Wald GR Chapter 6 Solutions: The Schwarzschild Solution

Source: Robert M. Wald, General Relativity, Chapter 6; 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 6, Problem 1: Isotropic Coordinates

Let $M$ be a three-dimensional manifold with a spherically symmetric Riemannian metric, and suppose $\nabla_a r\neq0$, where $r$ is defined in (6.1.3).

Show that one may introduce an isotropic radial coordinate $\widetilde r$ for which \[ds^2=H(\widetilde r) \left[d\widetilde r^2+\widetilde r^2d\Omega^2\right].\tag{1.1}\] Thus every spherically symmetric three-dimensional space is conformally flat.

Assuming the metric can be written in a conformally flat form, then we must have \[h(r)dr^2 = H(\tilde r)d\tilde r^2, \quad H(\tilde r)\tilde r^2 = r^2,\tag{1.2}\] which gives the differential equation \[\frac{\sqrt{h(r)}}{r}dr = \frac{d \tilde r}{\tilde r} \implies \tilde r = A \exp\left( \int^r dr' \frac{\sqrt{h(r')}}{r'}\right).\tag{1.3}\]
Express the Schwarzschild metric in these coordinates and obtain \[ds^2 =-\frac{(1-M/2\widetilde r)^2}{(1+M/2\widetilde r)^2}\,dt^2 +\left(1+\frac{M}{2\widetilde r}\right)^4 \left[d\widetilde r^2+\widetilde r^2d\Omega^2\right].\tag{1.4}\]

For the Schwarzschild metric, \[\tilde r = A \exp\left(\int ^r dr' \frac{1}{\sqrt{r'^2 - 2Gmr'}}\right) = 2A\left(r-GM + \sqrt{r(r-GM)}\right).\tag{1.5}\] $A$ can be fixed by imposing the metric to be asymptotically flat, that is as $r \to \infty$, $H(\tilde r ) \to 1$. Therefore, we must have $A = 1/4$. Hence, playing around with the algebra slightly, we find \[\tilde r = \tilde r\left(1 + \frac{GM}{2\tilde r}\right)^2,\tag{1.6}\] which gives \[ds^2 =-\frac{(1-M/2\widetilde r)^2}{(1+M/2\widetilde r)^2}\,dt^2 +\left(1+\frac{M}{2\widetilde r}\right)^4 \left[d\widetilde r^2+\widetilde r^2d\Omega^2\right].\tag{1.7}\]

2 Chapter 6, Problem 3: The Reissner--Nordström Solution

Consider the source-free Maxwell equations (4.3.12) and (4.3.13), with $j^a=0$, on the static, spherically symmetric spacetime (6.1.5).

Argue that a Maxwell tensor with the same static and spherical symmetries must have the form \[F_{ab} =2A(r)(e_0)_{[a}(e_1)_{b]} +2B(r)(e_2)_{[a}(e_3)_{b]},\tag{2.1}\] where the $(e_\mu)_a$ are defined in (6.1.6).

Since $F$ is a two form, it should in general take the form \[F = F_{ab} \hat \theta^a \wedge \hat \theta^b,\tag{2.2}\] where $\hat \theta^a$ is the dual basis of $\hat e_a$. Now, excluding the basis that are not invariant under $SO(3)$ transformation, we are left with \[F = A(r) \hat \theta^0 \wedge \hat \theta^1 + B(r) \hat{\theta}^2 \wedge\hat \theta^3,\tag{2.3}\] where $A, B$ are only functions of $r$ by the requirement of the universe being static.
Restrict to $B(r)=0$ and solve Maxwell's equations to find \[A(r)=-\frac{q}{r^2},\tag{2.4}\] interpreting $q$ as the total charge. The branch with $B(r)\neq0$ is a duality rotation of this electric solution and represents a magnetic monopole.

