# 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.

# Riemann and Ricci tensors in the weak field limit

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 22, Boxes 22.2-22.4.

The Einstein equation is

$\displaystyle G^{ij}+\Lambda g^{ij}=8\pi GT^{ij} \ \ \ \ \ (1)$

where ${\Lambda}$ is the cosmological constant and ${T^{ij}}$ is the stress-energy tensor. The Einstein tensor ${G^{ij}}$ is defined in terms of the Ricci tensor and curvature scalar as

$\displaystyle G^{ij}\equiv R^{ij}-\frac{1}{2}g^{ij}R \ \ \ \ \ (2)$

The Einstein equation can be written in the alternative form

$\displaystyle R^{ij}=8\pi G\left(T^{ij}-\frac{1}{2}g^{ij}T\right)+\Lambda g^{ij} \ \ \ \ \ (3)$

where

$\displaystyle T\equiv g_{ij}T^{ij} \ \ \ \ \ (4)$

In preparation for examining the behaviour of this equation in the weak field limit, we need to see how the Ricci tensor behaves for metrics that are nearly flat. Since the Ricci tensor is a contraction of the Riemann tensor, we need first to see how the Riemann tensor behaves in this limit.

The metric for a weak field can be written as

$\displaystyle g_{ij}=\eta_{ij}+h_{ij} \ \ \ \ \ (5)$

where ${\eta_{ij}}$ is the flat space metric and ${h_{ij}}$ is a small perturbation, where small means ${\left|h_{ij}\right|\ll1}$. Note that since ${g_{ij}}$ and ${\eta_{ij}}$ are both symmetric, we must have

$\displaystyle h_{ij}=h_{ji} \ \ \ \ \ (6)$

As an approximation, we’ll discard all terms of order ${\left|h_{ij}\right|^{2}}$ or higher. First, we’ll get an approximation for the inverse metric ${g^{ij}}$. Since ${\eta_{ij}}$ is diagonal with elements ${\left[-1,1,1,1\right]}$ it is its own inverse so

$\displaystyle \eta^{ij}=\eta_{ij} \ \ \ \ \ (7)$

Therefore it’s reasonable to assume that

$\displaystyle g^{ij}=\eta^{ij}+b^{ij} \ \ \ \ \ (8)$

where ${\left|b^{ij}\right|\ll1}$. We can find ${b^{ij}}$ from the condition that

$\displaystyle g^{ij}g_{jk}=\delta_{k}^{i} \ \ \ \ \ (9)$

Therefore

 $\displaystyle g^{ij}g_{jk}$ $\displaystyle =$ $\displaystyle \eta^{ij}\eta_{jk}+\eta^{ij}h_{jk}+b^{ij}\eta_{jk}+b^{ij}h_{jk}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle \delta_{k}^{i}+\eta^{ij}h_{jk}+b^{ij}\eta_{jk} \ \ \ \ \ (11)$

to first order (since we’re assuming both ${b^{ij}}$ and ${h_{jk}}$ are very small). Therefore

 $\displaystyle b^{ij}\eta_{jk}$ $\displaystyle =$ $\displaystyle -\eta^{ij}h_{jk}\ \ \ \ \ (12)$ $\displaystyle b^{ij}\eta^{kl}\eta_{jk}$ $\displaystyle =$ $\displaystyle -\eta^{kl}\eta^{ij}h_{jk}\ \ \ \ \ (13)$ $\displaystyle b^{ij}\delta_{j}^{l}$ $\displaystyle =$ $\displaystyle -\eta^{kl}\eta^{ij}h_{jk}\ \ \ \ \ (14)$ $\displaystyle b^{il}$ $\displaystyle =$ $\displaystyle -\eta^{kl}\eta^{ij}h_{jk}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -h^{il} \ \ \ \ \ (16)$

where

$\displaystyle h^{il}\equiv\eta^{kl}\eta^{ij}h_{jk} \ \ \ \ \ (17)$

Thus the perturbation ${h^{ij}}$ in the inverse metric is of the same order as the perturbation ${h_{jk}}$ in the original metric which means that, to first order, we can raise an index of a tensor whose components are all of order ${h_{jk}}$ by simply multiplying by the flat space metric ${\eta^{ij}}$ rather than the full metric ${g^{ij}}$. That is, for some tensor ${A_{kl}}$ where ${\left|A_{kl}\right|\ll1}$ we have, to first order

