# Christoffel symbols for a general diagonal metric

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Appendix.

In the appendix to Moore’s book, he gives some worksheets which can be used to calculate Christoffel symbols and the Ricci tensor for a general diagonal metric. I thought it would be useful to show where these formulas come from.

The key to calculating Christoffel symbols is to compare two forms of the geodesic equation. Suppose our diagonal metric is

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

where ${A,B,C}$ and ${D}$ are positive functions of the coordinates. Note the minus sign in the first term; this ensures that ${x^{0}}$ is a time coordinate.

The two forms of the geodesic equation are

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

and in terms of the Christoffel symbols:

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

For a diagonal metric, 2 becomes (no sum over ${a}$):

$\displaystyle g_{aa}\ddot{x}^{a}+\sum_{i}\left[\partial_{i}g_{aa}\dot{x}^{a}\dot{x}^{i}-\frac{1}{2}\partial_{a}g_{ii}\left(\dot{x}^{i}\right)^{2}\right]=0 \ \ \ \ \ (4)$

Therefore, for ${a=0}$ we have, using the notation ${A_{i}\equiv\partial_{i}A}$:

$\displaystyle -A\ddot{x}^{0}-\sum_{i}A_{i}\dot{x}^{a}\dot{x}^{i}-\frac{1}{2}\left[-A_{0}\left(\dot{x}^{0}\right)^{2}+B_{0}\left(\dot{x}^{1}\right)^{2}+C_{0}\left(\dot{x}^{2}\right)^{2}+D_{0}\left(\dot{x}^{3}\right)^{2}\right]=0 \ \ \ \ \ (5)$

Dividing through by ${-A}$ we get

$\displaystyle \ddot{x}^{0}+\frac{1}{A}\sum_{i}A_{i}\dot{x}^{0}\dot{x}^{i}+\frac{1}{2A}\left[-A_{0}\left(\dot{x}^{0}\right)^{2}+B_{0}\left(\dot{x}^{1}\right)^{2}+C_{0}\left(\dot{x}^{2}\right)^{2}+D_{0}\left(\dot{x}^{3}\right)^{2}\right]=0 \ \ \ \ \ (6)$

We can now compare this term by term with 3 to read off the Christoffel symbols ${\Gamma_{\; ij}^{0}}$:

 $\displaystyle \Gamma_{\;00}^{0}$ $\displaystyle =$ $\displaystyle \frac{A_{0}}{A}-\frac{A_{0}}{2A}=\frac{A_{0}}{2A}\ \ \ \ \ (7)$ $\displaystyle \Gamma_{\;01}^{0}$ $\displaystyle =$ $\displaystyle \Gamma_{\;10}^{0}=\frac{A_{0}}{2A}\ \ \ \ \ (8)$ $\displaystyle \Gamma_{\;11}^{0}$ $\displaystyle =$ $\displaystyle \frac{B_{0}}{2A} \ \ \ \ \ (9)$

and so on. The off-diagonal symbols such as ${\Gamma_{\;01}^{0}=\Gamma_{\;10}^{0}}$ are obtained from the ${\frac{1}{A}\sum_{i}A_{i}\dot{x}^{0}\dot{x}^{i}}$ term, remembering that each term in the sum contributes to two Christoffel symbols. For example

$\displaystyle \frac{1}{A}A_{1}\dot{x}^{0}\dot{x}^{1}=\left(\Gamma_{\;01}^{0}+\Gamma_{\;10}^{0}\right)\dot{x}^{0}\dot{x}^{1} \ \ \ \ \ (10)$

The complete worksheet for the Christoffel symbols and Ricci tensor is available as a PDF from Moore’s website.