Writing the field strength two form in terms of the coordinate basis, we have \[F = A(r) f^{1/2}h^{1/2}dt\wedge dr.\tag{2.5}\]\ Since $A, f ,h$ are all functions of $r$, $dF = 0$ gives no extra informations. The wedge of $F$ is given by \[\star F = A(r)r^2\sin\theta d\theta \wedge d\phi,\tag{2.6}\] up to some constant that would not affect our calculation here. Then, the source-free Maxwell equation gives \[\partial_r(A(r)r^2) = 0 \implies A(r) = -\frac{q}{r^2}.\tag{2.7}\]
Insert the electromagnetic stress-energy tensor from part (b) into \[G_{ab}=8\pi T_{ab},\tag{2.8}\] solve the resulting equations, and show that the general metric is \[ds^2 =-\left(1-\frac{2M}{r}+\frac{q^2}{r^2}\right)dt^2 +\left(1-\frac{2M}{r}+\frac{q^2}{r^2}\right)^{-1}dr^2 +r^2d\Omega^2.\tag{2.9}\]

The resulting energy momentum tensor in vierbein basis is \[T_{00} = \frac{1}{8\pi}\frac{q^2}{r^4},\tag{2.10}\] \[T_{11} = -\frac{1}{8\pi}\frac{q^2}{r^4},\tag{2.11}\] \[T_{22} = \frac{1}{8\pi}\frac{q^2}{r^4},\tag{2.12}\] and \[T_{33} = \frac{1}{8\pi}\frac{q^2}{r^4}.\tag{2.13}\] The Einstein equation is given by \[R_{00} +\frac{1}{2}R = 8\pi T_{00}, \tag{2.14}\] \[R_{11} - \frac{1}{2}R = 8\pi T_{11}, \tag{2.15}\] \[R_{22} - \frac{1}{2}R = 8\pi T_{22}, \tag{2.16}\] and \[R_{33} - \frac{1}{2}R = 8\pi T_{33}, \tag{2.17}\] where the Ricci tensor is given in (6.1.35-6.1.37). Combing (2.14) and (2.15) , we find \[R_{00} + R_{11} = 0,\tag{2.18}\] which is equivalent to (6.1.38), and hence \[f = h^{-1}.\tag{2.19}\] Note that (2.16) and (2.17) are completely identical, plugging $f=1/h$ in one of them, we find \[f'' + \frac{2f'}{r}= \frac{2q^2}{r^4},\tag{2.20}\] where we have used \[R = -R_{00} + R_{11} + 2R_{22}.\tag{2.21}\] This gives \[f = h^{-1} = 1- \frac{C}{r} -\frac{q^2}{r^2}.\tag{2.22}\]

3 Chapter 6, Problem 4: Local Force and Redshift

Let $(M,g_{ab})$ be stationary with timelike Killing field $\xi^a$, and define \[V^2=-\xi^a\xi_a.\tag{3.1}\]

For a stationary observer, show that the acceleration $a^b=u^a\nabla_a u^b$ is \[a^b=\nabla^b\ln V.\tag{3.2}\]

\[a^b = u^a \nabla_a u^b = \frac{u^a \nabla_a \xi^b}{V} - \frac{u^a \xi^b \nabla_a V}{V^2}.\tag{3.3}\] The second term vanishes by \[\frac{u^a \xi^b \nabla_a V}{V^2} = -\frac{2\xi^a\xi^b\xi^c\nabla_a\xi^c}{V^3} = 0\tag{3.4}\] by $\nabla_a \xi_b = - \nabla_b \xi_a$. Then, we have \[a^b = \frac{\xi^a \nabla_a \xi^b}{V^2} = -\frac{\xi^a\nabla^b \xi_a}{V^2} = \frac{1}{2}\nabla^b \log V^2 = \nabla^b \log V.\tag{3.5}\]
Suppose additionally that the spacetime is asymptotically flat. Thus there are coordinates $t,x,y,z$, with $\xi^a=(\partial/\partial t)^a$, in which $g_{ab}$ tends to $\operatorname{diag}(-1,1,1,1)$ as $r=(x^2+y^2+z^2)^{1/2}\to\infty$. See Chapter 11 for further discussion of asymptotic flatness.
A particle of mass $m$ has energy at infinity \[E=-m\xi^a u_a.\tag{3.6}\] Hold the particle stationary with a massless string whose far end is held by a stationary observer at large $r$. If $F$ is the force applied locally to the particle, part (a) gives \[F=mV^{-1}\left(\nabla_aV\nabla^aV\right)^{1/2}.\tag{3.7}\] Use conservation of energy to show that the force supplied at infinity is \[F_\infty=VF.\tag{3.8}\] Interpret the factor $V$ as the redshift relating the two force magnitudes.

