# Black hole with static charge; Reissner-Nordström solution

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

The derivation of the Schwarzschild metric can be enhanced to include a source, such as a black hole, with a static electric charge ${Q}$. The resulting metric is known as the Reissner-Nordstöm solution. In order to include the effects of the charge, we have to realize that even if there is no mass outside the source, the electric field carries energy so its contribution to the stress-energy tensor must be included. We’ll begin with a quick review of the electromagnetic stress-energy tensor. The electromagnetic field tensor is

$\displaystyle F^{\mu\nu}=\left[\begin{array}{cccc} 0 & E_{x} & E_{y} & E_{z}\\ -E_{x} & 0 & B_{z} & -B_{y}\\ -E_{y} & -B_{z} & 0 & B_{x}\\ -E_{z} & B_{y} & -B_{x} & 0 \end{array}\right] \ \ \ \ \ (1)$

where ${E_{i}}$ and ${B_{i}}$ are the spatial components of the electric and magnetic fields. We can just as well write this in spherical coordinates as

$\displaystyle F^{\mu\nu}=\left[\begin{array}{cccc} 0 & E_{r} & E_{\theta} & E_{\phi}\\ -E_{r} & 0 & B_{\phi} & -B_{\theta}\\ -E_{\theta} & -B_{\phi} & 0 & B_{r}\\ -E_{\phi} & B_{\theta} & -B_{r} & 0 \end{array}\right] \ \ \ \ \ (2)$

The electromagnetic stress-energy tensor can be written in terms of ${F^{\mu\nu}}$ (generalized to non-flat space with a metric ${g^{\mu\nu}}$):

 $\displaystyle T^{\mu\nu}$ $\displaystyle =$ $\displaystyle -\frac{1}{8\pi k}\left(2F^{\mu\kappa}F_{\kappa\lambda}g^{\lambda\nu}+\frac{1}{2}F_{\lambda\kappa}F^{\lambda\kappa}g^{\mu\nu}\right)\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{8\pi k}\left(2F^{\mu\kappa}F_{\kappa}^{\;\nu}+\frac{1}{2}F_{\lambda\kappa}F^{\lambda\kappa}g^{\mu\nu}\right)\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{8\pi k}\left(2F^{\mu\kappa}g_{\kappa\alpha}F^{\alpha\nu}+\frac{1}{2}F_{\lambda\kappa}F^{\lambda\kappa}g^{\mu\nu}\right)\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{8\pi k}\left(2F^{\mu\kappa}g_{\kappa\alpha}F^{\nu\alpha}-\frac{1}{2}F_{\lambda\kappa}F^{\lambda\kappa}g^{\mu\nu}\right) \ \ \ \ \ (6)$

where to get the last line, we used the anti-symmetry of ${F^{\alpha\nu}=-F^{\nu\alpha}}$. The constant ${k=1/4\pi\epsilon_{0}}$ in more conventional notation.

The Einstein equation (with ${\Lambda=0}$) is:

$\displaystyle R^{\mu\nu}=8\pi G\left(T^{\mu\nu}-\frac{1}{2}g^{\mu\nu}T\right) \ \ \ \ \ (7)$

so to work out the components of ${R^{\mu\nu}}$ we need the scalar ${T=T_{\;\mu}^{\mu}}$. For any metric ${g^{\mu\nu}}$ we have from 3

 $\displaystyle T=T_{\;\mu}^{\mu}$ $\displaystyle =$ $\displaystyle g_{\mu\nu}T^{\mu\nu}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{8\pi k}\left(2F^{\mu\kappa}g_{\kappa\alpha}g_{\mu\nu}F^{\nu\alpha}-\frac{1}{2}F_{\lambda\kappa}F^{\lambda\kappa}g^{\mu\nu}g_{\mu\nu}\right)\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{8\pi k}\left(2F_{\nu\alpha}F^{\nu\alpha}-2F_{\lambda\kappa}F^{\lambda\kappa}\right)\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (11)$

The double contraction in the second line is ${g^{\mu\nu}g_{\mu\nu}=4}$ (as can be seen by working it out in a local inertial frame where ${g^{\mu\nu}=\eta^{\mu\nu}}$), and in the third line we lowered the two indices of ${F^{\mu\kappa}}$ in the first term.

To proceed, we need to make a few assumptions. First, we’ll assume that a charged black hole gives rise to a spherically symmetric field tensor with only an electric field (the magnetic field is zero). We’ll also assume that the metric obeys Birkhoff’s theorem, so that it is independent of time. In that case, the only non-zero component of ${F_{\mu\nu}}$ is the radial electric field, which is an unknown function of ${r}$ only. That is

$\displaystyle F_{\mu\nu}=\left[\begin{array}{cccc} 0 & -E\left(r\right) & 0 & 0\\ E\left(r\right) & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{array}\right] \ \ \ \ \ (12)$

Note that we’re looking at the components of ${F_{\mu\nu}}$ with both indices lowered, so that the electric field is negative for ${F_{tr}}$ and positive for ${F_{rt}}$. [I should add that the original derivation of this assumed flat space, so I’m a bit hazy on how we can make this assumption for non-flat space. I suppose, given that we’re not specifying ${E}$ at this point, we can make this assumption.]

At this stage, we’ll take the metric to be

$\displaystyle ds^{2}=-Adt^{2}+Bdr^{2}+r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2} \ \ \ \ \ (13)$

with ${A}$ and ${B}$ to be determined. The components of ${F^{\mu\nu}}$ with raised indices are then since the metric is diagonal,

 $\displaystyle F_{\mu\nu}$ $\displaystyle =$ $\displaystyle g_{\mu\alpha}g_{\nu\beta}F^{\alpha\beta}\ \ \ \ \ (14)$ $\displaystyle F_{tr}$ $\displaystyle =$ $\displaystyle g_{tt}g_{rr}F^{tr}\ \ \ \ \ (15)$ $\displaystyle -E$ $\displaystyle =$ $\displaystyle -ABF^{tr}\ \ \ \ \ (16)$ $\displaystyle F^{tr}$ $\displaystyle =$ $\displaystyle \frac{E}{AB}=-F^{rt} \ \ \ \ \ (17)$

Given ${F^{\mu\nu}}$ and ${F_{\mu\nu}}$ we can now work out the RHS of 7, remembering that ${T=0}$ from 11:

 $\displaystyle 8\pi GT^{\mu\nu}$ $\displaystyle =$ $\displaystyle \frac{G}{k}\left(2F^{\mu\kappa}g_{\kappa\alpha}F^{\nu\alpha}-\frac{1}{2}F_{\lambda\kappa}F^{\lambda\kappa}g^{\mu\nu}\right)\ \ \ \ \ (18)$ $\displaystyle 8\pi GT_{\mu\nu}$ $\displaystyle =$ $\displaystyle \frac{G}{k}g_{\gamma\mu}g_{\delta\nu}\left(2F^{\gamma\kappa}g_{\kappa\alpha}F^{\delta\alpha}-\frac{1}{2}F_{\lambda\kappa}F^{\lambda\kappa}g^{\gamma\delta}\right) \ \ \ \ \ (19)$

From 12 and 17 we have

 $\displaystyle F_{\lambda\kappa}F^{\lambda\kappa}$ $\displaystyle =$ $\displaystyle -\frac{2E^{2}}{AB}\ \ \ \ \ (20)$ $\displaystyle 8\pi GT_{\mu\nu}$ $\displaystyle =$ $\displaystyle \frac{G}{k}g_{\gamma\mu}g_{\delta\nu}\left(2F^{\gamma\kappa}g_{\kappa\alpha}F^{\delta\alpha}+\frac{E^{2}}{AB}g^{\gamma\delta}\right)\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{G}{k}g_{\gamma\mu}g_{\delta\nu}2F^{\gamma\kappa}g_{\kappa\alpha}F^{\delta\alpha}+g_{\mu\nu}\frac{GE^{2}}{kAB} \ \ \ \ \ (22)$

For the individual components, we get, using ${g_{tt}=-A}$, ${g_{rr}=B}$ and ${g_{\theta\theta}=r^{2}}$

 $\displaystyle 8\pi GT_{tt}$ $\displaystyle =$ $\displaystyle \frac{G}{k}A^{2}\left(2F^{tr}BF^{tr}\right)-\frac{GE^{2}}{kB}\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{G}{k}\left[2A^{2}B\frac{E^{2}}{A^{2}B^{2}}-\frac{E^{2}}{B}\right]\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{GE^{2}}{kB}\ \ \ \ \ (25)$ $\displaystyle 8\pi GT_{rr}$ $\displaystyle =$ $\displaystyle \frac{G}{k}\left[2B^{2}F^{rt}\left(-A\right)F^{rt}+B\frac{E^{2}}{AB}\right]\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{G}{k}\left[-2AB^{2}\frac{E^{2}}{A^{2}B^{2}}+\frac{E^{2}}{A}\right]\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{GE^{2}}{kA}\ \ \ \ \ (28)$ $\displaystyle 8\pi GT_{\theta\theta}$ $\displaystyle =$ $\displaystyle 0+g_{\theta\theta}\frac{GE^{2}}{kAB}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{GE^{2}r^{2}}{kAB} \ \ \ \ \ (30)$

We can now plug these into 7 to get the equations that must be solved to find ${A}$ and ${B}$. We get

 $\displaystyle R_{tt}$ $\displaystyle =$ $\displaystyle \frac{GE^{2}}{kB}\ \ \ \ \ (31)$ $\displaystyle R_{rr}$ $\displaystyle =$ $\displaystyle -\frac{GE^{2}}{kA}\ \ \ \ \ (32)$ $\displaystyle BR_{tt}+AR_{rr}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (33)$

