Difference between revisions of "Evolution of Universe"

$$ \begin{gathered} \rho _r(t) = \rho _m(t);\\ \rho _{r0}\frac{a_0^4}{a^4} = \rho _{m0}\frac{a_0^3}{a^3};\\ \frac{\rho _{m0}}{\rho _{r0}} = \frac{\Omega _{m0}}{\Omega _{r0}} = \frac{a_0}{a} = 1 + z;\\ \Omega _{r0} = 5 \cdot {10^{ - 5}} \\ z = \frac{\Omega _{m0}}{\Omega _{r0}} - 1 \approx 5400\\ \end{gathered} $$

$$ V(x) = - \frac{1}{2}\sum\limits_i \Omega _{i0}x^{ - (1 + 3w_i)};\quad x \equiv \frac{a}{a_0} $$ In the SCM $$ V(x) = - \frac{1}{2}\Omega _{m0}x^{-1}-\frac{1}{2}\Omega _{\Lambda 0}x^2 \simeq - 0.135x^{-1}-0.365x^2 $$

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 \[\begin{array}{l} {A_4} = 1 - \frac{3^2}{2}\Omega _m;\\ {A_5} = 1 + 3\Omega _m + \frac{3^3}{2}\Omega _m^2;\\ {A_6} = 1 - \frac{3^3}{2}\Omega _m - 3^4\Omega _m^2 - \frac{3^4}{4}\Omega _m^3 \end{array}\]

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 $w_0,w_a:$ \[\begin{array}{l} q_0 = \frac{1}{2} + \frac{3}{2}\left( 1 - \Omega _m \right)w_0;\\ j_0 = 1 + \frac{3}{2}\left( 1 - \Omega _m \right)\left[3w_0\left( 1 + w_0 \right) + w_a \right];\\ s_0 = - \frac{7}{2} - \frac{33}{4}\left( 1 - \Omega _m \right)w_a - \frac{9}{4}\left( 1 - \Omega _m \right)\left[ 9 + \left(7 - \Omega _m \right)w_a \right]w_0 - \\ - \frac{9}{4}\left( 1 - \Omega _m \right)\left( 16 - 3\Omega _m \right)w_0^2 - \\ - \frac{27}{4}\left( 1 - \Omega _m \right)\left(3 - \Omega _m \right)w_0^3;\\ l_0 = \frac{35}{2} + \frac{1 - \Omega _m}{4}\left[ 213 + (7 - \Omega _m)w_a \right]w_a + \\ + \frac{1 - \Omega _m}{4}\left[ 489 + 9(82 - 21\Omega _m)w_a \right]w_0 + \\ + \frac{9}{2}\left( 1 - \Omega _m \right)\left[ 67 - 21\Omega _m + \frac{3}{2}(23 - 11\Omega _m)w_a \right]w_0^2 + \\ + \frac{27}{4}\left( {1 - {\Omega _m}} \right)(47 - 24{\Omega _m})w_0^3 + \\ + \frac{81}{2}\left( 1 - \Omega _m \right)(3 - 2\Omega _m)w_0^4 \end{array}\]

In the SCM $\left( {{w_0},{w_a}} \right) = \left( { - 1,0} \right)$ and therefore \[\begin{array}{l} {q_0} = \frac{1}{2} - \frac{3}{2}\left( {1 - {\Omega _m}} \right);\\ {j_0} = 1;\\ {s_0} = 1 - \frac{9}{2}{\Omega _m};\\ {l_0} = 1 + 3{\Omega _m} + \frac{27}{2}\Omega _m^2 \end{array}\]

$$ \begin{gathered} t_U=\int_z^\infty\frac{dz}{(1 + z)H(z)} ;\; t_UH_0=\int_z^\infty\frac{dz}{(1 + z)\sqrt{\Omega _{m0}(1 + z)^3 + \Omega _{\Lambda 0} } }; \\ t_{ph}H_0=\int_0^z \frac{dz}{(1 + z)\sqrt{\Omega _{m0}(1 + z)^3 +\Omega _{\Lambda 0}}}\\ \end{gathered} $$ At $$z = 0.1\,\,\, \to \,\,\,\,t_U H_0 = 0.899\,\,\,\,\,t_{ph} H_0 = 0.093$$ $$z = 1\,\, \to \,\,\,t_U H_0 = 0.431\,\,\,\,\,t_{ph} H_0 = 0.561$$ $$z = 100\,\,\, \to \,\,t_U H_0 = 1.264 \times 10^{ - 3} \,\,\,\,\,t_{ph} H_0 = 0.991$$ $$z = 1000\,\, \to \,\,\,t_U H_0 = 3.663 \times 10^{ - 5} \,\,\,\,\,t_{ph} H_0 = 0.993$$

