I was given the following metric for a sphere
$$g_{\mu\nu} = diag(1, r^2, r^2\sin^2\theta)$$
and tasked to calculate the Christoffel symbols. There are 2 ways that I know of to calculate them. One is from the equation
$$2\Gamma^{\alpha}_{\beta\gamma} = g^{\alpha\rho}(g_{\alpha\rho,~\beta} + g_{\beta\rho,~\alpha} -g_{\alpha\beta,~\rho})$$
and the other way is from the Lagrangian $L$.
What puzzles me:
For the 1st method, say if I want to calculate $\Gamma^{r}_{\theta\theta}$, I simply get
$$2\Gamma^{r}_{\theta\theta} = g^{rr}(g_{r\theta,~\theta} + g_{\theta r,~\theta} -g_{\theta\theta,~r})$$
and with only the last term surviving, I get $\Gamma^{r}_{\theta\theta} = -r$. Also, $\Gamma^{r}_{\phi\phi} = -r\sin^2\theta$
For the 2nd method I have $L = g_{\mu\nu}\dot{x}^\mu\dot{x}^\nu = \dot{r}^2 + r^2\dot{\theta}^2 + r^2\sin^2\theta\dot{\phi}^2$, and by Euler-Lagrange equation I get for the radial component
$$2\ddot{r} – (2r\dot{\theta}^2 + 2r\sin^2\theta\dot{\phi}^2) = 0$$
and here is where I do not get it. I have to compare the coefficients with the equation
$$\ddot{r} + \Gamma^{r}_{\theta\theta}\dot{\theta}^2 = 0$$
to extract out $\Gamma^{r}_{\theta\theta}$ but there is this $2r\sin^2\theta\dot{\phi}^2$ term which prevents me from direct comparison. Even if I assume that the equation may actually be a combination of 2 equations namely,
$$a\ddot{r} + 2\Gamma^{r}_{\theta\theta}\dot{\theta}^2 = 0$$
and
$$(2-a)\ddot{r} + 2\Gamma^{r}_{\phi\phi}\dot{\phi}^2 = 0$$
where $a$ is some constant. The only way to retrieve the correct $\Gamma^{r}_{\theta\theta}$ is for $a=2$. In this case I would not be able to retrieve $\Gamma^{r}_{\phi\phi}$. Any clues?
Best Answer
The geodesic equation is
$$ \frac{d^2x^\mu}{dt^2} + \Gamma^\mu_{\nu \rho} \frac{dx^\nu}{dt}\frac{dx^\rho}{dt} = 0$$
The coefficient of $\dot{\phi}^2$ you're seeing corresponds to $\Gamma^r_{\phi\phi}$.