# Cosmic strings

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 23; Problem 23.6.

Another example of a solution to the Einstein equation is a simple model of a cosmic string. Cosmic strings are postulated objects that are left over from the big bang. They are virtually one-dimensional structures with a radius much smaller than an atomic nucleus, but with lengths of hundreds of thousands of light years. As a model of a cosmic string, suppose we have an infinite, straight string stretching along the ${z}$ axis, and that the string is axially symmetric, that is, that its structure depends only on the radial coordinate ${r}$ measured from the ${z}$ axis. The metric describing the string is a generalization of the cylindrical coordinate system:

$\displaystyle ds^{2}=-dt^{2}+dr^{2}+f^{2}\left(r\right)d\phi^{2}+dz^{2} \ \ \ \ \ (1)$

For an ordinary cylindrical system in flat space, ${f\left(r\right)=r}$.

[The interpretation of the ${r}$ coordinate is qualitatively different from the Schwarzschild metric, where we assumed spherical symmetry and used this to write down the angular components of the metric as those that apply in flat space, that is ${r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2}}$. This choice results in the radial coordinate ${r}$ being a circumferential coordinate, in that the circumference of a circle with radial coordinate ${r}$ is ${2\pi r}$, but the distance from the origin to a point on the circle is not ${r}$. In the cylindrical metric here, ${r}$ is not a circumferential coordinate because the metric component ${g_{\phi\phi}=f^{2}\ne1}$, so the circumference of a circle of radius ${r}$ is ${2\pi f}$ (as you can verify by setting ${dt=dr=dz=0}$ and integrating over ${\phi}$ from 0 to ${2\pi}$ for a fixed ${r}$). However, because ${g_{rr}=1}$, the ${r}$ coordinate here does represent the actual distance from the ${z}$ axis to a point on a circle with coordinate ${r}$.]

We take the stress-energy tensor to be

$\displaystyle T_{\;t}^{t}=T_{\;z}^{z}=-\sigma\left(r\right) \ \ \ \ \ (2)$

[Moore doesn’t explain where these come from, but we’ll just accept this for now.] From the definition of the stress-energy tensor ${T^{tt}=-T_{\;t}^{t}=\sigma}$ is the energy density.

$\displaystyle R_{\mu\nu}=8\pi G\left(T_{\mu\nu}-\frac{1}{2}g_{\mu\nu}T\right) \ \ \ \ \ (3)$

The scalar ${T}$ is

$\displaystyle T=T_{\;\mu}^{\mu}=T_{\;t}^{t}+T_{\;z}^{z}=-2\sigma \ \ \ \ \ (4)$

The non-zero components of ${T_{\mu\nu}}$ can be found from 2 by lowering the first index:

 $\displaystyle T_{tt}$ $\displaystyle =$ $\displaystyle g_{tt}T_{\;t}^{t}=\sigma\ \ \ \ \ (5)$ $\displaystyle T_{zz}$ $\displaystyle =$ $\displaystyle g_{zz}T_{\;z}^{z}=-\sigma \ \ \ \ \ (6)$

From 3 we therefore have

 $\displaystyle R_{tt}$ $\displaystyle =$ $\displaystyle 8\pi G\left(T_{tt}-\frac{1}{2}g_{tt}T\right)\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 8\pi G\left(\sigma-\sigma\right)=0\ \ \ \ \ (8)$ $\displaystyle R_{rr}$ $\displaystyle =$ $\displaystyle 8\pi G\left(0+g_{rr}\sigma\right)\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 8\pi G\sigma\ \ \ \ \ (10)$ $\displaystyle R_{\phi\phi}$ $\displaystyle =$ $\displaystyle 8\pi G\left(0+g_{\phi\phi}\sigma\right)\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 8\pi Gf^{2}\sigma\ \ \ \ \ (12)$ $\displaystyle R_{zz}$ $\displaystyle =$ $\displaystyle 8\pi G\left(-\sigma+g_{zz}\sigma\right)=0 \ \ \ \ \ (13)$

All off-diagonal components of ${R_{\mu\nu}}$ are zero since both ${T_{\mu\nu}}$ and ${g_{\mu\nu}}$ are diagonal. Thus

$\displaystyle R_{rr}=\frac{R_{\phi\phi}}{f^{2}} \ \ \ \ \ (14)$

Using the Ricci tensor worksheet we can work out ${R_{\mu\nu}}$ in terms of ${g_{\mu\nu}}$. The only non-zero terms are those involving a derivative of ${g_{\phi\phi}}$ with respect to ${r}$ on its own (that is, not multiplied by some other derivative), or in terms of the notation of the worksheet, those terms involving either ${C_{1}}$ or ${C_{11}}$ on their own. We have (where a subscript 1 indicates a derivative with respect to ${r}$):

 $\displaystyle C$ $\displaystyle =$ $\displaystyle f^{2}\ \ \ \ \ (15)$ $\displaystyle C_{1}$ $\displaystyle =$ $\displaystyle \frac{d\left(f^{2}\right)}{dr}=2ff_{1}\ \ \ \ \ (16)$ $\displaystyle C_{11}$ $\displaystyle =$ $\displaystyle \frac{d\left(2ff_{1}\right)}{dr}=2f_{1}^{2}+2ff_{11} \ \ \ \ \ (17)$

The only components of ${R_{\mu\nu}}$ involving these two derivatives on their own are ${R_{rr}}$ and ${R_{\phi\phi}}$:

 $\displaystyle R_{rr}$ $\displaystyle =$ $\displaystyle -\frac{1}{2C}C_{11}+\frac{1}{4C^{2}}C_{1}^{2}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{f_{1}^{2}}{f^{2}}-\frac{f_{11}}{f}+\frac{4f^{2}f_{1}^{2}}{4f^{4}}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{f_{11}}{f}\ \ \ \ \ (20)$ $\displaystyle R_{\phi\phi}$ $\displaystyle =$ $\displaystyle -f_{1}^{2}-ff_{11}+\frac{4f^{2}f_{1}^{2}}{4f^{2}}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -ff_{11}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle f^{2}R_{rr} \ \ \ \ \ (23)$

Thus 14 is satisfied here as well. [Note that there are a couple of errors in Moore’s problem statement – see the errata list here.] Combining 10 and 20 gives

$\displaystyle f_{11}=\frac{d^{2}f}{dr^{2}}=-8\pi Gf\left(r\right)\sigma\left(r\right) \ \ \ \ \ (24)$

Moore now says that we require the metric to be non-singular at ${r=0}$ (actually he says ‘non-singular at the origin’ although I assume he means ‘non-singular at all points on the ${z}$ axis, since there’s nothing special about ${z=0}$ here). It’s not entirely clear to me why we would require this since the Schwarzschild metric is singular at ${r=0}$. He also says that this requirement leads to the metric reducing to the flat space metric as ${r\rightarrow0}$. Again, this isn’t exactly obvious; there are lots of metrics that are finite at ${r=0}$ so why choose flat space? Anyway, let’s plow onwards…

If we require ${f\left(r\right)\rightarrow r}$ as ${r\rightarrow0}$ to give us the flat, cylindrical metric near the ${z}$ axis, then ${f_{1}=\frac{df}{dr}\rightarrow1}$ as ${r\rightarrow0}$. We can then integrate 24 to give, for points outside the string’s radius ${r_{s}}$:

 $\displaystyle f_{1}$ $\displaystyle =$ $\displaystyle -4G\int_{0}^{r_{s}}2\pi f\left(r\right)\sigma\left(r\right)dr+A\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -4G\mu+A \ \ \ \ \ (26)$

where ${A}$ is a constant of integration, ${r_{s}}$ is the radius of the string and

$\displaystyle \mu\left(r_{s}\right)\equiv\int_{0}^{r_{s}}2\pi f\left(r\right)\sigma\left(r\right)dr \ \ \ \ \ (27)$

is the string’s energy density (energy per unit length). [Recall from above that the circumference of a circle of radius ${r}$ is ${2\pi f}$ and ${\sigma}$ is the energy density (per unit volume), so the energy in a cylindrical shell of radius ${r}$, thickness ${dr}$ and unit length is ${2\pi f\left(r\right)\sigma\left(r\right)dr}$.] In the limiting case of no string at all, ${r_{s}=0}$ and the metric reduces to flat space where ${f_{1}=1}$ so ${A=1}$ and we have for ${r>r_{s}}$:

$\displaystyle \frac{df}{dr}=1-4G\mu \ \ \ \ \ (28)$

Since ${\mu}$ is a constant for ${r>r_{s}}$, we can integrate this directly to get

$\displaystyle f\left(r\right)=\left(1-4G\mu\right)r+K \ \ \ \ \ (29)$

where ${K}$ is a constant of integration. For very small ${r}$ we should have ${f\rightarrow r}$ and since ${r_{s}}$ is very small, we’d expect ${\mu}$ to be small, so ${K}$ would be close to zero.

The resulting metric is

$\displaystyle ds^{2}=-dt^{2}+dr^{2}+\left(1-4G\mu\right)^{2}r^{2}d\phi^{2}+dz^{2} \ \ \ \ \ (30)$

We can redefine the angular coordinate ${\phi}$ (in a way similar to the redefinition of the time coordinate used in deriving Birkhoff’s theorem) by defining

$\displaystyle \tilde{\phi}\equiv\left(1-4G\mu\right)\phi \ \ \ \ \ (31)$

to get what appears to be a flat space metric:

$\displaystyle ds^{2}=-dt^{2}+dr^{2}+r^{2}d\tilde{\phi}^{2}+dz^{2} \ \ \ \ \ (32)$

However, remember that the radial coordinate ${r}$ is not the same as that used in flat space, since the circumference of a circle of radius ${r}$ is given by ${2\pi\left(1-4G\mu\right)r}$, so is actually slightly smaller than ${2\pi r}$. Also, the new axial coordinate ${\tilde{\phi}}$ covers ${2\pi\left(1-4G\mu\right)<2\pi}$ for a complete circle.