$\displaystyle A_{\; l}^{j}=g^{jk}A_{kl}=\left(\eta^{jk}-h^{jk}\right)A_{kl}=\eta^{jk}A_{kl}-\mathcal{O}\left(\left|h\right|^{2}\right)\approx\eta^{jk}A_{kl} \ \ \ \ \ (18)$

From 3 with ${\Lambda=0}$ and indices lowered, the Einstein equation is

$\displaystyle R_{ij}=8\pi G\left(T_{ij}-\frac{1}{2}g_{ij}T\right) \ \ \ \ \ (19)$

so we need to see what ${R_{ij}}$ looks like in the weak field limit. We’ll start with the Christoffel symbols

$\displaystyle \Gamma_{\; ij}^{m}=\frac{1}{2}g^{ml}\left(\partial_{j}g_{il}+\partial_{i}g_{lj}-\partial_{l}g_{ji}\right) \ \ \ \ \ (20)$

In the weak field limit

$\displaystyle \partial_{j}g_{il}=\partial_{j}\left(\eta_{il}+h_{il}\right)=\partial_{j}h_{il} \ \ \ \ \ (21)$

since ${\partial_{j}\eta_{il}=0}$. Therefore,

$\displaystyle \Gamma_{\; ij}^{m}=\frac{1}{2}g^{ml}\left(\partial_{j}h_{il}+\partial_{i}h_{lj}-\partial_{l}h_{ji}\right) \ \ \ \ \ (22)$

This result is exact so far. Now if we want only first order terms, we can replace ${g^{ml}=\eta^{ml}-h^{ml}}$ by just ${\eta^{ml}}$ since the ${h^{ml}}$ multiplied into the derivatives gives only second order terms. Therefore, to first order

$\displaystyle \Gamma_{\; ij}^{m}=\frac{1}{2}\eta^{ml}\left(\partial_{j}h_{il}+\partial_{i}h_{lj}-\partial_{l}h_{ji}\right) \ \ \ \ \ (23)$

Now the Riemann tensor is defined as

$\displaystyle R_{\; jlm}^{i}\equiv-\left[\partial_{m}\Gamma_{\; lj}^{i}-\partial_{l}\Gamma_{\; mj}^{i}+\Gamma_{\; lj}^{k}\Gamma_{\; km}^{i}-\Gamma_{\; mj}^{k}\Gamma_{\; lk}^{i}\right] \ \ \ \ \ (24)$

Since the Christoffel symbols are all of order ${h_{ij}}$ the last two terms in the Riemann tensor are of order ${\left|h_{ij}\right|^{2}}$ so we can drop them. With the first index lowered, we get

 $\displaystyle R_{njlm}$ $\displaystyle =$ $\displaystyle -g_{ni}\left[\partial_{m}\Gamma_{\; lj}^{i}-\partial_{l}\Gamma_{\; mj}^{i}\right]\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{2}\eta_{ni}\left[\eta^{ik}\partial_{m}\left(\partial_{j}h_{lk}+\partial_{l}h_{kj}-\partial_{k}h_{jl}\right)-\eta^{ik}\partial_{l}\left(\partial_{j}h_{mk}+\partial_{m}h_{kj}-\partial_{k}h_{jm}\right)\right]\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{2}\left[\partial_{m}\left(\partial_{j}h_{ln}+\partial_{l}h_{nj}-\partial_{n}h_{jl}\right)-\partial_{l}\left(\partial_{j}h_{mn}+\partial_{m}h_{nj}-\partial_{n}h_{jm}\right)\right]\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\partial_{j}\partial_{l}h_{nm}+\partial_{n}\partial_{m}h_{jl}-\partial_{n}\partial_{l}h_{jm}-\partial_{j}\partial_{m}h_{nl}\right] \ \ \ \ \ (28)$

where we used 6 to get the last line.

