\Lambda 0} = \Omega _{m0}(1 + z^*)^3; \\
z^* = \left( \frac{2\Omega_{\Lambda 0}}
{\Omega _{m0}}\right)^{1/3} - 1; \\
\Omega _{\Lambda 0}\simeq 0.73;\quad \Omega _{m0}\simeq 0.27; \\
z^* \simeq 0.745 \\
\end{gathered}
$$

Let us now determine the time moment $t^*$ of the transition
$$
\begin{gathered}
a^* = \frac{1}
{1 + z^*} = \left(\frac{\Omega _{m0}}
{2\Omega _{\Lambda 0}} \right)^{1/3}; \\
t(a) = \frac{2}{3\sqrt{\Omega_{\Lambda 0}}H_0}Arsh\left(\sqrt{ \frac{\Omega_{\Lambda 0}}
{\Omega_{m0}}}a^{3/2} \right); \\
t^* = \frac{2}
{3\sqrt {\Omega _{\Lambda 0}}}H_0Arsh\left(\sqrt {1/2} \right) \\
\end{gathered}
$$
There is an alternative way to solve the problem.
In the multi-component Universe with the corresponding state equations $p_i = w_i\rho _i$ the deceleration parameter $q \equiv - \frac{\ddot aa}{\dot a^2}$ equals to
$$
q = \frac{\Omega }
{2} + \frac{3}
{2}\sum\limits_i w_i\Omega _i
$$
In the case of the SCM $\Omega =1$ and
$$
q = \frac{1}
{2} - \frac{3}
{2}\frac{\Omega _{\Lambda 0}}
{\left[\left(1 + z\right)^3\Omega _{m0}+\Omega_{\Lambda 0}\right]}
$$

From the condition $q=0$ it follows that
$$
z^* =\left(\frac{2\Omega_{\Lambda 0}}
{\Omega _{m0}} \right)^{1/3}- 1
$$
For $\Omega _{\Lambda 0} = 0.73,\;\Omega _{m0} = 0.27$ one obtains that
$$
z^* = 0.745
$$
</p>
</div>
</div></div>

<div id="SCM31"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 32 ===
Solve the previous problem using the derivative $d\eta/d\ln a$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\frac{d\eta}
{d\ln a} = a\frac{dt}{ada} = \frac{1}{aH} = \frac{1}{\dot a}
$$

If the quantity $aH = \dot a$ grows, then $\ddot a > 0,$ which corresponds to the accelerated expansion (inflation) of the Universe. In the SCM the quantity
$$
\frac{aH}{H_0} \simeq \sqrt{ \frac{0.27 + 0.73a^3}{a}}
$$
starts to grow at $a=0.573,$ which corresponds to $z=0.745$
</p>
</div>
</div></div>

<div id="SCM32"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 33 ===
Is dark energy domination necessary for transition to the accelerated expansion?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
No. The transition to the accelerated expansion takes place at $(z=0.75)$, which is considerably earlier than the transition to the dark energy domination stage at $(z=0.45).$
</p>
</div>
</div></div>

<div id="SCM33"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 34 ===
Consider flat Universe composed of matter and dark energy in form of cosmological constant. Find the redshift value corresponding to equality of densities of the both components $\rho_m(z_{eq})=\rho_\Lambda(z_{eq})$ and the one corresponding to beginning of the accelerated expansion $q\left(z_{accel} \right) = 0$. Obtain relation between $z_{eq}$ and $z_{accel}$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
z_{eq} = \left(\frac{\Omega _\Lambda}
{\Omega _{m0}} \right)^{1/3}- 1; \\
z_{accel}= \left(\frac{2\Omega_\Lambda }
{\Omega _{m0}}\right)^{1/3}- 1; \\
z_{accel} = 2^{1/3}\left(z_{eq}+ 1\right) - 1 \\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM34"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 35 ===
Show that density perturbations stop to grow after the transition from dust to $\Lambda$-dominated era.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
On the $\Lambda$-dominated stage the non-relativistic matter density perturbation size depends on conformal time as
\begin{equation}\label{1SCM30}
a(\eta)=-\frac{1}{H\eta}, \,\, \eta=-const\cdotp e^{-Ht},\,\,\eta<0
\end{equation}
where the Hubble parameter $H$ is constant, $H^2=(8\pi/3)G\rho_\Lambda$.
The linearized Einstein equations for the case of single-component medium take the following form in the momentum representation:
\begin{equation}\label{3SCM30}
k^2\Phi+3\frac{a^{'}}{a}+3\frac{a^{'2}}{a^2}\Phi=-4\pi Ga^2\delta\rho,
\end{equation}
\begin{equation}\label{2SCM30}
\Phi^{''}+3\frac{a^{'}}{a}\Phi^{'}+\left(2\frac{a^{''}}{a} - \frac{a^{'2}}{a^2}\right)\Phi=4\pi Ga^2\delta p.
\end{equation}
Assuming that the $\Lambda$-term is constant (which is exactly true for the vacuum energy density) one obtains $\delta\rho_\Lambda=\delta p_\Lambda=0$.
Using \eqref{2SCM30} and taking into account \eqref{1SCM30} we come up with the equation: $${\Phi}^{''}-\frac{3}{\eta}{\Phi}^{'}-\frac{3}{\eta^2}\Phi=0,$$
Solutions of the equation behave as
$$\Phi\propto\eta\propto\frac 1a$$ and $$\Phi\propto\eta^3\propto\frac{1}{a^3}.$$
Use now \eqref{3SCM30} for the modes beyond the horizon and take into account that for dust density $\rho_M\propto a^{-3}.$ Keeping only the first solution for the gravitational potential (the second decays too fast), one can see that the density perturbations do not grow: for the modes under horizon the relation
$$\delta_M\propto a^3\frac{k^2}{a^2}\Phi=const$$ holds.

Therefore the perturbations stop to grow at the transition from the dust-dominated to the $\Lambda$-dominated stage. If the present accelerated expansion of the Universe is actually caused by non-zero cosmological constant, then the structure formation from small primary perturbations is eventually close to end and it will never resume again: the structures larger than the presently observed ones will never appear.
</p>
</div>
</div></div>

<div id="SCM35"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 36 ===
What happens to the velocity fluctuations of non-relativistic matter and radiation with respect to Hubble flow in the epoch of cosmological constant domination?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM36"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 37 ===
Find the ratio of baryon to non-baryon components in the galactic halo.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM37"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 38 ===
Imagine that in the Universe described by SCM the dark energy was instantly switched off. Analyze further dynamics of the Universe.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="local-group"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 39 ===
Estimate density of dark energy in form of cosmological constant using the Hubble diagram for the neighborhood of the Local group.
<gallery widths=600px heights=500px>
File:im_eps.JPG|Hubble diagram for the neighborhood of the Local group.
</gallery>
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
An elegant and simple method was recently proposed (see A.D.Chernin, UFN 178, 267 (2008)), which allows not only to become convinced in the existence of the dark energy but also to estimate its density. The method is based on the assumption, that the dark energy (in form of the cosmological constant) governs the Hubble flow even on comparably small distances $\left( \sim \,Mpc \right)$. Consider the resulting force $\vec{F}$ (per unit mass), acting on a galaxy from the picture \ref{hubble_diagram}. The force is a sum of two contributions:
$$
\vec{F}=\vec{F}_N+\vec{F}_\Lambda
$$
Here $\vec{F}_N$ is the force produced by the total mass of the Local group, and $F_{\Lambda }$ is the force created by the dark energy. For a crude estimate let us neglect the non-spherical mass distribution in the Local group. Then the force acting on a satellite galaxy in a given point $R$ outside the local Group equals to
$$
F=-\frac{GM_{LG}}{R^2}+\frac{8\pi G}{3}{\rho_{\Lambda }}R,
$$
where $M_{LG}\simeq (1\div 3) \times 10^{12}M_{\odot }$ is the total mass of the Local group. As can be seen from the picture, the transition from domination of the local group gravity to the vacuum domination takes place on the distance of order $R_{0}=1.8\div 2\; \, Mpc.$ It allows to make a rough estimate of the dark energy density from the condition $F\left( R=R_0 \right)=0.$ The estimate results in the following
$$
(0.1\pm 0.03)<\rho_V<(1\pm 0.3)\times 10^{-26} \, kg/m^3.
$$
The proximity of the obtained crude estimate to the actual value of the dark energy density in the SCM is striking.
$$
\rho_{V}\simeq (0.75\pm 0.05)\times 10^{-26}\, kg/m^3.
$$
</p>
</div>
</div></div>

<div id="SCM38"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 40 ===
Estimate the Local group mass by methods used in the previous problem.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM39"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 41 ===
Find "weak" points in the argumentation of the two preceding problems.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM40"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 42 ===
The product of the age of the Universe and current Hubble parameter (the Hubble's constant) is a very important test (Sandage consistency test) of internal consistency for any model of Universe. Analyze the
parameters on which the product $H_0 t_0$ depends in the Big Bang model and in the SCM.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM41"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 43 ===
Show that for a fixed source of radiation the luminosity distance for high redshift values in flat Universe is greater for the dark energy dominated case compared to the non-relativistic matter dominated one.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the case of Universe filled by dark energy and non-relativistic matter, the photometric distance is equal to
$$
d_L = \frac{1 + z}
{H_0}\int_0^z \frac{dz'}
{\sqrt{\Omega _{m0}\left(1 + z'\right)^3+ \Omega _{DE0}\left(1 + z' \right)^{3\left( 1 + w_{DE}\right)}}}
$$

In the case $\Omega _{m0} = 1,\;\Omega _{DE0} = 0$
one obtains
$$
d_L = \frac{2}
{H_0}\left[1 + z -(1 + z)^{1/2}\right]
$$
and in the case $\Omega _{m0} = 0,\;\Omega _{DE0} = 1$
$$
d_L = \frac{1}{H_0}z(1 + z)
$$
instead.

In means that the supernova appear dimmer in the dark energy dominated Universe.
</p>
</div>
</div></div>

<div id="SCM42"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 44 ===
In the observations that discovered the accelerated expansion of the Universe the researchers in particular detected two $Ia$ type supernovae: $1992P$, $z=0.026$, $m=16.08$ and $1997ap$, $z=0.83$,
$m=24.32$. Show that these observed parameters are in accordance with the SCM.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The apparent $m$ and absolute $M$ stellar magnitudes are related with the photometric distance $d_L$ by the expression
\begin{equation}\label{p(1)}
m - M = 5\log _{10}\left(\frac{d_L}
{\mbox{Mpc}} \right) + 25
\end{equation}
The photometric distance in the SCM is equal to
\begin{equation}\label{p(2)}
d_L = \frac{1 + z}
{H_0}\int_0^z \frac{dz'}
{\sqrt {\Omega _{m0}(1 + z)^3 + \Omega _\Lambda}}
\end{equation}
For small $z$ $d_L(z) \simeq z/H_0.$ Using the supernova $1992P$ with small $z=0.026$ one can find $M = - 19.09.$ Assuming that the $1a$-type supernova are standard candles (they have equal absolute magnitudes), one obtains from the relation \ref{p(1)} supernova with $z=0.83$
$$H_0d_L\simeq 1.16$$
From the other hand it follows from the relation \ref{p(1)} that in the case
$\Omega _{m0} = 1,\;\Omega_\Lambda = 0$
$$
H_0d_L\simeq 0.95
$$
In the SCM one gets much better agreement:
$$
H_0d_L \simeq 1.2
$$
</p>
</div>
</div></div>

<div id="SCM43"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 45 ===
Find the redshift value, at which a source of linear dimension $d$ has minimum visible angular
size.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The apparent angular size equals to $$\delta \theta = {{(1 + z)^2 d } \over {d_L (z)}},$$
where $d_L (z)$ is the photometric distance from a source with the redshift $z$. In the case of cosmological constant domination the photometric distance is:
$$d_L (z) = - {z \over {H_0 \Omega _\Lambda ^{1/2} }},$$
In the SCM (see the previous problem)
$$
d_L = \frac{1 + z}
{H_0}\int_0^z \frac{dz'}
{\sqrt {\Omega _{m0}(1 + z)^3 + \Omega _\Lambda}}
$$
then the angular size is
$$\delta \theta = - {{H_0 \Omega _\Lambda ^{1/2} d} \over 2}{{(1 + z)^2 } \over z},$$
and $z = - 1$. It corresponds to infinitely far future.
</p>
</div>
</div></div>

<div id="SCM44"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 46 ===
Compare the observed value of the dark energy density with the
one expected from the dimensionality considerations (the cosmological constant problem).
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The dimensionality considerations suggest the following value for density of the dark energy in form of the cosmological constant
\[\rho_\Lambda^{(dim)}\sim M_{Pl}^4\approx10^{76}GeV^4.\]
At the same time the ''observed'' value fixed by the SCM is
\[\rho_\Lambda^{(obs)}\approx10^{-48}GeV^4.\]
This huge disagreement
\[\frac{\rho_\Lambda^{(dim)}}{\rho_\Lambda^{(obs)}}>10^{120}\]
is called the ''cosmological constant problem''. It presents a fundamental problem for elementary particles theory and physics in general, rather than for cosmology itself. (Look for details in R.Bousso, [http://arxiv.org/abs/0708.4231 TASI Lectures on the cosmological constant])
</p>
</div>
</div></div>

<div id="SCM45"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 47 ===
Determine the density of vacuum energy using the Planck scale as cutoff parameter.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For a quantum field with a given mode of frequency $\omega$, the zero-point energy is $\pm\omega/2$ with the plus sign for a bosonic field and the minus sign for a fermionic field. The total zero-point energy is then $\pm\sum\limits_i\omega_i/2$. In the continuum limit, we have for a free field (bosonic or fermionic)
\[\frac12\sum\limits_i\omega_i=\frac12\int\frac{d^3xd^3k}{(2\pi)^3}(k^2+m^2)^{1/2}=V\int\frac{k^2dk}{4\pi^2}(k^2+m^2)^{1/2}.\] This integral diverges, thus we need to introduce the physical cutoff. In cosmological context this cutoff must be the Planck mass at largest. With the cutoff $k_{max}\sim M_{Pl}$ one obtains
\[\rho_\Lambda=\int_0^{k_{max}}\frac{k^2dk}{4\pi^2}(k^2+m^2)^{1/2} \approx\frac{k_{max}^4}{16\pi^2}=\frac{M_{Pl}^4}{16\pi^2} \approx10^{74}GeV^4.\]
This result only slightly modifies the dimensional considerations given in the previous problem.
</p>
</div>
</div></div>

<div id="SCM46"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 48 ===
Identifying the vacuum fluctuations density with the observable dark energy value in SCM, find the required frequency cutoff magnitude in the fluctuation spectrum.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[\nu_{max}=10^{12}Hz.\]
</p>
</div>
</div></div>

<div id="SCM47"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 49 ===
What purely cosmological problem originates from the divergence of the zero-point energy density?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
If the zero-point energy density is infinite, then our Universe is infinitely curved and the space size is infinitely small.
</p>
</div>
</div></div>

<div id="SCM48"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 50 ===
With exact supersymmetry, the bosonic contribution to cosmological constant is canceled by its fermionic counterpart. However, we know that our world looks like not supersymmetric. Supersymmetry, if exists, has to be broken above or around $100GeV$ scale. Compare the observed value of the dark energy density with the one expected from broken supersymmetry.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
If we take the $k_{max}^{(SUSY)}\sim100GeV$ cutoff, then
\[\rho_\Lambda^{(SUSY)}=\frac{\left(k_{max}^{(SUSY)}\right)^4}{16\pi^2}\approx10^6GeV^4.\]
\[\frac{\rho_\Lambda^{(SUSY)}}{\rho_\Lambda^{(obs)}}>10^{54}.\]
Thus supersymmetry does not solve the problem of the cosmological constant.
</p>
</div>
</div></div>

<div id="SCM49"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 51 ===
Determine duration of the inflation period.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Let us see what happens with the relative density during all the history of Universe. From the first Friedmann equation it follows that in the matter-dominated epoch $\left| \Omega - 1 \right| \sim t^{2/3},$ and in the radiation-dominated epoch $\left| \Omega - 1 \right| \sim t^{2/3}.$ Then assume that inflation started at $t_i,$ and finished at $t_f.$ Then in the period of $t_f < t < t_{eq} = 50000$ years the radiation dominated and the range $t_{eq} < t < t_0 $ corresponds to the matter domination epoch. Then the presently observed difference
$$
\left|\Omega (t_0 ) - 1 \right| = \left| \Omega (t_i ) - 1 \right|e^{ - 2H(t_f - t_i )} \left( \frac{t_{eq} }
{t_f } \right)\left( \frac{t_0}
{t_{eq} }\right)^{2/3}
$$
According the SCM $\left|\Omega \left( t_0\right) - 1 \right| < 0.04.$ If one assumes that before the beginning of the inflation there was $\left|\Omega \left( t_i \right) - 1 \right| \sim 1,$ then
$$
N > \frac{1}{2}\ln \left[\frac{\Omega \left( t_i \right) - 1}{\Omega \left(t_0\right) - 1}\left( \frac{t_{eq}}{t_f } \right)\left( \frac{t_0}{t_{eq}} \right)^{2/3} \right] \approx 60
$$

Assume that radiation dominated before the inflation started. Then $H = \frac{1}{2t}.$ If at the moment of time $t_i$ the vacuum dark energy started to dominate and then the expansion continued with constant rate, then $H\approx 1/t_i$ during the inflation. And finally one obtains that
$$
N = H\left(t_f - t_i\right) \approx \frac{t_f - t_i }
{t_i }
$$
it is commonly assumed that the inflation finishing time is of order of the Grand Unification time $t_f\approx 10^{-35}.$ Therefore $t_f\approx 10^{-38}.$ Evidently the inflation could last longer as we used only lower estimate for $N.$
</p>
</div>
</div></div>

<div id="SCM50"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 52 ===
Plot the dependence of luminosity distance $d_L$ (in units of $H_0^{-1}$) on the redshift $z$ for the two-component flat Universe with non-relativistic liquid ($w=0$) and cosmological constant
($w=-1$). Consider the following cases:

a) $\Omega_\Lambda^0=0$;

b) $\Omega_\Lambda^0=0.3$;

c) $\Omega_\Lambda^0=0.7$;

d) $\Omega_\Lambda^0=1$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM51"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 53 ===
For a Universe filled by dark energy with state equation $p_{DE}=w_{DE}\rho_{DE}$ and non-relativistic matter obtain the Taylor series for $d_L$ in terms of $z$ near the observation point
$z_0=0$. Explain the obtained result.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the universe filled by dark energy with the state equation $p_{DE}=w_{DE}\rho_{DE}$ and non-relativistic dark matter the photometric distance takes on the form:
\begin{equation}
\label{d_L}
d_L = \frac{1 + z}
{H_0}\int_0^z \frac{dz'}
{\sqrt{\Omega _{m0}\left(1 + z'\right)^3+ \Omega _{DE0}\left(1 + z' \right)^{3\left( 1 + w_{DE}\right)}}}
\end{equation}
Expand the integrand in Taylor series in $z$ in vicinity of the observation point $z_0=0$ to obtain
$$
\begin{gathered}
\label{tey}
\frac{1}{\sqrt{\Omega _{m0}\left(1 + z'\right)^3+ \Omega _{DE0}\left(1 + z' \right)^{3\left( 1 + w_{DE}\right)}}}\simeq \\
\simeq 1 -\frac{3}{2}\left(1+w_{DE}\Omega _{DE0}\right)z+\frac{3}{4}\left(3w_{DE}\Omega _{DE0}-1\right)z^2.\nonumber
\end{gathered}
$$
Substitution of the decomposition (\ref{tey}) into the integral (\ref{d_L}) results in the following:
\begin{equation}
\label{d_L1}
d_L = \frac{1 + z}
{H_0}\left(z -\frac{3}{4}\left(1+w_{DE}\Omega _{DE0}\right)z^2+\frac{1}{4}\left(3w_{DE}\Omega _{DE0}-1\right)z^3\right).
\end{equation}
</p>
</div>
</div></div>

<div id="SCM_45"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 54 ===
Determine position of the first acoustic peak in the CMB power spectrum produced by baryon oscillations on the surface of the last scattering.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Early Universe was filled by photon-baryon plasma, which can be treated as a single-component liquid. The baryons were trapped by the potential walls generated by density fluctuations, and they were eventually compressed. The contraction lead to heating of the plasma and therefore to increasing radiative pressure of the photons which is directed outside. Finally the radiative pressure stop the compression and lead to expansion of the plasma. During the expansion the plasma cools down and its radiative pressure decreases. The gravity starts to dominate again leading to repeated compression. The concurrence between the gravity and the pressure lead to longitudinal (acoustic) oscillations in the photon-baryon fluid. When the matter and the radiation get decoupled in the recombination process, the picture of the acoustic oscillations remains frozen into the CMB. Today we detect the evidence of the primordial acoustic waves (dense and diluted regions) in form of the primary anisotropy in the CMB.

It is well known that any acoustic wave, whatever complicatedly shaped, can be represented in form of superposition of modes with different wave numbers $k$, $k\propto 1/\lambda $. Every mode $\lambda $ corresponds to certain angular scale $\theta $ on the sky. Therefore in order to facilitate the comparison of the theory with observations, one should use the angular (multipole) decomposition in terms of the Legendre polynomials $P_{l} (\cos \theta )$, instead of the Fourier transform in terms of sines and cosines. Order of the polynomial $l$ plays the same role as the index $k$ in the Fourier decomposition. For $l\ge 2$ the Legendre polynomials are oscillating functions in the interval $\left[1,-1\right]$. Number of the oscillations increases with growth of $l$. Therefore
\begin{equation} \label{acust1_}
l\propto \frac{1}{\theta }
\end{equation}
The temperature fluctuations can be observed and qualitatively analyzed with help of paired measurements in the directions $\stackrel{\frown}{n},\stackrel{\frown}{n}'$, so that $\stackrel{\frown}{n}\cdot \stackrel{\frown}{n}'=\cos \theta $. Taking average over all the pairs in the assumptions of Gaussian fluctuations, one obtains the two-point correlator $C(\theta )$, which can be presented in terms of the multipole decomposition
\begin{equation} \label{acust2_}
\left\langle \delta T(\stackrel{\frown}{n})\cdot \delta T(\stackrel{\frown}{n}')\right\rangle \equiv C(\theta )=\sum _{l}\frac{2l+1}{4\pi } C_{l} P_{l} (\cos \theta )
\end{equation}
where $C_{l} $ are the coefficients.

As we already mentioned, the analysis of the temperature fluctuations enables us to clarify the structure of the longitudinal oscillations. The mode with the maximum wavelength corresponds to the maximum angular size of the primary anisotropy. This fundamental mode was detected the first. There are reliable evidences of the fact that the second and the third modes were also already detected. The distance $r_{s} $, passed by the acoustic wave during the time period before the recombination, is called the sound horizon. The sound horizon is fixed by (or rather fixes) the physical scale o the last scattering surface. Size of the sound horizon depends on values of the physical parameters. Distance to the last scattering surface $d_{sls} $ depends on the cosmological parameters too. Together they determine the angular size of the sound horizon$\theta _{s} $ (see Fig.)

<gallery widths=600px heights=500px>
File:acustic_pic.JPG|Dependence of the deceleration parameter on the redshift.
</gallery>

\begin{equation} \label{acust3_}
\theta _{s} \approx \frac{r_{s} }{d_{sls} }
\end{equation}
Analysis of the temperature anisotropy allows to determine $\theta _{s} $. Varying the cosmological parameters in $r_{s} ,d_{sls} ,$ one achieves the best agreement with the observations and thus fixes the cosmological parameters.

We can estimate the sound horizon size as the length passed by sound from the moment $t=0$ to the recombination time $t_{*} $
\begin{equation} \label{acust4_}
r_{s} (z_{*} ;\Omega _{b} ,\Omega _{r} )\approx \int _{0}^{t_{*} }c_{s} dt
\end{equation}
where $z_{*} $ is the redshift of the recombination moment $\left(z_{*} \approx 1100\right)$, $c_{s} $ is the sound speed in the photon-baryon medium
\begin{equation} \label{acust5_}
c_{s} \approx c\left[3\left(1+\frac{3\Omega _{b} }{\Omega _{r} } \right)\right]^{-1/2}
\end{equation}
To determine $dt$ one should use the first Friedmann equation in the form
\begin{equation} \label{acust6_}
\left(\frac{da}{dt} \right)^{2} =H_{0}^{2} \left[\Omega _{r0} \left(\frac{a_{0} }{a} \right)^{2} +\Omega _{m0} \frac{a_{0} }{a} +\Omega _{k0} +\Omega _{\Lambda 0} \left(\frac{a}{a_{0} } \right)^{2} \right]
\end{equation}
Making use of the relations $\frac{a_{0} }{a} =1+z$ and $\Omega _{r} +\Omega _{m} +\Omega _{\Lambda } +\Omega _{k} =1$, one obtains
\begin{equation} \label{acust7_}
dt=H_{0}^{-1} (1+z)^{-1} \left\{\left(1+z\right)^{2} \left(1+\Omega _{m,0} z\right)+z(z+2)\left[\left(1+z)^{2} \Omega _{r.0} -\Omega _{\Lambda ,0} \right)\right]\right\}^{-1/2} dz
\end{equation}
Distance to the last scattering surface, which corresponds to its angular size, is determined by the quantity called the angular diameter distance. It is connected by the photometric distance $d$ by the simple relation
\begin{equation} \label{acust8_}
d_{sls} =\frac{d}{(1+z_{*} )^{2} }
\end{equation}
The position of the first acoustic peak is determined by
\begin{equation} \label{acust9_}
l\approx \frac{d_{sls} }{r_{s} }.
\end{equation}
Let us now predict the position of the first peak for the case of flat Universe. To the leading order the sound speed is
\[c_{s} =\frac{c}{\sqrt{3} }. \]
Let us assume also that in the recombination epoch the Universe was matter-dominated. Under such assumptions one gets
\begin{equation} \label{acust10_}
r_{s} =\frac{c_{s} }{H_{0} \sqrt{\Omega _{m} } } \int _{z_{*} }^{\infty }(1+z)^{-5/2} dz =r_{s} =\frac{2c_{s} }{3H_{0} \sqrt{\Omega _{m} } } \left(1+z_{*} \right)^{-3/2}
\end{equation}
Using the standard procedure (see section 4), we obtain for the comoving radial coordinate $r_{sls} $ of the last scattering surface the following
\begin{equation} \label{acust11_}
r_{lsr} =\frac{c}{H_{0} } \int _{0}^{z_{*} }\left[\Omega _{m} (1+z)^{^{3} } +\Omega _{\Lambda } \right] ^{-1/2} dz
\end{equation}
Replacing the integrand by its binomial decomposition
\[\Omega _{m}^{-1/2} (1+z)^{-3/2} -\left(\Omega _{\Lambda } /2\Omega _{m}^{3/2} \right)\left(1+z\right)^{-9/2} \]
and taking into account that
\[d_{sls} =r_{sls} /(1+z_{*} ),\]
we obtain
\begin{equation} \label{acust12_}
d_{sls} =\frac{2c}{7H_{0} (1+z_{*} )} \left\{7\Omega _{m}^{-1/2} -2\Omega _{\Lambda } \Omega _{m}^{-3/2} +{\rm O} \left[\left(1+z_{*} \right)^{-1/2} \right]\right\}.
\end{equation}
Using the relation $\Omega _{\Lambda } =1-\Omega _{m} $ and neglecting the higher order terms, one obtains
\begin{equation} \label{acust13_}
d_{sls} \approx \frac{2c\Omega _{m}^{-1/2} }{7H_{0} (1+z_{*} )} \left(9-2\Omega _{m}^{3} \right)
\end{equation}
Combination of the relations \eqref{acust10_} and \eqref{acust13_} gives for the first acoustic peak the following
\begin{equation} \label{acust14_}
l\approx \frac{d_{sls} }{r_{s} } \approx 0.74\sqrt{1+z_{*} } \left(9-2\Omega _{m}^{3} \right)\simeq 220.
\end{equation}
This result agrees very well with numerous observations.
</p>
</div>
</div></div>

<div id="SCM52"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 55 ===
\bf Compare the asymptotes of time dependence of the scale factor
$a(t)$ for the SCM and de Sitter models. Explain physical reasons of
their distinction.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM53"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 56 ===
Redshift for any object slowly changes due to the acceleration (or deceleration) of the Universe expansion. Estimate change of velocity in one year.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

$$
\Delta z = \dot z \Delta t = H_0 \left(1 + z - {H(z)\over H_0} \right)\Delta t
$$

$$H(z)=H_0 \left[\Omega_{m0}(1+z)^3+\Omega_{\Lambda 0}\right]^{1/2}$$
$$\Omega_{m0}+\Omega_{\Lambda 0}=1,~\Omega_{m0}\approx 0.3$$
$$\Delta z \approx 2\cdot 10^{-11}$$
$$\Delta v = c{\Delta z\over 1+z}\approx 0.25~\mbox{cm/s}$$

</p>
</div>
</div></div>

<div id="SCM54"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 57 ===
Determine the lower limit of ratio of the total volume of the Universe to the observed one?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
If the Universe is the 3-sphere, then its radius is $R_0=a_0$
$$
R_0 = \frac{1}{H_0 \sqrt {\left| \Omega _{curv} \right|} }
$$
For the SCM parameters the cosmological horizon equals to
$$
L_{p,0} = a_0 \int_0^{t_0 } \frac{dt}{a(t)} = \frac{3.6}{H_0}
$$
We assumed that $a(t)\sim t^{2/3}.$ Therefore
$$
\frac{R_0}
{L_{p,0} } = \frac{1}
{3.6\sqrt {\left| \Omega _{curv}\right|} }
$$
Taking into account that $\left|\Omega _{curv} \right| < 0.02,$ one obtains $\frac{a_0}{L_{p0} } > 2$.
Number of the regions similar to the one inside our horizon equals to the ratio of the 3-sphere volume $2\pi ^2 a_0^3 $ to the volume of the observed Universe
$$
N = \frac{2\pi ^2 a_0^3}
{(4\pi /3)L_{p0}^3 } \simeq 4.7\left(\frac{a_0 }
{L_{p0} } \right)^3 > 38
$$
Thus we can see not more than $3 \%$ of the Universe volume.
</p>
</div>
</div></div>

<div id="SCM55"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 58 ===
What is the difference between the inflationary expansion in the early Universe and the present accelerated expansion?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The difference is in the fact that in the present time there is about $27 \%$ of non-relativistic matter which was absent in the inflation period. The numerical values of Hubble parameter in the beginning of the inflation was very different from the one at the beginning of the current period of accelerated expansion of Universe.
</p>
</div>
</div></div>

<div id="SCM56"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 59 ===
Compare the values of Hubble parameter at the beginning of the
inflation period and at the beginning of the present accelerated
expansion of Universe.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
h^2 (x) = \Omega _{m0} x^3 + (1 - \Omega _{m0} )x^\alpha ;\quad \alpha = 3(1 + w)
$$
Assume for simplicity that the Universe is spatially flat. From the definition $O(x)$
it follows that
$$
O(x) = \Omega _{m0} + \left( 1 - \Omega _{m0}\right)\frac{x^\alpha - 1}
{x^3 - 1}
$$
The required result follows from the fact that $\alpha =0$ for cosmological constant, $\alpha >0$ for quintessence and $\alpha <0$ for phantom energy.
</p>
</div>
</div></div>

<div id="SCM57"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 60 ===
Plot the dependencies $h(x)=H(x)/H_0(x)$, where $x=1+z$, in the SCM, for quintessence and for phantom energy cases.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Plots of the dependencies $
h(x) = H(x)/H_0 (x);\quad x = 1 + z
$ in the SCM are presented on Fig. \ref{fig:52} for phantom energy and quintessence.
<gallery widths=600px heights=500px>
File:12_52.JPG|Dependence of the deceleration parameter on the redshift.
</gallery>
</p>
</div>
</div></div>

<div id="SCM58"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 61 ===
Find the constraints imposed by the Weak Energy Condition (WEC) for the dark energy on the redshift-dependent Hubble parameter. ([http://arxiv.org/abs/astro-ph/0703416 Inspired by A.Sen, R.Scherrer])
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The redshift-dependent Hubble parameter $H(z)$ is defined by
\[H^2(z)=\frac{8\pi G}{3}(\rho_m+\rho_{DE})\]
This equation can be rewritten as
\begin{equation}\label{SCM_eq_76_1}
E^2(z)\equiv\frac{H^2(z)}{H^2_0(z)}=\Omega_{m0}(1+z)^3+(1-\Omega_{m0})\frac{\rho_{DE}}{\rho_{DE0}}
\end{equation}
The WEC forces \[\frac{\rho_{DE}}{\rho_{DE0}}\ge1 \ \left(\frac{d\rho}{dz}\ge0\right)\] for $z>1$, so equation (\ref{SCM_eq_76_1}) becomes
\begin{equation}\label{SCM_eq_76_2}
E^2(z)\ge\Omega_{m0}(1+z)^3+(1-\Omega_{m0})
\end{equation}
Going back to equation (\ref{SCM_eq_76_2}) and taking the derivative with respect to $z$ (which we denote throughout with a prime) gives:
\[
2E(z)\frac{dE}{dz}=3\Omega_{m0}(1+z)^2+\frac{1-\Omega_{m0}}{\rho_{DE0}}\frac{d\rho_{DE}}{dz}.
\]
Now the WEC implies that \[\frac{d\rho_{DE}}{dz}\ge0,\] so we get
\begin{equation}\label{SCM_eq_76_3}
E(z)\frac{dE}{dz}\ge\frac32\Omega_{m0}(1+z)^2.
\end{equation}
Equations (\ref{SCM_eq_76_2}) and (\ref{SCM_eq_76_3}) give the constraints that the WEC for the dark energy places on the redshift-dependent Hubble parameter.
</p>
</div>
</div></div>

<div id="SCM59"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 62 ===
Using the statefinders, show that the power law cosmology mimics SCM model at (see Chapters 3 and 9)
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Recall that the statefinders are defined as
\[r\equiv\frac{\dddot a}{aH^3},\ s\equiv\frac{r-1}{3(q-1/2)}, \ q\ne\frac12.\]
The SCM model corresponds to the point $r=1,\, s=0$ in the $(s,r)$ plane. In power-law cosmology (see Chapter 3) the statefinders are given by
\[r=2q^2+q,\, s=\frac23(q+1).\]
We note that $r=1,\, s=0$ for $q=-1$. Thus the power law cosmology mimics SCM model with $q=-1$.
</p>
</div>
</div></div>

Evolution of Universe

2012-10-04T17:56:02Z

Den: /* Problem 40 */

[[Category:Standard Cosmological Model|2]]

__NOTOC__
<div id="SCM-10"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 1 ===
Rewrite the first Friedman equation in terms of redshift and analyze the contributions of separate components at different stages of Universe evolution.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM11"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2 ===
Find the time dependence for the scale factor and analyze asymptotes of the dependence. Plot $a(t)$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM12"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 3 ===
Determine the redshift value corresponding to equality of radiation and matter densities.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
\rho _r(t) = \rho _m(t);\\
\rho _{r0}\frac{a_0^4}{a^4} = \rho _{m0}\frac{a_0^3}{a^3};\\
\frac{\rho _{m0}}{\rho _{r0}} = \frac{\Omega _{m0}}{\Omega _{r0}} = \frac{a_0}{a} = 1 + z;\\
\Omega _{r0} = 5 \cdot {10^{ - 5}} \\
z = \frac{\Omega _{m0}}{\Omega _{r0}} - 1 \approx 5400\\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM_9"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 4 ===
Construct effective one-dimensional potential (see [[Solutions_of_Friedman_equations_in_the_Big_Bang_model#dyn15|problem]] of Chapter 3)
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
<div id="V(x)"></div>
$$
V(x) = - \frac{1}{2}\sum\limits_i \Omega _{i0}x^{ - (1 + 3w_i)};\quad x \equiv \frac{a}{a_0}
$$
In the SCM
$$
V(x) = - \frac{1}{2}\Omega _{m0}x^{-1}-\frac{1}{2}\Omega _{\Lambda 0}x^2 \simeq - 0.135x^{-1}-0.365x^2
$$

<gallery widths=600px heights=500px>
File:12_9.JPG|
</gallery>
</p>
</div>
</div></div>

<div id="SCM13"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 5 ===
Show that the following holds: $\dot H= -4\pi G\rho_m$ and $\ddot H= 12\pi G\rho_m H$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM_18"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 6 ===
Expand the scale factor in Taylor series near the time moment
$t_0$:
\[\frac{a(t)}{a(t_0)}=1+\sum\limits_{n=1}^\infty\frac{A_n(t_0)}{n!}[H_0(t-t_0)]^n;\quad A_n\equiv\frac{1}{aH^n}\frac{d^na}{dt^n}\]
and calculate values for few first coefficients $A_n$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The coefficients $A_1$ and $A_2$ equal to:
$$A_1 = \frac{\dot a}{aH} = 1;\quad A_2 = \frac{\ddot a}{aH^2} = - q = 1 - \frac{3}{2}{\Omega _m}$$
The parameters $A_n(n > 2)$ can be calculated by consequent differentiation of the relation
$$\ddot a = a\left( \dot H + H^2 \right)$$
The time derivatives of the scale factor can be determined making use of the fact that in SCM $\dot H = - 4\pi G\rho _m$, and $\rho _m = \rho _0a^{ - 3}$. For example,
$$A_3 = 1 + \frac{\ddot H}{H^3} + 3\frac{\dot H}{H^2}$$
Using $\ddot H = 12\pi G{\rho _m}H,\;\dot H = - 4\pi G\rho _m$, one finds that
$$A_3 = 1$$
Let us present the expressions for some consequent expansion coefficients
\[\begin{array}{l}
{A_4} = 1 - \frac{3^2}{2}\Omega _m;\\
{A_5} = 1 + 3\Omega _m + \frac{3^3}{2}\Omega _m^2;\\
{A_6} = 1 - \frac{3^3}{2}\Omega _m - 3^4\Omega _m^2 - \frac{3^4}{4}\Omega _m^3
\end{array}\]
</p>
</div>
</div></div>

<div id="SCM14"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 7 ===
Show that all the coefficients $A_n$ can be expressed through
elementary functions of the deceleration parameter $q$ or the
density parameter \[\Omega_m=\frac23(1+q).\]
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM15"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 8 ===
Consider the case of flat Universe filled by non-relativistic matter and dark
energy with state equation $p_X=w\rho_X$, where the state parameter $w$ is parameterized as the following \[w=w_0+ w_a(1-a)=w_0 + w_a \frac{z}{1+z}.\]
Express current values of cosmographic parameters through $w_0$ and
$w_a$. % (See [http://arxiv.org/abs/0802.1583 Cosmography of f(R) gravity])
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the considered case
\[\frac{H^2(z)}{H_0^2} = \Omega _m(1 + z)^3 + \Omega _X(1 + z)^{3\left( 1 + w_0 + w_a \right)}e^{ - \frac{3w_az}{1 + z}}\]
Calculation of the cosmographic parameters requires in general to integrate $H(z)$ in order to obtain $a(t).$ One can avoid this procedure with help of the relation
\[\frac{d}{dt} = - (1 + z)H(z)\frac{d}{dz}.\]
We can, given $H(z)$, apply that relation to calculate $\dot H,\ddot H, \dddot H, \ddddot H $ and so on. Using the expressions for time derivatives of Hubble parameter obtained in the problem (\ref{equ61_6}), Chapter 2,
\[\begin{array}{l}
\dot H = - H^2(1 + q);\\
\ddot H = H^3\left( j + 3q + 2 \right);\\
\dddot H = H^4\left[ s - 4j - 3q(q + 4) - 6 \right];\\
\ddddot H = H^5\left[ l - 5s + 10\left( q + 2 \right)j + 30(q + 2)q + 24 \right]
\end{array}\]
one can express the current values of the cosmological parameters (for $z = 0$) in terms of $w_0,w_a:$
\[\begin{array}{l}
q_0 = \frac{1}{2} + \frac{3}{2}\left( 1 - \Omega _m \right)w_0;\\
j_0 = 1 + \frac{3}{2}\left( 1 - \Omega _m \right)\left[3w_0\left( 1 + w_0 \right) + w_a \right];\\
s_0 = - \frac{7}{2} - \frac{33}{4}\left( 1 - \Omega _m \right)w_a - \frac{9}{4}\left( 1 - \Omega _m \right)\left[ 9 + \left(7 - \Omega _m \right)w_a \right]w_0 - \\
- \frac{9}{4}\left( 1 - \Omega _m \right)\left( 16 - 3\Omega _m \right)w_0^2 - \\
- \frac{27}{4}\left( 1 - \Omega _m \right)\left(3 - \Omega _m \right)w_0^3;\\
l_0 = \frac{35}{2} + \frac{1 - \Omega _m}{4}\left[ 213 + (7 - \Omega _m)w_a \right]w_a + \\
+ \frac{1 - \Omega _m}{4}\left[ 489 + 9(82 - 21\Omega _m)w_a \right]w_0 + \\
+ \frac{9}{2}\left( 1 - \Omega _m \right)\left[ 67 - 21\Omega _m + \frac{3}{2}(23 - 11\Omega _m)w_a \right]w_0^2 + \\
+ \frac{{27}}{4}\left( {1 - {\Omega _m}} \right)(47 - 24{\Omega _m})w_0^3 + \\
+ \frac{81}{2}\left( 1 - \Omega _m \right)(3 - 2\Omega _m)w_0^4
\end{array}\]
</p>
</div>
</div></div>

<div id="SCM16"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 9 ===
Show that the results of the previous problem applied to SCM coincide with
the ones obtained in the [[#SCM_18|problem]].
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the SCM $\left( {{w_0},{w_a}} \right) = \left( { - 1,0} \right)$ and therefore
\[\begin{array}{l}
{q_0} = \frac{1}{2} - \frac{3}{2}\left( {1 - {\Omega _m}} \right);\\
{j_0} = 1;\\
{s_0} = 1 - \frac{9}{2}{\Omega _m};\\
{l_0} = 1 + 3{\Omega _m} + \frac{{27}}{2}\Omega _m^2
\end{array}\]
</p>
</div>
</div></div>

<div id="SCM10"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 10 ===
Photons with $z=0.1,\ 1,\ 100,\
1000$ are registered. What was the Universe age $t_U$ in the moment of
their emission? What period of time $t_{ph}$ were the photons on the
way? Plot $t_U(z)$ and $t_{ph}(z)$
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
t_U=\int_z^\infty\frac{dz}{(1 + z)H(z)} ;\; t_UH_0=\int_z^\infty\frac{dz}{(1 + z)\sqrt{\Omega _{m0}(1 + z)^3 + \Omega _{\Lambda 0} } }; \\
t_{ph}H_0=\int_0^z \frac{dz}{(1 + z)\sqrt{\Omega _{m0}(1 + z)^3 +\Omega _{\Lambda 0}}}\\
\end{gathered}
$$
At $$z = 0.1\,\,\, \to \,\,\,\,t_U H_0 = 0.899\,\,\,\,\,t_{ph} H_0 = 0.093$$
$$z = 1\,\, \to \,\,\,t_U H_0 = 0.431\,\,\,\,\,t_{ph} H_0 = 0.561$$
$$z = 100\,\,\, \to \,\,t_U H_0 = 1.264 \times 10^{ - 3} \,\,\,\,\,t_{ph} H_0 = 0.991$$
$$z = 1000\,\, \to \,\,\,t_U H_0 = 3.663 \times 10^{ - 5} \,\,\,\,\,t_{ph} H_0 = 0.993$$

<gallery widths=600px heights=500px>
File:12_10(1).JPG|
</gallery>

<gallery widths=600px heights=500px>
File:12_10(2).JPG|
</gallery>
</p>
</div>
</div></div>

<div id="SCM17"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 11 ===
Determine the present physical distance to the object that emitted light with current redshift $z$
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The physical distance to the object emitted a photon is related to the redshift $z$ by
$$
R(z) = \frac{1}{1 + z}\int_0^z\frac{dz'}{H(z')}
$$
In the SCM
$$
H =H_0\sqrt {\Omega _{r0}(1 + z)^4 + \Omega _{m0}(1 + z)^3+\Omega _{\Lambda 0}}
$$
The parameters ${H_0},\;\Omega _{r0},\Omega _{m0},\Omega _{\Lambda 0}$ are fixed by the model.
</p>
</div>
</div></div>

<div id="SCM_15"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 12 ===
A photon was emitted at time $t$ and registered at time $t_0$ with red
shift $z$. Find and plot the dependence of emission time on the redshift $t(z)$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
a = \frac{1}{1 + z}; \\
a(t) = A^{1/3}\sinh ^{2/3}\left(t/{t_\Lambda } \right);\quad A \equiv \frac{\Omega_{ m0}}{\Omega _{\Lambda 0}}; \\
\frac{t_0}{t_\Lambda }= Arth\sqrt {\Omega _{\Lambda 0}} ; \\
\frac{t}{t_0} = \frac{1}{Arth\sqrt{\Omega _{\Lambda 0}}}Arsh\left[\sqrt {\frac{\Omega _{\Lambda 0}}{\Omega _{m0}}} \frac{1}{\left(1 + z\right)^{3/2}} \right] \\
\end{gathered}
$$
<gallery widths=600px heights=500px>
File:12_13.JPG|
</gallery>
</p>
</div>
</div></div>

<div id="SCM18"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 13 ===
Find the time dependence for the scale factor and analyze its asymptotes. Plot $a(t)$
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the SCM
$$
\left(\frac{\dot a}{a}\right)^2 = H_0^2\left[ \Omega _{m0}\left(\frac{a_0}{a} \right)^3 + \Omega _{\Lambda 0} \right]
$$
Solution of the equation in the case $\Omega _{\Lambda 0} > 0$ reads
$$
\begin{gathered}
a(t) = a_0\left( \frac{\Omega _{m0}}{\Omega _{\Lambda 0}}\right)^{1/3}\left[ sh\left( \frac{3}{2}\sqrt {\Omega _{\Lambda 0}}H_0t \right) \right]^{2/3}; \\
a\left( t \ll H_0^{ - 1}\right) \propto t^{2/3};\; a\left( t \gg H_0^{ - 1} \right) \sim e^{H_0t} \\
\end{gathered}
$$
Below it is convenient to use the expression for time dependence of the scale factor in the following form
$$
a(t) = A^{1/3}\sinh ^{2/3}\left( t/{t_\Lambda }\right);\quad
A \equiv \frac{\Omega_{m0}}{\Omega _{\Lambda 0}} =
\frac{1 - \Omega _{\Lambda 0}}{\Omega _{\Lambda 0}};\quad t_\Lambda \equiv \frac{2}{3}\left(H_0 \right)^{-1}\left( \Omega _{\Lambda 0}\right)^{ - 1/2}
$$
</p>
</div>
</div></div>

<div id="SCM19"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 14 ===
Using the explicit solution for scale factor $a(t)$, obtained in the previous problem, find the cosmic horizon $R_h$ (see Chapter 3)
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
As we have seen above
\[a(t)=A\sinh^{2/3}{(t/t_\Lambda)},\]
where \[A=\frac{\Omega_{m0}}{\Omega_{\Lambda0}},\ t_\Lambda=\frac{2}{3H_\infty}.\]
Consider \[f(t)\equiv\ln{a(t)}=\ln A + \frac23\ln\sinh\frac{t}{t_\Lambda},\] then
\[\dot f=\frac{2}{3t_\Lambda\tanh(t/t_\Lambda)}=\frac{H_\infty}{\tanh(3tH_\infty/2)},\]
and finally
\[R_h=\frac{c}{\dot f}=\frac{c}{H_\infty}\tanh\left(\frac32tH_\infty\right).\]
</p>
</div>
</div></div>

<div id="SCM20"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 15 ===
Show that the Universe becomes $R_h$-delimited only after the cosmological constant starts to dominate.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
According to the result of the previous problem
\[\dot R_h=\frac32\left[1-\tanh^2\left(\frac32tH_\infty\right)\right]c,\]
so for early times $\dot R_h>c$ and an observer does not experience a divergent redshift with increasing $R$. Then after some ''transition'' time $t^*$, estimated by the condition
\[ct^*=R_h(t^*),\]
he (or she) will begin to encounter an observational limit at a finite radius $R_h$. Numerical solution of the equation gives $t^*\approx0.86H_\infty^{-1}$. This time is roughly the point at which the Universe transits from matter-dominated to $\Lambda$-dominated epoch. Eventually, the Universe becomes de Sitter one and therefore $\dot R_h\rightarrow0$ with $R_h$ settling at the fixed value $c/H_\infty$.
</p>
</div>
</div></div>

<div id="SCM21"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 16 ===
Analyze the stability of Friedmann equations.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Friedmann equations are given by
\[H^2=\frac{8\pi G}{3}\rho_m+\frac\Lambda3,\]
\[\dot H=-4\pi G\rho_m,\]
\[\dot\rho_m+3H\rho_m=0\]
Let us introduce the following dimensionless variables (recall that $\rho_m,\ \Lambda>0$):
\[x=\frac{\sqrt{8\pi G\rho_m}}{\sqrt3H},\ y=\frac{\sqrt\Lambda}{\sqrt3H}\]
Equations for $x$ and $y$ read (with $N\equiv\ln a$)
\[x'=\frac{dx}{dN}=\frac32x(x^2-1),\]
\[y'=\frac{dy}{dN}=\frac32yx^2.\]
The critical points, i.e. the solutions corresponding to $x'=0,\ y'=0$ and their eigenvalues are given by
\[x_{cr}=0,\ y_{cr}=1 (\Lambda-dominant\, case,)\ \mu_1=0, \mu_2=-\frac32;\]
\[x_{cr}=1,\ y_{cr}=0 (matter-dominant\, case,)\ \mu_1=3, \mu_2=\frac32.\]
From the eigenvalues we see that the matter dominant phase is unstable $\mu_{1,2}>0$ while the cosmological constant dominant phase has $\mu_2<0$ being stable. The existence of a zero eigenvalue $\mu_1=0$ in the first case originates from the fact that the two variables x and y are connected by the relation $x^2+y^2=1$. Therefore in this case one can reduce the dynamics to one-dimensional space.
</p>
</div>
</div></div>

<div id="SCM22"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 17 ===
Determine solutions for the perturbations $\delta\rho_m$ and $\delta H$. Make sure that the solutions are stable.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The variation of Friedmann equations yields
\[2H\delta H=\frac{\kappa^2}{3}\delta\rho_m,\]
\[\delta\dot H=-\frac{\kappa^2}{2}\delta\rho_m,\]
\[\delta\dot\rho_m+3\rho_m\delta H+ 3H\delta\rho_m=0,\]
where $\kappa^2\equiv8\pi G$.
Solution for the above equation is
\[\delta H=\frac{t_\Lambda}{4\kappa^2}C\tanh(\tau) e^{-f(\tau)},\ \delta\rho_m= \frac{C}{4\kappa^4}e^{-f(\tau)},\]
where \[\tau\equiv\frac{t}{t_\Lambda},\]
\[f(\tau)=-2\tau +\ln\left(-1+e^{4\tau}\right)+2\ln\tanh\tau.\]
Here $C$ is an arbitrary constant. We see that $f(\tau)$ approaches $2\tau$ as $t\rightarrow\infty$ and both $\delta H$ and $\delta\rho_m$ decay which implies that the considered solution is stable.
</p>
</div>
</div></div>

<div id="SCM23"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 18 ===
Rewrite the first Friedman equation in terms of conformal time.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[\frac{1}{a^4}\left( \frac{da}{d\eta } \right)^2=H_0^2\left[ \Omega_m a(\eta )^{-3}+\Omega_{\Lambda} \right]\]
</p>
</div>
</div></div>

<div id="SCM24"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 19 ===
Find relation between the scale factor and conformal time
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[\eta =\frac{1}{H_0}\int\limits_0^a\frac{dx}{\sqrt{x}{{\left( {{\Omega }_{m}}+{{\Omega }_{\Lambda }}{{x}^{3}} \right)}^{1/2}}}\]
</p>
</div>
</div></div>

<div id="SCM25"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 20 ===
Find explicit dependence of the scale factor on the conformal
time.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The integral obtained in the previous problem can be expressed in terms of elliptic integral of the first kind $F\left( \varphi ,k \right)$. For that purpose we rewrite it in the form
$$
\Omega^{1/6}_{\Lambda} \Omega^{1/3}_m H_{0} \eta = \int\limits_0^{u} \dfrac{dx}{\sqrt{x}\sqrt{1 + x^3}}.
$$
The upper limit equals to $u=\left(\dfrac{\Omega_{\Lambda}}{\Omega_m}\right)^{1/3}a.$ The integral can be presented in the form
$$3^{1/4} \Omega^{1/6}_{\Lambda} \Omega^{1/3}_m H_{0} \eta = F\left(\arccos\dfrac{1+(1-\sqrt{3})u}{1+(1+\sqrt{3})u}, \dfrac{\sqrt{2+\sqrt{3}}}{2}\right).$$

Here $y={{3}^{1/4}}\Omega _{\Lambda }^{1/6}\Omega _{m}^{1/3}{{H}_{0}}\eta ,\quad k=\frac{\sqrt{2+\sqrt{3}}}{2}\approx 0.97.$
</p>
</div>
</div></div>

<div id="SCM25"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 21 ===
Find relative density of dark energy $10^9$ years later.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
t = t_\Lambda Arcth\sqrt {\Omega _\Lambda} ;\quad \Omega _{\Lambda } = th^2\frac{t}{t_\Lambda };\\
t_\Lambda \simeq 10.768 \cdot 10^{9}\;\mbox{years}\;\;t = 14.7\cdot 10^{9}\;\mbox{years}\\
\Omega _{\Lambda 0} \simeq 0.77\\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM_17"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 22 ===
Find variation rate of relative density of dark energy. What are its asymptotic values? Plot its time dependence.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
\frac{d\Omega _\Lambda }{dt} = \frac{2}{t_\Lambda }\frac{sh(t/{t_\Lambda })}
{ch^3(t/{t_\Lambda })}; \\
x \equiv \frac{t}{t_\Lambda };\quad \frac{d\Omega _\Lambda }{dx} = 2\frac{sh(x)}{ch^3(x)} \\
\end{gathered}
$$

<gallery widths=600px heights=500px>
File:12_15.JPG|
</gallery>
</p>
</div>
</div></div>

<div id="SCM26"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 23 ===
Estimate size of the cosmological horizon.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The cosmological horizon is defined as
$$
L_{H0} = a_0\int\limits_0^{t_0}\frac{dt}{a\left( t \right)}.
$$
Using the results of the problem \ref{A_t_lambda}, one obtains
$$
L_{H0}= A^{-1/3}\int\limits_0^{t_0}\frac{dt}{sh^{2/3}\left( t/{t_\Lambda } \right)};
$$
</p>
</div>
</div></div>

<div id="SCM_19"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 24 ===
Find time dependence of Hubble parameter and plot it.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
H = \frac{\dot a}{a} =\frac{2}{3}\left(t_\Lambda \right)^{ - 1}cth\left( {\frac{t}{t_\Lambda }} \right)
$$

In the SCM ${t_0} \simeq 1.2t_\Lambda .$ Therefore
$$
H{t_0} \simeq 0.8cth\left( {1.2t/{t_0}} \right)
$$

<gallery widths=600px heights=500px>
File:12_16.JPG|Time dependence of the Hubble parameter (in units of the age of the Universe).
</gallery>

</p>
</div>
</div></div>

<div id="SCM27"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 25 ===
Find time dependence of dark matter density.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Present the first Friedmann equation in the form
\[\rho_m=\frac{1}{8\pi G}(3h^2-\Lambda)\]
and make use of results of the previous problem to obtain
\[\rho_m=\frac{\Lambda}{8\pi G}\frac{1}{\sinh^2(t/t_\Lambda)}.\]

\item {\bf Find the asymptotic (in time) value of the Hubble parameter.}

The required asymptote can be obtained in two ways: first, from the time dependence of the Hubble parameter (see the previous problem), and second, immediately from the first Friedmann equation in the SCM. Taking into account that $\rho_{m} \to 0$ at $t\to \infty$, one gets
$$
H_{\infty} = \frac{2}{3t_\Lambda} = \sqrt {\frac{\Lambda}{3}}.
$$
</p>
</div>
</div></div>

<div id="SCM28"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 26 ===
At present the age of the Universe $t_0\simeq13.7\cdot10^9$ years is close to the Hubble time $t_H=H_0^{-1}\simeq14\cdot10^9$ years.
Does the relation $t^*\simeq t_H(t^*)=H^{-1}(t^*)$ for the age $t^*$
of the Universe hold for any moment of its evolution
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
This relation cannot be satisfied in arbitrary moment of time, as the Hubble parameter has limit value in the SCM (see the previous problem):
$$
\frac{t^*}{t_H(t^*)} = H\left( t^*\right)t^* = \frac{2}{3}\frac{Arcth\sqrt {\Omega _\Lambda(t^*)}}{\sqrt{\Omega_\Lambda (t^*)}}.
$$
For $t\to \infty,\;\Omega _\Lambda\left( t^* \right) \to 1,$ and $$\frac{t^*}{t_H(t^*)} \to \infty. $$

\item {\bf Determine values of state parameter for the dark energy, that provide accelerated expansion of Universe in the present time.}

$$
\begin{gathered}
\sum\limits_i (\rho _{i0} + 3p_{i0}) < 0\; \Rightarrow \rho _{m0} + \rho _{DE0} + 3w_{DE}\rho _{DE0} < 0; \\
w_{DE} < - \frac{1}{3}\left( 1 + \frac{\Omega _{m0}}{\Omega _{DE0}}\right); \\
w_{DE} < - \frac{1}{3}\Omega _{DE0}^{-1};\; w_{DE} < - 0.46\\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM29"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 27 ===
Find current value of the deceleration parameter.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For a multi-component system the deceleration parameter equals to
$$
q = \frac{\Omega}{2} + \frac{3}{2}\sum\limits_i {w_i}\Omega _i
$$
In the SCM $\Omega = 1,w_m = 0,\;w_\Lambda= - 1,$ and
$$
\begin{gathered}
q = \frac{1 - 3\Omega_{\Lambda 0}}{2};\\
q_0 \simeq - 0.6 \\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM_24"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 28 ===
Find the redshift dependence of the deceleration parameter. Analyze the limiting cases.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
q \equiv - \frac{a\ddot a}{\dot a^2} = - \frac{\ddot a}{a}\frac{1}{H^2}; \\
\frac{\ddot a}{a} = - \frac{4\pi G}{3}\sum\limits_i \left( \rho _i + 3p_i \right) = - \frac{1}{2}H_0^2\sum\limits_i \Omega _{i0}\left(1 + 3w_i\right)(1 + z)^{3\left(1 + w_i\right)} ; \\
H^2 = H_0^2\sum\limits_i \Omega _{i0}(1 + z)^{3\left( 1 + w_i \right)}; \\
q = \frac{3}{2}\frac{\sum\limits_i \Omega _{i0}\left( 1 + w_i \right)(1 + z)^{3\left( 1 + w_i \right)}}
{\sum\limits_i \Omega _{i0}(1 + z)^{3\left( 1 + w_i\right)} } - 1 \\
\end{gathered}
$$
In the SCM
$$
q = \frac{1}{2}\frac{\Omega _{m0}(1 + z)^3- 2\Omega _{\Lambda 0} }
{\Omega _{m0}(1 + z)^3+\Omega_{\Lambda 0}}
$$

$$
q(z \to \infty ) = \frac{1}{2},\;q(z \to - 1) = - 1
$$

<gallery widths=600px heights=500px>
File:12_24.JPG|Dependence of the deceleration parameter on the redshift.
</gallery>
</p>
</div>
</div></div>

<div id="SCM_25"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 29 ===
Find and plot the time dependence of the deceleration parameter.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
q = - \frac{\ddot aa}
{\dot a^2} = - \frac{\ddot a}
{aH^2}; \\
a(t) = A^{1/3}sh^{2/3}\left(t/{t_\Lambda}\right);\quad A\equiv \frac{\Omega_ {m0}}
{\Omega _{\Lambda 0}}; \\
\frac{t_0}
{t_\Lambda } = Arth\sqrt {\Omega _{\Lambda 0}} ;\quad H = \frac{2}
{3}\frac{1}
{t_\Lambda }cth\left( \frac{t}
{t_\Lambda }\right); \\
q = \frac{1}
{2}\left[ 1 - 3th^2\left( \frac{t}
{t_\Lambda }\right)\right] \\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM29"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 30 ===
Find the moment of time when the dark energy started to dominate over dark matter. What redshift did it correspond to?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
\left( \frac{\dot a}{a} \right)^2 = \frac{8\pi G}
{3}\rho _{cr}\left( \Omega _{m0}(1 + z)^3+\Omega_{\Lambda 0}\right); \\
\Omega _{m0}(1 + z^*)^3 =\Omega _{\Lambda 0};\\
z^* = \left(\frac{\Omega _{\Lambda 0}}
{\Omega _{m0}}\right)^{1/3}-1; \\
\Omega _{\Lambda 0}\simeq 0.73;\quad \Omega _{m0}\simeq 0.27; \\
z^* \simeq 0.39; \\
\Omega _\Lambda = th^2\left(\frac{t}
{t_\Lambda }\right);\quad \Omega _m = 1 - \Omega _\Lambda = ch^{-2}\left( \frac{t}
{t_\Lambda } \right); \\
\Omega _\Lambda(\tilde t) = \Omega _m(\tilde t) \to \tilde t = \frac{Arsh(1)}{Arth(\sqrt {\Omega _{\Lambda 0} })}t_0 \simeq 9.5\cdot10^9 \,years \\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM30"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 31 ===
Determine the moment of time and redshift value corresponding to the transition from decelerated
expansion of the Universe to the accelerated one.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
\frac{\ddot a}
{a} = \frac{4\pi G}
{3}\rho_{cr}\left[2\Omega_{\Lambda 0} - \Omega _{m0}(1 + z)^3\right]; \\
2\Omega _{\Lambda 0} = \Omega _{m0}(1 + z^*)^3; \\
z^* = \left( \frac{2\Omega_{\Lambda 0}}
{\Omega _{m0}}\right)^{1/3} - 1; \\
\Omega _{\Lambda 0}\simeq 0.73;\quad \Omega _{m0}\simeq 0.27; \\
z^* \simeq 0.745 \\
\end{gathered}
$$

Let us now determine the time moment $t^*$ of the transition
$$
\begin{gathered}
a^* = \frac{1}
{1 + z^*} = \left(\frac{\Omega _{m0}}
{2\Omega _{\Lambda 0}} \right)^{1/3}; \\
t(a) = \frac{2}{3\sqrt{\Omega_{\Lambda 0}}H_0}Arsh\left(\sqrt{ \frac{\Omega_{\Lambda 0}}
{\Omega_{m0}}}a^{3/2} \right); \\
t^* = \frac{2}
{3\sqrt {\Omega _{\Lambda 0}}}H_0Arsh\left(\sqrt {1/2} \right) \\
\end{gathered}
$$
There is an alternative way to solve the problem.
In the multi-component Universe with the corresponding state equations $p_i = w_i\rho _i$ the deceleration parameter $q \equiv - \frac{\ddot aa}{\dot a^2}$ equals to
$$
q = \frac{\Omega }
{2} + \frac{3}
{2}\sum\limits_i w_i\Omega _i
$$
In the case of the SCM $\Omega =1$ and
$$
q = \frac{1}
{2} - \frac{3}
{2}\frac{\Omega _{\Lambda 0}}
{\left[\left(1 + z\right)^3\Omega _{m0}+\Omega_{\Lambda 0}\right]}
$$

From the condition $q=0$ it follows that
$$
z^* =\left(\frac{2\Omega_{\Lambda 0}}
{\Omega _{m0}} \right)^{1/3}- 1
$$
For $\Omega _{\Lambda 0} = 0.73,\;\Omega _{m0} = 0.27$ one obtains that
$$
z^* = 0.745
$$
</p>
</div>
</div></div>

<div id="SCM31"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 32 ===
Solve the previous problem using the derivative $d\eta/d\ln a$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\frac{d\eta}
{d\ln a} = a\frac{dt}{ada} = \frac{1}{aH} = \frac{1}{\dot a}
$$

If the quantity $aH = \dot a$ grows, then $\ddot a > 0,$ which corresponds to the accelerated expansion (inflation) of the Universe. In the SCM the quantity
$$
\frac{aH}{H_0} \simeq \sqrt{ \frac{0.27 + 0.73a^3}{a}}
$$
starts to grow at $a=0.573,$ which corresponds to $z=0.745$
</p>
</div>
</div></div>

<div id="SCM32"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 33 ===
Is dark energy domination necessary for transition to the accelerated expansion?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
No. The transition to the accelerated expansion takes place at $(z=0.75)$, which is considerably earlier than the transition to the dark energy domination stage at $(z=0.45).$
</p>
</div>
</div></div>

<div id="SCM33"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 34 ===
Consider flat Universe composed of matter and dark energy in form of cosmological constant. Find the redshift value corresponding to equality of densities of the both components $\rho_m(z_{eq})=\rho_\Lambda(z_{eq})$ and the one corresponding to beginning of the accelerated expansion $q\left(z_{accel} \right) = 0$. Obtain relation between $z_{eq}$ and $z_{accel}$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
z_{eq} = \left(\frac{\Omega _\Lambda}
{\Omega _{m0}} \right)^{1/3}- 1; \\
z_{accel}= \left(\frac{2\Omega_\Lambda }
{\Omega _{m0}}\right)^{1/3}- 1; \\
z_{accel} = 2^{1/3}\left(z_{eq}+ 1\right) - 1 \\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM34"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 35 ===
Show that density perturbations stop to grow after the transition from dust to $\Lambda$-dominated era.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
On the $\Lambda$-dominated stage the non-relativistic matter density perturbation size depends on conformal time as
\begin{equation}\label{1SCM30}
a(\eta)=-\frac{1}{H\eta}, \,\, \eta=-const\cdotp e^{-Ht},\,\,\eta<0
\end{equation}
where the Hubble parameter $H$ is constant, $H^2=(8\pi/3)G\rho_\Lambda$.
The linearized Einstein equations for the case of single-component medium take the following form in the momentum representation:
\begin{equation}\label{3SCM30}
k^2\Phi+3\frac{a^{'}}{a}+3\frac{a^{'2}}{a^2}\Phi=-4\pi Ga^2\delta\rho,
\end{equation}
\begin{equation}\label{2SCM30}
\Phi^{''}+3\frac{a^{'}}{a}\Phi^{'}+\left(2\frac{a^{''}}{a} - \frac{a^{'2}}{a^2}\right)\Phi=4\pi Ga^2\delta p.
\end{equation}
Assuming that the $\Lambda$-term is constant (which is exactly true for the vacuum energy density) one obtains $\delta\rho_\Lambda=\delta p_\Lambda=0$.
Using \eqref{2SCM30} and taking into account \eqref{1SCM30} we come up with the equation: $${\Phi}^{''}-\frac{3}{\eta}{\Phi}^{'}-\frac{3}{\eta^2}\Phi=0,$$
Solutions of the equation behave as
$$\Phi\propto\eta\propto\frac 1a$$ and $$\Phi\propto\eta^3\propto\frac{1}{a^3}.$$
Use now \eqref{3SCM30} for the modes beyond the horizon and take into account that for dust density $\rho_M\propto a^{-3}.$ Keeping only the first solution for the gravitational potential (the second decays too fast), one can see that the density perturbations do not grow: for the modes under horizon the relation
$$\delta_M\propto a^3\frac{k^2}{a^2}\Phi=const$$ holds.

