Wald GR Chapter 5 Solutions: Homogeneous, Isotropic Cosmology

Source: Robert M. Wald, General Relativity, Chapter 5; selected worked problems and related solutions.

Problem prompts are paraphrased from Robert M. Wald, General Relativity (University of Chicago Press, 1984); the equations, notation, numbering, and guidance follow the source. See the official publisher page.

1 Chapter 5, Problem 1: Robertson--Walker Coordinates

Show that the Robertson--Walker metric (5.1.11) can be rewritten as \[ds^2=-d\tau^2+a^2(\tau)\left[ \frac{dr^2}{1-kr^2} +r^2\left(d\theta^2+\sin^2\theta\,d\phi^2\right) \right].\tag{1.1}\] For $k=+1$, determine which part of the three-sphere is covered by these coordinates.

2 Chapter 5, Problem 2: The Friedmann Equations for Curved Spatial Sections

Derive Einstein's equations (5.2.14) and (5.2.15) for the three-sphere $(k=+1)$ and hyperboloid $(k=-1)$ geometries.

The metric tensor is \[g_{\mu\nu} = \operatorname{diag}(-1,a^2/(1-kr^2),a^2r^2,a^2r^2\sin^2\theta).\tag{2.1}\] The Levi-Civita connection is given by \[\Gamma^\rho{}_{\mu\nu} = \frac{1}{2}g^{\rho\sigma}(\partial_\mu g_{\nu\sigma} + \partial_\nu g_{\mu\sigma} - \partial_\sigma g_{\mu\nu}).\tag{2.2}\] We start with \[\Gamma^i{}_{i\tau} = \Gamma^i{}_{\tau i} = \frac{1}{2}g^{ii}(\partial_ig_{ii}) = \frac{\dot a}{a}.\tag{2.3}\] Meanwhile, \[\Gamma^\tau{}_{rr} = -\frac{1}{2}g^{\tau\tau}\partial_\tau g_{rr} = \frac{a\dot a}{1-kr^2},\quad \Gamma^\tau{}_{\theta\theta} = r^2a\dot a, \quad \Gamma^\tau{}_{\phi\phi} = r^2\sin^2\theta a\dot a.\tag{2.4}\] On top of that, we also have \[\Gamma^r{}_{rr} = \frac{1}{2}g^{rr}\partial_rg_{rr} = \frac{kr}{1-kr^2}, \quad \Gamma^r{}_{\theta\theta} = - \frac{1}{2}g^{rr}\partial_\theta g_{rr} = -(1-kr^2)a^2r, \quad \Gamma^r{}_{\phi\phi} = -(1-kr^2)a^2r\sin^2\theta.\tag{2.5}\] Finally, we also have \[\Gamma^\theta{}_{\phi\phi} = -\frac{1}{2}g^{\theta\theta}\partial_\theta g_{\phi\phi} = -\sin\theta\cos\theta, \quad \Gamma^\theta{}_{\theta r} = \Gamma^\theta{}_{r\theta } = \frac{1}{2}g^{\theta\theta}\partial_rg_{\theta\theta} = \frac{1}{r}, \quad \Gamma^\phi{}_{\phi r} = \Gamma^\phi{}_{r\phi } = \frac{1}{r},\tag{2.6}\] and \[\Gamma^\phi_{\theta\phi} = \Gamma^\phi_{\phi\theta} = \cot\theta\tag{2.7}\] with all the rest of the components vanishes.