Now to get the Ricci tensor in the weak field limit. We have, to first order

 $\displaystyle R_{jm}$ $\displaystyle =$ $\displaystyle g^{nl}R_{njlm}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\eta^{nl}\left[\partial_{j}\partial_{l}h_{nm}+\partial_{n}\partial_{m}h_{jl}-\partial_{n}\partial_{l}h_{jm}-\partial_{j}\partial_{m}h_{nl}\right]\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\eta^{nl}\left(\partial_{j}\partial_{l}h_{nm}-\frac{1}{2}\partial_{j}\partial_{m}h_{nl}\right)+\eta^{nl}\left(\partial_{n}\partial_{m}h_{jl}-\frac{1}{2}\partial_{j}\partial_{m}h_{nl}\right)\right]-\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{1}{2}\eta^{nl}\partial_{n}\partial_{l}h_{jm}\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\eta^{nl}\partial_{j}\left(\partial_{l}h_{nm}-\frac{1}{2}\partial_{m}h_{nl}\right)+\eta^{nl}\partial_{m}\left(\partial_{n}h_{jl}-\frac{1}{2}\partial_{j}h_{nl}\right)-\eta^{nl}\partial_{n}\partial_{l}h_{jm}\right]\ \ \ \ \ (32)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\eta^{ln}\partial_{j}\left(\partial_{l}h_{nm}-\frac{1}{2}\partial_{m}h_{nl}\right)+\eta^{nl}\partial_{m}\left(\partial_{n}h_{lj}-\frac{1}{2}\partial_{j}h_{nl}\right)-\eta^{nl}\partial_{n}\partial_{l}h_{jm}\right]\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\eta^{nl}\partial_{j}\left(\partial_{n}h_{lm}-\frac{1}{2}\partial_{m}h_{nl}\right)+\eta^{nl}\partial_{m}\left(\partial_{n}h_{lj}-\frac{1}{2}\partial_{j}h_{nl}\right)-\eta^{nl}\partial_{n}\partial_{l}h_{jm}\right]\ \ \ \ \ (34)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(\partial_{j}H_{m}+\partial_{m}H_{j}-\eta^{nl}\partial_{n}\partial_{l}h_{jm}\right) \ \ \ \ \ (35)$

where

$\displaystyle H_{m}\equiv\eta^{nl}\left(\partial_{n}h_{lm}-\frac{1}{2}\partial_{m}h_{nl}\right) \ \ \ \ \ (36)$

The index magic we performed above includes using ${\eta^{nl}=\eta^{ln}}$ and ${h_{nl}=h_{ln}}$ in 33 and then swapping the bound indices ${l}$ and ${n}$ and then using ${h_{ln}=h_{nl}}$ again in the first term in 34. The Ricci tensor in the weak field limit is thus

$\displaystyle R_{jm}=\frac{1}{2}\left(\partial_{j}H_{m}+\partial_{m}H_{j}-\eta^{nl}\partial_{n}\partial_{l}h_{jm}\right) \ \ \ \ \ (37)$

# 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).

# Ricci tensor and curvature scalar for a sphere

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 19; Box 19.6.

As an example of calculating the Ricci tensor and curvature scalar we’ll find them for the 2-d surface of a sphere. The Ricci tensor is calculated from the Riemann tensor, and that in turn depends on the Christoffel symbols, so we’ll need them first. It’s easiest to find them from the geodesic equation

$\displaystyle g_{aj}\ddot{x}^{j}+\left(\partial_{i}g_{aj}-\frac{1}{2}\partial_{a}g_{ij}\right)\dot{x}^{j}\dot{x}^{i}=0 \ \ \ \ \ (1)$

which is formally equivalent to

$\displaystyle \ddot{x}^{m}+\Gamma_{\; ij}^{m}\dot{x}^{j}\dot{x}^{i}=0 \ \ \ \ \ (2)$

The metric for a sphere is

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

where ${r}$ is the constant radius of the sphere, so

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

We have two equations arising from 1. For ${a=\theta}$

$\displaystyle r^{2}\ddot{\theta}-r^{2}\sin\theta\cos\theta\dot{\phi}^{2}=0 \ \ \ \ \ (6)$

Comparing with 2 we get, after dividing out the ${r^{2}}$:

$\displaystyle \Gamma_{\;\phi\phi}^{\theta}=-\sin\theta\cos\theta=-\frac{1}{2}\sin2\theta \ \ \ \ \ (7)$

For ${a=\phi}$:

$\displaystyle r^{2}\sin^{2}\theta\ddot{\phi}^{2}+2r^{2}\sin\theta\cos\theta\dot{\theta}\dot{\phi}=0 \ \ \ \ \ (8)$

Dividing through by ${r^{2}\sin^{2}\theta}$ and comparing with 2 we get (remember that the second term is ${\left(\Gamma_{\;\theta\phi}^{\phi}+\Gamma_{\;\phi\theta}^{\phi}\right)\dot{\theta}\dot{\phi}}$ and that ${\Gamma_{\;\theta\phi}^{\phi}=\Gamma_{\;\phi\theta}^{\phi}}$):

$\displaystyle \Gamma_{\;\theta\phi}^{\phi}=\Gamma_{\;\phi\theta}^{\phi}=\cot\theta \ \ \ \ \ (9)$

All other Christoffel symbols are zero.