Therefore the perturbations stop to grow at the transition from the dust-dominated to the $\Lambda$-dominated stage. If the present accelerated expansion of the Universe is actually caused by non-zero cosmological constant, then the structure formation from small primary perturbations is eventually close to end and it will never resume again: the structures larger than the presently observed ones will never appear.
</p>
</div>
</div></div>

<div id="SCM35"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 36 ===
What happens to the velocity fluctuations of non-relativistic matter and radiation with respect to Hubble flow in the epoch of cosmological constant domination?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM36"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 37 ===
Find the ratio of baryon to non-baryon components in the galactic halo.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM37"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 38 ===
Imagine that in the Universe described by SCM the dark energy was instantly switched off. Analyze further dynamics of the Universe.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="local-group"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 39 ===
Estimate density of dark energy in form of cosmological constant using the Hubble diagram for the neighborhood of the Local group.
<gallery widths=600px heights=500px>
File:im_eps.JPG|Hubble diagram for the neighborhood of the Local group.
</gallery>
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
An elegant and simple method was recently proposed (see A.D.Chernin, UFN 178, 267 (2008)), which allows not only to become convinced in the existence of the dark energy but also to estimate its density. The method is based on the assumption, that the dark energy (in form of the cosmological constant) governs the Hubble flow even on comparably small distances $\left( \sim \,Mpc \right)$. Consider the resulting force $\vec{F}$ (per unit mass), acting on a galaxy from the picture \ref{hubble_diagram}. The force is a sum of two contributions:
$$
\vec{F}=\vec{F}_N+\vec{F}_\Lambda
$$
Here $\vec{F}_N$ is the force produced by the total mass of the Local group, and $F_{\Lambda }$ is the force created by the dark energy. For a crude estimate let us neglect the non-spherical mass distribution in the Local group. Then the force acting on a satellite galaxy in a given point $R$ outside the local Group equals to
$$
F=-\frac{GM_{LG}}{R^2}+\frac{8\pi G}{3}{\rho_{\Lambda }}R,
$$
where $M_{LG}\simeq (1\div 3) \times 10^{12}M_{\odot }$ is the total mass of the Local group. As can be seen from the picture, the transition from domination of the local group gravity to the vacuum domination takes place on the distance of order $R_{0}=1.8\div 2\; \, Mpc.$ It allows to make a rough estimate of the dark energy density from the condition $F\left( R=R_0 \right)=0.$ The estimate results in the following
$$
(0.1\pm 0.03)<\rho_V<(1\pm 0.3)\times 10^{-26} \, kg/m^3.
$$
The proximity of the obtained crude estimate to the actual value of the dark energy density in the SCM is striking.
$$
\rho_{V}\simeq (0.75\pm 0.05)\times 10^{-26}\, kg/m^3.
$$
</p>
</div>
</div></div>

<div id="SCM38"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 40 ===
Estimate the Local group mass by methods used in the previous problem.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM39"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 41 ===
Find "weak" points in the argumentation of the two preceding problems.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM40"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 42 ===
The product of the age of the Universe and current Hubble parameter (the Hubble's constant) is a very important test (Sandage consistency test) of internal consistency for any model of Universe. Analyze the
parameters on which the product $H_0 t_0$ depends in the Big Bang model and in the SCM.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM41"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 43 ===
Show that for a fixed source of radiation the luminosity distance for high redshift values in flat Universe is greater for the dark energy dominated case compared to the non-relativistic matter dominated one.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the case of Universe filled by dark energy and non-relativistic matter, the photometric distance is equal to
$$
d_L = \frac{1 + z}
{H_0}\int_0^z \frac{dz'}
{\sqrt{\Omega _{m0}\left(1 + z'\right)^3+ \Omega _{DE0}\left(1 + z' \right)^{3\left( 1 + w_{DE}\right)}}}
$$

In the case $\Omega _{m0} = 1,\;\Omega _{DE0} = 0$
one obtains
$$
d_L = \frac{2}
{H_0}\left[1 + z -(1 + z)^{1/2}\right]
$$
and in the case $\Omega _{m0} = 0,\;\Omega _{DE0} = 1$
$$
d_L = \frac{1}{H_0}z(1 + z)
$$
instead.

In means that the supernova appear dimmer in the dark energy dominated Universe.
</p>
</div>
</div></div>

<div id="SCM42"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 44 ===
In the observations that discovered the accelerated expansion of the Universe the researchers in particular detected two $Ia$ type supernovae: $1992P$, $z=0.026$, $m=16.08$ and $1997ap$, $z=0.83$,
$m=24.32$. Show that these observed parameters are in accordance with the SCM.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The apparent $m$ and absolute $M$ stellar magnitudes are related with the photometric distance $d_L$ by the expression
\begin{equation}\label{p(1)}
m - M = 5\log _{10}\left(\frac{d_L}
{\mbox{Mpc}} \right) + 25
\end{equation}
The photometric distance in the SCM is equal to
\begin{equation}\label{p(2)}
d_L = \frac{1 + z}
{H_0}\int_0^z \frac{dz'}
{\sqrt {\Omega _{m0}(1 + z)^3 + \Omega _\Lambda}}
\end{equation}
For small $z$ $d_L(z) \simeq z/H_0.$ Using the supernova $1992P$ with small $z=0.026$ one can find $M = - 19.09.$ Assuming that the $1a$-type supernova are standard candles (they have equal absolute magnitudes), one obtains from the relation \ref{p(1)} supernova with $z=0.83$
$$H_0d_L\simeq 1.16$$
From the other hand it follows from the relation \ref{p(1)} that in the case
$\Omega _{m0} = 1,\;\Omega_\Lambda = 0$
$$
H_0d_L\simeq 0.95
$$
In the SCM one gets much better agreement:
$$
H_0d_L \simeq 1.2
$$
</p>
</div>
</div></div>

<div id="SCM43"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 45 ===
Find the redshift value, at which a source of linear dimension $d$ has minimum visible angular
size.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The apparent angular size equals to $$\delta \theta = {{(1 + z)^2 d } \over {d_L (z)}},$$
where $d_L (z)$ is the photometric distance from a source with the redshift $z$. In the case of cosmological constant domination the photometric distance is:
$$d_L (z) = - {z \over {H_0 \Omega _\Lambda ^{1/2} }},$$
In the SCM (see the previous problem)
$$
d_L = \frac{1 + z}
{H_0}\int_0^z \frac{dz'}
{\sqrt {\Omega _{m0}(1 + z)^3 + \Omega _\Lambda}}
$$
then the angular size is
$$\delta \theta = - {{H_0 \Omega _\Lambda ^{1/2} d} \over 2}{{(1 + z)^2 } \over z},$$
and $z = - 1$. It corresponds to infinitely far future.
</p>
</div>
</div></div>

<div id="SCM44"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 46 ===
Compare the observed value of the dark energy density with the
one expected from the dimensionality considerations (the cosmological constant problem).
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The dimensionality considerations suggest the following value for density of the dark energy in form of the cosmological constant
\[\rho_\Lambda^{(dim)}\sim M_{Pl}^4\approx10^{76}GeV^4.\]
At the same time the ''observed'' value fixed by the SCM is
\[\rho_\Lambda^{(obs)}\approx10^{-48}GeV^4.\]
This huge disagreement
\[\frac{\rho_\Lambda^{(dim)}}{\rho_\Lambda^{(obs)}}>10^{120}\]
is called the ''cosmological constant problem''. It presents a fundamental problem for elementary particles theory and physics in general, rather than for cosmology itself. (Look for details in R.Bousso, [http://arxiv.org/abs/0708.4231 TASI Lectures on the cosmological constant])
</p>
</div>
</div></div>

<div id="SCM45"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 47 ===
Determine the density of vacuum energy using the Planck scale as cutoff parameter.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For a quantum field with a given mode of frequency $\omega$, the zero-point energy is $\pm\omega/2$ with the plus sign for a bosonic field and the minus sign for a fermionic field. The total zero-point energy is then $\pm\sum\limits_i\omega_i/2$. In the continuum limit, we have for a free field (bosonic or fermionic)
\[\frac12\sum\limits_i\omega_i=\frac12\int\frac{d^3xd^3k}{(2\pi)^3}(k^2+m^2)^{1/2}=V\int\frac{k^2dk}{4\pi^2}(k^2+m^2)^{1/2}.\] This integral diverges, thus we need to introduce the physical cutoff. In cosmological context this cutoff must be the Planck mass at largest. With the cutoff $k_{max}\sim M_{Pl}$ one obtains
\[\rho_\Lambda=\int_0^{k_{max}}\frac{k^2dk}{4\pi^2}(k^2+m^2)^{1/2} \approx\frac{k_{max}^4}{16\pi^2}=\frac{M_{Pl}^4}{16\pi^2} \approx10^{74}GeV^4.\]
This result only slightly modifies the dimensional considerations given in the previous problem.
</p>
</div>
</div></div>

<div id="SCM46"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 48 ===
Identifying the vacuum fluctuations density with the observable dark energy value in SCM, find the required frequency cutoff magnitude in the fluctuation spectrum.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[\nu_{max}=10^{12}Hz.\]
</p>
</div>
</div></div>

<div id="SCM47"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 49 ===
What purely cosmological problem originates from the divergence of the zero-point energy density?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
If the zero-point energy density is infinite, then our Universe is infinitely curved and the space size is infinitely small.
</p>
</div>
</div></div>

<div id="SCM48"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 50 ===
With exact supersymmetry, the bosonic contribution to cosmological constant is canceled by its fermionic counterpart. However, we know that our world looks like not supersymmetric. Supersymmetry, if exists, has to be broken above or around $100GeV$ scale. Compare the observed value of the dark energy density with the one expected from broken supersymmetry.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
If we take the $k_{max}^{(SUSY)}\sim100GeV$ cutoff, then
\[\rho_\Lambda^{(SUSY)}=\frac{\left(k_{max}^{(SUSY)}\right)^4}{16\pi^2}\approx10^6GeV^4.\]
\[\frac{\rho_\Lambda^{(SUSY)}}{\rho_\Lambda^{(obs)}}>10^{54}.\]
Thus supersymmetry does not solve the problem of the cosmological constant.
</p>
</div>
</div></div>

<div id="SCM49"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 51 ===
Determine duration of the inflation period.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Let us see what happens with the relative density during all the history of Universe. From the first Friedmann equation it follows that in the matter-dominated epoch $\left| \Omega - 1 \right| \sim t^{2/3},$ and in the radiation-dominated epoch $\left| \Omega - 1 \right| \sim t^{2/3}.$ Then assume that inflation started at $t_i,$ and finished at $t_f.$ Then in the period of $t_f < t < t_{eq} = 50000$ years the radiation dominated and the range $t_{eq} < t < t_0 $ corresponds to the matter domination epoch. Then the presently observed difference
$$
\left|\Omega (t_0 ) - 1 \right| = \left| \Omega (t_i ) - 1 \right|e^{ - 2H(t_f - t_i )} \left( \frac{t_{eq} }
{t_f } \right)\left( \frac{t_0}
{t_{eq} }\right)^{2/3}
$$
According the SCM $\left|\Omega \left( t_0\right) - 1 \right| < 0.04.$ If one assumes that before the beginning of the inflation there was $\left|\Omega \left( t_i \right) - 1 \right| \sim 1,$ then
$$
N > \frac{1}{2}\ln \left[\frac{\Omega \left( t_i \right) - 1}{\Omega \left(t_0\right) - 1}\left( \frac{t_{eq}}{t_f } \right)\left( \frac{t_0}{t_{eq}} \right)^{2/3} \right] \approx 60
$$

Assume that radiation dominated before the inflation started. Then $H = \frac{1}{2t}.$ If at the moment of time $t_i$ the vacuum dark energy started to dominate and then the expansion continued with constant rate, then $H\approx 1/t_i$ during the inflation. And finally one obtains that
$$
N = H\left(t_f - t_i\right) \approx \frac{t_f - t_i }
{t_i }
$$
it is commonly assumed that the inflation finishing time is of order of the Grand Unification time $t_f\approx 10^{-35}.$ Therefore $t_f\approx 10^{-38}.$ Evidently the inflation could last longer as we used only lower estimate for $N.$
</p>
</div>
</div></div>

<div id="SCM50"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 52 ===
Plot the dependence of luminosity distance $d_L$ (in units of $H_0^{-1}$) on the redshift $z$ for the two-component flat Universe with non-relativistic liquid ($w=0$) and cosmological constant
($w=-1$). Consider the following cases:

a) $\Omega_\Lambda^0=0$;

b) $\Omega_\Lambda^0=0.3$;

c) $\Omega_\Lambda^0=0.7$;

d) $\Omega_\Lambda^0=1$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM51"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 53 ===
For a Universe filled by dark energy with state equation $p_{DE}=w_{DE}\rho_{DE}$ and non-relativistic matter obtain the Taylor series for $d_L$ in terms of $z$ near the observation point
$z_0=0$. Explain the obtained result.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the universe filled by dark energy with the state equation $p_{DE}=w_{DE}\rho_{DE}$ and non-relativistic dark matter the photometric distance takes on the form:
\begin{equation}
\label{d_L}
d_L = \frac{1 + z}
{H_0}\int_0^z \frac{dz'}
{\sqrt{\Omega _{m0}\left(1 + z'\right)^3+ \Omega _{DE0}\left(1 + z' \right)^{3\left( 1 + w_{DE}\right)}}}
\end{equation}
Expand the integrand in Taylor series in $z$ in vicinity of the observation point $z_0=0$ to obtain
$$
\begin{gathered}
\label{tey}
\frac{1}{\sqrt{\Omega _{m0}\left(1 + z'\right)^3+ \Omega _{DE0}\left(1 + z' \right)^{3\left( 1 + w_{DE}\right)}}}\simeq \\
\simeq 1 -\frac{3}{2}\left(1+w_{DE}\Omega _{DE0}\right)z+\frac{3}{4}\left(3w_{DE}\Omega _{DE0}-1\right)z^2.\nonumber
\end{gathered}
$$
Substitution of the decomposition (\ref{tey}) into the integral (\ref{d_L}) results in the following:
\begin{equation}
\label{d_L}
d_L = \frac{1 + z}
{H_0}\left(z -\frac{3}{4}\left(1+w_{DE}\Omega _{DE0}\right)z^2+\frac{1}{4}\left(3w_{DE}\Omega _{DE0}-1\right)z^3\right).
\end{equation}
</p>
</div>
</div></div>

<div id="SCM_45"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 54 ===
Determine position of the first acoustic peak in the CMB power spectrum produced by baryon oscillations on the surface of the last scattering.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Early Universe was filled by photon-baryon plasma, which can be treated as a single-component liquid. The baryons were trapped by the potential walls generated by density fluctuations, and they were eventually compressed. The contraction lead to heating of the plasma and therefore to increasing radiative pressure of the photons which is directed outside. Finally the radiative pressure stop the compression and lead to expansion of the plasma. During the expansion the plasma cools down and its radiative pressure decreases. The gravity starts to dominate again leading to repeated compression. The concurrence between the gravity and the pressure lead to longitudinal (acoustic) oscillations in the photon-baryon fluid. When the matter and the radiation get decoupled in the recombination process, the picture of the acoustic oscillations remains frozen into the CMB. Today we detect the evidence of the primordial acoustic waves (dense and diluted regions) in form of the primary anisotropy in the CMB.

It is well known that any acoustic wave, whatever complicatedly shaped, can be represented in form of superposition of modes with different wave numbers $k$, $k\propto 1/\lambda $. Every mode $\lambda $ corresponds to certain angular scale $\theta $ on the sky. Therefore in order to facilitate the comparison of the theory with observations, one should use the angular (multipole) decomposition in terms of the Legendre polynomials $P_{l} (\cos \theta )$, instead of the Fourier transform in terms of sines and cosines. Order of the polynomial $l$ plays the same role as the index $k$ in the Fourier decomposition. For $l\ge 2$ the Legendre polynomials are oscillating functions in the interval $\left[1,-1\right]$. Number of the oscillations increases with growth of $l$. Therefore
\begin{equation} \label{acust1_}
l\propto \frac{1}{\theta }
\end{equation}
The temperature fluctuations can be observed and qualitatively analyzed with help of paired measurements in the directions $\stackrel{\frown}{n},\stackrel{\frown}{n}'$, so that $\stackrel{\frown}{n}\cdot \stackrel{\frown}{n}'=\cos \theta $. Taking average over all the pairs in the assumptions of Gaussian fluctuations, one obtains the two-point correlator $C(\theta )$, which can be presented in terms of the multipole decomposition
\begin{equation} \label{acust2_}
\left\langle \delta T(\stackrel{\frown}{n})\cdot \delta T(\stackrel{\frown}{n}')\right\rangle \equiv C(\theta )=\sum _{l}\frac{2l+1}{4\pi } C_{l} P_{l} (\cos \theta )
\end{equation}
where $C_{l} $ are the coefficients.

As we already mentioned, the analysis of the temperature fluctuations enables us to clarify the structure of the longitudinal oscillations. The mode with the maximum wavelength corresponds to the maximum angular size of the primary anisotropy. This fundamental mode was detected the first. There are reliable evidences of the fact that the second and the third modes were also already detected. The distance $r_{s} $, passed by the acoustic wave during the time period before the recombination, is called the sound horizon. The sound horizon is fixed by (or rather fixes) the physical scale o the last scattering surface. Size of the sound horizon depends on values of the physical parameters. Distance to the last scattering surface $d_{sls} $ depends on the cosmological parameters too. Together they determine the angular size of the sound horizon$\theta _{s} $ (see Fig.)

<gallery widths=600px heights=500px>
File:acustic_pic.JPG|Dependence of the deceleration parameter on the redshift.
</gallery>

\begin{equation} \label{acust3_}
\theta _{s} \approx \frac{r_{s} }{d_{sls} }
\end{equation}
Analysis of the temperature anisotropy allows to determine $\theta _{s} $. Varying the cosmological parameters in $r_{s} ,d_{sls} ,$ one achieves the best agreement with the observations and thus fixes the cosmological parameters.

We can estimate the sound horizon size as the length passed by sound from the moment $t=0$ to the recombination time $t_{*} $
\begin{equation} \label{acust4_}
r_{s} (z_{*} ;\Omega _{b} ,\Omega _{r} )\approx \int _{0}^{t_{*} }c_{s} dt
\end{equation}
where $z_{*} $ is the redshift of the recombination moment $\left(z_{*} \approx 1100\right)$, $c_{s} $ is the sound speed in the photon-baryon medium
\begin{equation} \label{acust5_}
c_{s} \approx c\left[3\left(1+\frac{3\Omega _{b} }{\Omega _{r} } \right)\right]^{-1/2}
\end{equation}
To determine $dt$ one should use the first Friedmann equation in the form
\begin{equation} \label{acust6_}
\left(\frac{da}{dt} \right)^{2} =H_{0}^{2} \left[\Omega _{r0} \left(\frac{a_{0} }{a} \right)^{2} +\Omega _{m0} \frac{a_{0} }{a} +\Omega _{k0} +\Omega _{\Lambda 0} \left(\frac{a}{a_{0} } \right)^{2} \right]
\end{equation}
Making use of the relations $\frac{a_{0} }{a} =1+z$ and $\Omega _{r} +\Omega _{m} +\Omega _{\Lambda } +\Omega _{k} =1$, one obtains
\begin{equation} \label{acust7_}
dt=H_{0}^{-1} (1+z)^{-1} \left\{\left(1+z\right)^{2} \left(1+\Omega _{m,0} z\right)+z(z+2)\left[\left(1+z)^{2} \Omega _{r.0} -\Omega _{\Lambda ,0} \right)\right]\right\}^{-1/2} dz
\end{equation}
Distance to the last scattering surface, which corresponds to its angular size, is determined by the quantity called the angular diameter distance. It is connected by the photometric distance $d$ by the simple relation
\begin{equation} \label{acust8_}
d_{sls} =\frac{d}{(1+z_{*} )^{2} }
\end{equation}
The position of the first acoustic peak is determined by
\begin{equation} \label{acust9_}
l\approx \frac{d_{sls} }{r_{s} }.
\end{equation}
Let us now predict the position of the first peak for the case of flat Universe. To the leading order the sound speed is
\[c_{s} =\frac{c}{\sqrt{3} }. \]
Let us assume also that in the recombination epoch the Universe was matter-dominated. Under such assumptions one gets
\begin{equation} \label{acust10_}
r_{s} =\frac{c_{s} }{H_{0} \sqrt{\Omega _{m} } } \int _{z_{*} }^{\infty }(1+z)^{-5/2} dz =r_{s} =\frac{2c_{s} }{3H_{0} \sqrt{\Omega _{m} } } \left(1+z_{*} \right)^{-3/2}
\end{equation}
Using the standard procedure (see section 4), we obtain for the comoving radial coordinate $r_{sls} $ of the last scattering surface the following
\begin{equation} \label{acust11_}
r_{lsr} =\frac{c}{H_{0} } \int _{0}^{z_{*} }\left[\Omega _{m} (1+z)^{^{3} } +\Omega _{\Lambda } \right] ^{-1/2} dz
\end{equation}
Replacing the integrand by its binomial decomposition
\[\Omega _{m}^{-1/2} (1+z)^{-3/2} -\left(\Omega _{\Lambda } /2\Omega _{m}^{3/2} \right)\left(1+z\right)^{-9/2} \]
and taking into account that
\[d_{sls} =r_{sls} /(1+z_{*} ),\]
we obtain
\begin{equation} \label{acust12_}
d_{sls} =\frac{2c}{7H_{0} (1+z_{*} )} \left\{7\Omega _{m}^{-1/2} -2\Omega _{\Lambda } \Omega _{m}^{-3/2} +{\rm O} \left[\left(1+z_{*} \right)^{-1/2} \right]\right\}.
\end{equation}
Using the relation $\Omega _{\Lambda } =1-\Omega _{m} $ and neglecting the higher order terms, one obtains
\begin{equation} \label{acust13_}
d_{sls} \approx \frac{2c\Omega _{m}^{-1/2} }{7H_{0} (1+z_{*} )} \left(9-2\Omega _{m}^{3} \right)
\end{equation}
Combination of the relations \eqref{acust10_} and \eqref{acust13_} gives for the first acoustic peak the following
\begin{equation} \label{acust14_}
l\approx \frac{d_{sls} }{r_{s} } \approx 0.74\sqrt{1+z_{*} } \left(9-2\Omega _{m}^{3} \right)\simeq 220.
\end{equation}
This result agrees very well with numerous observations.
</p>
</div>
</div></div>

<div id="SCM52"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 55 ===
\bf Compare the asymptotes of time dependence of the scale factor
$a(t)$ for the SCM and de Sitter models. Explain physical reasons of
their distinction.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM53"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 56 ===
Redshift for any object slowly changes due to the acceleration (or deceleration) of the Universe expansion. Estimate change of velocity in one year.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

$$
\Delta z = \dot z \Delta t = H_0 \left(1 + z - {H(z)\over H_0} \right)\Delta t
$$

$$H(z)=H_0 \left[\Omega_{m0}(1+z)^3+\Omega_{\Lambda 0}\right]^{1/2}$$
$$\Omega_{m0}+\Omega_{\Lambda 0}=1,~\Omega_{m0}\approx 0.3$$
$$\Delta z \approx 2\cdot 10^{-11}$$
$$\Delta v = c{\Delta z\over 1+z}\approx 0.25~\mbox{cm/s}$$

</p>
</div>
</div></div>

<div id="SCM54"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 57 ===
Determine the lower limit of ratio of the total volume of the Universe to the observed one?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
If the Universe is the 3-sphere, then its radius is $R_0=a_0$
$$
R_0 = \frac{1}{H_0 \sqrt {\left| \Omega _{curv} \right|} }
$$
For the SCM parameters the cosmological horizon equals to
$$
L_{p,0} = a_0 \int_0^{t_0 } \frac{dt}{a(t)} = \frac{3.6}{H_0}
$$
We assumed that $a(t)\sim t^{2/3}.$ Therefore
$$
\frac{R_0}
{L_{p,0} } = \frac{1}
{3.6\sqrt {\left| \Omega _{curv}\right|} }
$$
Taking into account that $\left|\Omega _{curv} \right| < 0.02,$ one obtains $\frac{a_0}{L_{p0} } > 2$.
Number of the regions similar to the one inside our horizon equals to the ratio of the 3-sphere volume $2\pi ^2 a_0^3 $ to the volume of the observed Universe
$$
N = \frac{2\pi ^2 a_0^3}
{(4\pi /3)L_{p0}^3 } \simeq 4.7\left(\frac{a_0 }
{L_{p0} } \right)^3 > 38
$$
Thus we can see not more than $3 \%$ of the Universe volume.
</p>
</div>
</div></div>

<div id="SCM55"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 58 ===
What is the difference between the inflationary expansion in the early Universe and the present accelerated expansion?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The difference is in the fact that in the present time there is about $27 \%$ of non-relativistic matter which was absent in the inflation period. The numerical values of Hubble parameter in the beginning of the inflation was very different from the one at the beginning of the current period of accelerated expansion of Universe.
</p>
</div>
</div></div>

<div id="SCM56"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 59 ===
Compare the values of Hubble parameter at the beginning of the
inflation period and at the beginning of the present accelerated
expansion of Universe.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
h^2 (x) = \Omega _{m0} x^3 + (1 - \Omega _{m0} )x^\alpha ;\quad \alpha = 3(1 + w)
$$
Assume for simplicity that the Universe is spatially flat. From the definition $O(x)$
it follows that
$$
O(x) = \Omega _{m0} + \left( 1 - \Omega _{m0}\right)\frac{x^\alpha - 1}
{x^3 - 1}
$$
The required result follows from the fact that $\alpha =0$ for cosmological constant, $\alpha >0$ for quintessence and $\alpha <0$ for phantom energy.
</p>
</div>
</div></div>

<div id="SCM57"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 60 ===
Plot the dependencies $h(x)=H(x)/H_0(x)$, where $x=1+z$, in the SCM, for quintessence and for phantom energy cases.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Plots of the dependencies $
h(x) = H(x)/H_0 (x);\quad x = 1 + z
$ in the SCM are presented on Fig. \ref{fig:52} for phantom energy and quintessence.
<gallery widths=600px heights=500px>
File:12_52.JPG|Dependence of the deceleration parameter on the redshift.
</gallery>
</p>
</div>
</div></div>

<div id="SCM58"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 61 ===
Find the constraints imposed by the Weak Energy Condition (WEC) for the dark energy on the redshift-dependent Hubble parameter. ([http://arxiv.org/abs/astro-ph/0703416 Inspired by A.Sen, R.Scherrer])
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The redshift-dependent Hubble parameter $H(z)$ is defined by
\[H^2(z)=\frac{8\pi G}{3}(\rho_m+\rho_{DE})\]
This equation can be rewritten as
\begin{equation}\label{SCM_eq_76_1}
E^2(z)\equiv\frac{H^2(z)}{H^2_0(z)}=\Omega_{m0}(1+z)^3+(1-\Omega_{m0})\frac{\rho_{DE}}{\rho_{DE0}}
\end{equation}
The WEC forces \[\frac{\rho_{DE}}{\rho_{DE0}}\ge1 \ \left(\frac{d\rho}{dz}\ge0\right)\] for $z>1$, so equation (\ref{SCM_eq_76_1}) becomes
\begin{equation}\label{SCM_eq_76_2}
E^2(z)\ge\Omega_{m0}(1+z)^3+(1-\Omega_{m0})
\end{equation}
Going back to equation (\ref{SCM_eq_76_2}) and taking the derivative with respect to $z$ (which we denote throughout with a prime) gives:
\[
2E(z)\frac{dE}{dz}=3\Omega_{m0}(1+z)^2+\frac{1-\Omega_{m0}}{\rho_{DE0}}\frac{d\rho_{DE}}{dz}.
\]
Now the WEC implies that \[\frac{d\rho_{DE}}{dz}\ge0,\] so we get
\begin{equation}\label{SCM_eq_76_3}
E(z)\frac{dE}{dz}\ge\frac32\Omega_{m0}(1+z)^2.
\end{equation}
Equations (\ref{SCM_eq_76_2}) and (\ref{SCM_eq_76_3}) give the constraints that the WEC for the dark energy places on the redshift-dependent Hubble parameter.
</p>
</div>
</div></div>

<div id="SCM59"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 62 ===
Using the statefinders, show that the power law cosmology mimics SCM model at (see Chapters 3 and 9)
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Recall that the statefinders are defined as
\[r\equiv\frac{\dddot a}{aH^3},\ s\equiv\frac{r-1}{3(q-1/2)}, \ q\ne\frac12.\]
The SCM model corresponds to the point $r=1,\, s=0$ in the $(s,r)$ plane. In power-law cosmology (see Chapter 3) the statefinders are given by
\[r=2q^2+q,\, s=\frac23(q+1).\]
We note that $r=1,\, s=0$ for $q=-1$. Thus the power law cosmology mimics SCM model with $q=-1$.
</p>
</div>
</div></div>

Evolution of Universe

2012-10-04T17:53:05Z

Den: /* Problem 12 */

[[Category:Standard Cosmological Model|2]]

__NOTOC__
<div id="SCM-10"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 1 ===
Rewrite the first Friedman equation in terms of redshift and analyze the contributions of separate components at different stages of Universe evolution.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM11"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2 ===
Find the time dependence for the scale factor and analyze asymptotes of the dependence. Plot $a(t)$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM12"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 3 ===
Determine the redshift value corresponding to equality of radiation and matter densities.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
\rho _r(t) = \rho _m(t);\\
\rho _{r0}\frac{a_0^4}{a^4} = \rho _{m0}\frac{a_0^3}{a^3};\\
\frac{\rho _{m0}}{\rho _{r0}} = \frac{\Omega _{m0}}{\Omega _{r0}} = \frac{a_0}{a} = 1 + z;\\
\Omega _{r0} = 5 \cdot {10^{ - 5}} \\
z = \frac{\Omega _{m0}}{\Omega _{r0}} - 1 \approx 5400\\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM_9"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 4 ===
Construct effective one-dimensional potential (see [[Solutions_of_Friedman_equations_in_the_Big_Bang_model#dyn15|problem]] of Chapter 3)
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
<div id="V(x)"></div>
$$
V(x) = - \frac{1}{2}\sum\limits_i \Omega _{i0}x^{ - (1 + 3w_i)};\quad x \equiv \frac{a}{a_0}
$$
In the SCM
$$
V(x) = - \frac{1}{2}\Omega _{m0}x^{-1}-\frac{1}{2}\Omega _{\Lambda 0}x^2 \simeq - 0.135x^{-1}-0.365x^2
$$

<gallery widths=600px heights=500px>
File:12_9.JPG|
</gallery>
</p>
</div>
</div></div>

<div id="SCM13"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 5 ===
Show that the following holds: $\dot H= -4\pi G\rho_m$ and $\ddot H= 12\pi G\rho_m H$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM_18"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 6 ===
Expand the scale factor in Taylor series near the time moment
$t_0$:
\[\frac{a(t)}{a(t_0)}=1+\sum\limits_{n=1}^\infty\frac{A_n(t_0)}{n!}[H_0(t-t_0)]^n;\quad A_n\equiv\frac{1}{aH^n}\frac{d^na}{dt^n}\]
and calculate values for few first coefficients $A_n$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The coefficients $A_1$ and $A_2$ equal to:
$$A_1 = \frac{\dot a}{aH} = 1;\quad A_2 = \frac{\ddot a}{aH^2} = - q = 1 - \frac{3}{2}{\Omega _m}$$
The parameters $A_n(n > 2)$ can be calculated by consequent differentiation of the relation
$$\ddot a = a\left( \dot H + H^2 \right)$$
The time derivatives of the scale factor can be determined making use of the fact that in SCM $\dot H = - 4\pi G\rho _m$, and $\rho _m = \rho _0a^{ - 3}$. For example,
$$A_3 = 1 + \frac{\ddot H}{H^3} + 3\frac{\dot H}{H^2}$$
Using $\ddot H = 12\pi G{\rho _m}H,\;\dot H = - 4\pi G\rho _m$, one finds that
$$A_3 = 1$$
Let us present the expressions for some consequent expansion coefficients
\[\begin{array}{l}
{A_4} = 1 - \frac{3^2}{2}\Omega _m;\\
{A_5} = 1 + 3\Omega _m + \frac{3^3}{2}\Omega _m^2;\\
{A_6} = 1 - \frac{3^3}{2}\Omega _m - 3^4\Omega _m^2 - \frac{3^4}{4}\Omega _m^3
\end{array}\]
</p>
</div>
</div></div>

<div id="SCM14"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 7 ===
Show that all the coefficients $A_n$ can be expressed through
elementary functions of the deceleration parameter $q$ or the
density parameter \[\Omega_m=\frac23(1+q).\]
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM15"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 8 ===
Consider the case of flat Universe filled by non-relativistic matter and dark
energy with state equation $p_X=w\rho_X$, where the state parameter $w$ is parameterized as the following \[w=w_0+ w_a(1-a)=w_0 + w_a \frac{z}{1+z}.\]
Express current values of cosmographic parameters through $w_0$ and
$w_a$. % (See [http://arxiv.org/abs/0802.1583 Cosmography of f(R) gravity])
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the considered case
\[\frac{H^2(z)}{H_0^2} = \Omega _m(1 + z)^3 + \Omega _X(1 + z)^{3\left( 1 + w_0 + w_a \right)}e^{ - \frac{3w_az}{1 + z}}\]
Calculation of the cosmographic parameters requires in general to integrate $H(z)$ in order to obtain $a(t).$ One can avoid this procedure with help of the relation
\[\frac{d}{dt} = - (1 + z)H(z)\frac{d}{dz}.\]
We can, given $H(z)$, apply that relation to calculate $\dot H,\ddot H, \dddot H, \ddddot H $ and so on. Using the expressions for time derivatives of Hubble parameter obtained in the problem (\ref{equ61_6}), Chapter 2,
\[\begin{array}{l}
\dot H = - H^2(1 + q);\\
\ddot H = H^3\left( j + 3q + 2 \right);\\
\dddot H = H^4\left[ s - 4j - 3q(q + 4) - 6 \right];\\
\ddddot H = H^5\left[ l - 5s + 10\left( q + 2 \right)j + 30(q + 2)q + 24 \right]
\end{array}\]
one can express the current values of the cosmological parameters (for $z = 0$) in terms of $w_0,w_a:$
\[\begin{array}{l}
q_0 = \frac{1}{2} + \frac{3}{2}\left( 1 - \Omega _m \right)w_0;\\
j_0 = 1 + \frac{3}{2}\left( 1 - \Omega _m \right)\left[3w_0\left( 1 + w_0 \right) + w_a \right];\\
s_0 = - \frac{7}{2} - \frac{33}{4}\left( 1 - \Omega _m \right)w_a - \frac{9}{4}\left( 1 - \Omega _m \right)\left[ 9 + \left(7 - \Omega _m \right)w_a \right]w_0 - \\
- \frac{9}{4}\left( 1 - \Omega _m \right)\left( 16 - 3\Omega _m \right)w_0^2 - \\
- \frac{27}{4}\left( 1 - \Omega _m \right)\left(3 - \Omega _m \right)w_0^3;\\
l_0 = \frac{35}{2} + \frac{1 - \Omega _m}{4}\left[ 213 + (7 - \Omega _m)w_a \right]w_a + \\
+ \frac{1 - \Omega _m}{4}\left[ 489 + 9(82 - 21\Omega _m)w_a \right]w_0 + \\
+ \frac{9}{2}\left( 1 - \Omega _m \right)\left[ 67 - 21\Omega _m + \frac{3}{2}(23 - 11\Omega _m)w_a \right]w_0^2 + \\
+ \frac{{27}}{4}\left( {1 - {\Omega _m}} \right)(47 - 24{\Omega _m})w_0^3 + \\
+ \frac{81}{2}\left( 1 - \Omega _m \right)(3 - 2\Omega _m)w_0^4
\end{array}\]
</p>
</div>
</div></div>

<div id="SCM16"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 9 ===
Show that the results of the previous problem applied to SCM coincide with
the ones obtained in the [[#SCM_18|problem]].
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the SCM $\left( {{w_0},{w_a}} \right) = \left( { - 1,0} \right)$ and therefore
\[\begin{array}{l}
{q_0} = \frac{1}{2} - \frac{3}{2}\left( {1 - {\Omega _m}} \right);\\
{j_0} = 1;\\
{s_0} = 1 - \frac{9}{2}{\Omega _m};\\
{l_0} = 1 + 3{\Omega _m} + \frac{{27}}{2}\Omega _m^2
\end{array}\]
</p>
</div>
</div></div>

<div id="SCM10"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 10 ===
Photons with $z=0.1,\ 1,\ 100,\
1000$ are registered. What was the Universe age $t_U$ in the moment of
their emission? What period of time $t_{ph}$ were the photons on the
way? Plot $t_U(z)$ and $t_{ph}(z)$
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
t_U=\int_z^\infty\frac{dz}{(1 + z)H(z)} ;\; t_UH_0=\int_z^\infty\frac{dz}{(1 + z)\sqrt{\Omega _{m0}(1 + z)^3 + \Omega _{\Lambda 0} } }; \\
t_{ph}H_0=\int_0^z \frac{dz}{(1 + z)\sqrt{\Omega _{m0}(1 + z)^3 +\Omega _{\Lambda 0}}}\\
\end{gathered}
$$
At $$z = 0.1\,\,\, \to \,\,\,\,t_U H_0 = 0.899\,\,\,\,\,t_{ph} H_0 = 0.093$$
$$z = 1\,\, \to \,\,\,t_U H_0 = 0.431\,\,\,\,\,t_{ph} H_0 = 0.561$$
$$z = 100\,\,\, \to \,\,t_U H_0 = 1.264 \times 10^{ - 3} \,\,\,\,\,t_{ph} H_0 = 0.991$$
$$z = 1000\,\, \to \,\,\,t_U H_0 = 3.663 \times 10^{ - 5} \,\,\,\,\,t_{ph} H_0 = 0.993$$

<gallery widths=600px heights=500px>
File:12_10(1).JPG|
</gallery>

<gallery widths=600px heights=500px>
File:12_10(2).JPG|
</gallery>
</p>
</div>
</div></div>

<div id="SCM17"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 11 ===
Determine the present physical distance to the object that emitted light with current redshift $z$
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The physical distance to the object emitted a photon is related to the redshift $z$ by
$$
R(z) = \frac{1}{1 + z}\int_0^z\frac{dz'}{H(z')}
$$
In the SCM
$$
H =H_0\sqrt {\Omega _{r0}(1 + z)^4 + \Omega _{m0}(1 + z)^3+\Omega _{\Lambda 0}}
$$
The parameters ${H_0},\;\Omega _{r0},\Omega _{m0},\Omega _{\Lambda 0}$ are fixed by the model.
</p>
</div>
</div></div>

<div id="SCM_15"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 12 ===
A photon was emitted at time $t$ and registered at time $t_0$ with red
shift $z$. Find and plot the dependence of emission time on the redshift $t(z)$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
a = \frac{1}{1 + z}; \\
a(t) = A^{1/3}\sinh ^{2/3}\left(t/{t_\Lambda } \right);\quad A \equiv \frac{\Omega_{ m0}}{\Omega _{\Lambda 0}}; \\
\frac{t_0}{t_\Lambda }= Arth\sqrt {\Omega _{\Lambda 0}} ; \\
\frac{t}{t_0} = \frac{1}{Arth\sqrt{\Omega _{\Lambda 0}}}Arsh\left[\sqrt {\frac{\Omega _{\Lambda 0}}{\Omega _{m0}}} \frac{1}{\left(1 + z\right)^{3/2}} \right] \\
\end{gathered}
$$
<gallery widths=600px heights=500px>
File:12_13.JPG|
</gallery>
</p>
</div>
</div></div>

<div id="SCM18"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 13 ===
Find the time dependence for the scale factor and analyze its asymptotes. Plot $a(t)$
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the SCM
$$
\left(\frac{\dot a}{a}\right)^2 = H_0^2\left[ \Omega _{m0}\left(\frac{a_0}{a} \right)^3 + \Omega _{\Lambda 0} \right]
$$
Solution of the equation in the case $\Omega _{\Lambda 0} > 0$ reads
$$
\begin{gathered}
a(t) = a_0\left( \frac{\Omega _{m0}}{\Omega _{\Lambda 0}}\right)^{1/3}\left[ sh\left( \frac{3}{2}\sqrt {\Omega _{\Lambda 0}}H_0t \right) \right]^{2/3}; \\
a\left( t \ll H_0^{ - 1}\right) \propto t^{2/3};\; a\left( t \gg H_0^{ - 1} \right) \sim e^{H_0t} \\
\end{gathered}
$$
Below it is convenient to use the expression for time dependence of the scale factor in the following form
$$
a(t) = A^{1/3}\sinh ^{2/3}\left( t/{t_\Lambda }\right);\quad
A \equiv \frac{\Omega_{m0}}{\Omega _{\Lambda 0}} =
\frac{1 - \Omega _{\Lambda 0}}{\Omega _{\Lambda 0}};\quad t_\Lambda \equiv \frac{2}{3}\left(H_0 \right)^{-1}\left( \Omega _{\Lambda 0}\right)^{ - 1/2}
$$
</p>
</div>
</div></div>

<div id="SCM19"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 14 ===
Using the explicit solution for scale factor $a(t)$, obtained in the previous problem, find the cosmic horizon $R_h$ (see Chapter 3)
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
As we have seen above
\[a(t)=A\sinh^{2/3}{(t/t_\Lambda)},\]
where \[A=\frac{\Omega_{m0}}{\Omega_{\Lambda0}},\ t_\Lambda=\frac{2}{3H_\infty}.\]
Consider \[f(t)\equiv\ln{a(t)}=\ln A + \frac23\ln\sinh\frac{t}{t_\Lambda},\] then
\[\dot f=\frac{2}{3t_\Lambda\tanh(t/t_\Lambda)}=\frac{H_\infty}{\tanh(3tH_\infty/2)},\]
and finally
\[R_h=\frac{c}{\dot f}=\frac{c}{H_\infty}\tanh\left(\frac32tH_\infty\right).\]
</p>
</div>
</div></div>

<div id="SCM20"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 15 ===
Show that the Universe becomes $R_h$-delimited only after the cosmological constant starts to dominate.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
According to the result of the previous problem
\[\dot R_h=\frac32\left[1-\tanh^2\left(\frac32tH_\infty\right)\right]c,\]
so for early times $\dot R_h>c$ and an observer does not experience a divergent redshift with increasing $R$. Then after some ''transition'' time $t^*$, estimated by the condition
\[ct^*=R_h(t^*),\]
he (or she) will begin to encounter an observational limit at a finite radius $R_h$. Numerical solution of the equation gives $t^*\approx0.86H_\infty^{-1}$. This time is roughly the point at which the Universe transits from matter-dominated to $\Lambda$-dominated epoch. Eventually, the Universe becomes de Sitter one and therefore $\dot R_h\rightarrow0$ with $R_h$ settling at the fixed value $c/H_\infty$.
</p>
</div>
</div></div>

<div id="SCM21"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 16 ===
Analyze the stability of Friedmann equations.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Friedmann equations are given by
\[H^2=\frac{8\pi G}{3}\rho_m+\frac\Lambda3,\]
\[\dot H=-4\pi G\rho_m,\]
\[\dot\rho_m+3H\rho_m=0\]
Let us introduce the following dimensionless variables (recall that $\rho_m,\ \Lambda>0$):
\[x=\frac{\sqrt{8\pi G\rho_m}}{\sqrt3H},\ y=\frac{\sqrt\Lambda}{\sqrt3H}\]
Equations for $x$ and $y$ read (with $N\equiv\ln a$)
\[x'=\frac{dx}{dN}=\frac32x(x^2-1),\]
\[y'=\frac{dy}{dN}=\frac32yx^2.\]
The critical points, i.e. the solutions corresponding to $x'=0,\ y'=0$ and their eigenvalues are given by
\[x_{cr}=0,\ y_{cr}=1 (\Lambda-dominant\, case,)\ \mu_1=0, \mu_2=-\frac32;\]
\[x_{cr}=1,\ y_{cr}=0 (matter-dominant\, case,)\ \mu_1=3, \mu_2=\frac32.\]
From the eigenvalues we see that the matter dominant phase is unstable $\mu_{1,2}>0$ while the cosmological constant dominant phase has $\mu_2<0$ being stable. The existence of a zero eigenvalue $\mu_1=0$ in the first case originates from the fact that the two variables x and y are connected by the relation $x^2+y^2=1$. Therefore in this case one can reduce the dynamics to one-dimensional space.
</p>
</div>
</div></div>

<div id="SCM22"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 17 ===
Determine solutions for the perturbations $\delta\rho_m$ and $\delta H$. Make sure that the solutions are stable.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The variation of Friedmann equations yields
\[2H\delta H=\frac{\kappa^2}{3}\delta\rho_m,\]
\[\delta\dot H=-\frac{\kappa^2}{2}\delta\rho_m,\]
\[\delta\dot\rho_m+3\rho_m\delta H+ 3H\delta\rho_m=0,\]
where $\kappa^2\equiv8\pi G$.
Solution for the above equation is
\[\delta H=\frac{t_\Lambda}{4\kappa^2}C\tanh(\tau) e^{-f(\tau)},\ \delta\rho_m= \frac{C}{4\kappa^4}e^{-f(\tau)},\]
where \[\tau\equiv\frac{t}{t_\Lambda},\]
\[f(\tau)=-2\tau +\ln\left(-1+e^{4\tau}\right)+2\ln\tanh\tau.\]
Here $C$ is an arbitrary constant. We see that $f(\tau)$ approaches $2\tau$ as $t\rightarrow\infty$ and both $\delta H$ and $\delta\rho_m$ decay which implies that the considered solution is stable.
</p>
</div>
</div></div>

<div id="SCM23"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 18 ===
Rewrite the first Friedman equation in terms of conformal time.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[\frac{1}{a^4}\left( \frac{da}{d\eta } \right)^2=H_0^2\left[ \Omega_m a(\eta )^{-3}+\Omega_{\Lambda} \right]\]
</p>
</div>
</div></div>

<div id="SCM24"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 19 ===
Find relation between the scale factor and conformal time
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[\eta =\frac{1}{H_0}\int\limits_0^a\frac{dx}{\sqrt{x}{{\left( {{\Omega }_{m}}+{{\Omega }_{\Lambda }}{{x}^{3}} \right)}^{1/2}}}\]
</p>
</div>
</div></div>

<div id="SCM25"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 20 ===
Find explicit dependence of the scale factor on the conformal
time.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The integral obtained in the previous problem can be expressed in terms of elliptic integral of the first kind $F\left( \varphi ,k \right)$. For that purpose we rewrite it in the form
$$
\Omega^{1/6}_{\Lambda} \Omega^{1/3}_m H_{0} \eta = \int\limits_0^{u} \dfrac{dx}{\sqrt{x}\sqrt{1 + x^3}}.
$$
The upper limit equals to $u=\left(\dfrac{\Omega_{\Lambda}}{\Omega_m}\right)^{1/3}a.$ The integral can be presented in the form
$$3^{1/4} \Omega^{1/6}_{\Lambda} \Omega^{1/3}_m H_{0} \eta = F\left(\arccos\dfrac{1+(1-\sqrt{3})u}{1+(1+\sqrt{3})u}, \dfrac{\sqrt{2+\sqrt{3}}}{2}\right).$$

Here $y={{3}^{1/4}}\Omega _{\Lambda }^{1/6}\Omega _{m}^{1/3}{{H}_{0}}\eta ,\quad k=\frac{\sqrt{2+\sqrt{3}}}{2}\approx 0.97.$
</p>
</div>
</div></div>

<div id="SCM25"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 21 ===
Find relative density of dark energy $10^9$ years later.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
t = t_\Lambda Arcth\sqrt {\Omega _\Lambda} ;\quad \Omega _{\Lambda } = th^2\frac{t}{t_\Lambda };\\
t_\Lambda \simeq 10.768 \cdot 10^{9}\;\mbox{years}\;\;t = 14.7\cdot 10^{9}\;\mbox{years}\\
\Omega _{\Lambda 0} \simeq 0.77\\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM_17"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 22 ===
Find variation rate of relative density of dark energy. What are its asymptotic values? Plot its time dependence.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
\frac{d\Omega _\Lambda }{dt} = \frac{2}{t_\Lambda }\frac{sh(t/{t_\Lambda })}
{ch^3(t/{t_\Lambda })}; \\
x \equiv \frac{t}{t_\Lambda };\quad \frac{d\Omega _\Lambda }{dx} = 2\frac{sh(x)}{ch^3(x)} \\
\end{gathered}
$$

<gallery widths=600px heights=500px>
File:12_15.JPG|
</gallery>
</p>
</div>
</div></div>

<div id="SCM26"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 23 ===
Estimate size of the cosmological horizon.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The cosmological horizon is defined as
$$
L_{H0} = a_0\int\limits_0^{t_0}\frac{dt}{a\left( t \right)}.
$$
Using the results of the problem \ref{A_t_lambda}, one obtains
$$
L_{H0}= A^{-1/3}\int\limits_0^{t_0}\frac{dt}{sh^{2/3}\left( t/{t_\Lambda } \right)};
$$
</p>
</div>
</div></div>

<div id="SCM_19"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 24 ===
Find time dependence of Hubble parameter and plot it.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
H = \frac{\dot a}{a} =\frac{2}{3}\left(t_\Lambda \right)^{ - 1}cth\left( {\frac{t}{t_\Lambda }} \right)
$$

In the SCM ${t_0} \simeq 1.2t_\Lambda .$ Therefore
$$
H{t_0} \simeq 0.8cth\left( {1.2t/{t_0}} \right)
$$

<gallery widths=600px heights=500px>
File:12_16.JPG|Time dependence of the Hubble parameter (in units of the age of the Universe).
</gallery>

</p>
</div>
</div></div>

<div id="SCM27"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 25 ===
Find time dependence of dark matter density.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Present the first Friedmann equation in the form
\[\rho_m=\frac{1}{8\pi G}(3h^2-\Lambda)\]
and make use of results of the previous problem to obtain
\[\rho_m=\frac{\Lambda}{8\pi G}\frac{1}{\sinh^2(t/t_\Lambda)}.\]

\item {\bf Find the asymptotic (in time) value of the Hubble parameter.}

The required asymptote can be obtained in two ways: first, from the time dependence of the Hubble parameter (see the previous problem), and second, immediately from the first Friedmann equation in the SCM. Taking into account that $\rho_{m} \to 0$ at $t\to \infty$, one gets
$$
H_{\infty} = \frac{2}{3t_\Lambda} = \sqrt {\frac{\Lambda}{3}}.
$$
</p>
</div>
</div></div>

<div id="SCM28"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 26 ===
At present the age of the Universe $t_0\simeq13.7\cdot10^9$ years is close to the Hubble time $t_H=H_0^{-1}\simeq14\cdot10^9$ years.
Does the relation $t^*\simeq t_H(t^*)=H^{-1}(t^*)$ for the age $t^*$
of the Universe hold for any moment of its evolution
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
This relation cannot be satisfied in arbitrary moment of time, as the Hubble parameter has limit value in the SCM (see the previous problem):
$$
\frac{t^*}{t_H(t^*)} = H\left( t^*\right)t^* = \frac{2}{3}\frac{Arcth\sqrt {\Omega _\Lambda(t^*)}}{\sqrt{\Omega_\Lambda (t^*)}}.
$$
For $t\to \infty,\;\Omega _\Lambda\left( t^* \right) \to 1,$ and $$\frac{t^*}{t_H(t^*)} \to \infty. $$

\item {\bf Determine values of state parameter for the dark energy, that provide accelerated expansion of Universe in the present time.}

$$
\begin{gathered}
\sum\limits_i (\rho _{i0} + 3p_{i0}) < 0\; \Rightarrow \rho _{m0} + \rho _{DE0} + 3w_{DE}\rho _{DE0} < 0; \\
w_{DE} < - \frac{1}{3}\left( 1 + \frac{\Omega _{m0}}{\Omega _{DE0}}\right); \\
w_{DE} < - \frac{1}{3}\Omega _{DE0}^{-1};\; w_{DE} < - 0.46\\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM29"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 27 ===
Find current value of the deceleration parameter.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For a multi-component system the deceleration parameter equals to
$$
q = \frac{\Omega}{2} + \frac{3}{2}\sum\limits_i {w_i}\Omega _i
$$
In the SCM $\Omega = 1,w_m = 0,\;w_\Lambda= - 1,$ and
$$
\begin{gathered}
q = \frac{1 - 3\Omega_{\Lambda 0}}{2};\\
q_0 \simeq - 0.6 \\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM_24"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 28 ===
Find the redshift dependence of the deceleration parameter. Analyze the limiting cases.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
q \equiv - \frac{a\ddot a}{\dot a^2} = - \frac{\ddot a}{a}\frac{1}{H^2}; \\
\frac{\ddot a}{a} = - \frac{4\pi G}{3}\sum\limits_i \left( \rho _i + 3p_i \right) = - \frac{1}{2}H_0^2\sum\limits_i \Omega _{i0}\left(1 + 3w_i\right)(1 + z)^{3\left(1 + w_i\right)} ; \\
H^2 = H_0^2\sum\limits_i \Omega _{i0}(1 + z)^{3\left( 1 + w_i \right)}; \\
q = \frac{3}{2}\frac{\sum\limits_i \Omega _{i0}\left( 1 + w_i \right)(1 + z)^{3\left( 1 + w_i \right)}}
{\sum\limits_i \Omega _{i0}(1 + z)^{3\left( 1 + w_i\right)} } - 1 \\
\end{gathered}
$$
In the SCM
$$
q = \frac{1}{2}\frac{\Omega _{m0}(1 + z)^3- 2\Omega _{\Lambda 0} }
{\Omega _{m0}(1 + z)^3+\Omega_{\Lambda 0}}
$$

$$
q(z \to \infty ) = \frac{1}{2},\;q(z \to - 1) = - 1
$$

<gallery widths=600px heights=500px>
File:12_24.JPG|Dependence of the deceleration parameter on the redshift.
</gallery>
</p>
</div>
</div></div>

<div id="SCM_25"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 29 ===
Find and plot the time dependence of the deceleration parameter.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
q = - \frac{\ddot aa}
{\dot a^2} = - \frac{\ddot a}
{aH^2}; \\
a(t) = A^{1/3}sh^{2/3}\left(t/{t_\Lambda}\right);\quad A\equiv \frac{\Omega_ {m0}}
{\Omega _{\Lambda 0}}; \\
\frac{t_0}
{t_\Lambda } = Arth\sqrt {\Omega _{\Lambda 0}} ;\quad H = \frac{2}
{3}\frac{1}
{t_\Lambda }cth\left( \frac{t}
{t_\Lambda }\right); \\
q = \frac{1}
{2}\left[ 1 - 3th^2\left( \frac{t}
{t_\Lambda }\right)\right] \\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM29"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 30 ===
Find the moment of time when the dark energy started to dominate over dark matter. What redshift did it correspond to?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
\left( \frac{\dot a}{a} \right)^2 = \frac{8\pi G}
{3}\rho _{cr}\left( \Omega _{m0}(1 + z)^3+\Omega_{\Lambda 0}\right); \\
\Omega _{m0}(1 + z^*)^3 =\Omega _{\Lambda 0};\\
z^* = \left(\frac{\Omega _{\Lambda 0}}
{\Omega _{m0}}\right)^{1/3}-1; \\
\Omega _{\Lambda 0}\simeq 0.73;\quad \Omega _{m0}\simeq 0.27; \\
z^* \simeq 0.39; \\
\Omega _\Lambda = th^2\left(\frac{t}
{t_\Lambda }\right);\quad \Omega _m = 1 - \Omega _\Lambda = ch^{-2}\left( \frac{t}
{t_\Lambda } \right); \\
\Omega _\Lambda(\tilde t) = \Omega _m(\tilde t) \to \tilde t = \frac{Arsh(1)}{Arth(\sqrt {\Omega _{\Lambda 0} })}t_0 \simeq 9.5\cdot10^9 \,years \\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM30"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 31 ===
Determine the moment of time and redshift value corresponding to the transition from decelerated
expansion of the Universe to the accelerated one.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
\frac{\ddot a}
{a} = \frac{4\pi G}
{3}\rho_{cr}\left[2\Omega_{\Lambda 0} - \Omega _{m0}(1 + z)^3\right]; \\
2\Omega _{\Lambda 0} = \Omega _{m0}(1 + z^*)^3; \\
z^* = \left( \frac{2\Omega_{\Lambda 0}}
{\Omega _{m0}}\right)^{1/3} - 1; \\
\Omega _{\Lambda 0}\simeq 0.73;\quad \Omega _{m0}\simeq 0.27; \\
z^* \simeq 0.745 \\
\end{gathered}
$$

Let us now determine the time moment $t^*$ of the transition
$$
\begin{gathered}
a^* = \frac{1}
{1 + z^*} = \left(\frac{\Omega _{m0}}
{2\Omega _{\Lambda 0}} \right)^{1/3}; \\
t(a) = \frac{2}{3\sqrt{\Omega_{\Lambda 0}}H_0}Arsh\left(\sqrt{ \frac{\Omega_{\Lambda 0}}
{\Omega_{m0}}}a^{3/2} \right); \\
t^* = \frac{2}
{3\sqrt {\Omega _{\Lambda 0}}}H_0Arsh\left(\sqrt {1/2} \right) \\
\end{gathered}
$$
There is an alternative way to solve the problem.
In the multi-component Universe with the corresponding state equations $p_i = w_i\rho _i$ the deceleration parameter $q \equiv - \frac{\ddot aa}{\dot a^2}$ equals to
$$
q = \frac{\Omega }
{2} + \frac{3}
{2}\sum\limits_i w_i\Omega _i
$$
In the case of the SCM $\Omega =1$ and
$$
q = \frac{1}
{2} - \frac{3}
{2}\frac{\Omega _{\Lambda 0}}
{\left[\left(1 + z\right)^3\Omega _{m0}+\Omega_{\Lambda 0}\right]}
$$

From the condition $q=0$ it follows that
$$
z^* =\left(\frac{2\Omega_{\Lambda 0}}
{\Omega _{m0}} \right)^{1/3}- 1
$$
For $\Omega _{\Lambda 0} = 0.73,\;\Omega _{m0} = 0.27$ one obtains that
$$
z^* = 0.745
$$
</p>
</div>
</div></div>

<div id="SCM31"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 32 ===
Solve the previous problem using the derivative $d\eta/d\ln a$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\frac{d\eta}
{d\ln a} = a\frac{dt}{ada} = \frac{1}{aH} = \frac{1}{\dot a}
$$

If the quantity $aH = \dot a$ grows, then $\ddot a > 0,$ which corresponds to the accelerated expansion (inflation) of the Universe. In the SCM the quantity
$$
\frac{aH}{H_0} \simeq \sqrt{ \frac{0.27 + 0.73a^3}{a}}
$$
starts to grow at $a=0.573,$ which corresponds to $z=0.745$
</p>
</div>
</div></div>

<div id="SCM32"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 33 ===
Is dark energy domination necessary for transition to the accelerated expansion?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
No. The transition to the accelerated expansion takes place at $(z=0.75)$, which is considerably earlier than the transition to the dark energy domination stage at $(z=0.45).$
</p>
</div>
</div></div>

<div id="SCM33"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 34 ===
Consider flat Universe composed of matter and dark energy in form of cosmological constant. Find the redshift value corresponding to equality of densities of the both components $\rho_m(z_{eq})=\rho_\Lambda(z_{eq})$ and the one corresponding to beginning of the accelerated expansion $q\left(z_{accel} \right) = 0$. Obtain relation between $z_{eq}$ and $z_{accel}$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
z_{eq} = \left(\frac{\Omega _\Lambda}
{\Omega _{m0}} \right)^{1/3}- 1; \\
z_{accel}= \left(\frac{2\Omega_\Lambda }
{\Omega _{m0}}\right)^{1/3}- 1; \\
z_{accel} = 2^{1/3}\left(z_{eq}+ 1\right) - 1 \\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM34"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 35 ===
Show that density perturbations stop to grow after the transition from dust to $\Lambda$-dominated era.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
On the $\Lambda$-dominated stage the non-relativistic matter density perturbation size depends on conformal time as
\begin{equation}\label{1SCM30}
a(\eta)=-\frac{1}{H\eta}, \,\, \eta=-const\cdotp e^{-Ht},\,\,\eta<0
\end{equation}
where the Hubble parameter $H$ is constant, $H^2=(8\pi/3)G\rho_\Lambda$.
The linearized Einstein equations for the case of single-component medium take the following form in the momentum representation:
\begin{equation}\label{3SCM30}
k^2\Phi+3\frac{a^{'}}{a}+3\frac{a^{'2}}{a^2}\Phi=-4\pi Ga^2\delta\rho,
\end{equation}
\begin{equation}\label{2SCM30}
\Phi^{''}+3\frac{a^{'}}{a}\Phi^{'}+\left(2\frac{a^{''}}{a} - \frac{a^{'2}}{a^2}\right)\Phi=4\pi Ga^2\delta p.
\end{equation}
Assuming that the $\Lambda$-term is constant (which is exactly true for the vacuum energy density) one obtains $\delta\rho_\Lambda=\delta p_\Lambda=0$.
Using \eqref{2SCM30} and taking into account \eqref{1SCM30} we come up with the equation: $${\Phi}^{''}-\frac{3}{\eta}{\Phi}^{'}-\frac{3}{\eta^2}\Phi=0,$$
Solutions of the equation behave as
$$\Phi\propto\eta\propto\frac 1a$$ and $$\Phi\propto\eta^3\propto\frac{1}{a^3}.$$
Use now \eqref{3SCM30} for the modes beyond the horizon and take into account that for dust density $\rho_M\propto a^{-3}.$ Keeping only the first solution for the gravitational potential (the second decays too fast), one can see that the density perturbations do not grow: for the modes under horizon the relation
$$\delta_M\propto a^3\frac{k^2}{a^2}\Phi=const$$ holds.

