New from march

New from March-2015

1

Problem 1

problem id: 2501_01

Show that the luminosity distance can be in general presented in the following form \[d_L=\frac{1+z}{H_0\sqrt{\Omega_{k0}}}\sinh\left({H_0\sqrt{\Omega_{k0}}}\int\limits_0^z\frac{dz}{H(z)}\right),\] where $\Omega_{k0}$ is relative contribution of the space curvature.

solution

Use the relation \[\rho_{curv}=-\frac3{8\pi G}\frac{k}{R^2},\] where $R$ is the (comoving) radius of curvature of the open or closed Universe, to find \[\Omega_{k0}=-\frac{k}{RH_0}.\] It is easy to see that the expression \[d_L=\frac{1+z}{H_0\sqrt{\Omega_{k0}}}\sinh\left({H_0\sqrt{\Omega_{k0}}}\int\limits_0^z\frac{dz}{H(z)}\right)\] reproduces the commonly known expression for the luminosity distance \[ d_L(z)=(1+z)\left\{ \begin{array}{lr} R\sinh\left[\frac1{H_0R}\int\limits_0^z\frac{dz'}{E(z')}\right], & open\\ H_0^{-1}\int\limits_0^z\frac{dz'}{E(z')}, & flat\\ R\sin\left[\frac1{H_0R}\int\limits_0^z\frac{dz'}{E(z')}\right], & closed \end{array} \right.\]

2

Problem 2

problem id: 2501_02

Give physical interpretation of the conservation equation \[\dot{\rho}+3(\rho+p)H=0.\]

solution

The energy density has time dependence determined by the conservation equation \[\dot{\rho}=-3H\rho-3Hp.\] The two terms determine behavior of the homogeneous fluid which contains the energy in a dynamic Universe. The $H$ term provides the "friction", while the density term tracks the reduction in density due to the volume increase during expansion and the pressure term tracks the reduction in pressure energy during expansion.

3

Problem 3

problem id: 2501_03

Show that for both case of matter and radiation domination, the acceleration $\ddot a$ slows as the scale factor grows.

solution

From the second Friedmann equation \[\frac{\ddot a}{a}=-\frac{4\pi G}{3}(\rho+3p)\] clearly shows that both for matter and for radiation domination, the acceleration is negative; an expanding Universe dominated by the energy density of matter or radiation will decelerate. The acceleration in a matter dominated phase goes as $a^{-2}$ while in a radiation dominated phase it goes as $a^{-3}$. In either case the deceleration is large when $a(t)$ is small and then slows as the scale factor grows.

4

Problem 4

problem id: 2501_04

Obtain relation between the cosmological and conformal time for the Universe dominated by matter, radiation and cosmological constant respectively.}

solution

	$\rho(a)$	$a(t)$	$\eta$	$H$
matter	$1/a^3$	$t^{2/3}$	$t^{1/3}$	$(2/3)/t$
radiation	$1/a^4$	$t^{1/2}$	$t^{1/2}$	$(1/2)/t$
$\Lambda$	constant	$e^{Ht}$	$C(e^{Ht}-1)/(aH)$	constant

5

Problem 5

problem id: 2501_05

Show that for the power-paw expansion $a(t)\propto t^\alpha$ with $\alpha<1$ (decelerated expansion) the Hubble radius grows faster than the Universe expands.

solution

Indeed, in the case of the power-paw expansion with $\alpha<1$ (decelerated expansion) the Hubble radius grows faster than the Universe expands: $R_H=H^{-1}\propto t$ while $a(t)\propto t^\alpha$. In power law situations the Hubble radius has expansion velocity greater than the light speed \[\frac{d}{dt}\left(\frac1H\right)=\frac1\alpha.\] This behavior is true only for a decelerating Universe composed of matter and radiation. This is not a physical velocity, violating special relativity, but the velocity of expansion of the metric itself.

6

Problem 6

problem id: 2501_06

Why the cosmological constant cannot be used as a source for inflation in the inflation model?

solution

The cosmological constant do provide the accelerated expansion of Universe needed to realize the inflation: $a(t)\propto e^{Ht}$, $q=-1$. However, in approximation of the cosmological constant, $H$ is constant for all time. Therefore a dynamical mechanism for the limited time of inflation is needed. The physical mechanism for the existence of an approximately constant value of $H$ which lasts for a limited time is given by a scalar field. For a large initial potential energy of scalar field the state equation parameter $w\approx-1$ and the scalar field in process of the "slow roll" imitates the cosmological constant for sufficiently long period of time to solve the flatness and causal problems. Then due to shape of the potential the scalar field exits from the slow-roll regime, oscillates about its' potential minimum decaying into less massive particles insuring that inflation time is finite.