In 2D, the Riemann tensor has only one independent component, which we can take to be ${R_{\theta\phi\theta\phi}}$, which can be calculated from

$\displaystyle R_{\; j\ell m}^{i}\equiv\partial_{\ell}\Gamma_{\; mj}^{i}-\partial_{m}\Gamma_{\;\ell j}^{i}+\Gamma_{\; mj}^{k}\Gamma_{\;\ell k}^{i}-\Gamma_{\;\ell j}^{k}\Gamma_{\; km}^{i} \ \ \ \ \ (10)$

Lowering the first index, we have

 $\displaystyle R_{\theta\phi\theta\phi}$ $\displaystyle =$ $\displaystyle g_{\theta i}R_{\;\phi\theta\phi}^{i}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g_{\theta\theta}R_{\;\phi\theta\phi}^{\theta}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}\left(\partial_{\theta}\Gamma_{\;\phi\phi}^{\theta}-\partial_{\phi}\Gamma_{\;\theta\phi}^{\theta}+\Gamma_{\;\phi\phi}^{k}\Gamma_{\;\theta k}^{\theta}-\Gamma_{\;\theta\phi}^{k}\Gamma_{\; k\phi}^{\theta}\right)\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}\left(-\cos2\theta-0+0-\Gamma_{\;\theta\phi}^{\phi}\Gamma_{\;\phi\phi}^{\theta}\right)\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}\left(\sin^{2}\theta-\cos^{2}\theta+\cos^{2}\theta\right)\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}\sin^{2}\theta \ \ \ \ \ (16)$

We can now find the Ricci tensor.

$\displaystyle R_{ij}=g^{ab}R_{aibj} \ \ \ \ \ (17)$

Since the metric is diagonal and ${g^{ab}}$ is the inverse of ${g_{ab}}$, we have

 $\displaystyle g^{\theta\theta}$ $\displaystyle =$ $\displaystyle \frac{1}{g^{\theta\theta}}=\frac{1}{r^{2}}\ \ \ \ \ (18)$ $\displaystyle g^{\phi\phi}$ $\displaystyle =$ $\displaystyle \frac{1}{g^{\phi\phi}}=\frac{1}{r^{2}\sin^{2}\theta} \ \ \ \ \ (19)$

so, using the symmetries of the Riemann tensor,

 $\displaystyle R_{\theta\theta}$ $\displaystyle =$ $\displaystyle g^{ab}R_{a\theta b\theta}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g^{\phi\phi}R_{\phi\theta\phi\theta}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{r^{2}\sin^{2}\theta}r^{2}\sin^{2}\theta\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (23)$ $\displaystyle R_{\phi\phi}$ $\displaystyle =$ $\displaystyle g^{ab}R_{a\phi b\phi}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g^{\theta\theta}R_{\theta\phi\theta\phi}\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sin^{2}\theta\ \ \ \ \ (26)$ $\displaystyle R_{\theta\phi}$ $\displaystyle =$ $\displaystyle R_{\phi\theta}=0 \ \ \ \ \ (27)$

We can get the upstairs version of the Ricci tensor as well:

 $\displaystyle R^{\theta\theta}$ $\displaystyle =$ $\displaystyle g^{\theta i}g^{\theta j}R_{ij}\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g^{\theta\theta}g^{\theta\theta}R_{\theta\theta}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{r^{4}}\ \ \ \ \ (30)$ $\displaystyle R^{\phi\phi}$ $\displaystyle =$ $\displaystyle g^{\phi\phi}g^{\phi\phi}R_{\phi\phi}\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{r^{4}\sin^{2}\theta} \ \ \ \ \ (32)$

The curvature scalar is

 $\displaystyle R$ $\displaystyle =$ $\displaystyle g_{ij}R^{ij}\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}\frac{1}{r^{4}}+r^{2}\sin^{2}\theta\frac{1}{r^{4}\sin^{2}\theta}\ \ \ \ \ (34)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{r^{2}} \ \ \ \ \ (35)$

As we would expect, the curvature of a sphere decreases as its radius gets larger.