Therefore the perturbations stop to grow at the transition from the dust-dominated to the $\Lambda$-dominated stage. If the present accelerated expansion of the Universe is actually caused by non-zero cosmological constant, then the structure formation from small primary perturbations is eventually close to end and it will never resume again: the structures larger than the presently observed ones will never appear.
</p>
</div>
</div></div>

<div id="SCM35"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 36 ===
What happens to the velocity fluctuations of non-relativistic matter and radiation with respect to Hubble flow in the epoch of cosmological constant domination?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM36"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 37 ===
Find the ratio of baryon to non-baryon components in the galactic halo.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM37"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 38 ===
Imagine that in the Universe described by SCM the dark energy was instantly switched off. Analyze further dynamics of the Universe.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="local-group"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 39 ===
Estimate density of dark energy in form of cosmological constant using the Hubble diagram for the neighborhood of the Local group.
<gallery widths=600px heights=500px>
File:im_eps.JPG|Hubble diagram for the neighborhood of the Local group.
</gallery>
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
An elegant and simple method was recently proposed (see A.D.Chernin, UFN 178, 267 (2008)), which allows not only to become convinced in the existence of the dark energy but also to estimate its density. The method is based on the assumption, that the dark energy (in form of the cosmological constant) governs the Hubble flow even on comparably small distances $\left( \sim \,Mpc \right)$. Consider the resulting force $\vec{F}$ (per unit mass), acting on a galaxy from the picture \ref{hubble_diagram}. The force is a sum of two contributions:
$$
\vec{F}=\vec{F}_N+\vec{F}_\Lambda
$$
Here $\vec{F}_N$ is the force produced by the total mass of the Local group, and $F_{\Lambda }$ is the force created by the dark energy. For a crude estimate let us neglect the non-spherical mass distribution in the Local group. Then the force acting on a satellite galaxy in a given point $R$ outside the local Group equals to
$$
F=-\frac{GM_{LG}}{R^2}+\frac{8\pi G}{3}{\rho_{\Lambda }}R,
$$
where $M_{LG}\simeq (1\div 3) \times 10^{12}M_{\odot }$ is the total mass of the Local group. As can be seen from the picture, the transition from domination of the local group gravity to the vacuum domination takes place on the distance of order $R_{0}=1.8\div 2\; \, Mpc.$ It allows to make a rough estimate of the dark energy density from the condition $F\left( R=R_0 \right)=0.$ The estimate results in the following
$$
(0.1\pm 0.03)<\rho_V<(1\pm 0.3)\times 10^{-26} \, kg/m^3.
$$
The proximity of the obtained crude estimate to the actual value of the dark energy density in the SCM is striking.
$$
\rho_{V}\simeq (0.75\pm 0.05)\times 10^{-26}\, kg/m^3.
$$
</p>
</div>
</div></div>

<div id="SCM38"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 40 ===
Estimate the Local group mass by methods used in the previous problem.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM39"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 41 ===
Find "weak" points in the argumentation of the two preceding problems.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM40"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 42 ===
The product of the age of the Universe and current Hubble parameter (the Hubble's constant) is a very important test (Sandage consistency test) of internal consistency for any model of Universe. Analyze the
parameters on which the product $H_0 t_0$ depends in the Big Bang model and in the SCM.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM41"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 43 ===
Show that for a fixed source of radiation the luminosity distance for high redshift values in flat Universe is greater for the dark energy dominated case compared to the non-relativistic matter dominated one.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the case of Universe filled by dark energy and non-relativistic matter, the photometric distance is equal to
$$
d_L = \frac{1 + z}
{H_0}\int_0^z \frac{dz'}
{\sqrt{\Omega _{m0}\left(1 + z'\right)^3+ \Omega _{DE0}\left(1 + z' \right)^{3\left( 1 + w_{DE}\right)}}}
$$

In the case $\Omega _{m0} = 1,\;\Omega _{DE0} = 0$
one obtains
$$
d_L = \frac{2}
{H_0}\left[1 + z -(1 + z)^{1/2}\right]
$$
and in the case $\Omega _{m0} = 0,\;\Omega _{DE0} = 1$
$$
d_L = \frac{1}{H_0}z(1 + z)
$$
instead.

In means that the supernova appear dimmer in the dark energy dominated Universe.
</p>
</div>
</div></div>

<div id="SCM42"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 44 ===
In the observations that discovered the accelerated expansion of the Universe the researchers in particular detected two $Ia$ type supernovae: $1992P$, $z=0.026$, $m=16.08$ and $1997ap$, $z=0.83$,
$m=24.32$. Show that these observed parameters are in accordance with the SCM.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The apparent $m$ and absolute $M$ stellar magnitudes are related with the photometric distance $d_L$ by the expression
\begin{equation}\label{p(1)}
m - M = 5\log _{10}\left(\frac{d_L}
{\mbox{Mpc}} \right) + 25
\end{equation}
The photometric distance in the SCM is equal to
\begin{equation}\label{p(2)}
d_L = \frac{1 + z}
{H_0}\int_0^z \frac{dz'}
{\sqrt {\Omega _{m0}(1 + z)^3 + \Omega _\Lambda}}
\end{equation}
For small $z$ $d_L(z) \simeq z/H_0.$ Using the supernova $1992P$ with small $z=0.026$ one can find $M = - 19.09.$ Assuming that the $1a$-type supernova are standard candles (they have equal absolute magnitudes), one obtains from the relation \ref{p(1)} supernova with $z=0.83$
$$H_0d_L\simeq 1.16$$
From the other hand it follows from the relation \ref{p(1)} that in the case
$\Omega _{m0} = 1,\;\Omega_\Lambda = 0$
$$
H_0d_L\simeq 0.95
$$
In the SCM one gets much better agreement:
$$
H_0d_L \simeq 1.2
$$
</p>
</div>
</div></div>

<div id="SCM43"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 45 ===
Find the redshift value, at which a source of linear dimension $d$ has minimum visible angular
size.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The apparent angular size equals to $$\delta \theta = {{(1 + z)^2 d } \over {d_L (z)}},$$
where $d_L (z)$ is the photometric distance from a source with the redshift $z$. In the case of cosmological constant domination the photometric distance is:
$$d_L (z) = - {z \over {H_0 \Omega _\Lambda ^{1/2} }},$$
In the SCM (see the previous problem)
$$
d_L = \frac{1 + z}
{H_0}\int_0^z \frac{dz'}
{\sqrt {\Omega _{m0}(1 + z)^3 + \Omega _\Lambda}}
$$
then the angular size is
$$\delta \theta = - {{H_0 \Omega _\Lambda ^{1/2} d} \over 2}{{(1 + z)^2 } \over z},$$
and $z = - 1$. It corresponds to infinitely far future.
</p>
</div>
</div></div>

<div id="SCM44"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 46 ===
Compare the observed value of the dark energy density with the
one expected from the dimensionality considerations (the cosmological constant problem).
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The dimensionality considerations suggest the following value for density of the dark energy in form of the cosmological constant
\[\rho_\Lambda^{(dim)}\sim M_{Pl}^4\approx10^{76}GeV^4.\]
At the same time the ''observed'' value fixed by the SCM is
\[\rho_\Lambda^{(obs)}\approx10^{-48}GeV^4.\]
This huge disagreement
\[\frac{\rho_\Lambda^{(dim)}}{\rho_\Lambda^{(obs)}}>10^{120}\]
is called the ''cosmological constant problem''. It presents a fundamental problem for elementary particles theory and physics in general, rather than for cosmology itself. (Look for details in R.Bousso, [http://arxiv.org/abs/0708.4231 TASI Lectures on the cosmological constant])
</p>
</div>
</div></div>

<div id="SCM45"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 47 ===
Determine the density of vacuum energy using the Planck scale as cutoff parameter.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For a quantum field with a given mode of frequency $\omega$, the zero-point energy is $\pm\omega/2$ with the plus sign for a bosonic field and the minus sign for a fermionic field. The total zero-point energy is then $\pm\sum\limits_i\omega_i/2$. In the continuum limit, we have for a free field (bosonic or fermionic)
\[\frac12\sum\limits_i\omega_i=\frac12\int\frac{d^3xd^3k}{(2\pi)^3}(k^2+m^2)^{1/2}=V\int\frac{k^2dk}{4\pi^2}(k^2+m^2)^{1/2}.\] This integral diverges, thus we need to introduce the physical cutoff. In cosmological context this cutoff must be the Planck mass at largest. With the cutoff $k_{max}\sim M_{Pl}$ one obtains
\[\rho_\Lambda=\int_0^{k_{max}}\frac{k^2dk}{4\pi^2}(k^2+m^2)^{1/2} \approx\frac{k_{max}^4}{16\pi^2}=\frac{M_{Pl}^4}{16\pi^2} \approx10^{74}GeV^4.\]
This result only slightly modifies the dimensional considerations given in the previous problem.
</p>
</div>
</div></div>

<div id="SCM46"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 48 ===
Identifying the vacuum fluctuations density with the observable dark energy value in SCM, find the required frequency cutoff magnitude in the fluctuation spectrum.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[\nu_{max}=10^{12}Hz.\]
</p>
</div>
</div></div>

<div id="SCM47"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 49 ===
What purely cosmological problem originates from the divergence of the zero-point energy density?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
If the zero-point energy density is infinite, then our Universe is infinitely curved and the space size is infinitely small.
</p>
</div>
</div></div>

<div id="SCM48"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 50 ===
With exact supersymmetry, the bosonic contribution to cosmological constant is canceled by its fermionic counterpart. However, we know that our world looks like not supersymmetric. Supersymmetry, if exists, has to be broken above or around $100GeV$ scale. Compare the observed value of the dark energy density with the one expected from broken supersymmetry.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
If we take the $k_{max}^{(SUSY)}\sim100GeV$ cutoff, then
\[\rho_\Lambda^{(SUSY)}=\frac{\left(k_{max}^{(SUSY)}\right)^4}{16\pi^2}\approx10^6GeV^4.\]
\[\frac{\rho_\Lambda^{(SUSY)}}{\rho_\Lambda^{(obs)}}>10^{54}.\]
Thus supersymmetry does not solve the problem of the cosmological constant.
</p>
</div>
</div></div>

<div id="SCM49"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 51 ===
Determine duration of the inflation period.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Let us see what happens with the relative density during all the history of Universe. From the first Friedmann equation it follows that in the matter-dominated epoch $\left| \Omega - 1 \right| \sim t^{2/3},$ and in the radiation-dominated epoch $\left| \Omega - 1 \right| \sim t^{2/3}.$ Then assume that inflation started at $t_i,$ and finished at $t_f.$ Then in the period of $t_f < t < t_{eq} = 50000$ years the radiation dominated and the range $t_{eq} < t < t_0 $ corresponds to the matter domination epoch. Then the presently observed difference
$$
\left|\Omega (t_0 ) - 1 \right| = \left| \Omega (t_i ) - 1 \right|e^{ - 2H(t_f - t_i )} \left( \frac{t_{eq} }
{t_f } \right)\left( \frac{t_0}
{t_{eq} }\right)^{2/3}
$$
According the SCM $\left|\Omega \left( t_0\right) - 1 \right| < 0.04.$ If one assumes that before the beginning of the inflation there was $\left|\Omega \left( t_i \right) - 1 \right| \sim 1,$ then
$$
N > \frac{1}{2}\ln \left[\frac{\Omega \left( t_i \right) - 1}{\Omega \left(t_0\right) - 1}\left( \frac{t_{eq}}{t_f } \right)\left( \frac{t_0}{t_{eq}} \right)^{2/3} \right] \approx 60
$$

Assume that radiation dominated before the inflation started. Then $H = \frac{1}{2t}.$ If at the moment of time $t_i$ the vacuum dark energy started to dominate and then the expansion continued with constant rate, then $H\approx 1/t_i$ during the inflation. And finally one obtains that
$$
N = H\left(t_f - t_i\right) \approx \frac{t_f - t_i }
{t_i }
$$
it is commonly assumed that the inflation finishing time is of order of the Grand Unification time $t_f\approx 10^{-35}.$ Therefore $t_f\approx 10^{-38}.$ Evidently the inflation could last longer as we used only lower estimate for $N.$
</p>
</div>
</div></div>

<div id="SCM50"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 52 ===
Plot the dependence of luminosity distance $d_L$ (in units of $H_0^{-1}$) on the redshift $z$ for the two-component flat Universe with non-relativistic liquid ($w=0$) and cosmological constant
($w=-1$). Consider the following cases:

a) $\Omega_\Lambda^0=0$;

b) $\Omega_\Lambda^0=0.3$;

c) $\Omega_\Lambda^0=0.7$;

d) $\Omega_\Lambda^0=1$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM51"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 53 ===
For a Universe filled by dark energy with state equation $p_{DE}=w_{DE}\rho_{DE}$ and non-relativistic matter obtain the Taylor series for $d_L$ in terms of $z$ near the observation point
$z_0=0$. Explain the obtained result.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the universe filled by dark energy with the state equation $p_{DE}=w_{DE}\rho_{DE}$ and non-relativistic dark matter the photometric distance takes on the form:
\begin{equation}
\label{d_L}
d_L = \frac{1 + z}
{H_0}\int_0^z \frac{dz'}
{\sqrt{\Omega _{m0}\left(1 + z'\right)^3+ \Omega _{DE0}\left(1 + z' \right)^{3\left( 1 + w_{DE}\right)}}}
\end{equation}
Expand the integrand in Taylor series in $z$ in vicinity of the observation point $z_0=0$ to obtain
$$
\begin{gathered}
\label{tey}
\frac{1}{\sqrt{\Omega _{m0}\left(1 + z'\right)^3+ \Omega _{DE0}\left(1 + z' \right)^{3\left( 1 + w_{DE}\right)}}}\simeq \\
\simeq 1 -\frac{3}{2}\left(1+w_{DE}\Omega _{DE0}\right)z+\frac{3}{4}\left(3w_{DE}\Omega _{DE0}-1\right)z^2.\nonumber
\end{gathered}
$$
Substitution of the decomposition (\ref{tey}) into the integral (\ref{d_L}) results in the following:
\begin{equation}
\label{d_L}
d_L = \frac{1 + z}
{H_0}\left(z -\frac{3}{4}\left(1+w_{DE}\Omega _{DE0}\right)z^2+\frac{1}{4}\left(3w_{DE}\Omega _{DE0}-1\right)z^3\right).
\end{equation}
</p>
</div>
</div></div>

<div id="SCM_45"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 54 ===
Determine position of the first acoustic peak in the CMB power spectrum produced by baryon oscillations on the surface of the last scattering.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Early Universe was filled by photon-baryon plasma, which can be treated as a single-component liquid. The baryons were trapped by the potential walls generated by density fluctuations, and they were eventually compressed. The contraction lead to heating of the plasma and therefore to increasing radiative pressure of the photons which is directed outside. Finally the radiative pressure stop the compression and lead to expansion of the plasma. During the expansion the plasma cools down and its radiative pressure decreases. The gravity starts to dominate again leading to repeated compression. The concurrence between the gravity and the pressure lead to longitudinal (acoustic) oscillations in the photon-baryon fluid. When the matter and the radiation get decoupled in the recombination process, the picture of the acoustic oscillations remains frozen into the CMB. Today we detect the evidence of the primordial acoustic waves (dense and diluted regions) in form of the primary anisotropy in the CMB.

It is well known that any acoustic wave, whatever complicatedly shaped, can be represented in form of superposition of modes with different wave numbers $k$, $k\propto 1/\lambda $. Every mode $\lambda $ corresponds to certain angular scale $\theta $ on the sky. Therefore in order to facilitate the comparison of the theory with observations, one should use the angular (multipole) decomposition in terms of the Legendre polynomials $P_{l} (\cos \theta )$, instead of the Fourier transform in terms of sines and cosines. Order of the polynomial $l$ plays the same role as the index $k$ in the Fourier decomposition. For $l\ge 2$ the Legendre polynomials are oscillating functions in the interval $\left[1,-1\right]$. Number of the oscillations increases with growth of $l$. Therefore
\begin{equation} \label{acust1_}
l\propto \frac{1}{\theta }
\end{equation}
The temperature fluctuations can be observed and qualitatively analyzed with help of paired measurements in the directions $\stackrel{\frown}{n},\stackrel{\frown}{n}'$, so that $\stackrel{\frown}{n}\cdot \stackrel{\frown}{n}'=\cos \theta $. Taking average over all the pairs in the assumptions of Gaussian fluctuations, one obtains the two-point correlator $C(\theta )$, which can be presented in terms of the multipole decomposition
\begin{equation} \label{acust2_}
\left\langle \delta T(\stackrel{\frown}{n})\cdot \delta T(\stackrel{\frown}{n}')\right\rangle \equiv C(\theta )=\sum _{l}\frac{2l+1}{4\pi } C_{l} P_{l} (\cos \theta )
\end{equation}
where $C_{l} $ are the coefficients.

As we already mentioned, the analysis of the temperature fluctuations enables us to clarify the structure of the longitudinal oscillations. The mode with the maximum wavelength corresponds to the maximum angular size of the primary anisotropy. This fundamental mode was detected the first. There are reliable evidences of the fact that the second and the third modes were also already detected. The distance $r_{s} $, passed by the acoustic wave during the time period before the recombination, is called the sound horizon. The sound horizon is fixed by (or rather fixes) the physical scale o the last scattering surface. Size of the sound horizon depends on values of the physical parameters. Distance to the last scattering surface $d_{sls} $ depends on the cosmological parameters too. Together they determine the angular size of the sound horizon$\theta _{s} $ (see Fig.)

<gallery widths=600px heights=500px>
File:acustic_pic.JPG|Dependence of the deceleration parameter on the redshift.
</gallery>

\begin{equation} \label{acust3_}
\theta _{s} \approx \frac{r_{s} }{d_{sls} }
\end{equation}
Analysis of the temperature anisotropy allows to determine $\theta _{s} $. Varying the cosmological parameters in $r_{s} ,d_{sls} ,$ one achieves the best agreement with the observations and thus fixes the cosmological parameters.

We can estimate the sound horizon size as the length passed by sound from the moment $t=0$ to the recombination time $t_{*} $
\begin{equation} \label{acust4_}
r_{s} (z_{*} ;\Omega _{b} ,\Omega _{r} )\approx \int _{0}^{t_{*} }c_{s} dt
\end{equation}
where $z_{*} $ is the redshift of the recombination moment $\left(z_{*} \approx 1100\right)$, $c_{s} $ is the sound speed in the photon-baryon medium
\begin{equation} \label{acust5_}
c_{s} \approx c\left[3\left(1+\frac{3\Omega _{b} }{\Omega _{r} } \right)\right]^{-1/2}
\end{equation}
To determine $dt$ one should use the first Friedmann equation in the form
\begin{equation} \label{acust6_}
\left(\frac{da}{dt} \right)^{2} =H_{0}^{2} \left[\Omega _{r0} \left(\frac{a_{0} }{a} \right)^{2} +\Omega _{m0} \frac{a_{0} }{a} +\Omega _{k0} +\Omega _{\Lambda 0} \left(\frac{a}{a_{0} } \right)^{2} \right]
\end{equation}
Making use of the relations $\frac{a_{0} }{a} =1+z$ and $\Omega _{r} +\Omega _{m} +\Omega _{\Lambda } +\Omega _{k} =1$, one obtains
\begin{equation} \label{acust7_}
dt=H_{0}^{-1} (1+z)^{-1} \left\{\left(1+z\right)^{2} \left(1+\Omega _{m,0} z\right)+z(z+2)\left[\left(1+z)^{2} \Omega _{r.0} -\Omega _{\Lambda ,0} \right)\right]\right\}^{-1/2} dz
\end{equation}
Distance to the last scattering surface, which corresponds to its angular size, is determined by the quantity called the angular diameter distance. It is connected by the photometric distance $d$ by the simple relation
\begin{equation} \label{acust8_}
d_{sls} =\frac{d}{(1+z_{*} )^{2} }
\end{equation}
The position of the first acoustic peak is determined by
\begin{equation} \label{acust9_}
l\approx \frac{d_{sls} }{r_{s} }.
\end{equation}
Let us now predict the position of the first peak for the case of flat Universe. To the leading order the sound speed is
\[c_{s} =\frac{c}{\sqrt{3} }. \]
Let us assume also that in the recombination epoch the Universe was matter-dominated. Under such assumptions one gets
\begin{equation} \label{acust10_}
r_{s} =\frac{c_{s} }{H_{0} \sqrt{\Omega _{m} } } \int _{z_{*} }^{\infty }(1+z)^{-5/2} dz =r_{s} =\frac{2c_{s} }{3H_{0} \sqrt{\Omega _{m} } } \left(1+z_{*} \right)^{-3/2}
\end{equation}
Using the standard procedure (see section 4), we obtain for the comoving radial coordinate $r_{sls} $ of the last scattering surface the following
\begin{equation} \label{acust11_}
r_{lsr} =\frac{c}{H_{0} } \int _{0}^{z_{*} }\left[\Omega _{m} (1+z)^{^{3} } +\Omega _{\Lambda } \right] ^{-1/2} dz
\end{equation}
Replacing the integrand by its binomial decomposition
\[\Omega _{m}^{-1/2} (1+z)^{-3/2} -\left(\Omega _{\Lambda } /2\Omega _{m}^{3/2} \right)\left(1+z\right)^{-9/2} \]
and taking into account that
\[d_{sls} =r_{sls} /(1+z_{*} ),\]
we obtain
\begin{equation} \label{acust12_}
d_{sls} =\frac{2c}{7H_{0} (1+z_{*} )} \left\{7\Omega _{m}^{-1/2} -2\Omega _{\Lambda } \Omega _{m}^{-3/2} +{\rm O} \left[\left(1+z_{*} \right)^{-1/2} \right]\right\}.
\end{equation}
Using the relation $\Omega _{\Lambda } =1-\Omega _{m} $ and neglecting the higher order terms, one obtains
\begin{equation} \label{acust13_}
d_{sls} \approx \frac{2c\Omega _{m}^{-1/2} }{7H_{0} (1+z_{*} )} \left(9-2\Omega _{m}^{3} \right)
\end{equation}
Combination of the relations \eqref{acust10_} and \eqref{acust13_} gives for the first acoustic peak the following
\begin{equation} \label{acust14_}
l\approx \frac{d_{sls} }{r_{s} } \approx 0.74\sqrt{1+z_{*} } \left(9-2\Omega _{m}^{3} \right)\simeq 220.
\end{equation}
This result agrees very well with numerous observations.
</p>
</div>
</div></div>

<div id="SCM52"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 55 ===
\bf Compare the asymptotes of time dependence of the scale factor
$a(t)$ for the SCM and de Sitter models. Explain physical reasons of
their distinction.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM53"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 56 ===
Redshift for any object slowly changes due to the acceleration (or deceleration) of the Universe expansion. Estimate change of velocity in one year.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

$$
\Delta z = \dot z \Delta t = H_0 \left(1 + z - {H(z)\over H_0} \right)\Delta t
$$

$$H(z)=H_0 \left[\Omega_{m0}(1+z)^3+\Omega_{\Lambda 0}\right]^{1/2}$$
$$\Omega_{m0}+\Omega_{\Lambda 0}=1,~\Omega_{m0}\approx 0.3$$
$$\Delta z \approx 2\cdot 10^{-11}$$
$$\Delta v = c{\Delta z\over 1+z}\approx 0.25~\mbox{cm/s}$$

</p>
</div>
</div></div>

<div id="SCM54"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 57 ===
Determine the lower limit of ratio of the total volume of the Universe to the observed one?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
If the Universe is the 3-sphere, then its radius is $R_0=a_0$
$$
R_0 = \frac{1}{H_0 \sqrt {\left| \Omega _{curv} \right|} }
$$
For the SCM parameters the cosmological horizon equals to
$$
L_{p,0} = a_0 \int_0^{t_0 } \frac{dt}{a(t)} = \frac{3.6}{H_0}
$$
We assumed that $a(t)\sim t^{2/3}.$ Therefore
$$
\frac{R_0}
{L_{p,0} } = \frac{1}
{3.6\sqrt {\left| \Omega _{curv}\right|} }
$$
Taking into account that $\left|\Omega _{curv} \right| < 0.02,$ one obtains $\frac{a_0}{L_{p0} } > 2$.
Number of the regions similar to the one inside our horizon equals to the ratio of the 3-sphere volume $2\pi ^2 a_0^3 $ to the volume of the observed Universe
$$
N = \frac{2\pi ^2 a_0^3}
{(4\pi /3)L_{p0}^3 } \simeq 4.7\left(\frac{a_0 }
{L_{p0} } \right)^3 > 38
$$
Thus we can see not more than $3 \%$ of the Universe volume.
</p>
</div>
</div></div>

<div id="SCM55"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 58 ===
What is the difference between the inflationary expansion in the early Universe and the present accelerated expansion?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The difference is in the fact that in the present time there is about $27 \%$ of non-relativistic matter which was absent in the inflation period. The numerical values of Hubble parameter in the beginning of the inflation was very different from the one at the beginning of the current period of accelerated expansion of Universe.
</p>
</div>
</div></div>

<div id="SCM56"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 59 ===
Compare the values of Hubble parameter at the beginning of the
inflation period and at the beginning of the present accelerated
expansion of Universe.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
h^2 (x) = \Omega _{m0} x^3 + (1 - \Omega _{m0} )x^\alpha ;\quad \alpha = 3(1 + w)
$$
Assume for simplicity that the Universe is spatially flat. From the definition $O(x)$
it follows that
$$
O(x) = \Omega _{m0} + \left( 1 - \Omega _{m0}\right)\frac{x^\alpha - 1}
{x^3 - 1}
$$
The required result follows from the fact that $\alpha =0$ for cosmological constant, $\alpha >0$ for quintessence and $\alpha <0$ for phantom energy.
</p>
</div>
</div></div>

<div id="SCM57"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 60 ===
Plot the dependencies $h(x)=H(x)/H_0(x)$, where $x=1+z$, in the SCM, for quintessence and for phantom energy cases.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Plots of the dependencies $
h(x) = H(x)/H_0 (x);\quad x = 1 + z
$ in the SCM are presented on Fig. \ref{fig:52} for phantom energy and quintessence.
<gallery widths=600px heights=500px>
File:12_52.JPG|Dependence of the deceleration parameter on the redshift.
</gallery>
</p>
</div>
</div></div>

<div id="SCM58"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 61 ===
Find the constraints imposed by the Weak Energy Condition (WEC) for the dark energy on the redshift-dependent Hubble parameter. ([http://arxiv.org/abs/astro-ph/0703416 Inspired by A.Sen, R.Scherrer])
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The redshift-dependent Hubble parameter $H(z)$ is defined by
\[H^2(z)=\frac{8\pi G}{3}(\rho_m+\rho_{DE})\]
This equation can be rewritten as
\begin{equation}\label{SCM_eq_76_1}
E^2(z)\equiv\frac{H^2(z)}{H^2_0(z)}=\Omega_{m0}(1+z)^3+(1-\Omega_{m0})\frac{\rho_{DE}}{\rho_{DE0}}
\end{equation}
The WEC forces \[\frac{\rho_{DE}}{\rho_{DE0}}\ge1 \ \left(\frac{d\rho}{dz}\ge0\right)\] for $z>1$, so equation (\ref{SCM_eq_76_1}) becomes
\begin{equation}\label{SCM_eq_76_2}
E^2(z)\ge\Omega_{m0}(1+z)^3+(1-\Omega_{m0})
\end{equation}
Going back to equation (\ref{SCM_eq_76_2}) and taking the derivative with respect to $z$ (which we denote throughout with a prime) gives:
\[
2E(z)\frac{dE}{dz}=3\Omega_{m0}(1+z)^2+\frac{1-\Omega_{m0}}{\rho_{DE0}}\frac{d\rho_{DE}}{dz}.
\]
Now the WEC implies that \[\frac{d\rho_{DE}}{dz}\ge0,\] so we get
\begin{equation}\label{SCM_eq_76_3}
E(z)\frac{dE}{dz}\ge\frac32\Omega_{m0}(1+z)^2.
\end{equation}
Equations (\ref{SCM_eq_76_2}) and (\ref{SCM_eq_76_3}) give the constraints that the WEC for the dark energy places on the redshift-dependent Hubble parameter.
</p>
</div>
</div></div>

<div id="SCM59"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 62 ===
Using the statefinders, show that the power law cosmology mimics SCM model at (see Chapters 3 and 9)
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Recall that the statefinders are defined as
\[r\equiv\frac{\dddot a}{aH^3},\ s\equiv\frac{r-1}{3(q-1/2)}, \ q\ne\frac12.\]
The SCM model corresponds to the point $r=1,\, s=0$ in the $(s,r)$ plane. In power-law cosmology (see Chapter 3) the statefinders are given by
\[r=2q^2+q,\, s=\frac23(q+1).\]
We note that $r=1,\, s=0$ for $q=-1$. Thus the power law cosmology mimics SCM model with $q=-1$.
</p>
</div>
</div></div>

Evolution of Universe

2012-10-04T17:50:47Z

Den: /* Problem 4 */

[[Category:Standard Cosmological Model|2]]

__NOTOC__
<div id="SCM-10"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 1 ===
Rewrite the first Friedman equation in terms of redshift and analyze the contributions of separate components at different stages of Universe evolution.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM11"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2 ===
Find the time dependence for the scale factor and analyze asymptotes of the dependence. Plot $a(t)$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM12"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 3 ===
Determine the redshift value corresponding to equality of radiation and matter densities.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
\rho _r(t) = \rho _m(t);\\
\rho _{r0}\frac{a_0^4}{a^4} = \rho _{m0}\frac{a_0^3}{a^3};\\
\frac{\rho _{m0}}{\rho _{r0}} = \frac{\Omega _{m0}}{\Omega _{r0}} = \frac{a_0}{a} = 1 + z;\\
\Omega _{r0} = 5 \cdot {10^{ - 5}} \\
z = \frac{\Omega _{m0}}{\Omega _{r0}} - 1 \approx 5400\\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM_9"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 4 ===
Construct effective one-dimensional potential (see [[Solutions_of_Friedman_equations_in_the_Big_Bang_model#dyn15|problem]] of Chapter 3)
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
<div id="V(x)"></div>
$$
V(x) = - \frac{1}{2}\sum\limits_i \Omega _{i0}x^{ - (1 + 3w_i)};\quad x \equiv \frac{a}{a_0}
$$
In the SCM
$$
V(x) = - \frac{1}{2}\Omega _{m0}x^{-1}-\frac{1}{2}\Omega _{\Lambda 0}x^2 \simeq - 0.135x^{-1}-0.365x^2
$$

<gallery widths=600px heights=500px>
File:12_9.JPG|
</gallery>
</p>
</div>
</div></div>

<div id="SCM13"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 5 ===
Show that the following holds: $\dot H= -4\pi G\rho_m$ and $\ddot H= 12\pi G\rho_m H$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM_18"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 6 ===
Expand the scale factor in Taylor series near the time moment
$t_0$:
\[\frac{a(t)}{a(t_0)}=1+\sum\limits_{n=1}^\infty\frac{A_n(t_0)}{n!}[H_0(t-t_0)]^n;\quad A_n\equiv\frac{1}{aH^n}\frac{d^na}{dt^n}\]
and calculate values for few first coefficients $A_n$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The coefficients $A_1$ and $A_2$ equal to:
$$A_1 = \frac{\dot a}{aH} = 1;\quad A_2 = \frac{\ddot a}{aH^2} = - q = 1 - \frac{3}{2}{\Omega _m}$$
The parameters $A_n(n > 2)$ can be calculated by consequent differentiation of the relation
$$\ddot a = a\left( \dot H + H^2 \right)$$
The time derivatives of the scale factor can be determined making use of the fact that in SCM $\dot H = - 4\pi G\rho _m$, and $\rho _m = \rho _0a^{ - 3}$. For example,
$$A_3 = 1 + \frac{\ddot H}{H^3} + 3\frac{\dot H}{H^2}$$
Using $\ddot H = 12\pi G{\rho _m}H,\;\dot H = - 4\pi G\rho _m$, one finds that
$$A_3 = 1$$
Let us present the expressions for some consequent expansion coefficients
\[\begin{array}{l}
{A_4} = 1 - \frac{3^2}{2}\Omega _m;\\
{A_5} = 1 + 3\Omega _m + \frac{3^3}{2}\Omega _m^2;\\
{A_6} = 1 - \frac{3^3}{2}\Omega _m - 3^4\Omega _m^2 - \frac{3^4}{4}\Omega _m^3
\end{array}\]
</p>
</div>
</div></div>

<div id="SCM14"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 7 ===
Show that all the coefficients $A_n$ can be expressed through
elementary functions of the deceleration parameter $q$ or the
density parameter \[\Omega_m=\frac23(1+q).\]
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM15"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 8 ===
Consider the case of flat Universe filled by non-relativistic matter and dark
energy with state equation $p_X=w\rho_X$, where the state parameter $w$ is parameterized as the following \[w=w_0+ w_a(1-a)=w_0 + w_a \frac{z}{1+z}.\]
Express current values of cosmographic parameters through $w_0$ and
$w_a$. % (See [http://arxiv.org/abs/0802.1583 Cosmography of f(R) gravity])
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the considered case
\[\frac{H^2(z)}{H_0^2} = \Omega _m(1 + z)^3 + \Omega _X(1 + z)^{3\left( 1 + w_0 + w_a \right)}e^{ - \frac{3w_az}{1 + z}}\]
Calculation of the cosmographic parameters requires in general to integrate $H(z)$ in order to obtain $a(t).$ One can avoid this procedure with help of the relation
\[\frac{d}{dt} = - (1 + z)H(z)\frac{d}{dz}.\]
We can, given $H(z)$, apply that relation to calculate $\dot H,\ddot H, \dddot H, \ddddot H $ and so on. Using the expressions for time derivatives of Hubble parameter obtained in the problem (\ref{equ61_6}), Chapter 2,
\[\begin{array}{l}
\dot H = - H^2(1 + q);\\
\ddot H = H^3\left( j + 3q + 2 \right);\\
\dddot H = H^4\left[ s - 4j - 3q(q + 4) - 6 \right];\\
\ddddot H = H^5\left[ l - 5s + 10\left( q + 2 \right)j + 30(q + 2)q + 24 \right]
\end{array}\]
one can express the current values of the cosmological parameters (for $z = 0$) in terms of $w_0,w_a:$
\[\begin{array}{l}
q_0 = \frac{1}{2} + \frac{3}{2}\left( 1 - \Omega _m \right)w_0;\\
j_0 = 1 + \frac{3}{2}\left( 1 - \Omega _m \right)\left[3w_0\left( 1 + w_0 \right) + w_a \right];\\
s_0 = - \frac{7}{2} - \frac{33}{4}\left( 1 - \Omega _m \right)w_a - \frac{9}{4}\left( 1 - \Omega _m \right)\left[ 9 + \left(7 - \Omega _m \right)w_a \right]w_0 - \\
- \frac{9}{4}\left( 1 - \Omega _m \right)\left( 16 - 3\Omega _m \right)w_0^2 - \\
- \frac{27}{4}\left( 1 - \Omega _m \right)\left(3 - \Omega _m \right)w_0^3;\\
l_0 = \frac{35}{2} + \frac{1 - \Omega _m}{4}\left[ 213 + (7 - \Omega _m)w_a \right]w_a + \\
+ \frac{1 - \Omega _m}{4}\left[ 489 + 9(82 - 21\Omega _m)w_a \right]w_0 + \\
+ \frac{9}{2}\left( 1 - \Omega _m \right)\left[ 67 - 21\Omega _m + \frac{3}{2}(23 - 11\Omega _m)w_a \right]w_0^2 + \\
+ \frac{{27}}{4}\left( {1 - {\Omega _m}} \right)(47 - 24{\Omega _m})w_0^3 + \\
+ \frac{81}{2}\left( 1 - \Omega _m \right)(3 - 2\Omega _m)w_0^4
\end{array}\]
</p>
</div>
</div></div>

<div id="SCM16"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 9 ===
Show that the results of the previous problem applied to SCM coincide with
the ones obtained in the [[#SCM_18|problem]].
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the SCM $\left( {{w_0},{w_a}} \right) = \left( { - 1,0} \right)$ and therefore
\[\begin{array}{l}
{q_0} = \frac{1}{2} - \frac{3}{2}\left( {1 - {\Omega _m}} \right);\\
{j_0} = 1;\\
{s_0} = 1 - \frac{9}{2}{\Omega _m};\\
{l_0} = 1 + 3{\Omega _m} + \frac{{27}}{2}\Omega _m^2
\end{array}\]
</p>
</div>
</div></div>

<div id="SCM10"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 10 ===
Photons with $z=0.1,\ 1,\ 100,\
1000$ are registered. What was the Universe age $t_U$ in the moment of
their emission? What period of time $t_{ph}$ were the photons on the
way? Plot $t_U(z)$ and $t_{ph}(z)$
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
t_U=\int_z^\infty\frac{dz}{(1 + z)H(z)} ;\; t_UH_0=\int_z^\infty\frac{dz}{(1 + z)\sqrt{\Omega _{m0}(1 + z)^3 + \Omega _{\Lambda 0} } }; \\
t_{ph}H_0=\int_0^z \frac{dz}{(1 + z)\sqrt{\Omega _{m0}(1 + z)^3 +\Omega _{\Lambda 0}}}\\
\end{gathered}
$$
At $$z = 0.1\,\,\, \to \,\,\,\,t_U H_0 = 0.899\,\,\,\,\,t_{ph} H_0 = 0.093$$
$$z = 1\,\, \to \,\,\,t_U H_0 = 0.431\,\,\,\,\,t_{ph} H_0 = 0.561$$
$$z = 100\,\,\, \to \,\,t_U H_0 = 1.264 \times 10^{ - 3} \,\,\,\,\,t_{ph} H_0 = 0.991$$
$$z = 1000\,\, \to \,\,\,t_U H_0 = 3.663 \times 10^{ - 5} \,\,\,\,\,t_{ph} H_0 = 0.993$$

<gallery widths=600px heights=500px>
File:12_10(1).JPG|
</gallery>

<gallery widths=600px heights=500px>
File:12_10(2).JPG|
</gallery>
</p>
</div>
</div></div>

<div id="SCM17"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 11 ===
Determine the present physical distance to the object that emitted light with current redshift $z$
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The physical distance to the object emitted a photon is related to the redshift $z$ by
$$
R(z) = \frac{1}{1 + z}\int_0^z\frac{dz'}{H(z')}
$$
In the SCM
$$
H =H_0\sqrt {\Omega _{r0}(1 + z)^4 + \Omega _{m0}(1 + z)^3+\Omega _{\Lambda 0}}
$$
The parameters ${H_0},\;\Omega _{r0},\Omega _{m0},\Omega _{\Lambda 0}$ are fixed by the model.
</p>
</div>
</div></div>

<div id="SCM_15"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 12 ===
A photon was emitted at time $t$ and registered at time $t_0$ with red
shift $z$. Find and plot the dependence of emission time on the redshift $t(z)$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
a = \frac{1}{1 + z}; \\
a(t) = A^{1/3}\sh ^{2/3}\left(t/{t_\Lambda } \right);\quad A \equiv \frac{\Omega_{ m0}}{\Omega _{\Lambda 0}}; \\
\frac{t_0}{t_\Lambda }= Arth\sqrt {\Omega _{\Lambda 0}} ; \\
\frac{t}{t_0} = \frac{1}{Arth\sqrt{\Omega _{\Lambda 0}}}Arsh\left[\sqrt {\frac{\Omega _{\Lambda 0}}{\Omega _{m0}}} \frac{1}{\left(1 + z\right)^{3/2}} \right] \\
\end{gathered}
$$
<gallery widths=600px heights=500px>
File:12_13.JPG|
</gallery>
</p>
</div>
</div></div>

<div id="SCM18"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 13 ===
Find the time dependence for the scale factor and analyze its asymptotes. Plot $a(t)$
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the SCM
$$
\left(\frac{\dot a}{a}\right)^2 = H_0^2\left[ \Omega _{m0}\left(\frac{a_0}{a} \right)^3 + \Omega _{\Lambda 0} \right]
$$
Solution of the equation in the case $\Omega _{\Lambda 0} > 0$ reads
$$
\begin{gathered}
a(t) = a_0\left( \frac{\Omega _{m0}}{\Omega _{\Lambda 0}}\right)^{1/3}\left[ sh\left( \frac{3}{2}\sqrt {\Omega _{\Lambda 0}}H_0t \right) \right]^{2/3}; \\
a\left( t \ll H_0^{ - 1}\right) \propto t^{2/3};\; a\left( t \gg H_0^{ - 1} \right) \sim e^{H_0t} \\
\end{gathered}
$$
Below it is convenient to use the expression for time dependence of the scale factor in the following form
$$
a(t) = A^{1/3}\sinh ^{2/3}\left( t/{t_\Lambda }\right);\quad
A \equiv \frac{\Omega_{m0}}{\Omega _{\Lambda 0}} =
\frac{1 - \Omega _{\Lambda 0}}{\Omega _{\Lambda 0}};\quad t_\Lambda \equiv \frac{2}{3}\left(H_0 \right)^{-1}\left( \Omega _{\Lambda 0}\right)^{ - 1/2}
$$
</p>
</div>
</div></div>

<div id="SCM19"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 14 ===
Using the explicit solution for scale factor $a(t)$, obtained in the previous problem, find the cosmic horizon $R_h$ (see Chapter 3)
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
As we have seen above
\[a(t)=A\sinh^{2/3}{(t/t_\Lambda)},\]
where \[A=\frac{\Omega_{m0}}{\Omega_{\Lambda0}},\ t_\Lambda=\frac{2}{3H_\infty}.\]
Consider \[f(t)\equiv\ln{a(t)}=\ln A + \frac23\ln\sinh\frac{t}{t_\Lambda},\] then
\[\dot f=\frac{2}{3t_\Lambda\tanh(t/t_\Lambda)}=\frac{H_\infty}{\tanh(3tH_\infty/2)},\]
and finally
\[R_h=\frac{c}{\dot f}=\frac{c}{H_\infty}\tanh\left(\frac32tH_\infty\right).\]
</p>
</div>
</div></div>

<div id="SCM20"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 15 ===
Show that the Universe becomes $R_h$-delimited only after the cosmological constant starts to dominate.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
According to the result of the previous problem
\[\dot R_h=\frac32\left[1-\tanh^2\left(\frac32tH_\infty\right)\right]c,\]
so for early times $\dot R_h>c$ and an observer does not experience a divergent redshift with increasing $R$. Then after some ''transition'' time $t^*$, estimated by the condition
\[ct^*=R_h(t^*),\]
he (or she) will begin to encounter an observational limit at a finite radius $R_h$. Numerical solution of the equation gives $t^*\approx0.86H_\infty^{-1}$. This time is roughly the point at which the Universe transits from matter-dominated to $\Lambda$-dominated epoch. Eventually, the Universe becomes de Sitter one and therefore $\dot R_h\rightarrow0$ with $R_h$ settling at the fixed value $c/H_\infty$.
</p>
</div>
</div></div>

<div id="SCM21"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 16 ===
Analyze the stability of Friedmann equations.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Friedmann equations are given by
\[H^2=\frac{8\pi G}{3}\rho_m+\frac\Lambda3,\]
\[\dot H=-4\pi G\rho_m,\]
\[\dot\rho_m+3H\rho_m=0\]
Let us introduce the following dimensionless variables (recall that $\rho_m,\ \Lambda>0$):
\[x=\frac{\sqrt{8\pi G\rho_m}}{\sqrt3H},\ y=\frac{\sqrt\Lambda}{\sqrt3H}\]
Equations for $x$ and $y$ read (with $N\equiv\ln a$)
\[x'=\frac{dx}{dN}=\frac32x(x^2-1),\]
\[y'=\frac{dy}{dN}=\frac32yx^2.\]
The critical points, i.e. the solutions corresponding to $x'=0,\ y'=0$ and their eigenvalues are given by
\[x_{cr}=0,\ y_{cr}=1 (\Lambda-dominant\, case,)\ \mu_1=0, \mu_2=-\frac32;\]
\[x_{cr}=1,\ y_{cr}=0 (matter-dominant\, case,)\ \mu_1=3, \mu_2=\frac32.\]
From the eigenvalues we see that the matter dominant phase is unstable $\mu_{1,2}>0$ while the cosmological constant dominant phase has $\mu_2<0$ being stable. The existence of a zero eigenvalue $\mu_1=0$ in the first case originates from the fact that the two variables x and y are connected by the relation $x^2+y^2=1$. Therefore in this case one can reduce the dynamics to one-dimensional space.
</p>
</div>
</div></div>

<div id="SCM22"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 17 ===
Determine solutions for the perturbations $\delta\rho_m$ and $\delta H$. Make sure that the solutions are stable.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The variation of Friedmann equations yields
\[2H\delta H=\frac{\kappa^2}{3}\delta\rho_m,\]
\[\delta\dot H=-\frac{\kappa^2}{2}\delta\rho_m,\]
\[\delta\dot\rho_m+3\rho_m\delta H+ 3H\delta\rho_m=0,\]
where $\kappa^2\equiv8\pi G$.
Solution for the above equation is
\[\delta H=\frac{t_\Lambda}{4\kappa^2}C\tanh(\tau) e^{-f(\tau)},\ \delta\rho_m= \frac{C}{4\kappa^4}e^{-f(\tau)},\]
where \[\tau\equiv\frac{t}{t_\Lambda},\]
\[f(\tau)=-2\tau +\ln\left(-1+e^{4\tau}\right)+2\ln\tanh\tau.\]
Here $C$ is an arbitrary constant. We see that $f(\tau)$ approaches $2\tau$ as $t\rightarrow\infty$ and both $\delta H$ and $\delta\rho_m$ decay which implies that the considered solution is stable.
</p>
</div>
</div></div>

<div id="SCM23"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 18 ===
Rewrite the first Friedman equation in terms of conformal time.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[\frac{1}{a^4}\left( \frac{da}{d\eta } \right)^2=H_0^2\left[ \Omega_m a(\eta )^{-3}+\Omega_{\Lambda} \right]\]
</p>
</div>
</div></div>

<div id="SCM24"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 19 ===
Find relation between the scale factor and conformal time
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[\eta =\frac{1}{H_0}\int\limits_0^a\frac{dx}{\sqrt{x}{{\left( {{\Omega }_{m}}+{{\Omega }_{\Lambda }}{{x}^{3}} \right)}^{1/2}}}\]
</p>
</div>
</div></div>

<div id="SCM25"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 20 ===
Find explicit dependence of the scale factor on the conformal
time.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The integral obtained in the previous problem can be expressed in terms of elliptic integral of the first kind $F\left( \varphi ,k \right)$. For that purpose we rewrite it in the form
$$
\Omega^{1/6}_{\Lambda} \Omega^{1/3}_m H_{0} \eta = \int\limits_0^{u} \dfrac{dx}{\sqrt{x}\sqrt{1 + x^3}}.
$$
The upper limit equals to $u=\left(\dfrac{\Omega_{\Lambda}}{\Omega_m}\right)^{1/3}a.$ The integral can be presented in the form
$$3^{1/4} \Omega^{1/6}_{\Lambda} \Omega^{1/3}_m H_{0} \eta = F\left(\arccos\dfrac{1+(1-\sqrt{3})u}{1+(1+\sqrt{3})u}, \dfrac{\sqrt{2+\sqrt{3}}}{2}\right).$$

Here $y={{3}^{1/4}}\Omega _{\Lambda }^{1/6}\Omega _{m}^{1/3}{{H}_{0}}\eta ,\quad k=\frac{\sqrt{2+\sqrt{3}}}{2}\approx 0.97.$
</p>
</div>
</div></div>

<div id="SCM25"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 21 ===
Find relative density of dark energy $10^9$ years later.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
t = t_\Lambda Arcth\sqrt {\Omega _\Lambda} ;\quad \Omega _{\Lambda } = th^2\frac{t}{t_\Lambda };\\
t_\Lambda \simeq 10.768 \cdot 10^{9}\;\mbox{years}\;\;t = 14.7\cdot 10^{9}\;\mbox{years}\\
\Omega _{\Lambda 0} \simeq 0.77\\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM_17"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 22 ===
Find variation rate of relative density of dark energy. What are its asymptotic values? Plot its time dependence.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
\frac{d\Omega _\Lambda }{dt} = \frac{2}{t_\Lambda }\frac{sh(t/{t_\Lambda })}
{ch^3(t/{t_\Lambda })}; \\
x \equiv \frac{t}{t_\Lambda };\quad \frac{d\Omega _\Lambda }{dx} = 2\frac{sh(x)}{ch^3(x)} \\
\end{gathered}
$$

<gallery widths=600px heights=500px>
File:12_15.JPG|
</gallery>
</p>
</div>
</div></div>

<div id="SCM26"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 23 ===
Estimate size of the cosmological horizon.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The cosmological horizon is defined as
$$
L_{H0} = a_0\int\limits_0^{t_0}\frac{dt}{a\left( t \right)}.
$$
Using the results of the problem \ref{A_t_lambda}, one obtains
$$
L_{H0}= A^{-1/3}\int\limits_0^{t_0}\frac{dt}{sh^{2/3}\left( t/{t_\Lambda } \right)};
$$
</p>
</div>
</div></div>

<div id="SCM_19"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 24 ===
Find time dependence of Hubble parameter and plot it.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
H = \frac{\dot a}{a} =\frac{2}{3}\left(t_\Lambda \right)^{ - 1}cth\left( {\frac{t}{t_\Lambda }} \right)
$$

In the SCM ${t_0} \simeq 1.2t_\Lambda .$ Therefore
$$
H{t_0} \simeq 0.8cth\left( {1.2t/{t_0}} \right)
$$

<gallery widths=600px heights=500px>
File:12_16.JPG|Time dependence of the Hubble parameter (in units of the age of the Universe).
</gallery>

</p>
</div>
</div></div>

<div id="SCM27"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 25 ===
Find time dependence of dark matter density.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Present the first Friedmann equation in the form
\[\rho_m=\frac{1}{8\pi G}(3h^2-\Lambda)\]
and make use of results of the previous problem to obtain
\[\rho_m=\frac{\Lambda}{8\pi G}\frac{1}{\sinh^2(t/t_\Lambda)}.\]

\item {\bf Find the asymptotic (in time) value of the Hubble parameter.}

The required asymptote can be obtained in two ways: first, from the time dependence of the Hubble parameter (see the previous problem), and second, immediately from the first Friedmann equation in the SCM. Taking into account that $\rho_{m} \to 0$ at $t\to \infty$, one gets
$$
H_{\infty} = \frac{2}{3t_\Lambda} = \sqrt {\frac{\Lambda}{3}}.
$$
</p>
</div>
</div></div>

<div id="SCM28"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 26 ===
At present the age of the Universe $t_0\simeq13.7\cdot10^9$ years is close to the Hubble time $t_H=H_0^{-1}\simeq14\cdot10^9$ years.
Does the relation $t^*\simeq t_H(t^*)=H^{-1}(t^*)$ for the age $t^*$
of the Universe hold for any moment of its evolution
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
This relation cannot be satisfied in arbitrary moment of time, as the Hubble parameter has limit value in the SCM (see the previous problem):
$$
\frac{t^*}{t_H(t^*)} = H\left( t^*\right)t^* = \frac{2}{3}\frac{Arcth\sqrt {\Omega _\Lambda(t^*)}}{\sqrt{\Omega_\Lambda (t^*)}}.
$$
For $t\to \infty,\;\Omega _\Lambda\left( t^* \right) \to 1,$ and $$\frac{t^*}{t_H(t^*)} \to \infty. $$

\item {\bf Determine values of state parameter for the dark energy, that provide accelerated expansion of Universe in the present time.}

$$
\begin{gathered}
\sum\limits_i (\rho _{i0} + 3p_{i0}) < 0\; \Rightarrow \rho _{m0} + \rho _{DE0} + 3w_{DE}\rho _{DE0} < 0; \\
w_{DE} < - \frac{1}{3}\left( 1 + \frac{\Omega _{m0}}{\Omega _{DE0}}\right); \\
w_{DE} < - \frac{1}{3}\Omega _{DE0}^{-1};\; w_{DE} < - 0.46\\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM29"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 27 ===
Find current value of the deceleration parameter.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For a multi-component system the deceleration parameter equals to
$$
q = \frac{\Omega}{2} + \frac{3}{2}\sum\limits_i {w_i}\Omega _i
$$
In the SCM $\Omega = 1,w_m = 0,\;w_\Lambda= - 1,$ and
$$
\begin{gathered}
q = \frac{1 - 3\Omega_{\Lambda 0}}{2};\\
q_0 \simeq - 0.6 \\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM_24"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 28 ===
Find the redshift dependence of the deceleration parameter. Analyze the limiting cases.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
q \equiv - \frac{a\ddot a}{\dot a^2} = - \frac{\ddot a}{a}\frac{1}{H^2}; \\
\frac{\ddot a}{a} = - \frac{4\pi G}{3}\sum\limits_i \left( \rho _i + 3p_i \right) = - \frac{1}{2}H_0^2\sum\limits_i \Omega _{i0}\left(1 + 3w_i\right)(1 + z)^{3\left(1 + w_i\right)} ; \\
H^2 = H_0^2\sum\limits_i \Omega _{i0}(1 + z)^{3\left( 1 + w_i \right)}; \\
q = \frac{3}{2}\frac{\sum\limits_i \Omega _{i0}\left( 1 + w_i \right)(1 + z)^{3\left( 1 + w_i \right)}}
{\sum\limits_i \Omega _{i0}(1 + z)^{3\left( 1 + w_i\right)} } - 1 \\
\end{gathered}
$$
In the SCM
$$
q = \frac{1}{2}\frac{\Omega _{m0}(1 + z)^3- 2\Omega _{\Lambda 0} }
{\Omega _{m0}(1 + z)^3+\Omega_{\Lambda 0}}
$$

$$
q(z \to \infty ) = \frac{1}{2},\;q(z \to - 1) = - 1
$$

<gallery widths=600px heights=500px>
File:12_24.JPG|Dependence of the deceleration parameter on the redshift.
</gallery>
</p>
</div>
</div></div>

<div id="SCM_25"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 29 ===
Find and plot the time dependence of the deceleration parameter.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
q = - \frac{\ddot aa}
{\dot a^2} = - \frac{\ddot a}
{aH^2}; \\
a(t) = A^{1/3}sh^{2/3}\left(t/{t_\Lambda}\right);\quad A\equiv \frac{\Omega_ {m0}}
{\Omega _{\Lambda 0}}; \\
\frac{t_0}
{t_\Lambda } = Arth\sqrt {\Omega _{\Lambda 0}} ;\quad H = \frac{2}
{3}\frac{1}
{t_\Lambda }cth\left( \frac{t}
{t_\Lambda }\right); \\
q = \frac{1}
{2}\left[ 1 - 3th^2\left( \frac{t}
{t_\Lambda }\right)\right] \\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM29"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 30 ===
Find the moment of time when the dark energy started to dominate over dark matter. What redshift did it correspond to?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
\left( \frac{\dot a}{a} \right)^2 = \frac{8\pi G}
{3}\rho _{cr}\left( \Omega _{m0}(1 + z)^3+\Omega_{\Lambda 0}\right); \\
\Omega _{m0}(1 + z^*)^3 =\Omega _{\Lambda 0};\\
z^* = \left(\frac{\Omega _{\Lambda 0}}
{\Omega _{m0}}\right)^{1/3}-1; \\
\Omega _{\Lambda 0}\simeq 0.73;\quad \Omega _{m0}\simeq 0.27; \\
z^* \simeq 0.39; \\
\Omega _\Lambda = th^2\left(\frac{t}
{t_\Lambda }\right);\quad \Omega _m = 1 - \Omega _\Lambda = ch^{-2}\left( \frac{t}
{t_\Lambda } \right); \\
\Omega _\Lambda(\tilde t) = \Omega _m(\tilde t) \to \tilde t = \frac{Arsh(1)}{Arth(\sqrt {\Omega _{\Lambda 0} })}t_0 \simeq 9.5\cdot10^9 \,years \\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM30"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 31 ===
Determine the moment of time and redshift value corresponding to the transition from decelerated
expansion of the Universe to the accelerated one.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
\frac{\ddot a}
{a} = \frac{4\pi G}
{3}\rho_{cr}\left[2\Omega_{\Lambda 0} - \Omega _{m0}(1 + z)^3\right]; \\
2\Omega _{\Lambda 0} = \Omega _{m0}(1 + z^*)^3; \\
z^* = \left( \frac{2\Omega_{\Lambda 0}}
{\Omega _{m0}}\right)^{1/3} - 1; \\
\Omega _{\Lambda 0}\simeq 0.73;\quad \Omega _{m0}\simeq 0.27; \\
z^* \simeq 0.745 \\
\end{gathered}
$$

Let us now determine the time moment $t^*$ of the transition
$$
\begin{gathered}
a^* = \frac{1}
{1 + z^*} = \left(\frac{\Omega _{m0}}
{2\Omega _{\Lambda 0}} \right)^{1/3}; \\
t(a) = \frac{2}{3\sqrt{\Omega_{\Lambda 0}}H_0}Arsh\left(\sqrt{ \frac{\Omega_{\Lambda 0}}
{\Omega_{m0}}}a^{3/2} \right); \\
t^* = \frac{2}
{3\sqrt {\Omega _{\Lambda 0}}}H_0Arsh\left(\sqrt {1/2} \right) \\
\end{gathered}
$$
There is an alternative way to solve the problem.
In the multi-component Universe with the corresponding state equations $p_i = w_i\rho _i$ the deceleration parameter $q \equiv - \frac{\ddot aa}{\dot a^2}$ equals to
$$
q = \frac{\Omega }
{2} + \frac{3}
{2}\sum\limits_i w_i\Omega _i
$$
In the case of the SCM $\Omega =1$ and
$$
q = \frac{1}
{2} - \frac{3}
{2}\frac{\Omega _{\Lambda 0}}
{\left[\left(1 + z\right)^3\Omega _{m0}+\Omega_{\Lambda 0}\right]}
$$

From the condition $q=0$ it follows that
$$
z^* =\left(\frac{2\Omega_{\Lambda 0}}
{\Omega _{m0}} \right)^{1/3}- 1
$$
For $\Omega _{\Lambda 0} = 0.73,\;\Omega _{m0} = 0.27$ one obtains that
$$
z^* = 0.745
$$
</p>
</div>
</div></div>

<div id="SCM31"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 32 ===
Solve the previous problem using the derivative $d\eta/d\ln a$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\frac{d\eta}
{d\ln a} = a\frac{dt}{ada} = \frac{1}{aH} = \frac{1}{\dot a}
$$

If the quantity $aH = \dot a$ grows, then $\ddot a > 0,$ which corresponds to the accelerated expansion (inflation) of the Universe. In the SCM the quantity
$$
\frac{aH}{H_0} \simeq \sqrt{ \frac{0.27 + 0.73a^3}{a}}
$$
starts to grow at $a=0.573,$ which corresponds to $z=0.745$
</p>
</div>
</div></div>

<div id="SCM32"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 33 ===
Is dark energy domination necessary for transition to the accelerated expansion?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
No. The transition to the accelerated expansion takes place at $(z=0.75)$, which is considerably earlier than the transition to the dark energy domination stage at $(z=0.45).$
</p>
</div>
</div></div>

<div id="SCM33"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 34 ===
Consider flat Universe composed of matter and dark energy in form of cosmological constant. Find the redshift value corresponding to equality of densities of the both components $\rho_m(z_{eq})=\rho_\Lambda(z_{eq})$ and the one corresponding to beginning of the accelerated expansion $q\left(z_{accel} \right) = 0$. Obtain relation between $z_{eq}$ and $z_{accel}$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
z_{eq} = \left(\frac{\Omega _\Lambda}
{\Omega _{m0}} \right)^{1/3}- 1; \\
z_{accel}= \left(\frac{2\Omega_\Lambda }
{\Omega _{m0}}\right)^{1/3}- 1; \\
z_{accel} = 2^{1/3}\left(z_{eq}+ 1\right) - 1 \\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM34"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 35 ===
Show that density perturbations stop to grow after the transition from dust to $\Lambda$-dominated era.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
On the $\Lambda$-dominated stage the non-relativistic matter density perturbation size depends on conformal time as
\begin{equation}\label{1SCM30}
a(\eta)=-\frac{1}{H\eta}, \,\, \eta=-const\cdotp e^{-Ht},\,\,\eta<0
\end{equation}
where the Hubble parameter $H$ is constant, $H^2=(8\pi/3)G\rho_\Lambda$.
The linearized Einstein equations for the case of single-component medium take the following form in the momentum representation:
\begin{equation}\label{3SCM30}
k^2\Phi+3\frac{a^{'}}{a}+3\frac{a^{'2}}{a^2}\Phi=-4\pi Ga^2\delta\rho,
\end{equation}
\begin{equation}\label{2SCM30}
\Phi^{''}+3\frac{a^{'}}{a}\Phi^{'}+\left(2\frac{a^{''}}{a} - \frac{a^{'2}}{a^2}\right)\Phi=4\pi Ga^2\delta p.
\end{equation}
Assuming that the $\Lambda$-term is constant (which is exactly true for the vacuum energy density) one obtains $\delta\rho_\Lambda=\delta p_\Lambda=0$.
Using \eqref{2SCM30} and taking into account \eqref{1SCM30} we come up with the equation: $${\Phi}^{''}-\frac{3}{\eta}{\Phi}^{'}-\frac{3}{\eta^2}\Phi=0,$$
Solutions of the equation behave as
$$\Phi\propto\eta\propto\frac 1a$$ and $$\Phi\propto\eta^3\propto\frac{1}{a^3}.$$
Use now \eqref{3SCM30} for the modes beyond the horizon and take into account that for dust density $\rho_M\propto a^{-3}.$ Keeping only the first solution for the gravitational potential (the second decays too fast), one can see that the density perturbations do not grow: for the modes under horizon the relation
$$\delta_M\propto a^3\frac{k^2}{a^2}\Phi=const$$ holds.

