# Friedman equations

## Contents

- 1 Problem 1: derivation of Friedman equations
- 2 Problem 2: reformulation in terms of conformal time
- 3 Problem 3: relations between equations
- 4 Problem 4: source of gravity in GR in the weak field limit
- 5 Problem 5: expansion and pressure
- 6 Problem 6: second equation for $k=0$
- 7 Problem 7: Lorentz invariance
- 8 Problem 8: critical density
- 9 Problem 9: relative densities
- 10 Problem 10: scale factor via observables
- 11 Problem 11: Newtonian interpretation
- 12 Problem 12: $\dot H$
- 13 Problem 13: Raychaudhuri equation
- 14 Problem 14: conservation equation
- 15 Problem 15: conservation in terms of conformal time
- 16 Problem 16: conservation equations for the expanding Universe
- 17 Problem 17: relations between the three equations
- 18 Problem 18: e-foldings number
- 19 Problem 19: conservation in terms of e-foldings number
- 20 Problem 20: pressure in terms of H
- 21 Problem 21: EoS parameter in terms of H
- 22 Problem 22: hidden symmetry of Friedman equations
- 23 Problem 23: relation between the total pressure and the deceleration parameter
- 24 Problem 24: evolution of the state parameter
- 25 Problem 25: EoS via the expansion dynamics
- 26 Problem 26: upper bound on w
- 27 Problem 27: non-relativistic state parameter
- 28 Problem 28: condition $\dot X = 0$

### Problem 1: derivation of Friedman equations

Starting from the Einstein equations, derive the equations for the scale factor $a(t)$ of the FLRW metric-the Friedman equations: \begin{align}\label{FriedmanEqI} \Big(\frac{\dot a}{a}\Big)^2& =\; \frac{8\pi G}{3}\rho -\frac{k}{a^2};\\ \label{FriedmanEqII} \frac{\ddot a}{a} &=- \frac{4\pi G}3\big(\rho+3p). \end{align} Consider matter an ideal fluid with $T^{\mu}_{\nu}=diag(\rho,-p,-p,-p)$.

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 explicit expressions for the Ricci tensor components of of the FLRW metric, one obtains from the component $\binom{0}{0}$ the following equation: \[ - 3\frac{\ddot a}{a} = 8\pi G \cdot \frac{1}{2}(\rho + 3p),\] 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 the problem on cosmological energy-momentum tensor.

### Problem 2: reformulation in terms of conformal time

Derive the Friedman equations in terms of conformal time.

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*}

### Problem 3: relations between equations

Show that the first Friedman equation is the first integral of the second one.

### Problem 4: source of gravity in GR in the weak field limit

Show that in the weak field limit of General Relativity the source of gravity is the quantity $(\rho + 3p)$.

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.

### Problem 5: expansion and pressure

How does the magnitude of pressure affect the expansion rate?

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.

### Problem 6: second equation for $k=0$

Consider the case $k = 0$ and show that the second Friedman equation can be presented in the form \[HH' = - 4\pi G\left(\rho + p\right),\] where $H' \equiv \frac{dH}{d\ln a}.$

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 \[H'H = - 4\pi G(\rho + p).\]

### Problem 7: Lorentz invariance

Are solutions of Friedman equations Lorentz-invariant?

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.

### Problem 8: critical density

The critical density corresponds to the case of spatially-flat Universe $k=0$. Determine its actual value.

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

### Problem 9: relative densities

Show that the first Friedman equation can be presented in the form \[\sum\limits_i\Omega_i=1,\] where $\Omega_i$ are relative densities of the components, \[\Omega_i\equiv\frac{\rho_{i}}{\rho_{cr}}, \quad \rho_{cr}=\frac{3H^{2}}{8\pi G}, \quad \rho_{curv}=-\frac{3}{8\pi G}\frac{k}{a^2},\] and $\rho_{curv}$ describes the contribution to the total density of the spatial curvature.

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 \[1 = \sum\limits_i \frac{8\pi G}{3H^2} \left( \rho _i - \frac{3}{8\pi G}\frac{k}{a^2} \right) = \sum\limits_i \frac{8\pi G}{3H^2}\left( \rho _i + \rho _{curv} \right) = \sum\limits_i \Omega_i .\]

### Problem 10: scale factor via observables

Express the scale factor $a$ in a non-flat Universe through the Hubble's radius and total relative density $\rho/\rho_{cr}$.

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

### Problem 11: Newtonian interpretation

Show that the relative curvature density $\rho_{curv}/\rho_{cr}$ in a given region can be interpreted as a measure of difference between the average potential and kinetic energies in the region.

### Problem 12: $\dot H$

Prove that in the case of spatially flat Universe \[\dot{H} = - 4\pi G (\rho + p).\]

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.

