2

Given a sphere, find the christoffel symbols

$X(\theta,\phi)=(r \sin\phi \cos\theta,r \sin\phi \sin\theta, r \cos \phi)$

So $$(g_{ij})=\begin{pmatrix} r^2\sin^2\phi& 0\\ 0& r^2 \end{pmatrix}$$

$$(g^{ij})=\frac{1}{r^4 \sin^2 \phi}\begin{pmatrix} r^2& 0\\ 0& r^2\sin^2\phi\end{pmatrix}=\begin{pmatrix} {\frac{1}{r^2 \sin^2 \phi}} & 0\\ 0& \frac{1}{r^2}\end{pmatrix}$$

$$g_{\{ij;1\}}=\begin{pmatrix} 0& 0\\ 0& 0 \end{pmatrix}$$

$$g_{\{ij;2\}}=\begin{pmatrix} 2r^2\sin \phi \cos \phi& 0\\ 0& 0 \end{pmatrix}$$

So $$\Gamma^1_{22}=\frac{1}{2}\frac{1}{r^2\sin^2 \phi}(0+0-0)+\frac{1}{2}\cdot0(0+0-0)=0$$

But the answer is $$\Gamma^1_{22}=-\sin \phi \cos \phi$$

Community
  • 1
gbox
  • 12,867

1 Answers1

2

I think the Christoffel symbol you want is $\Gamma_{\theta\theta}^{\phi}$, which is $$\begin{align}\Gamma_{\theta\theta}^{\phi}&=\frac{1}{2}g^{\phi k}(\partial_{\theta}g_{\theta k}+\partial_{\theta}g_{\theta k}-\partial_{k}g_{\theta\theta})\\ &=\frac{1}{2}g^{\phi\phi}(\partial_{\theta}g_{\theta\phi}+\partial_{\theta}g_{\theta\phi}-\partial_{\phi}g_{\theta\theta})\\ &=\frac{1}{2}g^{\phi\phi}(0+0-\partial_{\phi}g_{\theta\theta})\\ &=-\frac{1}{2}\cdot\frac{1}{r^2}\cdot2r^2\sin\phi\cos\phi\\ &=-\sin\phi\cos\phi \end{align}$$

A. Goodier
  • 10,964