Therefore the perturbations stop to grow at the transition from the dust-dominated to the $\Lambda$-dominated stage. If the present accelerated expansion of the Universe is actually caused by non-zero cosmological constant, then the structure formation from small primary perturbations is eventually close to end and it will never resume again: the structures larger than the presently observed ones will never appear.
</p>
</div>
</div></div>

<div id="SCM35"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 36 ===
What happens to the velocity fluctuations of non-relativistic matter and radiation with respect to Hubble flow in the epoch of cosmological constant domination?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM36"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 37 ===
Find the ratio of baryon to non-baryon components in the galactic halo.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM37"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 38 ===
Imagine that in the Universe described by SCM the dark energy was instantly switched off. Analyze further dynamics of the Universe.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="local-group"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 39 ===
Estimate density of dark energy in form of cosmological constant using the Hubble diagram for the neighborhood of the Local group.
<gallery widths=600px heights=500px>
File:im_eps.JPG|Hubble diagram for the neighborhood of the Local group.
</gallery>
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
An elegant and simple method was recently proposed (see A.D.Chernin, UFN 178, 267 (2008)), which allows not only to become convinced in the existence of the dark energy but also to estimate its density. The method is based on the assumption, that the dark energy (in form of the cosmological constant) governs the Hubble flow even on comparably small distances $\left( \sim \,Mpc \right)$. Consider the resulting force $\vec{F}$ (per unit mass), acting on a galaxy from the picture \ref{hubble_diagram}. The force is a sum of two contributions:
$$
\vec{F}=\vec{F}_N+\vec{F}_\Lambda
$$
Here $\vec{F}_N$ is the force produced by the total mass of the Local group, and $F_{\Lambda }$ is the force created by the dark energy. For a crude estimate let us neglect the non-spherical mass distribution in the Local group. Then the force acting on a satellite galaxy in a given point $R$ outside the local Group equals to
$$
F=-\frac{GM_{LG}}{R^2}+\frac{8\pi G}{3}{\rho_{\Lambda }}R,
$$
where $M_{LG}\simeq (1\div 3) \times 10^{12}M_{\odot }$ is the total mass of the Local group. As can be seen from the picture, the transition from domination of the local group gravity to the vacuum domination takes place on the distance of order $R_{0}=1.8\div 2\; \, Mpc.$ It allows to make a rough estimate of the dark energy density from the condition $F\left( R=R_0 \right)=0.$ The estimate results in the following
$$
(0.1\pm 0.03)<\rho_V<(1\pm 0.3)\times 10^{-26} \, kg/m^3.
$$
The proximity of the obtained crude estimate to the actual value of the dark energy density in the SCM is striking.
$$
\rho_{V}\simeq (0.75\pm 0.05)\times 10^{-26}\, kg/m^3.
$$
</p>
</div>
</div></div>

<div id="SCM38"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 40 ===
Estimate the Local group mass by methods used in the previous problem.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM39"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 41 ===
Find "weak" points in the argumentation of the two preceding problems.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM40"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 42 ===
The product of the age of the Universe and current Hubble parameter (the Hubble's constant) is a very important test (Sandage consistency test) of internal consistency for any model of Universe. Analyze the
parameters on which the product $H_0 t_0$ depends in the Big Bang model and in the SCM.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM41"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 43 ===
Show that for a fixed source of radiation the luminosity distance for high redshift values in flat Universe is greater for the dark energy dominated case compared to the non-relativistic matter dominated one.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the case of Universe filled by dark energy and non-relativistic matter, the photometric distance is equal to
$$
d_L = \frac{1 + z}
{H_0}\int_0^z \frac{dz'}
{\sqrt{\Omega _{m0}\left(1 + z'\right)^3+ \Omega _{DE0}\left(1 + z' \right)^{3\left( 1 + w_{DE}\right)}}}
$$

In the case $\Omega _{m0} = 1,\;\Omega _{DE0} = 0$
one obtains
$$
d_L = \frac{2}
{H_0}\left[1 + z -(1 + z)^{1/2}\right]
$$
and in the case $\Omega _{m0} = 0,\;\Omega _{DE0} = 1$
$$
d_L = \frac{1}{H_0}z(1 + z)
$$
instead.

In means that the supernova appear dimmer in the dark energy dominated Universe.
</p>
</div>
</div></div>

<div id="SCM42"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 44 ===
In the observations that discovered the accelerated expansion of the Universe the researchers in particular detected two $Ia$ type supernovae: $1992P$, $z=0.026$, $m=16.08$ and $1997ap$, $z=0.83$,
$m=24.32$. Show that these observed parameters are in accordance with the SCM.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The apparent $m$ and absolute $M$ stellar magnitudes are related with the photometric distance $d_L$ by the expression
\begin{equation}\label{p(1)}
m - M = 5\log _{10}\left(\frac{d_L}
{\mbox{Mpc}} \right) + 25
\end{equation}
The photometric distance in the SCM is equal to
\begin{equation}\label{p(2)}
d_L = \frac{1 + z}
{H_0}\int_0^z \frac{dz'}
{\sqrt {\Omega _{m0}(1 + z)^3 + \Omega _\Lambda}}
\end{equation}
For small $z$ $d_L(z) \simeq z/H_0.$ Using the supernova $1992P$ with small $z=0.026$ one can find $M = - 19.09.$ Assuming that the $1a$-type supernova are standard candles (they have equal absolute magnitudes), one obtains from the relation \ref{p(1)} supernova with $z=0.83$
$$H_0d_L\simeq 1.16$$
From the other hand it follows from the relation \ref{p(1)} that in the case
$\Omega _{m0} = 1,\;\Omega_\Lambda = 0$
$$
H_0d_L\simeq 0.95
$$
In the SCM one gets much better agreement:
$$
H_0d_L \simeq 1.2
$$
</p>
</div>
</div></div>

<div id="SCM43"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 45 ===
Find the redshift value, at which a source of linear dimension $d$ has minimum visible angular
size.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The apparent angular size equals to $$\delta \theta = {{(1 + z)^2 d } \over {d_L (z)}},$$
where $d_L (z)$ is the photometric distance from a source with the redshift $z$. In the case of cosmological constant domination the photometric distance is:
$$d_L (z) = - {z \over {H_0 \Omega _\Lambda ^{1/2} }},$$
In the SCM (see the previous problem)
$$
d_L = \frac{1 + z}
{H_0}\int_0^z \frac{dz'}
{\sqrt {\Omega _{m0}(1 + z)^3 + \Omega _\Lambda}}
$$
then the angular size is
$$\delta \theta = - {{H_0 \Omega _\Lambda ^{1/2} d} \over 2}{{(1 + z)^2 } \over z},$$
and $z = - 1$. It corresponds to infinitely far future.
</p>
</div>
</div></div>

<div id="SCM44"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 46 ===
Compare the observed value of the dark energy density with the
one expected from the dimensionality considerations (the cosmological constant problem).
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The dimensionality considerations suggest the following value for density of the dark energy in form of the cosmological constant
\[\rho_\Lambda^{(dim)}\sim M_{Pl}^4\approx10^{76}GeV^4.\]
At the same time the ''observed'' value fixed by the SCM is
\[\rho_\Lambda^{(obs)}\approx10^{-48}GeV^4.\]
This huge disagreement
\[\frac{\rho_\Lambda^{(dim)}}{\rho_\Lambda^{(obs)}}>10^{120}\]
is called the ''cosmological constant problem''. It presents a fundamental problem for elementary particles theory and physics in general, rather than for cosmology itself. (Look for details in R.Bousso, [http://arxiv.org/abs/0708.4231 TASI Lectures on the cosmological constant])
</p>
</div>
</div></div>

<div id="SCM45"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 47 ===
Determine the density of vacuum energy using the Planck scale as cutoff parameter.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For a quantum field with a given mode of frequency $\omega$, the zero-point energy is $\pm\omega/2$ with the plus sign for a bosonic field and the minus sign for a fermionic field. The total zero-point energy is then $\pm\sum\limits_i\omega_i/2$. In the continuum limit, we have for a free field (bosonic or fermionic)
\[\frac12\sum\limits_i\omega_i=\frac12\int\frac{d^3xd^3k}{(2\pi)^3}(k^2+m^2)^{1/2}=V\int\frac{k^2dk}{4\pi^2}(k^2+m^2)^{1/2}.\] This integral diverges, thus we need to introduce the physical cutoff. In cosmological context this cutoff must be the Planck mass at largest. With the cutoff $k_{max}\sim M_{Pl}$ one obtains
\[\rho_\Lambda=\int_0^{k_{max}}\frac{k^2dk}{4\pi^2}(k^2+m^2)^{1/2} \approx\frac{k_{max}^4}{16\pi^2}=\frac{M_{Pl}^4}{16\pi^2} \approx10^{74}GeV^4.\]
This result only slightly modifies the dimensional considerations given in the previous problem.
</p>
</div>
</div></div>

<div id="SCM46"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 48 ===
Identifying the vacuum fluctuations density with the observable dark energy value in SCM, find the required frequency cutoff magnitude in the fluctuation spectrum.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[\nu_{max}=10^{12}Hz.\]
</p>
</div>
</div></div>

<div id="SCM47"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 49 ===
What purely cosmological problem originates from the divergence of the zero-point energy density?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
If the zero-point energy density is infinite, then our Universe is infinitely curved and the space size is infinitely small.
</p>
</div>
</div></div>

<div id="SCM48"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 50 ===
With exact supersymmetry, the bosonic contribution to cosmological constant is canceled by its fermionic counterpart. However, we know that our world looks like not supersymmetric. Supersymmetry, if exists, has to be broken above or around $100GeV$ scale. Compare the observed value of the dark energy density with the one expected from broken supersymmetry.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
If we take the $k_{max}^{(SUSY)}\sim100GeV$ cutoff, then
\[\rho_\Lambda^{(SUSY)}=\frac{\left(k_{max}^{(SUSY)}\right)^4}{16\pi^2}\approx10^6GeV^4.\]
\[\frac{\rho_\Lambda^{(SUSY)}}{\rho_\Lambda^{(obs)}}>10^{54}.\]
Thus supersymmetry does not solve the problem of the cosmological constant.
</p>
</div>
</div></div>

<div id="SCM49"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 51 ===
Determine duration of the inflation period.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Let us see what happens with the relative density during all the history of Universe. From the first Friedmann equation it follows that in the matter-dominated epoch $\left| \Omega - 1 \right| \sim t^{2/3},$ and in the radiation-dominated epoch $\left| \Omega - 1 \right| \sim t^{2/3}.$ Then assume that inflation started at $t_i,$ and finished at $t_f.$ Then in the period of $t_f < t < t_{eq} = 50000$ years the radiation dominated and the range $t_{eq} < t < t_0 $ corresponds to the matter domination epoch. Then the presently observed difference
$$
\left|\Omega (t_0 ) - 1 \right| = \left| \Omega (t_i ) - 1 \right|e^{ - 2H(t_f - t_i )} \left( \frac{t_{eq} }
{t_f } \right)\left( \frac{t_0}
{t_{eq} }\right)^{2/3}
$$
According the SCM $\left|\Omega \left( t_0\right) - 1 \right| < 0.04.$ If one assumes that before the beginning of the inflation there was $\left|\Omega \left( t_i \right) - 1 \right| \sim 1,$ then
$$
N > \frac{1}{2}\ln \left[\frac{\Omega \left( t_i \right) - 1}{\Omega \left(t_0\right) - 1}\left( \frac{t_{eq}}{t_f } \right)\left( \frac{t_0}{t_{eq}} \right)^{2/3} \right] \approx 60
$$

Assume that radiation dominated before the inflation started. Then $H = \frac{1}{2t}.$ If at the moment of time $t_i$ the vacuum dark energy started to dominate and then the expansion continued with constant rate, then $H\approx 1/t_i$ during the inflation. And finally one obtains that
$$
N = H\left(t_f - t_i\right) \approx \frac{t_f - t_i }
{t_i }
$$
it is commonly assumed that the inflation finishing time is of order of the Grand Unification time $t_f\approx 10^{-35}.$ Therefore $t_f\approx 10^{-38}.$ Evidently the inflation could last longer as we used only lower estimate for $N.$
</p>
</div>
</div></div>

<div id="SCM50"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 52 ===
Plot the dependence of luminosity distance $d_L$ (in units of $H_0^{-1}$) on the redshift $z$ for the two-component flat Universe with non-relativistic liquid ($w=0$) and cosmological constant
($w=-1$). Consider the following cases:

a) $\Omega_\Lambda^0=0$;

b) $\Omega_\Lambda^0=0.3$;

c) $\Omega_\Lambda^0=0.7$;

d) $\Omega_\Lambda^0=1$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM51"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 53 ===
For a Universe filled by dark energy with state equation $p_{DE}=w_{DE}\rho_{DE}$ and non-relativistic matter obtain the Taylor series for $d_L$ in terms of $z$ near the observation point
$z_0=0$. Explain the obtained result.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the universe filled by dark energy with the state equation $p_{DE}=w_{DE}\rho_{DE}$ and non-relativistic dark matter the photometric distance takes on the form:
\begin{equation}
\label{d_L}
d_L = \frac{1 + z}
{H_0}\int_0^z \frac{dz'}
{\sqrt{\Omega _{m0}\left(1 + z'\right)^3+ \Omega _{DE0}\left(1 + z' \right)^{3\left( 1 + w_{DE}\right)}}}
\end{equation}
Expand the integrand in Taylor series in $z$ in vicinity of the observation point $z_0=0$ to obtain
$$
\begin{gathered}
\label{tey}
\frac{1}{\sqrt{\Omega _{m0}\left(1 + z'\right)^3+ \Omega _{DE0}\left(1 + z' \right)^{3\left( 1 + w_{DE}\right)}}}\simeq \\
\simeq 1 -\frac{3}{2}\left(1+w_{DE}\Omega _{DE0}\right)z+\frac{3}{4}\left(3w_{DE}\Omega _{DE0}-1\right)z^2.\nonumber
\end{gathered}
$$
Substitution of the decomposition (\ref{tey}) into the integral (\ref{d_L}) results in the following:
\begin{equation}
\label{d_L}
d_L = \frac{1 + z}
{H_0}\left(z -\frac{3}{4}\left(1+w_{DE}\Omega _{DE0}\right)z^2+\frac{1}{4}\left(3w_{DE}\Omega _{DE0}-1\right)z^3\right).
\end{equation}
</p>
</div>
</div></div>

<div id="SCM_45"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 54 ===
Determine position of the first acoustic peak in the CMB power spectrum produced by baryon oscillations on the surface of the last scattering.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Early Universe was filled by photon-baryon plasma, which can be treated as a single-component liquid. The baryons were trapped by the potential walls generated by density fluctuations, and they were eventually compressed. The contraction lead to heating of the plasma and therefore to increasing radiative pressure of the photons which is directed outside. Finally the radiative pressure stop the compression and lead to expansion of the plasma. During the expansion the plasma cools down and its radiative pressure decreases. The gravity starts to dominate again leading to repeated compression. The concurrence between the gravity and the pressure lead to longitudinal (acoustic) oscillations in the photon-baryon fluid. When the matter and the radiation get decoupled in the recombination process, the picture of the acoustic oscillations remains frozen into the CMB. Today we detect the evidence of the primordial acoustic waves (dense and diluted regions) in form of the primary anisotropy in the CMB.

It is well known that any acoustic wave, whatever complicatedly shaped, can be represented in form of superposition of modes with different wave numbers $k$, $k\propto 1/\lambda $. Every mode $\lambda $ corresponds to certain angular scale $\theta $ on the sky. Therefore in order to facilitate the comparison of the theory with observations, one should use the angular (multipole) decomposition in terms of the Legendre polynomials $P_{l} (\cos \theta )$, instead of the Fourier transform in terms of sines and cosines. Order of the polynomial $l$ plays the same role as the index $k$ in the Fourier decomposition. For $l\ge 2$ the Legendre polynomials are oscillating functions in the interval $\left[1,-1\right]$. Number of the oscillations increases with growth of $l$. Therefore
\begin{equation} \label{acust1_}
l\propto \frac{1}{\theta }
\end{equation}
The temperature fluctuations can be observed and qualitatively analyzed with help of paired measurements in the directions $\stackrel{\frown}{n},\stackrel{\frown}{n}'$, so that $\stackrel{\frown}{n}\cdot \stackrel{\frown}{n}'=\cos \theta $. Taking average over all the pairs in the assumptions of Gaussian fluctuations, one obtains the two-point correlator $C(\theta )$, which can be presented in terms of the multipole decomposition
\begin{equation} \label{acust2_}
\left\langle \delta T(\stackrel{\frown}{n})\cdot \delta T(\stackrel{\frown}{n}')\right\rangle \equiv C(\theta )=\sum _{l}\frac{2l+1}{4\pi } C_{l} P_{l} (\cos \theta )
\end{equation}
where $C_{l} $ are the coefficients.

As we already mentioned, the analysis of the temperature fluctuations enables us to clarify the structure of the longitudinal oscillations. The mode with the maximum wavelength corresponds to the maximum angular size of the primary anisotropy. This fundamental mode was detected the first. There are reliable evidences of the fact that the second and the third modes were also already detected. The distance $r_{s} $, passed by the acoustic wave during the time period before the recombination, is called the sound horizon. The sound horizon is fixed by (or rather fixes) the physical scale o the last scattering surface. Size of the sound horizon depends on values of the physical parameters. Distance to the last scattering surface $d_{sls} $ depends on the cosmological parameters too. Together they determine the angular size of the sound horizon$\theta _{s} $ (see Fig.)

<gallery widths=600px heights=500px>
File:acustic_pic.JPG|Dependence of the deceleration parameter on the redshift.
</gallery>

\begin{equation} \label{acust3_}
\theta _{s} \approx \frac{r_{s} }{d_{sls} }
\end{equation}
Analysis of the temperature anisotropy allows to determine $\theta _{s} $. Varying the cosmological parameters in $r_{s} ,d_{sls} ,$ one achieves the best agreement with the observations and thus fixes the cosmological parameters.

We can estimate the sound horizon size as the length passed by sound from the moment $t=0$ to the recombination time $t_{*} $
\begin{equation} \label{acust4_}
r_{s} (z_{*} ;\Omega _{b} ,\Omega _{r} )\approx \int _{0}^{t_{*} }c_{s} dt
\end{equation}
where $z_{*} $ is the redshift of the recombination moment $\left(z_{*} \approx 1100\right)$, $c_{s} $ is the sound speed in the photon-baryon medium
\begin{equation} \label{acust5_}
c_{s} \approx c\left[3\left(1+\frac{3\Omega _{b} }{\Omega _{r} } \right)\right]^{-1/2}
\end{equation}
To determine $dt$ one should use the first Friedmann equation in the form
\begin{equation} \label{acust6_}
\left(\frac{da}{dt} \right)^{2} =H_{0}^{2} \left[\Omega _{r0} \left(\frac{a_{0} }{a} \right)^{2} +\Omega _{m0} \frac{a_{0} }{a} +\Omega _{k0} +\Omega _{\Lambda 0} \left(\frac{a}{a_{0} } \right)^{2} \right]
\end{equation}
Making use of the relations $\frac{a_{0} }{a} =1+z$ and $\Omega _{r} +\Omega _{m} +\Omega _{\Lambda } +\Omega _{k} =1$, one obtains
\begin{equation} \label{acust7_}
dt=H_{0}^{-1} (1+z)^{-1} \left\{\left(1+z\right)^{2} \left(1+\Omega _{m,0} z\right)+z(z+2)\left[\left(1+z)^{2} \Omega _{r.0} -\Omega _{\Lambda ,0} \right)\right]\right\}^{-1/2} dz
\end{equation}
Distance to the last scattering surface, which corresponds to its angular size, is determined by the quantity called the angular diameter distance. It is connected by the photometric distance $d$ by the simple relation
\begin{equation} \label{acust8_}
d_{sls} =\frac{d}{(1+z_{*} )^{2} }
\end{equation}
The position of the first acoustic peak is determined by
\begin{equation} \label{acust9_}
l\approx \frac{d_{sls} }{r_{s} }.
\end{equation}
Let us now predict the position of the first peak for the case of flat Universe. To the leading order the sound speed is
\[c_{s} =\frac{c}{\sqrt{3} }. \]
Let us assume also that in the recombination epoch the Universe was matter-dominated. Under such assumptions one gets
\begin{equation} \label{acust10_}
r_{s} =\frac{c_{s} }{H_{0} \sqrt{\Omega _{m} } } \int _{z_{*} }^{\infty }(1+z)^{-5/2} dz =r_{s} =\frac{2c_{s} }{3H_{0} \sqrt{\Omega _{m} } } \left(1+z_{*} \right)^{-3/2}
\end{equation}
Using the standard procedure (see section 4), we obtain for the comoving radial coordinate $r_{sls} $ of the last scattering surface the following
\begin{equation} \label{acust11_}
r_{lsr} =\frac{c}{H_{0} } \int _{0}^{z_{*} }\left[\Omega _{m} (1+z)^{^{3} } +\Omega _{\Lambda } \right] ^{-1/2} dz
\end{equation}
Replacing the integrand by its binomial decomposition
\[\Omega _{m}^{-1/2} (1+z)^{-3/2} -\left(\Omega _{\Lambda } /2\Omega _{m}^{3/2} \right)\left(1+z\right)^{-9/2} \]
and taking into account that
\[d_{sls} =r_{sls} /(1+z_{*} ),\]
we obtain
\begin{equation} \label{acust12_}
d_{sls} =\frac{2c}{7H_{0} (1+z_{*} )} \left\{7\Omega _{m}^{-1/2} -2\Omega _{\Lambda } \Omega _{m}^{-3/2} +{\rm O} \left[\left(1+z_{*} \right)^{-1/2} \right]\right\}.
\end{equation}
Using the relation $\Omega _{\Lambda } =1-\Omega _{m} $ and neglecting the higher order terms, one obtains
\begin{equation} \label{acust13_}
d_{sls} \approx \frac{2c\Omega _{m}^{-1/2} }{7H_{0} (1+z_{*} )} \left(9-2\Omega _{m}^{3} \right)
\end{equation}
Combination of the relations \eqref{acust10_} and \eqref{acust13_} gives for the first acoustic peak the following
\begin{equation} \label{acust14_}
l\approx \frac{d_{sls} }{r_{s} } \approx 0.74\sqrt{1+z_{*} } \left(9-2\Omega _{m}^{3} \right)\simeq 220.
\end{equation}
This result agrees very well with numerous observations.
</p>
</div>
</div></div>

<div id="SCM52"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 55 ===
\bf Compare the asymptotes of time dependence of the scale factor
$a(t)$ for the SCM and de Sitter models. Explain physical reasons of
their distinction.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM53"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 56 ===
Redshift for any object slowly changes due to the acceleration (or deceleration) of the Universe expansion. Estimate change of velocity in one year.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

$$
\Delta z = \dot z \Delta t = H_0 \left(1 + z - {H(z)\over H_0} \right)\Delta t
$$

$$H(z)=H_0 \left[\Omega_{m0}(1+z)^3+\Omega_{\Lambda 0}\right]^{1/2}$$
$$\Omega_{m0}+\Omega_{\Lambda 0}=1,~\Omega_{m0}\approx 0.3$$
$$\Delta z \approx 2\cdot 10^{-11}$$
$$\Delta v = c{\Delta z\over 1+z}\approx 0.25~\mbox{cm/s}$$

</p>
</div>
</div></div>

<div id="SCM54"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 57 ===
Determine the lower limit of ratio of the total volume of the Universe to the observed one?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
If the Universe is the 3-sphere, then its radius is $R_0=a_0$
$$
R_0 = \frac{1}{H_0 \sqrt {\left| \Omega _{curv} \right|} }
$$
For the SCM parameters the cosmological horizon equals to
$$
L_{p,0} = a_0 \int_0^{t_0 } \frac{dt}{a(t)} = \frac{3.6}{H_0}
$$
We assumed that $a(t)\sim t^{2/3}.$ Therefore
$$
\frac{R_0}
{L_{p,0} } = \frac{1}
{3.6\sqrt {\left| \Omega _{curv}\right|} }
$$
Taking into account that $\left|\Omega _{curv} \right| < 0.02,$ one obtains $\frac{a_0}{L_{p0} } > 2$.
Number of the regions similar to the one inside our horizon equals to the ratio of the 3-sphere volume $2\pi ^2 a_0^3 $ to the volume of the observed Universe
$$
N = \frac{2\pi ^2 a_0^3}
{(4\pi /3)L_{p0}^3 } \simeq 4.7\left(\frac{a_0 }
{L_{p0} } \right)^3 > 38
$$
Thus we can see not more than $3 \%$ of the Universe volume.
</p>
</div>
</div></div>

<div id="SCM55"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 58 ===
What is the difference between the inflationary expansion in the early Universe and the present accelerated expansion?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The difference is in the fact that in the present time there is about $27 \%$ of non-relativistic matter which was absent in the inflation period. The numerical values of Hubble parameter in the beginning of the inflation was very different from the one at the beginning of the current period of accelerated expansion of Universe.
</p>
</div>
</div></div>

<div id="SCM56"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 59 ===
Compare the values of Hubble parameter at the beginning of the
inflation period and at the beginning of the present accelerated
expansion of Universe.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
h^2 (x) = \Omega _{m0} x^3 + (1 - \Omega _{m0} )x^\alpha ;\quad \alpha = 3(1 + w)
$$
Assume for simplicity that the Universe is spatially flat. From the definition $O(x)$
it follows that
$$
O(x) = \Omega _{m0} + \left( 1 - \Omega _{m0}\right)\frac{x^\alpha - 1}
{x^3 - 1}
$$
The required result follows from the fact that $\alpha =0$ for cosmological constant, $\alpha >0$ for quintessence and $\alpha <0$ for phantom energy.
</p>
</div>
</div></div>

<div id="SCM57"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 60 ===
Plot the dependencies $h(x)=H(x)/H_0(x)$, where $x=1+z$, in the SCM, for quintessence and for phantom energy cases.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Plots of the dependencies $
h(x) = H(x)/H_0 (x);\quad x = 1 + z
$ in the SCM are presented on Fig. \ref{fig:52} for phantom energy and quintessence.
<gallery widths=600px heights=500px>
File:12_52.JPG|Dependence of the deceleration parameter on the redshift.
</gallery>
</p>
</div>
</div></div>

<div id="SCM58"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 61 ===
Find the constraints imposed by the Weak Energy Condition (WEC) for the dark energy on the redshift-dependent Hubble parameter. ([http://arxiv.org/abs/astro-ph/0703416 Inspired by A.Sen, R.Scherrer])
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The redshift-dependent Hubble parameter $H(z)$ is defined by
\[H^2(z)=\frac{8\pi G}{3}(\rho_m+\rho_{DE})\]
This equation can be rewritten as
\begin{equation}\label{SCM_eq_76_1}
E^2(z)\equiv\frac{H^2(z)}{H^2_0(z)}=\Omega_{m0}(1+z)^3+(1-\Omega_{m0})\frac{\rho_{DE}}{\rho_{DE0}}
\end{equation}
The WEC forces \[\frac{\rho_{DE}}{\rho_{DE0}}\ge1 \ \left(\frac{d\rho}{dz}\ge0\right)\] for $z>1$, so equation (\ref{SCM_eq_76_1}) becomes
\begin{equation}\label{SCM_eq_76_2}
E^2(z)\ge\Omega_{m0}(1+z)^3+(1-\Omega_{m0})
\end{equation}
Going back to equation (\ref{SCM_eq_76_2}) and taking the derivative with respect to $z$ (which we denote throughout with a prime) gives:
\[
2E(z)\frac{dE}{dz}=3\Omega_{m0}(1+z)^2+\frac{1-\Omega_{m0}}{\rho_{DE0}}\frac{d\rho_{DE}}{dz}.
\]
Now the WEC implies that \[\frac{d\rho_{DE}}{dz}\ge0,\] so we get
\begin{equation}\label{SCM_eq_76_3}
E(z)\frac{dE}{dz}\ge\frac32\Omega_{m0}(1+z)^2.
\end{equation}
Equations (\ref{SCM_eq_76_2}) and (\ref{SCM_eq_76_3}) give the constraints that the WEC for the dark energy places on the redshift-dependent Hubble parameter.
</p>
</div>
</div></div>

<div id="SCM59"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 62 ===
Using the statefinders, show that the power law cosmology mimics SCM model at (see Chapters 3 and 9)
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Recall that the statefinders are defined as
\[r\equiv\frac{\dddot a}{aH^3},\ s\equiv\frac{r-1}{3(q-1/2)}, \ q\ne\frac12.\]
The SCM model corresponds to the point $r=1,\, s=0$ in the $(s,r)$ plane. In power-law cosmology (see Chapter 3) the statefinders are given by
\[r=2q^2+q,\, s=\frac23(q+1).\]
We note that $r=1,\, s=0$ for $q=-1$. Thus the power law cosmology mimics SCM model with $q=-1$.
</p>
</div>
</div></div>

Evolution of Universe

2012-10-04T17:50:04Z

Den:

[[Category:Standard Cosmological Model|2]]

__NOTOC__
<div id="SCM-10"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 1 ===
Rewrite the first Friedman equation in terms of redshift and analyze the contributions of separate components at different stages of Universe evolution.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM11"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2 ===
Find the time dependence for the scale factor and analyze asymptotes of the dependence. Plot $a(t)$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM12"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 3 ===
Determine the redshift value corresponding to equality of radiation and matter densities.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
\rho _r(t) = \rho _m(t);\\
\rho _{r0}\frac{a_0^4}{a^4} = \rho _{m0}\frac{a_0^3}{a^3};\\
\frac{\rho _{m0}}{\rho _{r0}} = \frac{\Omega _{m0}}{\Omega _{r0}} = \frac{a_0}{a} = 1 + z;\\
\Omega _{r0} = 5 \cdot {10^{ - 5}} \\
z = \frac{\Omega _{m0}}{\Omega _{r0}} - 1 \approx 5400\\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM_9"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 4 ===
Construct effective one-dimensional potential (see [[Solutions_of_Friedman_equations_in_the_Big_Bang_model#dyn15|problem]] \ref{u_eff} of Chapter 3)
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
<div id="V(x)"></div>
$$
V(x) = - \frac{1}{2}\sum\limits_i \Omega _{i0}x^{ - (1 + 3w_i)};\quad x \equiv \frac{a}{a_0}
$$
In the SCM
$$
V(x) = - \frac{1}{2}\Omega _{m0}x^{-1}-\frac{1}{2}\Omega _{\Lambda 0}x^2 \simeq - 0.135x^{-1}-0.365x^2
$$

<gallery widths=600px heights=500px>
File:12_9.JPG|
</gallery>
</p>
</div>
</div></div>

<div id="SCM13"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 5 ===
Show that the following holds: $\dot H= -4\pi G\rho_m$ and $\ddot H= 12\pi G\rho_m H$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM_18"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 6 ===
Expand the scale factor in Taylor series near the time moment
$t_0$:
\[\frac{a(t)}{a(t_0)}=1+\sum\limits_{n=1}^\infty\frac{A_n(t_0)}{n!}[H_0(t-t_0)]^n;\quad A_n\equiv\frac{1}{aH^n}\frac{d^na}{dt^n}\]
and calculate values for few first coefficients $A_n$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The coefficients $A_1$ and $A_2$ equal to:
$$A_1 = \frac{\dot a}{aH} = 1;\quad A_2 = \frac{\ddot a}{aH^2} = - q = 1 - \frac{3}{2}{\Omega _m}$$
The parameters $A_n(n > 2)$ can be calculated by consequent differentiation of the relation
$$\ddot a = a\left( \dot H + H^2 \right)$$
The time derivatives of the scale factor can be determined making use of the fact that in SCM $\dot H = - 4\pi G\rho _m$, and $\rho _m = \rho _0a^{ - 3}$. For example,
$$A_3 = 1 + \frac{\ddot H}{H^3} + 3\frac{\dot H}{H^2}$$
Using $\ddot H = 12\pi G{\rho _m}H,\;\dot H = - 4\pi G\rho _m$, one finds that
$$A_3 = 1$$
Let us present the expressions for some consequent expansion coefficients
\[\begin{array}{l}
{A_4} = 1 - \frac{3^2}{2}\Omega _m;\\
{A_5} = 1 + 3\Omega _m + \frac{3^3}{2}\Omega _m^2;\\
{A_6} = 1 - \frac{3^3}{2}\Omega _m - 3^4\Omega _m^2 - \frac{3^4}{4}\Omega _m^3
\end{array}\]
</p>
</div>
</div></div>

<div id="SCM14"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 7 ===
Show that all the coefficients $A_n$ can be expressed through
elementary functions of the deceleration parameter $q$ or the
density parameter \[\Omega_m=\frac23(1+q).\]
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM15"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 8 ===
Consider the case of flat Universe filled by non-relativistic matter and dark
energy with state equation $p_X=w\rho_X$, where the state parameter $w$ is parameterized as the following \[w=w_0+ w_a(1-a)=w_0 + w_a \frac{z}{1+z}.\]
Express current values of cosmographic parameters through $w_0$ and
$w_a$. % (See [http://arxiv.org/abs/0802.1583 Cosmography of f(R) gravity])
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the considered case
\[\frac{H^2(z)}{H_0^2} = \Omega _m(1 + z)^3 + \Omega _X(1 + z)^{3\left( 1 + w_0 + w_a \right)}e^{ - \frac{3w_az}{1 + z}}\]
Calculation of the cosmographic parameters requires in general to integrate $H(z)$ in order to obtain $a(t).$ One can avoid this procedure with help of the relation
\[\frac{d}{dt} = - (1 + z)H(z)\frac{d}{dz}.\]
We can, given $H(z)$, apply that relation to calculate $\dot H,\ddot H, \dddot H, \ddddot H $ and so on. Using the expressions for time derivatives of Hubble parameter obtained in the problem (\ref{equ61_6}), Chapter 2,
\[\begin{array}{l}
\dot H = - H^2(1 + q);\\
\ddot H = H^3\left( j + 3q + 2 \right);\\
\dddot H = H^4\left[ s - 4j - 3q(q + 4) - 6 \right];\\
\ddddot H = H^5\left[ l - 5s + 10\left( q + 2 \right)j + 30(q + 2)q + 24 \right]
\end{array}\]
one can express the current values of the cosmological parameters (for $z = 0$) in terms of $w_0,w_a:$
\[\begin{array}{l}
q_0 = \frac{1}{2} + \frac{3}{2}\left( 1 - \Omega _m \right)w_0;\\
j_0 = 1 + \frac{3}{2}\left( 1 - \Omega _m \right)\left[3w_0\left( 1 + w_0 \right) + w_a \right];\\
s_0 = - \frac{7}{2} - \frac{33}{4}\left( 1 - \Omega _m \right)w_a - \frac{9}{4}\left( 1 - \Omega _m \right)\left[ 9 + \left(7 - \Omega _m \right)w_a \right]w_0 - \\
- \frac{9}{4}\left( 1 - \Omega _m \right)\left( 16 - 3\Omega _m \right)w_0^2 - \\
- \frac{27}{4}\left( 1 - \Omega _m \right)\left(3 - \Omega _m \right)w_0^3;\\
l_0 = \frac{35}{2} + \frac{1 - \Omega _m}{4}\left[ 213 + (7 - \Omega _m)w_a \right]w_a + \\
+ \frac{1 - \Omega _m}{4}\left[ 489 + 9(82 - 21\Omega _m)w_a \right]w_0 + \\
+ \frac{9}{2}\left( 1 - \Omega _m \right)\left[ 67 - 21\Omega _m + \frac{3}{2}(23 - 11\Omega _m)w_a \right]w_0^2 + \\
+ \frac{{27}}{4}\left( {1 - {\Omega _m}} \right)(47 - 24{\Omega _m})w_0^3 + \\
+ \frac{81}{2}\left( 1 - \Omega _m \right)(3 - 2\Omega _m)w_0^4
\end{array}\]
</p>
</div>
</div></div>

<div id="SCM16"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 9 ===
Show that the results of the previous problem applied to SCM coincide with
the ones obtained in the [[#SCM_18|problem]].
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the SCM $\left( {{w_0},{w_a}} \right) = \left( { - 1,0} \right)$ and therefore
\[\begin{array}{l}
{q_0} = \frac{1}{2} - \frac{3}{2}\left( {1 - {\Omega _m}} \right);\\
{j_0} = 1;\\
{s_0} = 1 - \frac{9}{2}{\Omega _m};\\
{l_0} = 1 + 3{\Omega _m} + \frac{{27}}{2}\Omega _m^2
\end{array}\]
</p>
</div>
</div></div>

<div id="SCM10"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 10 ===
Photons with $z=0.1,\ 1,\ 100,\
1000$ are registered. What was the Universe age $t_U$ in the moment of
their emission? What period of time $t_{ph}$ were the photons on the
way? Plot $t_U(z)$ and $t_{ph}(z)$
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
t_U=\int_z^\infty\frac{dz}{(1 + z)H(z)} ;\; t_UH_0=\int_z^\infty\frac{dz}{(1 + z)\sqrt{\Omega _{m0}(1 + z)^3 + \Omega _{\Lambda 0} } }; \\
t_{ph}H_0=\int_0^z \frac{dz}{(1 + z)\sqrt{\Omega _{m0}(1 + z)^3 +\Omega _{\Lambda 0}}}\\
\end{gathered}
$$
At $$z = 0.1\,\,\, \to \,\,\,\,t_U H_0 = 0.899\,\,\,\,\,t_{ph} H_0 = 0.093$$
$$z = 1\,\, \to \,\,\,t_U H_0 = 0.431\,\,\,\,\,t_{ph} H_0 = 0.561$$
$$z = 100\,\,\, \to \,\,t_U H_0 = 1.264 \times 10^{ - 3} \,\,\,\,\,t_{ph} H_0 = 0.991$$
$$z = 1000\,\, \to \,\,\,t_U H_0 = 3.663 \times 10^{ - 5} \,\,\,\,\,t_{ph} H_0 = 0.993$$

<gallery widths=600px heights=500px>
File:12_10(1).JPG|
</gallery>

<gallery widths=600px heights=500px>
File:12_10(2).JPG|
</gallery>
</p>
</div>
</div></div>

<div id="SCM17"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 11 ===
Determine the present physical distance to the object that emitted light with current redshift $z$
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The physical distance to the object emitted a photon is related to the redshift $z$ by
$$
R(z) = \frac{1}{1 + z}\int_0^z\frac{dz'}{H(z')}
$$
In the SCM
$$
H =H_0\sqrt {\Omega _{r0}(1 + z)^4 + \Omega _{m0}(1 + z)^3+\Omega _{\Lambda 0}}
$$
The parameters ${H_0},\;\Omega _{r0},\Omega _{m0},\Omega _{\Lambda 0}$ are fixed by the model.
</p>
</div>
</div></div>

<div id="SCM_15"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 12 ===
A photon was emitted at time $t$ and registered at time $t_0$ with red
shift $z$. Find and plot the dependence of emission time on the redshift $t(z)$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
a = \frac{1}{1 + z}; \\
a(t) = A^{1/3}\sh ^{2/3}\left(t/{t_\Lambda } \right);\quad A \equiv \frac{\Omega_{ m0}}{\Omega _{\Lambda 0}}; \\
\frac{t_0}{t_\Lambda }= Arth\sqrt {\Omega _{\Lambda 0}} ; \\
\frac{t}{t_0} = \frac{1}{Arth\sqrt{\Omega _{\Lambda 0}}}Arsh\left[\sqrt {\frac{\Omega _{\Lambda 0}}{\Omega _{m0}}} \frac{1}{\left(1 + z\right)^{3/2}} \right] \\
\end{gathered}
$$
<gallery widths=600px heights=500px>
File:12_13.JPG|
</gallery>
</p>
</div>
</div></div>

<div id="SCM18"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 13 ===
Find the time dependence for the scale factor and analyze its asymptotes. Plot $a(t)$
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the SCM
$$
\left(\frac{\dot a}{a}\right)^2 = H_0^2\left[ \Omega _{m0}\left(\frac{a_0}{a} \right)^3 + \Omega _{\Lambda 0} \right]
$$
Solution of the equation in the case $\Omega _{\Lambda 0} > 0$ reads
$$
\begin{gathered}
a(t) = a_0\left( \frac{\Omega _{m0}}{\Omega _{\Lambda 0}}\right)^{1/3}\left[ sh\left( \frac{3}{2}\sqrt {\Omega _{\Lambda 0}}H_0t \right) \right]^{2/3}; \\
a\left( t \ll H_0^{ - 1}\right) \propto t^{2/3};\; a\left( t \gg H_0^{ - 1} \right) \sim e^{H_0t} \\
\end{gathered}
$$
Below it is convenient to use the expression for time dependence of the scale factor in the following form
$$
a(t) = A^{1/3}\sinh ^{2/3}\left( t/{t_\Lambda }\right);\quad
A \equiv \frac{\Omega_{m0}}{\Omega _{\Lambda 0}} =
\frac{1 - \Omega _{\Lambda 0}}{\Omega _{\Lambda 0}};\quad t_\Lambda \equiv \frac{2}{3}\left(H_0 \right)^{-1}\left( \Omega _{\Lambda 0}\right)^{ - 1/2}
$$
</p>
</div>
</div></div>

<div id="SCM19"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 14 ===
Using the explicit solution for scale factor $a(t)$, obtained in the previous problem, find the cosmic horizon $R_h$ (see Chapter 3)
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
As we have seen above
\[a(t)=A\sinh^{2/3}{(t/t_\Lambda)},\]
where \[A=\frac{\Omega_{m0}}{\Omega_{\Lambda0}},\ t_\Lambda=\frac{2}{3H_\infty}.\]
Consider \[f(t)\equiv\ln{a(t)}=\ln A + \frac23\ln\sinh\frac{t}{t_\Lambda},\] then
\[\dot f=\frac{2}{3t_\Lambda\tanh(t/t_\Lambda)}=\frac{H_\infty}{\tanh(3tH_\infty/2)},\]
and finally
\[R_h=\frac{c}{\dot f}=\frac{c}{H_\infty}\tanh\left(\frac32tH_\infty\right).\]
</p>
</div>
</div></div>

<div id="SCM20"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 15 ===
Show that the Universe becomes $R_h$-delimited only after the cosmological constant starts to dominate.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
According to the result of the previous problem
\[\dot R_h=\frac32\left[1-\tanh^2\left(\frac32tH_\infty\right)\right]c,\]
so for early times $\dot R_h>c$ and an observer does not experience a divergent redshift with increasing $R$. Then after some ''transition'' time $t^*$, estimated by the condition
\[ct^*=R_h(t^*),\]
he (or she) will begin to encounter an observational limit at a finite radius $R_h$. Numerical solution of the equation gives $t^*\approx0.86H_\infty^{-1}$. This time is roughly the point at which the Universe transits from matter-dominated to $\Lambda$-dominated epoch. Eventually, the Universe becomes de Sitter one and therefore $\dot R_h\rightarrow0$ with $R_h$ settling at the fixed value $c/H_\infty$.
</p>
</div>
</div></div>

<div id="SCM21"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 16 ===
Analyze the stability of Friedmann equations.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Friedmann equations are given by
\[H^2=\frac{8\pi G}{3}\rho_m+\frac\Lambda3,\]
\[\dot H=-4\pi G\rho_m,\]
\[\dot\rho_m+3H\rho_m=0\]
Let us introduce the following dimensionless variables (recall that $\rho_m,\ \Lambda>0$):
\[x=\frac{\sqrt{8\pi G\rho_m}}{\sqrt3H},\ y=\frac{\sqrt\Lambda}{\sqrt3H}\]
Equations for $x$ and $y$ read (with $N\equiv\ln a$)
\[x'=\frac{dx}{dN}=\frac32x(x^2-1),\]
\[y'=\frac{dy}{dN}=\frac32yx^2.\]
The critical points, i.e. the solutions corresponding to $x'=0,\ y'=0$ and their eigenvalues are given by
\[x_{cr}=0,\ y_{cr}=1 (\Lambda-dominant\, case,)\ \mu_1=0, \mu_2=-\frac32;\]
\[x_{cr}=1,\ y_{cr}=0 (matter-dominant\, case,)\ \mu_1=3, \mu_2=\frac32.\]
From the eigenvalues we see that the matter dominant phase is unstable $\mu_{1,2}>0$ while the cosmological constant dominant phase has $\mu_2<0$ being stable. The existence of a zero eigenvalue $\mu_1=0$ in the first case originates from the fact that the two variables x and y are connected by the relation $x^2+y^2=1$. Therefore in this case one can reduce the dynamics to one-dimensional space.
</p>
</div>
</div></div>

<div id="SCM22"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 17 ===
Determine solutions for the perturbations $\delta\rho_m$ and $\delta H$. Make sure that the solutions are stable.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The variation of Friedmann equations yields
\[2H\delta H=\frac{\kappa^2}{3}\delta\rho_m,\]
\[\delta\dot H=-\frac{\kappa^2}{2}\delta\rho_m,\]
\[\delta\dot\rho_m+3\rho_m\delta H+ 3H\delta\rho_m=0,\]
where $\kappa^2\equiv8\pi G$.
Solution for the above equation is
\[\delta H=\frac{t_\Lambda}{4\kappa^2}C\tanh(\tau) e^{-f(\tau)},\ \delta\rho_m= \frac{C}{4\kappa^4}e^{-f(\tau)},\]
where \[\tau\equiv\frac{t}{t_\Lambda},\]
\[f(\tau)=-2\tau +\ln\left(-1+e^{4\tau}\right)+2\ln\tanh\tau.\]
Here $C$ is an arbitrary constant. We see that $f(\tau)$ approaches $2\tau$ as $t\rightarrow\infty$ and both $\delta H$ and $\delta\rho_m$ decay which implies that the considered solution is stable.
</p>
</div>
</div></div>

<div id="SCM23"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 18 ===
Rewrite the first Friedman equation in terms of conformal time.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[\frac{1}{a^4}\left( \frac{da}{d\eta } \right)^2=H_0^2\left[ \Omega_m a(\eta )^{-3}+\Omega_{\Lambda} \right]\]
</p>
</div>
</div></div>

<div id="SCM24"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 19 ===
Find relation between the scale factor and conformal time
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[\eta =\frac{1}{H_0}\int\limits_0^a\frac{dx}{\sqrt{x}{{\left( {{\Omega }_{m}}+{{\Omega }_{\Lambda }}{{x}^{3}} \right)}^{1/2}}}\]
</p>
</div>
</div></div>

<div id="SCM25"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 20 ===
Find explicit dependence of the scale factor on the conformal
time.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The integral obtained in the previous problem can be expressed in terms of elliptic integral of the first kind $F\left( \varphi ,k \right)$. For that purpose we rewrite it in the form
$$
\Omega^{1/6}_{\Lambda} \Omega^{1/3}_m H_{0} \eta = \int\limits_0^{u} \dfrac{dx}{\sqrt{x}\sqrt{1 + x^3}}.
$$
The upper limit equals to $u=\left(\dfrac{\Omega_{\Lambda}}{\Omega_m}\right)^{1/3}a.$ The integral can be presented in the form
$$3^{1/4} \Omega^{1/6}_{\Lambda} \Omega^{1/3}_m H_{0} \eta = F\left(\arccos\dfrac{1+(1-\sqrt{3})u}{1+(1+\sqrt{3})u}, \dfrac{\sqrt{2+\sqrt{3}}}{2}\right).$$

Here $y={{3}^{1/4}}\Omega _{\Lambda }^{1/6}\Omega _{m}^{1/3}{{H}_{0}}\eta ,\quad k=\frac{\sqrt{2+\sqrt{3}}}{2}\approx 0.97.$
</p>
</div>
</div></div>

<div id="SCM25"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 21 ===
Find relative density of dark energy $10^9$ years later.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
t = t_\Lambda Arcth\sqrt {\Omega _\Lambda} ;\quad \Omega _{\Lambda } = th^2\frac{t}{t_\Lambda };\\
t_\Lambda \simeq 10.768 \cdot 10^{9}\;\mbox{years}\;\;t = 14.7\cdot 10^{9}\;\mbox{years}\\
\Omega _{\Lambda 0} \simeq 0.77\\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM_17"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 22 ===
Find variation rate of relative density of dark energy. What are its asymptotic values? Plot its time dependence.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
\frac{d\Omega _\Lambda }{dt} = \frac{2}{t_\Lambda }\frac{sh(t/{t_\Lambda })}
{ch^3(t/{t_\Lambda })}; \\
x \equiv \frac{t}{t_\Lambda };\quad \frac{d\Omega _\Lambda }{dx} = 2\frac{sh(x)}{ch^3(x)} \\
\end{gathered}
$$

<gallery widths=600px heights=500px>
File:12_15.JPG|
</gallery>
</p>
</div>
</div></div>

<div id="SCM26"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 23 ===
Estimate size of the cosmological horizon.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The cosmological horizon is defined as
$$
L_{H0} = a_0\int\limits_0^{t_0}\frac{dt}{a\left( t \right)}.
$$
Using the results of the problem \ref{A_t_lambda}, one obtains
$$
L_{H0}= A^{-1/3}\int\limits_0^{t_0}\frac{dt}{sh^{2/3}\left( t/{t_\Lambda } \right)};
$$
</p>
</div>
</div></div>

<div id="SCM_19"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 24 ===
Find time dependence of Hubble parameter and plot it.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
H = \frac{\dot a}{a} =\frac{2}{3}\left(t_\Lambda \right)^{ - 1}cth\left( {\frac{t}{t_\Lambda }} \right)
$$

In the SCM ${t_0} \simeq 1.2t_\Lambda .$ Therefore
$$
H{t_0} \simeq 0.8cth\left( {1.2t/{t_0}} \right)
$$

<gallery widths=600px heights=500px>
File:12_16.JPG|Time dependence of the Hubble parameter (in units of the age of the Universe).
</gallery>

</p>
</div>
</div></div>

<div id="SCM27"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 25 ===
Find time dependence of dark matter density.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Present the first Friedmann equation in the form
\[\rho_m=\frac{1}{8\pi G}(3h^2-\Lambda)\]
and make use of results of the previous problem to obtain
\[\rho_m=\frac{\Lambda}{8\pi G}\frac{1}{\sinh^2(t/t_\Lambda)}.\]

\item {\bf Find the asymptotic (in time) value of the Hubble parameter.}

The required asymptote can be obtained in two ways: first, from the time dependence of the Hubble parameter (see the previous problem), and second, immediately from the first Friedmann equation in the SCM. Taking into account that $\rho_{m} \to 0$ at $t\to \infty$, one gets
$$
H_{\infty} = \frac{2}{3t_\Lambda} = \sqrt {\frac{\Lambda}{3}}.
$$
</p>
</div>
</div></div>

<div id="SCM28"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 26 ===
At present the age of the Universe $t_0\simeq13.7\cdot10^9$ years is close to the Hubble time $t_H=H_0^{-1}\simeq14\cdot10^9$ years.
Does the relation $t^*\simeq t_H(t^*)=H^{-1}(t^*)$ for the age $t^*$
of the Universe hold for any moment of its evolution
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
This relation cannot be satisfied in arbitrary moment of time, as the Hubble parameter has limit value in the SCM (see the previous problem):
$$
\frac{t^*}{t_H(t^*)} = H\left( t^*\right)t^* = \frac{2}{3}\frac{Arcth\sqrt {\Omega _\Lambda(t^*)}}{\sqrt{\Omega_\Lambda (t^*)}}.
$$
For $t\to \infty,\;\Omega _\Lambda\left( t^* \right) \to 1,$ and $$\frac{t^*}{t_H(t^*)} \to \infty. $$

\item {\bf Determine values of state parameter for the dark energy, that provide accelerated expansion of Universe in the present time.}

$$
\begin{gathered}
\sum\limits_i (\rho _{i0} + 3p_{i0}) < 0\; \Rightarrow \rho _{m0} + \rho _{DE0} + 3w_{DE}\rho _{DE0} < 0; \\
w_{DE} < - \frac{1}{3}\left( 1 + \frac{\Omega _{m0}}{\Omega _{DE0}}\right); \\
w_{DE} < - \frac{1}{3}\Omega _{DE0}^{-1};\; w_{DE} < - 0.46\\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM29"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 27 ===
Find current value of the deceleration parameter.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For a multi-component system the deceleration parameter equals to
$$
q = \frac{\Omega}{2} + \frac{3}{2}\sum\limits_i {w_i}\Omega _i
$$
In the SCM $\Omega = 1,w_m = 0,\;w_\Lambda= - 1,$ and
$$
\begin{gathered}
q = \frac{1 - 3\Omega_{\Lambda 0}}{2};\\
q_0 \simeq - 0.6 \\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM_24"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 28 ===
Find the redshift dependence of the deceleration parameter. Analyze the limiting cases.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
q \equiv - \frac{a\ddot a}{\dot a^2} = - \frac{\ddot a}{a}\frac{1}{H^2}; \\
\frac{\ddot a}{a} = - \frac{4\pi G}{3}\sum\limits_i \left( \rho _i + 3p_i \right) = - \frac{1}{2}H_0^2\sum\limits_i \Omega _{i0}\left(1 + 3w_i\right)(1 + z)^{3\left(1 + w_i\right)} ; \\
H^2 = H_0^2\sum\limits_i \Omega _{i0}(1 + z)^{3\left( 1 + w_i \right)}; \\
q = \frac{3}{2}\frac{\sum\limits_i \Omega _{i0}\left( 1 + w_i \right)(1 + z)^{3\left( 1 + w_i \right)}}
{\sum\limits_i \Omega _{i0}(1 + z)^{3\left( 1 + w_i\right)} } - 1 \\
\end{gathered}
$$
In the SCM
$$
q = \frac{1}{2}\frac{\Omega _{m0}(1 + z)^3- 2\Omega _{\Lambda 0} }
{\Omega _{m0}(1 + z)^3+\Omega_{\Lambda 0}}
$$

$$
q(z \to \infty ) = \frac{1}{2},\;q(z \to - 1) = - 1
$$

<gallery widths=600px heights=500px>
File:12_24.JPG|Dependence of the deceleration parameter on the redshift.
</gallery>
</p>
</div>
</div></div>

<div id="SCM_25"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 29 ===
Find and plot the time dependence of the deceleration parameter.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
q = - \frac{\ddot aa}
{\dot a^2} = - \frac{\ddot a}
{aH^2}; \\
a(t) = A^{1/3}sh^{2/3}\left(t/{t_\Lambda}\right);\quad A\equiv \frac{\Omega_ {m0}}
{\Omega _{\Lambda 0}}; \\
\frac{t_0}
{t_\Lambda } = Arth\sqrt {\Omega _{\Lambda 0}} ;\quad H = \frac{2}
{3}\frac{1}
{t_\Lambda }cth\left( \frac{t}
{t_\Lambda }\right); \\
q = \frac{1}
{2}\left[ 1 - 3th^2\left( \frac{t}
{t_\Lambda }\right)\right] \\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM29"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 30 ===
Find the moment of time when the dark energy started to dominate over dark matter. What redshift did it correspond to?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
\left( \frac{\dot a}{a} \right)^2 = \frac{8\pi G}
{3}\rho _{cr}\left( \Omega _{m0}(1 + z)^3+\Omega_{\Lambda 0}\right); \\
\Omega _{m0}(1 + z^*)^3 =\Omega _{\Lambda 0};\\
z^* = \left(\frac{\Omega _{\Lambda 0}}
{\Omega _{m0}}\right)^{1/3}-1; \\
\Omega _{\Lambda 0}\simeq 0.73;\quad \Omega _{m0}\simeq 0.27; \\
z^* \simeq 0.39; \\
\Omega _\Lambda = th^2\left(\frac{t}
{t_\Lambda }\right);\quad \Omega _m = 1 - \Omega _\Lambda = ch^{-2}\left( \frac{t}
{t_\Lambda } \right); \\
\Omega _\Lambda(\tilde t) = \Omega _m(\tilde t) \to \tilde t = \frac{Arsh(1)}{Arth(\sqrt {\Omega _{\Lambda 0} })}t_0 \simeq 9.5\cdot10^9 \,years \\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM30"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 31 ===
Determine the moment of time and redshift value corresponding to the transition from decelerated
expansion of the Universe to the accelerated one.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
\frac{\ddot a}
{a} = \frac{4\pi G}
{3}\rho_{cr}\left[2\Omega_{\Lambda 0} - \Omega _{m0}(1 + z)^3\right]; \\
2\Omega _{\Lambda 0} = \Omega _{m0}(1 + z^*)^3; \\
z^* = \left( \frac{2\Omega_{\Lambda 0}}
{\Omega _{m0}}\right)^{1/3} - 1; \\
\Omega _{\Lambda 0}\simeq 0.73;\quad \Omega _{m0}\simeq 0.27; \\
z^* \simeq 0.745 \\
\end{gathered}
$$

Let us now determine the time moment $t^*$ of the transition
$$
\begin{gathered}
a^* = \frac{1}
{1 + z^*} = \left(\frac{\Omega _{m0}}
{2\Omega _{\Lambda 0}} \right)^{1/3}; \\
t(a) = \frac{2}{3\sqrt{\Omega_{\Lambda 0}}H_0}Arsh\left(\sqrt{ \frac{\Omega_{\Lambda 0}}
{\Omega_{m0}}}a^{3/2} \right); \\
t^* = \frac{2}
{3\sqrt {\Omega _{\Lambda 0}}}H_0Arsh\left(\sqrt {1/2} \right) \\
\end{gathered}
$$
There is an alternative way to solve the problem.
In the multi-component Universe with the corresponding state equations $p_i = w_i\rho _i$ the deceleration parameter $q \equiv - \frac{\ddot aa}{\dot a^2}$ equals to
$$
q = \frac{\Omega }
{2} + \frac{3}
{2}\sum\limits_i w_i\Omega _i
$$
In the case of the SCM $\Omega =1$ and
$$
q = \frac{1}
{2} - \frac{3}
{2}\frac{\Omega _{\Lambda 0}}
{\left[\left(1 + z\right)^3\Omega _{m0}+\Omega_{\Lambda 0}\right]}
$$

From the condition $q=0$ it follows that
$$
z^* =\left(\frac{2\Omega_{\Lambda 0}}
{\Omega _{m0}} \right)^{1/3}- 1
$$
For $\Omega _{\Lambda 0} = 0.73,\;\Omega _{m0} = 0.27$ one obtains that
$$
z^* = 0.745
$$
</p>
</div>
</div></div>

<div id="SCM31"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 32 ===
Solve the previous problem using the derivative $d\eta/d\ln a$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\frac{d\eta}
{d\ln a} = a\frac{dt}{ada} = \frac{1}{aH} = \frac{1}{\dot a}
$$

If the quantity $aH = \dot a$ grows, then $\ddot a > 0,$ which corresponds to the accelerated expansion (inflation) of the Universe. In the SCM the quantity
$$
\frac{aH}{H_0} \simeq \sqrt{ \frac{0.27 + 0.73a^3}{a}}
$$
starts to grow at $a=0.573,$ which corresponds to $z=0.745$
</p>
</div>
</div></div>

<div id="SCM32"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 33 ===
Is dark energy domination necessary for transition to the accelerated expansion?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
No. The transition to the accelerated expansion takes place at $(z=0.75)$, which is considerably earlier than the transition to the dark energy domination stage at $(z=0.45).$
</p>
</div>
</div></div>

<div id="SCM33"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 34 ===
Consider flat Universe composed of matter and dark energy in form of cosmological constant. Find the redshift value corresponding to equality of densities of the both components $\rho_m(z_{eq})=\rho_\Lambda(z_{eq})$ and the one corresponding to beginning of the accelerated expansion $q\left(z_{accel} \right) = 0$. Obtain relation between $z_{eq}$ and $z_{accel}$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
z_{eq} = \left(\frac{\Omega _\Lambda}
{\Omega _{m0}} \right)^{1/3}- 1; \\
z_{accel}= \left(\frac{2\Omega_\Lambda }
{\Omega _{m0}}\right)^{1/3}- 1; \\
z_{accel} = 2^{1/3}\left(z_{eq}+ 1\right) - 1 \\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM34"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 35 ===
Show that density perturbations stop to grow after the transition from dust to $\Lambda$-dominated era.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
On the $\Lambda$-dominated stage the non-relativistic matter density perturbation size depends on conformal time as
\begin{equation}\label{1SCM30}
a(\eta)=-\frac{1}{H\eta}, \,\, \eta=-const\cdotp e^{-Ht},\,\,\eta<0
\end{equation}
where the Hubble parameter $H$ is constant, $H^2=(8\pi/3)G\rho_\Lambda$.
The linearized Einstein equations for the case of single-component medium take the following form in the momentum representation:
\begin{equation}\label{3SCM30}
k^2\Phi+3\frac{a^{'}}{a}+3\frac{a^{'2}}{a^2}\Phi=-4\pi Ga^2\delta\rho,
\end{equation}
\begin{equation}\label{2SCM30}
\Phi^{''}+3\frac{a^{'}}{a}\Phi^{'}+\left(2\frac{a^{''}}{a} - \frac{a^{'2}}{a^2}\right)\Phi=4\pi Ga^2\delta p.
\end{equation}
Assuming that the $\Lambda$-term is constant (which is exactly true for the vacuum energy density) one obtains $\delta\rho_\Lambda=\delta p_\Lambda=0$.
Using \eqref{2SCM30} and taking into account \eqref{1SCM30} we come up with the equation: $${\Phi}^{''}-\frac{3}{\eta}{\Phi}^{'}-\frac{3}{\eta^2}\Phi=0,$$
Solutions of the equation behave as
$$\Phi\propto\eta\propto\frac 1a$$ and $$\Phi\propto\eta^3\propto\frac{1}{a^3}.$$
Use now \eqref{3SCM30} for the modes beyond the horizon and take into account that for dust density $\rho_M\propto a^{-3}.$ Keeping only the first solution for the gravitational potential (the second decays too fast), one can see that the density perturbations do not grow: for the modes under horizon the relation
$$\delta_M\propto a^3\frac{k^2}{a^2}\Phi=const$$ holds.

Therefore the perturbations stop to grow at the transition from the dust-dominated to the $\Lambda$-dominated stage. If the present accelerated expansion of the Universe is actually caused by non-zero cosmological constant, then the structure formation from small primary perturbations is eventually close to end and it will never resume again: the structures larger than the presently observed ones will never appear.
</p>
</div>
</div></div>

<div id="SCM35"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 36 ===
What happens to the velocity fluctuations of non-relativistic matter and radiation with respect to Hubble flow in the epoch of cosmological constant domination?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM36"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 37 ===
Find the ratio of baryon to non-baryon components in the galactic halo.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM37"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 38 ===
Imagine that in the Universe described by SCM the dark energy was instantly switched off. Analyze further dynamics of the Universe.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="local-group"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 39 ===
Estimate density of dark energy in form of cosmological constant using the Hubble diagram for the neighborhood of the Local group.
<gallery widths=600px heights=500px>
File:im_eps.JPG|Hubble diagram for the neighborhood of the Local group.
</gallery>
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
An elegant and simple method was recently proposed (see A.D.Chernin, UFN 178, 267 (2008)), which allows not only to become convinced in the existence of the dark energy but also to estimate its density. The method is based on the assumption, that the dark energy (in form of the cosmological constant) governs the Hubble flow even on comparably small distances $\left( \sim \,Mpc \right)$. Consider the resulting force $\vec{F}$ (per unit mass), acting on a galaxy from the picture \ref{hubble_diagram}. The force is a sum of two contributions:
$$
\vec{F}=\vec{F}_N+\vec{F}_\Lambda
$$
Here $\vec{F}_N$ is the force produced by the total mass of the Local group, and $F_{\Lambda }$ is the force created by the dark energy. For a crude estimate let us neglect the non-spherical mass distribution in the Local group. Then the force acting on a satellite galaxy in a given point $R$ outside the local Group equals to
$$
F=-\frac{GM_{LG}}{R^2}+\frac{8\pi G}{3}{\rho_{\Lambda }}R,
$$
where $M_{LG}\simeq (1\div 3) \times 10^{12}M_{\odot }$ is the total mass of the Local group. As can be seen from the picture, the transition from domination of the local group gravity to the vacuum domination takes place on the distance of order $R_{0}=1.8\div 2\; \, Mpc.$ It allows to make a rough estimate of the dark energy density from the condition $F\left( R=R_0 \right)=0.$ The estimate results in the following
$$
(0.1\pm 0.03)<\rho_V<(1\pm 0.3)\times 10^{-26} \, kg/m^3.
$$
The proximity of the obtained crude estimate to the actual value of the dark energy density in the SCM is striking.
$$
\rho_{V}\simeq (0.75\pm 0.05)\times 10^{-26}\, kg/m^3.
$$
</p>
</div>
</div></div>

<div id="SCM38"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 40 ===
Estimate the Local group mass by methods used in the previous problem.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM39"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 41 ===
Find "weak" points in the argumentation of the two preceding problems.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM40"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 42 ===
The product of the age of the Universe and current Hubble parameter (the Hubble's constant) is a very important test (Sandage consistency test) of internal consistency for any model of Universe. Analyze the
parameters on which the product $H_0 t_0$ depends in the Big Bang model and in the SCM.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM41"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 43 ===
Show that for a fixed source of radiation the luminosity distance for high redshift values in flat Universe is greater for the dark energy dominated case compared to the non-relativistic matter dominated one.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the case of Universe filled by dark energy and non-relativistic matter, the photometric distance is equal to
$$
d_L = \frac{1 + z}
{H_0}\int_0^z \frac{dz'}
{\sqrt{\Omega _{m0}\left(1 + z'\right)^3+ \Omega _{DE0}\left(1 + z' \right)^{3\left( 1 + w_{DE}\right)}}}
$$

In the case $\Omega _{m0} = 1,\;\Omega _{DE0} = 0$
one obtains
$$
d_L = \frac{2}
{H_0}\left[1 + z -(1 + z)^{1/2}\right]
$$
and in the case $\Omega _{m0} = 0,\;\Omega _{DE0} = 1$
$$
d_L = \frac{1}{H_0}z(1 + z)
$$
instead.

