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

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  15. 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.

    Reply
  16. 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.

    Reply

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