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

 $\displaystyle \frac{1}{2B}\left[\partial_{rr}^{2}A-\partial_{tt}^{2}B+\frac{\left(\partial_{t}B\right)^{2}}{2B}+\frac{\left(\partial_{t}A\right)\left(\partial_{t}B\right)-\left(\partial_{r}A\right)^{2}}{2A}-\frac{\left(\partial_{r}A\right)\left(\partial_{r}B\right)}{2B}+\frac{2\partial_{r}A}{r}\right]$ $\displaystyle =$ $\displaystyle R_{tt}\ \ \ \ \ (34)$ $\displaystyle \frac{1}{2A}\left[\partial_{tt}^{2}B-\partial_{rr}^{2}A+\frac{\left(\partial_{r}A\right)^{2}-\left(\partial_{t}A\right)\left(\partial_{t}B\right)}{2A}+\frac{\left(\partial_{r}A\right)\left(\partial_{r}B\right)-\left(\partial_{t}B\right)^{2}}{2B}+\frac{2A\partial_{r}B}{rB}\right]$ $\displaystyle =$ $\displaystyle R_{rr}\ \ \ \ \ (35)$ $\displaystyle -\frac{r\partial_{r}A}{2AB}+\frac{r\partial_{r}B}{2B^{2}}+1-\frac{1}{B}$ $\displaystyle =$ $\displaystyle R_{\theta\theta}\ \ \ \ \ (36)$ $\displaystyle \frac{\partial_{t}B}{rB}$ $\displaystyle =$ $\displaystyle R_{tr} \ \ \ \ \ (37)$

Because of Birkhoff’s theorem, all time derivatives are zero, so these equations simplify to

 $\displaystyle \frac{1}{2B}\left[\partial_{rr}^{2}A-\frac{\left(\partial_{r}A\right)^{2}}{2A}-\frac{\left(\partial_{r}A\right)\left(\partial_{r}B\right)}{2B}+\frac{2\partial_{r}A}{r}\right]$ $\displaystyle =$ $\displaystyle R_{tt}\ \ \ \ \ (38)$ $\displaystyle \frac{1}{2A}\left[-\partial_{rr}^{2}A+\frac{\left(\partial_{r}A\right)^{2}}{2A}+\frac{\left(\partial_{r}A\right)\left(\partial_{r}B\right)}{2B}+\frac{2A\partial_{r}B}{rB}\right]$ $\displaystyle =$ $\displaystyle R_{rr}\ \ \ \ \ (39)$ $\displaystyle -\frac{r\partial_{r}A}{2AB}+\frac{r\partial_{r}B}{2B^{2}}+1-\frac{1}{B}$ $\displaystyle =$ $\displaystyle R_{\theta\theta}\ \ \ \ \ (40)$ $\displaystyle 0$ $\displaystyle =$ $\displaystyle R_{tr} \ \ \ \ \ (41)$

Applying 33 we get

 $\displaystyle \frac{\partial_{r}A}{r}+\frac{A\partial_{r}B}{rB}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (42)$ $\displaystyle \frac{\partial_{r}A}{A}+\frac{\partial_{r}B}{B}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (43)$

We can replace the partial derivatives by total derivatives since ${A}$ and ${B}$ depend only on ${r}$. Multiplying through by ${AB}$ we get

$\displaystyle B\frac{dA}{dr}+A\frac{dB}{dr}=\frac{d}{dr}\left(AB\right)=0 \ \ \ \ \ (44)$

so ${AB=\mbox{constant}}$. For very large ${r}$, the metric must reduce to ${\eta_{\mu\nu}}$ so both ${A\rightarrow1}$ and ${B\rightarrow1}$. Thus the product ${AB=1}$ everywhere, which means from 17 that ${F^{tr}=-F^{rt}=E}$ and thus ${F^{\mu\nu}=-F_{\mu\nu}}$.

So far, we have established that ${A=\frac{1}{B}}$ but to get the two components separately, we need to use the fact that we’re dealing a charged black hole. Maxwell’s equations can be written in tensor form as

$\displaystyle \nabla_{\nu}F^{\mu\nu}=4\pi kJ^{\mu} \ \ \ \ \ (45)$

where the absolute gradient is defined in terms of Christoffel symbols as

$\displaystyle \nabla_{\rho}F^{\mu\nu}=\partial_{\rho}F^{\mu\nu}+F^{\mu\alpha}\Gamma_{\alpha\rho}^{\nu}+F^{\alpha\nu}\Gamma_{\alpha\rho}^{\mu} \ \ \ \ \ (46)$

Contracting ${\rho}$ with ${\nu}$ we get

$\displaystyle \nabla_{\nu}F^{\mu\nu}=\partial_{\nu}F^{\mu\nu}+F^{\mu\alpha}\Gamma_{\alpha\nu}^{\nu}+F^{\alpha\nu}\Gamma_{\alpha\nu}^{\mu} \ \ \ \ \ (47)$

In empty space, ${J^{\mu}=0}$ since there is no charge or current, so for ${\mu=t}$ we have

$\displaystyle \nabla_{\nu}F^{t\nu}=\partial_{\nu}F^{t\nu}+F^{t\alpha}\Gamma_{\alpha\nu}^{\nu}+F^{\alpha\nu}\Gamma_{\alpha\nu}^{t} \ \ \ \ \ (48)$

Unfortunately, this means working out a few Christoffel symbols, but we can use the worksheet to make things easier. The only non-zero components of ${F^{\mu\nu}}$ are ${F^{tr}=-F^{rt}}$. Because ${\Gamma_{\alpha\nu}^{t}=\Gamma_{\nu\alpha}^{t}}$, the last term is zero after the sums are done, so

$\displaystyle \nabla_{\nu}F^{t\nu}=\partial_{r}F^{tr}+F^{tr}\Gamma_{r\nu}^{\nu} \ \ \ \ \ (49)$

In the notation of the worksheet, we have, using ${B=\frac{1}{A}}$, ${C=r^{2}}$ and ${D=r^{2}\sin^{2}\theta}$ and a subscript 1 means ‘take the derivative with respect to ${r}$‘:

 $\displaystyle \Gamma_{r\nu}^{\nu}$ $\displaystyle =$ $\displaystyle \Gamma_{1\nu}^{\nu}\ \ \ \ \ (50)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2A}A_{1}+\frac{1}{2B}B_{1}+\frac{1}{2C}C_{1}+\frac{1}{2D}D_{1}\ \ \ \ \ (51)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2A}A_{1}-\frac{1}{2A}A_{1}+\frac{1}{r}+\frac{1}{r}\ \ \ \ \ (52)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{r}\ \ \ \ \ (53)$ $\displaystyle \nabla_{\nu}F^{t\nu}$ $\displaystyle =$ $\displaystyle \partial_{r}E+\frac{2E}{r}=0\ \ \ \ \ (54)$ $\displaystyle r^{2}\partial_{r}E+2rE$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (55)$ $\displaystyle \frac{d}{dr}\left(r^{2}E\right)$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (56)$ $\displaystyle E\left(r\right)$ $\displaystyle =$ $\displaystyle \frac{b}{r^{2}} \ \ \ \ \ (57)$

where ${b}$ is a constant of integration. If this is to reduce to the Coulomb field at large ${r}$, then we require

$\displaystyle b=kQ=\frac{Q}{4\pi\epsilon_{0}} \ \ \ \ \ (58)$

From 40 and 30 with ${AB=1}$ we have

 $\displaystyle -\frac{r\partial_{r}A}{2}+\frac{r\partial_{r}B}{2B^{2}}+1-\frac{1}{B}$ $\displaystyle =$ $\displaystyle \frac{GE^{2}r^{2}}{k}\ \ \ \ \ (59)$ $\displaystyle -\frac{r\partial_{r}A}{2}+\frac{rA^{2}\partial_{r}\frac{1}{A}}{2}+1-A$ $\displaystyle =$ $\displaystyle \frac{GkQ^{2}}{r^{2}}\ \ \ \ \ (60)$ $\displaystyle -r\frac{dA}{dr}+1-A$ $\displaystyle =$ $\displaystyle \frac{GkQ^{2}}{r^{2}}\ \ \ \ \ (61)$ $\displaystyle -\frac{d\left(rA\right)}{dr}+1$ $\displaystyle =$ $\displaystyle \frac{GkQ^{2}}{r^{2}}\ \ \ \ \ (62)$ $\displaystyle \frac{d\left(rA\right)}{dr}$ $\displaystyle =$ $\displaystyle 1-\frac{GkQ^{2}}{r^{2}}\ \ \ \ \ (63)$ $\displaystyle A\left(r\right)$ $\displaystyle =$ $\displaystyle 1+\frac{GkQ^{2}}{r^{2}}+\frac{K}{r} \ \ \ \ \ (64)$

where ${K}$ is a constant of integration. In order for this to reduce to the Schwarzschild metric component ${-g_{tt}=\left(1-\frac{2GM}{r}\right)}$ when ${Q=0}$, we must have ${K=-2GM}$, so

 $\displaystyle A\left(r\right)$ $\displaystyle =$ $\displaystyle 1-\frac{2GM}{r}+\frac{GkQ^{2}}{r^{2}}\ \ \ \ \ (65)$ $\displaystyle B\left(r\right)$ $\displaystyle =$ $\displaystyle \left[1-\frac{2GM}{r}+\frac{GkQ^{2}}{r^{2}}\right]^{-1} \ \ \ \ \ (66)$

An event horizon occurs whenever ${g_{tt}=0}$. In this case, this gives rise to a quadratic equation in ${r}$:

 $\displaystyle r^{2}-2GMr+GkQ^{2}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (67)$ $\displaystyle r$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[2GM\pm\sqrt{4G^{2}M^{2}-4GkQ^{2}}\right]\ \ \ \ \ (68)$ $\displaystyle$ $\displaystyle =$ $\displaystyle GM\pm\sqrt{G^{2}M^{2}-GkQ^{2}} \ \ \ \ \ (69)$

For real solutions, we must have

 $\displaystyle G^{2}M^{2}$ $\displaystyle \ge$ $\displaystyle GkQ^{2}\ \ \ \ \ (70)$ $\displaystyle GM^{2}$ $\displaystyle \ge$ $\displaystyle kQ^{2} \ \ \ \ \ (71)$

If the charge ${Q}$ is large enough to violate this condition, there are no event horizons meaning that the singularity at ${r=0}$ becomes a naked singularity.