# Einstein equation solution for the interior of a spherically symmetric star

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 23; Problem 23.5.

The derivation of the Schwarzschild metric applies to the empty space outside a spherically symmetric source. We can apply a similar method to try to find a metric for the interior of a spherically symmetric, static source such as a non-rotating star. As usual, we can make a number of simplifying assumptions. Spherically symmetry implies that the metric has the general form

$\displaystyle ds^{2}=-A\left(r\right)dt^{2}+B\left(r\right)dr^{2}+r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2} \ \ \ \ \ (1)$

where the functions ${A}$ and ${B}$ are to be determined; they depend only on ${r}$ because we’re assuming that everything is independent of time. Another simplification is the assumption that the star’s matter is a perfect fluid. We’ve looked at the Einstein equation for a perfect fluid before, which has the general form (with ${\Lambda=0}$):

$\displaystyle R^{\mu\nu}=8\pi G\left(T^{\mu\nu}-\frac{1}{2}g^{\mu\nu}T\right) \ \ \ \ \ (2)$

The stress-energy tensor is

$\displaystyle T^{\mu\nu}=\left(\rho+P\right)u^{\mu}u^{\nu}+Pg^{\mu\nu} \ \ \ \ \ (3)$

where ${u^{\mu}}$ is the four-velocity, ${\rho}$ is the energy density and ${P}$ is the pressure, both of which can be functions of ${r}$.
We have for the stress-energy scalar:

 $\displaystyle T$ $\displaystyle =$ $\displaystyle \left(\rho+P\right)g_{\mu\nu}u^{\mu}u^{\nu}+Pg_{\mu\nu}g^{\mu\nu}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\left(\rho+P\right)+4P\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 3P-\rho \ \ \ \ \ (6)$

since ${g_{\mu\nu}g^{\mu\nu}=4}$ and ${g_{\mu\nu}u^{\mu}u^{\nu}=-1}$ in any coordinate system. This gives

$\displaystyle T^{\mu\nu}-\frac{1}{2}g^{\mu\nu}T=\left(\rho+P\right)u^{\mu}u^{\nu}+\frac{1}{2}g^{\mu\nu}\left(\rho-P\right) \ \ \ \ \ (7)$

Lowering the indices in 7, we get

$\displaystyle T_{\mu\nu}-\frac{1}{2}g_{\mu\nu}T=\left(\rho+P\right)u_{\mu}u_{\nu}+\frac{1}{2}g_{\mu\nu}\left(\rho-P\right) \ \ \ \ \ (8)$

Since the fluid is at rest, the spatial components of the four-velocity ${u_{i}}$ are all zero. From the condition ${g_{\mu\nu}u^{\mu}u^{\nu}=-1}$ we have

 $\displaystyle g_{\mu\nu}u^{\mu}u^{\nu}$ $\displaystyle =$ $\displaystyle g_{tt}u^{t}u^{t}=-A\left(u^{t}\right)^{2}=-1\ \ \ \ \ (9)$ $\displaystyle u^{t}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{A}}\ \ \ \ \ (10)$ $\displaystyle u_{t}$ $\displaystyle =$ $\displaystyle g_{t\nu}u^{\nu}=g_{tt}u^{t}=-Au^{t}=-\sqrt{A} \ \ \ \ \ (11)$

Equation 2 with indices lowered is

$\displaystyle R_{\mu\nu}=8\pi G\left(T_{\mu\nu}-\frac{1}{2}g_{\mu\nu}T\right) \ \ \ \ \ (12)$

so with the above results, we get

 $\displaystyle R_{tt}$ $\displaystyle =$ $\displaystyle 8\pi G\left(\left(\rho+P\right)u_{t}u_{t}+\frac{1}{2}g_{tt}\left(\rho-P\right)\right)\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 8\pi G\left[\left(\rho+P\right)A-\frac{A}{2}\left(\rho-P\right)\right]\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 4\pi GA\left(\rho+3P\right)\ \ \ \ \ (15)$ $\displaystyle R_{rr}$ $\displaystyle =$ $\displaystyle 8\pi G\left(\left(\rho+P\right)u_{r}u_{r}+\frac{1}{2}g_{rr}\left(\rho-P\right)\right)\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 4\pi G\left(0+B\left(\rho-P\right)\right)\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 4\pi GB\left(\rho-P\right)\ \ \ \ \ (18)$ $\displaystyle R_{\theta\theta}$ $\displaystyle =$ $\displaystyle 8\pi G\left(\left(\rho+P\right)u_{\theta}u_{\theta}+\frac{1}{2}g_{\theta\theta}\left(\rho-P\right)\right)\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 4\pi G\left(0+r^{2}\left(\rho-P\right)\right)\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 4\pi Gr^{2}\left(\rho-P\right) \ \ \ \ \ (21)$

We can combine these three equations to eliminate ${P}$:

$\displaystyle \frac{R_{tt}}{A}+\frac{R_{rr}}{B}+\frac{2R_{\theta\theta}}{r^{2}}=16\pi G\rho \ \ \ \ \ (22)$

When we worked out the Ricci tensor in terms of the metric, we got the equations

 $\displaystyle \frac{1}{2B}\left[\partial_{rr}^{2}A-\partial_{tt}^{2}B+\frac{\left(\partial_{t}B\right)^{2}}{2B}+\frac{\left(\partial_{t}A\right)\left(\partial_{t}B\right)-\left(\partial_{r}A\right)^{2}}{2A}-\frac{\left(\partial_{r}A\right)\left(\partial_{r}B\right)}{2B}+\frac{2\partial_{r}A}{r}\right]$ $\displaystyle =$ $\displaystyle R_{tt}\ \ \ \ \ (23)$ $\displaystyle \frac{1}{2A}\left[\partial_{tt}^{2}B-\partial_{rr}^{2}A+\frac{\left(\partial_{r}A\right)^{2}-\left(\partial_{t}A\right)\left(\partial_{t}B\right)}{2A}+\frac{\left(\partial_{r}A\right)\left(\partial_{r}B\right)-\left(\partial_{t}B\right)^{2}}{2B}+\frac{2A\partial_{r}B}{rB}\right]$ $\displaystyle =$ $\displaystyle R_{rr}\ \ \ \ \ (24)$ $\displaystyle -\frac{r\partial_{r}A}{2AB}+\frac{r\partial_{r}B}{2B^{2}}+1-\frac{1}{B}$ $\displaystyle =$ $\displaystyle R_{\theta\theta}\ \ \ \ \ (25)$ $\displaystyle \frac{\partial_{t}B}{rB}$ $\displaystyle =$ $\displaystyle R_{tr} \ \ \ \ \ (26)$

All time derivatives are zero, so these equations simplify to

 $\displaystyle \frac{1}{2B}\left[\partial_{rr}^{2}A-\frac{\left(\partial_{r}A\right)^{2}}{2A}-\frac{\left(\partial_{r}A\right)\left(\partial_{r}B\right)}{2B}+\frac{2\partial_{r}A}{r}\right]$ $\displaystyle =$ $\displaystyle R_{tt}\ \ \ \ \ (27)$ $\displaystyle \frac{1}{2A}\left[-\partial_{rr}^{2}A+\frac{\left(\partial_{r}A\right)^{2}}{2A}+\frac{\left(\partial_{r}A\right)\left(\partial_{r}B\right)}{2B}+\frac{2A\partial_{r}B}{rB}\right]$ $\displaystyle =$ $\displaystyle R_{rr}\ \ \ \ \ (28)$ $\displaystyle -\frac{r\partial_{r}A}{2AB}+\frac{r\partial_{r}B}{2B^{2}}+1-\frac{1}{B}$ $\displaystyle =$ $\displaystyle R_{\theta\theta}\ \ \ \ \ (29)$ $\displaystyle 0$ $\displaystyle =$ $\displaystyle R_{tr} \ \ \ \ \ (30)$

Applying 22 we get

 $\displaystyle \frac{\partial_{r}A}{ABr}+\frac{\partial_{r}B}{rB^{2}}-\frac{\partial_{r}A}{rAB}+\frac{\partial_{r}B}{rB^{2}}+\frac{2}{r^{2}}-\frac{2}{r^{2}B}$ $\displaystyle =$ $\displaystyle 16\pi G\rho\left(r\right)\ \ \ \ \ (31)$ $\displaystyle r\frac{\partial_{r}B}{B^{2}}+1-\frac{1}{B}$ $\displaystyle =$ $\displaystyle 8\pi Gr^{2}\rho\left(r\right)\ \ \ \ \ (32)$ $\displaystyle \frac{d}{dr}\left[r\left(1-\frac{1}{B}\right)\right]$ $\displaystyle =$ $\displaystyle 8\pi Gr^{2}\rho\left(r\right)\ \ \ \ \ (33)$ $\displaystyle r\left(1-\frac{1}{B}\right)$ $\displaystyle =$ $\displaystyle 2G\int_{0}^{r}4\pi\left(r'\right)^{2}\rho\left(r'\right)dr' \ \ \ \ \ (34)$

The integral on the last line resembles the mass of the star out to radius ${r}$, as it’s the integral of the density ${\rho}$ over a set of spherical shells of radius ${r'}$, surface area ${4\pi\left(r'\right)^{2}}$ and thickness ${dr'}$. However, the Schwarzschild radial coordinate ${r}$ isn’t equal to the actual radial distance (it’s a circumferential coordinate), so this isn’t quite accurate. With the definition

$\displaystyle m\left(r\right)\equiv\int_{0}^{r}4\pi\left(r'\right)^{2}\rho\left(r'\right)dr' \ \ \ \ \ (35)$

we therefore have

 $\displaystyle r\left(1-\frac{1}{B}\right)$ $\displaystyle =$ $\displaystyle 2Gm\left(r\right)\ \ \ \ \ (36)$ $\displaystyle B$ $\displaystyle =$ $\displaystyle \left[1-\frac{2Gm\left(r\right)}{r}\right]^{-1} \ \ \ \ \ (37)$

