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
This expression can be cumbersome to work with, since it involves calculating the inverse metric tensor 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 ):
The following equation is formally equivalent to this:
The method for calculating the Christoffel symbols is to work out the terms in 2, divide through by and then compare the result term by term with 3. By doing this we are able to read off the as the coefficients of in 2.
Example We can use this technique to work out the for the Schwarzschild (S) metric, which is
First, take in 2. Since the S metric doesn’t depend on for all elements. Further, since the S metric is diagonal, is restricted to , so the equation becomes
Since there are 2 non-zero derivatives, so this equation expands to
By comparing this with 3 we can read off the symbols:
Here, we’ve used the symmetry of the Christoffel symbols. Because no other terms appear in the equation, all the other are zero, so the complete set is, where the rows are labelled , , and from top to bottom, and the columns the same order from left to right:
Now consider . This time, one of the does depend on so we will get a contribution from the term. We get
Again, we can read off the symbols to get
For things get a bit messier since all four terms depend on . We get
Comparing terms, we get
Finally, for the metric is again independent of so the situation is a lot simpler:
The symbols are