In means that the supernova appear dimmer in the dark energy dominated Universe.
</p>
</div>
</div></div>

<div id="SCM42"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 44 ===
In the observations that discovered the accelerated expansion of the Universe the researchers in particular detected two $Ia$ type supernovae: $1992P$, $z=0.026$, $m=16.08$ and $1997ap$, $z=0.83$,
$m=24.32$. Show that these observed parameters are in accordance with the SCM.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The apparent $m$ and absolute $M$ stellar magnitudes are related with the photometric distance $d_L$ by the expression
\begin{equation}\label{p(1)}
m - M = 5\log _{10}\left(\frac{d_L}
{\mbox{Mpc}} \right) + 25
\end{equation}
The photometric distance in the SCM is equal to
\begin{equation}\label{p(2)}
d_L = \frac{1 + z}
{H_0}\int_0^z \frac{dz'}
{\sqrt {\Omega _{m0}(1 + z)^3 + \Omega _\Lambda}}
\end{equation}
For small $z$ $d_L(z) \simeq z/H_0.$ Using the supernova $1992P$ with small $z=0.026$ one can find $M = - 19.09.$ Assuming that the $1a$-type supernova are standard candles (they have equal absolute magnitudes), one obtains from the relation \ref{p(1)} supernova with $z=0.83$
$$H_0d_L\simeq 1.16$$
From the other hand it follows from the relation \ref{p(1)} that in the case
$\Omega _{m0} = 1,\;\Omega_\Lambda = 0$
$$
H_0d_L\simeq 0.95
$$
In the SCM one gets much better agreement:
$$
H_0d_L \simeq 1.2
$$
</p>
</div>
</div></div>

<div id="SCM43"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 45 ===
Find the redshift value, at which a source of linear dimension $d$ has minimum visible angular
size.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The apparent angular size equals to $$\delta \theta = {{(1 + z)^2 d } \over {d_L (z)}},$$
where $d_L (z)$ is the photometric distance from a source with the redshift $z$. In the case of cosmological constant domination the photometric distance is:
$$d_L (z) = - {z \over {H_0 \Omega _\Lambda ^{1/2} }},$$
In the SCM (see the previous problem)
$$
d_L = \frac{1 + z}
{H_0}\int_0^z \frac{dz'}
{\sqrt {\Omega _{m0}(1 + z)^3 + \Omega _\Lambda}}
$$
then the angular size is
$$\delta \theta = - {{H_0 \Omega _\Lambda ^{1/2} d} \over 2}{{(1 + z)^2 } \over z},$$
and $z = - 1$. It corresponds to infinitely far future.
</p>
</div>
</div></div>

<div id="SCM44"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 46 ===
Compare the observed value of the dark energy density with the
one expected from the dimensionality considerations (the cosmological constant problem).
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The dimensionality considerations suggest the following value for density of the dark energy in form of the cosmological constant
\[\rho_\Lambda^{(dim)}\sim M_{Pl}^4\approx10^{76}GeV^4.\]
At the same time the ''observed'' value fixed by the SCM is
\[\rho_\Lambda^{(obs)}\approx10^{-48}GeV^4.\]
This huge disagreement
\[\frac{\rho_\Lambda^{(dim)}}{\rho_\Lambda^{(obs)}}>10^{120}\]
is called the ''cosmological constant problem''. It presents a fundamental problem for elementary particles theory and physics in general, rather than for cosmology itself. (Look for details in R.Bousso, [http://arxiv.org/abs/0708.4231 TASI Lectures on the cosmological constant])
</p>
</div>
</div></div>

<div id="SCM45"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 47 ===
Determine the density of vacuum energy using the Planck scale as cutoff parameter.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For a quantum field with a given mode of frequency $\omega$, the zero-point energy is $\pm\omega/2$ with the plus sign for a bosonic field and the minus sign for a fermionic field. The total zero-point energy is then $\pm\sum\limits_i\omega_i/2$. In the continuum limit, we have for a free field (bosonic or fermionic)
\[\frac12\sum\limits_i\omega_i=\frac12\int\frac{d^3xd^3k}{(2\pi)^3}(k^2+m^2)^{1/2}=V\int\frac{k^2dk}{4\pi^2}(k^2+m^2)^{1/2}.\] This integral diverges, thus we need to introduce the physical cutoff. In cosmological context this cutoff must be the Planck mass at largest. With the cutoff $k_{max}\sim M_{Pl}$ one obtains
\[\rho_\Lambda=\int_0^{k_{max}}\frac{k^2dk}{4\pi^2}(k^2+m^2)^{1/2} \approx\frac{k_{max}^4}{16\pi^2}=\frac{M_{Pl}^4}{16\pi^2} \approx10^{74}GeV^4.\]
This result only slightly modifies the dimensional considerations given in the previous problem.
</p>
</div>
</div></div>

<div id="SCM46"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 48 ===
Identifying the vacuum fluctuations density with the observable dark energy value in SCM, find the required frequency cutoff magnitude in the fluctuation spectrum.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[\nu_{max}=10^{12}Hz.\]
</p>
</div>
</div></div>

<div id="SCM47"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 49 ===
What purely cosmological problem originates from the divergence of the zero-point energy density?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
If the zero-point energy density is infinite, then our Universe is infinitely curved and the space size is infinitely small.
</p>
</div>
</div></div>

<div id="SCM48"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 50 ===
With exact supersymmetry, the bosonic contribution to cosmological constant is canceled by its fermionic counterpart. However, we know that our world looks like not supersymmetric. Supersymmetry, if exists, has to be broken above or around $100GeV$ scale. Compare the observed value of the dark energy density with the one expected from broken supersymmetry.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
If we take the $k_{max}^{(SUSY)}\sim100GeV$ cutoff, then
\[\rho_\Lambda^{(SUSY)}=\frac{\left(k_{max}^{(SUSY)}\right)^4}{16\pi^2}\approx10^6GeV^4.\]
\[\frac{\rho_\Lambda^{(SUSY)}}{\rho_\Lambda^{(obs)}}>10^{54}.\]
Thus supersymmetry does not solve the problem of the cosmological constant.
</p>
</div>
</div></div>

<div id="SCM49"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 51 ===
Determine duration of the inflation period.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Let us see what happens with the relative density during all the history of Universe. From the first Friedmann equation it follows that in the matter-dominated epoch $\left| \Omega - 1 \right| \sim t^{2/3},$ and in the radiation-dominated epoch $\left| \Omega - 1 \right| \sim t^{2/3}.$ Then assume that inflation started at $t_i,$ and finished at $t_f.$ Then in the period of $t_f < t < t_{eq} = 50000$ years the radiation dominated and the range $t_{eq} < t < t_0 $ corresponds to the matter domination epoch. Then the presently observed difference
$$
\left|\Omega (t_0 ) - 1 \right| = \left| \Omega (t_i ) - 1 \right|e^{ - 2H(t_f - t_i )} \left( \frac{t_{eq} }
{t_f } \right)\left( \frac{t_0}
{t_{eq} }\right)^{2/3}
$$
According the SCM $\left|\Omega \left( t_0\right) - 1 \right| < 0.04.$ If one assumes that before the beginning of the inflation there was $\left|\Omega \left( t_i \right) - 1 \right| \sim 1,$ then
$$
N > \frac{1}{2}\ln \left[\frac{\Omega \left( t_i \right) - 1}{\Omega \left(t_0\right) - 1}\left( \frac{t_{eq}}{t_f } \right)\left( \frac{t_0}{t_{eq}} \right)^{2/3} \right] \approx 60
$$

Assume that radiation dominated before the inflation started. Then $H = \frac{1}{2t}.$ If at the moment of time $t_i$ the vacuum dark energy started to dominate and then the expansion continued with constant rate, then $H\approx 1/t_i$ during the inflation. And finally one obtains that
$$
N = H\left(t_f - t_i\right) \approx \frac{t_f - t_i }
{t_i }
$$
it is commonly assumed that the inflation finishing time is of order of the Grand Unification time $t_f\approx 10^{-35}.$ Therefore $t_f\approx 10^{-38}.$ Evidently the inflation could last longer as we used only lower estimate for $N.$
</p>
</div>
</div></div>

<div id="SCM50"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 52 ===
Plot the dependence of luminosity distance $d_L$ (in units of $H_0^{-1}$) on the redshift $z$ for the two-component flat Universe with non-relativistic liquid ($w=0$) and cosmological constant
($w=-1$). Consider the following cases:

a) $\Omega_\Lambda^0=0$;

b) $\Omega_\Lambda^0=0.3$;

c) $\Omega_\Lambda^0=0.7$;

d) $\Omega_\Lambda^0=1$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM51"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 53 ===
For a Universe filled by dark energy with state equation $p_{DE}=w_{DE}\rho_{DE}$ and non-relativistic matter obtain the Taylor series for $d_L$ in terms of $z$ near the observation point
$z_0=0$. Explain the obtained result.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the universe filled by dark energy with the state equation $p_{DE}=w_{DE}\rho_{DE}$ and non-relativistic dark matter the photometric distance takes on the form:
\begin{equation}
\label{d_L}
d_L = \frac{1 + z}
{H_0}\int_0^z \frac{dz'}
{\sqrt{\Omega _{m0}\left(1 + z'\right)^3+ \Omega _{DE0}\left(1 + z' \right)^{3\left( 1 + w_{DE}\right)}}}
\end{equation}
Expand the integrand in Taylor series in $z$ in vicinity of the observation point $z_0=0$ to obtain
$$
\begin{gathered}
\label{tey}
\frac{1}{\sqrt{\Omega _{m0}\left(1 + z'\right)^3+ \Omega _{DE0}\left(1 + z' \right)^{3\left( 1 + w_{DE}\right)}}}\simeq \\
\simeq 1 -\frac{3}{2}\left(1+w_{DE}\Omega _{DE0}\right)z+\frac{3}{4}\left(3w_{DE}\Omega _{DE0}-1\right)z^2.\nonumber
\end{gathered}
$$
Substitution of the decomposition (\ref{tey}) into the integral (\ref{d_L}) results in the following:
\begin{equation}
\label{d_L}
d_L = \frac{1 + z}
{H_0}\left(z -\frac{3}{4}\left(1+w_{DE}\Omega _{DE0}\right)z^2+\frac{1}{4}\left(3w_{DE}\Omega _{DE0}-1\right)z^3\right).
\end{equation}
</p>
</div>
</div></div>

<div id="SCM_45"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 54 ===
Determine position of the first acoustic peak in the CMB power spectrum produced by baryon oscillations on the surface of the last scattering.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Early Universe was filled by photon-baryon plasma, which can be treated as a single-component liquid. The baryons were trapped by the potential walls generated by density fluctuations, and they were eventually compressed. The contraction lead to heating of the plasma and therefore to increasing radiative pressure of the photons which is directed outside. Finally the radiative pressure stop the compression and lead to expansion of the plasma. During the expansion the plasma cools down and its radiative pressure decreases. The gravity starts to dominate again leading to repeated compression. The concurrence between the gravity and the pressure lead to longitudinal (acoustic) oscillations in the photon-baryon fluid. When the matter and the radiation get decoupled in the recombination process, the picture of the acoustic oscillations remains frozen into the CMB. Today we detect the evidence of the primordial acoustic waves (dense and diluted regions) in form of the primary anisotropy in the CMB.

It is well known that any acoustic wave, whatever complicatedly shaped, can be represented in form of superposition of modes with different wave numbers $k$, $k\propto 1/\lambda $. Every mode $\lambda $ corresponds to certain angular scale $\theta $ on the sky. Therefore in order to facilitate the comparison of the theory with observations, one should use the angular (multipole) decomposition in terms of the Legendre polynomials $P_{l} (\cos \theta )$, instead of the Fourier transform in terms of sines and cosines. Order of the polynomial $l$ plays the same role as the index $k$ in the Fourier decomposition. For $l\ge 2$ the Legendre polynomials are oscillating functions in the interval $\left[1,-1\right]$. Number of the oscillations increases with growth of $l$. Therefore
\begin{equation} \label{acust1_}
l\propto \frac{1}{\theta }
\end{equation}
The temperature fluctuations can be observed and qualitatively analyzed with help of paired measurements in the directions $\stackrel{\frown}{n},\stackrel{\frown}{n}'$, so that $\stackrel{\frown}{n}\cdot \stackrel{\frown}{n}'=\cos \theta $. Taking average over all the pairs in the assumptions of Gaussian fluctuations, one obtains the two-point correlator $C(\theta )$, which can be presented in terms of the multipole decomposition
\begin{equation} \label{acust2_}
\left\langle \delta T(\stackrel{\frown}{n})\cdot \delta T(\stackrel{\frown}{n}')\right\rangle \equiv C(\theta )=\sum _{l}\frac{2l+1}{4\pi } C_{l} P_{l} (\cos \theta )
\end{equation}
where $C_{l} $ are the coefficients.

As we already mentioned, the analysis of the temperature fluctuations enables us to clarify the structure of the longitudinal oscillations. The mode with the maximum wavelength corresponds to the maximum angular size of the primary anisotropy. This fundamental mode was detected the first. There are reliable evidences of the fact that the second and the third modes were also already detected. The distance $r_{s} $, passed by the acoustic wave during the time period before the recombination, is called the sound horizon. The sound horizon is fixed by (or rather fixes) the physical scale o the last scattering surface. Size of the sound horizon depends on values of the physical parameters. Distance to the last scattering surface $d_{sls} $ depends on the cosmological parameters too. Together they determine the angular size of the sound horizon$\theta _{s} $ (see Fig.)

<gallery widths=600px heights=500px>
File:acustic_pic.JPG|Dependence of the deceleration parameter on the redshift.
</gallery>

\begin{equation} \label{acust3_}
\theta _{s} \approx \frac{r_{s} }{d_{sls} }
\end{equation}
Analysis of the temperature anisotropy allows to determine $\theta _{s} $. Varying the cosmological parameters in $r_{s} ,d_{sls} ,$ one achieves the best agreement with the observations and thus fixes the cosmological parameters.

We can estimate the sound horizon size as the length passed by sound from the moment $t=0$ to the recombination time $t_{*} $
\begin{equation} \label{acust4_}
r_{s} (z_{*} ;\Omega _{b} ,\Omega _{r} )\approx \int _{0}^{t_{*} }c_{s} dt
\end{equation}
where $z_{*} $ is the redshift of the recombination moment $\left(z_{*} \approx 1100\right)$, $c_{s} $ is the sound speed in the photon-baryon medium
\begin{equation} \label{acust5_}
c_{s} \approx c\left[3\left(1+\frac{3\Omega _{b} }{\Omega _{r} } \right)\right]^{-1/2}
\end{equation}
To determine $dt$ one should use the first Friedmann equation in the form
\begin{equation} \label{acust6_}
\left(\frac{da}{dt} \right)^{2} =H_{0}^{2} \left[\Omega _{r0} \left(\frac{a_{0} }{a} \right)^{2} +\Omega _{m0} \frac{a_{0} }{a} +\Omega _{k0} +\Omega _{\Lambda 0} \left(\frac{a}{a_{0} } \right)^{2} \right]
\end{equation}
Making use of the relations $\frac{a_{0} }{a} =1+z$ and $\Omega _{r} +\Omega _{m} +\Omega _{\Lambda } +\Omega _{k} =1$, one obtains
\begin{equation} \label{acust7_}
dt=H_{0}^{-1} (1+z)^{-1} \left\{\left(1+z\right)^{2} \left(1+\Omega _{m,0} z\right)+z(z+2)\left[\left(1+z)^{2} \Omega _{r.0} -\Omega _{\Lambda ,0} \right)\right]\right\}^{-1/2} dz
\end{equation}
Distance to the last scattering surface, which corresponds to its angular size, is determined by the quantity called the angular diameter distance. It is connected by the photometric distance $d$ by the simple relation
\begin{equation} \label{acust8_}
d_{sls} =\frac{d}{(1+z_{*} )^{2} }
\end{equation}
The position of the first acoustic peak is determined by
\begin{equation} \label{acust9_}
l\approx \frac{d_{sls} }{r_{s} }.
\end{equation}
Let us now predict the position of the first peak for the case of flat Universe. To the leading order the sound speed is
\[c_{s} =\frac{c}{\sqrt{3} }. \]
Let us assume also that in the recombination epoch the Universe was matter-dominated. Under such assumptions one gets
\begin{equation} \label{acust10_}
r_{s} =\frac{c_{s} }{H_{0} \sqrt{\Omega _{m} } } \int _{z_{*} }^{\infty }(1+z)^{-5/2} dz =r_{s} =\frac{2c_{s} }{3H_{0} \sqrt{\Omega _{m} } } \left(1+z_{*} \right)^{-3/2}
\end{equation}
Using the standard procedure (see section 4), we obtain for the comoving radial coordinate $r_{sls} $ of the last scattering surface the following
\begin{equation} \label{acust11_}
r_{lsr} =\frac{c}{H_{0} } \int _{0}^{z_{*} }\left[\Omega _{m} (1+z)^{^{3} } +\Omega _{\Lambda } \right] ^{-1/2} dz
\end{equation}
Replacing the integrand by its binomial decomposition
\[\Omega _{m}^{-1/2} (1+z)^{-3/2} -\left(\Omega _{\Lambda } /2\Omega _{m}^{3/2} \right)\left(1+z\right)^{-9/2} \]
and taking into account that
\[d_{sls} =r_{sls} /(1+z_{*} ),\]
we obtain
\begin{equation} \label{acust12_}
d_{sls} =\frac{2c}{7H_{0} (1+z_{*} )} \left\{7\Omega _{m}^{-1/2} -2\Omega _{\Lambda } \Omega _{m}^{-3/2} +{\rm O} \left[\left(1+z_{*} \right)^{-1/2} \right]\right\}.
\end{equation}
Using the relation $\Omega _{\Lambda } =1-\Omega _{m} $ and neglecting the higher order terms, one obtains
\begin{equation} \label{acust13_}
d_{sls} \approx \frac{2c\Omega _{m}^{-1/2} }{7H_{0} (1+z_{*} )} \left(9-2\Omega _{m}^{3} \right)
\end{equation}
Combination of the relations \eqref{acust10_} and \eqref{acust13_} gives for the first acoustic peak the following
\begin{equation} \label{acust14_}
l\approx \frac{d_{sls} }{r_{s} } \approx 0.74\sqrt{1+z_{*} } \left(9-2\Omega _{m}^{3} \right)\simeq 220.
\end{equation}
This result agrees very well with numerous observations.
</p>
</div>
</div></div>

<div id="SCM52"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 55 ===
\bf Compare the asymptotes of time dependence of the scale factor
$a(t)$ for the SCM and de Sitter models. Explain physical reasons of
their distinction.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM53"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 56 ===
Redshift for any object slowly changes due to the acceleration (or deceleration) of the Universe expansion. Estimate change of velocity in one year.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

$$
\Delta z = \dot z \Delta t = H_0 \left(1 + z - {H(z)\over H_0} \right)\Delta t
$$

$$H(z)=H_0 \left[\Omega_{m0}(1+z)^3+\Omega_{\Lambda 0}\right]^{1/2}$$
$$\Omega_{m0}+\Omega_{\Lambda 0}=1,~\Omega_{m0}\approx 0.3$$
$$\Delta z \approx 2\cdot 10^{-11}$$
$$\Delta v = c{\Delta z\over 1+z}\approx 0.25~\mbox{cm/s}$$

</p>
</div>
</div></div>

<div id="SCM54"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 57 ===
Determine the lower limit of ratio of the total volume of the Universe to the observed one?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
If the Universe is the 3-sphere, then its radius is $R_0=a_0$
$$
R_0 = \frac{1}{H_0 \sqrt {\left| \Omega _{curv} \right|} }
$$
For the SCM parameters the cosmological horizon equals to
$$
L_{p,0} = a_0 \int_0^{t_0 } \frac{dt}{a(t)} = \frac{3.6}{H_0}
$$
We assumed that $a(t)\sim t^{2/3}.$ Therefore
$$
\frac{R_0}
{L_{p,0} } = \frac{1}
{3.6\sqrt {\left| \Omega _{curv}\right|} }
$$
Taking into account that $\left|\Omega _{curv} \right| < 0.02,$ one obtains $\frac{a_0}{L_{p0} } > 2$.
Number of the regions similar to the one inside our horizon equals to the ratio of the 3-sphere volume $2\pi ^2 a_0^3 $ to the volume of the observed Universe
$$
N = \frac{2\pi ^2 a_0^3}
{(4\pi /3)L_{p0}^3 } \simeq 4.7\left(\frac{a_0 }
{L_{p0} } \right)^3 > 38
$$
Thus we can see not more than $3 \%$ of the Universe volume.
</p>
</div>
</div></div>

<div id="SCM55"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 58 ===
What is the difference between the inflationary expansion in the early Universe and the present accelerated expansion?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The difference is in the fact that in the present time there is about $27 \%$ of non-relativistic matter which was absent in the inflation period. The numerical values of Hubble parameter in the beginning of the inflation was very different from the one at the beginning of the current period of accelerated expansion of Universe.
</p>
</div>
</div></div>

<div id="SCM56"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 59 ===
Compare the values of Hubble parameter at the beginning of the
inflation period and at the beginning of the present accelerated
expansion of Universe.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
h^2 (x) = \Omega _{m0} x^3 + (1 - \Omega _{m0} )x^\alpha ;\quad \alpha = 3(1 + w)
$$
Assume for simplicity that the Universe is spatially flat. From the definition $O(x)$
it follows that
$$
O(x) = \Omega _{m0} + \left( 1 - \Omega _{m0}\right)\frac{x^\alpha - 1}
{x^3 - 1}
$$
The required result follows from the fact that $\alpha =0$ for cosmological constant, $\alpha >0$ for quintessence and $\alpha <0$ for phantom energy.
</p>
</div>
</div></div>

<div id="SCM57"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 60 ===
Plot the dependencies $h(x)=H(x)/H_0(x)$, where $x=1+z$, in the SCM, for quintessence and for phantom energy cases.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Plots of the dependencies $
h(x) = H(x)/H_0 (x);\quad x = 1 + z
$ in the SCM are presented on Fig. \ref{fig:52} for phantom energy and quintessence.
<gallery widths=600px heights=500px>
File:12_52.JPG|Dependence of the deceleration parameter on the redshift.
</gallery>
</p>
</div>
</div></div>

<div id="SCM58"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 61 ===
Find the constraints imposed by the Weak Energy Condition (WEC) for the dark energy on the redshift-dependent Hubble parameter. ([http://arxiv.org/abs/astro-ph/0703416 Inspired by A.Sen, R.Scherrer])
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The redshift-dependent Hubble parameter $H(z)$ is defined by
\[H^2(z)=\frac{8\pi G}{3}(\rho_m+\rho_{DE})\]
This equation can be rewritten as
\begin{equation}\label{SCM_eq_76_1}
E^2(z)\equiv\frac{H^2(z)}{H^2_0(z)}=\Omega_{m0}(1+z)^3+(1-\Omega_{m0})\frac{\rho_{DE}}{\rho_{DE0}}
\end{equation}
The WEC forces \[\frac{\rho_{DE}}{\rho_{DE0}}\ge1 \ \left(\frac{d\rho}{dz}\ge0\right)\] for $z>1$, so equation (\ref{SCM_eq_76_1}) becomes
\begin{equation}\label{SCM_eq_76_2}
E^2(z)\ge\Omega_{m0}(1+z)^3+(1-\Omega_{m0})
\end{equation}
Going back to equation (\ref{SCM_eq_76_2}) and taking the derivative with respect to $z$ (which we denote throughout with a prime) gives:
\[
2E(z)\frac{dE}{dz}=3\Omega_{m0}(1+z)^2+\frac{1-\Omega_{m0}}{\rho_{DE0}}\frac{d\rho_{DE}}{dz}.
\]
Now the WEC implies that \[\frac{d\rho_{DE}}{dz}\ge0,\] so we get
\begin{equation}\label{SCM_eq_76_3}
E(z)\frac{dE}{dz}\ge\frac32\Omega_{m0}(1+z)^2.
\end{equation}
Equations (\ref{SCM_eq_76_2}) and (\ref{SCM_eq_76_3}) give the constraints that the WEC for the dark energy places on the redshift-dependent Hubble parameter.
</p>
</div>
</div></div>

<div id="SCM59"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 62 ===
Using the statefinders, show that the power law cosmology mimics SCM model at (see Chapters 3 and 9)
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Recall that the statefinders are defined as
\[r\equiv\frac{\dddot a}{aH^3},\ s\equiv\frac{r-1}{3(q-1/2)}, \ q\ne\frac12.\]
The SCM model corresponds to the point $r=1,\, s=0$ in the $(s,r)$ plane. In power-law cosmology (see Chapter 3) the statefinders are given by
\[r=2q^2+q,\, s=\frac23(q+1).\]
We note that $r=1,\, s=0$ for $q=-1$. Thus the power law cosmology mimics SCM model with $q=-1$.
</p>
</div>
</div></div>

Evolution of Universe

2012-10-04T17:48:58Z

Den:

<div id="SCM-10"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 1 ===
Rewrite the first Friedman equation in terms of redshift and analyze the contributions of separate components at different stages of Universe evolution.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM11"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2 ===
Find the time dependence for the scale factor and analyze asymptotes of the dependence. Plot $a(t)$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM12"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 3 ===
Determine the redshift value corresponding to equality of radiation and matter densities.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
\rho _r(t) = \rho _m(t);\\
\rho _{r0}\frac{a_0^4}{a^4} = \rho _{m0}\frac{a_0^3}{a^3};\\
\frac{\rho _{m0}}{\rho _{r0}} = \frac{\Omega _{m0}}{\Omega _{r0}} = \frac{a_0}{a} = 1 + z;\\
\Omega _{r0} = 5 \cdot {10^{ - 5}} \\
z = \frac{\Omega _{m0}}{\Omega _{r0}} - 1 \approx 5400\\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM_9"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 4 ===
Construct effective one-dimensional potential (see [[Solutions_of_Friedman_equations_in_the_Big_Bang_model#dyn15|problem]] \ref{u_eff} of Chapter 3)
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
<div id="V(x)"></div>
$$
V(x) = - \frac{1}{2}\sum\limits_i \Omega _{i0}x^{ - (1 + 3w_i)};\quad x \equiv \frac{a}{a_0}
$$
In the SCM
$$
V(x) = - \frac{1}{2}\Omega _{m0}x^{-1}-\frac{1}{2}\Omega _{\Lambda 0}x^2 \simeq - 0.135x^{-1}-0.365x^2
$$

<gallery widths=600px heights=500px>
File:12_9.JPG|
</gallery>
</p>
</div>
</div></div>

<div id="SCM13"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 5 ===
Show that the following holds: $\dot H= -4\pi G\rho_m$ and $\ddot H= 12\pi G\rho_m H$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM_18"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 6 ===
Expand the scale factor in Taylor series near the time moment
$t_0$:
\[\frac{a(t)}{a(t_0)}=1+\sum\limits_{n=1}^\infty\frac{A_n(t_0)}{n!}[H_0(t-t_0)]^n;\quad A_n\equiv\frac{1}{aH^n}\frac{d^na}{dt^n}\]
and calculate values for few first coefficients $A_n$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The coefficients $A_1$ and $A_2$ equal to:
$$A_1 = \frac{\dot a}{aH} = 1;\quad A_2 = \frac{\ddot a}{aH^2} = - q = 1 - \frac{3}{2}{\Omega _m}$$
The parameters $A_n(n > 2)$ can be calculated by consequent differentiation of the relation
$$\ddot a = a\left( \dot H + H^2 \right)$$
The time derivatives of the scale factor can be determined making use of the fact that in SCM $\dot H = - 4\pi G\rho _m$, and $\rho _m = \rho _0a^{ - 3}$. For example,
$$A_3 = 1 + \frac{\ddot H}{H^3} + 3\frac{\dot H}{H^2}$$
Using $\ddot H = 12\pi G{\rho _m}H,\;\dot H = - 4\pi G\rho _m$, one finds that
$$A_3 = 1$$
Let us present the expressions for some consequent expansion coefficients
\[\begin{array}{l}
{A_4} = 1 - \frac{3^2}{2}\Omega _m;\\
{A_5} = 1 + 3\Omega _m + \frac{3^3}{2}\Omega _m^2;\\
{A_6} = 1 - \frac{3^3}{2}\Omega _m - 3^4\Omega _m^2 - \frac{3^4}{4}\Omega _m^3
\end{array}\]
</p>
</div>
</div></div>

<div id="SCM14"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 7 ===
Show that all the coefficients $A_n$ can be expressed through
elementary functions of the deceleration parameter $q$ or the
density parameter \[\Omega_m=\frac23(1+q).\]
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM15"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 8 ===
Consider the case of flat Universe filled by non-relativistic matter and dark
energy with state equation $p_X=w\rho_X$, where the state parameter $w$ is parameterized as the following \[w=w_0+ w_a(1-a)=w_0 + w_a \frac{z}{1+z}.\]
Express current values of cosmographic parameters through $w_0$ and
$w_a$. % (See [http://arxiv.org/abs/0802.1583 Cosmography of f(R) gravity])
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the considered case
\[\frac{H^2(z)}{H_0^2} = \Omega _m(1 + z)^3 + \Omega _X(1 + z)^{3\left( 1 + w_0 + w_a \right)}e^{ - \frac{3w_az}{1 + z}}\]
Calculation of the cosmographic parameters requires in general to integrate $H(z)$ in order to obtain $a(t).$ One can avoid this procedure with help of the relation
\[\frac{d}{dt} = - (1 + z)H(z)\frac{d}{dz}.\]
We can, given $H(z)$, apply that relation to calculate $\dot H,\ddot H, \dddot H, \ddddot H $ and so on. Using the expressions for time derivatives of Hubble parameter obtained in the problem (\ref{equ61_6}), Chapter 2,
\[\begin{array}{l}
\dot H = - H^2(1 + q);\\
\ddot H = H^3\left( j + 3q + 2 \right);\\
\dddot H = H^4\left[ s - 4j - 3q(q + 4) - 6 \right];\\
\ddddot H = H^5\left[ l - 5s + 10\left( q + 2 \right)j + 30(q + 2)q + 24 \right]
\end{array}\]
one can express the current values of the cosmological parameters (for $z = 0$) in terms of $w_0,w_a:$
\[\begin{array}{l}
q_0 = \frac{1}{2} + \frac{3}{2}\left( 1 - \Omega _m \right)w_0;\\
j_0 = 1 + \frac{3}{2}\left( 1 - \Omega _m \right)\left[3w_0\left( 1 + w_0 \right) + w_a \right];\\
s_0 = - \frac{7}{2} - \frac{33}{4}\left( 1 - \Omega _m \right)w_a - \frac{9}{4}\left( 1 - \Omega _m \right)\left[ 9 + \left(7 - \Omega _m \right)w_a \right]w_0 - \\
- \frac{9}{4}\left( 1 - \Omega _m \right)\left( 16 - 3\Omega _m \right)w_0^2 - \\
- \frac{27}{4}\left( 1 - \Omega _m \right)\left(3 - \Omega _m \right)w_0^3;\\
l_0 = \frac{35}{2} + \frac{1 - \Omega _m}{4}\left[ 213 + (7 - \Omega _m)w_a \right]w_a + \\
+ \frac{1 - \Omega _m}{4}\left[ 489 + 9(82 - 21\Omega _m)w_a \right]w_0 + \\
+ \frac{9}{2}\left( 1 - \Omega _m \right)\left[ 67 - 21\Omega _m + \frac{3}{2}(23 - 11\Omega _m)w_a \right]w_0^2 + \\
+ \frac{{27}}{4}\left( {1 - {\Omega _m}} \right)(47 - 24{\Omega _m})w_0^3 + \\
+ \frac{81}{2}\left( 1 - \Omega _m \right)(3 - 2\Omega _m)w_0^4
\end{array}\]
</p>
</div>
</div></div>

<div id="SCM16"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 9 ===
Show that the results of the previous problem applied to SCM coincide with
the ones obtained in the [[#SCM_18|problem]].
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the SCM $\left( {{w_0},{w_a}} \right) = \left( { - 1,0} \right)$ and therefore
\[\begin{array}{l}
{q_0} = \frac{1}{2} - \frac{3}{2}\left( {1 - {\Omega _m}} \right);\\
{j_0} = 1;\\
{s_0} = 1 - \frac{9}{2}{\Omega _m};\\
{l_0} = 1 + 3{\Omega _m} + \frac{{27}}{2}\Omega _m^2
\end{array}\]
</p>
</div>
</div></div>

<div id="SCM10"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 10 ===
Photons with $z=0.1,\ 1,\ 100,\
1000$ are registered. What was the Universe age $t_U$ in the moment of
their emission? What period of time $t_{ph}$ were the photons on the
way? Plot $t_U(z)$ and $t_{ph}(z)$
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
t_U=\int_z^\infty\frac{dz}{(1 + z)H(z)} ;\; t_UH_0=\int_z^\infty\frac{dz}{(1 + z)\sqrt{\Omega _{m0}(1 + z)^3 + \Omega _{\Lambda 0} } }; \\
t_{ph}H_0=\int_0^z \frac{dz}{(1 + z)\sqrt{\Omega _{m0}(1 + z)^3 +\Omega _{\Lambda 0}}}\\
\end{gathered}
$$
At $$z = 0.1\,\,\, \to \,\,\,\,t_U H_0 = 0.899\,\,\,\,\,t_{ph} H_0 = 0.093$$
$$z = 1\,\, \to \,\,\,t_U H_0 = 0.431\,\,\,\,\,t_{ph} H_0 = 0.561$$
$$z = 100\,\,\, \to \,\,t_U H_0 = 1.264 \times 10^{ - 3} \,\,\,\,\,t_{ph} H_0 = 0.991$$
$$z = 1000\,\, \to \,\,\,t_U H_0 = 3.663 \times 10^{ - 5} \,\,\,\,\,t_{ph} H_0 = 0.993$$

<gallery widths=600px heights=500px>
File:12_10(1).JPG|
</gallery>

<gallery widths=600px heights=500px>
File:12_10(2).JPG|
</gallery>
</p>
</div>
</div></div>

<div id="SCM17"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 11 ===
Determine the present physical distance to the object that emitted light with current redshift $z$
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The physical distance to the object emitted a photon is related to the redshift $z$ by
$$
R(z) = \frac{1}{1 + z}\int_0^z\frac{dz'}{H(z')}
$$
In the SCM
$$
H =H_0\sqrt {\Omega _{r0}(1 + z)^4 + \Omega _{m0}(1 + z)^3+\Omega _{\Lambda 0}}
$$
The parameters ${H_0},\;\Omega _{r0},\Omega _{m0},\Omega _{\Lambda 0}$ are fixed by the model.
</p>
</div>
</div></div>

<div id="SCM_15"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 12 ===
A photon was emitted at time $t$ and registered at time $t_0$ with red
shift $z$. Find and plot the dependence of emission time on the redshift $t(z)$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
a = \frac{1}{1 + z}; \\
a(t) = A^{1/3}\sh ^{2/3}\left(t/{t_\Lambda } \right);\quad A \equiv \frac{\Omega_{ m0}}{\Omega _{\Lambda 0}}; \\
\frac{t_0}{t_\Lambda }= Arth\sqrt {\Omega _{\Lambda 0}} ; \\
\frac{t}{t_0} = \frac{1}{Arth\sqrt{\Omega _{\Lambda 0}}}Arsh\left[\sqrt {\frac{\Omega _{\Lambda 0}}{\Omega _{m0}}} \frac{1}{\left(1 + z\right)^{3/2}} \right] \\
\end{gathered}
$$
<gallery widths=600px heights=500px>
File:12_13.JPG|
</gallery>
</p>
</div>
</div></div>

<div id="SCM18"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 13 ===
Find the time dependence for the scale factor and analyze its asymptotes. Plot $a(t)$
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the SCM
$$
\left(\frac{\dot a}{a}\right)^2 = H_0^2\left[ \Omega _{m0}\left(\frac{a_0}{a} \right)^3 + \Omega _{\Lambda 0} \right]
$$
Solution of the equation in the case $\Omega _{\Lambda 0} > 0$ reads
$$
\begin{gathered}
a(t) = a_0\left( \frac{\Omega _{m0}}{\Omega _{\Lambda 0}}\right)^{1/3}\left[ sh\left( \frac{3}{2}\sqrt {\Omega _{\Lambda 0}}H_0t \right) \right]^{2/3}; \\
a\left( t \ll H_0^{ - 1}\right) \propto t^{2/3};\; a\left( t \gg H_0^{ - 1} \right) \sim e^{H_0t} \\
\end{gathered}
$$
Below it is convenient to use the expression for time dependence of the scale factor in the following form
$$
a(t) = A^{1/3}\sinh ^{2/3}\left( t/{t_\Lambda }\right);\quad
A \equiv \frac{\Omega_{m0}}{\Omega _{\Lambda 0}} =
\frac{1 - \Omega _{\Lambda 0}}{\Omega _{\Lambda 0}};\quad t_\Lambda \equiv \frac{2}{3}\left(H_0 \right)^{-1}\left( \Omega _{\Lambda 0}\right)^{ - 1/2}
$$
</p>
</div>
</div></div>

<div id="SCM19"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 14 ===
Using the explicit solution for scale factor $a(t)$, obtained in the previous problem, find the cosmic horizon $R_h$ (see Chapter 3)
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
As we have seen above
\[a(t)=A\sinh^{2/3}{(t/t_\Lambda)},\]
where \[A=\frac{\Omega_{m0}}{\Omega_{\Lambda0}},\ t_\Lambda=\frac{2}{3H_\infty}.\]
Consider \[f(t)\equiv\ln{a(t)}=\ln A + \frac23\ln\sinh\frac{t}{t_\Lambda},\] then
\[\dot f=\frac{2}{3t_\Lambda\tanh(t/t_\Lambda)}=\frac{H_\infty}{\tanh(3tH_\infty/2)},\]
and finally
\[R_h=\frac{c}{\dot f}=\frac{c}{H_\infty}\tanh\left(\frac32tH_\infty\right).\]
</p>
</div>
</div></div>

<div id="SCM20"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 15 ===
Show that the Universe becomes $R_h$-delimited only after the cosmological constant starts to dominate.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
According to the result of the previous problem
\[\dot R_h=\frac32\left[1-\tanh^2\left(\frac32tH_\infty\right)\right]c,\]
so for early times $\dot R_h>c$ and an observer does not experience a divergent redshift with increasing $R$. Then after some ''transition'' time $t^*$, estimated by the condition
\[ct^*=R_h(t^*),\]
he (or she) will begin to encounter an observational limit at a finite radius $R_h$. Numerical solution of the equation gives $t^*\approx0.86H_\infty^{-1}$. This time is roughly the point at which the Universe transits from matter-dominated to $\Lambda$-dominated epoch. Eventually, the Universe becomes de Sitter one and therefore $\dot R_h\rightarrow0$ with $R_h$ settling at the fixed value $c/H_\infty$.
</p>
</div>
</div></div>

<div id="SCM21"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 16 ===
Analyze the stability of Friedmann equations.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Friedmann equations are given by
\[H^2=\frac{8\pi G}{3}\rho_m+\frac\Lambda3,\]
\[\dot H=-4\pi G\rho_m,\]
\[\dot\rho_m+3H\rho_m=0\]
Let us introduce the following dimensionless variables (recall that $\rho_m,\ \Lambda>0$):
\[x=\frac{\sqrt{8\pi G\rho_m}}{\sqrt3H},\ y=\frac{\sqrt\Lambda}{\sqrt3H}\]
Equations for $x$ and $y$ read (with $N\equiv\ln a$)
\[x'=\frac{dx}{dN}=\frac32x(x^2-1),\]
\[y'=\frac{dy}{dN}=\frac32yx^2.\]
The critical points, i.e. the solutions corresponding to $x'=0,\ y'=0$ and their eigenvalues are given by
\[x_{cr}=0,\ y_{cr}=1 (\Lambda-dominant\, case,)\ \mu_1=0, \mu_2=-\frac32;\]
\[x_{cr}=1,\ y_{cr}=0 (matter-dominant\, case,)\ \mu_1=3, \mu_2=\frac32.\]
From the eigenvalues we see that the matter dominant phase is unstable $\mu_{1,2}>0$ while the cosmological constant dominant phase has $\mu_2<0$ being stable. The existence of a zero eigenvalue $\mu_1=0$ in the first case originates from the fact that the two variables x and y are connected by the relation $x^2+y^2=1$. Therefore in this case one can reduce the dynamics to one-dimensional space.
</p>
</div>
</div></div>

<div id="SCM22"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 17 ===
Determine solutions for the perturbations $\delta\rho_m$ and $\delta H$. Make sure that the solutions are stable.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The variation of Friedmann equations yields
\[2H\delta H=\frac{\kappa^2}{3}\delta\rho_m,\]
\[\delta\dot H=-\frac{\kappa^2}{2}\delta\rho_m,\]
\[\delta\dot\rho_m+3\rho_m\delta H+ 3H\delta\rho_m=0,\]
where $\kappa^2\equiv8\pi G$.
Solution for the above equation is
\[\delta H=\frac{t_\Lambda}{4\kappa^2}C\tanh(\tau) e^{-f(\tau)},\ \delta\rho_m= \frac{C}{4\kappa^4}e^{-f(\tau)},\]
where \[\tau\equiv\frac{t}{t_\Lambda},\]
\[f(\tau)=-2\tau +\ln\left(-1+e^{4\tau}\right)+2\ln\tanh\tau.\]
Here $C$ is an arbitrary constant. We see that $f(\tau)$ approaches $2\tau$ as $t\rightarrow\infty$ and both $\delta H$ and $\delta\rho_m$ decay which implies that the considered solution is stable.
</p>
</div>
</div></div>

<div id="SCM23"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 18 ===
Rewrite the first Friedman equation in terms of conformal time.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[\frac{1}{a^4}\left( \frac{da}{d\eta } \right)^2=H_0^2\left[ \Omega_m a(\eta )^{-3}+\Omega_{\Lambda} \right]\]
</p>
</div>
</div></div>

<div id="SCM24"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 19 ===
Find relation between the scale factor and conformal time
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[\eta =\frac{1}{H_0}\int\limits_0^a\frac{dx}{\sqrt{x}{{\left( {{\Omega }_{m}}+{{\Omega }_{\Lambda }}{{x}^{3}} \right)}^{1/2}}}\]
</p>
</div>
</div></div>

<div id="SCM25"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 20 ===
Find explicit dependence of the scale factor on the conformal
time.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The integral obtained in the previous problem can be expressed in terms of elliptic integral of the first kind $F\left( \varphi ,k \right)$. For that purpose we rewrite it in the form
$$
\Omega^{1/6}_{\Lambda} \Omega^{1/3}_m H_{0} \eta = \int\limits_0^{u} \dfrac{dx}{\sqrt{x}\sqrt{1 + x^3}}.
$$
The upper limit equals to $u=\left(\dfrac{\Omega_{\Lambda}}{\Omega_m}\right)^{1/3}a.$ The integral can be presented in the form
$$3^{1/4} \Omega^{1/6}_{\Lambda} \Omega^{1/3}_m H_{0} \eta = F\left(\arccos\dfrac{1+(1-\sqrt{3})u}{1+(1+\sqrt{3})u}, \dfrac{\sqrt{2+\sqrt{3}}}{2}\right).$$

Here $y={{3}^{1/4}}\Omega _{\Lambda }^{1/6}\Omega _{m}^{1/3}{{H}_{0}}\eta ,\quad k=\frac{\sqrt{2+\sqrt{3}}}{2}\approx 0.97.$
</p>
</div>
</div></div>

<div id="SCM25"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 21 ===
Find relative density of dark energy $10^9$ years later.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
t = t_\Lambda Arcth\sqrt {\Omega _\Lambda} ;\quad \Omega _{\Lambda } = th^2\frac{t}{t_\Lambda };\\
t_\Lambda \simeq 10.768 \cdot 10^{9}\;\mbox{years}\;\;t = 14.7\cdot 10^{9}\;\mbox{years}\\
\Omega _{\Lambda 0} \simeq 0.77\\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM_17"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 22 ===
Find variation rate of relative density of dark energy. What are its asymptotic values? Plot its time dependence.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
\frac{d\Omega _\Lambda }{dt} = \frac{2}{t_\Lambda }\frac{sh(t/{t_\Lambda })}
{ch^3(t/{t_\Lambda })}; \\
x \equiv \frac{t}{t_\Lambda };\quad \frac{d\Omega _\Lambda }{dx} = 2\frac{sh(x)}{ch^3(x)} \\
\end{gathered}
$$

<gallery widths=600px heights=500px>
File:12_15.JPG|
</gallery>
</p>
</div>
</div></div>

<div id="SCM26"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 23 ===
Estimate size of the cosmological horizon.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The cosmological horizon is defined as
$$
L_{H0} = a_0\int\limits_0^{t_0}\frac{dt}{a\left( t \right)}.
$$
Using the results of the problem \ref{A_t_lambda}, one obtains
$$
L_{H0}= A^{-1/3}\int\limits_0^{t_0}\frac{dt}{sh^{2/3}\left( t/{t_\Lambda } \right)};
$$
</p>
</div>
</div></div>

<div id="SCM_19"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 24 ===
Find time dependence of Hubble parameter and plot it.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
H = \frac{\dot a}{a} =\frac{2}{3}\left(t_\Lambda \right)^{ - 1}cth\left( {\frac{t}{t_\Lambda }} \right)
$$

In the SCM ${t_0} \simeq 1.2t_\Lambda .$ Therefore
$$
H{t_0} \simeq 0.8cth\left( {1.2t/{t_0}} \right)
$$

<gallery widths=600px heights=500px>
File:12_16.JPG|Time dependence of the Hubble parameter (in units of the age of the Universe).
</gallery>

</p>
</div>
</div></div>

<div id="SCM27"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 25 ===
Find time dependence of dark matter density.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Present the first Friedmann equation in the form
\[\rho_m=\frac{1}{8\pi G}(3h^2-\Lambda)\]
and make use of results of the previous problem to obtain
\[\rho_m=\frac{\Lambda}{8\pi G}\frac{1}{\sinh^2(t/t_\Lambda)}.\]

\item {\bf Find the asymptotic (in time) value of the Hubble parameter.}

The required asymptote can be obtained in two ways: first, from the time dependence of the Hubble parameter (see the previous problem), and second, immediately from the first Friedmann equation in the SCM. Taking into account that $\rho_{m} \to 0$ at $t\to \infty$, one gets
$$
H_{\infty} = \frac{2}{3t_\Lambda} = \sqrt {\frac{\Lambda}{3}}.
$$
</p>
</div>
</div></div>

<div id="SCM28"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 26 ===
At present the age of the Universe $t_0\simeq13.7\cdot10^9$ years is close to the Hubble time $t_H=H_0^{-1}\simeq14\cdot10^9$ years.
Does the relation $t^*\simeq t_H(t^*)=H^{-1}(t^*)$ for the age $t^*$
of the Universe hold for any moment of its evolution
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
This relation cannot be satisfied in arbitrary moment of time, as the Hubble parameter has limit value in the SCM (see the previous problem):
$$
\frac{t^*}{t_H(t^*)} = H\left( t^*\right)t^* = \frac{2}{3}\frac{Arcth\sqrt {\Omega _\Lambda(t^*)}}{\sqrt{\Omega_\Lambda (t^*)}}.
$$
For $t\to \infty,\;\Omega _\Lambda\left( t^* \right) \to 1,$ and $$\frac{t^*}{t_H(t^*)} \to \infty. $$

\item {\bf Determine values of state parameter for the dark energy, that provide accelerated expansion of Universe in the present time.}

$$
\begin{gathered}
\sum\limits_i (\rho _{i0} + 3p_{i0}) < 0\; \Rightarrow \rho _{m0} + \rho _{DE0} + 3w_{DE}\rho _{DE0} < 0; \\
w_{DE} < - \frac{1}{3}\left( 1 + \frac{\Omega _{m0}}{\Omega _{DE0}}\right); \\
w_{DE} < - \frac{1}{3}\Omega _{DE0}^{-1};\; w_{DE} < - 0.46\\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM29"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 27 ===
Find current value of the deceleration parameter.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For a multi-component system the deceleration parameter equals to
$$
q = \frac{\Omega}{2} + \frac{3}{2}\sum\limits_i {w_i}\Omega _i
$$
In the SCM $\Omega = 1,w_m = 0,\;w_\Lambda= - 1,$ and
$$
\begin{gathered}
q = \frac{1 - 3\Omega_{\Lambda 0}}{2};\\
q_0 \simeq - 0.6 \\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM_24"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 28 ===
Find the redshift dependence of the deceleration parameter. Analyze the limiting cases.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
q \equiv - \frac{a\ddot a}{\dot a^2} = - \frac{\ddot a}{a}\frac{1}{H^2}; \\
\frac{\ddot a}{a} = - \frac{4\pi G}{3}\sum\limits_i \left( \rho _i + 3p_i \right) = - \frac{1}{2}H_0^2\sum\limits_i \Omega _{i0}\left(1 + 3w_i\right)(1 + z)^{3\left(1 + w_i\right)} ; \\
H^2 = H_0^2\sum\limits_i \Omega _{i0}(1 + z)^{3\left( 1 + w_i \right)}; \\
q = \frac{3}{2}\frac{\sum\limits_i \Omega _{i0}\left( 1 + w_i \right)(1 + z)^{3\left( 1 + w_i \right)}}
{\sum\limits_i \Omega _{i0}(1 + z)^{3\left( 1 + w_i\right)} } - 1 \\
\end{gathered}
$$
In the SCM
$$
q = \frac{1}{2}\frac{\Omega _{m0}(1 + z)^3- 2\Omega _{\Lambda 0} }
{\Omega _{m0}(1 + z)^3+\Omega_{\Lambda 0}}
$$

$$
q(z \to \infty ) = \frac{1}{2},\;q(z \to - 1) = - 1
$$

<gallery widths=600px heights=500px>
File:12_24.JPG|Dependence of the deceleration parameter on the redshift.
</gallery>
</p>
</div>
</div></div>

<div id="SCM_25"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 29 ===
Find and plot the time dependence of the deceleration parameter.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
q = - \frac{\ddot aa}
{\dot a^2} = - \frac{\ddot a}
{aH^2}; \\
a(t) = A^{1/3}sh^{2/3}\left(t/{t_\Lambda}\right);\quad A\equiv \frac{\Omega_ {m0}}
{\Omega _{\Lambda 0}}; \\
\frac{t_0}
{t_\Lambda } = Arth\sqrt {\Omega _{\Lambda 0}} ;\quad H = \frac{2}
{3}\frac{1}
{t_\Lambda }cth\left( \frac{t}
{t_\Lambda }\right); \\
q = \frac{1}
{2}\left[ 1 - 3th^2\left( \frac{t}
{t_\Lambda }\right)\right] \\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM29"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 30 ===
Find the moment of time when the dark energy started to dominate over dark matter. What redshift did it correspond to?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
\left( \frac{\dot a}{a} \right)^2 = \frac{8\pi G}
{3}\rho _{cr}\left( \Omega _{m0}(1 + z)^3+\Omega_{\Lambda 0}\right); \\
\Omega _{m0}(1 + z^*)^3 =\Omega _{\Lambda 0};\\
z^* = \left(\frac{\Omega _{\Lambda 0}}
{\Omega _{m0}}\right)^{1/3}-1; \\
\Omega _{\Lambda 0}\simeq 0.73;\quad \Omega _{m0}\simeq 0.27; \\
z^* \simeq 0.39; \\
\Omega _\Lambda = th^2\left(\frac{t}
{t_\Lambda }\right);\quad \Omega _m = 1 - \Omega _\Lambda = ch^{-2}\left( \frac{t}
{t_\Lambda } \right); \\
\Omega _\Lambda(\tilde t) = \Omega _m(\tilde t) \to \tilde t = \frac{Arsh(1)}{Arth(\sqrt {\Omega _{\Lambda 0} })}t_0 \simeq 9.5\cdot10^9 \,years \\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM30"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 31 ===
Determine the moment of time and redshift value corresponding to the transition from decelerated
expansion of the Universe to the accelerated one.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
\frac{\ddot a}
{a} = \frac{4\pi G}
{3}\rho_{cr}\left[2\Omega_{\Lambda 0} - \Omega _{m0}(1 + z)^3\right]; \\
2\Omega _{\Lambda 0} = \Omega _{m0}(1 + z^*)^3; \\
z^* = \left( \frac{2\Omega_{\Lambda 0}}
{\Omega _{m0}}\right)^{1/3} - 1; \\
\Omega _{\Lambda 0}\simeq 0.73;\quad \Omega _{m0}\simeq 0.27; \\
z^* \simeq 0.745 \\
\end{gathered}
$$

Let us now determine the time moment $t^*$ of the transition
$$
\begin{gathered}
a^* = \frac{1}
{1 + z^*} = \left(\frac{\Omega _{m0}}
{2\Omega _{\Lambda 0}} \right)^{1/3}; \\
t(a) = \frac{2}{3\sqrt{\Omega_{\Lambda 0}}H_0}Arsh\left(\sqrt{ \frac{\Omega_{\Lambda 0}}
{\Omega_{m0}}}a^{3/2} \right); \\
t^* = \frac{2}
{3\sqrt {\Omega _{\Lambda 0}}}H_0Arsh\left(\sqrt {1/2} \right) \\
\end{gathered}
$$
There is an alternative way to solve the problem.
In the multi-component Universe with the corresponding state equations $p_i = w_i\rho _i$ the deceleration parameter $q \equiv - \frac{\ddot aa}{\dot a^2}$ equals to
$$
q = \frac{\Omega }
{2} + \frac{3}
{2}\sum\limits_i w_i\Omega _i
$$
In the case of the SCM $\Omega =1$ and
$$
q = \frac{1}
{2} - \frac{3}
{2}\frac{\Omega _{\Lambda 0}}
{\left[\left(1 + z\right)^3\Omega _{m0}+\Omega_{\Lambda 0}\right]}
$$

From the condition $q=0$ it follows that
$$
z^* =\left(\frac{2\Omega_{\Lambda 0}}
{\Omega _{m0}} \right)^{1/3}- 1
$$
For $\Omega _{\Lambda 0} = 0.73,\;\Omega _{m0} = 0.27$ one obtains that
$$
z^* = 0.745
$$
</p>
</div>
</div></div>

<div id="SCM31"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 32 ===
Solve the previous problem using the derivative $d\eta/d\ln a$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\frac{d\eta}
{d\ln a} = a\frac{dt}{ada} = \frac{1}{aH} = \frac{1}{\dot a}
$$

If the quantity $aH = \dot a$ grows, then $\ddot a > 0,$ which corresponds to the accelerated expansion (inflation) of the Universe. In the SCM the quantity
$$
\frac{aH}{H_0} \simeq \sqrt{ \frac{0.27 + 0.73a^3}{a}}
$$
starts to grow at $a=0.573,$ which corresponds to $z=0.745$
</p>
</div>
</div></div>

<div id="SCM32"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 33 ===
Is dark energy domination necessary for transition to the accelerated expansion?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
No. The transition to the accelerated expansion takes place at $(z=0.75)$, which is considerably earlier than the transition to the dark energy domination stage at $(z=0.45).$
</p>
</div>
</div></div>

<div id="SCM33"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 34 ===
Consider flat Universe composed of matter and dark energy in form of cosmological constant. Find the redshift value corresponding to equality of densities of the both components $\rho_m(z_{eq})=\rho_\Lambda(z_{eq})$ and the one corresponding to beginning of the accelerated expansion $q\left(z_{accel} \right) = 0$. Obtain relation between $z_{eq}$ and $z_{accel}$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
z_{eq} = \left(\frac{\Omega _\Lambda}
{\Omega _{m0}} \right)^{1/3}- 1; \\
z_{accel}= \left(\frac{2\Omega_\Lambda }
{\Omega _{m0}}\right)^{1/3}- 1; \\
z_{accel} = 2^{1/3}\left(z_{eq}+ 1\right) - 1 \\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM34"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 35 ===
Show that density perturbations stop to grow after the transition from dust to $\Lambda$-dominated era.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
On the $\Lambda$-dominated stage the non-relativistic matter density perturbation size depends on conformal time as
\begin{equation}\label{1SCM30}
a(\eta)=-\frac{1}{H\eta}, \,\, \eta=-const\cdotp e^{-Ht},\,\,\eta<0
\end{equation}
where the Hubble parameter $H$ is constant, $H^2=(8\pi/3)G\rho_\Lambda$.
The linearized Einstein equations for the case of single-component medium take the following form in the momentum representation:
\begin{equation}\label{3SCM30}
k^2\Phi+3\frac{a^{'}}{a}+3\frac{a^{'2}}{a^2}\Phi=-4\pi Ga^2\delta\rho,
\end{equation}
\begin{equation}\label{2SCM30}
\Phi^{''}+3\frac{a^{'}}{a}\Phi^{'}+\left(2\frac{a^{''}}{a} - \frac{a^{'2}}{a^2}\right)\Phi=4\pi Ga^2\delta p.
\end{equation}
Assuming that the $\Lambda$-term is constant (which is exactly true for the vacuum energy density) one obtains $\delta\rho_\Lambda=\delta p_\Lambda=0$.
Using \eqref{2SCM30} and taking into account \eqref{1SCM30} we come up with the equation: $${\Phi}^{''}-\frac{3}{\eta}{\Phi}^{'}-\frac{3}{\eta^2}\Phi=0,$$
Solutions of the equation behave as
$$\Phi\propto\eta\propto\frac 1a$$ and $$\Phi\propto\eta^3\propto\frac{1}{a^3}.$$
Use now \eqref{3SCM30} for the modes beyond the horizon and take into account that for dust density $\rho_M\propto a^{-3}.$ Keeping only the first solution for the gravitational potential (the second decays too fast), one can see that the density perturbations do not grow: for the modes under horizon the relation
$$\delta_M\propto a^3\frac{k^2}{a^2}\Phi=const$$ holds.

Therefore the perturbations stop to grow at the transition from the dust-dominated to the $\Lambda$-dominated stage. If the present accelerated expansion of the Universe is actually caused by non-zero cosmological constant, then the structure formation from small primary perturbations is eventually close to end and it will never resume again: the structures larger than the presently observed ones will never appear.
</p>
</div>
</div></div>

<div id="SCM35"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 36 ===
What happens to the velocity fluctuations of non-relativistic matter and radiation with respect to Hubble flow in the epoch of cosmological constant domination?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM36"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 37 ===
Find the ratio of baryon to non-baryon components in the galactic halo.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM37"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 38 ===
Imagine that in the Universe described by SCM the dark energy was instantly switched off. Analyze further dynamics of the Universe.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="local-group"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 39 ===
Estimate density of dark energy in form of cosmological constant using the Hubble diagram for the neighborhood of the Local group.
<gallery widths=600px heights=500px>
File:im_eps.JPG|Hubble diagram for the neighborhood of the Local group.
</gallery>
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
An elegant and simple method was recently proposed (see A.D.Chernin, UFN 178, 267 (2008)), which allows not only to become convinced in the existence of the dark energy but also to estimate its density. The method is based on the assumption, that the dark energy (in form of the cosmological constant) governs the Hubble flow even on comparably small distances $\left( \sim \,Mpc \right)$. Consider the resulting force $\vec{F}$ (per unit mass), acting on a galaxy from the picture \ref{hubble_diagram}. The force is a sum of two contributions:
$$
\vec{F}=\vec{F}_N+\vec{F}_\Lambda
$$
Here $\vec{F}_N$ is the force produced by the total mass of the Local group, and $F_{\Lambda }$ is the force created by the dark energy. For a crude estimate let us neglect the non-spherical mass distribution in the Local group. Then the force acting on a satellite galaxy in a given point $R$ outside the local Group equals to
$$
F=-\frac{GM_{LG}}{R^2}+\frac{8\pi G}{3}{\rho_{\Lambda }}R,
$$
where $M_{LG}\simeq (1\div 3) \times 10^{12}M_{\odot }$ is the total mass of the Local group. As can be seen from the picture, the transition from domination of the local group gravity to the vacuum domination takes place on the distance of order $R_{0}=1.8\div 2\; \, Mpc.$ It allows to make a rough estimate of the dark energy density from the condition $F\left( R=R_0 \right)=0.$ The estimate results in the following
$$
(0.1\pm 0.03)<\rho_V<(1\pm 0.3)\times 10^{-26} \, kg/m^3.
$$
The proximity of the obtained crude estimate to the actual value of the dark energy density in the SCM is striking.
$$
\rho_{V}\simeq (0.75\pm 0.05)\times 10^{-26}\, kg/m^3.
$$
</p>
</div>
</div></div>

<div id="SCM38"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 40 ===
Estimate the Local group mass by methods used in the previous problem.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM39"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 41 ===
Find "weak" points in the argumentation of the two preceding problems.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM40"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 42 ===
The product of the age of the Universe and current Hubble parameter (the Hubble's constant) is a very important test (Sandage consistency test) of internal consistency for any model of Universe. Analyze the
parameters on which the product $H_0 t_0$ depends in the Big Bang model and in the SCM.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM41"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 43 ===
Show that for a fixed source of radiation the luminosity distance for high redshift values in flat Universe is greater for the dark energy dominated case compared to the non-relativistic matter dominated one.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the case of Universe filled by dark energy and non-relativistic matter, the photometric distance is equal to
$$
d_L = \frac{1 + z}
{H_0}\int_0^z \frac{dz'}
{\sqrt{\Omega _{m0}\left(1 + z'\right)^3+ \Omega _{DE0}\left(1 + z' \right)^{3\left( 1 + w_{DE}\right)}}}
$$

In the case $\Omega _{m0} = 1,\;\Omega _{DE0} = 0$
one obtains
$$
d_L = \frac{2}
{H_0}\left[1 + z -(1 + z)^{1/2}\right]
$$
and in the case $\Omega _{m0} = 0,\;\Omega _{DE0} = 1$
$$
d_L = \frac{1}{H_0}z(1 + z)
$$
instead.

