What is the Lagrangian for geodesics in Schwarzschild coordinates?

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
3 replies · 4K views
Demon117
Messages
162
Reaction score
1
For pure interest I have been trying to solve for the geodesics of the Schwarzschild metric. To do so I know I need to find the explicit Lagrangian for the variational principle for geodesics in this spacetime in Schwarzschild coordinates. How do I derive this lagrangian?

I know that the proper time along a timelike world line between two points in spacetime is

[tex]\sqrt {[-g_{{\alpha \beta }} \left( x \right) {{\it dx}}^{\alpha}{{<br /> \it dx}}^{\beta}]} \left( B-A \right)[/tex]

But how do I use this and what does it end up telling me?
 
Physics news on Phys.org
You can certainly do it that way, but why not skip a few steps and write down the geodesic equations immediately using Christoffel symbols, as described in any GR textbook.
 
A good explanation of this variational process can be found in Carroll (free online), ch3 I think although not 100% sure. It's not a Lagrangian your after here, it's just that the geodesics are "shortest distance", or as Carroll does maximize proper time...so you want to use Calculus of variations techniques to extrematise the functional you have written. You often get the Christoffel symbols by doing this process too.
 
You can solve the euler lagrange equations for the geodesics as extremal paths using the line element with factors of 1/2 which give you the geodesic equations or you can just find the Christoffel symbols based on its definition in terms of the derivatives of the metric tensor. Its up to you really but the first method is less time consuming for me at least.