It is convenient to use the Einstein equations in the form \[R^\mu_\nu = 8\pi G\left( T^\mu_\nu - \frac{1}{2}\delta _\nu ^\mu T \right).\] Using the results of problems this, this and this, one obtains from the component $\binom{0}{0}$ the following equation: \[ - 3\frac[[:Template:\ddot a]]{a} = 8\pi G \cdot \frac{1}{2}(\rho + 3p),\] and from the $\binom{1}{1}$ component \[- \frac{2{\dot a}^2 + a\ddot a + 2k}{a^2} = 8\pi G \cdot \frac{1}{2}(p - \rho ).\] The components $\binom{2}{2}$ and $\binom{3}{3}$ give the same equations as $\binom{1}{1}$, and the off-diagonal ones reduce to the identity$^*$ $0=0$. Excluding the second derivative from the two independent equations one obtains the first Friedman equation, and on excluding the first derivative --- the second Friedman equation. $^*$This is why the stress-energy tensor was taken in the form of stress-energy tensor of ideal fluid; otherwise we would have come to contradiction. More on this see in problem .

In order to rewrite the Friedman equations in terms of conformal time $dt=a(\eta)d\eta$, we note that \[\frac{d}{dt} =\frac{d\eta}{dt}\frac{d}{d\eta} =\frac{1}{a}\frac{d}{d\eta},\] and therefore \[\left(\frac{\dot{a}}{a}\right)^2 =\frac{1}{a^2}\left(\frac{da}{d\eta} \frac{d\eta}{dt}\right)^2 =\frac{(a')^2}{a^4},\] where the prime denotes the derivative with respect to the conformal time $\eta$. Then one gets \begin{align*} \ddot{a}=&\frac{d}{dt} \frac{da}{dt}= \frac{d}{dt}\frac{da}{dt} =\frac{d}{dt}\frac{da}{d\eta}\frac{d\eta}{dt} =\frac{d}{dt}\frac{a'}{a}=\\ =&\frac{d\eta}{dt}\frac{d}{d\eta}\left( \frac{a'}{a}\right)= \frac{1}{a}\left(-\frac{a'^2}{a^2}+\frac{a''}{a}\right)= -\frac{a'^2}{a^3}+\frac{a''}{a^2}. \end{align*} After substitution of the derivatives into the Friedman equations one obtains the latter in the conformal time: \begin{align*} & (a')^2 +ka^2=\frac{8\pi G}{3}\rho a^4,\\ &a'' +ka=\frac{4\pi G}{3}\left(\rho-3p\right)a^3. \end{align*}

The density distribution $\rho$ enters the Poisson equation $\Delta \Phi = 4\pi G\rho$ and serves as the source of gravitational field in the Newtonian theory. In order to obtain the analogue of this equation in the frame of General Relativity, consider the Newtonian approximation of the Einstein theory: let $g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$, where $h_{\mu\nu}$ is a small perturbation to the Minkowski metric $\eta_{\mu\nu}$. In this limit $h_{00}=1+2\Phi$, where $\Phi$ is the Newtonian gravitational potential. Then we linearize the Einstein equations \[R_{\mu \nu } = 8\pi G\left( T_{\mu \nu } - \frac{1}{2}Tg_{\mu \nu } \right)\] with respect to $h_{\mu\nu}$. In the first order of approximation we can neglect terms quadratic on $\Gamma$ in the Ricci tensor \begin{equation}\label{2-53} R_{\nu \sigma } = \Gamma _{\nu \sigma ,\mu }^\mu -\Gamma _{\nu \mu ,\sigma }^\mu + \Gamma _{\nu \sigma }^\alpha \Gamma _{\alpha \mu }^\mu -\Gamma _{\nu \mu }^\alpha \Gamma _{\alpha \sigma}^\mu, \end{equation} Also, for stationary metrics the second term in the $00$ component equals to zero, so \[R_{00} = \partial _\alpha \Gamma _{00}^\alpha= \frac{1}{2}\Delta g_{00} = \Delta \Phi.\] It is easy to see that $T_{\mu \nu }$ is of the first order of smallness with respect to $\Phi$, so from the $00$ component of the Einstein equation we obtain \[\Delta \Phi = 8\pi G\left( T_{00} - \frac{1}{2}T \right) = 8\pi G\left( \rho - \frac{1}{2}(\rho - 3p) \right) = 4\pi G(\rho +3p).\] Therefore the source of gravitational field is not $\rho$, but the combination $\rho + 3p$. In particular, an object satisfying the condition $\rho + 3p < 0$, will repel, rather than attract, non-relativistic particles.