7

Problem 7

problem id: 2501_07

Derive the following formula \[\Delta v\equiv\frac{\Delta z}{1+z}=H_0\Delta t_o \left[1-\frac{E(z)}{1+z}\right].\]

solution

9

Problem 9

problem id: 2501_09

Show that inflation ends when the parameter \[\varepsilon\equiv\frac{M_P^2}{16\pi}\left(\frac{dV}{d\varphi}\frac1V\right)^2=1.\]

solution

Using definition of the parameter $\varepsilon$ one finds \[\varepsilon=-\frac{\dot H}{H^2}.\] In the slow-roll approximation \[H^2=\frac{8\pi}{3M_P^2V},\] \[3H\dot\varphi=-\frac{dV}{d\varphi}.\] Therefore \[\frac{\ddot a}a=H^2(1-\varepsilon).\] The condition $\varepsilon=1$ is equivalent to $\ddot a=0$. When the value $\varepsilon=1$ is reached due to variation of the potential shape the Universe exits the regime of the accelerated expansion (inflation). Around tend, the inflaton field(s) typically begin oscillating around the minimum of the potential. (see Inflation and the Higgs Scalar, DANIEL GREEN (1412.2107).)

10

Problem 10

problem id: 2501_10

How does the number of e-folds $N$ depend on the slow-roll parameter $\varepsilon$?

solution

\[N\propto\int\limits_{\varphi_{end}}^{\varphi_{initial}}d\varphi\frac{V}{dV/d\varphi} \propto\int\limits_{\varphi_{end}}^{\varphi_{initial}}d\varphi/\sqrt\varepsilon.\] The number of $e$-folds depends on how fast the field is $f$-decreasing. The number of e-folds, $N$ , is inversely proportional to the square root of the slow roll parameter $\varepsilon$ or proportional to the inverse fractional change of the potential with the field, $V/(dV/d\varphi)$.

11

Problem 11

problem id: 2501_11

The exponential increase in $a(t)$ (during the inflation) drastically reduces the temperature since $Ta$ is a constant. After the field disappears, the Universe will need to re-heat to the high temperatures needed to create the light nuclei whose relative abundance is predicted by BB cosmology and is a major success of the BB model.

solution

17

Problem 17

problem id: 2501_17

Suppose that $dq/dt=f(q)$. Find the Hubble parameter in terms of $q$.

solution

\[1+q=\frac{d}{dt}\left(\frac1H\right),\quad \frac{dq}{dt}=f(q)\to dt=\frac{dq}{f(q)},\] \[\frac1{H(q)}=\int\frac{1+q}{f(q)}dq.\] (see S. Carloni et al., A new approach to reconstruction methods in f(R) gravity, arXiv:1005.1840)

18

Problem 18

problem id: 2501_18

Show that derivative w.r.t. the cosmic time can be related to that w.r.t. redshift as follows: \[\frac1{f(t)}\frac{df(t)}{dt}=-(1+z)H(z)\frac1{f(z)}\frac{df(z)}{dz}.\]

solution

§ 1

New in Observational Cosmology

Problems 1b – 3b

1b

Problem 1b

problem id: 2501_01o

Obtain relations between velocity of cosmological expansion and redshift.

solution

The exact velocity-proper distance relation (Hubble law) reads \[V_{exp}=HR.\] For the observer on the Earth living at the time $t=t_0$ the observed redshift serves as a measure of distance at this epoch. For photons moving along the comoving coordinate $r$, $cdt=\pm adr$. Consequently \[a_0dr=\frac{cdz}{H(z)}.\] Integration over the redshift gives the proper distance-redshift relation \[R(t_0,z)\equiv R(z)=\int\limits_0^z\frac{cdz'}{H(z')}.\] Using the $R(z)$ relation one gets the exact velocity-redshift relation \[V_{exp}(z)=H_0 R(z)=H_0\int\limits_0^z\frac{cdz'}{H(z')}.\] where the expansion velocity is taken for an object with the redshift $z$ observed at the time $t=t_0$.

