2.5.1
Problem 2.5.1: 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 $T^{\mu}_{\nu}=diag(\rho,-p,-p,-p)$.
solution
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.
2.5.2
Derive the Friedman equations in terms of conformal time.
solution
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*}
2.5.3
Problem 2.5.3: Relations between equations
Show that the first Friedman equation is the first integral of the second one.
2.5.4
Problem 2.5.4: 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 $(\rho + 3p)$.
solution
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.
2.5.5
Problem 2.5.5: Expansion and pressure
How does the magnitude of pressure affect the expansion rate?
solution
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.
2.5.6
Problem 2.5.6: 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 $H' \equiv \frac{dH}{d\ln a}.$
solution
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
\[H'H = - 4\pi G(\rho + p).\]
2.5.7
Problem 2.5.7: Lorentz invariance
Are solutions of Friedman equations Lorentz-invariant?
solution
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.
2.5.8
Problem 2.5.8: Critical density
The critical density corresponds to the case of spatially-flat Universe $k=0$. Determine its actual value.
solution
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
\[{\rho_{cr0}} = \frac{3H_0^2}{8\pi G}
\approx 1.12 \cdot {10^{ - 29}}g/cm^3\]
2.5.9
Problem 2.5.9: 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.
solution
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
\[1 = \sum\limits_i \frac{8\pi G}{3H^2}
\left( \rho _i - \frac{3}{8\pi G}\frac{k}{a^2} \right) = \sum\limits_i \frac{8\pi G}{3H^2}\left(
\rho _i + \rho _{curv} \right) = \sum\limits_i \Omega_i .\]
2.5.10
Problem 2.5.10: Scale factor via observables
Express the scale factor $a$ in a non-flat Universe through the Hubble's radius and total relative density $\rho/\rho_{cr}$.
solution
Extracting the curvature term from the Friedman equations. we can write
\[\Omega=1+\frac{k}{a^2 H^2},\]
so
\[a= \frac{H^{-1}}{\big|\Omega- 1\big|^{1/2}}.\]
2.5.11
Problem 2.5.11: 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.
2.5.12
Problem 2.5.12: $\dot H$
Prove that in the case of spatially flat Universe
\[\dot{H} = - 4\pi G (\rho + p).\]
solution
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.
2.5.13
Problem 2.5.13: Raychaudhuri equation
Obtain the Raychaudhuri equation
\[H^2 + \dot H = - \frac{4\pi G}{3}(\rho + 3p).\]
solution
The required result is obtained by straightforward plugging of the second Friedman equations into $\frac{\ddot a}{a} = H^2 + \dot H$.
2.5.14
Problem 2.5.14: 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.
solution
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.
2.5.15
Obtain the conservation equation in terms of the conformal time.
solution
Let the prime denote the derivative with respect to conformal time $\eta,$ then for $i-$th component of energy density one obtains
\[\rho '_i(\eta ) +
3\mathcal{H}\left( 1 + w_i \right)\rho_{i} = 0.\]
2.5.16
Problem 2.5.16: 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.
solution
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,
\[\dot \rho + 3\frac{\dot a}{a}(\rho + p) = 0.\]
2.5.17
Problem 2.5.17: 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.
solution
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 $k=0,\pm1$.
2.5.18
Problem 2.5.18: E-foldings number
For a spatially flat Universe rewrite the Friedman equations in terms of the $e$-foldings number for the scale factor
\[N(t)=\ln\frac{a(t)}{a_0}.\]
solution
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
\[\dot{\rho}+3\dot{N}(\rho+p)=0\]
2.5.19
Problem 2.5.19: 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 $p=w\rho$.
solution
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
\[\rho = \rho _0e^{ - 3(1 + w)N}.\]
2.5.20
Problem 2.5.20: Pressure in terms of H
For the case of spatially flat Universe express pressure in terms of the Hubble parameter and its time derivative.
solution
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
\[p =- \frac{1}{8\pi G}\left( 2\dot H + 3H^2 \right).\]
2.5.21
Problem 2.5.21: EoS parameter in terms of H
Express the state parameter $w = p/\rho$ in terms of the Hubble parameter and its time derivative.
solution
Using the results of the previous problem, we obtain
\[w = \frac{p}{\rho } = - 1 - \frac{2}{3}\frac{\dot H}{H^2}.\]
2.5.22
Problem 2.5.22: Hidden symmetry of Friedman equations
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
solution
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.
2.5.23
Problem 2.5.23: Relation between the total pressure and the deceleration parameter
Find the relation between the total pressure and the deceleration parameter for a flat Universe.
solution
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 $$p = \frac{H^2 }{8\pi G}\left( {2q - 1} \right)$$
2.5.24
Problem 2.5.24: Evolution of the state parameter
Evaluate the derivatives of the state parameter $w$ with respect to cosmological and conformal times.
solution
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
\[w'=3\mathcal{H}(w+1)(w-c_{s}^{2}).\]
2.5.25
Problem 2.5.25: 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.
solution
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 $H^2=\frac{\kappa^2}{3}\rho$
\[p=-\rho -\frac{2}{\kappa^2}
\frac{d}{d\ln a}
f'\left(f^{-1}\left(
\kappa\sqrt{\frac{\rho}{3}}\right)\right).\]
2.5.26
Problem 2.5.26: Upper bound on w
Find the upper bound for the state parameter $w$.
solution
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 $c = 1$).
2.5.27
Problem 2.5.27: Non-relativistic state parameter
Show that for non-relativistic particles the state parameter $w$ is much less than unity.
solution
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
\[w=\frac{p}{\rho} =\frac{\left<v^2\right>}{3 c^2}\ll 1.\]
2.5.28
Problem 2.5.28: 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 $\dot X = 0$.
solution
\[
\dot X = 2H\left( {\dot H - \frac{k}{a^2}} \right) - \frac{8\pi G}{3}\dot \rho
\]
Using
\[\dot H = \frac{k}{a^2} - 4\pi G\left( {\rho + p} \right),\]
one obtains
\[\dot X = \dot \rho + 3H\left( {\rho + p} \right) = 0.\]