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.

$$ \begin{gathered} a = \frac{1}{1 + z}; \\ a(t) = A^{1/3}\sinh ^{2/3}\left(t/{t_\Lambda } \right);\quad A \equiv \frac{\Omega_{ m0}}{\Omega _{\Lambda 0}}; \\ \frac{t_0}{t_\Lambda }= Arth\sqrt {\Omega _{\Lambda 0}} ; \\ \frac{t}{t_0} = \frac{1}{Arth\sqrt{\Omega _{\Lambda 0}}}Arsh\left[\sqrt {\frac{\Omega _{\Lambda 0}}{\Omega _{m0}}} \frac{1}{\left(1 + z\right)^{3/2}} \right] \\ \end{gathered} $$

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 $$ a(t) = A^{1/3}\sinh ^{2/3}\left( t/{t_\Lambda }\right);\quad A \equiv \frac{\Omega_{m0}}{\Omega _{\Lambda 0}} = \frac{1 - \Omega _{\Lambda 0}}{\Omega _{\Lambda 0}};\quad t_\Lambda \equiv \frac{2}{3}\left(H_0 \right)^{-1}\left( \Omega _{\Lambda 0}\right)^{ - 1/2} $$

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 \[R_h=\frac{c}{\dot f}=\frac{c}{H_\infty}\tanh\left(\frac32tH_\infty\right).\]

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 $c/H_\infty$.

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.

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.

\[\frac{1}{a^4}\left( \frac{da}{d\eta } \right)^2=H_0^2\left[ \Omega_m a(\eta )^{-3}+\Omega_{\Lambda} \right]\]

\[\eta =\frac{1}{H_0}\int\limits_0^a\frac{dx}{\sqrt{x}{{\left( {{\Omega }_{m}}+{{\Omega }_{\Lambda }}{{x}^{3}} \right)}^{1/2}}}\]

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

$$ \begin{gathered} t = t_\Lambda Arcth\sqrt {\Omega _\Lambda} ;\quad \Omega _{\Lambda } = th^2\frac{t}{t_\Lambda };\\ t_\Lambda \simeq 10.768 \cdot 10^{9}\;\mbox{years}\;\;t = 14.7\cdot 10^{9}\;\mbox{years}\\ \Omega _{\Lambda 0} \simeq 0.77\\ \end{gathered} $$

$$ \begin{gathered} \frac{d\Omega _\Lambda }{dt} = \frac{2}{t_\Lambda }\frac{sh(t/{t_\Lambda })} {ch^3(t/{t_\Lambda })}; \\ x \equiv \frac{t}{t_\Lambda };\quad \frac{d\Omega _\Lambda }{dx} = 2\frac{sh(x)}{ch^3(x)} \\ \end{gathered} $$

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 $$ L_{H0}= A^{-1/3}\int\limits_0^{t_0}\frac{dt}{sh^{2/3}\left( t/{t_\Lambda } \right)}; $$

$$ H = \frac{\dot a}{a} =\frac{2}{3}\left(t_\Lambda \right)^{ - 1}cth\left( {\frac{t}{t_\Lambda }} \right) $$ In the SCM ${t_0} \simeq 1.2t_\Lambda .$ Therefore $$ H{t_0} \simeq 0.8cth\left( {1.2t/{t_0}} \right) $$

• 

  Time dependence of the Hubble parameter (in units of the age of the Universe).