### Problem 13: Raychaudhuri equation

Obtain the Raychaudhuri equation \[H^2 + \dot H = - \frac{4\pi G}{3}(\rho + 3p).\]

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

### Problem 14: conservation equation

Starting from the Friedman equations, obtain the conservation equation for matter in an expanding Universe: \[\dot{\rho}+3H(\rho+p)=0.\] Show that it can be presented in the form \[\frac{d\ln\rho}{d\ln a}+3(1+w)=0,\] where $w=p/\rho$ is the state parameter for matter.

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.

### Problem 15: conservation in terms of conformal time

Obtain the conservation equation in terms of the conformal time.

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

### Problem 16: conservation equations for the expanding Universe

Starting from the energy momentum conservation law $\nabla_{\mu}T^{\mu\nu}=0$, obtain the conservation equation for the expanding Universe.

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 explicit expressions for ${\Gamma^{i}}_{jk}$, one obtains \[a^3\frac{dp}{dt} = \frac d{dt}\left[ a^3(\rho(t) + p(t))\right],\] and collecting the terms, \[\dot \rho + 3\frac{\dot a}{a}(\rho + p) = 0.\]

### Problem 17: relations between the three equations

Show that among the three equations (two Friedman equations and the conservation equation) only two are independent, i.e. any of the three can be obtained from the two others.

The derivation of the conservation equation from the two Friedman equations was discussed above. 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 same derivation, 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 $k=0,\pm1$.

### Problem 18: e-foldings number

For a spatially flat Universe rewrite the Friedman equations in terms of the $e$-foldings number for the scale factor \[N(t)=\ln\frac{a(t)}{a_0}.\]

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 \[\dot{\rho}+3\dot{N}(\rho+p)=0\]

### Problem 19: conservation in terms of e-foldings number

Express the conservation equation in terms of $N$ and find $\rho(N)$ for a substance with equation of state $p=w\rho$.

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 \[\rho = \rho _0e^{ - 3(1 + w)N}.\]

### Problem 20: pressure in terms of H

For the case of spatially flat Universe express pressure in terms of the Hubble parameter and its time derivative.

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 \[p =- \frac{1}{8\pi G}\left( 2\dot H + 3H^2 \right).\]

### Problem 21: EoS parameter in terms of H

Express the state parameter $w = p/\rho$ in terms of the Hubble parameter and its time derivative.

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

Show that for a spatially flat Universe the Friedman equations are invariant$^*$ under the change of variables to a new scale factor \[a \to \alpha=\frac{1}{a}\] and to the new equation of state: \[(w+1)\to (\omega+1)=-(w+1).\] ${}^{*}$V. Faraoni. A symmetry of the spatially flat Friedmann equations with barotropic fluids. arXiv:1108.2102v1

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, is also unchanged.

This *duality* maps expanding universes into contracting ones (filled with somewhat peculiar matter) and vice-versa.

### Problem 23: relation between the total pressure and the deceleration parameter

Find the relation between the total pressure and the deceleration parameter for a flat Universe.

From the conservation equation follows that \[p = - \frac{\dot \rho }{3H}\frac{w}{1 + w}.\] Using \[\begin{array}{l} w = \frac{2q - 1}{3}; \\ \dot \rho = \frac{3}{4\pi G}H\dot H,\quad \dot H = - H^2 (1 + q) \\ \end{array}\] find $$p = \frac{H^2 }{8\pi G}\left( {2q - 1} \right)$$

### Problem 24: evolution of the state parameter

Evaluate the derivatives of the state parameter $w$ with respect to cosmological and conformal times.

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 \[w'=3\mathcal{H}(w+1)(w-c_{s}^{2}).\]

### Problem 25: EoS via the expansion dynamics

Consider an FLRW Universe dominated by a substance, such that Hubble parameter depends on time as $H=f(t)$, where $f(t)$ is an arbitrary differentiable function. Find the state equation for the considered substance.

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

### Problem 26: upper bound on w

Find the upper bound for the state parameter $w$.

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

### Problem 27: non-relativistic state parameter

Show that for non-relativistic particles the state parameter $w$ is much less than unity.

For 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 \[w=\frac{p}{\rho} =\frac{\left<v^2\right>}{3 c^2}\ll 1.\]

### Problem 28: condition $\dot X = 0$

The Friedman equation can be regarded as a constraint of the form \[ X \equiv H^2 + \frac{k}{a^2} - \frac{8\pi G}{3}\rho = 0 \] Show that the conservation equation $\dot \rho + 3H(\rho + p)$ represents the condition $\dot X = 0$.

\[ \dot X = 2H\left( {\dot H - \frac{k}{a^2}} \right) - \frac{8\pi G}{3}\dot \rho \] Using \[\dot H = \frac{k}{a^2} - 4\pi G\left( {\rho + p} \right),\] one obtains \[\dot X = \dot \rho + 3H\left( {\rho + p} \right) = 0.\]