# 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 with negative mass

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 23; Problem 23.2.

In the derivation of the Schwarzschild metric we got to the form

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

where ${C}$ is a constant of integration, and by requiring the metric to reduce to Newton’s gravitation law for large ${r}$, we found that ${C=-2GM}$. However, we can also consider the case where ${C>0}$, which is essentially considering the possibility of negative mass.

First, how will a test mass released at rest from a distance ${r}$ behave? From our previous derivation we have the acceleration given by

$\displaystyle \frac{d^{2}r}{d\tau^{2}}=\frac{C}{2r^{2}} \ \ \ \ \ (2)$

Since ${C>0}$, the quantity on the RHS is positive, so the particle will accelerate away from the origin. Thus, as we might expect, a negative mass repels a positive mass.

Since the object is repelled, we might expect that it could not exist in a closed orbit about the negative mass. For the case of circular orbits, we can see this is true by scanning through the derivation we did earlier for the ordinary Schwarzschild metric, but replacing ${-2GM}$ by ${C}$. In the original metric, the radius of a circular orbit was given by the solutions of the quadratic equation

$\displaystyle GMr^{2}-l^{2}r+3GMl^{2}=0 \ \ \ \ \ (3)$

where ${l}$ is the constant angular momentum. Replacing ${-2GM}$ by ${C}$ and multiplying through by ${-1}$ gives us the quadratic

$\displaystyle \frac{C}{2}r^{2}+l^{2}r+\frac{3C}{2}l^{2}=0 \ \ \ \ \ (4)$

Since all 3 terms in this equation are intrinsically positive, it has no real, positive roots for ${r}$, so there are no circular orbits possible.

The ordinary Schwarzschild metric has an event horizon at ${r=2GM}$, that is, a point at which ${g_{tt}=0}$ and if ${r}$ crosses this point, the signs of both ${g_{tt}}$ and ${g_{rr}}$ change, causing a swap between the time and radial coordinates. In order for an event horizon to exist if ${C>0}$, there must be a value of ${r}$ where ${g_{tt}=0}$, but since ${g_{tt}=-\left(1+\frac{C}{r}\right)}$, this does not happen.

The singularity at ${r=0}$ is still there, however, and is a geometric singularity (that is, one that is a result of the intrinsic geometry of the metric) rather than a coordinate singularity (one that is an artifact of the coordinate system). Because there is no event horizon, this is a naked singularity, which can be approached directly without time and space swapping round.

There is one aspect of this metric that still puzzles me however. If we follow through the derivation of the constants of the motion that we did earlier for the Schwarzschild metric but replacing ${-2GM}$ by ${C}$, we get

 $\displaystyle \frac{dt}{d\tau}$ $\displaystyle =$ $\displaystyle e\left(1+\frac{C}{r}\right)^{-1}\ \ \ \ \ (5)$ $\displaystyle \frac{d\theta}{d\tau}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (6)$ $\displaystyle \frac{d\phi}{d\tau}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (7)$ $\displaystyle \frac{dr}{d\tau}$ $\displaystyle =$ $\displaystyle \pm\sqrt{e^{2}-\left(1+\frac{C}{r}\right)\left(1+\frac{\ell^{2}}{r^{2}}\right)} \ \ \ \ \ (8)$

where ${e}$ and ${l}$ are constants of the motion.

For the Schwarzschild metric, in general ${dt/d\tau=u^{t}}$, the time component of the four-velocity. In this case, ${t}$ is the time as measured by an observer at rest at infinity, and ${\tau}$ is the proper time as measured by the object, which may be moving. The four-momentum‘s time component is the energy, and the four-velocity is the four-momentum per unit mass, so ${e}$ is the energy per unit mass of the object which remains constant as the object moves in from infinity.

If we use the same interpretation for the ${C>0}$ case and consider a particle released from rest at a distance ${r}$, then ${\ell=0}$, ${e=1}$ and we end up with ${\frac{dr}{d\tau}=\pm\sqrt{-C/r}}$ which makes ${\frac{dr}{d\tau}}$ imaginary. However, if we take the derivative of this, we get

$\displaystyle \frac{d^{2}r}{d\tau^{2}}=-\frac{\sqrt{-C}}{r^{3/2}}\frac{dr}{d\tau}=-\frac{\sqrt{-C}}{r^{3/2}}\frac{\sqrt{-C}}{r^{1/2}}=\frac{C}{2r^{2}} \ \ \ \ \ (9)$

which agrees with 2.

Something odd is happening with the time coordinate ${t}$ in this case, since at ${r\rightarrow\infty}$, ${\frac{dt}{d\tau}=e}$ and as ${r}$ decreases, so does ${\frac{dt}{d\tau}}$, reaching zero at ${r=0}$. This is the opposite behaviour to the ${t}$ coordinate in the ordinary Schwarzschild metric, where the proper time slows down as we get closer to the event horizon, eventually stopping when ${r=2GM}$. Because the time coordinate behaves differently in this case, is it still correct to identify it with the proper time of an observer at infinity? Comments welcome.

# Schwarzschild metric: the Newtonian limit & Christoffel symbol worksheet

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

In our derivation of the Schwarzschild metric, we got as far as finding the dependence of the metric on the spacetime coordinates, giving the form

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

The final task is to find the constant ${X}$, which we can do by considering the behaviour of the metric for large ${r}$ and requiring that it reduce to the Newtonian graviational force law in that limit. [I’ve renamed the constant ${C}$ in the original post to ${X}$ here to avoid confusion with the ${C}$ that turns up in the metric tensor below.]

For an object initially at rest, the spatial comopnents of its four-velocity are all zero: ${u^{i}=0}$. However, the contraction of ${\mathbf{u}}$ with itself gives the invariant ${\mathbf{u}\cdot\mathbf{u}=-1}$, so we have

 $\displaystyle \mathbf{u}\cdot\mathbf{u}$ $\displaystyle =$ $\displaystyle g_{\mu\nu}u^{\mu}u^{\nu}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g_{tt}\left(u^{t}\right)^{2}=-1\ \ \ \ \ (3)$ $\displaystyle u^{t}$ $\displaystyle =$ $\displaystyle \sqrt{-\frac{1}{g_{tt}}} \ \ \ \ \ (4)$

Any object’s trajectory obeys the geodesic equation which, in terms of Christoffel symbols, is

$\displaystyle \ddot{x}^{\mu}+\Gamma_{\;\nu\sigma}^{\mu}\dot{x}^{\nu}\dot{x}^{\sigma}=0 \ \ \ \ \ (5)$

where a dot denotes a derivative with respect to proper time ${\tau}$, so that ${\dot{x}^{\nu}=u^{\nu}}$.

In our case, this reduces to

 $\displaystyle \ddot{x}^{\mu}+\Gamma_{tt}^{\mu}\left(u^{t}\right)^{2}$ $\displaystyle =$ $\displaystyle \ddot{x}^{\mu}-\frac{\Gamma_{tt}^{\mu}}{g_{tt}}=0\ \ \ \ \ (6)$ $\displaystyle \ddot{x}^{\mu}$ $\displaystyle =$ $\displaystyle \frac{\Gamma_{tt}^{\mu}}{g_{tt}}=-\frac{1}{A}\Gamma_{tt}^{\mu} \ \ \ \ \ (7)$

where ${A=-g_{tt}}$. We therefore need to calculate the Christoffel symbols ${\Gamma_{tt}^{\mu}}$, which we can do from their expression in terms of ${g_{\mu\nu}}$:

$\displaystyle \Gamma_{\;\nu\sigma}^{\mu}=\frac{1}{2}g^{\mu\lambda}\left(\partial_{\sigma}g_{\nu\lambda}+\partial_{\nu}g_{\lambda\sigma}-\partial_{\lambda}g_{\sigma\nu}\right) \ \ \ \ \ (8)$

This can get quite tedious, but Moore provides a worksheet in the Appendix which simplifies the task. The notation is the same as that used for Ricci tensor worksheet for the generic diagonal metric, written as

$\displaystyle ds^{2}=-A\left(dx^{0}\right)^{2}+B\left(dx^{1}\right)^{2}+C\left(dx^{2}\right)^{2}+D\left(dx^{3}\right)^{2} \ \ \ \ \ (9)$

where ${x^{0}}$ is the time coordinate and the other three are space coordinates. Note the minus sign in the first term: this makes explicit the fact that the metric component for time should be negative. Thus we have ${g_{00}=-A}$, ${g_{11}=B}$, ${g_{22}=C}$ and ${g_{33}=D}$.

Derivatives with respect to coordinates are written as subscripts, so that ${A_{01}=\frac{\partial^{2}A}{\partial x^{0}\partial x^{1}}}$ and so on. It’s important not to confuse this notation with tensor notation; ${A_{01}}$ is not the 01 component of a tensor. Although we don’t need all the Christoffel symbols here, I’ve produced the table for reference.

