# Christoffel symbols for Schwarzschild metric

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 17; Box 17.6, Problem P17.2.

We’ve seen that the Christoffel symbols in terms of the metric are given by

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

This expression can be cumbersome to work with, since it involves calculating the inverse metric tensor ${g^{ml}}$ and doing a lot of sums to find each Christoffel symbol. Often an easier way is to exploit the relation between the Christoffel symbols and the geodesic equation.

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

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

The following equation is formally equivalent to this:

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

The method for calculating the Christoffel symbols is to work out the terms in 2, divide through by ${g_{aj}}$ and then compare the result term by term with 3. By doing this we are able to read off the ${\Gamma_{\;ij}^{m}}$ as the coefficients of ${\dot{x}^{j}\dot{x}^{i}}$ in 2.

Example We can use this technique to work out the ${\Gamma_{\;ij}^{m}}$ for the Schwarzschild (S) metric, which is

$\displaystyle ds^{2}=-\left(1-\frac{2GM}{r}\right)dt^{2}+\left(1-\frac{2GM}{r}\right)^{-1}dr^{2}+r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2} \ \ \ \ \ (4)$

First, take ${a=\phi}$ in 2. Since the S metric doesn’t depend on ${\phi}$ ${\partial_{\phi}g_{ij}=0}$ for all elements. Further, since the S metric is diagonal, ${g_{\phi j}}$ is restricted to ${g_{\phi\phi}}$, so the equation becomes

$\displaystyle g_{\phi\phi}\ddot{\phi}+\partial_{i}g_{\phi\phi}\dot{\phi}\dot{x}^{i}=0 \ \ \ \ \ (5)$

Since ${g_{\phi\phi}=r^{2}\sin^{2}\theta}$ there are 2 non-zero derivatives, so this equation expands to

 $\displaystyle r^{2}\sin^{2}\theta\ddot{\phi}+2r\sin^{2}\theta\dot{\phi}\dot{r}+2r^{2}\sin\theta\cos\theta\dot{\phi}\dot{\theta}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (6)$ $\displaystyle \ddot{\phi}+\frac{2}{r}\dot{\phi}\dot{r}+2\cot\theta\dot{\phi}\dot{\theta}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (7)$

By comparing this with 3 we can read off the symbols:

 $\displaystyle \Gamma_{\;r\phi}^{\phi}+\Gamma_{\;\phi r}^{\phi}$ $\displaystyle =$ $\displaystyle \frac{2}{r}\ \ \ \ \ (8)$ $\displaystyle \Gamma_{\;r\phi}^{\phi}=\Gamma_{\;\phi r}^{\phi}$ $\displaystyle =$ $\displaystyle \frac{1}{r}\ \ \ \ \ (9)$ $\displaystyle \Gamma_{\;\theta\phi}^{\phi}+\Gamma_{\;\phi\theta}^{\phi}$ $\displaystyle =$ $\displaystyle 2\cot\theta\ \ \ \ \ (10)$ $\displaystyle \Gamma_{\;\theta\phi}^{\phi}=\Gamma_{\;\phi\theta}^{\phi}$ $\displaystyle =$ $\displaystyle \cot\theta \ \ \ \ \ (11)$

Here, we’ve used the symmetry of the Christoffel symbols. Because no other terms appear in the equation, all the other ${\Gamma_{\;ij}^{\phi}}$ are zero, so the complete set is, where the rows are labelled ${t}$, ${r}$, ${\theta}$ and ${\phi}$ from top to bottom, and the columns the same order from left to right:

$\displaystyle \Gamma_{\;ij}^{\phi}=\left[\begin{array}{cccc} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & \frac{1}{r}\\ 0 & 0 & 0 & \cot\theta\\ 0 & \frac{1}{r} & \cot\theta & 0 \end{array}\right] \ \ \ \ \ (12)$

Now consider ${a=\theta}$. This time, one of the ${g_{ij}}$ does depend on ${\theta}$ so we will get a contribution from the ${\partial_{\theta}g_{\phi\phi}}$ term. We get

