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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?

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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 $\Gamma^r_{\phi\phi}$.

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  • $\begingroup$ The problem is that in $2\ddot{r} - (2r\dot{\theta}^2 + 2r\sin^2\theta\dot{\phi}^2) = 0$ there are $\dot{\theta}^2$ and $\dot{\phi}^2$. Can't seem to be able to make a straightforward comparison. $\endgroup$ – Spaceman Spiff May 10 '15 at 15:13
  • $\begingroup$ @SpacemanSpiff Remember that in my formula there's a sum over $\nu$ and $\rho$, that's why there are $\dot{\theta}^2$ and $\dot{\phi}^2$ terms. $\endgroup$ – Javier May 10 '15 at 15:39
  • $\begingroup$ I get it now. Once I divide my equation by 2 the coefficients of each first-order terms are simply the Christoffel symbols. Thanks! $\endgroup$ – Spaceman Spiff May 10 '15 at 16:16

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