Physics 7xx: Cosmic inflation-II. Friedmann cosmology

This is the second part of this three part discussion of inflationary re-heat. This portion is in fact a Physics 711 problem that I like to assign during our trip through Goldstein’s chapter on relativity.

The Einstein field equations (note the signature) are

$G_{\mu\nu}=R_{\mu\nu}-{1\over 2} R \, g_{\mu\nu}={8\pi G\over c^4} \, T_{\mu\nu}\quad \mbox{keeping the G's and c's}$

and these can lead to a cosmology in which the universe collapses. Supposedly this troubled Einstein, so he added an ad-hoc cosmological term

$R^{\mu\nu}-{1\over 2} R \, g^{\mu\nu}+\Lambda g^{\mu\nu}={8\pi G\over c^4} \, T^{\mu\nu}, \qquad T^{\mu\nu}=P(g^{\mu\nu}+u^\mu u^\nu)+\rho c^2 u^\mu u^\nu$

which today comprises the entirety of our understanding of whatever it is that is causing the universe to expand at an increasing rate. It would be great if you could fill in this gap in our knowledge, and let me know how it all turns out…

Flat Friedmann cosmology assumes a very simple metric

$ds^2=a^2(t)\Big((dx)^2+(dy)^2+(dz)^2\Big)-c^2 \, (dt)^2$

which describes an inflating universe. Recent evidence of the homogeneity of the CMB suggests that the curvature is vanishingly small (talk to Peter Timbie or attend his cosmology journal club meetings to get the full story).
$\bullet$ Find all components of $R_{\mu\nu\rho\sigma}$ preferentially by writing a program in some CAS (please do not use a canned library or module, write your own!) or by the Cartan calculus/differential forms method, or simply by good old brute force and dogged perserverance. Find $R_{1212}$ by hand. Then find all components of the Ricci tensor, and the scalar curvature. and the Einstein curvature $G_{\mu\nu}$ .
Finally work out the $00$ component and the $11$ components of the field equations for a perfect fluid (the $ii$ for $i=1,2,3$ components are the same) including the cosmological term. Evaluate $T_{\mu\nu}$ in the rest frame of the fluid.

I will compute everything by hand here … From the metric

$ds^2=a^2(t)\Big((dx)^2+(dy)^2+(dz)^2\Big)-c^2 \, (dt)^2$

The only nonzero Christofel symbols must have one time $"0=4"$ index, and two other indices the same because $g$ is symmetric, these fall into two categories (no sum on $i$ )

$\Gamma^i \, _{i0}={1\over 2}g^{ii}\Big(\partial_0 g_{ii}+\partial_i g_{i0}-\partial_i g_{i0}\Big)={1\over 2}a^{-2}{da^2\over dt}={\dot{a}\over a}$

$\Gamma^0 \, _{ii}={1\over 2}g^{00}\Big(\partial_i g_{i0}+\partial_i g_{0i}-\partial_0 g_{ii}\Big)={1\over 2}(-{1\over c^2})(-{da^2\over dt})={a\dot{a}\over c^2}$

From

$R^\mu \, _{\alpha\rho\beta}=\partial_\rho \Gamma^\mu \, _{\alpha\beta}-\partial_\beta \Gamma^\mu \, _{\alpha\rho}+\Gamma^\mu \, _{\rho\sigma}\Gamma^\sigma \, _{\alpha\beta}-\Gamma^\mu \, _{\beta\sigma}\Gamma^\sigma \, _{\alpha\rho}$

we get three classes of nonzero Riemann tensor coefficients (no sum on $i$ );

$R^i \, _{0i0}=\partial_i \Gamma^i \, _{00}-\partial_0 \Gamma^i \, _{0i}+\Gamma^i \, _{ik}\Gamma^k \, _{00}-\Gamma^i \, _{0k}\Gamma^k \, _{0i}=-\partial_0 \Gamma^i \, _{0i}-(\Gamma^i \, _{0i})^2=-{d\over dt}({\dot{a}\over a})-({\dot{a}\over a})^2=-{\ddot{a}\over a}$