2b

Problem 2b

problem id: 2501_02o

Why the Linear Distance-Redshift Law in Near Space?

solution

Consider a nearby galaxy and the change of the scale factor during the small time interval $dt$ when the light travels from this galaxy to the observer: \[a(t_{obs})=a(t_{em})+\left.\frac{da}{dt}\right|_{t=t_{em}}dt.\] When the distance is short and the expansion speed much less than the speed of light, then the distance $R$ measured by the astronomer using any of the available ways is practically equal to the metric distance $a(t)r$ and the time interval can be approximated as $dt\approx a(t)r/c=R/c$. From this it follows that \[a(t_{obs})/a(t_{em})=1+z=1+\frac{\dot a}a \frac R c.\] and finally \[cz=HR.\]

3b

Problem 3b

problem id: 2501_03o

Find the exact relativistic Doppler velocity-redshift relation.

solution

\[(1+z)_{Dop}=\sqrt{\frac{c+V}{c-V}}.\] The exact relativistic Doppler velocity-redshift relation is \[V_{Dop}(z)=c\frac{2z+z^2}{2+2z+z^2}.\] For $z\to\infty$, the velocity $V_{Dop}(z)\to c$ corresponding to the limit $V_{Dop}le c$. Show, that $V_{Dop}$ and $V_{exp}$ give the same results only in the first order of $V/c$. Find $V_{exp}(z)$ for the three cosmological models: Einstein-de Sitter, Milne and de Sitter. For $\Omega_m=1$, $\Omega_\Lambda=0$: \[V_{exp}(z)=H_0 R(z)=H_0\int\limits_0^z\frac{cdz'}{H(z')}=\frac{2(\sqrt{1+z}-1)}{\sqrt{1+z}}c.\] For $\Omega_m=0$, $\Omega_\Lambda=0$: \[V_{exp}(z)=\frac{z(1+z/2)}{(1+z)}c.\] For $\Omega_m=0$, $\Omega_\Lambda=1$: \[V_{exp}(z)=zc.\]

§ 2

NEW 2

Problems 1c – 3f

1c

Problem 1c

problem id: new2015_3

The lookback time is defined as the difference between the present day age of the Universe and its age at redshift $z$, i.e. the difference between the age of the Universe at observation $t_0$ and the age of the Universe, $t$, when the photons were emitted. Find the lookback time for the Universe filled with non-relativistic matter, radiation and a component with state equation $p=w(z)\rho$.

solution

From the definitions of redshift $1+z=1/a$ we have \[\frac{dz}{dt}=-\frac{dot a}{a^2}=-H(z)(1+z)\] or \[dt=-\frac{dt}{H(z)(1+z)}.\] The lookback time is defined as \[t_0-t=\int\limits_t^{t_0}dt=\int\limits_0^z\frac{dz'}{H(z')(1+z')}=\frac1{H_0}\int\limits_0^z\frac{dz'}{E(z')(1+z')}\] where \[E(z)=\sqrt{\Omega_r(1+z)^4+\Omega_m(1+z)^3+\Omega_k(1+z)^2+\Omega_w\exp\left( 3\int\limits_0^zdz'\frac{1+w(z')}{1+z'}\right)}.\] From the definition of lookback time it is clear that the cosmological time or the time back to the Big Bang, is given by \[t(z)=\int\limits_z^\infty\frac{dz'}{H(z')(1+z')}.\]

2c

Problem 2c

problem id: new2015_4

Find the solutions corrected by LQC Friedmann equations for a matter dominated Universe.

solution

Solving the corrected first Friedmann equation and the conservation equation for a matter dominated Universe \[H^2=\frac\rho3\left(1-\frac\rho{\rho_c}\right),\quad \dot\rho+3H\rho=0;\] one obtains the following quantities \[a(t)=\left(\frac34\rho_ct^2+1\right)^{1/3},\quad \rho(t)=\frac{\rho_c}{\frac34\rho_ct^2+1},\quad H(t)=\frac{\frac12\rho_ct}{\frac34\rho_ct^2+1}.\] For small values of the energy density ($\rho\ll\rho_c$) the solution of standard Friedmann equations is recovered.