This gives us ${B}$, but what about ${A}$? If we use the stress-energy tensor’s conservation equation

$\displaystyle \nabla_{\nu}T^{\mu\nu}=0 \ \ \ \ \ (38)$

we can look at ${\mu=r}$ component, starting from 3. The absolute gradient is defined in terms of Christoffel symbols as

$\displaystyle \nabla_{\rho}T^{\mu\nu}=\partial_{\rho}T^{\mu\nu}+T^{\mu\alpha}\Gamma_{\alpha\rho}^{\nu}+T^{\alpha\nu}\Gamma_{\alpha\rho}^{\mu} \ \ \ \ \ (39)$

so for ${\mu=r}$ and ${\rho=\nu}$ we have

$\displaystyle \nabla_{\nu}T^{r\nu}=\partial_{\nu}T^{r\nu}+T^{\alpha\nu}\Gamma_{\alpha\nu}^{r}+T^{r\alpha}\Gamma_{\alpha\nu}^{\nu} \ \ \ \ \ (40)$

Since the covariant derivative of the metric tensor is zero and ${P}$ is a scalar (so its covariant derivative is the same as its ordinary derivative), we have

 $\displaystyle \nabla_{\nu}T^{r\nu}$ $\displaystyle =$ $\displaystyle \nabla_{\nu}\left[\left(\rho+P\right)u^{r}u^{\nu}\right]+\nabla_{\nu}\left(Pg^{r\nu}\right)\ \ \ \ \ (41)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \nabla_{\nu}\left[\left(\rho+P\right)u^{r}u^{\nu}\right]+g^{rr}\partial_{r}P\ \ \ \ \ (42)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \partial_{\nu}\left[\left(\rho+P\right)u^{r}u^{\nu}\right]+\left(\rho+P\right)\left[u^{\alpha}u^{\nu}\Gamma_{\alpha\nu}^{r}+u^{r}u^{\alpha}\Gamma_{\alpha\nu}^{\nu}\right]+g^{rr}\partial_{r}P\ \ \ \ \ (43)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (44)$

Because the fluid is at rest, ${u^{r}=u^{\theta}=u^{\phi}=0}$ so this reduces to

$\displaystyle \left(\rho+P\right)u^{t}u^{t}\Gamma_{tt}^{r}+g^{rr}\partial_{r}P=0 \ \ \ \ \ (45)$

For a diagonal metric ${g^{\mu\nu}=\frac{1}{g_{\mu\nu}}}$ so ${g^{rr}=\frac{1}{B}}$, from 10 ${u^{t}=\frac{1}{\sqrt{A}}}$, and from the Christoffel symbol worksheet ${\Gamma_{tt}^{r}=\Gamma_{00}^{1}=\frac{1}{2B}\partial_{r}A}$ so we have

 $\displaystyle \left(\rho+P\right)\frac{\partial_{r}A}{2AB}+\frac{\partial_{r}P}{B}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (46)$ $\displaystyle \frac{1}{A}\frac{dA}{dr}$ $\displaystyle =$ $\displaystyle -\frac{2}{\rho+P}\frac{dP}{dr} \ \ \ \ \ (47)$

To solve this, we need to know both ${\rho}$ and ${P}$ as functions of ${r}$. We can make some headway by using ${R_{\theta\theta}}$ in 21 and 25. We have

$\displaystyle -\frac{r\partial_{r}A}{2AB}+\frac{r\partial_{r}B}{2B^{2}}+1-\frac{1}{B}=4\pi Gr^{2}\left(\rho-P\right) \ \ \ \ \ (48)$

From 32 and 36

 $\displaystyle r\frac{\partial_{r}B}{B^{2}}$ $\displaystyle =$ $\displaystyle 8\pi Gr^{2}\rho-\left(1-\frac{1}{B}\right)\ \ \ \ \ (49)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 8\pi Gr^{2}\rho-\frac{2Gm}{r} \ \ \ \ \ (50)$

Plugging this and 37 into 48:

 $\displaystyle -\frac{r}{2B}\left(-\frac{2}{\rho+P}\frac{dP}{dr}\right)+4\pi Gr^{2}\rho-\frac{Gm}{r}+\frac{2Gm}{r}$ $\displaystyle =$ $\displaystyle 4\pi Gr^{2}\left(\rho-P\right)\ \ \ \ \ (51)$ $\displaystyle \frac{r^{2}}{\rho+P}\left(1-\frac{2Gm}{r}\right)\frac{dP}{dr}$ $\displaystyle =$ $\displaystyle -\left(4\pi Gr^{3}P+Gm\right)\ \ \ \ \ (52)$ $\displaystyle \frac{dP}{dr}$ $\displaystyle =$ $\displaystyle -\frac{\left(\rho+P\right)}{r^{2}}\frac{\left(4\pi Gr^{3}P+Gm\right)}{1-2Gm/r} \ \ \ \ \ (53)$

This is the Oppenheimer-Volkoff equation for stellar structure. In order to solve it, we need a couple of other equations involving ${m}$ and ${\rho}$, but we’ll leave that for later. [By the way, for anyone interested in trivia, the Volkoff after whom this equation is named is George Volkoff, and he was the Dean of Science at the University of British Columbia in Vancouver when I was doing my undergraduate degree in physics and astronomy there in the 1970s. Sadly, I never had him as a professor.]

# Schwarzschild metric with non-zero cosmological constant

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 23; Problem 23.3.

In the derivation of the Schwarzschild metric we used the Einstein equation in the form

$\displaystyle R^{\mu\nu}=\kappa\left(T^{\mu\nu}-\frac{1}{2}g^{\mu\nu}T\right)+\Lambda g^{\mu\nu} \ \ \ \ \ (1)$

and took the cosmological constant to be ${\Lambda=0}$, giving the condition that ${R^{\mu\nu}=0}$ in empty space. If we take ${\Lambda\ne0}$ (but still very small), then in empty space the Ricci tensor becomes

$\displaystyle R^{\mu\nu}=\Lambda g^{\mu\nu} \ \ \ \ \ (2)$

If we follow through the original derivation of the Schwarzschild metric with this condition, using the general diagonal metric

$\displaystyle ds^{2}=-Adt^{2}+Bdr^{2}+r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2} \ \ \ \ \ (3)$

as a starting point, then we get the equations

 $\displaystyle \partial_{t}B$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (4)$ $\displaystyle \frac{B}{A}R_{tt}+R_{rr}$ $\displaystyle =$ $\displaystyle \frac{2\partial_{r}A}{r}+\frac{2A\partial_{r}B}{rB}\ \ \ \ \ (5)$ $\displaystyle R_{\theta\theta}$ $\displaystyle =$ $\displaystyle r^{2}\Lambda\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{r\partial_{r}A}{2AB}+\frac{r\partial_{r}B}{2B^{2}}+1-\frac{1}{B} \ \ \ \ \ (7)$

Substituting 2 into 5, we get

 $\displaystyle \frac{B}{A}\left(-A\Lambda\right)+B\Lambda$ $\displaystyle =$ $\displaystyle \frac{2\partial_{r}A}{r}+\frac{2A\partial_{r}B}{rB}\ \ \ \ \ (8)$ $\displaystyle 0$ $\displaystyle =$ $\displaystyle \frac{2\partial_{r}A}{r}+\frac{2A\partial_{r}B}{rB}\ \ \ \ \ (9)$ $\displaystyle \frac{\partial_{r}A}{A}$ $\displaystyle =$ $\displaystyle -\frac{\partial_{r}B}{B} \ \ \ \ \ (10)$

Thus a non-zero ${\Lambda}$ doesn’t change this equation. The only difference comes from 7. Substituting 10 into 7 and rearranging terms gives

 $\displaystyle \partial_{r}\left(\frac{r}{B}\right)$ $\displaystyle =$ $\displaystyle 1-r^{2}\Lambda\ \ \ \ \ (11)$ $\displaystyle \frac{r}{B}$ $\displaystyle =$ $\displaystyle r-\frac{r^{3}\Lambda}{3}+C\ \ \ \ \ (12)$ $\displaystyle A=\frac{1}{B}$ $\displaystyle =$ $\displaystyle 1+\frac{C}{r}-\frac{r^{2}\Lambda}{3} \ \ \ \ \ (13)$

[The relation ${A=\frac{1}{B}}$ comes from rescaling the time coordinate.]

If ${\Lambda}$ is very small, so small that ${\frac{C}{r}\gg r^{2}\Lambda}$ for values of ${r}$ on the scale of intergalactic distances (millions of light years), then for these distances the cosmological constant term can be neglected and the requirement that we reclaim Newton’s law of gravity for these distances gives us the condition ${C=-2GM}$ as in the original derivation, in which we showed that

 $\displaystyle \ddot{x}^{\mu}$ $\displaystyle =$ $\displaystyle \frac{\Gamma_{tt}^{\mu}}{g_{tt}}=-\frac{1}{A}\Gamma_{tt}^{\mu}\ \ \ \ \ (14)$ $\displaystyle \frac{d^{2}r}{d\tau^{2}}$ $\displaystyle =$ $\displaystyle -\frac{1}{A}\Gamma_{tt}^{r} \ \ \ \ \ (15)$

The Christoffel symbol comes out to

$\displaystyle \Gamma_{tt}^{r}=\frac{1}{2B}\partial_{r}A \ \ \ \ \ (16)$

For our more general case, we therefore have

 $\displaystyle \frac{d^{2}r}{d\tau^{2}}$ $\displaystyle =$ $\displaystyle -\frac{1}{A}\Gamma_{tt}^{r}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{2AB}\partial_{r}A\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{2}\partial_{r}A\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{C}{2r^{2}}+\frac{r\Lambda}{3} \ \ \ \ \ (20)$

Thus for ${r}$ large, but not large enough that ${r\Lambda}$ becomes significant, we just have the Newtonian law of gravity with ${C=-2GM}$. For really large ${r}$, however, the ${\frac{r\Lambda}{3}}$ term will eventually dominate and change the sign of the radial acceleration from negative (attractive force) to positive (repulsive force).