In means that the supernova appear dimmer in the dark energy dominated Universe.
</p>
</div>
</div></div>

<div id="SCM42"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 44 ===
In the observations that discovered the accelerated expansion of the Universe the researchers in particular detected two $Ia$ type supernovae: $1992P$, $z=0.026$, $m=16.08$ and $1997ap$, $z=0.83$,
$m=24.32$. Show that these observed parameters are in accordance with the SCM.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The apparent $m$ and absolute $M$ stellar magnitudes are related with the photometric distance $d_L$ by the expression
\begin{equation}\label{p(1)}
m - M = 5\log _{10}\left(\frac{d_L}
{\mbox{Mpc}} \right) + 25
\end{equation}
The photometric distance in the SCM is equal to
\begin{equation}\label{p(2)}
d_L = \frac{1 + z}
{H_0}\int_0^z \frac{dz'}
{\sqrt {\Omega _{m0}(1 + z)^3 + \Omega _\Lambda}}
\end{equation}
For small $z$ $d_L(z) \simeq z/H_0.$ Using the supernova $1992P$ with small $z=0.026$ one can find $M = - 19.09.$ Assuming that the $1a$-type supernova are standard candles (they have equal absolute magnitudes), one obtains from the relation \ref{p(1)} supernova with $z=0.83$
$$H_0d_L\simeq 1.16$$
From the other hand it follows from the relation \ref{p(1)} that in the case
$\Omega _{m0} = 1,\;\Omega_\Lambda = 0$
$$
H_0d_L\simeq 0.95
$$
In the SCM one gets much better agreement:
$$
H_0d_L \simeq 1.2
$$
</p>
</div>
</div></div>

<div id="SCM43"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 45 ===
Find the redshift value, at which a source of linear dimension $d$ has minimum visible angular
size.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The apparent angular size equals to $$\delta \theta = {{(1 + z)^2 d } \over {d_L (z)}},$$
where $d_L (z)$ is the photometric distance from a source with the redshift $z$. In the case of cosmological constant domination the photometric distance is:
$$d_L (z) = - {z \over {H_0 \Omega _\Lambda ^{1/2} }},$$
In the SCM (see the previous problem)
$$
d_L = \frac{1 + z}
{H_0}\int_0^z \frac{dz'}
{\sqrt {\Omega _{m0}(1 + z)^3 + \Omega _\Lambda}}
$$
then the angular size is
$$\delta \theta = - {{H_0 \Omega _\Lambda ^{1/2} d} \over 2}{{(1 + z)^2 } \over z},$$
and $z = - 1$. It corresponds to infinitely far future.
</p>
</div>
</div></div>

<div id="SCM44"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 46 ===
Compare the observed value of the dark energy density with the
one expected from the dimensionality considerations (the cosmological constant problem).
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The dimensionality considerations suggest the following value for density of the dark energy in form of the cosmological constant
\[\rho_\Lambda^{(dim)}\sim M_{Pl}^4\approx10^{76}GeV^4.\]
At the same time the ''observed'' value fixed by the SCM is
\[\rho_\Lambda^{(obs)}\approx10^{-48}GeV^4.\]
This huge disagreement
\[\frac{\rho_\Lambda^{(dim)}}{\rho_\Lambda^{(obs)}}>10^{120}\]
is called the ''cosmological constant problem''. It presents a fundamental problem for elementary particles theory and physics in general, rather than for cosmology itself. (Look for details in R.Bousso, [http://arxiv.org/abs/0708.4231 TASI Lectures on the cosmological constant])
</p>
</div>
</div></div>

<div id="SCM45"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 47 ===
Determine the density of vacuum energy using the Planck scale as cutoff parameter.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For a quantum field with a given mode of frequency $\omega$, the zero-point energy is $\pm\omega/2$ with the plus sign for a bosonic field and the minus sign for a fermionic field. The total zero-point energy is then $\pm\sum\limits_i\omega_i/2$. In the continuum limit, we have for a free field (bosonic or fermionic)
\[\frac12\sum\limits_i\omega_i=\frac12\int\frac{d^3xd^3k}{(2\pi)^3}(k^2+m^2)^{1/2}=V\int\frac{k^2dk}{4\pi^2}(k^2+m^2)^{1/2}.\] This integral diverges, thus we need to introduce the physical cutoff. In cosmological context this cutoff must be the Planck mass at largest. With the cutoff $k_{max}\sim M_{Pl}$ one obtains
\[\rho_\Lambda=\int_0^{k_{max}}\frac{k^2dk}{4\pi^2}(k^2+m^2)^{1/2} \approx\frac{k_{max}^4}{16\pi^2}=\frac{M_{Pl}^4}{16\pi^2} \approx10^{74}GeV^4.\]
This result only slightly modifies the dimensional considerations given in the previous problem.
</p>
</div>
</div></div>

<div id="SCM46"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 48 ===
Identifying the vacuum fluctuations density with the observable dark energy value in SCM, find the required frequency cutoff magnitude in the fluctuation spectrum.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[\nu_{max}=10^{12}Hz.\]
</p>
</div>
</div></div>

<div id="SCM47"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 49 ===
What purely cosmological problem originates from the divergence of the zero-point energy density?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
If the zero-point energy density is infinite, then our Universe is infinitely curved and the space size is infinitely small.
</p>
</div>
</div></div>

<div id="SCM48"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 50 ===
With exact supersymmetry, the bosonic contribution to cosmological constant is canceled by its fermionic counterpart. However, we know that our world looks like not supersymmetric. Supersymmetry, if exists, has to be broken above or around $100GeV$ scale. Compare the observed value of the dark energy density with the one expected from broken supersymmetry.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
If we take the $k_{max}^{(SUSY)}\sim100GeV$ cutoff, then
\[\rho_\Lambda^{(SUSY)}=\frac{\left(k_{max}^{(SUSY)}\right)^4}{16\pi^2}\approx10^6GeV^4.\]
\[\frac{\rho_\Lambda^{(SUSY)}}{\rho_\Lambda^{(obs)}}>10^{54}.\]
Thus supersymmetry does not solve the problem of the cosmological constant.
</p>
</div>
</div></div>

<div id="SCM49"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 51 ===
Determine duration of the inflation period.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Let us see what happens with the relative density during all the history of Universe. From the first Friedmann equation it follows that in the matter-dominated epoch $\left| \Omega - 1 \right| \sim t^{2/3},$ and in the radiation-dominated epoch $\left| \Omega - 1 \right| \sim t^{2/3}.$ Then assume that inflation started at $t_i,$ and finished at $t_f.$ Then in the period of $t_f < t < t_{eq} = 50000$ years the radiation dominated and the range $t_{eq} < t < t_0 $ corresponds to the matter domination epoch. Then the presently observed difference
$$
\left|\Omega (t_0 ) - 1 \right| = \left| \Omega (t_i ) - 1 \right|e^{ - 2H(t_f - t_i )} \left( \frac{t_{eq} }
{t_f } \right)\left( \frac{t_0}
{t_{eq} }\right)^{2/3}
$$
According the SCM $\left|\Omega \left( t_0\right) - 1 \right| < 0.04.$ If one assumes that before the beginning of the inflation there was $\left|\Omega \left( t_i \right) - 1 \right| \sim 1,$ then
$$
N > \frac{1}{2}\ln \left[\frac{\Omega \left( t_i \right) - 1}{\Omega \left(t_0\right) - 1}\left( \frac{t_{eq}}{t_f } \right)\left( \frac{t_0}{t_{eq}} \right)^{2/3} \right] \approx 60
$$

Assume that radiation dominated before the inflation started. Then $H = \frac{1}{2t}.$ If at the moment of time $t_i$ the vacuum dark energy started to dominate and then the expansion continued with constant rate, then $H\approx 1/t_i$ during the inflation. And finally one obtains that
$$
N = H\left(t_f - t_i\right) \approx \frac{t_f - t_i }
{t_i }
$$
it is commonly assumed that the inflation finishing time is of order of the Grand Unification time $t_f\approx 10^{-35}.$ Therefore $t_f\approx 10^{-38}.$ Evidently the inflation could last longer as we used only lower estimate for $N.$
</p>
</div>
</div></div>

<div id="SCM50"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 52 ===
Plot the dependence of luminosity distance $d_L$ (in units of $H_0^{-1}$) on the redshift $z$ for the two-component flat Universe with non-relativistic liquid ($w=0$) and cosmological constant
($w=-1$). Consider the following cases:

a) $\Omega_\Lambda^0=0$;

b) $\Omega_\Lambda^0=0.3$;

c) $\Omega_\Lambda^0=0.7$;

d) $\Omega_\Lambda^0=1$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM51"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 53 ===
For a Universe filled by dark energy with state equation $p_{DE}=w_{DE}\rho_{DE}$ and non-relativistic matter obtain the Taylor series for $d_L$ in terms of $z$ near the observation point
$z_0=0$. Explain the obtained result.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the universe filled by dark energy with the state equation $p_{DE}=w_{DE}\rho_{DE}$ and non-relativistic dark matter the photometric distance takes on the form:
\begin{equation}
\label{d_L}
d_L = \frac{1 + z}
{H_0}\int_0^z \frac{dz'}
{\sqrt{\Omega _{m0}\left(1 + z'\right)^3+ \Omega _{DE0}\left(1 + z' \right)^{3\left( 1 + w_{DE}\right)}}}
\end{equation}
Expand the integrand in Taylor series in $z$ in vicinity of the observation point $z_0=0$ to obtain
$$
\begin{gathered}
\label{tey}
\frac{1}{\sqrt{\Omega _{m0}\left(1 + z'\right)^3+ \Omega _{DE0}\left(1 + z' \right)^{3\left( 1 + w_{DE}\right)}}}\simeq \\
\simeq 1 -\frac{3}{2}\left(1+w_{DE}\Omega _{DE0}\right)z+\frac{3}{4}\left(3w_{DE}\Omega _{DE0}-1\right)z^2.\nonumber
\end{gathered}
$$
Substitution of the decomposition (\ref{tey}) into the integral (\ref{d_L}) results in the following:
\begin{equation}
\label{d_L}
d_L = \frac{1 + z}
{H_0}\left(z -\frac{3}{4}\left(1+w_{DE}\Omega _{DE0}\right)z^2+\frac{1}{4}\left(3w_{DE}\Omega _{DE0}-1\right)z^3\right).
\end{equation}
</p>
</div>
</div></div>

<div id="SCM_45"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 54 ===
Determine position of the first acoustic peak in the CMB power spectrum produced by baryon oscillations on the surface of the last scattering.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Early Universe was filled by photon-baryon plasma, which can be treated as a single-component liquid. The baryons were trapped by the potential walls generated by density fluctuations, and they were eventually compressed. The contraction lead to heating of the plasma and therefore to increasing radiative pressure of the photons which is directed outside. Finally the radiative pressure stop the compression and lead to expansion of the plasma. During the expansion the plasma cools down and its radiative pressure decreases. The gravity starts to dominate again leading to repeated compression. The concurrence between the gravity and the pressure lead to longitudinal (acoustic) oscillations in the photon-baryon fluid. When the matter and the radiation get decoupled in the recombination process, the picture of the acoustic oscillations remains frozen into the CMB. Today we detect the evidence of the primordial acoustic waves (dense and diluted regions) in form of the primary anisotropy in the CMB.

It is well known that any acoustic wave, whatever complicatedly shaped, can be represented in form of superposition of modes with different wave numbers $k$, $k\propto 1/\lambda $. Every mode $\lambda $ corresponds to certain angular scale $\theta $ on the sky. Therefore in order to facilitate the comparison of the theory with observations, one should use the angular (multipole) decomposition in terms of the Legendre polynomials $P_{l} (\cos \theta )$, instead of the Fourier transform in terms of sines and cosines. Order of the polynomial $l$ plays the same role as the index $k$ in the Fourier decomposition. For $l\ge 2$ the Legendre polynomials are oscillating functions in the interval $\left[1,-1\right]$. Number of the oscillations increases with growth of $l$. Therefore
\begin{equation} \label{acust1_}
l\propto \frac{1}{\theta }
\end{equation}
The temperature fluctuations can be observed and qualitatively analyzed with help of paired measurements in the directions $\stackrel{\frown}{n},\stackrel{\frown}{n}'$, so that $\stackrel{\frown}{n}\cdot \stackrel{\frown}{n}'=\cos \theta $. Taking average over all the pairs in the assumptions of Gaussian fluctuations, one obtains the two-point correlator $C(\theta )$, which can be presented in terms of the multipole decomposition
\begin{equation} \label{acust2_}
\left\langle \delta T(\stackrel{\frown}{n})\cdot \delta T(\stackrel{\frown}{n}')\right\rangle \equiv C(\theta )=\sum _{l}\frac{2l+1}{4\pi } C_{l} P_{l} (\cos \theta )
\end{equation}
where $C_{l} $ are the coefficients.

As we already mentioned, the analysis of the temperature fluctuations enables us to clarify the structure of the longitudinal oscillations. The mode with the maximum wavelength corresponds to the maximum angular size of the primary anisotropy. This fundamental mode was detected the first. There are reliable evidences of the fact that the second and the third modes were also already detected. The distance $r_{s} $, passed by the acoustic wave during the time period before the recombination, is called the sound horizon. The sound horizon is fixed by (or rather fixes) the physical scale o the last scattering surface. Size of the sound horizon depends on values of the physical parameters. Distance to the last scattering surface $d_{sls} $ depends on the cosmological parameters too. Together they determine the angular size of the sound horizon$\theta _{s} $ (see Fig.)

<gallery widths=600px heights=500px>
File:acustic_pic.JPG|Dependence of the deceleration parameter on the redshift.
</gallery>

\begin{equation} \label{acust3_}
\theta _{s} \approx \frac{r_{s} }{d_{sls} }
\end{equation}
Analysis of the temperature anisotropy allows to determine $\theta _{s} $. Varying the cosmological parameters in $r_{s} ,d_{sls} ,$ one achieves the best agreement with the observations and thus fixes the cosmological parameters.

We can estimate the sound horizon size as the length passed by sound from the moment $t=0$ to the recombination time $t_{*} $
\begin{equation} \label{acust4_}
r_{s} (z_{*} ;\Omega _{b} ,\Omega _{r} )\approx \int _{0}^{t_{*} }c_{s} dt
\end{equation}
where $z_{*} $ is the redshift of the recombination moment $\left(z_{*} \approx 1100\right)$, $c_{s} $ is the sound speed in the photon-baryon medium
\begin{equation} \label{acust5_}
c_{s} \approx c\left[3\left(1+\frac{3\Omega _{b} }{\Omega _{r} } \right)\right]^{-1/2}
\end{equation}
To determine $dt$ one should use the first Friedmann equation in the form
\begin{equation} \label{acust6_}
\left(\frac{da}{dt} \right)^{2} =H_{0}^{2} \left[\Omega _{r0} \left(\frac{a_{0} }{a} \right)^{2} +\Omega _{m0} \frac{a_{0} }{a} +\Omega _{k0} +\Omega _{\Lambda 0} \left(\frac{a}{a_{0} } \right)^{2} \right]
\end{equation}
Making use of the relations $\frac{a_{0} }{a} =1+z$ and $\Omega _{r} +\Omega _{m} +\Omega _{\Lambda } +\Omega _{k} =1$, one obtains
\begin{equation} \label{acust7_}
dt=H_{0}^{-1} (1+z)^{-1} \left\{\left(1+z\right)^{2} \left(1+\Omega _{m,0} z\right)+z(z+2)\left[\left(1+z)^{2} \Omega _{r.0} -\Omega _{\Lambda ,0} \right)\right]\right\}^{-1/2} dz
\end{equation}
Distance to the last scattering surface, which corresponds to its angular size, is determined by the quantity called the angular diameter distance. It is connected by the photometric distance $d$ by the simple relation
\begin{equation} \label{acust8_}
d_{sls} =\frac{d}{(1+z_{*} )^{2} }
\end{equation}
The position of the first acoustic peak is determined by
\begin{equation} \label{acust9_}
l\approx \frac{d_{sls} }{r_{s} }.
\end{equation}
Let us now predict the position of the first peak for the case of flat Universe. To the leading order the sound speed is
\[c_{s} =\frac{c}{\sqrt{3} }. \]
Let us assume also that in the recombination epoch the Universe was matter-dominated. Under such assumptions one gets
\begin{equation} \label{acust10_}
r_{s} =\frac{c_{s} }{H_{0} \sqrt{\Omega _{m} } } \int _{z_{*} }^{\infty }(1+z)^{-5/2} dz =r_{s} =\frac{2c_{s} }{3H_{0} \sqrt{\Omega _{m} } } \left(1+z_{*} \right)^{-3/2}
\end{equation}
Using the standard procedure (see section 4), we obtain for the comoving radial coordinate $r_{sls} $ of the last scattering surface the following
\begin{equation} \label{acust11_}
r_{lsr} =\frac{c}{H_{0} } \int _{0}^{z_{*} }\left[\Omega _{m} (1+z)^{^{3} } +\Omega _{\Lambda } \right] ^{-1/2} dz
\end{equation}
Replacing the integrand by its binomial decomposition
\[\Omega _{m}^{-1/2} (1+z)^{-3/2} -\left(\Omega _{\Lambda } /2\Omega _{m}^{3/2} \right)\left(1+z\right)^{-9/2} \]
and taking into account that
\[d_{sls} =r_{sls} /(1+z_{*} ),\]
we obtain
\begin{equation} \label{acust12_}
d_{sls} =\frac{2c}{7H_{0} (1+z_{*} )} \left\{7\Omega _{m}^{-1/2} -2\Omega _{\Lambda } \Omega _{m}^{-3/2} +{\rm O} \left[\left(1+z_{*} \right)^{-1/2} \right]\right\}.
\end{equation}
Using the relation $\Omega _{\Lambda } =1-\Omega _{m} $ and neglecting the higher order terms, one obtains
\begin{equation} \label{acust13_}
d_{sls} \approx \frac{2c\Omega _{m}^{-1/2} }{7H_{0} (1+z_{*} )} \left(9-2\Omega _{m}^{3} \right)
\end{equation}
Combination of the relations \eqref{acust10_} and \eqref{acust13_} gives for the first acoustic peak the following
\begin{equation} \label{acust14_}
l\approx \frac{d_{sls} }{r_{s} } \approx 0.74\sqrt{1+z_{*} } \left(9-2\Omega _{m}^{3} \right)\simeq 220.
\end{equation}
This result agrees very well with numerous observations.
</p>
</div>
</div></div>

<div id="SCM52"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 55 ===
\bf Compare the asymptotes of time dependence of the scale factor
$a(t)$ for the SCM and de Sitter models. Explain physical reasons of
their distinction.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="SCM53"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 56 ===
Redshift for any object slowly changes due to the acceleration (or deceleration) of the Universe expansion. Estimate change of velocity in one year.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

$$
\Delta z = \dot z \Delta t = H_0 \left(1 + z - {H(z)\over H_0} \right)\Delta t
$$

$$H(z)=H_0 \left[\Omega_{m0}(1+z)^3+\Omega_{\Lambda 0}\right]^{1/2}$$
$$\Omega_{m0}+\Omega_{\Lambda 0}=1,~\Omega_{m0}\approx 0.3$$
$$\Delta z \approx 2\cdot 10^{-11}$$
$$\Delta v = c{\Delta z\over 1+z}\approx 0.25~\mbox{cm/s}$$

</p>
</div>
</div></div>

<div id="SCM54"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 57 ===
Determine the lower limit of ratio of the total volume of the Universe to the observed one?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
If the Universe is the 3-sphere, then its radius is $R_0=a_0$
$$
R_0 = \frac{1}{H_0 \sqrt {\left| \Omega _{curv} \right|} }
$$
For the SCM parameters the cosmological horizon equals to
$$
L_{p,0} = a_0 \int_0^{t_0 } \frac{dt}{a(t)} = \frac{3.6}{H_0}
$$
We assumed that $a(t)\sim t^{2/3}.$ Therefore
$$
\frac{R_0}
{L_{p,0} } = \frac{1}
{3.6\sqrt {\left| \Omega _{curv}\right|} }
$$
Taking into account that $\left|\Omega _{curv} \right| < 0.02,$ one obtains $\frac{a_0}{L_{p0} } > 2$.
Number of the regions similar to the one inside our horizon equals to the ratio of the 3-sphere volume $2\pi ^2 a_0^3 $ to the volume of the observed Universe
$$
N = \frac{2\pi ^2 a_0^3}
{(4\pi /3)L_{p0}^3 } \simeq 4.7\left(\frac{a_0 }
{L_{p0} } \right)^3 > 38
$$
Thus we can see not more than $3 \%$ of the Universe volume.
</p>
</div>
</div></div>

<div id="SCM55"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 58 ===
What is the difference between the inflationary expansion in the early Universe and the present accelerated expansion?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The difference is in the fact that in the present time there is about $27 \%$ of non-relativistic matter which was absent in the inflation period. The numerical values of Hubble parameter in the beginning of the inflation was very different from the one at the beginning of the current period of accelerated expansion of Universe.
</p>
</div>
</div></div>

<div id="SCM56"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 59 ===
Compare the values of Hubble parameter at the beginning of the
inflation period and at the beginning of the present accelerated
expansion of Universe.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
h^2 (x) = \Omega _{m0} x^3 + (1 - \Omega _{m0} )x^\alpha ;\quad \alpha = 3(1 + w)
$$
Assume for simplicity that the Universe is spatially flat. From the definition $O(x)$
it follows that
$$
O(x) = \Omega _{m0} + \left( 1 - \Omega _{m0}\right)\frac{x^\alpha - 1}
{x^3 - 1}
$$
The required result follows from the fact that $\alpha =0$ for cosmological constant, $\alpha >0$ for quintessence and $\alpha <0$ for phantom energy.
</p>
</div>
</div></div>

<div id="SCM57"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 60 ===
Plot the dependencies $h(x)=H(x)/H_0(x)$, where $x=1+z$, in the SCM, for quintessence and for phantom energy cases.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Plots of the dependencies $
h(x) = H(x)/H_0 (x);\quad x = 1 + z
$ in the SCM are presented on Fig. \ref{fig:52} for phantom energy and quintessence.
<gallery widths=600px heights=500px>
File:12_52.JPG|Dependence of the deceleration parameter on the redshift.
</gallery>
</p>
</div>
</div></div>

<div id="SCM58"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 61 ===
Find the constraints imposed by the Weak Energy Condition (WEC) for the dark energy on the redshift-dependent Hubble parameter. ([http://arxiv.org/abs/astro-ph/0703416 Inspired by A.Sen, R.Scherrer])
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The redshift-dependent Hubble parameter $H(z)$ is defined by
\[H^2(z)=\frac{8\pi G}{3}(\rho_m+\rho_{DE})\]
This equation can be rewritten as
\begin{equation}\label{SCM_eq_76_1}
E^2(z)\equiv\frac{H^2(z)}{H^2_0(z)}=\Omega_{m0}(1+z)^3+(1-\Omega_{m0})\frac{\rho_{DE}}{\rho_{DE0}}
\end{equation}
The WEC forces \[\frac{\rho_{DE}}{\rho_{DE0}}\ge1 \ \left(\frac{d\rho}{dz}\ge0\right)\] for $z>1$, so equation (\ref{SCM_eq_76_1}) becomes
\begin{equation}\label{SCM_eq_76_2}
E^2(z)\ge\Omega_{m0}(1+z)^3+(1-\Omega_{m0})
\end{equation}
Going back to equation (\ref{SCM_eq_76_2}) and taking the derivative with respect to $z$ (which we denote throughout with a prime) gives:
\[
2E(z)\frac{dE}{dz}=3\Omega_{m0}(1+z)^2+\frac{1-\Omega_{m0}}{\rho_{DE0}}\frac{d\rho_{DE}}{dz}.
\]
Now the WEC implies that \[\frac{d\rho_{DE}}{dz}\ge0,\] so we get
\begin{equation}\label{SCM_eq_76_3}
E(z)\frac{dE}{dz}\ge\frac32\Omega_{m0}(1+z)^2.
\end{equation}
Equations (\ref{SCM_eq_76_2}) and (\ref{SCM_eq_76_3}) give the constraints that the WEC for the dark energy places on the redshift-dependent Hubble parameter.
</p>
</div>
</div></div>

<div id="SCM59"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 62 ===
Using the statefinders, show that the power law cosmology mimics SCM model at (see Chapters 3 and 9)
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Recall that the statefinders are defined as
\[r\equiv\frac{\dddot a}{aH^3},\ s\equiv\frac{r-1}{3(q-1/2)}, \ q\ne\frac12.\]
The SCM model corresponds to the point $r=1,\, s=0$ in the $(s,r)$ plane. In power-law cosmology (see Chapter 3) the statefinders are given by
\[r=2q^2+q,\, s=\frac23(q+1).\]
We note that $r=1,\, s=0$ for $q=-1$. Thus the power law cosmology mimics SCM model with $q=-1$.
</p>
</div>
</div></div>

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Characteristic Parameters and Scales

2012-10-04T16:40:58Z

Den: /* Problem 2 */

[[Category:Standard Cosmological Model|1]]

__NOTOC__

<div id="SCM1"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 1 ===
Calculate the dark energy density and the cosmological constant value.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">For quark one obtains the following:
SCM assumes $\Omega _\Lambda = 0.73,$ and the Hubble constant value is generally accepted to be $H_0 = 70\, km\cdot s^{-1}\cdot Mpc^{- 1} = 2.26 \cdot 10^{ - 18}s^{-1}.$ Then the critical density corresponds to the following value
\[
\rho_{cr} = \frac{3H_0^2}{8\pi G} = 0.92 \cdot 10^{-29} g/cm^3
\]
and
\[\rho _\Lambda = \Omega _\Lambda \rho_{cr} = 0.67 \cdot 10^{-29} g/cm^3.\]</p>
</div>
</div></div>

<div id="SCM2"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2 ===
Estimate total number of baryons in the Universe.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
N = \frac{M}{m_n}=\frac{\frac{4}{3}\pi \left( \frac{c}{H_0} \right)^3\rho }{m_n}=\frac{\frac{4}{3}\pi \alpha \left( \frac{c}{H_0}\right)^3\frac{3H_0^2}{8\pi G}}{m_n}= \frac{1}{2}\alpha \frac{H_0^2}{Gm_n}\left( \frac{c}{H_0} \right)^3=\frac{1}{2}\alpha \frac{M_{Pl}}{m_n}\frac{M_{Pl}c^2}{\hbar H_0}\simeq 10^{80}\quad (\alpha \simeq 0.04)
$$
</p>
</div>
</div></div>

<div id="SCM_2"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 3 ===
Find dependence of relative density of dark energy $\Omega_\Lambda$ on the redshift. Plot $\Omega_\Lambda(z)$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Neglecting the contribution of radiation ($\rho _{r0} \ll \rho _{m0} < \rho _\Lambda$) one obtains
$$
\Omega _\Lambda (z) = \frac{\rho _\Lambda }{\rho _{cr}(z)} = \frac{\rho _\Lambda}{\rho _{m0}(1 + z)^3 + \rho _\Lambda } = \frac{\Omega _{\Lambda 0}}
{\Omega _{m0}(1 + z)^3 + \Omega _{\Lambda 0}};
$$
$$
\Omega _\Lambda (z) \simeq \frac{1}{\frac{1}{3}(1 + z)^3+ 1}
$$
<gallery widths=600px heights=500px>
File:12_1.jpg|
</gallery>
</p>
</div>
</div></div>

<div id="SCM3"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 4 ===
Estimate total number of stars in the Universe.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
\rho_{cr} \equiv \frac{3H_0^2}{8\pi G} \simeq 10^{-29}\mbox{g/cm}^3; \\
\rho_{bar}\simeq 0.04\rho _{cr}; \\
R_H \approx 10^{28}\mbox{cm}; \\
M_{bar} \approx 0.2 \times 10^{55}\,cm; \\
M_ \odot\simeq 10^{33}\,cm; \\
N_ \odot \approx 10^{21}. \\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM4"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 5 ===
Find the ratio of dark energy density to the energy density of electric field of intensity $1\,V/m$. Compare the dark energy density with gravitational field energy density on the Earth surface.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Transit from the relative density $\Omega_{DE}$ to th usual one $\rho_{DE}=\Omega_{DE}\rho_{cr}.$ The present value of the critical density of Universe is
$$\rho_{cr}\simeq 9\times 10^{-10} \;\mbox{J/m}^3,$$
therefore
$$
\rho_{DE}\simeq 6.3\times 10^{-10} \;J/m^3
$$
which is equivalent to the mass density
$$
\rho _{DE}^{(m)}\simeq 0.7\times 10^{-26}\;kg/m^3\simeq 3.9\;GeV/s^2\,m^3\simeq 4 \;proton mass/m^3
$$

Compare $\rho_{DE}$ with the energy density of the electric field $\rho_{E}$ with intensity $E=1\;V/m$ (such electric fields are easily available for laboratory measurements). For this field one gets
$$
\rho_E = \frac{\varepsilon_0E^2}{2} = 4.4 \times 10^{- 12}\mbox{J/m}^3.
$$

As one can see, it is possible to measure fields with energy density which is orders of magnitude less than the dark energy density. However the measurement of dark energy is far from being trivial. Unlike the the electric field, the dark energy cannot be switched on and off. The dark energy, at least in the form of cosmological constant, is homogeneously distributed in space, and we presently have no means to create local regions of increased dark energy density, similar to a capacitor in the case of the electric field. At last let us compare the dark energy density under interest to that of gravity field on the Earth surface where the laboratory measurements are performed. The latter equals to
$$
\rho _G = \frac{g^2}{8\pi G} = 5.7 \times 10^{+10} \mbox{J/m}^3.
$$
It is clear that if the background exceeds the measured quantity in twenty orders of magnitude, then reliable identification of this quantity requires a super-precision experiment.
</p>
</div>
</div></div>

<div id="SCM5"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 6 ===
Estimate the distance between two neutral hydrogen atoms at which the gravitational force of their attraction is balanced by the repulsion force generated by dark energy in the form of cosmological constant. Make the same estimates for the
Sun-Earth system.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\ddot R = \frac{\ddot a}{a}R.
$$
The acceleration generated by the cosmological constant is
$$\begin{gathered}
\ddot R = \frac{\ddot a}{a}R = - \frac{4\pi G}{3}(\rho _\Lambda + 3p_\Lambda )R = \frac{8\pi G}{3}\rho _\Lambda R; \\
\frac{\rho _{\Lambda 0}}{\rho _{cr0}} = \Omega _{\Lambda 0};\quad \rho _{cr0} = \frac{3H_0^2}{8\pi G} \\
\end{gathered}
$$
Then the contribution of the cosmological constant into the current acceleration is $\ddot R = \Omega _{\Lambda 0}H_0^2R.$ This quantity is to be compared with the usual gravity acceleration generated by the mass $M,$
$$
g = G\frac{M}{R^2}
$$
Then for the case of hydrogen atom one obtains
\[R = \left(\frac{Gm_p}{H_0^2\Omega _{\Lambda 0}} \right)^{1/3}\simeq 0.3 \, m.\]

For the Sun-Earth system the analogues estimate gives
$$
R=\frac{2\pi R_{a.u.}}{H_0\sqrt{\Omega_{\Lambda 0}}}\simeq 0.5 \,Mpc,
$$
where $R_{a.u.}-$ is the mean Sun-Earth distance.
</p>
</div>
</div></div>

<div id="SCM6"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 7 ===
Calculate magnitude of physical acceleration.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
\dot V = (\dot H + H^2)R; \\
H \equiv \frac{\dot a}{a};\;\dot H = \frac{\ddot aa - \dot a^2}{a^2} =\frac{\ddot a}{a} - H^2; \\
\dot H + H^2 = - \frac{4\pi G}{3}(\rho + 3p) = - \frac{4\pi G}{3}( - 2\rho _\Lambda + \rho _m) = \\
= \frac{8\pi G}{3}\left( \rho _\Lambda - \frac{1}{2}\rho _m \right) = H_0^2\left( \Omega _{\Lambda 0} - \frac{1}{2}\Omega _{m 0} \right); \\
\dot V = H_0^2R\left( \Omega _{\Lambda 0 } - \frac{1}{2}\Omega _{m 0} \right) \\
\end{gathered}
$$

$$
\begin{gathered}
R = 1\, Mpc;\\
\dot V \simeq 10^{ - 11}\, sm/s^2 \\
\end{gathered}
$$

</p>
</div>
</div></div>

<div id="SCM7"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 8 ===
How far can one see in the Universe?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The particle horizon limits the causally connected region of the Universe, which is in principle observable at given time moment. by definition it is
$$
L_{p} = a\int\limits_{0}^t\frac{dt}{a\left( t \right)}.
$$
where
$a(t) = A^{1/3}\sinh ^{2/3}\left( t/{t_\Lambda }\right)$
and
$t_\Lambda \equiv \frac{2}{3}\left(H_0 \right)^{-1}\left( \Omega _{\Lambda 0}\right)^{ - 1/2}\simeq 10.768 \cdot 10^{9}$ years.
$t_0 = t_\Lambda Arth\sqrt {\Omega _{\Lambda 0}}\simeq 13.7\cdot10^9 $ years.
\[A \equiv \frac{\Omega_{m0}}{\Omega _{\Lambda 0}} =
\frac{1 - \Omega _{\Lambda 0}}{\Omega _{\Lambda 0}} = 0.37.\]
After integration one finally obtains
$L_{p} = 47.563\cdot 10^9$ light years.
</p>
</div>
</div></div>

<div id="SCM8"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 9 ===
Find age of the Universe.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Using the result of the previous problem, one obtains the age of the Universe from the condition $a(t_0) = a_0 = 1$
$$
t_0 = \frac{2}{3}\left( H_0\right)^{-1}\left(\Omega _{\Lambda 0}\right)^{-1/2}Arsh \sqrt{\frac{\Omega _{\Lambda 0}}{\Omega _{m0}}}
$$
or
$$
t_0 = t_\Lambda Arsh\sqrt {\frac{\Omega _{\Lambda 0}}{\Omega _{m0}}}
$$
Using the relation $Arth(x) = Arsh\left(x/\sqrt {1 - x^2}\right)$ one gets
\[
t_0 = t_\Lambda Arth\sqrt {\Omega _{\Lambda 0}}
\simeq 13.7\cdot 10^9 years.\]
</p>
</div>
</div></div>

<div id="SCM9"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 10 ===
Give a qualitative explanation why the age of Universe in SCM is considerably greater than the age of matter dominated Universe (Einstein-de Sitter model).
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the Einstein-de Sitter model the expansion is always decelerating, while in the SCM the decelerated expansion turns to the accelerated one. Therefore for a given Hubble constant $H_0$ the preceding value in the Einstein-de Sitter model was greater than that in the SCM, and the average expansion rate was also greater, so the observed size of the Universe was achieved for a shorter time.
</p>
</div>
</div></div>

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Evolution of Universe

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[[Category:Standard Cosmological Model|2]]

<div id="SCM_24"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 28 ===
Find the redshift dependence of the deceleration parameter. Analyze the limiting cases.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
q \equiv - \frac{a\ddot a}{\dot a^2} = - \frac{\ddot a}{a}\frac{1}{H^2}; \\
\frac{\ddot a}{a} = - \frac{4\pi G}{3}\sum\limits_i \left( \rho _i + 3p_i \right) = - \frac{1}{2}H_0^2\sum\limits_i \Omega _{i0}\left(1 + 3w_i\right)(1 + z)^{3\left(1 + w_i\right)} ; \\
H^2 = H_0^2\sum\limits_i \Omega _{i0}(1 + z)^{3\left( 1 + w_i \right)}; \\
q = \frac{3}{2}\frac{\sum\limits_i \Omega _{i0}\left( 1 + w_i \right)(1 + z)^{3\left( 1 + w_i \right)}}
{\sum\limits_i \Omega _{i0}(1 + z)^{3\left( 1 + w_i\right)} } - 1\\
\end{gathered}
$$
In the SCM
$$
q = \frac{1}{2}\frac{\Omega _{m0}(1 + z)^3- 2\Omega _{\Lambda 0} }
{\Omega _{m0}(1 + z)^3+\Omega_{\Lambda 0}}
$$

$$
q(z \to \infty ) = \frac{1}{2},\;q(z \to - 1) = - 1
$$

<gallery widths=600px heights=500px>
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</gallery>
Dependence of the deceleration parameter on the redshift.
</p>
</div>
</div></div>

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Characteristic Parameters and Scales

2012-10-03T20:15:48Z

Den:

[[Category:Standard Cosmological Model|1]]

__NOTOC__

<div id="SCM1"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 1 ===
Calculate the dark energy density and the cosmological constant value.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">For quark one obtains the following:
SCM assumes $\Omega _\Lambda = 0.73,$ and the Hubble constant value is generally accepted to be $H_0 = 70\, km\cdot s^{-1}\cdot Mpc^{- 1} = 2.26 \cdot 10^{ - 18}s^{-1}.$ Then the critical density corresponds to the following value
\[
\rho_{cr} = \frac{3H_0^2}{8\pi G} = 0.92 \cdot 10^{-29} g/cm^3
\]
and
\[\rho _\Lambda = \Omega _\Lambda \rho_{cr} = 0.67 \cdot 10^{-29} g/cm^3.\]</p>
</div>
</div></div>

<div id="SCM2"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2 ===
Estimate total number of baryons in the Universe.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\begin{eqnarray}
N &=& \frac{M}{m_n}=\frac{\frac{4}{3}\pi \left( \frac{c}{H_0} \right)^3\rho }{m_n}=\frac{\frac{4}{3}\pi \alpha \left( \frac{c}{H_0}\right)^3\frac{3H_0^2}{8\pi G}}{m_n}= \\
&=& \frac{1}{2}\alpha \frac{H_0^2}{Gm_n}\left( \frac{c}{H_0} \right)^3=\frac{1}{2}\alpha \frac{M_{Pl}}{m_n}\frac{M_{Pl}c^2}{\hbar H_0}\simeq 10^{80}\quad (\alpha \simeq 0.04)
\end{eqnarray}
</p>
</div>
</div></div>

<div id="SCM_2"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 3 ===
Find dependence of relative density of dark energy $\Omega_\Lambda$ on the redshift. Plot $\Omega_\Lambda(z)$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Neglecting the contribution of radiation ($\rho _{r0} \ll \rho _{m0} < \rho _\Lambda$) one obtains
$$
\Omega _\Lambda (z) = \frac{\rho _\Lambda }{\rho _{cr}(z)} = \frac{\rho _\Lambda}{\rho _{m0}(1 + z)^3 + \rho _\Lambda } = \frac{\Omega _{\Lambda 0}}
{\Omega _{m0}(1 + z)^3 + \Omega _{\Lambda 0}};
$$
$$
\Omega _\Lambda (z) \simeq \frac{1}{\frac{1}{3}(1 + z)^3+ 1}
$$
<gallery widths=600px heights=500px>
File:12_1.jpg|
</gallery>
</p>
</div>
</div></div>

<div id="SCM3"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 4 ===
Estimate total number of stars in the Universe.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
\rho_{cr} \equiv \frac{3H_0^2}{8\pi G} \simeq 10^{-29}\mbox{g/cm}^3; \\
\rho_{bar}\simeq 0.04\rho _{cr}; \\
R_H \approx 10^{28}\mbox{cm}; \\
M_{bar} \approx 0.2 \times 10^{55}\,cm; \\
M_ \odot\simeq 10^{33}\,cm; \\
N_ \odot \approx 10^{21}. \\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM4"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 5 ===
Find the ratio of dark energy density to the energy density of electric field of intensity $1\,V/m$. Compare the dark energy density with gravitational field energy density on the Earth surface.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Transit from the relative density $\Omega_{DE}$ to th usual one $\rho_{DE}=\Omega_{DE}\rho_{cr}.$ The present value of the critical density of Universe is
$$\rho_{cr}\simeq 9\times 10^{-10} \;\mbox{J/m}^3,$$
therefore
$$
\rho_{DE}\simeq 6.3\times 10^{-10} \;J/m^3
$$
which is equivalent to the mass density
$$
\rho _{DE}^{(m)}\simeq 0.7\times 10^{-26}\;kg/m^3\simeq 3.9\;GeV/s^2\,m^3\simeq 4 \;proton mass/m^3
$$

Compare $\rho_{DE}$ with the energy density of the electric field $\rho_{E}$ with intensity $E=1\;V/m$ (such electric fields are easily available for laboratory measurements). For this field one gets
$$
\rho_E = \frac{\varepsilon_0E^2}{2} = 4.4 \times 10^{- 12}\mbox{J/m}^3.
$$

As one can see, it is possible to measure fields with energy density which is orders of magnitude less than the dark energy density. However the measurement of dark energy is far from being trivial. Unlike the the electric field, the dark energy cannot be switched on and off. The dark energy, at least in the form of cosmological constant, is homogeneously distributed in space, and we presently have no means to create local regions of increased dark energy density, similar to a capacitor in the case of the electric field. At last let us compare the dark energy density under interest to that of gravity field on the Earth surface where the laboratory measurements are performed. The latter equals to
$$
\rho _G = \frac{g^2}{8\pi G} = 5.7 \times 10^{+10} \mbox{J/m}^3.
$$
It is clear that if the background exceeds the measured quantity in twenty orders of magnitude, then reliable identification of this quantity requires a super-precision experiment.
</p>
</div>
</div></div>

<div id="SCM5"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 6 ===
Estimate the distance between two neutral hydrogen atoms at which the gravitational force of their attraction is balanced by the repulsion force generated by dark energy in the form of cosmological constant. Make the same estimates for the
Sun-Earth system.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\ddot R = \frac{\ddot a}{a}R.
$$
The acceleration generated by the cosmological constant is
$$\begin{gathered}
\ddot R = \frac{\ddot a}{a}R = - \frac{4\pi G}{3}(\rho _\Lambda + 3p_\Lambda )R = \frac{8\pi G}{3}\rho _\Lambda R; \\
\frac{\rho _{\Lambda 0}}{\rho _{cr0}} = \Omega _{\Lambda 0};\quad \rho _{cr0} = \frac{3H_0^2}{8\pi G} \\
\end{gathered}
$$
Then the contribution of the cosmological constant into the current acceleration is $\ddot R = \Omega _{\Lambda 0}H_0^2R.$ This quantity is to be compared with the usual gravity acceleration generated by the mass $M,$
$$
g = G\frac{M}{R^2}
$$
Then for the case of hydrogen atom one obtains
\[R = \left(\frac{Gm_p}{H_0^2\Omega _{\Lambda 0}} \right)^{1/3}\simeq 0.3 \, m.\]

For the Sun-Earth system the analogues estimate gives
$$
R=\frac{2\pi R_{a.u.}}{H_0\sqrt{\Omega_{\Lambda 0}}}\simeq 0.5 \,Mpc,
$$
where $R_{a.u.}-$ is the mean Sun-Earth distance.
</p>
</div>
</div></div>

<div id="SCM6"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 7 ===
Calculate magnitude of physical acceleration.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
\dot V = (\dot H + H^2)R; \\
H \equiv \frac{\dot a}{a};\;\dot H = \frac{\ddot aa - \dot a^2}{a^2} =\frac{\ddot a}{a} - H^2; \\
\dot H + H^2 = - \frac{4\pi G}{3}(\rho + 3p) = - \frac{4\pi G}{3}( - 2\rho _\Lambda + \rho _m) = \\
= \frac{8\pi G}{3}\left( \rho _\Lambda - \frac{1}{2}\rho _m \right) = H_0^2\left( \Omega _{\Lambda 0} - \frac{1}{2}\Omega _{m 0} \right); \\
\dot V = H_0^2R\left( \Omega _{\Lambda 0 } - \frac{1}{2}\Omega _{m 0} \right) \\
\end{gathered}
$$

$$
\begin{gathered}
R = 1\, Mpc;\\
\dot V \simeq 10^{ - 11}\, sm/s^2 \\
\end{gathered}
$$

</p>
</div>
</div></div>

<div id="SCM7"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 8 ===
How far can one see in the Universe?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The particle horizon limits the causally connected region of the Universe, which is in principle observable at given time moment. by definition it is
$$
L_{p} = a\int\limits_{0}^t\frac{dt}{a\left( t \right)}.
$$
where
$a(t) = A^{1/3}\sinh ^{2/3}\left( t/{t_\Lambda }\right)$
and
$t_\Lambda \equiv \frac{2}{3}\left(H_0 \right)^{-1}\left( \Omega _{\Lambda 0}\right)^{ - 1/2}\simeq 10.768 \cdot 10^{9}$ years.
$t_0 = t_\Lambda Arth\sqrt {\Omega _{\Lambda 0}}\simeq 13.7\cdot10^9 $ years.
\[A \equiv \frac{\Omega_{m0}}{\Omega _{\Lambda 0}} =
\frac{1 - \Omega _{\Lambda 0}}{\Omega _{\Lambda 0}} = 0.37.\]
After integration one finally obtains
$L_{p} = 47.563\cdot 10^9$ light years.
</p>
</div>
</div></div>

<div id="SCM8"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 9 ===
Find age of the Universe.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Using the result of the previous problem, one obtains the age of the Universe from the condition $a(t_0) = a_0 = 1$
$$
t_0 = \frac{2}{3}\left( H_0\right)^{-1}\left(\Omega _{\Lambda 0}\right)^{-1/2}Arsh \sqrt{\frac{\Omega _{\Lambda 0}}{\Omega _{m0}}}
$$
or
$$
t_0 = t_\Lambda Arsh\sqrt {\frac{\Omega _{\Lambda 0}}{\Omega _{m0}}}
$$
Using the relation $Arth(x) = Arsh\left(x/\sqrt {1 - x^2}\right)$ one gets
\[
t_0 = t_\Lambda Arth\sqrt {\Omega _{\Lambda 0}}
\simeq 13.7\cdot 10^9 years.\]
</p>
</div>
</div></div>

<div id="SCM9"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 10 ===
Give a qualitative explanation why the age of Universe in SCM is considerably greater than the age of matter dominated Universe (Einstein-de Sitter model).
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the Einstein-de Sitter model the expansion is always decelerating, while in the SCM the decelerated expansion turns to the accelerated one. Therefore for a given Hubble constant $H_0$ the preceding value in the Einstein-de Sitter model was greater than that in the SCM, and the average expansion rate was also greater, so the observed size of the Universe was achieved for a shorter time.
</p>
</div>
</div></div>

Characteristic Parameters and Scales

2012-10-03T20:14:15Z

Den:

[[Category:Standard Cosmological Model|1]]

__NOTOC__

<div id="SCM1"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 1 ===
Calculate the dark energy density and the cosmological constant value.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">For quark one obtains the following:
SCM assumes $\Omega _\Lambda = 0.73,$ and the Hubble constant value is generally accepted to be $H_0 = 70\, km\cdot s^{-1}\cdot Mpc^{- 1} = 2.26 \cdot 10^{ - 18}s^{-1}.$ Then the critical density corresponds to the following value
\[
\rho_{cr} = \frac{3H_0^2}{8\pi G} = 0.92 \cdot 10^{-29} g/cm^3
\]
and
\[\rho _\Lambda = \Omega _\Lambda \rho_{cr} = 0.67 \cdot 10^{-29} g/cm^3.\]</p>
</div>
</div></div>

<div id="SCM2"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2 ===
Estimate total number of baryons in the Universe.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\begin{eqnarray}
N &=& \frac{M}{m_n}=\frac{\frac{4}{3}\pi \left( \frac{c}{H_0} \right)^3\rho }{m_n}=\frac{\frac{4}{3}\pi \alpha \left( \frac{c}{H_0}\right)^3\frac{3H_0^2}{8\pi G}}{m_n}= \\
&=& \frac{1}{2}\alpha \frac{H_0^2}{Gm_n}\left( \frac{c}{H_0} \right)^3=\frac{1}{2}\alpha \frac{M_{Pl}}{m_n}\frac{M_{Pl}c^2}{\hbar H_0}\simeq 10^{80}\quad (\alpha \simeq 0.04)
\end{eqnarray}
</p>
</div>
</div></div>

<div id="SCM_2"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 3 ===
Find dependence of relative density of dark energy $\Omega_\Lambda$ on the redshift. Plot $\Omega_\Lambda(z)$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Neglecting the contribution of radiation ($\rho _{r0} \ll \rho _{m0} < \rho _\Lambda$) one obtains
$$
\Omega _\Lambda (z) = \frac{\rho _\Lambda }{\rho _{cr}(z)} = \frac{\rho _\Lambda}{\rho _{m0}(1 + z)^3 + \rho _\Lambda } = \frac{\Omega _{\Lambda 0}}
{\Omega _{m0}(1 + z)^3 + \Omega _{\Lambda 0}};
$$
$$
\Omega _\Lambda (z) \simeq \frac{1}{\frac{1}{3}(1 + z)^3+ 1}
$$
<gallery widths=600px heights=500px>
File:12_1.jpg|
</gallery>
</p>
</div>
</div></div>

<div id="SCM3"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 4 ===
Estimate total number of stars in the Universe.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
\rho_{cr} \equiv \frac{3H_0^2}{8\pi G} \simeq 10^{-29}\mbox{g/cm}^3; \\
\rho_{bar}\simeq 0.04\rho _{cr}; \\
R_H \approx 10^{28}\mbox{cm}; \\
M_{bar} \approx 0.2 \times 10^{55}\,cm; \\
M_ \odot\simeq 10^{33}\,cm; \\
N_ \odot \approx 10^{21}. \\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="SCM4"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 5 ===
Find the ratio of dark energy density to the energy density of electric field of intensity $1\,V/m$. Compare the dark energy density with gravitational field energy density on the Earth surface.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Transit from the relative density $\Omega_{DE}$ to th usual one $\rho_{DE}=\Omega_{DE}\rho_{cr}.$ The present value of the critical density of Universe is
$$\rho_{cr}\simeq 9\times 10^{-10} \;\mbox{J/m}^3,$$
therefore
$$
\rho_{DE}\simeq 6.3\times 10^{-10} \;J/m^3
$$
which is equivalent to the mass density
$$
\rho _{DE}^{(m)}\simeq 0.7\times 10^{-26}\;kg/m^3\simeq 3.9\;GeV/s^2\,m^3\simeq 4 \;proton mass/m^3
$$

Compare $\rho_{DE}$ with the energy density of the electric field $\rho_{E}$ with intensity $E=1\;V/m$ (such electric fields are easily available for laboratory measurements). For this field one gets
$$
\rho_E = \frac{\varepsilon_0E^2}{2} = 4.4 \times 10^{- 12}\mbox{J/m}^3.
$$

As one can see, it is possible to measure fields with energy density which is orders of magnitude less than the dark energy density. However the measurement of dark energy is far from being trivial. Unlike the the electric field, the dark energy cannot be switched on and off. The dark energy, at least in the form of cosmological constant, is homogeneously distributed in space, and we presently have no means to create local regions of increased dark energy density, similar to a capacitor in the case of the electric field. At last let us compare the dark energy density under interest to that of gravity field on the Earth surface where the laboratory measurements are performed. The latter equals to
$$
\rho _G = \frac{g^2}{8\pi G} = 5.7 \times 10^{+10} \mbox{J/m}^3.
$$
It is clear that if the background exceeds the measured quantity in twenty orders of magnitude, then reliable identification of this quantity requires a super-precision experiment.
</p>
</div>
</div></div>

<div id="SCM5"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 6 ===
Estimate the distance between two neutral hydrogen atoms at which the gravitational force of their attraction is balanced by the repulsion force generated by dark energy in the form of cosmological constant. Make the same estimates for the
Sun-Earth system.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\ddot R = \frac{\ddot a}{a}R.
$$
The acceleration generated by the cosmological constant is
$$\begin{gathered}
\ddot R = \frac{\ddot a}{a}R = - \frac{4\pi G}{3}(\rho _\Lambda + 3p_\Lambda )R = \frac{8\pi G}{3}\rho _\Lambda R; \\
\frac{\rho _{\Lambda 0}}{\rho _{cr0}} = \Omega _{\Lambda 0};\quad \rho _{cr0} = \frac{3H_0^2}{8\pi G} \\
\end{gathered}
$$
Then the contribution of the cosmological constant into the current acceleration is $\ddot R = \Omega _{\Lambda 0}H_0^2R.$ This quantity is to be compared with the usual gravity acceleration generated by the mass $M,$
$$
g = G\frac{M}{R^2}
$$
Then for the case of hydrogen atom one obtains
\[R = \left(\frac{Gm_p}{H_0^2\Omega _{\Lambda 0}} \right)^{1/3}\simeq 0.3 \, m.\]

For the Sun-Earth system the analogues estimate gives
$$
R=\frac{2\pi R_{a.u.}}{H_0\sqrt{\Omega_{\Lambda 0}}}\simeq 0.5 \,Mpc,
$$
where $R_{a.u.}-$ is the mean Sun-Earth distance.
</p>
</div>
</div></div>

<div id="SCM6"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 7 ===
Calculate magnitude of physical acceleration.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\begin{gathered}
\dot V = (\dot H + H^2)R; \\
H \equiv \frac{\dot a}{a};\;\dot H = \frac{\ddot aa - \dot a^2}{a^2} =\frac{\ddot a}{a} - H^2; \\
\dot H + H^2 = - \frac{4\pi G}{3}(\rho + 3p) = - \frac{4\pi G}{3}( - 2\rho _\Lambda + \rho _m) = \\
= \frac{8\pi G}{3}\left( \rho _\Lambda - \frac{1}{2}\rho _m \right) = H_0^2\left( \Omega _{\Lambda 0} - \frac{1}{2}\Omega _{m 0} \right); \\
\dot V = H_0^2R\left( \Omega _{\Lambda 0 } - \frac{1}{2}\Omega _{m 0} \right) \\
\end{gathered}
$$

$$
\begin{gathered}
R = 1\, Mpc;\\
\dot V \simeq 10^{ - 11}\, sm/s^2 \\
\end{gathered}
$$

</p>
</div>
</div></div>

<div id="SCM7"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 8 ===
How far can one see in the Universe?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The particle horizon limits the causally connected region of the Universe, which is in principle observable at given time moment. by definition it is
$$
L_{p} = a\int\limits_{0}^t\frac{dt}{a\left( t \right)}.
$$
where
$a(t) = A^{1/3}\sinh ^{2/3}\left( t/{t_\Lambda }\right)$
and
$t_\Lambda \equiv \frac{2}{3}\left(H_0 \right)^{-1}\left( \Omega _{\Lambda 0}\right)^{ - 1/2}\simeq 10.768 \cdot 10^{9}$ years.
$t_0 = t_\Lambda Arth\sqrt {\Omega _{\Lambda 0}}\simeq 13.7\cdot10^9 $ years.
\[A \equiv \frac{\Omega_{m0}}{\Omega _{\Lambda 0}} =
\frac{1 - \Omega _{\Lambda 0}}{\Omega _{\Lambda 0}} = 0.37.\]
After integration one finally obtains
$L_{p} = 47.563\cdot 10^9$ light years.
</p>
</div>
</div></div>

<div id="SCM8"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 9 ===
Find age of the Universe.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Using the result of the previous problem, one obtains the age of the Universe from the condition $a(t_0) = a_0 = 1$
$$
t_0 = \frac{2}{3}\left( H_0\right)^{-1}\left(\Omega _{\Lambda 0}\right)^{-1/2}Arsh \sqrt{\frac{\Omega _{\Lambda 0}}{\Omega _{m0}}}
$$
or
$$
t_0 = t_\Lambda Arsh\sqrt {\frac{\Omega _{\Lambda 0}}{\Omega _{m0}}}
$$
Using the relation $Arth(x) = Arsh\left(x/\sqrt {1 - x^2}\right)$ one gets
\[
t_0 = t_\Lambda Arth\sqrt {\Omega _{\Lambda 0}}
\simeq 13.7\cdot 10^9 years.\]
</p>
</div>
</div></div>

<div id="SCM9"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 10 ===
Give a qualitative explanation why the age of Universe in SCM is considerably greater than the age of matter dominated Universe (Einstein-de Sitter model).
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the Einstein-de Sitter model the expansion is always decelerating, while in the SCM the decelerated expansion turns to the accelerated one. Therefore for a given Hubble constant $H_0$ the preceding value in the Einstein-de Sitter model was greater than that in the SCM, and the average expansion rate was also greater, so the observed size of the Universe was achieved for a shorter time.
</p>
</div>
</div></div>

File:12 1.jpg

2012-10-02T11:33:41Z

Den:

6 Extras

2012-10-02T00:27:06Z

Den:

[[Category:Thermodynamics of Universe|8]]

__NOTOC__

<div id="razm39n"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 1 ===
What is more important for the open space suit: heating or cooling function?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In spite of wide-spread opinion that the pace is very cool place (the temperature is lower than $3\,K$), the cold as a cooling rate can be considered in different aspects. The heat conductivity in vacuum is also close to zero, therefore the heat flow from a warm body placed into open space can go only due to the radiation. The radiation intensity is proportional to fourth degree of the temperature. For example, if an astronaut gets into the open space (and far from near stars, so that heating from extra sources can be neglected) and cannot return to the spacecraft, he (or she) will not be covered by icy crust and there is no danger of fast icy death. His (or her) temperature (about $310 \, K$) is sufficient to stay in comfortable temperature conditions some time, at least to arrival of space rescue team. Under assumption that there is no heat production in astronaut's body and precipitation of water from the skin is also absent (the astronaut is inside the hermetic space suit without the heat insulation), then it will coll down in one degree approximately every forty minutes, even in absolutely black space suit, which emits energy the most efficiently. Along with decreasing of temperature the cooling rate will also decrease in accordance with the Stephan-Boltzmann law. Actually an astronaut in vacuum is in danger of overheat rather than cold, because the heat production rate in human body equals approximately $100 \,W$; efficient heat removal represent one of the most challenging problems faced by the space suit developers.
</p>
</div>
</div></div>

<div id="therm42"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2 ===
If the Universe is electrically neutral then how many electrons are there for each baryon?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$8/9.$
</p>
</div>
</div></div>

<div id="therm45"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 3 ===
Explain why the thermonuclear processes in the first stars considerably influenced the evolution of the Universe as a whole?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
A nuclear fusion reaction yields approximately $7\cdot 10^6\, eV$ of energy per hydrogen atom, while the hydrogen atom ionization energy is just $13.6\,\mbox{\it eV}.$ Therefore it is sufficient to convert quite small fraction ($\sim 10^{- 5}$) of total baryon mass of stars in order to ionize the rest of Universe. In fact one needs a fraction at lest one order of magnitude greater. The reason is that

1) only part of photons have energy above the ionization threshold;

2) due to the redshift each hydrogen atom suffers multiple recombinations.

</p>
</div>
</div></div>

<div id="therm46"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 4 ===
Why is the present relative abundance of chemical elements approximately the same as right after the creation of the Solar system?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Energy required for nuclear transformations must be of order of nuclei bound energy. Putting aside the lightest nuclei, a rough estimate for the bound energy per nucleon is of order of
$8\, MeV.$ Therefore the energy needed for nuclear transformations is million times grater than the one obtained in chemical reactions. Such energies (or temperatures) cannot be achieved in any natural process on Earth. Thus the presently observed relative abundance of elements is approximately the same that was $4.6\cdot 10^9$ years ago when the Solar System was formed. The radioactive elements make exclusion. Changes of their abundance present one of the methods to determine the age of Universe.
</p>
</div>
</div></div>

Primary Nucleosynthesis

2012-10-02T00:24:46Z

Den:

[[Category:Thermodynamics of Universe|7]]

__NOTOC__

<div id="therm38"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 1 ===
Find the ratio of neutrons to protons number densities in the case of thermal equilibrium between them.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Though total number of baryons is conserved, they can transform to each other in reactions of the type $n{\nu _e} \leftrightarrow p{e^-}, n{e^+} \leftrightarrow p{\bar \nu}_e.$ If their rates are sufficiently high to support the thermal equilibrium in the expanding Universe, then
$$
\frac{n_n}{n_p} = \left(\frac{m_n}{m_p}\right)^{3/2}e^{-\left(m_n - m_p\right)/T} \approx e^{-\Delta m/T},
$$
where $m_p = 938.272\,MeV,\;m_n = 939.565\,MeV;\Delta m = m_n - m_p = 1.293\,MeV.$
</p>
</div>
</div></div>

<div id="therm39"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2 ===
Up to what temperature can the reaction $n\nu_e\leftrightarrow pe^-$ support thermal equilibrium between protons and neutrons in the expanding Universe
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In order to know whether the thermal equilibrium is supported one needs to compare the rate of Universe expansion with that of the reactions, required to provide the equilibrium. The reaction rate
(per neutron) is given by the expression \[\Gamma = n\left\langle \sigma v\right\rangle,\] where $n$ is the number density of the target particles (neutrinos or electrons in the case under consideration),
$$
n_{\nu ,e} = \frac{3} {4}\frac{\zeta (3)}{\pi ^2}g_{\nu ,e}T^3,
$$
and the cross-section $\sigma$ can be found in frames of the Standard Model. To estimate the thermal mean $\left\langle \sigma v\right\rangle $ one can use the expression \[ \left\langle {\sigma
v} \right\rangle \approx G_F^2{T^2},\] where $G_F \approx 1.166 \times
10^{ - 5} \,GeV^{-2}$ is the Fermi constant of weak interaction. Omitting all the factors of order of unity one obtains $ \Gamma \approx G_F^2T^5.$ The expression for the Hubble parameter in radiation-dominated Universe was obtained in [[Peculiarities_of_Thermodynamics_in_Early_Universe#ter_15|problem]] of the present Chapter:
$$
H = 1.66\sqrt {g^*} \frac{T^2}{M_{Pl}}.
$$
The point, where $\Gamma = H$, determines the temperature ${T_f}$ of ''freeze-out'' (fixation) for the ratio ${n_n}/{n_p}.$ Equating the expressions for
$H$ and $\Gamma,$ one obtains
$$
G_F^2T^5 = 1.66\sqrt {g^*} \frac{T^2}{M_{Pl}},
$$
and then
$$
T_f = \left(\frac{1.66} {G_F^2M_{Pl}} \right)^{1/3}(g^*)^{1/6}.
$$
Using that ${g^*} = 10.75,$ one can obtain that $T_f\simeq1.2\,MeV$.
More accurate calculations give the value ${T_f} \approx 0.7\,MeV.$
</p>
</div>
</div></div>

<div id="therm40"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 3 ===
Determine the ratio $n_n/n_p$ at the temperature of freeze-out.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
\frac{n_n}{n_p} = e^{ - \left( m_n - m_p \right)/T_f} \approx e^{ -
1.3/0.7} \approx 0.16.
$$
</p>
</div>
</div></div>

<div id="therm41"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 4 ===
Determine the age of Universe when it reached the temperature of freeze-out.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In the region $T \sim 1\,MeV,\;{g^*} =
10.75\;$ the relation between time and temperature can be presented in the following form (see [[Peculiarities_of_Thermodynamics_in_Early_Universe#ter_16|problem]] and [[Peculiarities_of_Thermodynamics_in_Early_Universe#ter_20|problem]]) $tT^2
\approx 0.75\ \mbox{s}\cdot\,MeV^2$, and it follows for $T \approx
0.7\,\,MeV$ that $t \approx 1.5\,s$.
</p>
</div>
</div></div>

<div id="therm41n"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 5 ===
At what temperature and at what time did efficient deuterium synthesis start?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="therm42n"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 6 ===
Determine the time period during which the synthesis of light elements took place.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="therm43n"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 7 ===
Determine the ratio of neutrons to protons number densities at temperature interval from the freeze-out to the
creation of deuterium.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Even after the fixation (freezing-out) of the neutron concentration at $T_f \approx 0.7\, \, MeV $
(see [[#therm39|problem]] of the present Chapter) the neutrons can decay up to the deuterium creation, which starts near $T = 0.085\, \,MeV$. In order to determine the corresponding moment of time use the relation (see [[Peculiarities_of_Thermodynamics_in_Early_Universe#ter_15|problem]])
\[
H \approx 1.66\sqrt{g^*}\frac{T^2}{M_{Pl}}.
\]
Calculation effective number of degrees of freedom $g^* $ at the temperature corresponding to the deuteron creation threshold should be performed with due care. Firstly, at the temperature under consideration only photons and neutrinos remain relativistic, which is trivially taken into account. Secondly one should remember that in the considered energy range the temperature of neutrino background differs from that of microwave. Therefore in order to calculate the effective number of internal degrees of freedom the following expression is to be used (see [[Thermodynamical_Properties_of_Elementary_Particles#therm5|problem]])
\[
g^* = \sum\limits_{i = bosons} g_i \left(\frac{T_i}{T}\right)^4 + \frac{7}{8}\sum\limits_{j = fermions}g_i \left(\frac{T_j}{T}\right)^4 .
\]
For $T \gg m_e \approx 0.5\,\, MeV$ the reaction $e^ + e^ -
\leftrightarrow \gamma \gamma $ proceeds with the same rates in both directions. As long as temperature falls down below the electron mass, the photons are already unable to support the reaction $\gamma \gamma \to e^ + e^ - $. The fact that the reaction $e^
+ e^ - \leftrightarrow \gamma \gamma $ presents an additional source of photons, leads to result that the photon temperature appears higher than that of neutrino. Using solely thermodynamic considerations, one can shove that \[
\frac{T_\nu}{T_\gamma} = \left(\frac{4}{11}\right)^{1/3} \approx 0.714,\]
and therefore
\[
g_* = g_\nu + \frac{7}{8}2 \times 3\frac{T_\nu ^4}{T^4} =2 + \frac{7}{8}6\left(\frac{4}{11}\right)^{4/3} \approx 3.36.
\]
In the considered energy range (see [[Peculiarities_of_Thermodynamics_in_Early_Universe#ter_16|problem]] and [[Peculiarities_of_Thermodynamics_in_Early_Universe#ter_20|problem]])
\[
tT^2 \approx \frac{{0.301}} {{\sqrt {g^* } }}M_{Pl} = 1.32\ \,s\cdot \,MeV^2.
\]
and it follows that $t \approx 180\,s$. As a rough estimate one can take that the neutrons can decay during that time and they are afterwards absorbed due to creation of deuterium, and later helium. Therefore from the point of neutron number density fixation to starting point of deuterium creation one obtains
\[
\frac{n_n}{n_p} = e^{-\left(m_n - m_p \right)/T_f }e^{ - t/\tau _n }.
\]
Here $\tau _n \approx 885.7\, s$ is the mean lifetime of neutron. At $t \approx 180\,s$
\[
\frac{n_p}{n_n} \approx 0.13.
\]
Recall that at the beginning of the latter process the considered ratio was equal to $0.16$ (see [[#therm40|problem]]).
</p>
</div>
</div></div>

<div id="therm44n"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 8 ===
Determine the relative abundance of ${}^4He$ in the Universe
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="therm45n"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 9 ===
How many helium atoms are there for each hydrogen atom?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="therm46n"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 10 ===
What changes in relative ${}^4 He$ abundance would be caused by

a) decreasing of average neutron lifetime $\tau_n$;

b) decreasing or increasing of the temperature of
freeze-out $T_f$?