Now, we are ready to compute the Ricci tensor, which is given by \[R_{\mu\rho} = \partial_\nu \Gamma^\nu{}_{\mu\rho} - \partial_\mu \Gamma^\nu {}_{\nu\rho} + \Gamma^\alpha{}_{\mu\rho}\Gamma^\nu{}_{\alpha\nu} - \Gamma^\alpha{}_{\nu\rho}\Gamma^\nu{}_{\alpha\mu}.\tag{2.8}\] \[R_{\tau\tau} = -3\partial_\tau\Gamma^{r}{}_{r\tau} - 3\Gamma^r{}_{r\tau}\Gamma^r{}_{r\tau} = -3\frac{\ddot a}{a}.\tag{2.9}\] Note that $R_{**} = \frac{1-kr^2}{a^2}R_{rr}$, and \[\begin{aligned} R_{rr} &= \partial_\tau\Gamma^\tau{}_{rr} +\partial_r\Gamma^r{}_{rr} -\partial_r\Gamma^r{}_{rr} - 2\partial_r\Gamma^\theta{}_{\theta r}\\ &\qquad+ (\Gamma^r{}_{rr})^2 + 2\Gamma^r{}_{rr}\Gamma^\theta{}_{r\theta}+3\Gamma^\tau{}_{rr}\Gamma^r{}_{r\tau}- (\Gamma^r{}_{rr})^2 - 2\Gamma^\tau{}_{rr}\Gamma^r{}_{\tau r} - 2(\Gamma^\theta{}_{\theta r})^2\\ &=\frac{\dot a^2 + a \ddot a}{1-kr^2} + \frac{2}{r^2} +\frac{\dot a^2}{1-kr^2} +2\frac{k}{1-kr^2}+ \frac{2}{r^2}\\&= \frac{2\dot a^2 + a \ddot a}{1-kr^2} + \frac{2k}{1-kr^2}, \end{aligned}\tag{2.10}\] which gives \[R_{**} = \left(\frac{\ddot a}{a} + 2\frac{\dot a^2+k}{a^2}\right).\tag{2.11}\] The Ricci scalar is given by \[R = -R_{\tau\tau} + 3R_{**} = 6\left(\frac{\ddot a}{a} + \frac{\dot a^2+k}{a^2}\right).\tag{2.12}\] Therefore, we find Einstein's equations to be \[G_{\tau\tau} = 3\frac{\dot a^2+k}{a^2} = 8\pi \rho,\tag{2.13}\] \[G_{**} = -2\frac{\ddot a}{a} - \frac{\dot a^2+k}{a^2} = 8\pi P.\tag{2.14}\] Manipulating a bit, we find \[\frac{3\dot a ^2}{a^2} = 8\pi \rho - \frac{3k}{a^2},\tag{2.15}\] \[\frac{3\ddot a}{a} = -4\pi(\rho+3P).\tag{2.16}\]

3 Chapter 5, Problem 3: Cosmological Constant Solutions

Use the modified Einstein equation (5.2.17) with cosmological constant $\Lambda$ to write the counterparts of equations (5.2.14) and (5.2.15), including their $\Lambda$ terms. Show that a static solution exists precisely when $k=+1$ and $\Lambda>0$. For a dust-filled Einstein static universe $(P=0)$, relate its radius $a$ to the density $\rho$. Perturb $a$ slightly away from equilibrium and show that the static solution is unstable.

Einstein's equations become \[G_{\tau\tau} -\Lambda = 8\pi \rho,\tag{3.1}\] \[G_{**} + \Lambda = 8\pi P,\tag{3.2}\] which gives \[\frac{3\dot a ^2}{a^2} -\Lambda = 8\pi \rho - \frac{3k}{a^2},\tag{3.3}\] \[\frac{3\ddot a}{a} - \Lambda= -4\pi(\rho+3P).\tag{3.4}\] A static universe would have $\dot a = \ddot a = 0$, which gives \[\frac{3k}{a^2} = 8\pi \rho + \Lambda,\tag{3.5}\] \[\Lambda = 4\pi(\rho+3P).\tag{3.6}\] It is easy to see that we must have $k = +1$ for the static universe solution to exist. In the case of a dust-filled static universe, we have $a_0 = \sqrt{1/4\pi\rho}$ and $\Lambda = 4\pi \rho$.

Consider perturbing the universe to $a = a_0 + \delta a$ with $\delta a / a_0 \ll 1$, and release it ($\dot{\delta a} = 0$). Through mass conservation, we have $\rho_0a_0^3 = \rho(a_0 + \delta a)^3$. We find \[\ddot{\delta a} = \frac{\delta a}{a_0^2},\tag{3.7}\] which means that the static universe is not stable.
Set $\Lambda>0$ and $T_{ab}=0$ in the modified Einstein equation. Find the spatially homogeneous and isotropic solution for $k=0$. This gives de Sitter spacetime; the $k=\pm1$ solutions correspond to other choices of spacelike hypersurfaces in the same spacetime.

The equations become \[\frac{3\dot a^2}{a^2} = \Lambda,\quad \frac{3\ddot a}{a} = \Lambda.\tag{3.8}\] This has a solution \[a(\tau) = Ae^{\sqrt{\Lambda/3}\tau}.\tag{3.9}\]

4 Chapter 5, Problem 4: Cosmological Redshift

Derive the redshift relation (5.3.6) through the following steps.

Show that \[\nabla_a u_b=\frac{\dot a}{a}h_{ab}, \qquad \dot a=\frac{da}{d\tau},\tag{4.1}\] where $h_{ab}$ is defined in equation (5.1.10).