3c

Problem 3c

problem id: new2015_5

Show that in the case of the flat Friedmann metric, the third power of the scale factor $\varphi=a^3$ satisfies the equation \[\frac{d^2\varphi}{dt^2}=\frac32(\rho-p)\varphi,\quad 8\pi G=1.\] Check validity of this equation for different cosmological components: non-relativistic matter, cosmological constant, a component with the state equation $p=w\rho$.

solution

\[\frac{d^2a^3}{dt^2}=3\left(2H^2+\frac{\ddot a}a\right)a^3=3\left[\frac23\rho-\frac16(\rho+3p)\right]a^3=\frac32(\rho-p)\varphi.\]

---

$f(R)$ gravity theory is built by direct generalization of the Einstein-Hilbert action with the substitution $R\to f(R)$. The new action is \[S=\frac1{2\kappa}\int d^4x\sqrt{-g}f(R)+S_m(g_{\mu\nu},\psi);\quad \kappa\equiv8\pi G.\] Here $\psi$ is general notion for the matter fields. The chosen generalization contains function $f(R)$ which depends solely on the Ricci scalar $R$, but it does not include other invariants such as $R_{\mu\nu}R^{\mu\nu}$. The reason of that is the following: the action $f(R)$ is sufficiently general to reflect basic features of gravity, and at the same time it is simple enough so that the calculations present no technical difficulty. The function $f(R)$ must satisfy the stability conditions \[f'(R)>0,\quad f''(R)>0.\]

1d

Problem 1d

problem id: f_r_1

Obtain field equations for the $f(R)$ gravity.

solution

\[f'(R)R_{\mu\nu}-\frac12f(R)g_{\mu\nu}-\left(\nabla_\mu\nabla_\nu-g_{\mu\nu}\square\right)f'(R)=\kappa T_{\mu\nu}.\] Here \[T_{\mu\nu}=-\frac{2}{\sqrt{-g}}\frac{\delta S_m}{\delta g^{\mu\nu}},\] $\nabla_\mu$ is covariant derivative associated with the Levi-Civita connection of the metric, and $\square=\nabla_\mu\nabla^\mu$.

2d

Problem 2d

problem id: f_r_2

Obtain equation relating the scalar curvature $R$ with trace of the stress-energy tensor.

solution

\begin{equation}\label{f_r_2_e_1} f'(R)R-2f(R)+3\square f'(R)=\kappa T, \end{equation} where $T=g^{\mu\nu}T_{\mu\nu}$. $R$ and $T$ are related in differential rather in algebraic way like $R=-kT$ in general relativity. Note that equality $T=0$ does not imply that $R$ equals to zero or a constant.

3d

Problem 3d

problem id: f_r_3

Solutions with $R=const$ are called the maximally symmetric. Show that in the case $R=0$, $T_{\mu\nu}=0$ the maximally symmetric solution is the Minkowski space, and in the case $R=const\equiv C$, $T_{\mu\nu}=0$ the maximally symmetric solution coincides with the de- Sitter or anti-de Sitter depending on sign of $C$.

solution

For $R=const$ and $T_{\mu\nu}=0$ the equation \ref{f_r_2_e_1} obtained in the previous problem takes on the form \[f'(R)R-2f(R)=0.\] For given function $f(R)$ this equation represents an algebraic equation for $R$. If $R=0$ is a root of this equation then it immediately follows from the field equations that $R_{\mu\nu}=0$ and the maximally symmetric solution is the Minkowski space-time. If the equation has a root $R=const=C$ then \[R_{\mu\nu}=\frac C4 g_{\mu\nu}\] and the maximally symmetric solution corresponds to the de- Sitter or anti-de Sitter (or to the cosmological constant in usual General Relativity) depending on sign of $C$.

4b

Problem 4b

problem id: f_r_4

Use the FLRW metrics and the stress-energy tensor for an ideal liquid to obtain an analogue of the Friedmann equations for the $f(R)$ cosmology.

solution

\[H^2=\frac\kappa{3f'}\left[\rho+\frac12(Rf'-f)-3H\dot Rf''\right];\] \[2\dot H+H^2=-\frac\kappa{f'}\left[p+\dot R^2f'''+2H\dot Rf''+\ddot Rf''+\frac12(f-Rf')\right].\]

5b

Problem 5b

problem id: f_r_5

Show that introduction of effective energy density \[\rho_{eff}=\frac{Rf'-f}{2f'}-\frac{3H\dot Rf''}{f'}\] and effective pressure \[p_{eff}=\frac{\dot R^2f'''+2H\dot Rf''+\ddot Rf''+\frac12(f-Rf')}{f'}\] allows to represent the equations obtained in the previous problem in form of the standard Friedmann equations \begin{align} \nonumber H^2&=\frac\kappa3\rho_{eff};\\ \nonumber \frac{\ddot a}a&=-\frac\kappa6\left(\rho_{eff}+3p_{eff}\right). \end{align}

solution

6b

Problem 6b

problem id: f_r_6

What condition must the function $f(R)$ satisfy to in order to make \[w_{eff}\equiv\frac{p_{eff}}{\rho_{eff}}=-1?\]

solution

To reproduce (imitate) the de Sitter equation of state (i.e. the cosmological constant) $w_{eff}=-1$ the following condition is required \[\frac{f'''}{f''}=\frac{\dot RH-\ddot R}{\dot R^2}.\]