The addition of ${\Lambda}$ also has an effect on particle orbits. If we follow through the same derivation we did earlier for the ordinary Schwarzschild metric, but with ${\Lambda\ne0}$ then from 13 we have

 $\displaystyle g_{tt}$ $\displaystyle =$ $\displaystyle -\left(1-\frac{2GM}{r}-\frac{r^{2}\Lambda}{3}\right)\ \ \ \ \ (21)$ $\displaystyle g_{rr}$ $\displaystyle =$ $\displaystyle \left(1-\frac{2GM}{r}-\frac{r^{2}\Lambda}{3}\right)^{-1} \ \ \ \ \ (22)$

From the ${t}$ component of the geodesic equation we get the conserved quantity ${e}$:

$\displaystyle e=\left(1-\frac{2GM}{r}-\frac{r^{2}\Lambda}{3}\right)\frac{dt}{d\tau} \ \ \ \ \ (23)$

The angular components are unchanged, so we still have the conserved quantity ${\ell}$

$\displaystyle \ell=r^{2}\sin^{2}\theta\frac{d\phi}{d\tau} \ \ \ \ \ (24)$

Following through the earlier derivation for the ${r}$ component, we get

 $\displaystyle -1$ $\displaystyle =$ $\displaystyle -\left(1-\frac{2GM}{r}-\frac{r^{2}\Lambda}{3}\right)\left[e\left(1-\frac{2GM}{r}-\frac{r^{2}\Lambda}{3}\right)^{-1}\right]^{2}+\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \left(1-\frac{2GM}{r}-\frac{r^{2}\Lambda}{3}\right)^{-1}\left(\frac{dr}{d\tau}\right)^{2}+r^{2}\left(\frac{\ell}{r^{2}}\right)^{2}\ \ \ \ \ (25)$ $\displaystyle -\left(1-\frac{2GM}{r}-\frac{r^{2}\Lambda}{3}\right)$ $\displaystyle =$ $\displaystyle -e^{2}+\left(\frac{dr}{d\tau}\right)^{2}+\left(1-\frac{2GM}{r}-\frac{r^{2}\Lambda}{3}\right)\frac{\ell^{2}}{r^{2}}\ \ \ \ \ (26)$ $\displaystyle \frac{1}{2}\left(e^{2}-1\right)$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(\frac{dr}{d\tau}\right)^{2}+\frac{1}{2}\frac{\ell^{2}}{r^{2}}-GM\left(\frac{1}{r}+\frac{l^{2}}{r^{3}}\right)-\frac{\Lambda}{6}\left(\ell^{2}+r^{2}\right) \ \ \ \ \ (27)$

If we write this equation in the more suggestive notation we have

$\displaystyle \tilde{K}+\tilde{V}\left(r\right)=\tilde{E} \ \ \ \ \ (28)$

where

 $\displaystyle \tilde{K}$ $\displaystyle \equiv$ $\displaystyle \frac{1}{2}\left(\frac{dr}{d\tau}\right)^{2}\ \ \ \ \ (29)$ $\displaystyle \tilde{V}$ $\displaystyle \equiv$ $\displaystyle \frac{1}{2}\frac{\ell^{2}}{r^{2}}-GM\left(\frac{1}{r}+\frac{\ell^{2}}{r^{3}}\right)-\frac{\Lambda}{6}\left(\ell^{2}+r^{2}\right)\ \ \ \ \ (30)$ $\displaystyle \tilde{E}$ $\displaystyle \equiv$ $\displaystyle \frac{1}{2}\left(e^{2}-1\right) \ \ \ \ \ (31)$

If ${\Lambda=0}$, the ‘potential energy’ curve ${\tilde{V}}$ has a maximum at

$\displaystyle r=\frac{\ell^{2}-\sqrt{\ell^{4}-12G^{2}M^{2}\ell^{2}}}{2GM} \ \ \ \ \ (32)$

and a minimum at

$\displaystyle r=\frac{\ell^{2}+\sqrt{\ell^{4}-12G^{2}M^{2}\ell^{2}}}{2GM} \ \ \ \ \ (33)$

provided that ${\ell\ge\sqrt{12}GM}$. These two distances allow circular orbits (the former being unstable and the latter stable). As ${r\rightarrow\infty}$, ${\tilde{V}\rightarrow0}$ if ${\Lambda=0}$. However, if ${\Lambda>0}$, then, from 30, for very large ${r}$, ${\tilde{V}\rightarrow-\infty}$ so that an object at this great distance will recede from the central mass ${M}$, as we’d expect if the force becomes repulsive for very large distances.

# Schwarzschild metric: finding the metric; Birkhoff’s theorem

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 23; Boxes 23.3 – 23.4.

The expressions for the components of the Ricci tensor for a spherically symmetric source look quite frightening as differential equations, and in the general case would be impossible to solve exactly. However, if we restrict ourselves to the vacuum, that is, to the region outside the source, things simplify a lot. In that case, because the stress-energy tensor ${T_{ij}=0}$, it follows from the Einstein equation that all components of the Ricci tensor must also be zero:

$\displaystyle R_{ij}=8\pi G\left(T_{ij}-\frac{1}{2}g_{ij}T\right)=0 \ \ \ \ \ (1)$

The metric has the form

$\displaystyle ds^{2}=-Adt^{2}+Bdr^{2}+r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2} \ \ \ \ \ (2)$

and the Ricci components therefore give the PDEs:

 $\displaystyle \frac{1}{2B}\left[\partial_{rr}^{2}A-\partial_{tt}^{2}B+\frac{\left(\partial_{t}B\right)^{2}}{2B}+\frac{\left(\partial_{t}A\right)\left(\partial_{t}B\right)-\left(\partial_{r}A\right)^{2}}{2A}-\frac{\left(\partial_{r}A\right)\left(\partial_{r}B\right)}{2B}+\frac{2\partial_{r}A}{r}\right]$ $\displaystyle =$ $\displaystyle R_{tt}=0\ \ \ \ \ (3)$ $\displaystyle \frac{1}{2A}\left[\partial_{tt}^{2}B-\partial_{rr}^{2}A+\frac{\left(\partial_{r}A\right)^{2}-\left(\partial_{t}A\right)\left(\partial_{t}B\right)}{2A}+\frac{\left(\partial_{r}A\right)\left(\partial_{r}B\right)-\left(\partial_{t}B\right)^{2}}{2B}+\frac{2A\partial_{r}B}{rB}\right]$ $\displaystyle =$ $\displaystyle R_{rr}=0\ \ \ \ \ (4)$ $\displaystyle -\frac{r\partial_{r}A}{2AB}+\frac{r\partial_{r}B}{2B^{2}}+1-\frac{1}{B}$ $\displaystyle =$ $\displaystyle R_{\theta\theta}=0\ \ \ \ \ (5)$ $\displaystyle \frac{\partial_{t}B}{rB}$ $\displaystyle =$ $\displaystyle R_{tr}=0 \ \ \ \ \ (6)$

The ${R_{tr}}$ equation says

 $\displaystyle \partial_{t}B$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (7)$ $\displaystyle B$ $\displaystyle =$ $\displaystyle B\left(r\right) \ \ \ \ \ (8)$

That is, ${B}$ can depend on ${r}$ only.

Next, notice that the terms in the brackets for ${R_{tt}}$ and ${R_{rr}}$ cancel in pairs except for a couple of terms, so we have

 $\displaystyle 2BR_{tt}+2AR_{rr}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2\partial_{r}A}{r}+\frac{2A\partial_{r}B}{rB}\ \ \ \ \ (10)$ $\displaystyle \frac{\partial_{r}A}{A}$ $\displaystyle =$ $\displaystyle -\frac{\partial_{r}B}{B} \ \ \ \ \ (11)$

Plugging this into 5 we get

 $\displaystyle \frac{r\partial_{r}B}{B^{2}}+1-\frac{1}{B}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (12)$ $\displaystyle \frac{1}{B}-\frac{r\partial_{r}B}{B^{2}}$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (13)$ $\displaystyle \partial_{r}\left(\frac{r}{B}\right)$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (14)$ $\displaystyle \frac{r}{B}$ $\displaystyle =$ $\displaystyle r+C\ \ \ \ \ (15)$ $\displaystyle \frac{1}{B}$ $\displaystyle =$ $\displaystyle 1+\frac{C}{r} \ \ \ \ \ (16)$

where ${C}$ is a constant of integration.

Now, from 11 and given that ${B}$ does not depend on ${t}$, we must have ${\partial_{r}A/A}$ independent of ${t}$ also. This can happen only if any dependence ${A}$ has on ${t}$ cancels out when we take the quotient ${\partial_{r}A/A}$, and this can happen only if ${A\left(t,r\right)=f\left(t\right)a\left(r\right)}$ for some functions ${f}$ and ${a}$. In that case,

 $\displaystyle \frac{\partial_{r}A}{A}$ $\displaystyle =$ $\displaystyle -\frac{\partial_{r}B}{B}\ \ \ \ \ (17)$ $\displaystyle \frac{1}{a}\frac{da}{dr}$ $\displaystyle =$ $\displaystyle -\frac{1}{B}\frac{dB}{dr}\ \ \ \ \ (18)$ $\displaystyle \ln a$ $\displaystyle =$ $\displaystyle -\ln B+\ln K\ \ \ \ \ (19)$ $\displaystyle a$ $\displaystyle =$ $\displaystyle \frac{K}{B}=K\left(1+\frac{C}{r}\right)\ \ \ \ \ (20)$ $\displaystyle A$ $\displaystyle =$ $\displaystyle Kf\left(t\right)\left(1+\frac{C}{r}\right) \ \ \ \ \ (21)$

where we use total rather than partial derivatives in 18 because both ${a}$ and ${B}$ depend only on ${r}$, and ${K}$ is another constant of integration.

The metric now looks like this:

$\displaystyle ds^{2}=-Kf\left(t\right)\left(1+\frac{C}{r}\right)dt^{2}+\left(1+\frac{C}{r}\right)^{-1}dr^{2}+r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2} \ \ \ \ \ (22)$

In order for this metric to contain exactly one time coordinate, the coefficient of ${dt^{2}}$ must be negative (giving the time coordinate), while the coefficients of the other three coordinates must be positive. Therefore ${1+\frac{C}{r}>0}$ and ${Kf\left(t\right)>0}$.

At this stage, we can transform the time coordinate so that

$\displaystyle dt'=\sqrt{Kf\left(t\right)}dt \ \ \ \ \ (23)$

then replace ${t}$ by ${t'}$ and drop the prime to get

$\displaystyle ds^{2}=-\left(1+\frac{C}{r}\right)dt^{2}+\left(1+\frac{C}{r}\right)^{-1}dr^{2}+r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2} \ \ \ \ \ (24)$

We thus arrive (almost; we still have to find ${C}$) at the Schwarzschild metric. Note that in this form, the metric is independent of time, even though we haven’t assumed that the mass-energy of the source is independent of time, only that it is always spherically symmetric. Thus a star that expands or contracts while maintaining spherical symmetry would always give rise to the same metric. This is called Birkhoff’s theorem.

This choice of ${t}$ is the time measured by an observer at rest at infinity (${r\rightarrow\infty}$), since to such an observer ${ds^{2}=-\left(1+\frac{C}{r}\right)dt^{2}\rightarrow-dt^{2}}$. This might look like a bit of a fudge, since we hid the time dependence of ${g_{tt}}$ by sweeping it under the carpet with the rescaling of time in 23. However, on reflection, I think it does actually make sense, since in a more general case (if ${T_{ij}\ne0}$, say, or if the metric were non-diagonal), it wouldn’t be possible to find any time coordinate that gives a time-independent metric.

# Spherically symmetric solution to the Einstein equation

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 23; Box 23.1.

We’ll return now to general relativity, and build up to a derivation of the Schwarzschild metric. As a quick review, the problem is to find a solution to the Einstein equation in the form

$\displaystyle R^{ij}=8\pi G\left(T^{ij}-\frac{1}{2}g^{ij}T\right) \ \ \ \ \ (1)$

where ${R^{ij}}$ is the Ricci tensor, itself a contraction of the Riemann tensor, ${T^{ij}}$ is the stress-energy tensor and ${T=g_{ij}T^{ij}}$ is the stress-energy scalar.

In a practical problem, ${T^{ij}}$ will be given, and the problem is to determine the metric ${g^{ij}}$ from the Ricci tensor. The Ricci tensor is specified in terms of Christoffel symbols, which are in turn defined in terms of the metric and its derivatives, so the Einstein equation becomes a system of coupled, non-linear partial differential equations in the components of the metric tensor.

A good starting point is to look at spacetime around a source with spherical symmetry. We can picture this spacetime as a set of nested spherical shells, on the surface of which the usual 2-d spherical metric applies:

$\displaystyle ds^{2}=r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2} \ \ \ \ \ (2)$

This gives us several of the metric components already, in that

 $\displaystyle g_{\theta\theta}$ $\displaystyle =$ $\displaystyle r^{2}\ \ \ \ \ (3)$ $\displaystyle g_{\phi\phi}$ $\displaystyle =$ $\displaystyle r^{2}\sin^{2}\theta\ \ \ \ \ (4)$ $\displaystyle g_{\theta\phi}$ $\displaystyle =$ $\displaystyle g_{\phi\theta}=0 \ \ \ \ \ (5)$

We can set up the coordinates on each shell such that any line with fixed values of ${\theta}$ and ${\phi}$ is perpendicular to all the surfaces. This means that the basis vectors ${\mathbf{e}_{\theta}}$ and ${\mathbf{e}_{\phi}}$ are perpendicular to the third spatial basis vector ${\mathbf{e}_{r}}$, so that ${g_{r\theta}=g_{r\phi}=0}$. With spatial symmetry, there should be no difference in the way the metric treats motions in different directions of ${\theta}$ or ${\phi}$, so we’d expect the terms ${g_{r\theta}drd\theta}$, ${g_{r\phi}drd\phi}$, ${g_{t\theta}dtd\theta}$ and ${g_{t\phi}dtd\phi}$ to all be zero, which gives us four more (well, eight, actually, since ${g_{ij}=g_{ji}}$) metric components.

We’re left with ${g_{tt}}$, ${g_{rr}}$ and ${g_{rt}=g_{tr}}$. If ${t}$ is a time coordinate, we must have ${g_{tt}<0}$ and likewise, if ${r}$ is a spatial coordinate, then ${g_{rr}>0}$. We can, in fact, eliminate ${g_{rt}}$ by making a coordinate transformation as follows:

$\displaystyle t'=t+f\left(r,t\right) \ \ \ \ \ (6)$

where ${f}$ is some function of the original ${r}$ and ${t}$ coordinates (unknown at present). We can, in principle, always determine ${f}$ so that ${g_{rt'}=0}$ and then use ${t'}$ as our new time coordinate. [Note that we can’t use the symmetry argument to claim that ${g_{rt}=0}$, since in a spherically symmetric situation, it does make a difference whether you are travelling in the plus or minus ${r}$ direction, so it isn’t necessarily so that ${g_{rt}=0}$ in all cases.]

Take the differential of this equation to get

 $\displaystyle dt'$ $\displaystyle =$ $\displaystyle dt+\partial_{r}f\; dr+\partial_{t}f\; dt\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(1+\partial_{t}f\right)dt+\partial_{r}f\; dr\ \ \ \ \ (8)$ $\displaystyle dt$ $\displaystyle =$ $\displaystyle \frac{dt'-\partial_{r}f\; dr}{1+\partial_{t}f}\equiv\alpha\left(dt'-\partial_{r}f\; dr\right) \ \ \ \ \ (9)$

With the deductions above, our original metric equation is

$\displaystyle ds^{2}=g_{tt}dt^{2}+2g_{rt}dr\; dt+g_{rr}dr^{2}+r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2} \ \ \ \ \ (10)$

Substituting 9 into the first two terms on the RHS, we get

 $\displaystyle g_{tt}dt^{2}+2g_{rt}dr\; dt$ $\displaystyle =$ $\displaystyle g_{tt}\alpha^{2}\left(dt'-\partial_{r}f\; dr\right)^{2}+2\alpha g_{rt}dr\left(dt'-\partial_{r}f\; dr\right)\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle dr^{2}\left(g_{tt}\alpha^{2}\left(\partial_{r}f\right)^{2}-2\alpha g_{rt}\partial_{r}f\right)+\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle$ $\displaystyle dr\; dt'\left(2\alpha g_{rt}-2\alpha^{2}g_{tt}\partial_{r}f\right)+\left(dt'\right)^{2}g_{tt}\alpha^{2}\nonumber$

We can now set the coefficient of ${dr\; dt'}$ to zero to get

 $\displaystyle g_{rt}$ $\displaystyle =$ $\displaystyle \alpha g_{tt}\partial_{r}f\ \ \ \ \ (13)$ $\displaystyle \frac{g_{rt}}{g_{tt}}\left(1+\partial_{t}f\right)$ $\displaystyle =$ $\displaystyle \partial_{r}f \ \ \ \ \ (14)$

Assuming this partial differential equation for ${f\left(r,t\right)}$ can be solved (which we won’t be able to do a priori, since we don’t know ${g_{rt}}$ or ${g_{tt}}$, but in principle, the equation can be solved), it is always possible to find a time coordinate ${t'}$ such that ${g_{rt'}=0}$, so we might as well use that time coordinate from the start. Relabelling this time coordinate from ${t'}$ back to ${t}$, the spherically symmetric metric is then

$\displaystyle ds^{2}=g_{tt}dt^{2}+g_{rr}dr^{2}+r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2} \ \ \ \ \ (15)$

We therefore have only two metric components that need to be found by solving the Einstein equation 1, which we’ll get to in the next post.

# Wave solution of the weak-field Einstein equation

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 22, Problem 22.4.

In the weak-field limit, the Einstein equation becomes

$\displaystyle -\frac{1}{2}\square^{2}h_{jm}=8\pi G\left(T_{jm}-\frac{1}{2}\eta_{jm}T\right) \ \ \ \ \ (1)$

In empty space, the stress-energy tensor is zero (${T_{jm}=0}$) so the equation becomes

$\displaystyle \square^{2}h_{jm}=0 \ \ \ \ \ (2)$

This is just the 3-d wave equation for a wave with speed ${v=1}$:

$\displaystyle \nabla^{2}h_{jm}=\partial_{t}^{2}h_{jm} \ \ \ \ \ (3)$

so one solution is a wave travelling in the ${x}$ direction:

$\displaystyle h_{jm}=A_{jm}\cos\left(\omega t-kx\right) \ \ \ \ \ (4)$

where ${A_{jm}}$ is a constant matrix.

Substituting into 3 we get

$\displaystyle -A_{jm}k^{2}\cos\left(\omega t-kx\right)=-A_{jm}\omega^{2}\cos\left(\omega t-kx\right) \ \ \ \ \ (5)$

which gives the condition ${\omega=k}$. The speed ${v=1}$ means that the wave travels at the speed of light.

In the weak field limit, the Ricci tensor becomes

$\displaystyle R_{jm}=\frac{1}{2}\left(\partial_{j}H_{m}+\partial_{m}H_{j}-\eta^{nl}\partial_{n}\partial_{l}h_{jm}\right) \ \ \ \ \ (6)$

where

$\displaystyle H_{m}\equiv\eta^{nl}\left(\partial_{n}h_{lm}-\frac{1}{2}\partial_{m}h_{nl}\right) \ \ \ \ \ (7)$

We derived the wave equation above by taking all ${H_{m}=0}$, so we can see what conditions this imposes on ${A_{jm}}$. Since ${\omega=k}$ and ${h_{ij}}$ is independent of ${y}$ and ${z}$ we have

 $\displaystyle \partial_{t}h_{ij}$ $\displaystyle =$ $\displaystyle -A_{ij}\omega\sin\left(\omega\left(t-x\right)\right)\ \ \ \ \ (8)$ $\displaystyle \partial_{x}h_{ij}$ $\displaystyle =$ $\displaystyle A_{ij}\omega\sin\left(\omega\left(t-x\right)\right) \ \ \ \ \ (9)$

Thus requiring each ${H_{m}=0}$ allows us to cancel off ${\omega\sin\left(\omega\left(t-x\right)\right)}$ from all terms, giving for ${m=t}$

 $\displaystyle \eta^{tt}\left(-\frac{1}{2}A_{tt}\right)+\eta^{xx}\left(A_{xt}+\frac{1}{2}A_{xx}\right)+\eta^{yy}\left(\frac{1}{2}A_{yy}\right)\eta^{zz}\left(\frac{1}{2}A_{zz}\right)$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (10)$ $\displaystyle A_{tt}+A_{xx}+A_{yy}+A_{zz}+2A_{xt}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (11)$

The other three ${H_{m}}$s give us three more conditions:

 $\displaystyle A_{tt}+A_{xx}-A_{yy}-A_{zz}+2A_{tx}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (12)$ $\displaystyle A_{ty}+A_{xy}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (13)$ $\displaystyle A_{tz}+A_{xz}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (14)$

Since ${h_{ij}=h_{ji}}$ we must also have ${A_{ij}=A_{ji}}$ so from 11 and 12 we get

 $\displaystyle A_{zz}$ $\displaystyle =$ $\displaystyle -A_{yy}\ \ \ \ \ (15)$ $\displaystyle A_{tx}$ $\displaystyle =$ $\displaystyle -\frac{1}{2}\left(A_{tt}+A_{xx}\right) \ \ \ \ \ (16)$

Allowing for the symmetry of ${A_{ij}}$ leaves 10 components to determine, and the four conditions here impose another 4 constraints, leaving 6 independent components, which we can take as

 $\displaystyle A_{tt}$ $\displaystyle =$ $\displaystyle a\ \ \ \ \ (17)$ $\displaystyle A_{xx}$ $\displaystyle =$ $\displaystyle b\ \ \ \ \ (18)$ $\displaystyle A_{yy}$ $\displaystyle =$ $\displaystyle c\ \ \ \ \ (19)$ $\displaystyle A_{ty}$ $\displaystyle =$ $\displaystyle d\ \ \ \ \ (20)$ $\displaystyle A_{tz}$ $\displaystyle =$ $\displaystyle e\ \ \ \ \ (21)$ $\displaystyle A_{yz}$ $\displaystyle =$ $\displaystyle f \ \ \ \ \ (22)$

Using the conditions above, the matrix has the form

$\displaystyle A_{ij}=\left[\begin{array}{cccc} a & -\frac{1}{2}\left(a+b\right) & d & e\\ -\frac{1}{2}\left(a+b\right) & b & -d & -e\\ d & -d & c & f\\ e & -e & f & -c \end{array}\right] \ \ \ \ \ (23)$

# Stress-energy tensor in the weak field, low velocity limit

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 22, Box 22.5.

In the weak field limit, the Ricci tensor becomes

$\displaystyle R_{jm}=\frac{1}{2}\left(\partial_{j}H_{m}+\partial_{m}H_{j}-\eta^{nl}\partial_{n}\partial_{l}h_{jm}\right) \ \ \ \ \ (1)$

where

$\displaystyle H_{m}\equiv\eta^{nl}\left(\partial_{n}h_{lm}-\frac{1}{2}\partial_{m}h_{nl}\right) \ \ \ \ \ (2)$

and the ${h_{ij}}$ is the perturbation on the flat metric, so that

$\displaystyle g_{ij}=\eta_{ij}+h_{ij} \ \ \ \ \ (3)$

Because we can introduce a coordinate transformation for the four coordinates in the form

$\displaystyle \left(x'\right)^{i}=f^{i}\left(x^{j}\right) \ \ \ \ \ (4)$

there are four degrees of freedom that we can play with in specifying the form of ${H_{i}}$. It turns out (we may get around to a proof in some future post) that it is always possible to find a coordinate system in which all ${H_{i}=0}$. If we use such a coordinate system then the Ricci tensor is

$\displaystyle R_{jm}=-\frac{1}{2}\eta^{nl}\partial_{n}\partial_{l}h_{jm} \ \ \ \ \ (5)$

so the Einstein equation becomes

$\displaystyle -\frac{1}{2}\eta^{nl}\partial_{n}\partial_{l}h_{jm}=8\pi G\left(T_{jm}-\frac{1}{2}\eta_{jm}T\right) \ \ \ \ \ (6)$

Introducing the d’Alembertian operator

$\displaystyle \square^{2}\equiv\eta^{nl}\partial_{n}\partial_{l}=-\frac{d^{2}}{dt^{2}}+\nabla^{2} \ \ \ \ \ (7)$

we can write the Einstein equation as

$\displaystyle -\frac{1}{2}\square^{2}h_{jm}=8\pi G\left(T_{jm}-\frac{1}{2}\eta_{jm}T\right) \ \ \ \ \ (8)$

As it stands, this equation is a set of uncoupled differential equations for the ${h_{jm}}$ so in principle they can be solved. However, we can invoke another approximation by assuming that the system is in a steady state so that all time derivatives are zero. This doesn’t necessarily mean that the masses are all stationary, since we might have a star rotating at a constant angular velocity. In such cases, the stress-energy tensor ${T_{jm}}$ is constant in time so we would expect ${h_{jm}}$ to be independent of time as well. In that case, we get

$\displaystyle \nabla^{2}h_{jm}=-16\pi G\left(T_{jm}-\frac{1}{2}\eta_{jm}T\right) \ \ \ \ \ (9)$

This equation should look familiar from electrodynamics, where it is formally equivalent to Poisson’s equation for the electrostatic potential in terms of the charge distribution. In that case we had

$\displaystyle \nabla^{2}V=-\nabla\cdot\mathbf{E}=-\frac{\rho}{\epsilon_{0}} \ \ \ \ \ (10)$

where ${\rho}$ here is the charge density. The solution of this equation is

$\displaystyle V\left(\mathbf{r}\right)=\frac{1}{4\pi\epsilon_{0}}\int\frac{1}{\left|\mathbf{r}-\mathbf{r}'\right|}\rho\left(\mathbf{r}'\right)d^{3}\mathbf{r}' \ \ \ \ \ (11)$

By replacing ${\rho/\epsilon_{0}}$ with ${16\pi G\left(T_{jm}-\frac{1}{2}\eta_{jm}T\right)}$ we can find ${h_{jm}}$ as

$\displaystyle h_{jm}=2G\int\frac{2T_{jm}-\eta_{jm}T}{\left|\mathbf{r}-\mathbf{r}'\right|}d^{3}\mathbf{r}' \ \ \ \ \ (12)$

Thus if we know the stress-energy tensor as a function of position, we can work out the perturbations on the flat metric ${h_{jm}}$.

Example We can look at this equation for the case of a perfect fluid, where the stress-energy tensor is

$\displaystyle T_{ij}=\left(\rho_{0}+P_{0}\right)u_{i}u_{j}+P_{0}g_{ij} \ \ \ \ \ (13)$

where ${\rho_{0}}$ and ${P_{0}}$ are the fluid’s density and pressure in its rest frame, and ${u_{i}}$ is the fluid’s four-velocity in the observer’s frame. We need to find the numerator of the integrand in 12 for each component. First, we work out the stress-energy scalar ${T}$:

 $\displaystyle T$ $\displaystyle =$ $\displaystyle g^{ij}T_{ij}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\rho_{0}+P_{0}\right)g^{ij}u_{i}u_{j}+P_{0}g^{ij}g_{ij}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\left(\rho_{0}+P_{0}\right)+4P_{0}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\rho_{0}+3P_{0} \ \ \ \ \ (17)$

since ${g^{ij}u_{i}u_{j}=-1}$ and ${g^{ij}g_{ij}=\delta_{i}^{i}=4}$. Since ${T}$ is a scalar, this result is valid in all coordinate systems. Also, we haven’t yet used any approximations, so this result is valid for all perfect fluids, even ones where the density and pressure are large.

Now let’s assume that ${\rho_{0}}$ and ${P_{0}}$ are small and so we keep only up to first order terms, so any product of ${\rho_{0}}$ or ${P_{0}}$ with ${h^{ij}}$ can be ignored. Thus

 $\displaystyle T_{ij}-\frac{1}{2}g_{ij}T$ $\displaystyle \approx$ $\displaystyle T_{ij}-\frac{1}{2}\eta_{ij}T\ \ \ \ \ (18)$ $\displaystyle 2T_{ij}-g_{ij}T$ $\displaystyle \approx$ $\displaystyle 2T_{ij}-\eta_{ij}T\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle 2\left(\rho_{0}+P_{0}\right)u_{i}u_{j}+2P_{0}\eta_{ij}-\eta_{ij}\left(-\rho_{0}+3P_{0}\right)\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\left(\rho_{0}+P_{0}\right)u_{i}u_{j}+\eta_{ij}\left(\rho_{0}-P_{0}\right) \ \ \ \ \ (21)$

There are 3 cases to consider. First, ${i=j=t}$ and use ${\eta_{tt}=-1}$. Further, we’ll assume that the spatial velocity components are all small, so ${u_{t}\approx-1}$ and ${u_{i}u_{j}\approx0}$ if ${i}$ and ${j}$ are both spatial indices.

 $\displaystyle 2T_{tt}-\eta_{tt}T$ $\displaystyle =$ $\displaystyle 2\left(\rho_{0}+P_{0}\right)u_{t}u_{t}+\eta_{tt}\left(\rho_{0}-P_{0}\right)\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\left(\rho_{0}+P_{0}\right)-\left(\rho_{0}-P_{0}\right)\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \rho_{0}+3P_{0} \ \ \ \ \ (24)$

Now suppose ${i=t}$ and ${j}$ is a spatial index (or vice versa; since everything is symmetric it makes no difference). Then

 $\displaystyle 2T_{tj}-\eta_{tj}T$ $\displaystyle =$ $\displaystyle 2T_{tj}\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\left(\rho_{0}+P_{0}\right)u_{t}u_{j}\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -2\left(\rho_{0}+P_{0}\right)u_{j} \ \ \ \ \ (27)$

Finally, if both ${i}$ and ${j}$ are spatial indices we get, if ${i\ne j}$

 $\displaystyle 2T_{ij}-g_{ij}T$ $\displaystyle =$ $\displaystyle 2\left(\rho_{0}+P_{0}\right)u_{i}u_{j}+\eta_{ij}\left(\rho_{0}-P_{0}\right)\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle 0 \ \ \ \ \ (29)$

since the first term involves the second order term ${u_{i}u_{j}}$ and in the second term ${\eta_{ij}=0}$ if ${i\ne j}$.

Now if ${i=j}$ we get

 $\displaystyle 2T_{ii}-g_{ii}T$ $\displaystyle =$ $\displaystyle 2\left(\rho_{0}+P_{0}\right)u_{i}u_{i}+\eta_{ii}\left(\rho_{0}-P_{0}\right)\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle \rho_{0}-P_{0} \ \ \ \ \ (31)$

again, since the first term has the second order factor ${u_{i}^{2}}$ and in the second term ${\eta_{ii}=+1}$ if ${i}$ is a spatial index.

Therefore, if we know ${\rho_{0}}$ and ${P_{0}}$ as functions of position we can work out the perturbations ${h_{ij}}$ to the metric.

# Vacuum stress-energy and the cosmological constant

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 21, Problem 21.9.

The Einstein equation is

$\displaystyle R^{ij}=8\pi G\left(T^{ij}-\frac{1}{2}g^{ij}T\right)+\Lambda g^{ij} \ \ \ \ \ (1)$

Up to now, we’ve usually taken ${\Lambda=0}$, since we know from the Newtonian limit that ${\Lambda}$ must be very small. If ${\Lambda\ne0}$, the Newtonian limit becomes

$\displaystyle \nabla^{2}\Phi=4\pi G\rho-\Lambda \ \ \ \ \ (2)$

so ${\Lambda}$ acts as a negative mass density, that is, it adds a repulsive term into the gravitational force. Einstein originally introduced it to counter the attractive force of gravity on a cosmological scale, since at the time it was believed that the universe was static (neither expanding nor contracting) and if gravity were purely attractive, the universe would be contracting.

At the moment, the universe is believed to be expanding so ${\Lambda}$ is believed to be non-zero and positive, although still small enough that its effects are not noticeable on the scale of the solar system (or indeed on a galactic scale). Because of this, ${\Lambda}$ is called the cosmological constant.

We can include ${\Lambda}$ within the stress-energy tensor by defining a vacuum stress-energy as

$\displaystyle T_{vac}^{ij}=-\frac{\Lambda}{8\pi G}g^{ij} \ \ \ \ \ (3)$

We can define a vacuum stress-energy scalar:

 $\displaystyle T_{vac}$ $\displaystyle \equiv$ $\displaystyle g_{ij}T_{vac}^{ij}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\Lambda}{8\pi G}g_{ij}g^{ij}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{4\Lambda}{8\pi G} \ \ \ \ \ (6)$

Therefore

 $\displaystyle T_{vac}^{ij}-\frac{1}{2}g^{ij}T_{vac}$ $\displaystyle =$ $\displaystyle -\frac{\Lambda}{8\pi G}g^{ij}+\frac{2\Lambda}{8\pi G}g^{ij}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\Lambda}{8\pi G}g^{ij} \ \ \ \ \ (8)$

and we can write 1 as

 $\displaystyle R^{ij}$ $\displaystyle =$ $\displaystyle 8\pi G\left(T^{ij}-\frac{1}{2}g^{ij}T+T_{vac}^{ij}-\frac{1}{2}g^{ij}T_{vac}\right)\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle \equiv$ $\displaystyle 8\pi G\left(T_{all}^{ij}-\frac{1}{2}g^{ij}T_{all}\right) \ \ \ \ \ (10)$

where ${T_{all}^{ij}}$ includes the stress-energy from the mass-energy density and the vacuum.

The dominant energy condition is a constraint placed on the stress-energy tensor so that observers in any local orthogonal frame will measure the fluid’s speed to be less than the speed of light. The condition is that if ${a^{i}}$ is any four-vector that is causal, that is, it satisfies the conditions

 $\displaystyle \mathbf{a}\cdot\mathbf{a}$ $\displaystyle \le$ $\displaystyle 0\ \ \ \ \ (11)$ $\displaystyle a^{t}$ $\displaystyle >$ $\displaystyle 0 \ \ \ \ \ (12)$

then we require the stress-energy tensor ${T^{ij}}$ to satisfy the condition that if

$\displaystyle b^{i}=-T^{ij}g_{jk}a^{k} \ \ \ \ \ (13)$

then ${\mathbf{b}}$ is also a causal four-vector. For the vacuum stress-energy this condition says

 $\displaystyle b^{i}$ $\displaystyle =$ $\displaystyle -T_{vac}^{ij}g_{jk}a^{k}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\Lambda}{8\pi G}g^{ij}g_{jk}a^{k}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\Lambda}{8\pi G}\delta_{\; k}^{i}a^{k}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\Lambda}{8\pi G}a^{i} \ \ \ \ \ (17)$

That is, ${b^{i}}$ is just a positive (if ${\Lambda>0}$) constant multiplied by ${a^{i}}$, so if ${a^{i}}$ is causal, then ${b^{i}}$ must also be causal. Thus ${T_{vac}^{ij}}$ satisfies the dominant energy condition.

# Einstein equation for an exponential metric

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 21, Problem 21.8.

Consider the metric:

$\displaystyle ds^{2}=-e^{2gx}dt^{2}+dx^{2}+dy^{2}+dz^{2} \ \ \ \ \ (1)$

We’ll have a look at what the Einstein equation has to say about gravity in a spacetime using this metric. First, we’ll find the Christoffel symbols using the method of comparing the two forms of the geodesic equation. The geodesic equation is

$\displaystyle \frac{d}{d\tau}\left(g_{aj}\frac{dx^{j}}{d\tau}\right)-\frac{1}{2}\frac{\partial g_{ij}}{\partial x^{a}}\frac{dx^{i}}{d\tau}\frac{dx^{j}}{d\tau}=0 \ \ \ \ \ (2)$

The following equation is formally equivalent to this (where a dot above a symbol means the derivative with respect to ${\tau}$):

$\displaystyle \ddot{x}^{m}+\Gamma_{\;ij}^{m}\dot{x}^{j}\dot{x}^{i}=0 \ \ \ \ \ (3)$

Since the metric is diagonal and none of its components depends on ${y}$ or ${z}$, the LHS of 2 is identically zero for ${a=y}$ or ${a=z}$, so all Christoffel symbols with any index being ${y}$ or ${z}$ is zero.

Now look at ${a=t}$. We get from 2, since ${g_{ij}}$ doesn’t depend on ${t}$:

 $\displaystyle \frac{d}{d\tau}\left(g_{tj}\frac{dx^{j}}{d\tau}\right)$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (4)$ $\displaystyle -2ge^{2gx}\dot{x}\dot{t}-e^{2gx}\ddot{t}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (5)$ $\displaystyle \ddot{t}+2g\dot{x}\dot{t}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (6)$

Comparing with 3 we see that

$\displaystyle \Gamma_{\;tx}^{t}=\Gamma_{\;xt}^{t}=g \ \ \ \ \ (7)$

Now for ${a=x}$:

 $\displaystyle \ddot{x}-\frac{1}{2}\left(\partial_{x}g_{tt}\right)\dot{t}^{2}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (8)$ $\displaystyle \ddot{x}+ge^{2gx}\dot{t}^{2}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (9)$ $\displaystyle \Gamma_{\;tt}^{x}$ $\displaystyle =$ $\displaystyle ge^{2gx} \ \ \ \ \ (10)$

These are the only non-zero Christoffel symbols. We can get an expression for the acceleration of a particle in its rest frame by noting that

$\displaystyle \ddot{x}=-ge^{2gx}\dot{t}^{2} \ \ \ \ \ (11)$

The four velocity of a particle at rest (so ${dx=dy=dz=0}$) must satisfy

 $\displaystyle u^{i}u_{i}$ $\displaystyle =$ $\displaystyle -1\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g_{ij}u^{i}u^{j}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g_{tt}\dot{t}^{2}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -e^{2gx}\dot{t}^{2}\ \ \ \ \ (15)$ $\displaystyle \dot{t}^{2}$ $\displaystyle =$ $\displaystyle e^{-2gx} \ \ \ \ \ (16)$

Plugging into 11 we get

$\displaystyle \ddot{x}=-g \ \ \ \ \ (17)$

That is, the acceleration in the ${x}$ direction is a constant, so this would seem to indicate a uniform gravitational field. However, let’s apply the Einstein equation and see what we get. We’ll need the Riemann tensor in order to get the Ricci tensor, so we’ll use the definition of the former:

$\displaystyle R_{\;j\ell m}^{i}\equiv-\left[\partial_{m}\Gamma_{\;\ell j}^{i}-\partial_{\ell}\Gamma_{\;mj}^{i}+\Gamma_{\;\ell j}^{k}\Gamma_{\;km}^{i}-\Gamma_{\;mj}^{k}\Gamma_{\;\ell k}^{i}\right] \ \ \ \ \ (18)$

Since all Christoffel symbols with a ${y}$ or ${z}$ index are zero, the only possibly non-zero components of ${R_{ij\ell m}}$ are those containing only ${x}$ or ${t}$, and due to the symmetry relations, the only independent component is

 $\displaystyle R_{txtx}$ $\displaystyle =$ $\displaystyle g_{ti}R_{\;xtx}^{i}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{2gx}\left[\partial_{x}\Gamma_{\;tx}^{t}-\partial_{t}\Gamma_{\;xx}^{t}+\Gamma_{\;tx}^{k}\Gamma_{\;kx}^{t}-\Gamma_{\;xx}^{k}\Gamma_{\;tk}^{t}\right]\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{2gx}\left[0-0+\Gamma_{\;tx}^{t}\Gamma_{\;tx}^{t}-0\right]\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g^{2}e^{2gx} \ \ \ \ \ (22)$

Now for the Ricci tensor. We have

 $\displaystyle R_{ab}$ $\displaystyle =$ $\displaystyle R_{\;aib}^{i}\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g^{ij}R_{jaib} \ \ \ \ \ (24)$

Since the metric is diagonal, the upstairs components are just the reciprocals of the downstairs ones, so

 $\displaystyle g^{tt}$ $\displaystyle =$ $\displaystyle -e^{-2gx}\ \ \ \ \ (25)$ $\displaystyle g^{xx}$ $\displaystyle =$ $\displaystyle g^{yy}=g^{zz}=1 \ \ \ \ \ (26)$

and we get

 $\displaystyle R_{tt}$ $\displaystyle =$ $\displaystyle g^{xx}R_{xtxt}=g^{2}e^{2gx}\ \ \ \ \ (27)$ $\displaystyle R_{xx}$ $\displaystyle =$ $\displaystyle g^{tt}R_{txtx}=g^{tt}R_{xtxt}=-g^{2}\ \ \ \ \ (28)$ $\displaystyle R_{xt}=R_{tx}$ $\displaystyle =$ $\displaystyle g^{ab}R_{bxat}=0 \ \ \ \ \ (29)$

where in the last line we see that ${R_{bxat}}$ can be non-zero only if ${b=t}$ and ${a=x}$ but since ${g^{ab}}$ is diagonal, ${g^{xt}=0}$. All components of ${R_{ab}}$ involving ${y}$ or ${z}$ indices are zero because all components of ${R_{abcd}}$ involving ${y}$ or ${z}$ indices are zero. To use the Ricci tensor in the Einstein equation, we need the upstairs version, which is

 $\displaystyle R^{tt}$ $\displaystyle =$ $\displaystyle g^{ta}g^{tb}R_{ab}=e^{-4gx}\left(g^{2}e^{2gx}\right)=g^{2}e^{-2gx}\ \ \ \ \ (30)$ $\displaystyle R^{xx}$ $\displaystyle =$ $\displaystyle g^{xa}g^{xb}R_{ab}=R_{xx}=-g^{2} \ \ \ \ \ (31)$

In a vacuum (assuming ${\Lambda=0}$) the stress-energy tensor ${T^{ij}=0}$ and the Einstein equation says that

$\displaystyle R^{ab}=0 \ \ \ \ \ (32)$

The only way this can be true is if ${g=0}$, meaning that there is no gravitational field.

More generally, suppose that there is some fluid in the region so that ${T^{ij}\ne0}$. In that case

 $\displaystyle R^{tt}$ $\displaystyle =$ $\displaystyle g^{2}e^{-2gx}=8\pi G\left(T^{tt}+\frac{1}{2}e^{-2gx}T\right)\ \ \ \ \ (33)$ $\displaystyle R^{xx}$ $\displaystyle =$ $\displaystyle -g^{2}=8\pi G\left(T^{xx}-\frac{1}{2}T\right)\ \ \ \ \ (34)$ $\displaystyle R^{yy}$ $\displaystyle =$ $\displaystyle 0=8\pi G\left(T^{yy}-\frac{1}{2}T\right)\ \ \ \ \ (35)$ $\displaystyle R^{zz}$ $\displaystyle =$ $\displaystyle 0=8\pi G\left(T^{zz}-\frac{1}{2}T\right) \ \ \ \ \ (36)$

where the stress-energy scalar is

 $\displaystyle T$ $\displaystyle =$ $\displaystyle g_{ij}T^{ij}\ \ \ \ \ (37)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -e^{2gx}T^{tt}+T^{xx}+T^{yy}+T^{zz} \ \ \ \ \ (38)$

From 35 and 36 we have

$\displaystyle T^{yy}=T^{zz}=\frac{T}{2} \ \ \ \ \ (39)$

so from 38 we have

$\displaystyle -e^{2gx}T^{tt}+T^{xx}=0 \ \ \ \ \ (40)$

However, if we multiply 33 by ${e^{2gx}}$ and add it to 34 we get

$\displaystyle 0=8\pi G\left(T^{tt}e^{2gx}+T^{xx}\right) \ \ \ \ \ (41)$

Combining the last two equations we get

$\displaystyle T^{tt}=T^{xx}=0 \ \ \ \ \ (42)$

Plugging 39 into 34 we get

$\displaystyle T^{yy}=T^{zz}=\frac{g^{2}}{8\pi G} \ \ \ \ \ (43)$

This result agrees with the correction to Moore’s problem 21.8 given in the errata.

In a perfect fluid at rest, ${T^{tt}}$ is the energy density and the diagonal components ${T^{ii}}$ for ${i=x,y,z}$ are the pressures in each of the 3 directions. If these results are to be believed, the fluid has zero energy (${T^{tt}=0}$) and no pressure in the ${x}$ direction, but a non-zero pressure in the ${y}$ and ${z}$ directions. It’s hard to imagine how a fluid could have zero energy and yet still exert pressure, so this would appear to be why the results are absurd, as stated by Moore.

# Einstein equation on the surface of a sphere

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 21, Problem 21.7.

According to the Einstein equation, the Riemann tensor in 2D must be zero in empty space, implying that gravitational fields cannot exist in 2D. Another consequence of the Einstein equation is that the stress-energy must be zero on the surface of a sphere. That is, even though a 2D surface is manifestly curved, the curvature is not the result of any mass or energy. This is another example of how general relativity breaks down in two dimensions.

The Einstein equation is

$\displaystyle R^{ij}=\kappa\left(T^{ij}-\frac{1}{2}g^{ij}T\right)+\Lambda g^{ij} \ \ \ \ \ (1)$

$\displaystyle R^{ij}=\left[\begin{array}{cc} \frac{1}{r^{4}} & 0\\ 0 & \frac{1}{r^{4}\sin^{2}\theta} \end{array}\right] \ \ \ \ \ (2)$

The metric for a sphere is (in both forms):

 $\displaystyle g^{ij}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} \frac{1}{r^{2}} & 0\\ 0 & \frac{1}{r^{2}\sin^{2}\theta} \end{array}\right]\ \ \ \ \ (3)$ $\displaystyle g_{ij}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} r^{2} & 0\\ 0 & r^{2}\sin^{2}\theta \end{array}\right] \ \ \ \ \ (4)$

Since the off-diagonal elements of ${g^{ij}}$ and ${R^{ij}}$ are all zero, 1 tells us that

$\displaystyle T^{\theta\phi}=T^{\phi\theta}=0 \ \ \ \ \ (5)$

To deal with the diagonal elements, we first need the stress-energy scalar.

 $\displaystyle T$ $\displaystyle =$ $\displaystyle g_{ij}T^{ij}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}T^{\theta\theta}+r^{2}\sin^{2}\theta T^{\phi\phi} \ \ \ \ \ (7)$

We have

 $\displaystyle R^{\theta\theta}$ $\displaystyle =$ $\displaystyle \frac{1}{r^{4}}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \kappa\left(T^{\theta\theta}-\frac{1}{2r^{2}}T\right)+\frac{\Lambda}{r^{2}}\ \ \ \ \ (9)$ $\displaystyle R^{\phi\phi}$ $\displaystyle =$ $\displaystyle \frac{1}{r^{4}\sin^{2}\theta}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \kappa\left(T^{\phi\phi}-\frac{1}{2r^{2}\sin^{2}\theta}T\right)+\frac{\Lambda}{r^{2}\sin^{2}\theta} \ \ \ \ \ (11)$

Combining these we get

 $\displaystyle \frac{R^{\theta\theta}}{\sin^{2}\theta}-R^{\phi\phi}$ $\displaystyle =$ $\displaystyle \kappa\left(\frac{T^{\theta\theta}}{\sin^{2}\theta}-T^{\phi\phi}\right)=0\ \ \ \ \ (12)$ $\displaystyle \frac{T^{\theta\theta}}{\sin^{2}\theta}-T^{\phi\phi}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (13)$ $\displaystyle T^{\theta\theta}$ $\displaystyle =$ $\displaystyle T^{\phi\phi}\sin^{2}\theta\ \ \ \ \ (14)$ $\displaystyle T$ $\displaystyle =$ $\displaystyle 2r^{2}\sin^{2}\theta T^{\phi\phi} \ \ \ \ \ (15)$

Therefore

 $\displaystyle T^{\theta\theta}-\frac{1}{2r^{2}}T$ $\displaystyle =$ $\displaystyle T^{\phi\phi}-\frac{1}{2r^{2}\sin^{2}\theta}T=0 \ \ \ \ \ (16)$

holds identically. Thus the stress-energy contribution to the Einstein equation is always zero on a sphere (although the stress-energy tensor may have two non-zero components, these two components always combine to give zero contribution to the Einstein equation).