 $\displaystyle r^{2}\ddot{\theta}+2r\dot{r}\dot{\theta}-\frac{1}{2}r^{2}\left(2\sin\theta\cos\theta\right)\dot{\phi}^{2}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (13)$ $\displaystyle \ddot{\theta}+\frac{2}{r}\dot{r}\dot{\theta}-\sin\theta\cos\theta\dot{\phi}^{2}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (14)$

Again, we can read off the symbols to get

$\displaystyle \Gamma_{\;ij}^{\theta}=\left[\begin{array}{cccc} 0 & 0 & 0 & 0\\ 0 & 0 & \frac{1}{r} & 0\\ 0 & \frac{1}{r} & 0 & 0\\ 0 & 0 & 0 & -\sin\theta\cos\theta \end{array}\right] \ \ \ \ \ (15)$

For ${a=r}$ things get a bit messier since all four ${g_{ij}}$ terms depend on ${r}$. We get

 $\displaystyle 0$ $\displaystyle =$ $\displaystyle \left(1-\frac{2GM}{r}\right)^{-1}\ddot{r}-\frac{2GM}{r^{2}}\left(1-\frac{2GM}{r}\right)^{-2}\dot{r}^{2}-\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{1}{2}\left[-\frac{2GM}{r^{2}}\dot{t}^{2}-\frac{2GM}{r^{2}}\left(1-\frac{2GM}{r}\right)^{-2}\dot{r}^{2}+2r\dot{\theta}^{2}+2r\sin^{2}\theta\dot{\phi}^{2}\right]\nonumber$ $\displaystyle 0$ $\displaystyle =$ $\displaystyle \ddot{r}+\frac{GM}{r^{2}}\left(1-\frac{2GM}{r}\right)\dot{t}^{2}-\frac{GM}{r^{2}}\left(1-\frac{2GM}{r}\right)^{-1}\dot{r}^{2}-\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle$ $\displaystyle r\left(1-\frac{2GM}{r}\right)\dot{\theta}^{2}-r\sin^{2}\theta\left(1-\frac{2GM}{r}\right)\dot{\phi}^{2}\nonumber$

Comparing terms, we get

$\displaystyle \Gamma_{\;ij}^{r}=\left[\begin{array}{cccc} \frac{GM}{r^{2}}\left(1-\frac{2GM}{r}\right) & 0 & 0 & 0\\ 0 & -\frac{GM}{r^{2}}\left(1-\frac{2GM}{r}\right)^{-1} & 0 & 0\\ 0 & 0 & -r\left(1-\frac{2GM}{r}\right) & 0\\ 0 & 0 & 0 & -r\sin^{2}\theta\left(1-\frac{2GM}{r}\right) \end{array}\right] \ \ \ \ \ (18)$

Finally, for ${a=t}$ the metric is again independent of ${t}$ so the situation is a lot simpler:

 $\displaystyle -\left(1-\frac{2GM}{r}\right)\ddot{t}-\frac{2GM}{r^{2}}\dot{r}\dot{t}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (19)$ $\displaystyle \ddot{t}+\frac{2GM}{r^{2}}\left(1-\frac{2GM}{r}\right)^{-1}\dot{r}\dot{t}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (20)$

The symbols are

$\displaystyle \Gamma_{\;ij}^{t}=\left[\begin{array}{cccc} 0 & \frac{GM}{r^{2}}\left(1-\frac{2GM}{r}\right)^{-1} & 0 & 0\\ \frac{GM}{r^{2}}\left(1-\frac{2GM}{r}\right)^{-1} & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{array}\right] \ \ \ \ \ (21)$

## 20 thoughts on “Christoffel symbols for Schwarzschild metric”

1. Avishek Dusoye

I find this technique pretty cool. At the same time, i found a typo, in equation (17), the coefficient of r_dot|r_dot should be of power of -1, instead of -2, as we multiply equation (16) by (1-2GM/(r^2)) throughout on both sides. Cheers take care.

2. Enrique

This method is very interesting. By the way, in equations 10 and 11 there’s a Christoffel symbol that has r, theta, I think it should be theta,phi instead. Cheers.