# Christoffel symbol for Schwarzschild metric

I know that the christoffel (second kind) can be defined like this:

$$\Gamma^m_{ij} = \frac{1}{2} g^{mk}(\frac{\partial g_{ki}}{\partial U^j}+\frac{\partial g_{jk}}{\partial U^i}-\frac{\partial g_{ij}}{\partial U^k})$$

but I don't know how $U^i$ is defined (specifically for the Schwarzschild metric.

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I'm not sure what you're asking. $U^0 = t$, $U^1 = r$, $U^2 = \theta$, and $U^3 = \phi$ in the typical order with conventional coordinate labels in the Schwarzschild metric. Well, you can permute them any way you like, as long as you keep the metric tensor components consistent. – Stan Liou May 10 '13 at 1:30
So in that case, I can just differentiate with respect to t, r theta and phi and get the christoffel symbol? – user912 May 10 '13 at 1:32
Yes, just remember the Einstein summation convention when doing so. To be clear: the Schwarzschild metric is diagonal, in $-+++$ convention $g_{00} = -(1-2M/r)$, ..., $g_{33} = r^2\sin^2\theta$, the contravariant components $g^{mk}$ can be found through matrix inverse (so in this particular case they're just straight inverses because of diagonality). – Stan Liou May 10 '13 at 1:36
So $g^{00} = -1/(1-2M/r)$? – user912 May 10 '13 at 1:38
Yup. Diagonal metrics are nice that way. – Stan Liou May 10 '13 at 1:40

The Schwarzschild metric is, in $-+++$ sign convention and units of $c = 1$ is $$\mathrm{d}s^2 = -\left(1-\frac{2M}{r}\right)\mathrm{d}t^2 + \frac{\mathrm{d}r^2}{1-\frac{2M}{r}} + r^2\left(\mathrm{d}\theta^2 + \sin^2\theta\,\mathrm{d}\phi^2\right)\text{.}$$ We can index the coordinates arbitrarily, but let's take them in the typical order: $(U^0,U^1,U^2,U^3) = (t,r,\theta,\phi)$. In the metric, terms like $\mathrm{d}t^2$ are shorthand for the tensor product $\mathrm{d}t\otimes\mathrm{d}t$ and cross-terms like $\mathrm{d}t\,\mathrm{d}r$ for $\frac{1}{2}\left(\mathrm{d}t\otimes\mathrm{d}r+\mathrm{d}r\otimes\mathrm{d}t\right)$, since the metric must be symmetric. But we don't have any cross-terms, so the covariant metric components form a diagonal matrix: $$g_{ij} = \begin{bmatrix}-\left(1-\frac{2M}{r}\right) &0 &0 &0\\ 0&\left(1-\frac{2M}{r}\right)^{-1} &0&0\\0&0&r^2&0\\0&0&0&r^2\sin^2\theta\\ \end{bmatrix}\text{,}$$ while the contravariant matrix components $g^{ij}$ form the matrix inverse of the above, which in the case of diagonality just simplifies to $g^{ij} = 1/g_{ij}$ if $i = j$ and $g^{ij} = 0$ otherwise.
Write $_{,n}$ for the partial derivative with respect to $U^n$. Then the connection coefficients / Christoffel symbols $$\Gamma^m_{ij} = \frac{1}{2}g^{mk}\left[g_{ki,j} + g_{kj,i} - g_{ij,k}\right]$$ simplify in the diagonal case to just $$\Gamma^m_{ij} = \frac{1}{2}g^{mm}\left[g_{mi,j} + g_{mj,i} - g_{ij,m}\right]\text{,}$$ since terms with $m\neq k$ have $g^{mk} = 0$, and in the Schwarzschild case the metric components are independent of either $t$ or $\phi$, so: $$\begin{eqnarray*} g_{ij,0} = \frac{\partial g_{ij}}{\partial t} = 0&\quad\text{and}\quad& g_{ij,3} = \frac{\partial g_{ij}}{\partial \phi} = 0\text{.} \end{eqnarray*}$$ Thus for the Schwarszchild case, any off-diagonal term is zero and all partials by $t$ or $\phi$ are also zero. For example, $$\begin{eqnarray*} \Gamma^\phi_{\theta\phi} = \Gamma^3_{23} &= \frac{1}{2}\underbrace{\left(r^2\sin^2\theta\right)^{-1}}_{g^{33}}\left(\underbrace{g_{32,3}}_0 + g_{33,2} - \underbrace{g_{23,3}}_0\right)\\ &= \frac{1}{2}\left(r^2\sin^2\theta\right)^{-1}\underbrace{\left(r^2\cdot2\sin\theta\cos\theta\right)}_{\partial_\theta(r^2\sin^2\theta)}= \cot\theta\text{.}\end{eqnarray*}$$ $$\Gamma^\phi_{r\phi} = \Gamma^3_{13} = \ldots = r^{-1}\text{,}$$ and other than terms required by symmetry of lower indices ($\Gamma^\phi_{\phi\theta} = \Gamma^\phi_{\theta\phi}$, &c.), those are the only nonzero $\Gamma^\phi_{ij}$. You should be able to find the rest yourself.
Out of curiosity, would this give me the components that I'd need to calculate, or should I still check if they're nonzero myself (I noticed that the matrix for $k = 0$ has a bunch of terms that aren't 0). – user912 May 11 '13 at 11:29
I meant to say $m = 0$ – user912 May 11 '13 at 11:40