7b

Problem 7b

problem id: f_r_7

Let $f(R)\propto R^n$. Find $w_{eff}$ as a function of $n$ assuming the power law dependence for the scale factor $a(t)=a_0(t/t_0)^\alpha$.

solution

The result for $n\ne1$ is \[w_{eff}=-\frac{6n^2-7n+1}{6n^2-9n+3},\] and $\alpha$ in terms of $n$ reads \[\alpha=\frac{-2n^2+3n-1}{n-2}.\] Appropriate choice of $n$ leads to desired value of $w_{eff}$. for example $n=2$ leads to $w_{eff}=-1$, $\alpha=\infty$. Arbitrary dependence $a(t)$ would result in time dependence of $w_{eff}$.

8

Problem 8

problem id: f_r_8

Let the function \[f(R)=R-\frac{\mu^{2(n+1)}}{R^n}.\] Find $w_{eff}$ as a function of $n$ assuming the power law dependence for the scale factor $a(t)=a_0(t/t_0)^\alpha$.

solution

\[w_{eff}=-1+\frac{2(n+2)}{3(2n+1)(n+1).}\] In particular, for $n=1$ one finds that $w_{eff}=-2/3$ (accelerated expansion). Note that in this case positive values of $n$ imply presence of terms inversely proportional to $R$, in contrast to the case considered in the previous problem.

9b

Problem 9b

problem id: f_r_9

Show that \[\frac{dq}{d\ln(1+z)}=j-q(2q+1).\]

solution

13

Problem 13

problem id: f_r_13

Show that characteristic size of the large scale structures is set by actual value of the cosmological constant.

solution

Let us show an important consequence of the formula \begin{equation}\label{12_1} F=-\frac m{R^2}-q(t)H^2(t)R. \end{equation} Consider a Universe, which contains no matter (or radiation), but only dark energy in the form of a non-zero cosmological constant $\Lambda$. In this case, the Hubble parameter and, hence, the deceleration parameter become time-independent and are given by $H=\sqrt{\Lambda/3}$ and $q=-1$. Thus, the force (\ref{12_1}) also becomes time-independent, \begin{equation}\label{12_3} F=-\frac m{r^2}+\frac13\Lambda r. \end{equation} For the case of spatially finite (i.e. non-pointlike) spherically-symmetric massive objects (12.3) is replaced by \begin{equation}\label{12_4} F=-\frac{M(r)}{r^2}+\frac13\Lambda r, \end{equation} where $M(r)$ is the total mass of the object contained within the radius $r$. If the object has the radial density $\rho(r)$ then \[M(r)=\int\limits_0^r4\pi r'^2\rho(r')dr'.\] Although the de Sitter background is not an accurate representation of our Universe, the SCM is dominated by dark-energy in a form consistent with a simple cosmological constant. Even in the simple Newtonian case (\ref{12_4}), we see immediately that there is an obvious, but profound, difference between the cases $\Lambda=0$ and $\Lambda\ne0$. In the former, the force on a constituent particle of a galaxy or cluster (say) is attractive for all values of $r$ and tends gradually to zero as $r\to\infty$ (for any sensible radial density profile). In the latter case, however, the force on a constituent particle (or equivalently its radial acceleration) vanishes at the finite radius $r_F$ which satisfies $r_F=[3M(R_F)/\Lambda]^{1/3}$, beyond which the net force becomes repulsive. This suggests that a non-zero $\Lambda$ should set a maximum size, dependent on mass, for galaxies and clusters.