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 \[\rho_m=\frac{\Lambda}{8\pi G}\frac{1}{\sinh^2(t/t_\Lambda)}.\]

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 $$ H_{\infty} = \frac{2}{3t_\Lambda} = \sqrt {\frac{\Lambda}{3}}. $$

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

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 $$ \begin{gathered} q = \frac{1 - 3\Omega_{\Lambda 0}}{2};\\ q_0 \simeq - 0.6 \\ \end{gathered} $$

$$ \begin{gathered} q \equiv - \frac{a\ddot a}{\dot a^2} = - \frac{\ddot a}{a}\frac{1}{H^2}; \\ \frac{\ddot a}{a} = - \frac{4\pi G}{3}\sum\limits_i \left( \rho _i + 3p_i \right) = - \frac{1}{2}H_0^2\sum\limits_i \Omega _{i0}\left(1 + 3w_i\right)(1 + z)^{3\left(1 + w_i\right)} ; \\ H^2 = H_0^2\sum\limits_i \Omega _{i0}(1 + z)^{3\left( 1 + w_i \right)}; \\ q = \frac{3}{2}\frac{\sum\limits_i \Omega _{i0}\left( 1 + w_i \right)(1 + z)^{3\left( 1 + w_i \right)}} {\sum\limits_i \Omega _{i0}(1 + z)^{3\left( 1 + w_i\right)} } - 1 \\ \end{gathered} $$ In the SCM $$ q = \frac{1}{2}\frac{\Omega _{m0}(1 + z)^3- 2\Omega _{\Lambda 0} } {\Omega _{m0}(1 + z)^3+\Omega_{\Lambda 0}} $$ $$ q(z \to \infty ) = \frac{1}{2},\;q(z \to - 1) = - 1 $$

• 

  Dependence of the deceleration parameter on the redshift.

$$ \begin{gathered} q = - \frac{\ddot aa} {\dot a^2} = - \frac{\ddot a} {aH^2}; \\ a(t) = A^{1/3}sh^{2/3}\left(t/{t_\Lambda}\right);\quad A\equiv \frac{\Omega_ {m0}} {\Omega _{\Lambda 0}}; \\ \frac{t_0} {t_\Lambda } = Arth\sqrt {\Omega _{\Lambda 0}} ;\quad H = \frac{2} {3}\frac{1} {t_\Lambda }cth\left( \frac{t} {t_\Lambda }\right); \\ q = \frac{1} {2}\left[ 1 - 3th^2\left( \frac{t} {t_\Lambda }\right)\right] \\ \end{gathered} $$

$$ \begin{gathered} \left( \frac{\dot a}{a} \right)^2 = \frac{8\pi G} {3}\rho _{cr}\left( \Omega _{m0}(1 + z)^3+\Omega_{\Lambda 0}\right); \\ \Omega _{m0}(1 + z^*)^3 =\Omega _{\Lambda 0};\\ z^* = \left(\frac{\Omega _{\Lambda 0}} {\Omega _{m0}}\right)^{1/3}-1; \\ \Omega _{\Lambda 0}\simeq 0.73;\quad \Omega _{m0}\simeq 0.27; \\ z^* \simeq 0.39; \\ \Omega _\Lambda = th^2\left(\frac{t} {t_\Lambda }\right);\quad \Omega _m = 1 - \Omega _\Lambda = ch^{-2}\left( \frac{t} {t_\Lambda } \right); \\ \Omega _\Lambda(\tilde t) = \Omega _m(\tilde t) \to \tilde t = \frac{Arsh(1)}{Arth(\sqrt {\Omega _{\Lambda 0} })}t_0 \simeq 9.5\cdot10^9 \,years \\ \end{gathered} $$

$$ \begin{gathered} \frac{\ddot a} {a} = \frac{4\pi G} {3}\rho_{cr}\left[2\Omega_{\Lambda 0} - \Omega _{m0}(1 + z)^3\right]; \\ 2\Omega _{\Lambda 0} = \Omega _{m0}(1 + z^*)^3; \\ z^* = \left( \frac{2\Omega_{\Lambda 0}} {\Omega _{m0}}\right)^{1/3} - 1; \\ \Omega _{\Lambda 0}\simeq 0.73;\quad \Omega _{m0}\simeq 0.27; \\ z^* \simeq 0.745 \\ \end{gathered} $$ 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 $$ z^* = 0.745 $$

$$ \frac{d\eta} {d\ln a} = a\frac{dt}{ada} = \frac{1}{aH} = \frac{1}{\dot a} $$ 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 $z=0.745$

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 $(z=0.45).$

$$ \begin{gathered} z_{eq} = \left(\frac{\Omega _\Lambda} {\Omega _{m0}} \right)^{1/3}- 1; \\ z_{accel}= \left(\frac{2\Omega_\Lambda } {\Omega _{m0}}\right)^{1/3}- 1; \\ z_{accel} = 2^{1/3}\left(z_{eq}+ 1\right) - 1 \\ \end{gathered} $$

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.