The local magnitude of force is given by \[F = m \sqrt{a^a a_a} = mV^{-1}\left(\nabla_aV\nabla^aV\right)^{1/2},\tag{3.9}\] while the energy of a stationary particle measured by a local stationary observer is given by \[E = m.\tag{3.10}\] Assuming the stationary particle is pushed through a proper distance $\delta s$ adiabatically with a constant force $F$. Then, the work done locally is equal to the local change of energy is given by \[F\delta s = \delta E.\tag{3.11}\] For an observer at spatial infinity, he agrees on the proper distance, while the force seen by this observer, $F_\infty$, is different. Therefore, the change of energy seen by this observer is \[\delta E_\infty = F_\infty \delta s.\tag{3.12}\] By the energy redshift, we have \[V\delta E = \delta E_\infty.\tag{3.13}\] Therefore, we have \[F_\infty =VF.\tag{3.14}\]

4 Chapter 6, Problem 5: Relativistic Time Delay

Derive the general-relativistic time-delay formula (6.3.45). Draft solution: the derivation below does not yet display the final time-delay formula (6.3.45).

By time translational symmetry, the time delay from the Earth to the planet in coordinate time must be equivalent to \[\Delta t = \int^{R_E}_{R_0} dr\, \frac{1}{(1-2GM/r)\sqrt{1-(1-2GM/r)b^2/r^2}} + \int^{R_p}_{R_0} dr\, \frac{1}{(1-2GM/r)\sqrt{1-(1-2GM/r)b^2/r^2}}.\tag{4.1}\] To first order in $G$, this is \[\Delta t = \Delta t \bigg|_{G=0} + \frac{\partial (\Delta t)}{\partial G}\bigg|_{G=0}G + \mathcal{O}(G^2).\tag{4.2}\] The first term is simply the time elapsed when the star is absent. Therefore, it must be \[\Delta t \bigg|_{G=0} = (R_E^2-R_0^2)^{1/2} + (R_p^2 - R_0^2)^{1/2}\tag{4.3}\] by simple geometry. Recall that \[R_0^3 - b^2(R_0-2GM) = 0 \implies b = R_0 + GM + \mathcal{O}(G^2).\tag{4.4}\] Then, for the second term, we find \[\begin{aligned} \frac{\partial (\Delta t)}{\partial G}\bigg|_{G=0}G &= \frac{2GM}{R_0}\log\left(\frac{r}{R_0} + \frac{\sqrt{r^2-R_0^2}}{R_0}\right)\bigg|^{R_E}_{R_0}+\frac{2GM}{R_0}\log\left(\frac{r}{R_0} + \frac{\sqrt{r^2-R_0^2}}{R_0}\right)\bigg|^{R_p}_{R_0}\\ \qquad &-\frac{2GMr}{\sqrt{r^2 -b^2}}\bigg|^{R_E}_{R_0}-\frac{2GMr}{\sqrt{r^2 -b^2}}\bigg|^{R_p}_{R_0}. \end{aligned}\tag{4.5}\] Note that special attention is required when substituting $R_0$ into the last two terms. Carefully keeping terms of first order in $G$, one would eventually find the residue terms of (6.3.45).

5 Chapter 6, Problem 6: Lifetime Inside the Schwarzschild Horizon

Show that every particle in region II, $r<2M$, of the extended Schwarzschild spacetime in Figure 6.9, whether geodesic or accelerated, must obey \[\left|\frac{dr}{d\tau}\right| \geq\left(\frac{2M}{r}-1\right)^{1/2}.\tag{5.1}\] Deduce that no observer in region II can have a remaining proper lifetime greater than \[\tau=\pi M \sim 10^{-5}\left(\frac{M}{M_\odot}\right)\mathrm{s}\tag{5.2}\] before reaching $r=0$. Finally, show that this bound is approached by free fall from $r=2M$ in the limit $E\to0$.