14

Problem 14

problem id: f_r_14

Inverse problem to the previous one: find upper bound for the cosmological constant $\Lambda9r)$.

solution

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1e

Problem 1e

problem id: 1301_38

(Inspired by 1. M. Dunajski, G. Gibbons, Cosmic Jerk, Snap and Beyond, arXiv: 0807.0207)

Express the curvature parameter $k$ terms of the cosmographic parameters for the case of Universe filled with non-interacting cosmological constant and non-relativistic matter.

solution

Using the relation $\rho_m=M/a^3$ ($M=const$) we rewrite the first Friedmann equation in the form \begin{equation}\label{1301_38_e1} \dot a^2+k=\frac13\frac M a+\frac13\Lambda a,\quad 8\pi G=1.\end{equation} Then we differentiate the latter equation two times to find \begin{align} \label{1301_38_e2} \ddot a=&-\frac16\frac M{a^2}=\frac13\Lambda a,\\ \nonumber \dddot a=&-\frac13\frac{M\dot a}{a^3}=\frac13\Lambda\dot a. \end{align} Using the definitions of the cosmographic parameters \[H=\frac{\dot a}{a},\quad q=-a\frac{\ddot a}{\dot a^2},\quad j=a^2\frac{\dddot a}{\dot a^3},\] we represent (\ref{1301_38_e1}) in the form \begin{align} \nonumber q &= \frac12A-B;\\ \nonumber j &= A+B;\\ \nonumber A &\equiv\frac13 \frac{M}{a^3H^2};\\ \nonumber B &\equiv\frac13 \frac{\Lambda}{H^2}. \end{align} Then we find \begin{align} \nonumber A &=\frac23(j+q);\\ \nonumber B &=\frac23(\frac12j-q). \end{align} The Friedmann equation (\ref{1301_38_e1}) in terms of the above introduced variables $A$ and $B$ takes on the form \[\frac k{a^2}=(A+B-1)H^2\] or \[k=a^2H^2(j-1).\]

2e

Problem 2e

problem id: 1301_39

Do the same as in the previous problem for the case of Universe filled with non-interacting non-relativistic matter $\rho_m=M_m/a^3$ and radiation $\rho_r=M_r/a^4$.

solution

We represent the first Friedmann equation in the form \begin{equation}\label{1301_39_e1} \frac{\dot a^2}{a^2}+\frac k{a^2}=\frac{M_m}{a^3}+\frac{M_r}{a^4},\quad \frac{8\pi G}{3}=1.\end{equation} We twice differentiate the latter equation to find \begin{align} \label{1301_39_e2} \ddot a=&-\frac12\frac{M_m}{a^2}-\frac{M_r}{a^3},\\ \nonumber \dddot a=&\frac{M_m}{a^3}\dot a+3\frac{M_r}{a^4}\dot a. \end{align} Using definitions of the cosmographic parameters $q$ and $j$, one obtains \begin{align} \nonumber q &= \frac12A+B;\\ \nonumber j &= A+3B;\\ \nonumber A &\equiv\frac{M_m}{a^3H^2};\\ \nonumber B &\equiv\frac{M_r}{a^4H^2}. \end{align} Then we find \begin{align} \nonumber A &=-2j+6q;\\ \nonumber B &=j-2q. \end{align} The Friedmann equation (\ref{1301_39_e1}) in terms of the above introduced variables $A$ and $B$ takes on the form \[\frac k{a^2}=(A+B-1)H^2\] or in terms of the cosmographic parameters \[k=a^2H^2(4q-j-1).\]

3e

Problem 3e

problem id: 1301_40

Check the expressions for the curvature $k$ obtained in the previous problem for two cases: a) a flat Universe solely filled with non-relativistic matter; b) a flat Universe solely filled with radiation.

solution

In the first case $B=0$, $q=1/2$. Then $A=1$ and $k=0$. Note that in this case $j=1$. In the second case $A=0$, $q=1$. Then $B=1$, $k=0$. In that case $j=3$.

4c

Problem 4c

problem id: 1301_41

Find relation between the cosmographic parameters free of any cosmological parameter for the case of Universe considered in the problem #1301_38.

solution

Using expression for $\dddot a$ obtained in the problem 1, one finds \[\ddddot a=\frac M3\frac{\ddot a}{a^3}-M\frac{\dot a^2}{a^4}+\frac\Lambda3\ddot a.\] For the snap parameter \[s\equiv a^3\frac{\ddddot a}{\dot a^4}\] one obtains \[s=-3(A+B)q-3A.\] Substitution of the parameters \begin{align} \nonumber A &=\frac23(j+q);\\ \nonumber B &=\frac23(\frac12j-q). \end{align} introduced in the problem 1, one finally finds \[s+2(q+j)+qj=0.\] This fourth order ODE is equivalent to the Friedmann equation and has an advantage that it appears as a constraint on directly measurable quantities.

