Christoffel symbols in terms of the metric tensor

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 17; Box 17.4.

It’s time to find out how to calculate the Christoffel symbols. We start with their definition in terms of the basis vectors in some coordinate system:

\displaystyle  \boxed{\partial_{j}\mathbf{e}_{i}=\Gamma_{\;ij}^{k}\mathbf{e}_{k}} \ \ \ \ \ (1)

We take the scalar product of this equation with another basis vector {\mathbf{e}_{l}} and use the definition of the metric tensor as {g_{ij}=\mathbf{e}_{i}\cdot\mathbf{e}_{j}}:

\displaystyle   \Gamma_{\;ij}^{k}\mathbf{e}_{k}\cdot\mathbf{e}_{l} \displaystyle  = \displaystyle  \left(\partial_{j}\mathbf{e}_{i}\right)\cdot\mathbf{e}_{l}\ \ \ \ \ (2)
\displaystyle  \Gamma_{\;ij}^{k}g_{kl} \displaystyle  = \displaystyle  \partial_{j}\left(\mathbf{e}_{i}\cdot\mathbf{e}_{l}\right)-\left(\partial_{j}\mathbf{e}_{l}\right)\cdot\mathbf{e}_{i}\ \ \ \ \ (3)
\displaystyle  \displaystyle  = \displaystyle  \partial_{j}g_{il}-\Gamma_{\;lj}^{k}\mathbf{e}_{k}\cdot\mathbf{e}_{i}\ \ \ \ \ (4)
\displaystyle  \displaystyle  = \displaystyle  \partial_{j}g_{il}-\Gamma_{\;lj}^{k}g_{ki}\ \ \ \ \ (5)
\displaystyle  \Gamma_{\;ij}^{k}g_{kl}+\Gamma_{\;lj}^{k}g_{ki} \displaystyle  = \displaystyle  \partial_{j}g_{il} \ \ \ \ \ (6)

In this equation the index {k} is a dummy (being summed over), so only the indices {i}, {j} and {l} are specified. We can cyclically permute these indices to generate two more equations:

\displaystyle   \Gamma_{\;jl}^{k}g_{ki}+\Gamma_{\;il}^{k}g_{kj} \displaystyle  = \displaystyle  \partial_{l}g_{ji}\ \ \ \ \ (7)
\displaystyle  \Gamma_{\;li}^{k}g_{kj}+\Gamma_{\;ji}^{k}g_{kl} \displaystyle  = \displaystyle  \partial_{i}g_{lj} \ \ \ \ \ (8)

We can now use the symmetry of the Christoffel symbols to solve for {\Gamma_{\;ij}^{k}} by swapping indices in 7 and 8 to get

\displaystyle   \Gamma_{\;lj}^{k}g_{ki}+\Gamma_{\;il}^{k}g_{kj} \displaystyle  = \displaystyle  \partial_{l}g_{ji}\ \ \ \ \ (9)
\displaystyle  \Gamma_{\;il}^{k}g_{kj}+\Gamma_{\;ij}^{k}g_{kl} \displaystyle  = \displaystyle  \partial_{i}g_{lj} \ \ \ \ \ (10)

We can now add 6 to 10 and subtract 9 to get

\displaystyle  2\Gamma_{\;ij}^{k}g_{kl}=\partial_{j}g_{il}+\partial_{i}g_{lj}-\partial_{l}g_{ji} \ \ \ \ \ (11)

Finally we can use the fact that

\displaystyle  g^{ij}g_{jk}=\delta_{\;\;k}^{i} \ \ \ \ \ (12)

and multiply both sides of 11 by {g^{ml}} to get

\displaystyle   2\Gamma_{\;ij}^{k}g_{kl}g^{ml} \displaystyle  = \displaystyle  g^{ml}\left(\partial_{j}g_{il}+\partial_{i}g_{lj}-\partial_{l}g_{ji}\right)\ \ \ \ \ (13)
\displaystyle  \Gamma_{\;ij}^{k}\delta_{\;\;k}^{m} \displaystyle  = \displaystyle  \frac{1}{2}g^{ml}\left(\partial_{j}g_{il}+\partial_{i}g_{lj}-\partial_{l}g_{ji}\right)\ \ \ \ \ (14)
\displaystyle  \Gamma_{\;ij}^{m} \displaystyle  = \displaystyle  \frac{1}{2}g^{ml}\left(\partial_{j}g_{il}+\partial_{i}g_{lj}-\partial_{l}g_{ji}\right) \ \ \ \ \ (15)

This gives us a formula for explicitly evaluating Christoffel symbols:

\displaystyle  \boxed{\Gamma_{\;ij}^{m}=\frac{1}{2}g^{ml}\left(\partial_{j}g_{il}+\partial_{i}g_{lj}-\partial_{l}g_{ji}\right)} \ \ \ \ \ (16)

This is a bit cumbersome to use as it requires finding the inverse metric tensor {g^{ml}} and has 3 sums over different derivatives.

Example As an example, we’ll work out {\Gamma_{\;ij}^{m}} for 2-D polar coordinates. The metric tensor and its inverse here are:

\displaystyle   g_{ij} \displaystyle  = \displaystyle  \left[\begin{array}{cc} 1 & 0\\ 0 & r^{2} \end{array}\right]\ \ \ \ \ (17)
\displaystyle  g^{ij} \displaystyle  = \displaystyle  \left[\begin{array}{cc} 1 & 0\\ 0 & r^{-2} \end{array}\right] \ \ \ \ \ (18)

so the derivatives are

\displaystyle   \partial_{r}g_{ij} \displaystyle  = \displaystyle  \left[\begin{array}{cc} 0 & 0\\ 0 & 2r \end{array}\right]\ \ \ \ \ (19)
\displaystyle  \partial_{\theta}g_{ij} \displaystyle  = \displaystyle  \left[\begin{array}{cc} 0 & 0\\ 0 & 0 \end{array}\right] \ \ \ \ \ (20)

Having only one non-zero derivative helps a lot, since the only non-zero term on the RHS of 16 is {\partial_{r}g_{\theta\theta}=2r}. We’ll work out a couple of the symbols explicitly and then give the final result.

\displaystyle   \Gamma_{\;r\theta}^{\theta} \displaystyle  = \displaystyle  \frac{1}{2}g^{\theta l}\left(\partial_{\theta}g_{rl}+\partial_{r}g_{l\theta}-\partial_{l}g_{\theta r}\right)\ \ \ \ \ (21)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{2}g^{\theta\theta}\left(\partial_{\theta}g_{r\theta}+\partial_{r}g_{\theta\theta}-\partial_{\theta}g_{\theta r}\right)\ \ \ \ \ (22)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{2r^{2}}\left(2r\right)\ \ \ \ \ (23)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{r}\ \ \ \ \ (24)
\displaystyle  \displaystyle  = \displaystyle  \Gamma_{\;\theta r}^{\theta}\ \ \ \ \ (25)
\displaystyle  \Gamma_{\;\theta\theta}^{r} \displaystyle  = \displaystyle  \frac{1}{2}g^{rl}\left(\partial_{\theta}g_{\theta l}+\partial_{\theta}g_{l\theta}-\partial_{l}g_{\theta\theta}\right)\ \ \ \ \ (26)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{2}g^{rr}\left(\partial_{\theta}g_{\theta r}+\partial_{\theta}g_{r\theta}-\partial_{r}g_{\theta\theta}\right)\ \ \ \ \ (27)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{2}(-2r)\ \ \ \ \ (28)
\displaystyle  \displaystyle  = \displaystyle  -r \ \ \ \ \ (29)

All of the other symbols are zero. The final results are

\displaystyle   \Gamma_{\;ij}^{r} \displaystyle  = \displaystyle  \left[\begin{array}{cc} 0 & 0\\ 0 & -r \end{array}\right]\ \ \ \ \ (30)
\displaystyle  \Gamma_{\;ij}^{\theta} \displaystyle  = \displaystyle  \left[\begin{array}{cc} 0 & r^{-1}\\ r^{-1} & 0 \end{array}\right] \ \ \ \ \ (31)

Using these in 1 gives the 4 derivatives of the basis vectors:

\displaystyle   \partial_{r}\mathbf{e}_{r} \displaystyle  = \displaystyle  \Gamma_{\;rr}^{i}\mathbf{e}_{i}=0\ \ \ \ \ (32)
\displaystyle  \partial_{\theta}\mathbf{e}_{r} \displaystyle  = \displaystyle  \Gamma_{\;r\theta}^{i}\mathbf{e}_{i}=\frac{1}{r}\mathbf{e}_{\theta}\ \ \ \ \ (33)
\displaystyle  \partial_{r}\mathbf{e}_{\theta} \displaystyle  = \displaystyle  \Gamma_{\;\theta r}^{i}\mathbf{e}_{i}=\frac{1}{r}\mathbf{e}_{\theta}\ \ \ \ \ (34)
\displaystyle  \partial_{\theta}\mathbf{e}_{\theta} \displaystyle  = \displaystyle  \Gamma_{\;\theta\theta}^{i}\mathbf{e}_{i}=-r\mathbf{e}_{r} \ \ \ \ \ (35)

12 thoughts on “Christoffel symbols in terms of the metric tensor

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  8. Collin Witt

    Can you please clarify step (3). What happened from step to (2) to (3). Looks like a derivative? The funny part is I understand the rest from that point.

    Reply
      1. Collin Witt

        Yea, I figured out that it’s the derivative on the first term of the RHS side and then if you subtract you get what is above in step (2). Thanks for clarifying, it was not immediately clear.

        Reply

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