$R^0 \, _{ii0}=\partial_i \Gamma^0 \, _{i0}-\partial_0 \Gamma^0 \, _{ii}+\Gamma^0 \, _{ij}\Gamma^j \, _{i0}-\Gamma^0 \, _{0j} \Gamma^j \, _{ii}=-\partial_0 \Gamma^0 \, _{ii}+\Gamma^0 \, _{ii}\Gamma^i \, _{i0}=-{\dot{a}^2\over c^2}-{a\ddot{a}\over c^2}+({\dot{a}\over a})({a\dot{a}\over c^2})=-{a\ddot{a}\over c^2}$

and here no sum on $i,j$ ;

$R^i \, _{jij}=\Gamma^i\, _{ik}\Gamma^k \, _{jj}-\Gamma^i \, _{jk}\Gamma^k \, _{ji}=\Gamma^i \, _{i0}\Gamma^0 \, _{jj}=({\dot{a}\over a}({a\dot{a}\over c^2})={\dot{a}^2\over c^2}, \quad i\ne j$

Remember $R_{abcd}=-R_{abdc}$ and so forth. Ricci…

$R_{jj}=\sum_{i\ne j} R^i \, _{jij}+R^0 \, _{j0j}=2{\dot{a}^2\over c^2}+{a\ddot{a}\over c^2}, \qquad R_{00}=\sum_i R^i \, _{0i0}=-3{\ddot{a}\over a}$

Scalar curvature…

$R=g^{00}R_{00}+3g^{11}R_{11}=-{1\over c^2}({-3\ddot{a}\over a})+3({1\over a^2})(2{\dot{a}^2\over c^2}+{a\ddot{a}\over c^2})={6(\dot{a}^2+a\ddot{a})\over a^2c^2}$

Einstein tensor…no sum on $i$

$G_{ii}=2{\dot{a}^2\over c^2}+{a\ddot{a}\over c^2}-{1\over 2}a^2\Big({6(\dot{a}^2+a\ddot{a})\over a^2c^2}\Big)=-{\dot{a}^2\over c^2}-2{a\ddot{a}\over c^2}, \qquad G_{00}=-3{\ddot{a}\over a}-{1\over 2}(-c^2)\Big({6(\dot{a}^2+a\ddot{a})\over a^2c^2}\Big)=3{\dot{a}^2\over a^2}$

Field equations in the rest frame

$G^{00}=(-{1\over c^2})^2\, 3{\dot{a}^2\over a^2}+\Lambda (-{1\over c^2})={8\pi G\over c^4}\rho$

$G^{ii}+\Lambda a^{-2}=(a^{-2})^2\Big(-{\dot{a}^2\over c^2}-2{a\ddot{a}\over c^2}\Big)+\Lambda \, a^{-2}={8\pi G\over c^4}Pa^{-2}$

$\mbox{Finally...!}\quad 3{\dot{a}^2\over c^2 a^2}-\Lambda={8\pi G\over c^4} \, \rho \, c^2, \qquad -{\dot{a}^2\over a^2 c^2}-2{\ddot{a}\over ac^2}+\Lambda={8\pi G\over c^4} \, P$

An old qualifier question
Before the 2018 restructuring of the qualifier it was common to see problems such as this, from our undergrad relativity/cosmology course…
The first field equation

$\dot{a}=\sqrt{{8\pi G \rho+\Lambda c^2\over 3}} \, a$

can be differentiated with respect to $t$ to get a form

${\ddot{a}\over a}=\alpha \, {\dot{a}\over a}+\beta$

(do this, find $\alpha,\beta$ ) which can be simplified using both field equations to a form

$-\Big(\rho+{P\over c^2}\Big)\gamma=-\delta{\dot{a}\over a}=\dot{\rho}$

Do this, find $\gamma,\delta$ . This equation represents energy conservation in an expanding universe.
The Hubble “constant” $H$ (certainly $t$ -dependent) is taken to be $H={\dot{a}\over a}$ and the first field equation can be written as

${\dot{a}\over a}=H=\sqrt{{8\pi G\rho\over 3}+{\Lambda c^2\over 3}}=\sqrt{{8\pi G\rho\over 3}+\Omega_\Lambda}$

Explain why the density (rest energy density) of ordinary matter will contribute to the right-hand side a term such as the first one appearing here

$H^2=\Big({\dot{a}\over a}\Big)^2=H_0^2\Big(\Omega_{m}a^{-3}+\Omega_r a^{-4}+\Omega_\Lambda\Big)$

and why radiation contributes one such as the second. You should review adiabatic relations and equations of state of a photon gas in order to get this part right.