# Einstein equation in the Newtonian limit

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 21; Problem 21.1.

The Einstein equation is

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

where we have yet to determine the constant ${\kappa}$. To do this, we need to show that the Einstein equation reduces to Newton’s law of gravity for weak gravitational fields. Actually, there are three conditions that should hold in the Newtonian limit. First, as we’ve said, the gravitational field is weak, meaning that spacetime is nearly flat. Second, objects should travel with a speed much less than the speed of light (the spatial four-velocity components ${u^{i}\ll1}$ for ${i=x,y,z}$). The second condition implies that the only non-negligible component of the stress-energy tensor ${T^{ij}}$ is ${T^{tt}}$. For example, for a perfect fluid, we can assume that it’s effectively at rest, so the off-diagonal elements are all zero. For the diagonal spatial compoments, we have (for ${T^{zz}}$; the other 2 components have the analogous formulas):

$\displaystyle T^{zz}=\frac{1}{L^{3}}\int dp^{x}\int dp^{y}\int\left(p^{z}\right)^{2}\frac{N\left(p\right)}{p^{t}}dp^{z} \ \ \ \ \ (2)$

This equation is for a cubic volume of side length ${L}$ containing ${N\left(p\right)}$ particles of momentum ${p}$. Since ${p^{z}=mu^{z}}$ and ${u^{z}\ll1}$, ${T^{zz}\approx0}$ (the requirement of a weak gravitational field means that ${N\left(p\right)}$ can’t be very large, since we can’t have that much mass). As the spatial diagonal elements ${T^{ii}=P}$, the pressure and ${T^{tt}=\rho}$, the energy density, this condition translates to ${\rho\gg P}$.

With these assumptions, we can try to show that the relativistic equation of geodesic deviation reduces to the Newtonian version. That is, we want to show that

$\displaystyle \ddot{\mathbf{n}}^{i}=-R_{\;j\ell m}^{i}u^{m}u^{j}n^{\ell} \ \ \ \ \ (3)$

reduces to

$\displaystyle \ddot{n}^{i}=-\eta^{ij}\left(\partial_{k}\partial_{j}\Phi\right)n^{k}\mbox{ (for }i,j,k=x,y,z\mbox{)} \ \ \ \ \ (4)$

where ${n^{i}}$ is the separation of two infinitesimally close geodesics (this is the tidal force.) and ${\Phi}$ is the Newtonian gravitational potential.

Starting with 3, we can eliminate terms where ${m\ne t}$ or ${j\ne t}$ (because ${u^{i}\approx0}$ for ${i\ne t}$) to get

$\displaystyle \ddot{n}^{i}=-R_{\;t\ell t}^{i}n^{\ell} \ \ \ \ \ (5)$

where we’re now considering only the spatial components: ${i,\ell=x,y,z}$. Note that summing ${\ell}$ over ${x,y,z}$ is the same as summing it over ${t,x,y,z}$ since due to the anti-symmetry of the Riemann tensor under interchange of its last two indices, ${R_{\;ttt}^{i}=-R_{\;ttt}^{i}=0}$. Also, because we’re in the non-relativistic limit, the proper time and coordinate time are essentially the same thing: ${\tau\approx t}$, so the time derivative is with respect to ${t}$.

Comparing this to 4, we have (renaming ${\ell}$ to ${k}$ in the last equation):

$\displaystyle R_{\;tkt}^{i}\approx\eta^{ij}\left(\partial_{k}\partial_{j}\Phi\right) \ \ \ \ \ (6)$

Newton’s law of gravity in differential form is

 $\displaystyle \nabla^{2}\Phi$ $\displaystyle =$ $\displaystyle 4\pi G\rho\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \eta^{ij}\partial_{i}\partial_{j}\Phi\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle R_{\;tit}^{i}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle R_{tt} \ \ \ \ \ (10)$

(Again, the contraction over index ${i}$ in the second to last line can be taken over 3 or 4 coordinates, since ${R_{\;ttt}^{i}=0}$.) We can raise both indices on the Ricci tensor in the usual way:

 $\displaystyle R^{tt}$ $\displaystyle =$ $\displaystyle g^{ti}g^{tj}R_{ij}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle \eta^{ti}\eta^{tj}R_{ij}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(-1\right)^{2}R_{tt}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle R_{tt} \ \ \ \ \ (14)$

so we can write 10 in upper index form as

$\displaystyle \nabla^{2}\Phi\approx R^{tt} \ \ \ \ \ (15)$

Now looking back at 1 and using the condition that ${T^{tt}=\rho}$ is the only significant entry in the stress-energy tensor, we have

