# Metric Connection from Geodesic Equation

1. Mar 31, 2012

### alex3

For the following two-dimensional metric

$$ds^2 = a^2(d\theta^2 + \sin^2{\theta}d\phi^2)$$

using the Euler-Lagrange equations reveal the following equations of motion

$$\ddot{\phi} + 2\frac{\cos{\theta}}{\sin{\theta}}\dot{\theta}\dot{\phi} = 0$$
$$\ddot{\theta} - \sin{\theta}\cos{\theta}\dot{\phi}^2 = 0$$

Using the general geodesic equation form $\ddot{x}^{\alpha} + \Gamma^{\alpha}_{\beta\gamma}\dot{x}^\beta\dot{x}^{\gamma}=0$, we infer that the equations derived describe geodesics. This shows that the only non-zero terms of the metric connection $\Gamma^{\alpha}_{\beta\gamma}$ are

$$\Gamma^{\theta}_{\phi\phi} = -\sin{\theta}\cos{\theta},\quad \Gamma^{\phi}_{\theta\phi} = \Gamma^{\phi}_{\phi\theta} = \frac{\cos{\theta}}{\sin{\theta}}$$

My problem is comprehending where the factor of two has gone for the $\Gamma^{\phi}_{\phi\theta}$ term. Is it due to fact that it's a coefficient of a mixed derivative, why is that?

2. Mar 31, 2012

### tom.stoer

You have to sum over theta and phi, so the geodesic equation contains the two terms

$$\Gamma^{\phi}_{\theta\phi}\dot{\theta}\dot{\phi} + \Gamma^{\phi}_{\phi\theta}\dot{\phi}\dot{\theta}$$

3. Mar 31, 2012

### alex3

Of course! Thank you, it's very clear now.