The solution: The first equation

$\dot{a}=\sqrt{{8\pi G \rho+\Lambda c^2\over 3}} \, a$

can be differentiated

$\ddot{a}=\sqrt{{8\pi G \rho+\Lambda c^2\over 3}} \, \dot{a}+{1\over 2}{8\pi G \dot{\rho}\over \sqrt{3(8\pi G \rho+\Lambda c^2)}} \, a$

$\quad\mbox{or}\quad {\ddot{a}\over a}=\sqrt{{8\pi G \rho+\Lambda c^2\over 3}} \, {\dot{a}\over a}+{1\over 2}{8\pi G \dot{\rho}\over \sqrt{3(8\pi G \rho+\Lambda c^2)}}$

which can be simplified using both equations

$-{4\pi G\over 3}\Big(\rho+{3P\over c^2}\Big)+{\Lambda c^2\over 3}={8\pi G \rho+\Lambda c^2\over 3}+{1\over 2}{8\pi G \dot{\rho}\over \sqrt{3(8\pi G \rho+\Lambda c^2)}}$

$-\Big(\rho+{P\over c^2}\Big)\sqrt{3(8\pi G \rho+\Lambda c^2)}=-3\Big(\rho+{P\over c^2}\Big){\dot{a}\over a}=\dot{\rho}$

which an energy conservation statement.
The Hubble constant $H$ is taken to be $H={\dot{a}\over a}$ and the first equation can be written as

${\dot{a}\over a}=H=\sqrt{{8\pi G\rho\over 3}+{\Lambda c^2\over 3}}$

Matter density should scale with $a$ as all dimensions $x,y,z$ inflate by scale $a$

${U\over V}\rightarrow {U\over a^3V}={\Omega_m\over a^3}$

For a photon gas $U\sim VT^4$ , and Gibbs-Duhem says $U=TS-PV$ so $S\sim VT^3$ , if the photon gas expands with inflation adiabatically

${U\over V}\sim T^4\sim\Big({S\over V}\Big)^{4/3}\sim \Big({S\over a^3V}\Big)^{4/3}\sim {\Omega_r\over a^4}$

so that the Hubble equation becomes

$H^2=\Big({\dot{a}\over a}\Big)^2=H_0^2\Big(\Omega_{m}a^{-3}+\Omega_r a^{-4}+\Omega_\Lambda\Big)$

Another old qualifier question
Solve this for a matter-dominated universe and find the present age of the universe $t_0$ given that $a(t_0)=1$ . Your answer depends on $H_0,\Omega_m$ .

This one is short…

$a^{1/2} \, da=H_0\sqrt{\Omega_m} dt, \qquad a(t)=\Big({3\over 2}H_0\sqrt{\Omega_m} \, t\Big)^{2/3}$

$t_0={2\over 3H_0\sqrt{\Omega_m}}$

…and another (spring 2011)
Hubble’s law says that the velocity with which every point in space recedes from any other point selected as an origin is $\dot{a}=H(t) \, a$ , or ${v_r\over r}=H(t)$ where $H(t)$ is the same everywhere in space but may be changing over time. It is thought to be around $500 km/s/Mpc$ . Consider the earth and draw a sphere of radius $r_0$ around it. The average density of matter within the sphere is $\rho_0=\rho(0)={3M\over 4\pi r_0^3}$ . You can think of any bit of matter within $r_0$ to be on a sphere centered on the earth of radius $r<r_0$ . As the universe expands, spheres of different radii grow in proportion to the radius, so all matter within $r_0$ will be within the sphere once it expands to $r(t)>r_0$ , $r(0)=r_0$ , and so the total matter within any expanding volume remains the same as every volume grows, therefore

$\rho(t)=\rho(0) {r_0^3\over r^3(t)}$

and the matter density drops over time. What must $\rho_0$ be in order that at some point in the distant future the expansion stops?

This can be answered using escape velocity ideas from Physics 201/207. Select a bit of matter $m$ on the initial sphere $r_0$ ; at $t\rightarrow\infty$ it will have slowed and come to rest at $r(\infty)>>r_0$ . Apply Gauss’s law to find the gravitational PE of $m$ ; the total mass within $r_0$ is the same as the total mass within $r(\infty)$ , $M(r_0)=M(r(\infty))$