$\displaystyle T=g_{ij}T^{ij}=\eta_{ij}T^{ij}=-\rho \ \ \ \ \ (16)$

so

 $\displaystyle R^{tt}$ $\displaystyle =$ $\displaystyle \kappa\left(T^{tt}-\frac{1}{2}\eta^{tt}T\right)+\Lambda\eta^{tt}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\kappa}{2}\rho-\Lambda\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle \nabla^{2}\Phi\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 4\pi G\rho \ \ \ \ \ (20)$

Comparing the second and fourth lines, we see that ${\Lambda\approx0}$ and

$\displaystyle \kappa=8\pi G \ \ \ \ \ (21)$

and the Einstein equation becomes

$\displaystyle R^{ij}=8\pi G\left(T^{ij}-\frac{1}{2}g^{ij}T\right) \ \ \ \ \ (22)$

or in its original form

$\displaystyle G^{ij}=8\pi GT^{ij} \ \ \ \ \ (23)$

Actually, we can’t take ${\Lambda=0}$; all we can say is that for gravitational systems on the scale of the solar system (where Newtonian theory works well, except in the case of the orbit of Mercury) ${\Lambda\ll4\pi G\rho}$. To get a feel for how small ${\Lambda}$ needs to be, suppose we have a spherical gravitational potential in empty space (${\rho=0}$). Then in spherical coordinates

 $\displaystyle \nabla^{2}\Phi$ $\displaystyle =$ $\displaystyle \frac{1}{r^{2}}\frac{d}{dr}\left(r^{2}\frac{d\Phi}{dr}\right)\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle -\Lambda\ \ \ \ \ (25)$ $\displaystyle \frac{d}{dr}\left(r^{2}\frac{d\Phi}{dr}\right)$ $\displaystyle \approx$ $\displaystyle -\Lambda r^{2}\ \ \ \ \ (26)$ $\displaystyle \frac{d\Phi}{dr}$ $\displaystyle \approx$ $\displaystyle -\frac{\Lambda}{3}r \ \ \ \ \ (27)$

This is the radial component (the only non-zero component) of the gradient of the potential, and the negative gradient of the gravitational potential is the gravitational field, which is the acceleration of gravity. In the solar system, Newton’s theory says that the acceleration due to the sun is

$\displaystyle g=\frac{GM_{s}}{r^{2}} \ \ \ \ \ (28)$

so if ${\Lambda\ne0}$ but its effect is not felt in Newton’s theory, we must have

$\displaystyle \frac{\Lambda}{3}r\ll\frac{GM_{s}}{r^{2}} \ \ \ \ \ (29)$

in order for Newton’s theory to be valid in the solar system. Distances in the solar system are around ${r\approx10^{12}\mbox{ m}}$, ${GM_{s}\approx1500\mbox{ m}}$ and ${G}$ in general relativistic units is ${7.426\times10^{-28}\mbox{ m kg}^{-1}}$ so this means

$\displaystyle \frac{\Lambda}{8\pi G}\ll\frac{3\times1500}{8\pi\left(7.426\times10^{-28}\right)\left(10^{12}\right)^{3}}=2.4\times10^{-7}\mbox{ kg m}^{-3} \ \ \ \ \ (30)$

# Einstein tensor of zero implies a zero Ricci tensor

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 21; 2.

We wish to prove that ${G^{ij}=0}$ if and only if ${R^{ij}=0}$. The Einstein tensor is defined as

$\displaystyle G^{ij}\equiv R^{ij}-\frac{1}{2}g^{ij}R \ \ \ \ \ (1)$

Clearly if the Ricci tensor ${R^{ij}=0}$ then ${G^{ij}=0}$ (since the curvature scalar is the contraction of the Ricci tensor: ${R=g_{ij}R^{ij}}$). To prove the converse, suppose ${G^{ij}=0}$. Then we can multiply both sides by ${g_{ij}}$ to get

 $\displaystyle 0$ $\displaystyle =$ $\displaystyle g_{ij}R^{ij}-\frac{1}{2}g_{ij}g^{ij}R\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle R-2R\ \ \ \ \ (3)$ $\displaystyle R$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (4)$

since ${g_{ij}g^{ij}=4}$. With ${G^{ij}=0}$ and ${R=0}$, 1 tells us that ${R^{ij}=0}$. QED.