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="therm43"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 11 ===
What nuclear reactions provided the ${}^4 He$ synthesis in the early Universe?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Synthesis of deuterium presents a critically important step to synthesis of $^4He$, because the direct synthesis of helium from two protons and two neutrons is highly improbable event. After creation of deuterium the helium synthesis proceeds along the following reactions:
$$
\begin{gathered}
d + p \to ^3He + \gamma ;\\
d + d \to ^3He + n; \\
d + d \to t + p; \\
^3He + d \to ^4He + p;\\
t + d \to ^4He + n.\\
\end{gathered}
$$
</p>
</div>
</div></div>

<div id="therm44"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 12 ===
Why is synthesis of elements heavier than ${}^7 Li$ suppressed in the early Universe?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
There are no natural stable nuclei with $A=5$, therefore one should consider only fusion of $^4He$ with tritium and $^3He$:
$$
\begin{gathered}
t + ^4He \to ^7Li + \gamma ; \\
^3He + ^4He \to ^7Be + \gamma \to ^7Be + e^ - \to ^7Li + \nu _e. \\
\end{gathered}
$$
Synthesis of two nuclei $^4He$ leads to unstable nucleus $^8Be.$ The reaction
$$
^8Be + ^4He \to ^{12}C + \gamma
$$
is inefficient: low density of the reactants leads to the fact that the mean time between the collisions of the nuclei considerable exceeds the lifetime of the unstable nucleus $^8Be.$ This reaction becomes important in stars, but it does not make importance in early Universe. Thus the absence of stable elements with $A=5$ and $A=8$ makes it impossible to proceed beyond the $^7Li$ in primary nucleosynthesis.
</p>
</div>
</div></div>

<div id="therm47n"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 13 ===
In our Universe the neutron half-value period (the life-time) approximately equals to 600 seconds. What would the relative helium abundance be if the neutron life-time decreased down to 100 seconds?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="therm48nn"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 14 ===
At what temperature in Universe did the synthesis reactions stop?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
At $t\sim 1000\, s$ the temperature falls down to $T \approx 0.03\,MeV$. After that moment the kinetic energy of nuclei is insufficient to overcome the Coulomb barrier and the synthesis processes stop.
</p>
</div>
</div></div>

The Saha equation

2012-10-01T23:21:43Z

Den:

[[Category:Thermodynamics of Universe|6]]

__NOTOC__

Degree of ionization of atomic hydrogen in thermal equilibrium
can be described by the Saha equation
\[\frac{1-X}{X^2}=n\lambda_{Te}^3e^{\frac{I}{kT}},\] where $X=n_e/n$ is the degree of ionization,
$n_e$ and $n$ are concentrations of electrons and atoms (both
neutral and ionized) respectively,
\[\lambda_{Te}^2=\frac{2\pi\hbar^2}{m_e kT}\] is the electron's
thermal de Broglie wave length and $I=13.6eV$ is the ionization
energy for hydrogen. It is often used in astrophysics for
description of stellar dynamics.

<div id="cmb23"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 1 ===
Using the Saha equation, determine the hydrogen ionization degree

a) 100 seconds after the Big Bang;

b) at the epoch of recombination;

c) at present time.

Assume for simplicity $\Omega=1$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
</p>
</div>
</div></div>

<div id="therm33"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 2 ===
Assuming that the ionization degree at the last scattering was equal to $10\%$, determine the decoupling
temperature using the Saha equation.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The Saha equation for hydrogen can be presented in the following form:
$$
\frac{1 - X}{X^2} = n\lambda _{Te}^3\exp \left( \frac{I_0}{kT}\right),
$$
where $X = {n_e}/n$ is equilibrium ionization degree, $n_e$ and $n$ are concentrations of electrons and atoms (both neutral and ionized) respectively, \[\lambda _{Te}^2 = \frac{2\pi \hbar ^2}{m_ekT}\] is the electron's
thermal de Broglie wave length, $I_0 = 13.6\,\mbox{eV}$ ionization energy for hydrogen. Concentration of hydrogen on the last scattering surface is approximately equal to to concentration of baryons, which in turn is connected to concentration of photons in the Universe: $n_{_B} \simeq 6 \cdot 10^{ -10}n_\gamma.$ Explicit form of the latter relation can be found from the Planck distribution:
$$
n_\gamma = \frac{\rho _\gamma }{E_\gamma} = \frac{\alpha T^4}{3kT}= \frac{\pi ^2}{45}{\left( \frac{kT}{\hbar c}\right)^3}.
$$
Introducing the dimensionless variable \[y = \frac{I_0}{kT},\] one finally obtains:
$$
\frac{1 - X}{X^2} = Ay^{ - 3/2}e^y.
$$
where
$$
A = \frac{2^{3/2}\pi ^{7/2}}{45}
\frac{n_{_B}}{n_\gamma }\left(\frac{I_0}{m_ec^2} \right)^{3/2} \approx 3 \cdot 10^{ - 16}.
$$
Numerical solution of the obtained equation for $X = 0.1$ leads to $y \approx 46,$ which corresponds to temperature value $T \approx 3431\,K.$
</p>
</div>
</div></div>

<div id="therm34"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 3 ===
How many iterations in the Saha equation are needed in order to obtain the decoupling temperature with accuracy $1K$? Write down analytically the approximate result.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The Saha equation can be conveniently rewritten in the form:
$$
y = \frac{3}{2}\ln y + \ln \frac{1 - X}{AX^2}.
$$
Consecutive iterations of the latter equation for $X = 0.1$ and $A = 3
\cdot 10^{ - 16}$ give the following results:
$$
y_0 = \ln \frac{1 - X}{AX^2} \approx 40.24; \Rightarrow T_0 \approx 3920\,K;
$$
$$
y_1 = \frac{3}{2}\ln y_0 + \ln \frac{1 - X}{AX^2} \approx 45.8; \Rightarrow T_1 \approx 3446\,K;
$$
$$
y_2 = \frac{3}{2}\ln y_1 + \ln \frac{1 - X}{AX^2} \approx 46.0;\Rightarrow T_2 \approx 3431\,K.
$$
Result of the third iteration already coincides with the exact one with accuracy of one degree. It can be analytically presented in the following form:
$$
kT \approx \frac{I_0} {\ln \left( \frac{1 - X} {AX^2}{\left\{ \ln
\left[ \frac{1 - X} {AX^2}{\left(\ln \frac{1 - X} {AX^2}
\right)}^{3/2}\right]\right\}}^{3/2}\right)}.$$
</p>
</div>
</div></div>

<div id="therm35"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 4 ===
Estimate the duration of the epoch of recombination: how long did it take for hydrogen ionization
degree to change from $90\%$ to $10\%$ according to the Saha equation?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Numerical solution of the Saha equation for $X=0.9,$ gives the corresponding temperature $T_1 = 4\ 029\,K,$ and full recombination (corresponding to $X = 0.1$) takes place at $T_2 = 3\ 431\,K,$ which corresponds to the scale factor values $a_1 = 6.8 \cdot 10^{ - 4}$ and $a_2 = 7.9 \cdot 10^{ - 4}$ respectively (setting $a_0=1$ in the present time). The recombination took place in the matter dominated epoch ( $a = \left( t/{t_0} \right)^{2/3},\;t_0 = 14 \cdot 10^9 $ years), therefore $t_1 \approx 250\ 000$ and $t_2 \approx 310\ 000$ years, i.e. the recombination endured approximately $60\cdot 10^3$ years.
</p>
</div>
</div></div>

<div id="therm36"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 5 ===
Using the Saha equation, determine the hydrogen ionization degree in the center of the Sun ($\rho=100g/cm^3$, $T=1.5\cdot10^7K$).
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The Saha equation with given concentration of the gas can be conveniently presented in the following form
$$ \frac{1 - X} {X^2} = n\lambda_0^3y^{3/2}e^y,$$
where
$$
\lambda _0 \equiv \lambda _{Te} \left.\right|_{kT = I_0} = \frac{\sqrt {2\pi } \hbar c}{\sqrt m_ec^2I_0}\simeq 1.877 \cdot 10^{ - 8}\,cm.
$$
Thus for hydrogen density in the center of the Sun one obtains $ n \simeq \rho /m_p \approx 6 \cdot 10^{25}\,cm^{-3}.$ It then follows
$$
X = \frac{1}{\sqrt{\frac{1}{4} + n\lambda _0^3\left( \frac{I_0}{kT} \right)^{3/2}\exp \left( \frac{I_0}{kT} \right)} + \frac{1}{2}}\approx 0.754
$$
The unreasonable result (almost 25\% of neutral hydrogen in the center of the Sun) appeared because of inapplicability of the Saha equation in such dense and highly ionized plasma: in the considered case the Debye radius and interatomic spacing are of the same order $r_D \approx a \approx 10^{- 9}\,cm.$
</p>
</div>
</div></div>

The Saha equation

2012-10-01T23:17:34Z

Den: /* Problem 1 */

[[Category:Thermodynamics of Universe|6]]

__NOTOC__

Degree of ionization of atomic hydrogen in thermal equilibrium
can be described by the Saha equation
\[\frac{1-X}{X^2}=n\lambda_{Te}^3e^{\frac{I}{kT}},\] where $X=n_e/n$ is the degree of ionization,
$n_e$ and $n$ are concentrations of electrons and atoms (both
neutral and ionized) respectively,
\[\lambda_{Te}^2=\frac{2\pi\hbar^2}{m_e kT}\] is the electron's
thermal de Broglie wave length and $I=13.6eV$ is the ionization
energy for hydrogen. It is often used in astrophysics for
description of stellar dynamics.

<div id="cmb23"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 1 ===
Using the Saha equation, determine the hydrogen ionization degree

a) 100 seconds after the Big Bang;
b) at the epoch of recombination;
c) at present time.

Assume for simplicity $\Omega=1$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
</p>
</div>
</div></div>

<div id="therm33"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 2 ===
Assuming that the ionization degree at the last scattering was equal to $10\%$, determine the decoupling
temperature using the Saha equation.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The Saha equation for hydrogen can be presented in the following form:
$$
\frac{1 - X}{X^2} = n\lambda _{Te}^3\exp \left( \frac{I_0}{kT}\right),
$$
where $X = {n_e}/n$ is equilibrium ionization degree, $n_e$ and $n$ are concentrations of electrons and atoms (both neutral and ionized) respectively, \[\lambda _{Te}^2 = \frac{2\pi \hbar ^2}{m_ekT}\] is the electron's
thermal de Broglie wave length, $I_0 = 13.6\,\mbox{eV}$ ionization energy for hydrogen. Concentration of hydrogen on the last scattering surface is approximately equal to to concentration of baryons, which in turn is connected to concentration of photons in the Universe: $n_{_B} \simeq 6 \cdot 10^{ -10}n_\gamma.$ Explicit form of the latter relation can be found from the Planck distribution:
$$
n_\gamma = \frac{\rho _\gamma }{E_\gamma} = \frac{\alpha T^4}{3kT}= \frac{\pi ^2}{45}{\left( \frac{kT}{\hbar c}\right)^3}.
$$
Introducing the dimensionless variable \[y = \frac{I_0}{kT},\] one finally obtains:
$$
\frac{1 - X}{X^2} = Ay^{ - 3/2}e^y.
$$
where
$$
A = \frac{2^{3/2}\pi ^{7/2}}{45}
\frac{n_{_B}}{n_\gamma }\left(\frac{I_0}{m_ec^2} \right)^{3/2} \approx 3 \cdot 10^{ - 16}.
$$
Numerical solution of the obtained equation for $X = 0.1$ leads to $y \approx 46,$ which corresponds to temperature value $T \approx 3431\,K.$
</p>
</div>
</div></div>

<div id="therm34"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 3 ===
How many iterations in the Saha equation are needed in order to obtain the decoupling temperature with
accuracy $1K$? Write down analytically the approximate result.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The Saha equation can be conveniently rewritten in the form:
$$
y = \frac{3}{2}\ln y + \ln \frac{1 - X}{AX^2}.
$$
Consecutive iterations of the latter equation for $X = 0.1$ and $A = 3
\cdot 10^{ - 16}$ give the following results:
$$
y_0 = \ln \frac{1 - X}{AX^2} \approx 40.24; \Rightarrow T_0 \approx 3920\,K;
$$
$$
y_1 = \frac{3}{2}\ln y_0 + \ln \frac{1 - X}{AX^2} \approx 45.8; \Rightarrow T_1 \approx 3446\,K;
$$
$$
y_2 = \frac{3}{2}\ln y_1 + \ln \frac{1 - X}{AX^2} \approx 46.0;\Rightarrow T_2 \approx 3431\,K.
$$
Result of the third iteration already coincides with the exact one with accuracy of one degree. It can be analytically presented in the following form:
$$
kT \approx \frac{I_0}{\ln \left(\frac{1 - X}{AX^2}{\left\ \ln
\left[ \frac{1 - X} {AX^2}{\left(\ln \frac{1 - X} {AX^2}
\right)}^{3/2}\right]\right\}^{3/2}\right)}.
$$
</p>
</div>
</div></div>

<div id="therm35"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 4 ===
Estimate the duration of the epoch of recombination: how long did it take for hydrogen ionization
degree to change from $90\%$ to $10\%$ according to the Saha equation?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Numerical solution of the Saha equation for $X=0.9,$ gives the corresponding temperature $T_1 = 4\ 029\,K,$ and full recombination (corresponding to $X = 0.1$) takes place at $T_2 = 3\ 431\,K,$ which corresponds to the scale factor values $a_1 = 6.8 \cdot 10^{ - 4}$ and $a_2 = 7.9 \cdot 10^{ - 4}$ respectively (setting $a_0=1$ in the present time). The recombination took place in the matter dominated epoch ( $a = \left( t/{t_0} \right)^{2/3},\;t_0 = 14 \cdot 10^9 $ years), therefore $t_1 \approx 250\ 000$ and $t_2 \approx 310\ 000$ years, i.e. the recombination endured approximately $60\cdot 10^3$ years.
</p>
</div>
</div></div>

<div id="therm36"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 5 ===
Using the Saha equation, determine the hydrogen ionization degree in the center of the Sun ($\rho=100g/cm^3$, $T=1.5\cdot10^7K$).
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The Saha equation with given concentration of the gas can be conveniently presented in the following form
$$ \frac{1 - X} {X^2} = n\lambda_0^3y^{3/2}e^y,$$
where
$$
\lambda _0 \equiv \lambda _{Te} \left.\right|_{kT = I_0} = \frac{\sqrt {2\pi } \hbar c}{\sqrt m_ec^2I_0}\simeq 1.877 \cdot 10^{ - 8}\,cm.
$$
Thus for hydrogen density in the center of the Sun one obtains $ n \simeq \rho /m_p \approx 6 \cdot 10^{25}\,cm^{-3}.$ It then follows
$$
X = \frac{1}{\sqrt{\frac{1}{4} + n\lambda _0^3\left( \frac{I_0}{kT} \right)^{3/2}\exp \left( \frac{I_0}{kT} \right)} + \frac{1}{2}}\approx 0.754
$$
The unreasonable result (almost 25\% of neutral hydrogen in the center of the Sun) appeared because of inapplicability of the Saha equation in such dense and highly ionized plasma: in the considered case the Debye radius and interatomic spacing are of the same order $r_D \approx a \approx 10^{- 9}\,cm.$
</p>
</div>
</div></div>

The Saha equation

2012-10-01T23:17:20Z

Den:

[[Category:Thermodynamics of Universe|6]]

__NOTOC__

Degree of ionization of atomic hydrogen in thermal equilibrium
can be described by the Saha equation
\[\frac{1-X}{X^2}=n\lambda_{Te}^3e^{\frac{I}{kT}},\] where $X=n_e/n$ is the degree of ionization,
$n_e$ and $n$ are concentrations of electrons and atoms (both
neutral and ionized) respectively,
\[\lambda_{Te}^2=\frac{2\pi\hbar^2}{m_e kT}\] is the electron's
thermal de Broglie wave length and $I=13.6eV$ is the ionization
energy for hydrogen. It is often used in astrophysics for
description of stellar dynamics.

<div id="cmb23"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 1 ===
Using the Saha equation, determine the hydrogen ionization degree

a) 100 seconds after the Big Bang;
b) at the epoch of recombination;
c) at present time.

Assume for simplicity $\Omega=1$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
</p>
</div>
</div></div>

<div id="therm33"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2 ===
Assuming that the ionization degree at the last scattering was equal to $10\%$, determine the decoupling
temperature using the Saha equation.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The Saha equation for hydrogen can be presented in the following form:
$$
\frac{1 - X}{X^2} = n\lambda _{Te}^3\exp \left( \frac{I_0}{kT}\right),
$$
where $X = {n_e}/n$ is equilibrium ionization degree, $n_e$ and $n$ are concentrations of electrons and atoms (both neutral and ionized) respectively, \[\lambda _{Te}^2 = \frac{2\pi \hbar ^2}{m_ekT}\] is the electron's
thermal de Broglie wave length, $I_0 = 13.6\,\mbox{eV}$ ionization energy for hydrogen. Concentration of hydrogen on the last scattering surface is approximately equal to to concentration of baryons, which in turn is connected to concentration of photons in the Universe: $n_{_B} \simeq 6 \cdot 10^{ -10}n_\gamma.$ Explicit form of the latter relation can be found from the Planck distribution:
$$
n_\gamma = \frac{\rho _\gamma }{E_\gamma} = \frac{\alpha T^4}{3kT}= \frac{\pi ^2}{45}{\left( \frac{kT}{\hbar c}\right)^3}.
$$
Introducing the dimensionless variable \[y = \frac{I_0}{kT},\] one finally obtains:
$$
\frac{1 - X}{X^2} = Ay^{ - 3/2}e^y.
$$
where
$$
A = \frac{2^{3/2}\pi ^{7/2}}{45}
\frac{n_{_B}}{n_\gamma }\left(\frac{I_0}{m_ec^2} \right)^{3/2} \approx 3 \cdot 10^{ - 16}.
$$
Numerical solution of the obtained equation for $X = 0.1$ leads to $y \approx 46,$ which corresponds to temperature value $T \approx 3431\,K.$
</p>
</div>
</div></div>

<div id="therm34"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 3 ===
How many iterations in the Saha equation are needed in order to obtain the decoupling temperature with
accuracy $1K$? Write down analytically the approximate result.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The Saha equation can be conveniently rewritten in the form:
$$
y = \frac{3}{2}\ln y + \ln \frac{1 - X}{AX^2}.
$$
Consecutive iterations of the latter equation for $X = 0.1$ and $A = 3
\cdot 10^{ - 16}$ give the following results:
$$
y_0 = \ln \frac{1 - X}{AX^2} \approx 40.24; \Rightarrow T_0 \approx 3920\,K;
$$
$$
y_1 = \frac{3}{2}\ln y_0 + \ln \frac{1 - X}{AX^2} \approx 45.8; \Rightarrow T_1 \approx 3446\,K;
$$
$$
y_2 = \frac{3}{2}\ln y_1 + \ln \frac{1 - X}{AX^2} \approx 46.0;\Rightarrow T_2 \approx 3431\,K.
$$
Result of the third iteration already coincides with the exact one with accuracy of one degree. It can be analytically presented in the following form:
$$
kT \approx \frac{I_0}{\ln \left(\frac{1 - X}{AX^2}{\left\ \ln
\left[ \frac{1 - X} {AX^2}{\left(\ln \frac{1 - X} {AX^2}
\right)}^{3/2}\right]\right\}^{3/2}\right)}.
$$
</p>
</div>
</div></div>

<div id="therm35"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 4 ===
Estimate the duration of the epoch of recombination: how long did it take for hydrogen ionization
degree to change from $90\%$ to $10\%$ according to the Saha equation?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Numerical solution of the Saha equation for $X=0.9,$ gives the corresponding temperature $T_1 = 4\ 029\,K,$ and full recombination (corresponding to $X = 0.1$) takes place at $T_2 = 3\ 431\,K,$ which corresponds to the scale factor values $a_1 = 6.8 \cdot 10^{ - 4}$ and $a_2 = 7.9 \cdot 10^{ - 4}$ respectively (setting $a_0=1$ in the present time). The recombination took place in the matter dominated epoch ( $a = \left( t/{t_0} \right)^{2/3},\;t_0 = 14 \cdot 10^9 $ years), therefore $t_1 \approx 250\ 000$ and $t_2 \approx 310\ 000$ years, i.e. the recombination endured approximately $60\cdot 10^3$ years.
</p>
</div>
</div></div>

<div id="therm36"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 5 ===
Using the Saha equation, determine the hydrogen ionization degree in the center of the Sun ($\rho=100g/cm^3$, $T=1.5\cdot10^7K$).
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The Saha equation with given concentration of the gas can be conveniently presented in the following form
$$ \frac{1 - X} {X^2} = n\lambda_0^3y^{3/2}e^y,$$
where
$$
\lambda _0 \equiv \lambda _{Te} \left.\right|_{kT = I_0} = \frac{\sqrt {2\pi } \hbar c}{\sqrt m_ec^2I_0}\simeq 1.877 \cdot 10^{ - 8}\,cm.
$$
Thus for hydrogen density in the center of the Sun one obtains $ n \simeq \rho /m_p \approx 6 \cdot 10^{25}\,cm^{-3}.$ It then follows
$$
X = \frac{1}{\sqrt{\frac{1}{4} + n\lambda _0^3\left( \frac{I_0}{kT} \right)^{3/2}\exp \left( \frac{I_0}{kT} \right)} + \frac{1}{2}}\approx 0.754
$$
The unreasonable result (almost 25\% of neutral hydrogen in the center of the Sun) appeared because of inapplicability of the Saha equation in such dense and highly ionized plasma: in the considered case the Debye radius and interatomic spacing are of the same order $r_D \approx a \approx 10^{- 9}\,cm.$
</p>
</div>
</div></div>

Peculiarities of Thermodynamics in Early Universe

2012-10-01T22:48:04Z

Den: /* Problem 12 */

[[Category:Thermodynamics of Universe|5]]

__NOTOC__

<div id="time1"></div>
<div id="ter_15"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 1 ===
Find the temperature dependence for the Hubble parameter in the early flat Universe.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

In early Universe the total energy density is determined by by the relativistic particles. Therefore
$\left( \hbar = c = 1\right)$:
\[ H^2 = \frac{8\pi G}{3}\rho;\ \rho = \frac{\pi ^2}{30}g^* T^4;\ G = \frac{1}{M_{Pl}^2}\Rightarrow
H =(2\pi)^{3/2}\sqrt{\frac{g^*}{90}}\frac{T^2}{M_{Pl}}\approx 1.66\sqrt {g^* } \frac{T^2}{M_{Pl}}.\]
</p>
</div>
</div></div>

<div id="ter_16n"></div>
<div id="time2"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2 ===
Find the time dependence of the temperature of the early Universe by direct integration of the first Friedman equation.}
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[\left(\frac{\dot a}{a}\right)^2 = \frac{8\pi G}{3}\rho ;\quad \rho = \alpha T^4
;\quad aT = const = A\Rightarrow
a\dot a = A^2 \left(\frac{8\pi G\alpha}{3}\right)^{1/2}.\]
Integrating the latter equation with the initial condition $a(t = 0) = 0$
one obtains \[a(t)=A\left(\frac{32\piG\alpha}{3}\right)^{1/4}\sqrt{t}\Rightarrow T = \left( \frac{3}{32\pi G\alpha} \right)^{1/4} t^{-1/2}.\]</p>
</div>
</div></div>
<div id="ter_17n"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 3 ===
Prove that results of the problems [[#time1]] and [[#time1]] are equivalent.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">Use the relation
\[
\alpha = \frac{\pi ^2}{30}g^*,
\]
which follows from the comparison of the equations $\rho = \alpha T^4 $ and \[\rho
= \frac{\pi^2}{30}g^* T^4.\]</p>
</div>
</div></div>

<div id="ter_17"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 4 ===
Determine the energy density of the Universe at the Planck time.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[
\rho = \frac{3M_{Pl}^2}{8\pi}H^2 ;\quad H = \frac{1}{2t};
M_{Pl} = \frac{1}{t_{Pl}} \Rightarrow \rho \left( t = t_{Pl} \right) = \frac{3M_{Pl}^4}{32\pi} \approx 6 \cdot 10^{74} \,GeV^4.\]</p>
</div>
</div></div>

<div id="ter_18"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 5 ===
Show that at Planck time the energy density of the Universe corresponded to $10^{77}$ proton masses in one proton volume.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Taking into account that $0.2\,GeV \cdot \, fm = 1$ and using the result o previous problem one obtains $ \rho (t = t_{Pl} ) \approx 6 \cdot 10^{74} \, GeV^4 \approx 0.75 \cdot 10^{77} \,GeV/\,fm^3, $ which approximately corresponds to $10^{77}$ proton mass inside the volume of a proton.
</p>
</div>
</div></div>

<div id="ter_19"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 6 ===
What was the temperature of radiation-dominated Universe at the Planck time?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Using the result of the previous problem one obtains
\[T(t=t_{Pl})=\left(\frac{\rho(t=t_{Pl})}{\alpha}\right)^{1/4}\simeq 6.3\cdot10^{31}\, K.\]
</p>
</div>
</div></div>

<div id="ter_20"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 7 ===
Determine the age of the Universe when its temperature was equal to $1\ MeV$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In [[#ter_16|problem]] of the present Chapter it was obtained that
\[
t \simeq \frac{0.301}{\sqrt {g^*}}\frac{M_{Pl}}{T^2}.\]
At temperature $1\, MeV$ the relativistic particles are presented by photons, electrons, neutrinos of all three types and their antiparticles. Therefore
\[
g^* = 2 + \frac{7}{8}(4 + 2 \times 3) = 10.75.
\]
Taking into account that $1/\,GeV \approx 0.7 \times 10^{- 24} \,s $ one obtains
\[
t(T \approx 1\,MeV) \approx 0.75\, s.\]
</p>
</div>
</div></div>

<div id="ter_21"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 8 ===
In the first cyclic accelerator - the cyclotron (1931)- particles were accelerated up to energies of order $1MeV$. In the next generation accelerators - the bevatrons - energy was risen
to $1GeV$. In the last generation accelerator - the LHC -protons are accelerated to energy of $1\ TeV$. What times in the Universe history do those energies allow to investigate?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="ter_23"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 9 ===
Show that in the epoch when the energy density of the Universe was determined by ultra-relativistic matter and effective number of internal degrees of freedom did not change, held $\dot{T}/T\propto
-T^2$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For the ultra-relativistic component
\[
H \propto T^2 ,\quad T \propto a^{ - 1} ,\quad \dot T \propto - \frac{\dot a}{a^2}\Rightarrow
\frac{\dot T}{T} \propto - H \propto - T^2.\]
</p>
</div>
</div></div>

<div id="therm37"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 10 ===
Estimate the baryon-antibaryon asymmetry $A\equiv(n_b-n_{\bar{b}})/n_{\bar{b}}$ in the early Universe.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The early Universe means the period of evolution of Universe, after which there are no processes capable to violate the baryon number conservation law. Than the total baryon number in a comoving volume will be constant. Therefore
$$
\left(n_b-n_{\bar{b}}\right)a^3=\left(n_{b0}-n_{\bar{b}0}\right)a_0^3.
$$
However today the anti-baryons are practically absent, thus $n_{\bar{b}0}
\approx 0,$ and therefore
$$
A \equiv \frac{n_b-n_{\bar{b}}}{n_{\bar{b}}} = \frac{n_{\bar{b}0}}{n_b}\frac{a_0^3} {a^3}.
$$

Using the estimate $a \sim 1/T,$ represent the expression for asymmetry in the following form
$$
A \approx \frac{n_{\bar{b}0}}{n_b}\frac{T^3} {T_0^3}.
$$
The densities of photons and baryons are connected to temperature by the relations ${n_b} \approx {T^3}$ and $ n_{\gamma0} \approx T_0^3 $ (the numerical factors of order of unity are omitted). Using the above given ingredients, the asymmetry can be presented in the following form
$$
A \approx \frac{n_{b0}} {n_{\gamma0}}.
$$
Therefore the baryon-anti-baryon asymmetry equals to current ratio of the baryon number density to that of photon number. More rigorous analysis leads to the relation \[A \approx 6\frac{n_{b0}}{n_{\gamma0}}.\] The current photon number density is well defined by the CMB temperature and equals to $410.4\: cm^{ - 3}.$ The baryon number density can be estimated by several ways, for example, basing on relative abundance of hydrogen and deuterium. The ultimate result reads
$$
A \approx 3 \cdot 10^{ - 9}.
$$
The latter result can be interpreted in the following way: in early Universe there were three extra quarks per each billion of anti-quarks. presently observed matter is nothing that the result of that tiny asymmetry.
</p>
</div>
</div></div>

<div id="therm38n"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 11 ===
Determine the monopoles' number density and their contribution to the energy density of the Universe at the great Unification temperature. Compare the latter with the photons' energy density at the same temperature.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Consider two regions situated so far from each other that they are not causally connected. Thus the regions take generally speaking independent configurations. Therefore a so-called topological defect, analogous to a dislocation in ferromagnetic crystals, appears on the boundary between the regions. A simplest type of such a defect is analogous to a point dislocation. In typical Grand Unification Theories (GUT) such objet represents the magnetic monopole. It behaves as a particle with mass
$$
m_{mon}\approx \frac{M_X}{\alpha_U}.
$$
Here ${M_X}\approx 10^{16}\,GeV$ is the GUT energy scale and $\alpha_U\approx 1/40$ is the effective coupling constant. The magnetic monopoles could be created in the hot Universe at the phase transition connected to spontaneous symmetry breaking: when temperature of the Universe falls lower than $T_c\approx E_{GUT}\approx 10^{16}\,GeV,$ the Higgs field presented by $X$ and $Y$ massive bosons acquires non-zero vacuum mean.

Due to their giant mass, the monopoles become nonrelativistic component of the energy density in the Universe right after their creation. The monopole density is expected to roughly equal to unity in each isolated region. Size of such region is determined by the distance passed by light during the time period $t_c$ from the Big Bang to the phase transition. This distance simply equals to the particle horizon at time $t_c.$ If the universe was dominated by radiation up to that time, then $a\sim t^{1/2}$ and therefore the particle horizon equals to $a(t)\int_0^{t_c}\frac{dt'}{a(t')}=2t_c$. Then the predicted monopole number density reads
$$
n_{mon}\approx \frac{1}{\left( 2t_c \right)^3}
$$
The time period ${t_c}$ can be estimated from the relation
$$
t_c=\frac 14\sqrt{\frac{45}{\pi^3g^*}} M_{Pl}T_{c}^{-2}\approx 10^{-39}\, sec.
$$
As the monopoles are non-relativistic particles, then their contribution into the energy density is
$$
\rho_{mon}=n_{mon}m_{mon}\approx \frac{1}{\left( 2t_c \right)^3}\frac{M_X}{\alpha_U}
\approx 2\times 10^{57}\,GeV^4.
$$
Compare this value to the photon concentration at the same time
$$
\rho_{\gamma }=\frac{\pi^2}{15}T_c^4\approx 2\times 10^{63}\,GeV^4.
$$
</p>
</div>
</div></div>

<div id="therm39n"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 12 ===
At what temperature and time does the contribution of
monopoles into the Universe energy density become comparable to the
contribution of photons?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
As it was shown in the previous problem, initially the photon energy density considerably exceeded that of monopoles: $\rho_{\gamma }/\rho _{mon}\approx 10^6$. However as the photons are relativistic particles then $\rho _\gamma\sim 1/a^4,$ while the monopoles are non-relativistic and $\rho_{mon}\sim 1/a^3.$ The two energy densities become equal when the scale factor increases in $10^6$ times. Correspondingly the temperature decreases in $10^6$ times, as $a\sim 1/T.$ Using the relation $t\sim T^{-2},$ one obtains that the equality between $\rho_\gamma$ and $\rho_{mon}$ takes places after time increases by factor $10^{12}$. Thus, starting from the GUT scale, $T\approx 10^{16}\,GeV$ at times of order of $t\approx 10^{-39}\,sec,$ the equality $\rho_{\gamma }=\rho_{mon}$ is expected to take place at temperature $T\approx 10^{10}\,GeV$ and time $t\approx 10^{-27}\,sec.$
</p>
</div>
</div></div>

Peculiarities of Thermodynamics in Early Universe

2012-10-01T22:47:51Z

Den: /* Problem 9 */

Peculiarities of Thermodynamics in Early Universe

2012-10-01T22:47:28Z

Den:

[[Category:Thermodynamics of Universe|5]]

__NOTOC__

<div id="time1"></div>
<div id="ter_15"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 1 ===
Find the temperature dependence for the Hubble parameter in the early flat Universe.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

In early Universe the total energy density is determined by by the relativistic particles. Therefore
$\left( \hbar = c = 1\right)$:
\[ H^2 = \frac{8\pi G}{3}\rho;\ \rho = \frac{\pi ^2}{30}g^* T^4;\ G = \frac{1}{M_{Pl}^2}\Rightarrow
H =(2\pi)^{3/2}\sqrt{\frac{g^*}{90}}\frac{T^2}{M_{Pl}}\approx 1.66\sqrt {g^* } \frac{T^2}{M_{Pl}}.\]
</p>
</div>
</div></div>

<div id="ter_16n"></div>
<div id="time2"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2 ===
Find the time dependence of the temperature of the early Universe by direct integration of the first Friedman equation.}
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[\left(\frac{\dot a}{a}\right)^2 = \frac{8\pi G}{3}\rho ;\quad \rho = \alpha T^4
;\quad aT = const = A\Rightarrow
a\dot a = A^2 \left(\frac{8\pi G\alpha}{3}\right)^{1/2}.\]
Integrating the latter equation with the initial condition $a(t = 0) = 0$
one obtains \[a(t)=A\left(\frac{32\piG\alpha}{3}\right)^{1/4}\sqrt{t}\Rightarrow T = \left( \frac{3}{32\pi G\alpha} \right)^{1/4} t^{-1/2}.\]</p>
</div>
</div></div>
<div id="ter_17n"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 3 ===
Prove that results of the problems [[#time1]] and [[#time1]] are equivalent.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">Use the relation
\[
\alpha = \frac{\pi ^2}{30}g^*,
\]
which follows from the comparison of the equations $\rho = \alpha T^4 $ and \[\rho
= \frac{\pi^2}{30}g^* T^4.\]</p>
</div>
</div></div>

<div id="ter_17"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 4 ===
Determine the energy density of the Universe at the Planck time.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[
\rho = \frac{3M_{Pl}^2}{8\pi}H^2 ;\quad H = \frac{1}{2t};
M_{Pl} = \frac{1}{t_{Pl}} \Rightarrow \rho \left( t = t_{Pl} \right) = \frac{3M_{Pl}^4}{32\pi} \approx 6 \cdot 10^{74} \,GeV^4.\]</p>
</div>
</div></div>

<div id="ter_18"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 5 ===
Show that at Planck time the energy density of the Universe corresponded to $10^{77}$ proton masses in one proton volume.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Taking into account that $0.2\,GeV \cdot \, fm = 1$ and using the result o previous problem one obtains $ \rho (t = t_{Pl} ) \approx 6 \cdot 10^{74} \, GeV^4 \approx 0.75 \cdot 10^{77} \,GeV/\,fm^3, $ which approximately corresponds to $10^{77}$ proton mass inside the volume of a proton.
</p>
</div>
</div></div>

<div id="ter_19"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 6 ===
What was the temperature of radiation-dominated Universe at the Planck time?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Using the result of the previous problem one obtains
\[T(t=t_{Pl})=\left(\frac{\rho(t=t_{Pl})}{\alpha}\right)^{1/4}\simeq 6.3\cdot10^{31}\, K.\]
</p>
</div>
</div></div>

<div id="ter_20"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 7 ===
Determine the age of the Universe when its temperature was equal to $1\ MeV$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
In [[#ter_16|problem]] of the present Chapter it was obtained that
\[
t \simeq \frac{0.301}{\sqrt {g^*}}\frac{M_{Pl}}{T^2}.\]
At temperature $1\, MeV$ the relativistic particles are presented by photons, electrons, neutrinos of all three types and their antiparticles. Therefore
\[
g^* = 2 + \frac{7}{8}(4 + 2 \times 3) = 10.75.
\]
Taking into account that $1/\,GeV \approx 0.7 \times 10^{- 24} \,s $ one obtains
\[
t(T \approx 1\,MeV) \approx 0.75\, s.\]
</p>
</div>
</div></div>

<div id="ter_21"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 8 ===
In the first cyclic accelerator - the cyclotron (1931)- particles were accelerated up to energies of order $1MeV$. In the next generation accelerators - the bevatrons - energy was risen
to $1GeV$. In the last generation accelerator - the LHC -protons are accelerated to energy of $1\ TeV$. What times in the Universe history do those energies allow to investigate?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="ter_23"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 9 ===
Show that in the epoch when the energy density of the Universe was determined by ultra-relativistic matter and effective number of internal degrees of freedom did not change, held $\dot{T}/T\propto
-T^2$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For the ultra-relativistic component
\[
H \propto T^2 ,\quad T \propto a^{ - 1} ,\quad \dot T \propto - \frac{\dot a}{a^2}\Rightarrow
\frac{\dot T}{T} \propto - H \propto - T^2.\]
</p>
</div>
</div></div>

<div id="therm37"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 10 ===
Estimate the baryon-antibaryon asymmetry $A\equiv(n_b-n_{\bar{b}})/n_{\bar{b}}$ in the early Universe.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The early Universe means the period of evolution of Universe, after which there are no processes capable to violate the baryon number conservation law. Than the total baryon number in a comoving volume will be constant. Therefore
$$
\left(n_b-n_{\bar{b}}\right)a^3=\left(n_{b0}-n_{\bar{b}0}\right)a_0^3.
$$
However today the anti-baryons are practically absent, thus $n_{\bar{b}0}
\approx 0,$ and therefore
$$
A \equiv \frac{n_b-n_{\bar{b}}}{n_{\bar{b}}} = \frac{n_{\bar{b}0}}{n_b}\frac{a_0^3} {a^3}.
$$

Using the estimate $a \sim 1/T,$ represent the expression for asymmetry in the following form
$$
A \approx \frac{n_{\bar{b}0}}{n_b}\frac{T^3} {T_0^3}.
$$
The densities of photons and baryons are connected to temperature by the relations ${n_b} \approx {T^3}$ and $ n_{\gamma0} \approx T_0^3 $ (the numerical factors of order of unity are omitted). Using the above given ingredients, the asymmetry can be presented in the following form
$$
A \approx \frac{n_{b0}} {n_{\gamma0}}.
$$
Therefore the baryon-anti-baryon asymmetry equals to current ratio of the baryon number density to that of photon number. More rigorous analysis leads to the relation \[A \approx 6\frac{n_{b0}}{n_{\gamma0}}.\] The current photon number density is well defined by the CMB temperature and equals to $410.4\: cm^{ - 3}.$ The baryon number density can be estimated by several ways, for example, basing on relative abundance of hydrogen and deuterium. The ultimate result reads
$$
A \approx 3 \cdot 10^{ - 9}.
$$
The latter result can be interpreted in the following way: in early Universe there were three extra quarks per each billion of anti-quarks. presently observed matter is nothing that the result of that tiny asymmetry.
</p>
</div>
</div></div>

<div id="therm38n"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 11 ===
Determine the monopoles' number density and their contribution to the energy density of the Universe at the great Unification temperature. Compare the latter with the photons' energy density at the same temperature.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Consider two regions situated so far from each other that they are not causally connected. Thus the regions take generally speaking independent configurations. Therefore a so-called topological defect, analogous to a dislocation in ferromagnetic crystals, appears on the boundary between the regions. A simplest type of such a defect is analogous to a point dislocation. In typical Grand Unification Theories (GUT) such objet represents the magnetic monopole. It behaves as a particle with mass
$$
m_{mon}\approx \frac{M_X}{\alpha_U}.
$$
Here ${M_X}\approx 10^{16}\,GeV$ is the GUT energy scale and $\alpha_U\approx 1/40$ is the effective coupling constant. The magnetic monopoles could be created in the hot Universe at the phase transition connected to spontaneous symmetry breaking: when temperature of the Universe falls lower than $T_c\approx E_{GUT}\approx 10^{16}\,GeV,$ the Higgs field presented by $X$ and $Y$ massive bosons acquires non-zero vacuum mean.

Due to their giant mass, the monopoles become nonrelativistic component of the energy density in the Universe right after their creation. The monopole density is expected to roughly equal to unity in each isolated region. Size of such region is determined by the distance passed by light during the time period $t_c$ from the Big Bang to the phase transition. This distance simply equals to the particle horizon at time $t_c.$ If the universe was dominated by radiation up to that time, then $a\sim t^{1/2}$ and therefore the particle horizon equals to $a(t)\int_0^{t_c}\frac{dt'}{a(t')}=2t_c$. Then the predicted monopole number density reads
$$
n_{mon}\approx \frac{1}{\left( 2t_c \right)^3}
$$
The time period ${t_c}$ can be estimated from the relation
$$
t_c=\frac 14\sqrt{\frac{45}{\pi^3g^*}} M_{Pl}T_{c}^{-2}\approx 10^{-39}\, sec.
$$
As the monopoles are non-relativistic particles, then their contribution into the energy density is
$$
\rho_{mon}=n_{mon}m_{mon}\approx \frac{1}{\left( 2t_c \right)^3}\frac{M_X}{\alpha_U}
\approx 2\times 10^{57}\,GeV^4.
$$
Compare this value to the photon concentration at the same time
$$
\rho_{\gamma }=\frac{\pi^2}{15}T_c^4\approx 2\times 10^{63}\,GeV^4.
$$
</p>
</div>
</div></div>

<div id="therm39n"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 12 ===
At what temperature and time does the contribution of
monopoles into the Universe energy density become comparable to the
contribution of photons?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
As it was shown in the previous problem, initially the photon energy density considerably exceeded that of monopoles: $\rho_{\gamma }/\rho _{mon}\approx 10^6$. However as the photons are relativistic particles then $\rho _\gamma\sim 1/a^4,$ while the monopoles are non-relativistic and $\rho_{mon}\sim 1/a^3.$ The two energy densities become equal when the scale factor increases in $10^6$ times. Correspondingly the temperature decreases in $10^6$ times, as $a\sim 1/T.$ Using the relation $t\sim T^{-2},$ one obtains that the equality between $\rho_\gamma$ and $\rho_{mon}$ takes places after time increases by factor $10^{12}$. Thus, starting from the GUT scale, $T\approx 10^{16}\,GeV$ at times of order of $t\approx 10^{-39}\,sec,$ the equality $\rho_{\gamma }=\rho_{mon}$ is expected to take place at temperature $T\approx 10^{10}\,GeV$ and time $t\approx 10^{-27}\,sec.$
</p>
</div>
</div></div>

Connection between Temperature and Redshift

2012-10-01T21:52:03Z

Den: /* Problem 2 */

[[Category:Thermodynamics of Universe|4]]

__NOTOC__

<div id="therm23"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 1 ===
Find the dependence of radiation temperature on the redshift.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
T \propto a^{ - 1};\quad T = T_0(1 + z).
$$
</p>
</div>
</div></div>

<div id="therm24"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2 ===
Find the dependence of free non-relativistic gas temperature on the redshift.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
T \propto a^{ - 2};\quad T = T_0(1 + z)^2.
$$
</p>
</div>
</div></div>

<div id="therm25"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 3 ===
Estimate the time moment when the recombination started, i.e. when ionized plasma transited to the gas of neutral atoms, and determine the corresponding redshift value. The recombination temperature equals to $T_{rec} \approx 0.3\,eV.$
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
z_{rec} = \frac{T_{rec}}{T_0} - 1.
$$
As $1\mbox{\it eV} \approx 11600\,K$, then
$$
z_{rec} \approx \frac{0.3}{2.73}11600 \approx 1300.
$$
For a rough estimate assume that the transition $T_{rec} \to T_0$ occurs mainly in the matter-dominated epoch to obtain
$$
t_{rec} = t_0\left( \frac{a}{a_0} \right)^{3/2} =
t_0\left(\frac{T_0}{T_{rec}} \right)^{3/2}\approx
300\,000\,years.
$$
</p>
</div>
</div></div>

Connection between Temperature and Redshift

2012-10-01T21:50:36Z

Den:

[[Category:Thermodynamics of Universe|4]]

__NOTOC__

<div id="therm23"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 1 ===
Find the dependence of radiation temperature on the redshift.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
T \propto a^{ - 1};\quad T = T_0(1 + z).
$$
</p>
</div>
</div></div>

<div id="therm24"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2 ===
Find the dependence of free non-relativistic gas' temperature on the redshift.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
T \propto a^{ - 2};\quad T = T_0(1 + z)^2.
$$
</p>
</div>
</div></div>

<div id="therm25"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 3 ===
Estimate the time moment when the recombination started, i.e. when ionized plasma transited to the gas of neutral atoms, and determine the corresponding redshift value. The recombination temperature equals to $T_{rec} \approx 0.3\,eV.$
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
z_{rec} = \frac{T_{rec}}{T_0} - 1.
$$
As $1\mbox{\it eV} \approx 11600\,K$, then
$$
z_{rec} \approx \frac{0.3}{2.73}11600 \approx 1300.
$$
For a rough estimate assume that the transition $T_{rec} \to T_0$ occurs mainly in the matter-dominated epoch to obtain
$$
t_{rec} = t_0\left( \frac{a}{a_0} \right)^{3/2} =
t_0\left(\frac{T_0}{T_{rec}} \right)^{3/2}\approx
300\,000\,years.
$$
</p>
</div>
</div></div>

Entropy of Expanding Universe

2012-10-01T21:17:58Z

Den:

[[Category:Thermodynamics of Universe|3]]

__NOTOC__

<div id="therm23_1"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 1 ===
Transform the energy conditions for the flat Universe to
conditions for the entropy density (see [http://arxiv.org/abs/1009.4513])
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="therm29"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2 ===
Find the entropy density for the photon gas.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[
dS = \frac{dE}{T} + \frac{pdV}{T} = \frac{V}{T}\frac{d\rho}{dT}dT + \frac{\rho+ p}{T}dV \Rightarrow\] \[ \left( \frac{\partial S}{\partial V}
\right)_T = \frac{\rho + p} {T} \Rightarrow
S = \frac{\rho + p}{T}V + f(T).\]

As entropy is proportional to volume, then $f(T) =
0.$ For the photon gas $p = \rho/3$, therefore
$$
s = \frac{S}{V} = \frac{4}{3}\frac{\rho}{T} = \frac{4}{3}\alpha
{T^3}.
$$

</p>
</div>
</div></div>

<div id="therm30"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 3 ===
Use the result of the previous problem to derive an alternative proof of the fact that $aT=const$ in the adiabatically expanding Universe.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$S = \frac{4}{3}\alpha T^3V = const;\quad V \propto a^3 \Rightarrow aT = const.$$
</p>
</div>
</div></div>

<div id="therm31"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 4 ===
Find the adiabatic index for the CMB.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
p \propto \rho \propto a^{ - 4};\quad V \propto {a^3} \Rightarrow pV^\gamma = const,\quad \gamma = \frac{4}{3}.
$$
</p>
</div>
</div></div>

<div id="therm32"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 5 ===
Show that the ratio of CMB entropy density to the baryon number density $s_\gamma/n_b$ remains constant during the expansion of the Universe.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

$$s \propto T^3 \propto a^{ - 3};\quad n_b \propto a^{ - 3}.$$ Therefore their ratio remains constant during the expansion of Universe.
</p>
</div>
</div></div>

<div id="ter_6n"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 6 ===
Estimate the current entropy density of the Universe.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

The current entropy of the Universe is determined by the CMB photons. A rough estimate neglects the contribution of other particles, then one obtains
\[s = 2\frac{2\pi ^2}{45}T_0^3.\]
For $T_0 \approx 2.725K$
\[s \approx 1.5 \cdot 10^3 \, cm^{-3}.\]
</p>
</div>
</div></div>

<div id="ter_7n"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 7 ===
Estimate the entropy of the observable part of the Universe.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Size of the observable part of the Universe is $l_{H0} \sim 10^{28} \mbox{\it cm}$ and therefore
\[S = \frac{4\pi}{3}sl_{H0}^3 \sim 10^{88}.\]
</p>
</div>
</div></div>

<div id="ter_10"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 7 ===
Why is the expansion of the Universe described by the Friedman equations adiabatic?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
It is because the heat flow is absent in the homogeneous and isotropic Universe.
</p>
</div>
</div></div>

<div id="ter_11"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 8 ===
Show that entropy is conserved during the expansion of the Universe described by the Friedman equations.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The first law of thermodynamics
\[
dE + pdV = TdS
\]
can be transformed into the form
\[
a^3 \left[\frac{d\rho}{dt} + 3H(\rho + p)\right] =T\frac{dS}{dt}.\]
Then with the conservation equation
\[\frac{d\rho}{dt} + 3H(\rho + p) = 0,\] which follows from the Friedmann equation, one obtains \[ \frac{dS}{dt} = 0 \Rightarrow S = const.\]
</p>
</div>
</div></div>

<div id="ter_12"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 9 ===
Show that the entropy density behaves as $s\propto a^{-3}$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[S = const \Rightarrow sV = const \Rightarrow sV_0 a^3 = const \Rightarrow sa^3 = const \Rightarrow s \propto a^{ - 3}.\]
</p>
</div>
</div></div>

<div id="ter_13"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 10 ===
Using only thermodynamical considerations, show that if the energy density of some component is $\rho=const$
than the state equation for that component reads $p=-\rho$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
By definition,
\[
p = - \left( \frac{\partial E}{\partial V} \right)_S.
\]
If $\rho = const$, then \[\frac{\partial E}{\partial V} = \rho\Rightarrow p = - \rho.\]
</p>
</div>
</div></div>

<div id="ter_14"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 11 ===
Show that the product $aT$ is an approximate invariant in the Universe dominated by relativistic particles.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Consider a volume $V$, filled by photons and let it expands with the same rate as the whole Universe, i.e. $V \propto a^3 (t)$. For radiation $\rho = \alpha T^4 $, $p = \rho/3.$ The first law of thermodynamics applied to the expanding Universe with no heat flow reads
\[
\frac{dE}{dt} = - p\frac{dV}{dt}.
\]
As $E = \rho V$, then
\[
\frac{1}{T}\frac{dT}{dt} = - \frac{1}{3V}\frac{dV}{dt},\] or equivalently \[ \frac{d}{dt}(\ln T) = - \frac{d}{dt}(\ln a)\Rightarrow aT = const.
\]
</p>
</div>
</div></div>

Entropy of Expanding Universe

2012-10-01T21:15:02Z

Den:

[[Category:Thermodynamics of Universe|3]]

__NOTOC__

<div id="therm23_1"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 1 ===
Transform the energy conditions for the flat Universe to
conditions for the entropy density (see [http://arxiv.org/abs/1009.4513])
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="therm29"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2 ===
Find the entropy density for the photon gas.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

\[
dS = \frac{dE}
{T} + \frac{pdV} {T} = \frac{V}{T}\frac{d\rho}{dT}dT + \frac{\rho+ p} {T}dV \Rightarrow\] \[ \left( \frac{\partial S}{\partial V}
\right)_T = \frac{\rho + p} {T} \Rightarrow
S = \frac{\rho + p}{T}V + f(T).\]

As entropy is proportional to volume, then $f(T) =
0.$ For the photon gas $p = \rho/3$, therefore
$$
s = \frac{S}{V} = \frac{4}{3}\frac{\rho}{T} = \frac{4}{3}\alpha
{T^3}.
$$

</p>
</div>
</div></div>

<div id="Okun6"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 1 ===

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="therm30"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 3 ===
Use the result of the previous problem to derive an alternative proof of the fact that $aT=const$ in the adiabatically expanding Universe.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$S = \frac{4}{3}\alpha T^3V = const;\quad V \propto a^3 \Rightarrow aT = const.$$
</p>
</div>
</div></div>

<div id="therm31"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 4 ===
Find the adiabatic index for the CMB.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
$$
p \propto \rho \propto a^{ - 4};\quad V \propto {a^3} \Rightarrow pV^\gamma = const,\quad \gamma = \frac{4}{3}.
$$
</p>
</div>
</div></div>

<div id="therm32"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 5 ===
Show that the ratio of CMB entropy density to the baryon number density $s_\gamma/n_b$ remains constant during the expansion of the Universe.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

$$s \propto T^3 \propto a^{ - 3};\quad n_b \propto a^{ - 3}.$$ Therefore their ratio remains constant during the expansion of Universe.
</p>
</div>
</div></div>

<div id="ter_6n"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 6 ===
Estimate the current entropy density of the Universe.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

The current entropy of the Universe is determined by the CMB photons. A rough estimate neglects the contribution of other particles, then one obtains
\[s = 2\frac{2\pi ^2}{45}T_0^3.\]
For $T_0 \approx 2.725K$
\[s \approx 1.5 \cdot 10^3 \, cm^{-3}.\]
</p>
</div>
</div></div>

<div id="ter_7n"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 7 ===
Estimate the entropy of the observable part of the Universe.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Size of the observable part of the Universe is $l_{H0} \sim 10^{28} \mbox{\it cm}$ and therefore
\[S = \frac{4\pi}{3}sl_{H0}^3 \sim 10^{88}.\]
</p>
</div>
</div></div>

<div id="ter_10"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 7 ===
Why is the expansion of the Universe described by the Friedman equations adiabatic?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
It is because the heat flow is absent in the homogeneous and isotropic Universe.
</p>
</div>
</div></div>

<div id="ter_11"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 8 ===
Show that entropy is conserved during the expansion of the Universe described by the Friedman equations.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The first law of thermodynamics
\[
dE + pdV = TdS
\]
can be transformed into the form
\[
a^3 \left[\frac{d\rho}{dt} + 3H(\rho + p)\right] =T\frac{dS}{dt}.\]
Then with the conservation equation
\[\frac{d\rho}{dt} + 3H(\rho + p) = 0,\] which follows from the Friedmann equation, one obtains \[ \frac{dS}{dt} = 0 \Rightarrow S = const.\]
</p>
</div>
</div></div>

<div id="ter_12"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 9 ===
Show that the entropy density behaves as $s\propto a^{-3}$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[S = const \Rightarrow sV = const \Rightarrow sV_0 a^3 = const \Rightarrow sa^3 = const \Rightarrow s \propto a^{ - 3}.\]
</p>
</div>
</div></div>

<div id="ter_13"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 10 ===
Using only thermodynamical considerations, show that if the energy density of some component is $\rho=const$
than the state equation for that component reads $p=-\rho$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
By definition,
\[
p = - \left( \frac{\partial E}{\partial V} \right)_S.
\]
If $\rho = const$, then \[\frac{\partial E}{\partial V} = \rho\Rightarrow p = - \rho.\]
</p>
</div>
</div></div>

<div id="ter_14"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 11 ===
Show that the product $aT$ is an approximate invariant in the Universe dominated by relativistic particles.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Consider a volume $V$, filled by photons and let it expands with the same rate as the whole Universe, i.e. $V \propto a^3 (t)$. For radiation $\rho = \alpha T^4 $, $p = \rho/3.$ The first law of thermodynamics applied to the expanding Universe with no heat flow reads
\[
\frac{dE}{dt} = - p\frac{dV}{dt}.
\]
As $E = \rho V$, then
\[
\frac{1}{T}\frac{dT}{dt} = - \frac{1}{3V}\frac{dV}{dt},\] or equivalently \[ \frac{d}{dt}(\ln T) = - \frac{d}{dt}(\ln a)\Rightarrow aT = const.
\]
</p>
</div>
</div></div>

Thermodynamics of Non-Relativistic Gas

2012-10-01T20:32:13Z

Den:

[[Category:Thermodynamics of Universe|2]]

__NOTOC__

<div id="therm26"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 1 ===
Find the ratio of thermal capacities of matter in the form of monatomic gas and radiation.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

$$\frac{c_m}{c_\gamma}=\frac{1.5nk}{4\alpha T^3},$$
where $n$ is the concentration of the atoms.
</p>
</div>
</div></div>

<div id="therm27"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2 ===
Expansion of the Universe tends to violate thermal equilibrium
between the radiation ($T\propto a^{-1}$) and gas of
non-relativistic particles ($T\propto a^{-2}$). Which of these two
components determines the temperature of the Universe?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

It is the radiation. Temperature of the matter relaxes to that of radiation, because the latter has much greater thermal capacity.
</p>
</div>
</div></div>

<div id="therm28"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 3 ===
ow that in the non-relativistic limit ($kT\ll mc^2$) $p\ll\rho$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

Consider for example the case of non-relativistic gas. Then $p = nkT,\;\rho = mn,$ and therefore
$$
p = \frac{kT}{m}\rho \ll 1.
$$
</p>
</div>
</div></div>

<div id="ter_n_19"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 4 ===
Show that for a system of particles in thermal equilibrium the following holds
\[\frac{dp}{dT}=\frac1T(\rho+p).\]
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

Entropy of the system of particles in thermal equilibrium is the function $S=S(V,T)$ such that
\[dS=\frac1T(dE+pdV)=\frac1T[d(\rho V)+pdV]=\frac1T\left[V\frac{d\rho}{dT}dT+(p+\rho)dV\right].\]
Then \[\frac{\partial S}{\partial V}=\frac1T(\rho+p),\ \frac{\partial S}{\partial V}=\frac{\partial S}{\partial T}=\frac{V}{T}\frac{d\rho}{dT}.\]
The above given derivatives must satisfy the condition
\[\frac{\partial}{\partial T}\left[\frac1T(\rho+p)\right]=\frac{\partial}{\partial V}\left(\frac{V}{T}\frac{d\rho}{dT}.\right)\]
Finally one obtains
\[\frac{dp}{dT}=\frac1T(\rho+p).\]
</p>
</div>
</div></div>

<div id="ter_n_20"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 5 ===
Show that for a substance with the equation of state $p=w\rho$ the following holds
\[T\propto\rho^{\frac{w}{w+1}}.\]
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

\[\frac{dp}{dT}=\frac1T(\rho+p)\Rightarrow\frac{d\rho}{d T}=\frac{1+w}{w}\frac\rho T,\]
\[\frac{w}{1+w}\frac{d\rho}{\rho}=\frac{dT}{T}\Rightarrow T\propto\rho^{\frac{w}{w+1}}.\]

</p>
</div>
</div></div>

<div id="ter_n_21"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 6 ===
Show that for a substance with the equation of state $p=w\rho$ the following holds $T\propto a^{-3w}.$
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[T\propto\rho^{\frac{w}{w+1}},\ \rho\propto a^{-3(1+w)}\Rightarrow T\propto a^{-3w}.\]
</p>
</div>
</div></div>

<div id="ter_n_22"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 7 ===
Obtain a generalization of Stefan-Boltzmann law for cosmological fluid, described by generalized polytropic equation of state $p=w\rho+k\rho^{1+1/n},$ assuming that $1+w+k\rho^{1/n}>0$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For the polytropic equation of state the thermodynamical equation
\[\frac{dp}{dT}=\frac1T(\rho+p)\]
becomes
\[(w+k(1+1/n)\rho^{1/n})\frac{d\rho}{dT}=\frac1T(w+1+k\rho^{1/n})\rho.\]
This equation can be integrated into
\[T=T_*\left[1\pm(\rho/\rho_*)^{1/n}\right](w+n+1)/(w+1)(\rho/\rho_*)^{w/(w+1)},\]
where $T_*$ is constants of integration and $\rho_*\equiv[(1+w)/|k|]^n$ [see P-H. Chavanis, arXiv:1208.0797], the upper sign corresponds to $k>0$, and the lower sign corresponds to $k<0$. This relation can be viewed as a generalization of Stefan-Boltzmann law.

</p>
</div>
</div></div>

<div id="ter_n_22_1"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 8 ===
Show that the velocity of sound for cosmological fluid described by generalized polytropic equation of state $p=w\rho+k\rho^{1+1/n}$, where $1+w+k\rho^{1/n}>0$, vanishes at the point where the temperature is extremum.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For the generalized polytropic equation of state the velocity of sound is given by
\[c_s^2=\frac{d p}{d\rho}=w\pm(w+1)\frac{n+1}{n}\left(\frac\rho{\rho_*}\right)^{1/n},\]
\[\rho_*\equiv\left[\frac{w+1}{|k|}\right]^n.\]
The upper sign corresponds to $k>0$, and the lower sign corresponds to $k<0$. As we have seen in the previous problem
\[T=T_*\left[1\pm(\rho/\rho_*)^{1/n}\right](w+n+1)/(w+1)(\rho/\rho_*)^{w/(w+1)}.\]
The extremum of temperature (when it exists) is located at
\[\rho_e=\rho_*\left[\mp\frac{wn}{(w+1)(n+1)}\right]^n.\]
It is easy to see that the velocity of sound vanishes at the point where the temperature is extremum.

</p>
</div>
</div></div>

Thermodynamics of Non-Relativistic Gas

2012-10-01T20:30:14Z

Den:

[[Category:Thermodynamics of Universe|2]]

__TOC__

<div id="therm26"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 1 ===
Find the ratio of thermal capacities of matter in the form of monatomic gas and radiation.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

$$\frac{c_m}{c_\gamma}=\frac{1.5nk}{4\alpha T^3},$$
where $n$ is the concentration of the atoms.
</p>
</div>
</div></div>

<div id="therm27"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2 ===
Expansion of the Universe tends to violate thermal equilibrium
between the radiation ($T\propto a^{-1}$) and gas of
non-relativistic particles ($T\propto a^{-2}$). Which of these two
components determines the temperature of the Universe?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

It is the radiation. Temperature of the matter relaxes to that of radiation, because the latter has much greater thermal capacity.
</p>
</div>
</div></div>

<div id="therm28"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 3 ===
ow that in the non-relativistic limit ($kT\ll mc^2$) $p\ll\rho$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

Consider for example the case of non-relativistic gas. Then $p = nkT,\;\rho = mn,$ and therefore
$$
p = \frac{kT}{m}\rho \ll 1.
$$
</p>
</div>
</div></div>

<div id="ter_n_19"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 4 ===
Show that for a system of particles in thermal equilibrium the following holds
\[\frac{dp}{dT}=\frac1T(\rho+p).\]
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

Entropy of the system of particles in thermal equilibrium is the function $S=S(V,T)$ such that
\[dS=\frac1T(dE+pdV)=\frac1T[d(\rho V)+pdV]=\frac1T\left[V\frac{d\rho}{dT}dT+(p+\rho)dV\right].\]
Then \[\frac{\partial S}{\partial V}=\frac1T(\rho+p),\ \frac{\partial S}{\partial V}=\frac{\partial S}{\partial T}=\frac{V}{T}\frac{d\rho}{dT}.\]
The above given derivatives must satisfy the condition
\[\frac{\partial}{\partial T}\left[\frac1T(\rho+p)\right]=\frac{\partial}{\partial V}\left(\frac{V}{T}\frac{d\rho}{dT}.\right)\]
Finally one obtains
\[\frac{dp}{dT}=\frac1T(\rho+p).\]
</p>
</div>
</div></div>

<div id="ter_n_20"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 5 ===
Show that for a substance with the equation of state $p=w\rho$ the following holds
\[T\propto\rho^{\frac{w}{w+1}}.\]
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

\[\frac{dp}{dT}=\frac1T(\rho+p)\Rightarrow\frac{d\rho}{d T}=\frac{1+w}{w}\frac\rho T,\]
\[\frac{w}{1+w}\frac{d\rho}{\rho}=\frac{dT}{T}\Rightarrow T\propto\rho^{\frac{w}{w+1}}.\]

</p>
</div>
</div></div>

<div id="ter_n_21"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 6 ===
Show that for a substance with the equation of state $p=w\rho$ the following holds $T\propto a^{-3w}.$
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[T\propto\rho^{\frac{w}{w+1}},\ \rho\propto a^{-3(1+w)}\Rightarrow T\propto a^{-3w}.\]
</p>
</div>
</div></div>

<div id="ter_n_22"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 7 ===
Obtain a generalization of Stefan-Boltzmann law for cosmological fluid, described by generalized polytropic equation of state $p=w\rho+k\rho^{1+1/n},$ assuming that $1+w+k\rho^{1/n}>0$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For the polytropic equation of state the thermodynamical equation
\[\frac{dp}{dT}=\frac1T(\rho+p)\]
becomes
\[(w+k(1+1/n)\rho^{1/n})\frac{d\rho}{dT}=\frac1T(w+1+k\rho^{1/n})\rho.\]
This equation can be integrated into
\[T=T_*\left[1\pm(\rho/\rho_*)^{1/n}\right](w+n+1)/(w+1)(\rho/\rho_*)^{w/(w+1)},\]
where $T_*$ is constants of integration and $\rho_*\equiv[(1+w)/|k|]^n$ [see P-H. Chavanis, arXiv:1208.0797], the upper sign corresponds to $k>0$, and the lower sign corresponds to $k<0$. This relation can be viewed as a generalization of Stefan-Boltzmann law.

