Christoffel symbols of Schwarzschild metric with Lagrangian

[Stella.Physics]

So the Schwarzschild metric is given by

ds²= -(1-2M/r)dt² + (1-2M/r)^-1dr²+r²dθ²+r²sin²θ dφ²

and the Lagragian is ${\frac{d}{dσ}}[{\frac{1}{L}}{\frac{dx^α}{dσ}}] + {\frac{∂L}{∂x^α}}=0$

with L = dτ/dσ. So for each α=0,1,2,3 we have

${\frac{d^2 x^1}{dτ^2}}=0$ for Minkowski spacetime

also the geadesic equation is ${\frac{d^2 x^a}{dτ^2}}+Γ^a_{bc}{\frac{dx^b}{dτ}}{\frac{dx^c}{dτ}}=0$

So what happens in the Schwarzschild metric and how can I find the Christoffel symbols from the Schwarzschild metric and the geodesic equations?

 

[stevendaryl]

For slower-than-light geodesics, you can use an effective Lagrangian (which is basically the square of the one you use):

\mathcal{L} = \frac{1}{2} g_{\mu \nu} U^\mu U^\nu

where g_{\mu \nu} is a component of the metric tensor, and U^\mu = \frac{dx^\mu}{d\sigma}. This is slightly easier to work with than the one you are using (although it's a little work to see that they lead to the same equations of motion). The Lagrangian equations of motion for this lagrangian are:

\frac{d}{ds} \frac{\partial \mathcal{L}}{\partial U^\mu} - \frac{\partial \mathcal{L}}{\partial x^\mu} = 0

This gives the following form of the geodesic equation:

g_{\mu \nu} \frac{dU^\nu}{d\sigma} + [\frac{\partial g_{\mu \nu}}{\partial x^\lambda} - \frac{1}{2} \frac{\partial g_{\lambda \nu}}{\partial x^\mu}] U^\lambda U^\nu = 0

Unfortunately, the quantity in square brackets (call it Q_{\mu \nu \lambda}) is not exactly equal to the Christoffel symbol \Gamma_{\mu \nu \lambda} \equiv g_{\mu \alpha} \Gamma^\alpha_{\nu \lambda} = \frac{1}{2}(\frac{\partial g_{\mu\nu}}{\partial x^\lambda} + \frac{\partial g_{\mu\lambda}}{\partial x^\nu}- \frac{\partial g_{\lambda \nu}}{\partial x^\mu}). However, if you compute Q_{\mu \nu \lambda}, then you can find \Gamma_{\mu \nu \lambda} by symmetrizing:

\Gamma_{\mu \nu \lambda} = \frac{1}{2} [Q_{\mu \nu \lambda} + Q_{\mu \lambda \nu}]