The increase of positive pressure leads to additional deceleration of expansion of the Universe. It absolutely contradicts our intuitive conceptions: strongly pressured substance expands. However, cosmological medium with positive pressure acts in the opposite way. Our intuition fails because force is determined by the gradient of pressure, and in the uniform medium (such as our Universe on the considered scales) there are no gradients of pressure. The pressure term in the second Friedman equation is a purely relativistic effect, that can only be described by General Relativity.

First of all, \[\dot{H}=\frac{\ddot a}{a}-{H^2},\] so it follows from the Friedman equations that \[\dot H = - \frac{4\pi G}{3}\left( \rho + 3p\right) - \frac{8\pi G}{3}\rho =- 4\pi G(\rho + p).\] On the other hand, $\dot H = H'H$, and one obtains the first Friedman equation in the required form \[H'H = - 4\pi G(\rho + p).\]

Solution of the Friedman equations violate the Lorentz invariance. In a Friedmannian Universe there is a selected cosmological time $t,$ which is the proper time of the comoving observer, who sees the spatially homogeneous and isotropic Universe. The situation is typical for physics: equations of General relativity are invariant with respect to Lorentz transformations, but particular solutions of the equations are in general not.

Spatial flatness means $k=0$. Then making use of the definition \[H=\frac{\dot a }a, \quad H_0^2 = \frac[[:Template:8\pi G]]{3}{\rho_{cr0}},\] where $H_0$ is the present value of Hubble's parameter, we obtain from the first Friedman equation \[{\rho_{cr0}} = \frac[[:Template:3H 0^2]][[:Template:8\pi G]] \approx 1.12 \cdot {10^{ - 29}}\mbox{\it g/cm}^3\]

The first Friedman equation reads \[H^2 = \frac{8\pi G}{3}\sum\limits_i {\rho _i} - \frac{k}{a^2},\] where $\sum\limits_i \rho _i$ is the total density of all components contributing to the Universe' composition. Then \[1 = \sum\limits_i \frac{8\pi G}{3H^2} \left( \rho _i - \frac{3}{8\pi G}\frac{k}{a^2} \right) = \sum\limits_i \frac{8\pi G}{3H^2}\left( \rho _i + \rho _{curv} \right) = \sum\limits_i \Omega_i .\]

Extracting the curvature term from the Friedman equations. we can write \[\Omega=1+\frac{k}{a^2 H^2},\] so \[a= \frac{H^{-1}}{\big|\Omega- 1\big|^{1/2}}.\]

As $H = \frac{\dot a}{a}$, we have $\dot H = \frac{\ddot a}{a} - {H^2}$. Plugging the Friedman equation into the right hand side, we get the required equality.

The required result is obtained by straightforward plugging of the second Friedman equations into $\frac{\ddot a}{a} = H^2 + \dot H$.

First, we rewrite the first Friedman equation in the form \[\dot a^2 = \frac{8\pi G}{3}\rho a^2 - k.\] On differentiating it with respect to time, we get \[ 2\dot a\ddot a = \frac{8\pi G}{3}\left( \dot \rho a^2 + 2a\dot a\rho \right),\] and excluding $\ddot a$ with the help of the second Friedman equation, obtain \[\dot \rho + 3\frac{\dot a}{a}(\rho + p) = 0.\] Introducing the state parameter $w=p/\rho$, it is easy to rewrite it in terms of logarithms.

Let the prime denote the derivative with respect to conformal time $\eta,$ then for $i-$th component of energy density one obtains \[\rho '_i(\eta ) + 3\mathcal{H}\left( 1 + w_i \right)\rho_{i} = 0.\]

The energy-momentum tensor has the form: $T^{\mu}_{\nu}=diag\{\rho,- p,-p,-p\}$. The conservation law $\nabla_{\nu}T^{\mu\nu}=0$ can be rewritten in terms of Christoffel symbols as \[ 0=\nabla_{\nu}T^{\mu\nu} =\partial_{\nu}T_{\mu}^\nu - T_\alpha ^\nu \Gamma _{\mu \nu }^\alpha + T_\mu ^\alpha \Gamma _{\alpha \nu }^\nu,\] so that after substitution of explicit expressions for ${\Gamma^{i}}_{jk}$, obtained in problem, one obtains \[a^3\frac{dp}{dt} = \frac d{dt}\left[ a^3(\rho(t) + p(t))\right],\] and collecting the terms, \[\dot \rho + 3\frac[[:Template:\dot a]]{a}(\rho + p) = 0.\]