</p>
</div>
</div></div>

<div id="ter_n_22_1"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 8 ===
Show that the velocity of sound for cosmological fluid described by generalized polytropic equation of state $p=w\rho+k\rho^{1+1/n}$, where $1+w+k\rho^{1/n}>0$, vanishes at the point where the temperature is extremum.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For the generalized polytropic equation of state the velocity of sound is given by
\[c_s^2=\frac{d p}{d\rho}=w\pm(w+1)\frac{n+1}{n}\left(\frac\rho{\rho_*}\right)^{1/n},\]
\[\rho_*\equiv\left[\frac{w+1}{|k|}\right]^n.\]
The upper sign corresponds to $k>0$, and the lower sign corresponds to $k<0$. As we have seen in the previous problem
\[T=T_*\left[1\pm(\rho/\rho_*)^{1/n}\right](w+n+1)/(w+1)(\rho/\rho_*)^{w/(w+1)}.\]
The extremum of temperature (when it exists) is located at
\[\rho_e=\rho_*\left[\mp\frac{wn}{(w+1)(n+1)}\right]^n.\]
It is easy to see that the velocity of sound vanishes at the point where the temperature is extremum.

</p>
</div>
</div></div>

Thermodynamics of Non-Relativistic Gas

2012-10-01T20:29:07Z

Den:

Thermodynamics of Non-Relativistic Gas

2012-10-01T20:23:01Z

Den:

[[Category:Thermodynamics of Universe|2]]

__TOC__

<div id="therm26"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 1 ===
Find the ratio of thermal capacities of matter in the form of monatomic gas and radiation.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

$$\frac{c_m}{c_\gamma}=\frac{1.5nk}{4\alpha T^3},$$
where $n$ is the concentration of the atoms.
</p>
</div>
</div></div>

<div id="therm27"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2 ===
Expansion of the Universe tends to violate thermal equilibrium
between the radiation ($T\propto a^{-1}$) and gas of
non-relativistic particles ($T\propto a^{-2}$). Which of these two
components determines the temperature of the Universe?
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

It is the radiation. Temperature of the matter relaxes to that of radiation, because the latter has much greater thermal capacity.
</p>
</div>
</div></div>

<div id="therm28"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 3 ===
ow that in the non-relativistic limit ($kT\ll mc^2$) $p\ll\rho$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

Consider for example the case of non-relativistic gas. Then $p = nkT,\;\rho = mn,$ and therefore
$$
p = \frac{kT}{m}\rho \ll 1.
$$
</p>
</div>
</div></div>

<div id="ter_n_19"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 4 ===
Show that for a system of particles in thermal equilibrium the following holds
\[\frac{dp}{dT}=\frac1T(\rho+p).\]
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

Entropy of the system of particles in thermal equilibrium is the function $S=S(V,T)$ such that
\[dS=\frac1T(dE+pdV)=\frac1T[d(\rho V)+pdV]=\frac1T\left[V\frac{d\rho}{dT}dT+(p+\rho)dV\right].\]
Then \[\frac{\partial S}{\partial V}=\frac1T(\rho+p),\ \frac{\partial S}{\partial V}=\frac{\partial S}{\partial T}=\frac{V}{T}\frac{d\rho}{dT}.\]
The above given derivatives must satisfy the condition
\[\frac{\partial}{\partial T}\left[\frac1T(\rho+p)\right]=\frac{\partial}{\partial V}\left(\frac{V}{T}\frac{d\rho}{dT}.\right)\]
Finally one obtains
\[\frac{dp}{dT}=\frac1T(\rho+p).\]
</p>
</div>
</div></div>

<div id="ter_n_20"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 5 ===
Show that for a substance with the equation of state $p=w\rho$ the following holds
\[T\propto\rho^{\frac{w}{w+1}}.\]
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

\[\frac{dp}{dT}=\frac1T(\rho+p)\Rightarrow\frac{d\rho}{d T}=\frac{1+w}{w}\frac\rho T,\]
\[\frac{w}{1+w}\frac{d\rho}{\rho}=\frac{dT}{T}\Rightarrow T\propto\rho^{\frac{w}{w+1}}.\]

</p>
</div>
</div></div>

<div id="ter_n_21"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 6 ===
Show that for a substance with the equation of state $p=w\rho$ the following holds
\[T\propto a^{-3w}.\]
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

\[T\propto\rho^{\frac{w}{w+1}},\ \rho\propto a^{-3(1+w)}\Rightarrow T\propto a^{-3w}.\]
</p>
</div>
</div></div>

<div id="ter_n_22"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 7 ===
Obtain a generalization of Stefan-Boltzmann law for cosmological fluid, described by generalized polytropic equation of state \[p=w\rho+k\rho^{1+1/n},\]
assuming that $1+w+k\rho^{1/n}>0$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For the polytropic equation of state the thermodynamical equation
\[\frac{dp}{dT}=\frac1T(\rho+p)\]
becomes
\[(w+k(1+1/n)\rho^{1/n})\frac{d\rho}{dT}=\frac1T(w+1+k\rho^{1/n})\rho.\]
This equation can be integrated into
\[T=T_*\left[1\pm(\rho/\rho_*)^{1/n}\right](w+n+1)/(w+1)(\rho/\rho_*)^{w/(w+1)},\]
where $T_*$ is constants of integration and $\rho_*\equiv[(1+w)/|k|]^n$ [see P-H. Chavanis, arXiv:1208.0797], the upper sign corresponds to $k>0$, and the lower sign corresponds to $k<0$. This relation can be viewed as a generalization of Stefan-Boltzmann law.

</p>
</div>
</div></div>

<div id="ter_n_22_1"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 8 ===
Show that the velocity of sound for cosmological fluid described by generalized polytropic equation of state $p=w\rho+k\rho^{1+1/n}$, where $1+w+k\rho^{1/n}>0$, vanishes at the point where the temperature is extremum.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
For the generalized polytropic equation of state the velocity of sound is given by
\[c_s^2=\frac{d p}{d\rho}=w\pm(w+1)\frac{n+1}{n}\left(\frac\rho{\rho_*}\right)^{1/n},\]
\[\rho_*\equiv\left[\frac{w+1}{|k|}\right]^n.\]
The upper sign corresponds to $k>0$, and the lower sign corresponds to $k<0$. As we have seen in the previous problem
\[T=T_*\left[1\pm(\rho/\rho_*)^{1/n}\right](w+n+1)/(w+1)(\rho/\rho_*)^{w/(w+1)}.\]
The extremum of temperature (when it exists) is located at
\[\rho_e=\rho_*\left[\mp\frac{wn}{(w+1)(n+1)}\right]^n.\]
It is easy to see that the velocity of sound vanishes at the point where the temperature is extremum.

</p>
</div>
</div></div>

Thermodynamics of Non-Relativistic Gas

2012-10-01T18:35:45Z

Den:

[[Category:Thermodynamics of Universe|2]]

__TOC__

<div id="ter_9"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 11 ===

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="ter_9"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 11 ===

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="ter_9"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 11 ===

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="ter_9"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 11 ===

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="ter_9"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 11 ===

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="ter_9"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 11 ===

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="ter_9"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 11 ===

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

<div id="ter_9"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 11 ===

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

</p>
</div>
</div></div>

Thermodynamical Properties of Elementary Particles

2012-10-01T18:26:54Z

Den: /* Problem 11 */

[[Category:Thermodynamics of Universe|1]]

<pre style="width:70%">
In the mid-forties G.A.~Gamov proposed the idea
of the "hot" origin of the World. Therefore thermodynamics
was introduced into cosmology, and nuclear physics too.
Before him the science of the evolution of Universe
contained only dynamics and geometry of the World.
A.D.Chernin.
</pre>

<div id="tab:ter_sm">
{| cellpadding="5" cellspacing="0" border="1" align="center"
|+ |The Standard Model particles and their properties.
! rowspan="2" | Particles
!rowspan="2" |Mass
!colspan="2" |Number of states
! rowspan="2" width="100" |$g$ (particles and anti-particles)
|-
|spin
|color
|- align="center" |
|Photon ($\gamma$) || 0 || 2 || 1 || 2
|- align="center" |
|$W^+,W^-$ || $80.4\, GeV$ || 3 || 1 || 6
|- align="center" |
|$Z$ || $91.2\,GeV$ || 3 || 1 || 3
|- align="center" |
|Gluon ($g$) || 0||2||8||16
|- align="center" |
| Higgs boson ||$>114\, GeV$||1||1||1
|- align="center" |
|Bosons
| colspan="3" |
| 28
|- align="center" |
| $u,/,\bar{u}$ || $3\, MeV$ || 2||3||12
|- align="center" |
||$d,/,\bar{d}$||$6\, MeV$||2||3||12
|- align="center" |
|$s,/,\bar{s}$||$100\,MeV$||2||3||12
|- align="center" |
|$c,/,\bar{c}$||$1.2\,GeV$||2||3||12
|- align="center" |
|$b,/,\bar{b}$||$4.2\, GeV$||2||3||12
|- align="center" |
|$t,/,\bar{t}$||$175\,GeV$||2||3||12
|- align="center" |
|$e^+,\,e^-$|| $0.511\, MeV $||2||1||4
|- align="center" |
|$\mu^+,\,\mu^-$|| $105.7\,MeV$||2||1||4
|- align="center" |
|$\tau^+,\,\tau^-$|| $1.777\, GeV$||2||1||4
|- align="center" |
|$\nu_e,\,\bar{\nu_e}$|| $<3\,eV$||1||1||2
|- align="center" |
|$\nu_\mu,\,\bar{\nu_\mu}$|| $<0.19\,MeV $||1||1||2
|- align="center" |
|$\nu_\tau,\,\bar{\nu_\tau}$|| $<18.2\, MeV $||1||1||2
|- align="center" |
|Fermions
| colspan="3" |
| 90
|}
</div>

__TOC__

<div id="ter_1"></div>
<div style="border: 1px solid #AAA; padding:5px;">
<div id="ter_1"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 1 ===
Find the energy and number densities for bosons and fermions in the relativistic limit.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The energy density equals to
$$
\rho = \left\{ \begin{array}{c}
\frac{\pi^2}{30}gT^4 \quad \; for\;bosons\\
\frac{7}{8}\frac{\pi^2}{30}gT^4 \quad for \;fermions
\end{array}
\right.
$$
Here $g$ equals to the number of internal degrees of freedom for the particle. For photon (with spin $S=1$) $g=2$ due to absence of longitudinal polarization for the zero-mass particles.
The number density is the following:
\[
n = \left\{ \begin{array}{c}
\frac{\zeta (3)}{\pi ^2 }gT^3 \quad \;for\;bosons \\
\frac{3}
{4}\frac{\zeta (3)}{\pi ^2 }gT^3 \quad \; for\;fermions
\end{array} \right.
\]
Here $\zeta (x)$ is the Riemann $\zeta$-function; $\zeta (3) \approx
1.202$.</p>
</div>
</div></div>

<div id="ter_2"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2 ===
Find the number densities and energy densities, normalized to the photon energy density at the same temperature, for neutrinos, electrons, positrons, pions and nucleons in the relativistic limit.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[
\frac{n_i}{n_\gamma} = \left\{
\begin{gathered}
\frac{1}{2}g_i \quad for\; bosons\\
\frac{3}{8}g_i \quad for\; fermions\\
\end{gathered}
\right.;\quad
\frac{\rho_i}{\rho_\gamma} = \left\{
\begin{gathered}
\frac{1}{2}g_i \quad for\; bosons\\
\frac{7}{16}g_i \quad for\; fermions\\
\end{gathered} \right.
\]
\[
\begin{gathered}
\left(\frac{n_i }{n_\gamma},\frac{\rho_i}{\rho_\gamma} \right)_{\nu_{\mu ,e}} = \left( \frac{3}{8},\frac{7}{16} \right);\;\;\left( \frac{n_i}{n_\gamma},
\frac{\rho_i}{\rho_\gamma} \right)_{e^\pm} = \left(
\frac{3}{4},\frac{7}{8} \right);\\
\left( \frac{n_i}{n_\gamma},\frac{\rho_i}{\rho_\gamma} \right)_\pi = \left( \frac{1}{2},\frac{1}{2} \right);\;\;\left( \frac{n_i}{n_\gamma},\frac{\rho_i}{\rho_\gamma} \right)_{n,p} = \left( \frac{3}{4},\frac{7}{8}
\right) \\
\end{gathered}
\]
</p>
</div>
</div></div>

<div id="ter_3"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 3 ===
Calculate the average energy per particle in the relativistic and non-relativistic limits.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Average energy per particle reads$\left\langle \varepsilon \right\rangle = \rho /n$. For $T \gg m$
\[
\left\langle \varepsilon \right\rangle = \left\{
\begin{gathered}
\frac{\pi ^2}{30\zeta (3)}T\quad \; for\; bosons\\
\frac{7\pi ^4}{180\zeta(3)}T\quad for\; fermions\\
\end{gathered} \right.
\]
In the non-relativistic limit $(T \ll m)$ one obtains the following:
\[
\left\langle \varepsilon \right\rangle = m + \frac{3}{2}T \approx m
\]

</p>
</div>
</div></div>

<div id="ter_4"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 4 ===
Find the number of internal degrees of freedom for a quark.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

For quark one obtains the following:
\[
g = 2(\, spin) \times 3(\, particle) \times 2(\, particles + \, antiparticle) = 12
\]
</p>
</div>
</div></div>

<div id="ter_5n"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 5 ===
Find the entropy density for bosons end fermions with zero chemical potential.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

In the considered case one gets:
\[
s = \frac{\rho + p}{T};\quad p = \frac{1}{3}\rho
\]
Using the results of [[#ter_1|problem]] of the present Chapter one obtains the following:
\[
s = \frac{2\pi ^2}{45}g^* T^3,
\]
where $g^* = 1$ for bosons and $g^* = 7/8$ for fermions.
</p>
</div>
</div></div>

<div id="ter_5"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 6 ===

Find the effective number of internal degrees of freedom for a mixture of relativistic bosons and fermions.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

\[
\rho = \sum\limits_{i = bosons} g_i \frac{\pi^2}{30} T^4 +
\sum\limits_{j = fermions} \frac{7}{8}g_j\frac{\pi^2}{30}
T^4 \frac{\pi^2}{30}g^* T^4,
\]
where $g^* $ is the effective number of degrees of freedom:
\[
g^* = \sum\limits_{i = bosons} g_i + \frac{7}{8}\sum\limits_{j
= fermions}g_j.
\]
</p>
</div>
</div></div>

<div id="therm5"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 7 ===

Generalize the results of the previous problem for the case when some $i$-components have temperature $T_i$ different from the CMB temperature $T$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[
g^* = \sum\limits_{i = bosons}g_i \left(\frac{T_i}{T} \right)^4 + \frac{7}{8}\sum\limits_{j = fermions}g_i \left( \frac{T_j}{T}\right)^4
\] </p>
</div>
</div></div>

<div id="ter_6"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 8 ===
Find the effective number of internal degrees of freedom for the Standard Model particles at temperature $T>1TeV$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;"> At temperature $1\, TeV$ all the Standard Model
particles are relativistic (see [[#tab:ter_sm|table]]). Therefore
\[
g^* = 28 + \frac{7}{8} \times 90 = 106.75.
\] </p>
</div>
</div></div>

<div id="ter_7"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 9 ===
Find the change of the number of internal degrees of freedom due to the quark
hadronization process.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
At temperature $T \approx 0.2\, GeV$ quarks and gluons transform into colorless hadrons (protons and neutrons). Therefore
\[
\Delta g = 16\,(gluons) + 12 \times 6\,(quarks) -
8\,(nucleons) = 80.
\]
</p>
</div>
</div></div>

<div id="ter_8"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 10 ===
Find the relation between the energy density and temperature at
$10^{10}\, K$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Represent the expression for energy density of relativistic particles in the following form:
\[
\rho = \left( \sum\limits_{i = bosons} \frac{g_i}{2} +
\sum\limits_{j = fermions} \frac{7}{16}g_j \right)\rho
_\gamma.
\]
At the temperature under consideration the relativistic particles are presented by photons, electron, positrons and neutrinos. Therefore
\[
\rho = \left( 1 + \frac{7} {4} + \frac{7}{4} \right)\rho _\gamma
= \frac{9}{2}\rho _\gamma = \frac{9}{2}\alpha T^4,
\]
where the radiation constant \[\alpha =
\frac{\pi^2}{15}\frac{k^4}{(\hbar c)^3} \approx 7.56 \cdot
10^{-16}\frac{ J}{ m^3K^4}.\]
</p>
</div>
</div></div>

<div id="ter_9"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 11 ===
Find the ratio of the energy density at temperature $10^{10}\, K$ to that at $10^8\, K$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

Unlike the case considered in the previous problem, now only photons and neutrinos remain relativistic, because $10^8 K \sim 10^{ - 2} \, MeV$, and electron mass is $0.511 \, MeV$ is considerably greater. Therefore in the case under consideration one obtains the following:
\[
\rho = \left( 1 + \frac{7}{4} \right)\rho _\gamma =
\frac{11}{4}\rho _\gamma \Rightarrow
\frac{\rho\left(T_2 = 10^{10} K \right)}{\rho \left( T_1 = 10^8 K \right)} = \frac{18}
{11}\left(\frac{T_2}{T_1}\right)^4\simeq1.6\cdot10^8.\]
</p>
</div>
</div></div>

<div id="razm40"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 12 ===
Determine Fermi energy of cosmological neutrino background (CNB) (a gas).
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

For a relativistic Fermi-gas the Fermi energy reads the following
$E_F = \hbar c\left( 3\pi ^2 n \right)^{1/3} ;$ $\hbar c \simeq 197.3 \, MeV\cdot \, fm$, $n \simeq 300 \, cm^{-3}$, so
$$E_F \simeq 0.4 \times 10^{-3} \, eV.$$
</p>
</div>
</div></div>

Thermodynamical Properties of Elementary Particles

2012-10-01T18:25:39Z

Den: /* Problem 7 */

[[Category:Thermodynamics of Universe|1]]

<pre style="width:70%">
In the mid-forties G.A.~Gamov proposed the idea
of the "hot" origin of the World. Therefore thermodynamics
was introduced into cosmology, and nuclear physics too.
Before him the science of the evolution of Universe
contained only dynamics and geometry of the World.
A.D.Chernin.
</pre>

<div id="tab:ter_sm">
{| cellpadding="5" cellspacing="0" border="1" align="center"
|+ |The Standard Model particles and their properties.
! rowspan="2" | Particles
!rowspan="2" |Mass
!colspan="2" |Number of states
! rowspan="2" width="100" |$g$ (particles and anti-particles)
|-
|spin
|color
|- align="center" |
|Photon ($\gamma$) || 0 || 2 || 1 || 2
|- align="center" |
|$W^+,W^-$ || $80.4\, GeV$ || 3 || 1 || 6
|- align="center" |
|$Z$ || $91.2\,GeV$ || 3 || 1 || 3
|- align="center" |
|Gluon ($g$) || 0||2||8||16
|- align="center" |
| Higgs boson ||$>114\, GeV$||1||1||1
|- align="center" |
|Bosons
| colspan="3" |
| 28
|- align="center" |
| $u,/,\bar{u}$ || $3\, MeV$ || 2||3||12
|- align="center" |
||$d,/,\bar{d}$||$6\, MeV$||2||3||12
|- align="center" |
|$s,/,\bar{s}$||$100\,MeV$||2||3||12
|- align="center" |
|$c,/,\bar{c}$||$1.2\,GeV$||2||3||12
|- align="center" |
|$b,/,\bar{b}$||$4.2\, GeV$||2||3||12
|- align="center" |
|$t,/,\bar{t}$||$175\,GeV$||2||3||12
|- align="center" |
|$e^+,\,e^-$|| $0.511\, MeV $||2||1||4
|- align="center" |
|$\mu^+,\,\mu^-$|| $105.7\,MeV$||2||1||4
|- align="center" |
|$\tau^+,\,\tau^-$|| $1.777\, GeV$||2||1||4
|- align="center" |
|$\nu_e,\,\bar{\nu_e}$|| $<3\,eV$||1||1||2
|- align="center" |
|$\nu_\mu,\,\bar{\nu_\mu}$|| $<0.19\,MeV $||1||1||2
|- align="center" |
|$\nu_\tau,\,\bar{\nu_\tau}$|| $<18.2\, MeV $||1||1||2
|- align="center" |
|Fermions
| colspan="3" |
| 90
|}
</div>

__TOC__

<div id="ter_1"></div>
<div style="border: 1px solid #AAA; padding:5px;">
<div id="ter_1"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 1 ===
Find the energy and number densities for bosons and fermions in the relativistic limit.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The energy density equals to
$$
\rho = \left\{ \begin{array}{c}
\frac{\pi^2}{30}gT^4 \quad \; for\;bosons\\
\frac{7}{8}\frac{\pi^2}{30}gT^4 \quad for \;fermions
\end{array}
\right.
$$
Here $g$ equals to the number of internal degrees of freedom for the particle. For photon (with spin $S=1$) $g=2$ due to absence of longitudinal polarization for the zero-mass particles.
The number density is the following:
\[
n = \left\{ \begin{array}{c}
\frac{\zeta (3)}{\pi ^2 }gT^3 \quad \;for\;bosons \\
\frac{3}
{4}\frac{\zeta (3)}{\pi ^2 }gT^3 \quad \; for\;fermions
\end{array} \right.
\]
Here $\zeta (x)$ is the Riemann $\zeta$-function; $\zeta (3) \approx
1.202$.</p>
</div>
</div></div>

<div id="ter_2"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2 ===
Find the number densities and energy densities, normalized to the photon energy density at the same temperature, for neutrinos, electrons, positrons, pions and nucleons in the relativistic limit.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[
\frac{n_i}{n_\gamma} = \left\{
\begin{gathered}
\frac{1}{2}g_i \quad for\; bosons\\
\frac{3}{8}g_i \quad for\; fermions\\
\end{gathered}
\right.;\quad
\frac{\rho_i}{\rho_\gamma} = \left\{
\begin{gathered}
\frac{1}{2}g_i \quad for\; bosons\\
\frac{7}{16}g_i \quad for\; fermions\\
\end{gathered} \right.
\]
\[
\begin{gathered}
\left(\frac{n_i }{n_\gamma},\frac{\rho_i}{\rho_\gamma} \right)_{\nu_{\mu ,e}} = \left( \frac{3}{8},\frac{7}{16} \right);\;\;\left( \frac{n_i}{n_\gamma},
\frac{\rho_i}{\rho_\gamma} \right)_{e^\pm} = \left(
\frac{3}{4},\frac{7}{8} \right);\\
\left( \frac{n_i}{n_\gamma},\frac{\rho_i}{\rho_\gamma} \right)_\pi = \left( \frac{1}{2},\frac{1}{2} \right);\;\;\left( \frac{n_i}{n_\gamma},\frac{\rho_i}{\rho_\gamma} \right)_{n,p} = \left( \frac{3}{4},\frac{7}{8}
\right) \\
\end{gathered}
\]
</p>
</div>
</div></div>

<div id="ter_3"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 3 ===
Calculate the average energy per particle in the relativistic and non-relativistic limits.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Average energy per particle reads$\left\langle \varepsilon \right\rangle = \rho /n$. For $T \gg m$
\[
\left\langle \varepsilon \right\rangle = \left\{
\begin{gathered}
\frac{\pi ^2}{30\zeta (3)}T\quad \; for\; bosons\\
\frac{7\pi ^4}{180\zeta(3)}T\quad for\; fermions\\
\end{gathered} \right.
\]
In the non-relativistic limit $(T \ll m)$ one obtains the following:
\[
\left\langle \varepsilon \right\rangle = m + \frac{3}{2}T \approx m
\]

</p>
</div>
</div></div>

<div id="ter_4"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 4 ===
Find the number of internal degrees of freedom for a quark.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

For quark one obtains the following:
\[
g = 2(\, spin) \times 3(\, particle) \times 2(\, particles + \, antiparticle) = 12
\]
</p>
</div>
</div></div>

<div id="ter_5n"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 5 ===
Find the entropy density for bosons end fermions with zero chemical potential.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

In the considered case one gets:
\[
s = \frac{\rho + p}{T};\quad p = \frac{1}{3}\rho
\]
Using the results of [[#ter_1|problem]] of the present Chapter one obtains the following:
\[
s = \frac{2\pi ^2}{45}g^* T^3,
\]
where $g^* = 1$ for bosons and $g^* = 7/8$ for fermions.
</p>
</div>
</div></div>

<div id="ter_5"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 6 ===

Find the effective number of internal degrees of freedom for a mixture of relativistic bosons and fermions.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

\[
\rho = \sum\limits_{i = bosons} g_i \frac{\pi^2}{30} T^4 +
\sum\limits_{j = fermions} \frac{7}{8}g_j\frac{\pi^2}{30}
T^4 \frac{\pi^2}{30}g^* T^4,
\]
where $g^* $ is the effective number of degrees of freedom:
\[
g^* = \sum\limits_{i = bosons} g_i + \frac{7}{8}\sum\limits_{j
= fermions}g_j.
\]
</p>
</div>
</div></div>

<div id="therm5"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 7 ===

Generalize the results of the previous problem for the case when some $i$-components have temperature $T_i$ different from the CMB temperature $T$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[
g^* = \sum\limits_{i = bosons}g_i \left(\frac{T_i}{T} \right)^4 + \frac{7}{8}\sum\limits_{j = fermions}g_i \left( \frac{T_j}{T}\right)^4
\] </p>
</div>
</div></div>

<div id="ter_6"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 8 ===
Find the effective number of internal degrees of freedom for the Standard Model particles at temperature $T>1TeV$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;"> At temperature $1\, TeV$ all the Standard Model
particles are relativistic (see [[#tab:ter_sm|table]]). Therefore
\[
g^* = 28 + \frac{7}{8} \times 90 = 106.75.
\] </p>
</div>
</div></div>

<div id="ter_7"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 9 ===
Find the change of the number of internal degrees of freedom due to the quark
hadronization process.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
At temperature $T \approx 0.2\, GeV$ quarks and gluons transform into colorless hadrons (protons and neutrons). Therefore
\[
\Delta g = 16\,(gluons) + 12 \times 6\,(quarks) -
8\,(nucleons) = 80.
\]
</p>
</div>
</div></div>

<div id="ter_8"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 10 ===
Find the relation between the energy density and temperature at
$10^{10}\, K$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Represent the expression for energy density of relativistic particles in the following form:
\[
\rho = \left( \sum\limits_{i = bosons} \frac{g_i}{2} +
\sum\limits_{j = fermions} \frac{7}{16}g_j \right)\rho
_\gamma.
\]
At the temperature under consideration the relativistic particles are presented by photons, electron, positrons and neutrinos. Therefore
\[
\rho = \left( 1 + \frac{7} {4} + \frac{7}{4} \right)\rho _\gamma
= \frac{9}{2}\rho _\gamma = \frac{9}{2}\alpha T^4,
\]
where the radiation constant \[\alpha =
\frac{\pi^2}{15}\frac{k^4}{(\hbar c)^3} \approx 7.56 \cdot
10^{-16}\frac{ J}{ m^3K^4}.\]
</p>
</div>
</div></div>

<div id="ter_9"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 11 ===
Find the ratio of the energy density at temperature $10^{10}\ K$ to that at $10^8\ K$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

Unlike the case considered in the previous problem, now only photons and neutrinos remain relativistic, because $10^8 K \sim 10^{ - 2} \mbox{\it MeV}$, and electron mass is $0.511 \mbox{\it MeV}$ is considerably greater. Therefore in the case under consideration one obtains the following:
\[
\rho = \left( 1 + \frac{7}{4} \right)\rho _\gamma =
\frac{11}{4}\rho _\gamma \Rightarrow
\frac{\rho\left(T_2 = 10^{10} K \right)}{\rho \left( T_1 = 10^8 K \right)} = \frac{18}
{11}\left(\frac{T_2}{T_1}\right)^4\simeq1.6\cdot10^8.\]
</p>
</div>
</div></div>

<div id="razm40"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 12 ===
Determine Fermi energy of cosmological neutrino background (CNB) (a gas).
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

For a relativistic Fermi-gas the Fermi energy reads the following
$E_F = \hbar c\left( 3\pi ^2 n \right)^{1/3} ;$ $\hbar c \simeq 197.3 \, MeV\cdot \, fm$, $n \simeq 300 \, cm^{-3}$, so
$$E_F \simeq 0.4 \times 10^{-3} \, eV.$$
</p>
</div>
</div></div>

Thermodynamical Properties of Elementary Particles

2012-10-01T18:24:43Z

Den:

[[Category:Thermodynamics of Universe|1]]

<pre style="width:70%">
In the mid-forties G.A.~Gamov proposed the idea
of the "hot" origin of the World. Therefore thermodynamics
was introduced into cosmology, and nuclear physics too.
Before him the science of the evolution of Universe
contained only dynamics and geometry of the World.
A.D.Chernin.
</pre>

<div id="tab:ter_sm">
{| cellpadding="5" cellspacing="0" border="1" align="center"
|+ |The Standard Model particles and their properties.
! rowspan="2" | Particles
!rowspan="2" |Mass
!colspan="2" |Number of states
! rowspan="2" width="100" |$g$ (particles and anti-particles)
|-
|spin
|color
|- align="center" |
|Photon ($\gamma$) || 0 || 2 || 1 || 2
|- align="center" |
|$W^+,W^-$ || $80.4\, GeV$ || 3 || 1 || 6
|- align="center" |
|$Z$ || $91.2\,GeV$ || 3 || 1 || 3
|- align="center" |
|Gluon ($g$) || 0||2||8||16
|- align="center" |
| Higgs boson ||$>114\, GeV$||1||1||1
|- align="center" |
|Bosons
| colspan="3" |
| 28
|- align="center" |
| $u,/,\bar{u}$ || $3\, MeV$ || 2||3||12
|- align="center" |
||$d,/,\bar{d}$||$6\, MeV$||2||3||12
|- align="center" |
|$s,/,\bar{s}$||$100\,MeV$||2||3||12
|- align="center" |
|$c,/,\bar{c}$||$1.2\,GeV$||2||3||12
|- align="center" |
|$b,/,\bar{b}$||$4.2\, GeV$||2||3||12
|- align="center" |
|$t,/,\bar{t}$||$175\,GeV$||2||3||12
|- align="center" |
|$e^+,\,e^-$|| $0.511\, MeV $||2||1||4
|- align="center" |
|$\mu^+,\,\mu^-$|| $105.7\,MeV$||2||1||4
|- align="center" |
|$\tau^+,\,\tau^-$|| $1.777\, GeV$||2||1||4
|- align="center" |
|$\nu_e,\,\bar{\nu_e}$|| $<3\,eV$||1||1||2
|- align="center" |
|$\nu_\mu,\,\bar{\nu_\mu}$|| $<0.19\,MeV $||1||1||2
|- align="center" |
|$\nu_\tau,\,\bar{\nu_\tau}$|| $<18.2\, MeV $||1||1||2
|- align="center" |
|Fermions
| colspan="3" |
| 90
|}
</div>

__TOC__

<div id="ter_1"></div>
<div style="border: 1px solid #AAA; padding:5px;">
<div id="ter_1"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 1 ===
Find the energy and number densities for bosons and fermions in the relativistic limit.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The energy density equals to
$$
\rho = \left\{ \begin{array}{c}
\frac{\pi^2}{30}gT^4 \quad \; for\;bosons\\
\frac{7}{8}\frac{\pi^2}{30}gT^4 \quad for \;fermions
\end{array}
\right.
$$
Here $g$ equals to the number of internal degrees of freedom for the particle. For photon (with spin $S=1$) $g=2$ due to absence of longitudinal polarization for the zero-mass particles.
The number density is the following:
\[
n = \left\{ \begin{array}{c}
\frac{\zeta (3)}{\pi ^2 }gT^3 \quad \;for\;bosons \\
\frac{3}
{4}\frac{\zeta (3)}{\pi ^2 }gT^3 \quad \; for\;fermions
\end{array} \right.
\]
Here $\zeta (x)$ is the Riemann $\zeta$-function; $\zeta (3) \approx
1.202$.</p>
</div>
</div></div>

<div id="ter_2"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2 ===
Find the number densities and energy densities, normalized to the photon energy density at the same temperature, for neutrinos, electrons, positrons, pions and nucleons in the relativistic limit.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[
\frac{n_i}{n_\gamma} = \left\{
\begin{gathered}
\frac{1}{2}g_i \quad for\; bosons\\
\frac{3}{8}g_i \quad for\; fermions\\
\end{gathered}
\right.;\quad
\frac{\rho_i}{\rho_\gamma} = \left\{
\begin{gathered}
\frac{1}{2}g_i \quad for\; bosons\\
\frac{7}{16}g_i \quad for\; fermions\\
\end{gathered} \right.
\]
\[
\begin{gathered}
\left(\frac{n_i }{n_\gamma},\frac{\rho_i}{\rho_\gamma} \right)_{\nu_{\mu ,e}} = \left( \frac{3}{8},\frac{7}{16} \right);\;\;\left( \frac{n_i}{n_\gamma},
\frac{\rho_i}{\rho_\gamma} \right)_{e^\pm} = \left(
\frac{3}{4},\frac{7}{8} \right);\\
\left( \frac{n_i}{n_\gamma},\frac{\rho_i}{\rho_\gamma} \right)_\pi = \left( \frac{1}{2},\frac{1}{2} \right);\;\;\left( \frac{n_i}{n_\gamma},\frac{\rho_i}{\rho_\gamma} \right)_{n,p} = \left( \frac{3}{4},\frac{7}{8}
\right) \\
\end{gathered}
\]
</p>
</div>
</div></div>

<div id="ter_3"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 3 ===
Calculate the average energy per particle in the relativistic and non-relativistic limits.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Average energy per particle reads$\left\langle \varepsilon \right\rangle = \rho /n$. For $T \gg m$
\[
\left\langle \varepsilon \right\rangle = \left\{
\begin{gathered}
\frac{\pi ^2}{30\zeta (3)}T\quad \; for\; bosons\\
\frac{7\pi ^4}{180\zeta(3)}T\quad for\; fermions\\
\end{gathered} \right.
\]
In the non-relativistic limit $(T \ll m)$ one obtains the following:
\[
\left\langle \varepsilon \right\rangle = m + \frac{3}{2}T \approx m
\]

</p>
</div>
</div></div>

<div id="ter_4"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 4 ===
Find the number of internal degrees of freedom for a quark.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

For quark one obtains the following:
\[
g = 2(\, spin) \times 3(\, particle) \times 2(\, particles + \, antiparticle) = 12
\]
</p>
</div>
</div></div>

<div id="ter_5n"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 5 ===
Find the entropy density for bosons end fermions with zero chemical potential.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

In the considered case one gets:
\[
s = \frac{\rho + p}{T};\quad p = \frac{1}{3}\rho
\]
Using the results of [[#ter_1|problem]] of the present Chapter one obtains the following:
\[
s = \frac{2\pi ^2}{45}g^* T^3,
\]
where $g^* = 1$ for bosons and $g^* = 7/8$ for fermions.
</p>
</div>
</div></div>

<div id="ter_5"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 6 ===

Find the effective number of internal degrees of freedom for a mixture of relativistic bosons and fermions.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

\[
\rho = \sum\limits_{i = bosons} g_i \frac{\pi^2}{30} T^4 +
\sum\limits_{j = fermions} \frac{7}{8}g_j\frac{\pi^2}{30}
T^4 \frac{\pi^2}{30}g^* T^4,
\]
where $g^* $ is the effective number of degrees of freedom:
\[
g^* = \sum\limits_{i = bosons} g_i + \frac{7}{8}\sum\limits_{j
= fermions}g_j.
\]
</p>
</div>
</div></div>

<div id="therm5"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 7 ===

Generalize the results of the previous problem for the case when some $i$-components have temperature $T_i$ different from the CMB temperature $T$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[
g^* = \sum\limits_{i = bosons} {g_i \left(\frac{T_i}{T} \right)^4 + \frac{7}{8}\sum\limits_{j = fermions}g_i \left( \frac{T_j}{T}\right)^4
\] </p>
</div>
</div></div>

<div id="ter_6"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 8 ===
Find the effective number of internal degrees of freedom for the Standard Model particles at temperature $T>1TeV$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;"> At temperature $1\, TeV$ all the Standard Model
particles are relativistic (see [[#tab:ter_sm|table]]). Therefore
\[
g^* = 28 + \frac{7}{8} \times 90 = 106.75.
\] </p>
</div>
</div></div>

<div id="ter_7"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 9 ===
Find the change of the number of internal degrees of freedom due to the quark
hadronization process.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
At temperature $T \approx 0.2\, GeV$ quarks and gluons transform into colorless hadrons (protons and neutrons). Therefore
\[
\Delta g = 16\,(gluons) + 12 \times 6\,(quarks) -
8\,(nucleons) = 80.
\]
</p>
</div>
</div></div>

<div id="ter_8"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 10 ===
Find the relation between the energy density and temperature at
$10^{10}\, K$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Represent the expression for energy density of relativistic particles in the following form:
\[
\rho = \left( \sum\limits_{i = bosons} \frac{g_i}{2} +
\sum\limits_{j = fermions} \frac{7}{16}g_j \right)\rho
_\gamma.
\]
At the temperature under consideration the relativistic particles are presented by photons, electron, positrons and neutrinos. Therefore
\[
\rho = \left( 1 + \frac{7} {4} + \frac{7}{4} \right)\rho _\gamma
= \frac{9}{2}\rho _\gamma = \frac{9}{2}\alpha T^4,
\]
where the radiation constant \[\alpha =
\frac{\pi^2}{15}\frac{k^4}{(\hbar c)^3} \approx 7.56 \cdot
10^{-16}\frac{ J}{ m^3K^4}.\]
</p>
</div>
</div></div>

<div id="ter_9"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 11 ===
Find the ratio of the energy density at temperature $10^{10}\ K$ to that at $10^8\ K$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

Unlike the case considered in the previous problem, now only photons and neutrinos remain relativistic, because $10^8 K \sim 10^{ - 2} \mbox{\it MeV}$, and electron mass is $0.511 \mbox{\it MeV}$ is considerably greater. Therefore in the case under consideration one obtains the following:
\[
\rho = \left( 1 + \frac{7}{4} \right)\rho _\gamma =
\frac{11}{4}\rho _\gamma \Rightarrow
\frac{\rho\left(T_2 = 10^{10} K \right)}{\rho \left( T_1 = 10^8 K \right)} = \frac{18}
{11}\left(\frac{T_2}{T_1}\right)^4\simeq1.6\cdot10^8.\]
</p>
</div>
</div></div>

<div id="razm40"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 12 ===
Determine Fermi energy of cosmological neutrino background (CNB) (a gas).
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

For a relativistic Fermi-gas the Fermi energy reads the following
$E_F = \hbar c\left( 3\pi ^2 n \right)^{1/3} ;$ $\hbar c \simeq 197.3 \, MeV\cdot \, fm$, $n \simeq 300 \, cm^{-3}$, so
$$E_F \simeq 0.4 \times 10^{-3} \, eV.$$
</p>
</div>
</div></div>

Thermodynamical Properties of Elementary Particles

2012-10-01T18:07:32Z

Den:

[[Category:Thermodynamics of Universe|1]]

<pre style="width:70%">
In the mid-forties G.A.~Gamov proposed the idea
of the "hot" origin of the World. Therefore thermodynamics
was introduced into cosmology, and nuclear physics too.
Before him the science of the evolution of Universe
contained only dynamics and geometry of the World.
A.D.Chernin.
</pre>

<div id="tab:ter_sm">
{| cellpadding="5" cellspacing="0" border="1" align="center"
|+ |The Standard Model particles and their properties.==
! rowspan="2" | Particles
!rowspan="2" |Mass
!colspan="2" |Number of states
! rowspan="2" width="100" |$g$ (particles and anti-particles)
|-
|spin
|color
|- align="center" |
|Photon ($\gamma$) || 0 || 2 || 1 || 2
|- align="center" |
|$W^+,W^-$ || $80.4\, GeV$ || 3 || 1 || 6
|- align="center" |
|$Z$ || $91.2\,GeV$ || 3 || 1 || 3
|- align="center" |
|Gluon ($g$) || 0||2||8||16
|- align="center" |
| Higgs boson ||$>114\, GeV$||1||1||1
|- align="center" |
|Bosons
| colspan="3" |
| 28
|- align="center" |
| $u,/,\bar{u}$ || $3\, MeV$ || 2||3||12
|- align="center" |
||$d,/,\bar{d}$||$6\, MeV$||2||3||12
|- align="center" |
|$s,/,\bar{s}$||$100\,MeV$||2||3||12
|- align="center" |
|$c,/,\bar{c}$||$1.2\,GeV$||2||3||12
|- align="center" |
|$b,/,\bar{b}$||$4.2\, GeV$||2||3||12
|- align="center" |
|$t,/,\bar{t}$||$175\,GeV$||2||3||12
|- align="center" |
|$e^+,\,e^-$|| $0.511\, MeV $||2||1||4
|- align="center" |
|$\mu^+,\,\mu^-$|| $105.7\,MeV$||2||1||4
|- align="center" |
|$\tau^+,\,\tau^-$|| $1.777\, GeV$||2||1||4
|- align="center" |
|$\nu_e,\,\bar{\nu_e}$|| $<3\,eV$||1||1||2
|- align="center" |
|$\nu_\mu,\,\bar{\nu_\mu}$|| $<0.19\,MeV $||1||1||2
|- align="center" |
|$\nu_\tau,\,\bar{\nu_\tau}$|| $<18.2\, MeV $||1||1||2
|- align="center" |
|Fermions
| colspan="3" |
| 90
|}
</div>

__TOC__

<div id="лейбел"></div>
<div style="border: 1px solid #AAA; padding:5px;">
<div id="ter_1"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 1 ===
Find the energy and number densities for bosons and fermions in the relativistic limit.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The energy density equals to
$$
\rho = \left\{ \begin{array}{c}
\frac{\pi^2}{30}gT^4 \quad \; for\;bosons\\
\frac{7}{8}\frac{\pi^2}{30}gT^4 \quad for \;fermions
\end{array}
\right.
$$
Here $g$ equals to the number of internal degrees of freedom for the particle. For photon (with spin $S=1$) $g=2$ due to absence of longitudinal polarization for the zero-mass particles.
The number density is the following:
\[
n = \left\{ \begin{array}{c}
\frac{\zeta (3)}{\pi ^2 }gT^3 \quad \;for\;bosons \\
\frac{3}
{4}\frac{\zeta (3)}{\pi ^2 }gT^3 \quad \; for\;fermions
\end{array} \right.
\]
Here $\zeta (x)$ is the Riemann $\zeta$-function; $\zeta (3) \approx
1.202$.</p>
</div>
</div></div>

<div id="ter_2"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2 ===
Find the number densities and energy densities, normalized to the photon energy density at the same temperature, for neutrinos, electrons, positrons, pions and nucleons in the relativistic limit.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

\[
\frac{n_i}{n_\gamma} = \left\{
\begin{gathered}
\frac{1}{2}g_i \quad for\; bosons\\
\frac{3}{8}g_i \quad for\; fermions\\
\end{gathered}
\right.;\quad
\frac{\rho_i}{\rho_\gamma} = \left\{
\begin{gathered}
\frac{1}{2}g_i \quad for\; bosons\\
\frac{7}{16}g_i \quad for\; fermions\\
\end{gathered} \right.
\]

\[
\begin{gathered}
\left(\frac{n_i }{n_\gamma},\frac{\rho_i}{\rho_\gamma} \right)_{\nu_{\mu ,e}} = \left( \frac{3}{8},\frac{7}{16} \right);\;\;\left( \frac{n_i}{n_\gamma},
\frac{\rho_i}{\rho_\gamma} \right)_{e^\pm} = \left(
\frac{3}{4},\frac{7}{8} \right);\\
\left( \frac{n_i}{n_\gamma},\frac{\rho_i}{\rho_\gamma} \right)_\pi = \left( \frac{1}{2},\frac{1}{2} \right);\;\;\left( \frac{n_i}{n_\gamma},\frac{\rho_i}{\rho_\gamma} \right)_{n,p} = \left( \frac{3}{4},\frac{7}{8}
\right) \\
\end{gathered}
\]

</p>
</div>
</div></div>

<div id="ter_3"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 3 ===
Calculate the average energy per particle in the relativistic and non-relativistic limits.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Average energy per particle reads$\left\langle \varepsilon \right\rangle = \rho /n$. For $T \gg m$
\[
\left\langle \varepsilon \right\rangle = \left\{
\begin{gathered}
\frac{\pi ^2}{30\zeta (3)}T\quad \; for\; bosons\\
\frac{7\pi ^4}{180\zeta(3)}T\quad for\; fermions\\
\end{gathered} \right.
\]
In the non-relativistic limit $(T \ll m)$ one obtains the following:
\[
\left\langle \varepsilon \right\rangle = m + \frac{3}{2}T \approx m
\]

</p>
</div>
</div></div>

<div id="ter_4"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 4 ===
Find the number of internal degrees of freedom for a quark.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

For quark one obtains the following:
\[
g = 2(\, spin) \times 3(\, particle) \times 2(\, particles + \, antiparticle) = 12
\]
</p>
</div>
</div></div>

<div id="ter_5n"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 5 ===
Find the entropy density for bosons end fermions with zero chemical potential.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

In the considered case one gets:
\[
s = \frac{\rho + p}{T};\quad p = \frac{1}{3}\rho
\]
Using the results of [[#ter_1|problem]] of the present Chapter one obtains the following:
\[
s = \frac{2\pi ^2}{45}g^* T^3,
\]
where $g^* = 1$ for bosons and $g^* = 7/8$ for fermions.
</p>
</div>
</div></div>

<div id="ter_5"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 6 ===

Find the effective number of internal degrees of freedom for a mixture of relativistic bosons and fermions.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

\[
\rho = \sum\limits_{i = bosons} g_i \frac{\pi^2}{30} T^4 +
\sum\limits_{j = fermions} \frac{7}{8}g_j\frac{\pi^2}{30}
T^4 \frac{\pi^2}{30}g^* T^4,
\]
where $g^* $ is the effective number of degrees of freedom:
\[
g^* = \sum\limits_{i = bosons} g_i + \frac{7}{8}\sum\limits_{j
= fermions}g_j.
\]
</p>
</div>
</div></div>

<div id="therm5"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 7 ===

Generalize the results of the previous problem for the case when some $i$-components have temperature $T_i$ different from the CMB temperature $T$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
\[
g^* = \sum\limits_{i = bosons} {g_i \left(\frac{T_i}{T} \right)^4 + \frac{7}{8}\sum\limits_{j = fermions}g_i \left( \frac{T_j}{T}\right)^4
\] </p>
</div>
</div></div>

<div id="ter_6"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 8 ===
Find the effective number of internal degrees of freedom for the Standard Model particles at temperature $T>1TeV$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;"> At temperature $1\, TeV$ all the Standard Model
particles are relativistic (see [[#tab:ter_sm|table]]). Therefore
\[
g^* = 28 + \frac{7}{8} \times 90 = 106.75.
\] </p>
</div>
</div></div>

<div id="ter_1"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2 ===

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;"> </p>
</div>
</div></div>

<div id="ter_1"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2 ===

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;"> </p>
</div>
</div></div>

Thermodynamical Properties of Elementary Particles

2012-10-01T18:01:22Z

Den: /* Problem 2 */

[[Category:Thermodynamics of Universe|1]]

<pre style="width:70%">
In the mid-forties G.A.~Gamov proposed the idea
of the "hot" origin of the World. Therefore thermodynamics
was introduced into cosmology, and nuclear physics too.
Before him the science of the evolution of Universe
contained only dynamics and geometry of the World.
A.D.Chernin.
</pre>

{| cellpadding="5" cellspacing="0" border="1" align="center"
|+ |The Standard Model particles and their properties.==
! rowspan="2" | Particles
!rowspan="2" |Mass
!colspan="2" |Number of states
! rowspan="2" width="100" |$g$ (particles and anti-particles)
|-
|spin
|color
|- align="center" |
|Photon ($\gamma$) || 0 || 2 || 1 || 2
|- align="center" |
|$W^+,W^-$ || $80.4\, GeV$ || 3 || 1 || 6
|- align="center" |
|$Z$ || $91.2\,GeV$ || 3 || 1 || 3
|- align="center" |
|Gluon ($g$) || 0||2||8||16
|- align="center" |
| Higgs boson ||$>114\, GeV$||1||1||1
|- align="center" |
|Bosons
| colspan="3" |
| 28
|- align="center" |
| $u,/,\bar{u}$ || $3\, MeV$ || 2||3||12
|- align="center" |
||$d,/,\bar{d}$||$6\, MeV$||2||3||12
|- align="center" |
|$s,/,\bar{s}$||$100\,MeV$||2||3||12
|- align="center" |
|$c,/,\bar{c}$||$1.2\,GeV$||2||3||12
|- align="center" |
|$b,/,\bar{b}$||$4.2\, GeV$||2||3||12
|- align="center" |
|$t,/,\bar{t}$||$175\,GeV$||2||3||12
|- align="center" |
|$e^+,\,e^-$|| $0.511\, MeV $||2||1||4
|- align="center" |
|$\mu^+,\,\mu^-$|| $105.7\,MeV$||2||1||4
|- align="center" |
|$\tau^+,\,\tau^-$|| $1.777\, GeV$||2||1||4
|- align="center" |
|$\nu_e,\,\bar{\nu_e}$|| $<3\,eV$||1||1||2
|- align="center" |
|$\nu_\mu,\,\bar{\nu_\mu}$|| $<0.19\,MeV $||1||1||2
|- align="center" |
|$\nu_\tau,\,\bar{\nu_\tau}$|| $<18.2\, MeV $||1||1||2
|- align="center" |
|Fermions
| colspan="3" |
| 90
|}

__TOC__

<div id="лейбел"></div>
<div style="border: 1px solid #AAA; padding:5px;">
<div id="ter_1"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 1 ===
Find the energy and number densities for bosons and fermions in the relativistic limit.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The energy density equals to
$$
\rho = \left\{ \begin{array}{c}
\frac{\pi^2}{30}gT^4 \quad \; for\;bosons\\
\frac{7}{8}\frac{\pi^2}{30}gT^4 \quad for \;fermions
\end{array}
\right.
$$
Here $g$ equals to the number of internal degrees of freedom for the particle. For photon (with spin $S=1$) $g=2$ due to absence of longitudinal polarization for the zero-mass particles.
The number density is the following:
\[
n = \left\{ \begin{array}{c}
\frac{\zeta (3)}{\pi ^2 }gT^3 \quad \;for\;bosons \\
\frac{3}
{4}\frac{\zeta (3)}{\pi ^2 }gT^3 \quad \; for\;fermions
\end{array} \right.
\]
Here $\zeta (x)$ is the Riemann $\zeta$-function; $\zeta (3) \approx
1.202$.</p>
</div>
</div></div>

<div id="ter_2"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2 ===
Find the number densities and energy densities, normalized to the photon energy density at the same temperature, for neutrinos, electrons, positrons, pions and nucleons in the relativistic limit.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

\[
\frac{n_i}{n_\gamma} = \left\{
\begin{gathered}
\frac{1}{2}g_i \quad for\; bosons\\
\frac{3}{8}g_i \quad for\; fermions\\
\end{gathered}
\right.;\quad
\frac{\rho_i}{\rho_\gamma} = \left\{
\begin{gathered}
\frac{1}{2}g_i \quad for\; bosons\\
\frac{7}{16}g_i \quad for\; fermions\\
\end{gathered} \right.
\]

\[
\begin{gathered}
\left(\frac{n_i }{n_\gamma},\frac{\rho_i}{\rho_\gamma} \right)_{\nu_{\mu ,e}} = \left( \frac{3}{8},\frac{7}{16} \right);\;\;\left( \frac{n_i}{n_\gamma},
\frac{\rho_i}{\rho_\gamma} \right)_{e^\pm} = \left(
\frac{3}{4},\frac{7}{8} \right);\\
\left( \frac{n_i}{n_\gamma},\frac{\rho_i}{\rho_\gamma} \right)_\pi = \left( \frac{1}{2},\frac{1}{2} \right);\;\;\left( \frac{n_i}{n_\gamma},\frac{\rho_i}{\rho_\gamma} \right)_{n,p} = \left( \frac{3}{4},\frac{7}{8}
\right) \\
\end{gathered}
\]

</p>
</div>
</div></div>

<div id="ter_3"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 3 ===
Calculate the average energy per particle in the relativistic and non-relativistic limits.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Average energy per particle reads$\left\langle \varepsilon \right\rangle = \rho /n$. For $T \gg m$
\[
\left\langle \varepsilon \right\rangle = \left\{
\begin{gathered}
\frac{\pi ^2}{30\zeta (3)}T\quad \; for\; bosons\\
\frac{7\pi ^4}{180\zeta(3)}T\quad for\; fermions\\
\end{gathered} \right.
\]
In the non-relativistic limit $(T \ll m)$ one obtains the following:
\[
\left\langle \varepsilon \right\rangle = m + \frac{3}{2}T \approx m
\]

</p>
</div>
</div></div>

<div id="ter_4"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 4 ===
Find the number of internal degrees of freedom for a quark.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

For quark one obtains the following:
\[
g = 2(\, spin) \times 3(\, particle) \times 2(\, particles + \, antiparticle) = 12
\]
</p>
</div>
</div></div>

<div id="ter_5n"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 5 ===
Find the entropy density for bosons end fermions with zero chemical potential.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

In the considered case one gets:
\[
s = \frac{\rho + p}{T};\quad p = \frac{1}{3}\rho
\]
Using the results of [[#ter_1|problem]] of the present Chapter one obtains the following:
\[
s = \frac{2\pi ^2}{45}g^* T^3,
\]
where $g^* = 1$ for bosons and $g^* = 7/8$ for fermions.
</p>
</div>
</div></div>

<div id="ter_5"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 6 ===

Find the effective number of internal degrees of freedom for a mixture of relativistic bosons and fermions.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

\[
\rho = \sum\limits_{i = bosons} g_i \frac{\pi^2}{30} T^4 +
\sum\limits_{j = fermions} \frac{7}{8}g_j\frac{\pi^2}{30}
T^4 \frac{\pi^2}{30}g^* T^4,
\]
where $g^* $ is the effective number of degrees of freedom:
\[
g^* = \sum\limits_{i = bosons} g_i + \frac{7}{8}\sum\limits_{j
= fermions}g_j.
\]
</p>
</div>
</div></div>

<div id="ter_1"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2 ===

Generalize the results of the previous problem for the case when some $i$-components have temperature $T_i$ different from the CMB temperature $T$.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;"> </p>
</div>
</div></div>

<div id="therm5"></div>
<div style="border: 1px solid #AAA; padding:5px;">

=== Problem 2 ===

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;"> </p>
</div>
</div></div>

<div id="ter_1"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2 ===

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;"> </p>
</div>
</div></div>

<div id="ter_1"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2 ===

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;"> </p>
</div>
</div></div>

Thermodynamical Properties of Elementary Particles

2012-10-01T17:59:28Z

Den:

[[Category:Thermodynamics of Universe|1]]

<pre style="width:70%">
In the mid-forties G.A.~Gamov proposed the idea
of the "hot" origin of the World. Therefore thermodynamics
was introduced into cosmology, and nuclear physics too.
Before him the science of the evolution of Universe
contained only dynamics and geometry of the World.
A.D.Chernin.
</pre>

{| cellpadding="5" cellspacing="0" border="1" align="center"
|+ |The Standard Model particles and their properties.==
! rowspan="2" | Particles
!rowspan="2" |Mass
!colspan="2" |Number of states
! rowspan="2" width="100" |$g$ (particles and anti-particles)
|-
|spin
|color
|- align="center" |
|Photon ($\gamma$) || 0 || 2 || 1 || 2
|- align="center" |
|$W^+,W^-$ || $80.4\, GeV$ || 3 || 1 || 6
|- align="center" |
|$Z$ || $91.2\,GeV$ || 3 || 1 || 3
|- align="center" |
|Gluon ($g$) || 0||2||8||16
|- align="center" |
| Higgs boson ||$>114\, GeV$||1||1||1
|- align="center" |
|Bosons
| colspan="3" |
| 28
|- align="center" |
| $u,/,\bar{u}$ || $3\, MeV$ || 2||3||12
|- align="center" |
||$d,/,\bar{d}$||$6\, MeV$||2||3||12
|- align="center" |
|$s,/,\bar{s}$||$100\,MeV$||2||3||12
|- align="center" |
|$c,/,\bar{c}$||$1.2\,GeV$||2||3||12
|- align="center" |
|$b,/,\bar{b}$||$4.2\, GeV$||2||3||12
|- align="center" |
|$t,/,\bar{t}$||$175\,GeV$||2||3||12
|- align="center" |
|$e^+,\,e^-$|| $0.511\, MeV $||2||1||4
|- align="center" |
|$\mu^+,\,\mu^-$|| $105.7\,MeV$||2||1||4
|- align="center" |
|$\tau^+,\,\tau^-$|| $1.777\, GeV$||2||1||4
|- align="center" |
|$\nu_e,\,\bar{\nu_e}$|| $<3\,eV$||1||1||2
|- align="center" |
|$\nu_\mu,\,\bar{\nu_\mu}$|| $<0.19\,MeV $||1||1||2
|- align="center" |
|$\nu_\tau,\,\bar{\nu_\tau}$|| $<18.2\, MeV $||1||1||2
|- align="center" |
|Fermions
| colspan="3" |
| 90
|}

__TOC__

<div id="лейбел"></div>
<div style="border: 1px solid #AAA; padding:5px;">
<div id="ter_1"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 1 ===
Find the energy and number densities for bosons and fermions in the relativistic limit.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
The energy density equals to
$$
\rho = \left\{ \begin{array}{c}
\frac{\pi^2}{30}gT^4 \quad \; for\;bosons\\
\frac{7}{8}\frac{\pi^2}{30}gT^4 \quad for \;fermions
\end{array}
\right.
$$
Here $g$ equals to the number of internal degrees of freedom for the particle. For photon (with spin $S=1$) $g=2$ due to absence of longitudinal polarization for the zero-mass particles.
The number density is the following:
\[
n = \left\{ \begin{array}{c}
\frac{\zeta (3)}{\pi ^2 }gT^3 \quad \;for\;bosons \\
\frac{3}
{4}\frac{\zeta (3)}{\pi ^2 }gT^3 \quad \; for\;fermions
\end{array} \right.
\]
Here $\zeta (x)$ is the Riemann $\zeta$-function; $\zeta (3) \approx
1.202$.</p>
</div>
</div></div>

<div id="ter_2"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2 ===
Find the number densities and energy densities, normalized to the photon energy density at the same temperature, for neutrinos, electrons, positrons, pions and nucleons in the relativistic limit.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

\[
\frac{n_i}{n_\gamma} = \left\{
\begin{gathered}
\frac{1}{2}g_i \quad for\; bosons\\
\frac{3}{8}g_i \quad for\; fermions\\
\end{gathered}
\right.;\quad
\frac{\rho_i}{\rho_\gamma} = \left\{
\begin{gathered}
\frac{1}{2}g_i \quad for\; bosons\\
\frac{7}{16}g_i \quad for\; fermions\\
\end{gathered} \right.
\]

\[
\begin{gathered}
\left(\frac{n_i }{n_\gamma},\frac{\rho_i}{\rho_\gamma} \right)_{\nu_{\mu ,e}} = \left( \frac{3}{8},\frac{7}{16} \right);\;\;\left( \frac{n_i}{n_\gamma},
\frac{\rho_i}{\rho_\gamma} \right)_{e^\pm} = \left(
\frac{3}{4},\frac{7}{8} \right);\\
\left( \frac{n_i}{n_\gamma},\frac{\rho_i}{\rho_\gamma} \right)_\pi = \left( \frac{1}{2},\frac{1}{2} \right);\;\;\left( \frac{n_i}{n_\gamma},\frac{\rho_i}{\rho_\gamma} \right)_{n,p} = \left( \frac{3}{4},\frac{7}{8}
\right) \\
\end{gathered}
\]

</p>
</div>
</div></div>

<div id="ter_3"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 3 ===
Calculate the average energy per particle in the relativistic and non-relativistic limits.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">
Average energy per particle reads$\left\langle \varepsilon \right\rangle = \rho /n$. For $T \gg m$
\[
\left\langle \varepsilon \right\rangle = \left\{
\begin{gathered}
\frac{\pi ^2}{30\zeta (3)}T\quad \; for\; bosons\\
\frac{7\pi ^4}{180\zeta(3)}T\quad for\; fermions\\
\end{gathered} \right.
\]
In the non-relativistic limit $(T \ll m)$ one obtains the following:
\[
\left\langle \varepsilon \right\rangle = m + \frac{3}{2}T \approx m
\]

</p>
</div>
</div></div>

<div id="ter_4"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 4 ===
Find the number of internal degrees of freedom for a quark.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

For quark one obtains the following:
\[
g = 2(\, spin) \times 3(\, particle) \times 2(\, particles + \, antiparticle) = 12
\]
</p>
</div>
</div></div>

<div id="ter_5n"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 5 ===
Find the entropy density for bosons end fermions with zero chemical potential.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

In the considered case one gets:
\[
s = \frac{\rho + p}{T};\quad p = \frac{1}{3}\rho
\]
Using the results of [[#ter_1|problem]] of the present Chapter one obtains the following:
\[
s = \frac{2\pi ^2}{45}g^* T^3,
\]
where $g^* = 1$ for bosons and $g^* = 7/8$ for fermions.
</p>
</div>
</div></div>

<div id="ter_5"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 6 ===

Find the effective number of internal degrees of freedom for a mixture of relativistic bosons and fermions.

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">

\[
\rho = \sum\limits_{i = bosons} g_i \frac{\pi^2}{30} T^4 +
\sum\limits_{j = fermions} \frac{7}{8}g_j\frac{\pi^2}{30}
T^4 \frac{\pi^2}{30}g^* T^4,
\]
where $g^* $ is the effective number of degrees of freedom:
\[
g^* = \sum\limits_{i = bosons} g_i + \frac{7}{8}\sum\limits_{j
= fermions}g_j.
\]
</p>
</div>
</div></div>

<div id="ter_1"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2 ===

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;"> </p>
</div>
</div></div>

<div id="ter_1"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2 ===

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;"> </p>
</div>
</div></div>

<div id="ter_1"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2 ===

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;"> </p>
</div>
</div></div>

<div id="ter_1"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2 ===

<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;"> </p>
</div>
</div></div>

Thermodynamical Properties of Elementary Particles

2012-10-01T17:52:45Z

Den: /* Problem 1 */

Thermodynamical Properties of Elementary Particles

2012-10-01T13:53:15Z

Den:

Thermodynamical Properties of Elementary Particles

2012-10-01T13:52:49Z

Den:

Thermodynamical Properties of Elementary Particles

2012-10-01T13:51:57Z

Den: