Metric Connection from Geodesic Equation

[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?

 

[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}

 

[alex3]

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