# Einstein equation: alternative form

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

The general relativistic generalization of Newton’s law of gravity is

$\displaystyle G^{ij}+\Lambda g^{ij}=\kappa T^{ij} \ \ \ \ \ (1)$

where the Einstein tensor is defined in terms of the Ricci tensor and the curvature scalar as

$\displaystyle G^{ij}\equiv R^{ij}-\frac{1}{2}g^{ij}R \ \ \ \ \ (2)$

We can write this in a different form that is sometimes easier to use in calculations. Eliminating ${G^{ij}}$ we have

$\displaystyle R^{ij}-\frac{1}{2}g^{ij}R+\Lambda g^{ij}=\kappa T^{ij} \ \ \ \ \ (3)$

Multiplying both sides by ${g_{ij}}$ we get

$\displaystyle g_{ij}R^{ij}-\frac{1}{2}g_{ij}g^{ij}R+\Lambda g_{ij}g^{ij}=\kappa g_{ij}T^{ij} \ \ \ \ \ (4)$

Because the tensor ${g_{ij}}$ is the inverse of ${g^{ij}}$, their product gives the identity matrix of rank 4 (this can be seen by doing the calculation in a local inertial frame where ${g_{ij}=\eta_{ij}}$ and noting that since it’s a tensor equation, it’s valid in all coordinate systems). That is

$\displaystyle g_{ij}g^{jk}=\delta_{\; i}^{k} \ \ \ \ \ (5)$

so if we contract the ${\delta_{\; j}^{k}}$ tensor we just sum up its diagonal elements and since these are all 1 (and there are four rows), we get

$\displaystyle \delta_{\; k}^{k}=4 \ \ \ \ \ (6)$

Returning to 4 we get

 $\displaystyle g_{ij}R^{ij}-2R+4\Lambda$ $\displaystyle =$ $\displaystyle \kappa g_{ij}T^{ij} \ \ \ \ \ (7)$

Since the curvature scalar is given by

$\displaystyle R\equiv g_{ij}R^{ij} \ \ \ \ \ (8)$

and the stress-energy scalar is

$\displaystyle T\equiv g_{ij}T^{ij} \ \ \ \ \ (9)$

we get

$\displaystyle -R+4\Lambda=\kappa T \ \ \ \ \ (10)$

Multiplying this by ${-\frac{1}{2}g^{ij}}$ and subtracting from 3 we have

 $\displaystyle R^{ij}-\Lambda g^{ij}$ $\displaystyle =$ $\displaystyle \kappa\left(T^{ij}-\frac{1}{2}g^{ij}T\right) \ \ \ \ \ (11)$

Isolating the Ricci tensor gives us the alternative form of the Einstein equation:

$\displaystyle \boxed{R^{ij}=\kappa\left(T^{ij}-\frac{1}{2}g^{ij}T\right)+\Lambda g^{ij}} \ \ \ \ \ (12)$

# Einstein tensor and Einstein equation

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 21; Box 21.2.

Our first attempt at constructing a tensor generalization of Newton’s law of gravitation failed because the total derivative of the Ricci tensor is not, in general, zero. Recall that the general idea is to look for an equation of form

$\displaystyle G^{ij}=\kappa T^{ij} \ \ \ \ \ (1)$

where ${G^{ij}}$ depends on the Riemann tensor and the metric. At this stage, there’s no particular reason to select any special form for this tensor, but if possible, we’d like it to be linear in the Riemann tensor. Also remember that ${G^{ij}}$ must satisfy the conditions:

1. It must be symmetric: (${G^{ij}=G^{ji}}$).
2. It must be of rank 2.
3. Its total derivative must be zero: ${\nabla_{j}G^{ij}=0}$.

To that end, let’s try a formula as follows:

$\displaystyle R^{ij}+bg^{ij}R+\Lambda g^{ij} \ \ \ \ \ (2)$

where ${R}$ is the curvature scalar

$\displaystyle R=g^{ab}R_{ab} \ \ \ \ \ (3)$

and ${b}$ and ${\Lambda}$ are constants. Note that although ${R}$ is a scalar, it is not a constant, so we can’t just merge the last two terms.