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. $$ \rho_{V}\simeq (0.75\pm 0.05)\times 10^{-26}\, kg/m^3. $$

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.

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: $$ H_0d_L \simeq 1.2 $$

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.

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, TASI Lectures on the cosmological constant)

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.

\[\nu_{max}=10^{12}Hz.\]

If the zero-point energy density is infinite, then our Universe is infinitely curved and the space size is infinitely small.

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.

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 $N.$

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}

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.)

• 

  Dependence of the deceleration parameter on the redshift.

\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}

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

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.

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.

$$ h^2 (x) = \Omega _{m0} x^3 + (1 - \Omega _{m0} )x^\alpha ;\quad \alpha = 3(1 + w) $$ 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.

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.

• 

  Dependence of the deceleration parameter on the redshift.

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.

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 $q=-1$.

a) \begin{equation} H(z)=H_0 (1+z)^{3/2}\sqrt{\Omega_m+\Omega_{\Lambda}\exp\left[3\int_{0}^{z}\frac{w(z')}{(1+z')}dz'\right]}\,; \end{equation} b) \begin{equation} H(z)=H_0\left\{ \Omega_m (1+z)^3+\Omega_k(1+z)^2+\Omega_{\Lambda} (1+z)^{3(1+w_0+w_a})\exp[-3w_az/(1+z)]\right\}^{1/2}\,; \end{equation} c) \begin{equation} H(z)=H_0 \sqrt{\Omega_m(1+z)^3+(1-\Omega_m)}\,. \end{equation}

Excluding the $\ddot{a}$ between the equations (\ref{EoS1},\ref{EoS2}) rewrite the system as \begin{equation} \dot {a}^{2} = \frac{8\pi G}{3}\rho a^2 + \frac{\Lambda(t) }{3}a^2 - \frac{k}{R_0^2} \label{frm3} \end{equation} taking the derivative we obtain \begin{equation} 2\dot{a}\ddot{a}=\frac{8\pi G}{3}\dot{\rho}a^2+ \frac{8\pi G}{3}\rho 2 a \dot{a} + \frac{\dot \Lambda(t) }{3} a^2+\frac{\Lambda(t) }{3} 2 a \dot{a} \end{equation} further dividing both sides by $2 a \dot{a}$ we have \begin{equation} \frac{\ddot {a}}{a} = \frac{4\pi G}{3}\dot{\rho}\frac{a}{\dot {a}}+\frac{8\pi G}{3}{\rho}+\frac{\dot \Lambda(t) }{6} \frac{a}{\dot {a}}+\frac{\Lambda(t) }{3}. \label{frm5} \end{equation} Now substitute into the left side of the equation (\ref{frm5}) the right side of (\ref{EoS2}) we write \begin{equation} -\frac{4\pi G}{3}\rho -\frac{4\pi G}{3} 3p + \frac{\Lambda(t) }{3}=\frac{4\pi G}{3}\dot{\rho}\frac{a}{\dot {a}}+\frac{8\pi G}{3}{\rho}+\frac{\dot \Lambda(t) }{6} \frac{a}{\dot {a}}+\frac{\Lambda(t) }{3} \end{equation} resulting in similar we obtain conservation equation for the density of matter \begin{equation} \dot{\rho}=-3\frac{\dot {a}}{a}(\rho+p)-\frac{\dot \Lambda(t) }{8\pi G} \label{rho_der} \end{equation} Solving the differential equation \eqref{rho_der} finally obtain: \begin{equation} \rho=a(t)^{-3(1+w)}\left(-\int a(t)^{3(1+w)}\frac{\dot \Lambda(t) }{8\pi G}dt+C\right) \label{rho_t} \end{equation} where $C$ - constant of integration.