In a local chart, we have $u^\mu = (1,0,0,0)$. The FRW metric can be written as \[ds^2 = -d\tau^2 +h_{ij}(\tau)dx^i dx^j.\tag{4.2}\] The covariant derivative is therefore \[\nabla_\mu u_\nu = \partial_\mu u_\nu - \Gamma^\rho{}_{\mu\nu}u_\rho = \Gamma^\tau{}_{\mu\nu}.\tag{4.3}\] In problem 5.2, we found \[\Gamma^\tau{}_{ij} = \frac{\dot a}{a}h_{\mu\nu}(\tau),\tag{4.4}\] where \[h_{\mu\nu}= \begin{pmatrix} 0 &0\\0&h_{ij} \end{pmatrix}.\tag{4.5}\] Therefore, we find \[\nabla_\mu u_\nu = \frac{\dot a}{a}h_{\mu\nu}(\tau).\tag{4.6}\]
Along any null geodesic, prove \[\frac{d\omega}{d\lambda} =-k^ak^b\nabla_a u_b =-\frac{\dot a}{a}\omega^2,\tag{4.7}\] where $\lambda$ is an affine parameter.

Let the photon have coordinate $x^\mu(\lambda)$, then $k^\mu = dx^\mu/d\lambda$. In our case, \[\frac{d\omega}{d\lambda} = \frac{d x^\mu}{d\lambda}\partial_\mu \omega =k^\mu\nabla_\mu \omega = -k^\mu \nabla_\mu (k^\nu u_\nu).\tag{4.8}\] Since $k^\mu$ satisfies the geodesic equation, we have $k^\mu \nabla_\mu k^\nu =0$. Therefore, we have \[\frac{d\omega}{d\lambda} = - \frac{\dot a}{a}k^\mu k^\nu h_{\mu\nu} = - \frac{\dot a}{a}k^\mu k^\nu (u_\mu u_\nu + g_{\mu\nu}) = -\frac{\dot a}{a}\omega^2.\tag{4.9}\]
Integrate the result of part (b) to obtain equation (5.3.6).

Rearranging, we find \[\frac{d\omega}{\omega^2} = - \frac{d\lambda}{d\tau}\frac{da }{a}.\tag{4.10}\] Note that \[\omega = - g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\tau} = - g_{\mu\nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}\frac{d\tau}{d\lambda} = \frac{d\tau}{d\lambda}.\tag{4.11}\] Therefore, we find \[\frac{d\omega}{\omega} = -\frac{da}{a} \implies \frac{\omega_1}{\omega_2} = \frac{a(\tau_2)}{a(\tau_1)}.\tag{4.12}\]

5 Chapter 5, Problem 5: Radial Null Geodesics and Global Light Travel

Consider a radial null geodesic in a Robertson--Walker cosmology (5.1.11), so that \[\frac{d\theta}{d\lambda} =\frac{d\phi}{d\lambda}=0.\tag{5.1}\]

For each of the three spatial geometries, show that the change in the coordinate $\psi$ between times $\tau_1$ and $\tau_2$ is \[\Delta\psi=\int_{\tau_1}^{\tau_2}\frac{d\tau}{a(\tau)}.\tag{5.2}\] In the flat case $(k=0)$, $\psi$ is the usual radial coordinate, \[\psi=(x^2+y^2+z^2)^{1/2}.\tag{5.3}\]

For the null geodesic, $ds = 0$. Therefore, we have \[0 =-\frac{d\tau^2}{d\lambda^2} + a^2(\tau) \frac{d\psi^2}{d\lambda^2} \implies \Delta\psi = \int_{\tau_1}^{\tau_2}\frac{d\tau}{a(\tau)}.\tag{5.4}\]
In the dust-filled spherical model, prove that a ray emitted at the big bang completes exactly one circuit of the universe by the big crunch.

For the dust-filled spherical model, we have \[a =\frac{1}{2}C(1-\cos\eta), \quad \tau = \frac{1}{2}C(\eta - \sin \eta).\tag{5.5}\] The integral becomes \[\Delta\psi = \int_0^{2\pi} d\eta = 2\pi.\tag{5.6}\]
In the radiation-filled spherical model, prove instead that such a ray covers exactly half the universe before the big crunch.

In this case, we have \[a = \sqrt{C'}[1-(1-\tau/\sqrt{C'})^2]^{1/2},\tag{5.7}\] which gives \[\Delta \psi = \int ^{2\sqrt{C'}}_0 \frac{d\tau}{\sqrt{C'}[1-(1-\tau/\sqrt{C'})^2]^{1/2}} = \pi.\tag{5.8}\]