5c

Problem 5c

problem id: 1301_42

Perform the same procedure for the Chaplygin gas with the equation of state $p=-A/\rho$ and for the generalized Chaplygin gas with the equation of state $\rho=-A/\rho^\alpha$.

solution

For small values of $a(t)$ density and pressure of the Chaplygin gas reduces to that of dust $\rho\propto a^{-3}$, and for large $a$ one gets the de Sitter Universe: $\rho=const$, $p=-\rho$. In between these two regimes one can use the approximation \begin{equation}\label{scalar_5}\rho=\sqrt A+\frac{B}{\sqrt{2A}}a^{-6}.\end{equation} Thus $\sqrt A$ plays the role of a cosmological constant. We insert this to the Friedmann equation with $\Lambda$ and follow the procedure of eliminating the constants by differentiation. This leads to an approximate constraint \begin{equation}\label{scalar_6}s+5(q+j)+qj=0.\end{equation} Analogous procedure for the generalized Chaplygin gas leads to the equation \begin{equation}\label{scalar_7}s+(3\alpha+2)(q+j)+qj=0.\end{equation} Note that for $\alpha=1$ we reproduce the above obtained result for the Chaplygin gas, while $\alpha=0$ we return to the results obtained in the problem (4). If we want to exclude the parameter $\alpha$ from the latter equation we must take one more derivative of the Friedmann equation and introduce an additional cosmological parameter \[l=\frac1a\frac{d^5a}{dt^5}\left(\frac1a\frac{da}{dt}\right)^{-5}.\] As a result one obtains \begin{equation}\label{scalar_8} -2qs-2jq^2-lq-2sj-3sq^2-j^2q-lj+s^2-3q^2j-qsj+j^3-2j^2q^2=0. \end{equation} This constraint is again approximate and is valid only in the regime where the higher order terms in the expansion of $\rho$ can be dropped.

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1f

Problem 1f

problem id: new_22

It is of broad interest to understand better the nature of the early Universe especially the Big Bang. The discovery of the CMB in 1965 resolved a dichotomy then existing in theoretical cosmology between steady-state and Big Bang theories. The interpretation of the CMB as a relic of a Big Bang was compelling and the steady-state theory died. Actually at that time it was really a trichotomy being reduced to a dichotomy because a third theory is a cyclic cosmology model. Find the main argument again the latter theory.

solution

The main argument of the opponents to the cyclic model of Universe was based on the so-called Tolman Entropy Conundrum (R.C. Tolman, Relativity, Thermodynamics and Cosmology. Oxford University Press (1934)): the entropy of the Universe necessarily increases, due to the second law of thermodynamics, and therefore cycles become larger and longer in the future, smaller and shorter in the past, implying that a Big Bang must have occurred at A FINITE time in the past.

2f

Problem 2f

problem id: new_26

In 1881, Johnson Stoney proposed a set of fundamental units involving $c$, $G$ and $e$. Build the Stoney's natural units of length, mass and time. \begin{align} \nonumber L_S&=\left(\frac{Ge^2}{c^4}\right)^{1/2};\\ \nonumber M_S&=\left(\frac{e^2}{G}\right)^{1/2};\\ \nonumber T_S&=\left(\frac{Ge^2}{c^6}\right)^{1/2}; \end{align} Show that the fine structure constant $\alpha=e^2/(\hbar c)$ can be used to convert between Stoney and Planck units.

solution

\begin{align} \nonumber L_S&=L_{Pl}\alpha^{1/2};\\ \nonumber M_S&=M_{Pl}\alpha^{1/2};\\ \nonumber T_S&=T_{Pl}\alpha^{1/2}; \end{align}

3f

Problem 3f

problem id: new_27

Find dimensions of the Newtonian gravitational constant $G$ and charge $e$ in the $N$-dimensional space.

solution

By Gauss's theorem we have $[G]=L^NM^{-1}T^{-2}$; $[e^2]=L^NMT^{-2}$.