 ${\Gamma_{00}^{0}=\frac{1}{2A}A_{0}}$ ${\Gamma_{10}^{0}=\Gamma_{01}^{0}=\frac{1}{2A}A_{1}}$ ${\Gamma_{20}^{0}=\Gamma_{02}^{0}=\frac{1}{2A}A_{2}}$ ${\Gamma_{30}^{0}=\Gamma_{03}^{0}=\frac{1}{2A}A_{3}}$ ${\Gamma_{11}^{0}=\frac{1}{2A}B_{0}}$ ${\Gamma_{22}^{0}=\frac{1}{2A}C_{0}}$ ${\Gamma_{33}^{0}=\frac{1}{2A}D_{0}}$ other ${\Gamma_{\mu\nu}^{0}=0}$ ${\Gamma_{01}^{1}=\Gamma_{10}^{1}=\frac{1}{2B}B_{0}}$ ${\Gamma_{11}^{1}=\frac{1}{2B}B_{1}}$ ${\Gamma_{12}^{1}=\Gamma_{21}^{1}=\frac{1}{2B}B_{2}}$ ${\Gamma_{13}^{1}=\Gamma_{31}^{1}=\frac{1}{2B}B_{3}}$ ${\Gamma_{00}^{1}=\frac{1}{2B}A_{1}}$ ${\Gamma_{22}^{1}=-\frac{1}{2B}C_{1}}$ ${\Gamma_{33}^{1}=-\frac{1}{2B}D_{1}}$ other ${\Gamma_{\mu\nu}^{1}=0}$ ${\Gamma_{02}^{2}=\Gamma_{20}^{2}=\frac{1}{2C}C_{0}}$ ${\Gamma_{12}^{2}=\Gamma_{21}^{2}=\frac{1}{2C}C_{1}}$ ${\Gamma_{22}^{2}=\frac{1}{2C}C_{2}}$ ${\Gamma_{32}^{2}=\Gamma_{23}^{2}=\frac{1}{2C}C_{3}}$ ${\Gamma_{00}^{2}=\frac{1}{2C}A_{2}}$ ${\Gamma_{11}^{2}=-\frac{1}{2C}B_{2}}$ ${\Gamma_{33}^{2}=-\frac{1}{2C}D_{2}}$ other ${\Gamma_{\mu\nu}^{2}=0}$ ${\Gamma_{03}^{3}=\Gamma_{30}^{3}=\frac{1}{2D}D_{0}}$ ${\Gamma_{13}^{3}=\Gamma_{31}^{3}=\frac{1}{2D}D_{1}}$ ${\Gamma_{23}^{3}=\Gamma_{32}^{3}=\frac{1}{2D}D_{2}}$ ${\Gamma_{33}^{3}=\frac{1}{2D}D_{3}}$ ${\Gamma_{00}^{3}=\frac{1}{2D}A_{3}}$ ${\Gamma_{11}^{3}=-\frac{1}{2D}B_{3}}$ ${\Gamma_{22}^{3}=-\frac{1}{2D}C_{3}}$ other ${\Gamma_{\mu\nu}^{3}=0}$

In our case, we need only the ${\Gamma_{00}^{\mu}}$ terms, which occur in the first column. Since ${A=\left(1+\frac{X}{r}\right)}$ the derivatives with respect to ${t}$, ${\theta}$ and ${\phi}$ are all zero, and the only non-zero Christoffel symbol is

$\displaystyle \Gamma_{tt}^{r}=\Gamma_{00}^{1}=\frac{1}{2B}A_{1}=\frac{1}{2}\left(1+\frac{X}{r}\right)\frac{\partial A}{\partial r}=-\frac{X}{2r^{2}}\left(1+\frac{X}{r}\right) \ \ \ \ \ (10)$

Therefore from 7 we have

$\displaystyle \ddot{x}^{r}=\frac{d^{2}r}{d\tau^{2}}=-\frac{1}{A}\left(-\frac{X}{2r^{2}}\left(1+\frac{X}{r}\right)\right)=\frac{X}{2r^{2}} \ \ \ \ \ (11)$

For large ${r}$, the Schwarzschild metric reduces to flat space, so the radial coordinate becomes the Newtonian radial coordinate and the proper time ${\tau}$ becomes the Newtonian time ${t}$, so

$\displaystyle \frac{d^{2}r}{d\tau^{2}}\rightarrow\frac{d^{2}r}{dt^{2}}=\frac{X}{2r^{2}} \ \ \ \ \ (12)$

This is equivalent to Newton’s law of gravity for a mass a distance ${r}$ from a mass ${M}$ if

$\displaystyle X=-2GM \ \ \ \ \ (13)$

[The minus sign indicates that the test mass accelerates towards ${M}$, that is, in the direction of decreasing ${r}$.]

Making this substitution in 1 we get the final form of the Schwarzschild metric

$\displaystyle ds^{2}=-\left(1-\frac{2GM}{r}\right)dt^{2}+\left(1-\frac{2GM}{r}\right)^{-1}dr^{2}+r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2} \ \ \ \ \ (14)$

# 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.

# Ricci tensor for a spherically symmetric metric: the worksheet

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 23; Box 23.2.

In the last post, we developed the general form for the metric in a spherically symmetric situation:

$\displaystyle ds^{2}=g_{tt}dt^{2}+g_{rr}dr^{2}+r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2} \ \ \ \ \ (1)$

The next step is to use the Einstein equation in the form

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

to generate a system of differential equations that can be solved to find ${g_{tt}}$ and ${g_{rr}}$. This involves calculating the components of the Ricci tensor ${R^{ij}}$. Recall that the Ricci tensor is a contraction of the Riemann tensor:

$\displaystyle R_{ij}=R_{\;iaj}^{a} \ \ \ \ \ (3)$

and the Riemann tensor is defined in terms of Christoffel symbols:

$\displaystyle R_{\;j\ell m}^{i}\equiv-\left[\partial_{m}\Gamma_{\;\ell j}^{i}-\partial_{\ell}\Gamma_{\;mj}^{i}+\Gamma_{\;\ell j}^{k}\Gamma_{\;km}^{i}-\Gamma_{\;mj}^{k}\Gamma_{\;\ell k}^{i}\right] \ \ \ \ \ (4)$

The Christoffel symbols are, in turn, calculated from the metric tensor and its derivatives:

$\displaystyle \Gamma_{\;ij}^{m}=\frac{1}{2}g^{ml}\left(\partial_{j}g_{il}+\partial_{i}g_{lj}-\partial_{l}g_{ji}\right) \ \ \ \ \ (5)$

Each component of ${R_{ij}}$ is therefore ultimately a differential equation involving components of the metric tensor ${g_{ij}}$, so if we know the stress-energy tensor ${T_{ij}}$, 2 gives us a set of PDEs that can, in principle at least, be solved to find the metric tensor. Since ${R_{ij}}$ is symmetric, it has 10 independent components, each of which is a sum of terms involving the components of ${g_{ij}}$. For a general (non-diagonal) metric, this can get very messy and, even for a diagonal metric such as we have for the spherically symmetric case, things are bad enough. In the appendix to Moore’s book, he gives a worksheet for calculating the Christoffel symbols and the independent components of ${R_{ij}}$ for a general diagonal metric. In the worksheet, all the work of expanding ${R_{ij}}$ in terms of ${g_{ij}}$ has been done, so we just need to fill in the results for our specific metric, such as that given in 1.

The worksheets are given for the generic diagonal metric, written as

$\displaystyle ds^{2}=-A\left(dx^{0}\right)^{2}+B\left(dx^{1}\right)^{2}+C\left(dx^{2}\right)^{2}+D\left(dx^{3}\right)^{2} \ \ \ \ \ (6)$

where ${x^{0}}$ is the time coordinate and the other three are space coordinates. Note the minus sign in the first term: this makes explicit the fact that the metric component for time should be negative. Thus we have ${g_{00}=-A}$, ${g_{11}=B}$, ${g_{22}=C}$ and ${g_{33}=D}$.

Derivatives with respect to coordinates are written as subscripts, so that ${A_{01}=\frac{\partial^{2}A}{\partial x^{0}\partial x^{1}}}$ and so on. It’s important not to confuse this notation with tensor notation; ${A_{01}}$ is not the 01 component of a tensor.

The 10 independent components of ${R_{ij}}$ are:

 ${R_{00}=0}$ ${+\frac{1}{2B}A_{11}}$ ${+\frac{1}{2C}A_{22}}$ ${+\frac{1}{2D}A_{33}}$ ${+0}$ ${-\frac{1}{2B}B_{00}}$ ${-\frac{1}{2C}C_{00}}$ ${-\frac{1}{2D}D_{00}}$ ${+0}$ ${+\frac{1}{4B^{2}}B_{0}^{2}}$ ${+\frac{1}{4C^{2}}C_{0}^{2}}$ ${+\frac{1}{4D^{2}}D_{0}^{2}}$ ${+0}$ ${+\frac{1}{4AB}A_{0}B_{0}}$ ${+\frac{1}{4AC}A_{0}C_{0}}$ ${+\frac{1}{4AD}A_{0}D_{0}}$ ${-\frac{1}{4BA}A_{1}^{2}}$ ${-\frac{1}{4B^{2}}A_{1}B_{1}}$ ${+\frac{1}{4BC}A_{1}C_{1}}$ ${+\frac{1}{4BD}A_{1}D_{1}}$ ${-\frac{1}{4CA}A_{2}^{2}}$ ${+\frac{1}{4CB}A_{2}B_{2}}$ ${-\frac{1}{4C^{2}}A_{2}C_{2}}$ ${+\frac{1}{4CD}A_{2}D_{2}}$ ${-\frac{1}{4DA}A_{3}^{2}}$ ${+\frac{1}{4DB}A_{3}B_{3}}$ ${+\frac{1}{4DC}A_{3}C_{3}}$ ${-\frac{1}{4D^{2}}A_{3}D_{3}}$
 ${R_{11}=\frac{1}{2A}B_{00}}$ ${+0}$ ${-\frac{1}{2C}B_{22}}$ ${-\frac{1}{2D}B_{33}}$ ${-\frac{1}{2A}A_{11}}$ ${+0}$ ${-\frac{1}{2C}C_{11}}$ ${-\frac{1}{2D}D_{11}}$ ${+\frac{1}{4A^{2}}A_{1}^{2}}$ ${+0}$ ${+\frac{1}{4C^{2}}C_{1}^{2}}$ ${+\frac{1}{4D^{2}}D_{1}^{2}}$ ${-\frac{1}{4A^{2}}B_{0}A_{0}}$ ${-\frac{1}{4AB}B_{0}^{2}}$ ${+\frac{1}{4AC}B_{0}C_{0}}$ ${+\frac{1}{4AD}B_{0}D_{0}}$ ${+\frac{1}{4BA}B_{1}A_{1}}$ ${+0}$ ${+\frac{1}{4BC}B_{1}C_{1}}$ ${+\frac{1}{4BD}B_{1}D_{1}}$ ${-\frac{1}{4CA}B_{2}A_{2}}$ ${+\frac{1}{4CB}B_{2}^{2}}$ ${+\frac{1}{4C^{2}}B_{2}C_{2}}$ ${-\frac{1}{4CD}B_{2}D_{2}}$ ${-\frac{1}{4DA}B_{3}A_{3}}$ ${+\frac{1}{4DB}B_{3}^{2}}$ ${-\frac{1}{4DC}B_{3}C_{3}}$ ${+\frac{1}{4D^{2}}B_{3}D_{3}}$
 ${R_{22}=\frac{1}{2A}C_{00}}$ ${-\frac{1}{2B}C_{11}}$ ${+0}$ ${-\frac{1}{2D}C_{33}}$ ${-\frac{1}{2A}A_{22}}$ ${-\frac{1}{2B}B_{22}}$ ${+0}$ ${-\frac{1}{2D}D_{22}}$ ${+\frac{1}{4A^{2}}A_{2}^{2}}$ ${+\frac{1}{4B^{2}}B_{2}^{2}}$ ${+0}$ ${+\frac{1}{4D^{2}}D_{2}^{2}}$ ${-\frac{1}{4A^{2}}C_{0}A_{0}}$ ${+\frac{1}{4AB}C_{0}B_{0}}$ ${-\frac{1}{4AC}C_{0}^{2}}$ ${+\frac{1}{4AD}C_{0}D_{0}}$ ${-\frac{1}{4BA}C_{1}A_{1}}$ ${+\frac{1}{4B^{2}}C_{1}B_{1}}$ ${+\frac{1}{4BC}C_{1}^{2}}$ ${-\frac{1}{4BD}C_{1}D_{1}}$ ${+\frac{1}{4CA}C_{2}A_{2}}$ ${+\frac{1}{4CB}C_{2}B_{2}}$ ${+0}$ ${+\frac{1}{4CD}C_{2}D_{2}}$ ${-\frac{1}{4DA}C_{3}A_{3}}$ ${-\frac{1}{4DB}C_{3}B_{3}}$ ${+\frac{1}{4DC}C_{3}^{2}}$ ${+\frac{1}{4D^{2}}C_{3}D_{3}}$
 ${R_{33}=\frac{1}{2A}D_{00}}$ ${-\frac{1}{2B}D_{11}}$ ${-\frac{1}{2C}D_{22}}$ ${+0}$ ${-\frac{1}{2A}A_{33}}$ ${-\frac{1}{2B}B_{33}}$ ${-\frac{1}{2C}C_{33}}$ ${+0}$ ${+\frac{1}{4A^{2}}A_{3}^{2}}$ ${+\frac{1}{4B^{2}}B_{3}^{2}}$ ${+\frac{1}{4C^{2}}C_{3}^{2}}$ ${+0}$ ${-\frac{1}{4A^{2}}D_{0}A_{0}}$ ${+\frac{1}{4AB}D_{0}B_{0}}$ ${+\frac{1}{4AC}D_{0}C_{0}}$ ${-\frac{1}{4AD}D_{0}^{2}}$ ${-\frac{1}{4BA}D_{1}A_{1}}$ ${+\frac{1}{4B^{2}}D_{1}B_{1}}$ ${-\frac{1}{4BC}D_{1}C_{1}}$ ${+\frac{1}{4BD}D_{1}^{2}}$ ${-\frac{1}{4CA}D_{2}A_{2}}$ ${-\frac{1}{4CB}D_{2}B_{2}}$ ${+\frac{1}{4C^{2}}D_{2}C_{2}}$ ${+\frac{1}{4CD}D_{2}^{2}}$ ${+\frac{1}{4DA}D_{3}A_{3}}$ ${+\frac{1}{4DB}D_{3}B_{3}}$ ${+\frac{1}{4DC}D_{3}C_{3}}$ ${+0}$
 ${R_{01}=-\frac{1}{2C}C_{01}}$ –${\frac{1}{2D}D_{01}}$ ${+\frac{1}{4C^{2}}C_{0}C_{1}}$ ${+\frac{1}{4D^{2}}D_{0}D_{1}}$ ${+\frac{1}{4AC}A_{1}C_{0}}$ ${+\frac{1}{4AD}A_{1}D_{0}}$ ${+\frac{1}{4BC}B_{0}C_{1}}$ ${+\frac{1}{4BD}B_{0}D_{1}}$
 ${R_{02}=-\frac{1}{2B}B_{02}}$ –${\frac{1}{2D}D_{02}}$ ${+\frac{1}{4B^{2}}B_{0}B_{2}}$ ${+\frac{1}{4D^{2}}D_{0}D_{2}}$ ${+\frac{1}{4AB}A_{2}B_{0}}$ ${+\frac{1}{4AD}A_{2}D_{0}}$ ${+\frac{1}{4BC}B_{2}C_{0}}$ ${+\frac{1}{4CD}C_{0}D_{2}}$
 ${R_{03}=-\frac{1}{2B}B_{03}}$ –${\frac{1}{2C}C_{03}}$ ${+\frac{1}{4B^{2}}B_{0}B_{3}}$ ${+\frac{1}{4C^{2}}C_{0}C_{3}}$ ${+\frac{1}{4AB}A_{3}B_{0}}$ ${+\frac{1}{4AC}A_{3}C_{0}}$ ${+\frac{1}{4BD}B_{3}D_{0}}$ ${+\frac{1}{4CD}C_{3}D_{0}}$
 ${R_{12}=-\frac{1}{2A}A_{12}}$ –${\frac{1}{2D}D_{12}}$ ${+\frac{1}{4A^{2}}A_{1}A_{2}}$ ${+\frac{1}{4D^{2}}D_{1}D_{2}}$ ${+\frac{1}{4AB}A_{1}B_{2}}$ ${+\frac{1}{4BD}B_{2}D_{1}}$ ${+\frac{1}{4AC}A_{2}C_{1}}$ ${+\frac{1}{4CD}C_{1}D_{2}}$
 ${R_{13}=-\frac{1}{2A}A_{13}}$ –${\frac{1}{2C}C_{13}}$ ${+\frac{1}{4A^{2}}A_{1}A_{3}}$ ${+\frac{1}{4D^{2}}D_{1}D_{2}}$ ${+\frac{1}{4AB}A_{1}B_{3}}$ ${+\frac{1}{4BC}B_{3}C_{1}}$ ${+\frac{1}{4DA}D_{1}A_{3}}$ ${+\frac{1}{4CD}C_{3}D_{1}}$
 ${R_{23}=-\frac{1}{2A}A_{23}}$ –${\frac{1}{2B}B_{23}}$ ${+\frac{1}{4A^{2}}A_{2}A_{3}}$ ${+\frac{1}{4B^{2}}B_{2}B_{3}}$ ${+\frac{1}{4AC}A_{2}C_{3}}$ ${+\frac{1}{4BC}B_{2}C_{3}}$ ${+\frac{1}{4DA}D_{2}A_{3}}$ ${+\frac{1}{4BD}B_{3}D_{2}}$

To apply these tables to the specific metric 1, we observe that ${x^{3}=\phi}$ does not appear in any component of ${g_{ij}}$ so all terms with a subscript 3 are zero. Also, ${x^{2}=\theta}$ appears only in ${g_{\phi\phi}=D}$, so any subscript 2 on ${A}$, ${B}$ or ${C}$ also gives zero. Finally, ${x^{0}=t}$ doesn’t appear in ${C}$ or ${D}$, so a subscript 0 there also gives zero. After using these simplifications, we have:

 ${R_{tt}=0}$ ${+\frac{1}{2B}A_{11}}$ ${+\frac{1}{2C}A_{22}=0}$ ${+\frac{1}{2D}A_{33}=0}$ ${+0}$ ${-\frac{1}{2B}B_{00}}$ ${-\frac{1}{2C}C_{00}=0}$ ${-\frac{1}{2D}D_{00}=0}$ ${+0}$ ${+\frac{1}{4B^{2}}B_{0}^{2}}$ ${+\frac{1}{4C^{2}}C_{0}^{2}=0}$ ${+\frac{1}{4D^{2}}D_{0}^{2}=0}$ ${+0}$ ${+\frac{1}{4AB}A_{0}B_{0}}$ ${+\frac{1}{4AC}A_{0}C_{0}=0}$ ${+\frac{1}{4AD}A_{0}D_{0}=0}$ ${-\frac{1}{4BA}A_{1}^{2}}$ ${-\frac{1}{4B^{2}}A_{1}B_{1}}$ ${+\frac{1}{4BC}A_{1}C_{1}=\frac{1}{4r^{2}B}A_{1}\left(2r\right)}$ ${+\frac{1}{4BD}A_{1}D_{1}=\frac{1}{4Br^{2}\sin^{2}\theta}A_{1}\left(2r\sin^{2}\theta\right)}$ ${-\frac{1}{4CA}A_{2}^{2}=0}$ ${+\frac{1}{4CB}A_{2}B_{2}=0}$ ${-\frac{1}{4C^{2}}A_{2}C_{2}=0}$ ${+\frac{1}{4CD}A_{2}D_{2}=0}$ ${-\frac{1}{4DA}A_{3}^{2}=0}$ ${+\frac{1}{4DB}A_{3}B_{3}=0}$ ${+\frac{1}{4DC}A_{3}C_{3}=0}$ ${-\frac{1}{4D^{2}}A_{3}D_{3}=0}$

Collecting terms, we get

 $\displaystyle R_{tt}$ $\displaystyle =$ $\displaystyle \frac{1}{2B}A_{11}-\frac{1}{2B}B_{00}+\frac{1}{4B^{2}}B_{0}^{2}+\frac{1}{4AB}A_{0}B_{0}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle$ $\displaystyle -\frac{1}{4BA}A_{1}^{2}-\frac{1}{4B^{2}}A_{1}B_{1}+\frac{1}{2rB}A_{1}+\frac{1}{2rB}A_{1}\nonumber$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2B}\left[\partial_{rr}^{2}A-\partial_{tt}^{2}B+\frac{\left(\partial_{t}B\right)^{2}}{2B}+\frac{\left(\partial_{t}A\right)\left(\partial_{t}B\right)-\left(\partial_{r}A\right)^{2}}{2A}-\frac{\left(\partial_{r}A\right)\left(\partial_{r}B\right)}{2B}+\frac{2\partial_{r}A}{r}\right] \ \ \ \ \ (8)$
 ${R_{rr}=\frac{1}{2A}B_{00}}$ ${+0}$ ${-\frac{1}{2C}B_{22}=0}$ ${-\frac{1}{2D}B_{33}=0}$ ${-\frac{1}{2A}A_{11}}$ ${+0}$ ${-\frac{1}{2C}C_{11}=-\frac{1}{r^{2}}}$ ${-\frac{1}{2D}D_{11}=-\frac{1}{r^{2}}}$ ${+\frac{1}{4A^{2}}A_{1}^{2}}$ ${+0}$ ${+\frac{1}{4C^{2}}C_{1}^{2}=\frac{1}{r^{2}}}$ ${+\frac{1}{4D^{2}}D_{1}^{2}=\frac{1}{r^{2}}}$ ${-\frac{1}{4A^{2}}B_{0}A_{0}}$ ${-\frac{1}{4AB}B_{0}^{2}}$ ${+\frac{1}{4AC}B_{0}C_{0}=0}$ ${+\frac{1}{4AD}B_{0}D_{0}=0}$ ${+\frac{1}{4BA}B_{1}A_{1}}$ ${+0}$ ${+\frac{1}{4BC}B_{1}C_{1}=\frac{B_{1}}{2rB}}$ ${+\frac{1}{4BD}B_{1}D_{1}=\frac{B_{1}}{2rB}}$ ${-\frac{1}{4CA}B_{2}A_{2}=0}$ ${+\frac{1}{4CB}B_{2}^{2}=0}$ ${+\frac{1}{4C^{2}}B_{2}C_{2}=0}$ ${-\frac{1}{4CD}B_{2}D_{2}=0}$ ${-\frac{1}{4DA}B_{3}A_{3}=0}$ ${+\frac{1}{4DB}B_{3}^{2}=0}$ ${-\frac{1}{4DC}B_{3}C_{3}=0}$ ${+\frac{1}{4D^{2}}B_{3}D_{3}=0}$
 $\displaystyle R_{rr}$ $\displaystyle =$ $\displaystyle \frac{1}{2A}B_{00}-\frac{1}{2A}A_{11}+\frac{1}{4A^{2}}A_{1}^{2}-\frac{1}{4A^{2}}B_{0}A_{0}-\frac{1}{4AB}B_{0}^{2}+\frac{1}{4BA}B_{1}A_{1}+\frac{B_{1}}{rB}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2A}\left[\partial_{tt}^{2}B-\partial_{rr}^{2}A+\frac{\left(\partial_{r}A\right)^{2}-\left(\partial_{t}A\right)\left(\partial_{t}B\right)}{2A}+\frac{\left(\partial_{r}A\right)\left(\partial_{r}B\right)-\left(\partial_{t}B\right)^{2}}{2B}+\frac{2A\partial_{r}B}{rB}\right] \ \ \ \ \ (10)$
 ${R_{\theta\theta}=\frac{1}{2A}C_{00}=0}$ ${-\frac{1}{2B}C_{11}=-\frac{1}{B}}$ ${+0}$ ${-\frac{1}{2D}C_{33}=0}$ ${-\frac{1}{2A}A_{22}=0}$ ${-\frac{1}{2B}B_{22}=0}$ ${+0}$ ${-\frac{1}{2D}D_{22}=-\frac{2\cos2\theta}{2\sin^{2}\theta}=\frac{\sin^{2}\theta-\cos^{2}\theta}{\sin^{2}\theta}}$ ${+\frac{1}{4A^{2}}A_{2}^{2}=0}$ ${+\frac{1}{4B^{2}}B_{2}^{2}=0}$ ${+0}$ ${+\frac{1}{4D^{2}}D_{2}^{2}=\frac{\sin^{2}2\theta}{4\sin^{4}\theta}=\frac{\cos^{2}\theta}{\sin^{2}\theta}}$ ${-\frac{1}{4A^{2}}C_{0}A_{0}=0}$ ${+\frac{1}{4AB}C_{0}B_{0}=0}$ ${-\frac{1}{4AC}C_{0}^{2}=0}$ ${+\frac{1}{4AD}C_{0}D_{0}=0}$ ${-\frac{1}{4BA}C_{1}A_{1}=-\frac{A_{1}r}{2AB}}$ ${+\frac{1}{4B^{2}}C_{1}B_{1}=\frac{rB_{1}}{2B^{2}}}$ ${+\frac{1}{4BC}C_{1}^{2}=\frac{1}{B}}$ ${-\frac{1}{4BD}C_{1}D_{1}=-\frac{1}{B}}$ ${+\frac{1}{4CA}C_{2}A_{2}=0}$ ${+\frac{1}{4CB}C_{2}B_{2}=0}$ ${+0}$ ${+\frac{1}{4CD}C_{2}D_{2}=0}$ ${-\frac{1}{4DA}C_{3}A_{3}=0}$ ${-\frac{1}{4DB}C_{3}B_{3}=0}$ ${+\frac{1}{4DC}C_{3}^{2}=0}$ ${+\frac{1}{4D^{2}}C_{3}D_{3}=0}$
 $\displaystyle R_{\theta\theta}$ $\displaystyle =$ $\displaystyle -\frac{1}{B}+\frac{\sin^{2}\theta-\cos^{2}\theta}{\sin^{2}\theta}+\frac{\cos^{2}\theta}{\sin^{2}\theta}-\frac{A_{1}r}{2AB}+\frac{rB_{1}}{2B^{2}}+\frac{1}{B}-\frac{1}{B}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{A_{1}r}{2AB}+\frac{rB_{1}}{2B^{2}}+1-\frac{1}{B}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{r\partial_{r}A}{2AB}+\frac{r\partial_{r}B}{2B^{2}}+1-\frac{1}{B} \ \ \ \ \ (13)$
 ${R_{\phi\phi}=\frac{1}{2A}D_{00}=0}$ ${-\frac{1}{2B}D_{11}=-\frac{\sin^{2}\theta}{B}}$ ${-\frac{1}{2C}D_{22}=1-2\cos^{2}\theta}$ ${+0}$ ${-\frac{1}{2A}A_{33}=0}$ ${-\frac{1}{2B}B_{33}=0}$ ${-\frac{1}{2C}C_{33}=0}$ ${+0}$ ${+\frac{1}{4A^{2}}A_{3}^{2}=0}$ ${+\frac{1}{4B^{2}}B_{3}^{2}=0}$ ${+\frac{1}{4C^{2}}C_{3}^{2}=0}$ ${+0}$ ${-\frac{1}{4A^{2}}D_{0}A_{0}=0}$ ${+\frac{1}{4AB}D_{0}B_{0}=0}$ ${+\frac{1}{4AC}D_{0}C_{0}=0}$ ${-\frac{1}{4AD}D_{0}^{2}=0}$ ${-\frac{1}{4BA}D_{1}A_{1}=-\frac{r\sin^{2}\theta}{2BA}A_{1}}$ ${+\frac{1}{4B^{2}}D_{1}B_{1}=\frac{r\sin^{2}\theta}{2B^{2}}B_{1}}$ ${-\frac{1}{4BC}D_{1}C_{1}=-\frac{\sin^{2}\theta}{B}}$ ${+\frac{1}{4BD}D_{1}^{2}=\frac{\sin^{2}\theta}{B}}$ ${-\frac{1}{4CA}D_{2}A_{2}=0}$ ${-\frac{1}{4CB}D_{2}B_{2}=0}$ ${+\frac{1}{4C^{2}}D_{2}C_{2}=0}$ ${+\frac{1}{4CD}D_{2}^{2}=\cos^{2}\theta}$ ${+\frac{1}{4DA}D_{3}A_{3}=0}$ ${+\frac{1}{4DB}D_{3}B_{3}=0}$ ${+\frac{1}{4DC}D_{3}C_{3}=0}$ ${+0}$
 $\displaystyle R_{\phi\phi}$ $\displaystyle =$ $\displaystyle -\frac{\sin^{2}\theta}{B}+1-2\cos^{2}\theta-\frac{r\sin^{2}\theta}{2BA}A_{1}+\frac{r\sin^{2}\theta}{2B^{2}}B_{1}-\frac{\sin^{2}\theta}{B}+\frac{\sin^{2}\theta}{B}+\cos^{2}\theta\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sin^{2}\theta\left[-\frac{r\partial_{r}A}{2AB}+\frac{r\partial_{r}B}{2B^{2}}+1-\frac{1}{B}\right]\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle R_{\theta\theta}\sin^{2}\theta \ \ \ \ \ (16)$
 ${R_{tr}=-\frac{1}{2C}C_{01}=0}$ –${\frac{1}{2D}D_{01}=0}$ ${+\frac{1}{4C^{2}}C_{0}C_{1}=0}$ ${+\frac{1}{4D^{2}}D_{0}D_{1}=0}$ ${+\frac{1}{4AC}A_{1}C_{0}=0}$ ${+\frac{1}{4AD}A_{1}D_{0}=0}$ ${+\frac{1}{4BC}B_{0}C_{1}=\frac{B_{0}}{2rB}}$ ${+\frac{1}{4BD}B_{0}D_{1}=\frac{B_{0}}{2rB}}$

$\displaystyle R_{tr}=\frac{\partial_{t}B}{rB} \ \ \ \ \ (17)$

 ${R_{r\theta}=-\frac{1}{2A}A_{12}=0}$ –${\frac{1}{2D}D_{12}=-\frac{2\cos\theta}{r\sin\theta}}$ ${+\frac{1}{4A^{2}}A_{1}A_{2}=0}$ ${+\frac{1}{4D^{2}}D_{1}D_{2}=\frac{\cos\theta}{r\sin\theta}}$ ${+\frac{1}{4AB}A_{1}B_{2}=0}$ ${+\frac{1}{4BD}B_{2}D_{1}=0}$ ${+\frac{1}{4AC}A_{2}C_{1}=0}$ ${+\frac{1}{4CD}C_{1}D_{2}=\frac{\cos\theta}{r\sin\theta}}$

$\displaystyle R_{r\theta}=-\frac{2\cos\theta}{r\sin\theta}+\frac{\cos\theta}{r\sin\theta}+\frac{\cos\theta}{r\sin\theta}=0 \ \ \ \ \ (18)$

Looking at the worksheet tables above, we can see that applying the rules stated earlier makes all entries in ${R_{02}}$, ${R_{03}}$, ${R_{13}}$ and ${R_{23}}$ zero, so these components of the Ricci tensor are all zero.

# 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.

# Schwarzschild metric equivalent to weak field solution for spherical object

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 22, Problem 22.1.

In the weak field, low velocity limit, the perturbation to the flat space metric can be found in terms of the stress-energy tensor

$\displaystyle h_{jm}=2G\int\frac{2T_{jm}-\eta_{jm}T}{\left|\mathbf{r}-\mathbf{r}'\right|}d^{3}\mathbf{r}' \ \ \ \ \ (1)$

For a spherically symmetric object composed of a perfect fluid whose component particles are moving slowly (compared to light), the pressure is negligible and the density is approximately equal to the Newtonian mass density. Using our previous results, we have

 $\displaystyle \rho_{g}$ $\displaystyle =$ $\displaystyle 2T_{tt}-\eta_{tt}T=\rho_{0}+3P_{0}\approx\rho_{0}\ \ \ \ \ (2)$ $\displaystyle \rho_{c}$ $\displaystyle =$ $\displaystyle 2T_{ii}-\eta_{ii}T=\rho_{0}-P_{0}\approx\rho_{0} \ \ \ \ \ (3)$

where ${i}$ is a spatial index in the last line. Off-diagonal ${h_{jm}}$ components are proportional to the velocity so are all zero if the object is stationary.

Therefore from 1 we get

 $\displaystyle h_{aa}\left(\mathbf{r}\right)$ $\displaystyle =$ $\displaystyle 2G\rho_{0}\int_{\mathcal{V}}\frac{d^{3}\mathbf{r}'}{\left|\mathbf{r}-\mathbf{r}'\right|}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2G\rho_{0}\mathcal{V}}{r}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2GM}{r} \ \ \ \ \ (6)$

where ${\mathcal{V}}$ is the volume of the sphere. The metric for such an object is then

 $\displaystyle ds^{2}$ $\displaystyle =$ $\displaystyle g_{tt}dt^{2}+g_{xx}dx^{2}+g_{yy}dy^{2}+g_{zz}dz^{2}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\eta_{tt}+h_{tt}\right)dt^{2}+\left(\eta_{xx}+h_{xx}\right)dx^{2}+\left(\eta_{yy}+h_{yy}\right)dy^{2}+\left(\eta_{zz}+h_{zz}\right)dz^{2}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\left(1-\frac{2GM}{r}\right)dt^{2}+\left(1+\frac{2GM}{r}\right)\left(dx^{2}+dy^{2}+dz^{2}\right) \ \ \ \ \ (9)$

In spherical coordinates we have

$\displaystyle dx^{2}+dy^{2}+dz^{2}=dr^{2}+r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2} \ \ \ \ \ (10)$

so

$\displaystyle ds^{2}=-\left(1-\frac{2GM}{r}\right)dt^{2}+\left(1+\frac{2GM}{r}\right)\left(dr^{2}+r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2}\right) \ \ \ \ \ (11)$

The ${r}$ coordinate is not circumferential, in the sense that the distance around the equator of the sphere is not ${2\pi r}$. We can see this by noting that to find the equatorial circumference, we hold ${r}$ constant at ${r=R}$ and ${\theta=\frac{\pi}{2}}$, so that ${ds=\sqrt{1+\frac{2GM}{R}}Rd\phi}$ and

$\displaystyle C=\int_{0}^{2\pi}\left(1+\frac{2GM}{R}\right)Rd\phi=2\pi\sqrt{1+\frac{2GM}{R}}R \ \ \ \ \ (12)$

We can define an alternative radial coordinate as

$\displaystyle r_{c}\equiv\sqrt{1+\frac{2GM}{r}}r \ \ \ \ \ (13)$

The ${r_{c}}$ coordinate is circumferential, since the equatorial circumference is ${C=2\pi R_{c}}$.

Assuming that ${\frac{GM}{r}}$ is small (equivalent to assuming we’re a long way from the object so the field is weak), then

 $\displaystyle dr_{c}$ $\displaystyle =$ $\displaystyle \sqrt{1+\frac{2GM}{r}}dr+\left(1+\frac{2GM}{r}\right)^{-1/2}\left(-\frac{GM}{r^{2}}dr\right)r\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\sqrt{1+\frac{2GM}{r}}-\frac{GM}{r}\left(1+\frac{2GM}{r}\right)^{-1/2}\right]dr\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle \left[1+\frac{GM}{r}-\frac{GM}{r}\left(1-\frac{GM}{r}\right)\right]dr\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle dr+\mathcal{O}\left(\left(\frac{GM}{r}\right)^{2}\right) \ \ \ \ \ (17)$

Thus to first order ${dr=dr_{c}}$ so we can write 11 as

$\displaystyle ds^{2}=-\left(1-\frac{2GM}{r}\right)dt^{2}+\left(1+\frac{2GM}{r}\right)dr_{c}^{2}+r_{c}^{2}\left(d\theta^{2}+\sin^{2}\theta d\phi^{2}\right) \ \ \ \ \ (18)$

The Schwarzschild metric is

$\displaystyle ds^{2}=-\left(1-\frac{2GM}{r_{c}}\right)dt^{2}+\left(1-\frac{2GM}{r_{c}}\right)^{-1}dr_{c}^{2}+r_{c}^{2}d\theta^{2}+r_{c}^{2}\sin^{2}\theta d\phi^{2} \ \ \ \ \ (19)$

where we’ve used the circumferential coordinate ${r_{c}}$ for the radial component. We can show that, to first order, 18 and 19 are equivalent. Start with the second term in the Schwarzschild metric:

 $\displaystyle \left(1-\frac{2GM}{r_{c}}\right)^{-1}$ $\displaystyle \approx$ $\displaystyle 1+\frac{2GM}{r_{c}}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1+\frac{2GM}{\sqrt{1+\frac{2GM}{r}}r}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle 1+\frac{2GM}{r}\left(1-\frac{GM}{r}\right)\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1+\frac{2GM}{r}+\mathcal{O}\left(\left(\frac{GM}{r}\right)^{2}\right) \ \ \ \ \ (23)$

The time component also matches, since

 $\displaystyle 1-\frac{2GM}{r_{c}}$ $\displaystyle =$ $\displaystyle 1-\frac{2GM}{\sqrt{1+\frac{2GM}{r}}r}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle 1-\frac{2GM}{r}\left(1-\frac{GM}{r}\right)\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1-\frac{2GM}{r}+\mathcal{O}\left(\left(\frac{GM}{r}\right)^{2}\right) \ \ \ \ \ (26)$

Thus in the weak field limit, the approximate solution is equivalent to the Schwarzschild solution to first order.

# Riemann tensor in the Schwarzschild metric: observer’s view

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 19; Problem P19.8.

The component of a four-vector ${A}$ along a basis vector ${\mathbf{o}_{i}}$ is given by the dot product of the four-vector with the basis vector. Since the dot product is a scalar, it can be computed in any reference frame. In particular, if we’re dealing with a locally flat coordinate system with basis vectors ${\mathbf{o}_{i}}$ embedded in the Schwarzschild (S) metric, we can do the calculation in the global S metric since we know the components of ${\mathbf{o}_{i}}$ in the S frame. That is, the components of a four-vector ${A}$ along each of the basis vectors is

 $\displaystyle A_{i,obs}$ $\displaystyle =$ $\displaystyle g_{jm}\left(\mathbf{o}_{i}\right)^{j}A^{m}\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\mathbf{o}_{i}\right)^{j}A_{j} \ \ \ \ \ (2)$

The subscript ${obs}$ indicates that the vector component is that seen by an observer in the locally flat frame with basis vectors ${\mathbf{o}_{i}}$.

We can extend this idea to find the components of any tensor in the locally flat frame, since we just apply the same procedure to each index of the tensor. For the Riemann tensor ${R_{\; j\ell m}^{i}}$ we would get

$\displaystyle R_{ij\ell m,obs}=g_{ab}\left(\mathbf{o}_{i}\right)^{a}\left(\mathbf{o}_{j}\right)^{c}\left(\mathbf{o}_{\ell}\right)^{d}\left(\mathbf{o}_{m}\right)^{e}R_{\; cde}^{b} \ \ \ \ \ (3)$

To get the observer’s tensor with the first index raised, we need to use the metric to raise the index. The correct metric to use is the metric of the locally flat frame, which is ${\eta^{ij}}$. Therefore we get

 $\displaystyle R_{\; j\ell m,obs}^{i}$ $\displaystyle =$ $\displaystyle \eta^{if}R_{fj\ell m,obs}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \eta^{if}g_{ab}\left(\mathbf{o}_{f}\right)^{a}\left(\mathbf{o}_{j}\right)^{c}\left(\mathbf{o}_{\ell}\right)^{d}\left(\mathbf{o}_{m}\right)^{e}R_{\; cde}^{b} \ \ \ \ \ (5)$

For example, we can calculate the component ${R_{\; ztz,obs}^{t}}$ for a freely-falling observer by using the components for ${\mathbf{o}_{i}}$ that we worked out earlier.

$\displaystyle R_{\; ztz,obs}^{t}=\eta^{tf}g_{ab}\left(\mathbf{o}_{f}\right)^{a}\left(\mathbf{o}_{z}\right)^{c}\left(\mathbf{o}_{t}\right)^{d}\left(\mathbf{o}_{z}\right)^{e}R_{\; cde}^{b} \ \ \ \ \ (6)$

To save typing, I’ll write the unit vectors in normal type and without the parentheses, so this equation becomes

$\displaystyle R_{\; ztz,obs}^{t}=\eta^{tf}g_{ab}o_{f}^{a}o_{z}^{c}o_{t}^{d}o_{z}^{e}R_{\; cde}^{b} \ \ \ \ \ (7)$

where sums are implied over all repeated indices except ${t}$ and ${z}$.

Remember that a superscript on a basis vector refers to the component of that vector along the direction given by the superscript, and that this direction is one of those in the S metric. A subscript on a basis vector refers to the axis in the locally flat system along with that basis vector points. Thus ${o_{z}^{r}}$ is the component along the ${r}$ direction in the S metric of the vector ${\mathbf{o}_{z}}$ that points along the ${z}$ direction in the locally flat system.

We’ll reproduce here the basis vectors ${\mathbf{o}_{t}}$ and ${\mathbf{o}_{z}}$ for a freely falling observer (we won’t need the other two vectors):

 $\displaystyle \mathbf{o}_{t}$ $\displaystyle =$ $\displaystyle \left[\left(1-\frac{2GM}{r}\right)^{-1},-\sqrt{\frac{2GM}{r}},0,0\right]\ \ \ \ \ (8)$ $\displaystyle \mathbf{o}_{z}$ $\displaystyle =$ $\displaystyle \left[-\left(1-\frac{2GM}{r}\right)^{-1}\sqrt{\frac{2GM}{r}},1,0,0\right] \ \ \ \ \ (9)$

Unfortunately, since both of these vectors have two non-zero components, the sums in 7 give us more than one term. In order to make use of the symmetries of the Riemann tensor, we’ll rewrite 7 using ${R_{acde}=g_{ab}R_{\; cde}^{b}}$:

$\displaystyle R_{\; ztz,obs}^{t}=\eta^{tf}o_{f}^{a}o_{z}^{c}o_{t}^{d}o_{z}^{e}R_{acde} \ \ \ \ \ (10)$

Since ${\eta^{ij}}$ is diagonal, we must have ${f=t}$ and since ${\eta^{tt}=-1}$ we have

$\displaystyle R_{\; ztz,obs}^{t}=-o_{t}^{a}o_{z}^{c}o_{t}^{d}o_{z}^{e}R_{acde} \ \ \ \ \ (11)$

Because of the symmetries, we must have ${a\ne c}$ and ${d\ne e}$, so there are four possibilities for these four indices:

$\displaystyle \left[a,c,d,e\right]=\left[t,r,t,r\right],\left[r,t,r,t\right],\left[r,t,t,r\right],\left[t,r,r,t\right] \ \ \ \ \ (12)$

The sum in 11 thus expands to

$\displaystyle R_{\; ztz,obs}^{t}=-R_{trtr}o_{t}^{t}o_{z}^{r}o_{t}^{t}o_{z}^{r}-R_{rtrt}o_{t}^{r}o_{z}^{t}o_{t}^{r}o_{z}^{t}-R_{rttr}o_{t}^{r}o_{z}^{t}o_{t}^{t}o_{z}^{r}-R_{trrt}o_{t}^{t}o_{z}^{r}o_{t}^{r}o_{z}^{t} \ \ \ \ \ (13)$

Using the symmetries when swapping the first two or the last two indices in the Riemann tensor in this lowered form, we can rewrite this as

 $\displaystyle R_{\; ztz,obs}^{t}$ $\displaystyle =$ $\displaystyle -R_{trtr}\left[o_{t}^{t}o_{z}^{r}o_{t}^{t}o_{z}^{r}+o_{t}^{r}o_{z}^{t}o_{t}^{r}o_{z}^{t}-o_{t}^{r}o_{z}^{t}o_{t}^{t}o_{z}^{r}-o_{t}^{t}o_{z}^{r}o_{t}^{r}o_{z}^{t}\right]\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -R_{trtr}\left[\left(o_{t}^{t}o_{z}^{r}\right)^{2}+\left(o_{t}^{r}o_{z}^{t}\right)^{2}-2o_{t}^{r}o_{z}^{t}o_{t}^{t}o_{z}^{r}\right]\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -R_{trtr}\left(o_{t}^{t}o_{z}^{r}-o_{t}^{r}o_{z}^{t}\right)^{2}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -R_{trtr}\left(\left(1-\frac{2GM}{r}\right)^{-1}-\left[-\left(1-\frac{2GM}{r}\right)^{-1}\sqrt{\frac{2GM}{r}}\right]\left[-\sqrt{\frac{2GM}{r}}\right]\right)^{2}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -R_{trtr}\left[\left(1-\frac{2GM}{r}\right)^{-1}\left(1-\frac{2GM}{r}\right)\right]^{2}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -R_{trtr} \ \ \ \ \ (19)$

We now need to find ${R_{trtr}}$, which is

$\displaystyle R_{trtr}=g_{ta}R_{\; rtr}^{a}=g_{tt}R_{\; rtr}^{t} \ \ \ \ \ (20)$

We’ve worked out the last tensor component earlier, so plugging this in, we get

 $\displaystyle R_{\; ztz,obs}^{t}$ $\displaystyle =$ $\displaystyle -g_{tt}R_{\; rtr}^{t}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(1-\frac{2GM}{r}\right)\left[\frac{2GM}{r^{3}}\left(1-\frac{2GM}{r}\right)^{-1}\right]\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2GM}{r^{3}} \ \ \ \ \ (23)$

Thus this component of the Riemann tensor has no singularity at ${r=2GM}$ in the observer’s local frame.

# Riemann tensor in the Schwarzschild metric

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 18; Problem P18.6.

We’ll calculate one component of the Riemann tensor for the Schwarzschild metric. The tensor is

$\displaystyle R_{\; j\ell m}^{i}\equiv\partial_{\ell}\Gamma_{\; mj}^{i}-\partial_{m}\Gamma_{\;\ell j}^{i}+\Gamma_{\; mj}^{k}\Gamma_{\;\ell k}^{i}-\Gamma_{\;\ell j}^{k}\Gamma_{\; km}^{i} \ \ \ \ \ (1)$

As usual, we need the Christoffel symbols, but we’ve already worked these out.

 $\displaystyle \Gamma_{\; ij}^{t}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cccc} 0 & \frac{GM}{r^{2}}\left(1-\frac{2GM}{r}\right)^{-1} & 0 & 0\\ \frac{GM}{r^{2}}\left(1-\frac{2GM}{r}\right)^{-1} & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{array}\right]\ \ \ \ \ (2)$ $\displaystyle \Gamma_{\; ij}^{r}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cccc} \frac{GM}{r^{2}}\left(1-\frac{2GM}{r}\right) & 0 & 0 & 0\\ 0 & -\frac{GM}{r^{2}}\left(1-\frac{2GM}{r}\right)^{-1} & 0 & 0\\ 0 & 0 & -r\left(1-\frac{2GM}{r}\right) & 0\\ 0 & 0 & 0 & -r\sin^{2}\theta\left(1-\frac{2GM}{r}\right) \end{array}\right]\ \ \ \ \ (3)$ $\displaystyle \Gamma_{\; ij}^{\theta}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cccc} 0 & 0 & 0 & 0\\ 0 & 0 & \frac{1}{r} & 0\\ 0 & \frac{1}{r} & 0 & 0\\ 0 & 0 & 0 & -\sin\theta\cos\theta \end{array}\right]\ \ \ \ \ (4)$ $\displaystyle \Gamma_{\; ij}^{\phi}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cccc} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & \frac{1}{r}\\ 0 & 0 & 0 & \cot\theta\\ 0 & \frac{1}{r} & \cot\theta & 0 \end{array}\right] \ \ \ \ \ (5)$

We can plug these into the formula above to get ${R_{\; rtr}^{t}}$. We have

$\displaystyle R_{\; rtr}^{t}=\partial_{t}\Gamma_{\; rr}^{t}-\partial_{r}\Gamma_{\; tr}^{t}+\Gamma_{\; rr}^{k}\Gamma_{\; kt}^{t}-\Gamma_{\; tr}^{k}\Gamma_{\; rk}^{t} \ \ \ \ \ (6)$

We can work out these terms one at a time (only the index ${k}$ is summmed):

 $\displaystyle \partial_{t}\Gamma_{\; rr}^{t}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (7)$ $\displaystyle -\partial_{r}\Gamma_{\; tr}^{t}$ $\displaystyle =$ $\displaystyle \frac{2GM}{r^{3}}\left(1-\frac{2GM}{r}\right)^{-1}+\frac{GM}{r^{2}}\left(1-\frac{2GM}{r}\right)^{-2}\left(\frac{2GM}{r^{2}}\right)\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2GM}{r^{3}}\left(1-\frac{2GM}{r}\right)^{-1}+\frac{2G^{2}M^{2}}{r^{4}}\left(1-\frac{2GM}{r}\right)^{-2}\ \ \ \ \ (9)$ $\displaystyle \Gamma_{\; rr}^{k}\Gamma_{\; kt}^{t}$ $\displaystyle =$ $\displaystyle \Gamma_{\; rr}^{r}\Gamma_{\; rt}^{t}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{G^{2}M^{2}}{r^{4}}\left(1-\frac{2GM}{r}\right)^{-2}\ \ \ \ \ (11)$ $\displaystyle -\Gamma_{\; tr}^{k}\Gamma_{\; rk}^{t}$ $\displaystyle =$ $\displaystyle -\left(\Gamma_{\; rt}^{t}\right)^{2}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{G^{2}M^{2}}{r^{4}}\left(1-\frac{2GM}{r}\right)^{-2} \ \ \ \ \ (13)$

Adding these up we get

$\displaystyle R_{\; rtr}^{t}=\frac{2GM}{r^{3}}\left(1-\frac{2GM}{r}\right)^{-1} \ \ \ \ \ (14)$

Since this is never zero, Schwarzschild spacetime is curved everywhere, but as ${r\rightarrow\infty}$, ${R_{\; rtr}^{t}\rightarrow0}$ so the further we get from the mass, the less curved the spacetime becomes.