The derivation of the conservation equation from the two Friedman equations is discussed in problem. Here we consider the following two cases: $a)$ how to obtain the second Friedman equation from the first one plus the conservation law, and $b)$ how to obtain the first Friedman equation from the second one plus the conservation law. $a)$ We follow the idea of problem, but now exclude $\dot{\rho }$ with the help of conservation law to obtain: \[\ddot{a}=-\frac{4\pi G}{3}(\rho +3p)a\] $b)$ We use the conservation law to exclude pressure from the second Friedman equation and multiply both sides of the equation by $\dot{a}$: \[\ddot{a}\dot{a}=\frac{4\pi }{3}G\left( 2\rho a\dot{a}+\dot{\rho }{{a}^{2}} \right)\] Note that both sides of the equation are total derivatives: \[\frac{1}{2}\frac{d}{dt}{{\dot{a}}^{2}}=\frac{4\pi }{3}G\frac{d}{dt}\left( \rho {{a}^{2}} \right).\] Setting the integration constant equal to $k$ leads to \[{{\dot{a}}^{2}}=\frac{8\pi G}{3}\rho {{a}^{2}}-k.\] By rescaling of $a$ one can always set $k=0,\pm1$ (see problem).

In such variables $H=\dot{N}$, and if we additionally use the units in which $8\pi G=1$, we can bring them to \begin{align*} \rho&=3(\dot{N})^{2};\\ p&=-2\ddot{N}-3(\dot{N})^{2}. \end{align*} The conservation equation is \[\dot{\rho}+3\dot{N}(\rho+p)=0\]

From the previous problem we see that the conservation equation can be written in the form \[\frac{d\rho }{\rho} + 3(1 + w)dN = 0.\] Thus its solution is \[\rho = \rho _0e^{ - 3(1 + w)N}.\]

Excluding density $\rho$ from the Friedman equations, we obtain \[H^{2}+2\frac{\ddot{a}}{a}=-8\pi G \,p.\] Taking into account that $\dot{H}=\frac{\ddot{a}}{a}-H^2$, we have \[p =- \frac{1}{8\pi G}\left( 2\dot H + 3H^2 \right).\]

Using the results of the previous problem, we obtain \[w = \frac{p}{\rho } = - 1 - \frac{2}{3}\frac{\dot H}{H^2}.\]

As \[\frac{\dot{\alpha}}{\alpha}=-\frac{\dot{a}}{a},\] we have the new Hubble parameter \[\mathcal{H}=\frac{\dot{\alpha}}{\alpha}=-H\] and the first equation remains unchanged. The conservation equation can be then rewritten as \[0=\dot{\rho}+3H(1+w)\rho =\dot{\rho}+3\mathcal{H}(1+\omega)\rho,\] and thus is invariant too. Then the remaining equation, which can be derived from these two (see problem) is also unchanged. This duality maps expanding universes into contracting ones (filled with somewhat peculiar matter) and vice-versa.

Note that \[\dot{w} =\frac{d}{dt}\left( \frac{p}{\rho } \right) =\frac{\dot{p}\rho -p\dot{\rho }}{\rho ^2} =\frac{p}{\rho }\; \frac{\dot{\rho}}{\rho }\; \left( \frac{\dot{p}}{\dot{\rho}} \frac{\rho }{p}-1 \right).\] Using the definition of the adiabatic velocity of sound \[c_{s}^{2}=\frac{dp}{d\rho } =\frac{\dot{p}}{\dot{\rho }}\] and the conservation equation in the form \[\frac{{\dot{\rho }}}{\rho }=-3H(1+w),\] we obtain \[\dot{w}=3H(w+1)\left( w-c_{s}^{2} \right).\] From $\dot{w}=w'/a$ and $H=\mathcal{H}/a$ it is easy to come to \[w'=3\mathcal{H}(w+1)(w-c_{s}^{2}).\]

Including the curvature term in total density, we have (see problem) \[\rho_\mathrm{total} = \frac{3}{\kappa^2}H^2, \quad p_\mathrm{total}=-\frac{1}{\kappa^2} \left(2\dot H + 3H^2\right),\] where $\kappa^2 = 8\pi G$. Then from the first Friedman equation $H^2=\frac{\kappa^2}{3}\rho$ \[p=-\rho -\frac{2}{\kappa^2} \frac{d}{d\ln a} f'\left(f^{-1}\left( \kappa\sqrt{\frac{\rho}{3}}\right)\right).\]

Assuming that pressure $p$ depends only on density $\rho$ (the medium is barotropic), the velocity of sound in the liquid equals to: $$ c_s^2 = \frac[[:Template:Dp]][[:Template:D\rho]].$$ Due to the causality condition $c_s^2 = w \le 1$ (note that we use units with $c = 1$).

For a non-relativistic gas \[p=nkT=\frac{p}{mc^2}\cdot \frac{m\left<v^2\right>}{3} =\rho \frac{\left<v^2\right>}{3 c^2},\] so \[w=\frac{p}{\rho} =\frac{\left<v^2\right>}{3 c^2}\ll 1.\]