The value $c^4/(4G)$ of the force limit is the energy of a Schwarzschild black hole divided by twice its radius. The maximum power $c^5/(4G)$ is realized when such a black hole is radiated away in the time that light takes to travel along a length corresponding to twice the radius. It will become clear below why a Schwarzschild black hole, as an extremal case of general relativity, is necessary to realize these limit values. (General relativity and cosmology derived from principle of maximum power or force Christoph Schiller (0607090))

Problem

problem id: new_28

Consider n-dimensional homogeneous and isotropic Universe, filled with two non-interacting components: the cosmological constant and a component with the state equation $p_m=w\rho_m$. Express the curvature parameter $k$ through the cosmographic parameters.

solution

Evolution of the considered universe is described by the equations \[H^2=\frac2{n(n-1)}\rho-\frac k{a^2},\quad 8\pi G=1,\] \[\rho=\rho_m+\Lambda,\] \[\dot\rho_m+nH(\rho_m+p_m)=0.\] Using the conservation equation one finds \[\rho=\rho_0a^{-n(1+w)}+\Lambda.\] The first Friedmann equation then takes on the form \[\dot a^2=\frac2{n(n-1)}\rho_0a^{-n(1+w)+2}+\frac{2\Lambda}{n(n-1)}a^2-k.\] In the notions \[-n(1+w)\equiv N,\quad \frac2{n(n-1)}=M\] the Friedmann equation reads \begin{equation}\label{new_28_e_1} \dot a^2=M\rho_0a^{N+2}+M\Lambda a^2-k. \end{equation} Differentiation of the latter equation gives \begin{align}\label{new_28_e_2} \ddot a&=\frac12M\rho_0(N+2)a^{N+1}+M\Lambda a;\\ \nonumber\dddot a&=\frac{(N+1)(N+2)}2M\rho_0a^N\dot a+M\Lambda\dot a. \end{align} Transforming to the cosmographic parameters, one gets \begin{align}\label{new_28_e_3} q&\equiv-\frac{\ddot a a}{\dot a^2}=-\frac12M\rho_0(N+2)\frac{a^N}{H^2}-\frac{M\Lambda}{H^2},\\ \nonumber j&\equiv\frac{\dddot a a^2}{\dot a^3}=\frac{(N+1)(N+2)}2M\rho_0(N+2)\frac{a^N}{H^2}+\frac{M\Lambda}{H^2}. \end{align} Using the notions \[A\equiv-M\rho_0(N+2)\frac{a^N}{H^2},\quad B\equiv\frac{M\Lambda}{H^2},\] the equation (\ref{new_28_e_3}) can be presented in the form \begin{align}\label{new_28_e_4} q&=\frac12A-B,\\ \nonumber j&=-\frac{N+1}2A+B. \end{align} It then follows that \begin{equation}\label{new_28_e_5} A=-\frac2N(q+j),\quad B=-\frac1N[(N+1)q+j]. \end{equation} The first Friedmann equation in terms of the parameters $A$ and $B$ becomes \begin{equation}\label{new_28_e_6} \frac{k}{a^2H^2}=-\frac1{N+2}A+B-1, \end{equation} or transforming to the cosmographic parameters $q$ and $j$ \begin{equation}\label{new_28_e_7} k=a^2H^2\left\{\frac2{N(N+2)}(q+j)-\frac1N\left[(N+1)q+j\right]-1\right\}, \end{equation} For $n=3$, $w=0$ ($N=-3$) the obtained relation for spatial curvature reduces to \[k=a^2H^2(j-1),\] which reproduces result of the problem #1301_38.

Problem

problem id: new_29

Find relation between the cosmographic parameters free of any cosmological parameter for the case of Universe considered in the previous problem.

solution

Using the relation for $\dddot a$ obtained in the previous problem, one finds \[\ddddot a=\frac{N(N+1)(N+2)}2M\rho_0a^{N-1}\dot a^2+\frac{(N+1)(N+2)}2M\rho_0a^{N}\ddot a+M\Lambda\ddot a.\] Using the definition \[s=\frac{a^3}{\dot a^4}\ddddot a\] we obtain \[s=\left(\frac{N+1}2A-B\right)q-\frac{N(N+1)}{2}A,\] where $A$ and $B$ are defined in the previous problem. Transforming to the cosmographic parameters, the latter relation can be presented in the form \[s-\left\{\frac2N(q+j)+\frac1N[(N+1)q+j]q-\frac6N(q+j)\right\}=0.\] For $N=-3$ the obtained result reproduces that of the problem #1301_41.

Problem

problem id: new_30

Find the sound speed for the modified Chaplygin gas with the state equation \[p=B\rho-\frac A{\rho^\alpha}.\]

solution

\[c_s^2=\frac{\delta p}{\delta\rho}=\frac{\dot p}{\dot\rho}=-\alpha w+(1+\alpha)B,\quad w=\frac p\rho.\]