This form for ${G^{ij}}$ satisfies the first two conditions above, since everything on the RHS is of rank 2 and is also symmetric. So it just remains to show that the total derivative is zero. Since ${\nabla_{j}g^{ij}=0}$ always, the problem reduces to showing that

$\displaystyle \nabla_{j}\left(R^{ij}+bg^{ij}R\right)=0 \ \ \ \ \ (4)$

$\displaystyle \nabla_{s}R_{abmn}+\nabla_{n}R_{absm}+\nabla_{m}R_{abns}=0 \ \ \ \ \ (5)$

Multiplying through by ${g^{gs}g^{am}g^{bn}}$ we get (the metrics can be taken inside the derivatives since their derivatives are zero, so they act as constants):

$\displaystyle \nabla_{s}g^{gs}g^{am}g^{bn}R_{abmn}+\nabla_{n}g^{gs}g^{am}g^{bn}R_{absm}+\nabla_{m}g^{gs}g^{am}g^{bn}R_{abns}=0 \ \ \ \ \ (6)$

In the first term, we have

 $\displaystyle g^{am}g^{bn}R_{abmn}$ $\displaystyle =$ $\displaystyle g^{bn}R_{\;bmn}^{m}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g^{bn}R_{bn}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle R \ \ \ \ \ (9)$

so we get

$\displaystyle \nabla_{s}g^{gs}R+\nabla_{n}g^{gs}g^{am}g^{bn}R_{absm}+\nabla_{m}g^{gs}g^{am}g^{bn}R_{abns}=0 \ \ \ \ \ (10)$

We can now use the symmetries of the Riemann tensor to simplify the last two terms. In the second term, we use ${R_{absm}=R_{smab}=-R_{msab}}$ and in the third term we use ${R_{abns}=R_{nsab}=-R_{nsba}}$:

$\displaystyle \nabla_{s}g^{gs}R-\nabla_{n}g^{gs}g^{am}g^{bn}R_{msab}-\nabla_{m}g^{gs}g^{am}g^{bn}R_{nsba}=0 \ \ \ \ \ (11)$

The Ricci tensor with lowered indices is

$\displaystyle R_{ab}=g^{cd}R_{dacb} \ \ \ \ \ (12)$

so to raise both its indices we have

$\displaystyle R^{gs}=g^{ga}g^{sb}g^{mn}R_{namb} \ \ \ \ \ (13)$

Comparing this with the second term in 11 we can map the indices in that term as follows: ${m\rightarrow n}$, ${s\rightarrow a}$, ${a\rightarrow m}$, ${n\rightarrow s}$ and ${b\rightarrow b}$. Thus the second term is the same as

 $\displaystyle \nabla_{n}g^{gs}g^{am}g^{bn}R_{msab}$ $\displaystyle =$ $\displaystyle \nabla_{s}g^{ga}g^{sb}g^{mn}R_{namb}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \nabla_{s}R^{gs} \ \ \ \ \ (15)$

Similarly, in the third term we can map ${n\rightarrow n}$, ${s\rightarrow a}$, ${b\rightarrow m}$, ${a\rightarrow b}$ and ${m\rightarrow s}$ to get

 $\displaystyle \nabla_{m}g^{gs}g^{am}g^{bn}R_{nsba}$ $\displaystyle =$ $\displaystyle \nabla_{s}g^{ga}g^{sb}g^{mn}R_{namb}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \nabla_{s}R^{gs} \ \ \ \ \ (17)$

Thus 11 becomes

$\displaystyle \nabla_{s}g^{gs}R-2\nabla_{s}R^{gs}=0 \ \ \ \ \ (18)$

or

$\displaystyle \nabla_{s}\left(R^{gs}-\frac{1}{2}g^{gs}R\right)=0 \ \ \ \ \ (19)$

Thus from 2 we can satisfy ${\nabla_{j}\left(R^{ij}+bg^{ij}R+\Lambda g^{ij}\right)=0}$ if ${b=-\frac{1}{2}}$. In practice, ${G^{ij}}$ is defined as just the first two terms:

$\displaystyle \boxed{G^{ij}\equiv R^{ij}-\frac{1}{2}g^{ij}R} \ \ \ \ \ (20)$

This is known as the Einstein tensor and the equation

$\displaystyle \boxed{G^{ij}+\Lambda g^{ij}=\kappa T^{ij}} \ \ \ \ \ (21)$

is the Einstein equation. This is the general relativistic replacement for Newton’s law of gravity.