Timelike Geodesic and Christoffel Symbols

[wam_mi]

Homework Statement

How do I show the following metric have time-like geodesics, if \theta and R are constants

ds^{2} = R^{2} (-dt^{2} + (cosh(t))^{2} d\theta^{2})

Homework Equations

v^{a}v_{a} = -1 for time-like geodesic, where v^{a} is the tangent vector along the curve

The Attempt at a Solution

First, I write it as the Lagrangian

L = -R^{2}\dot{t}^{2} + (cosh(t))^{2} \dot{\theta}^{2} = -R^{2}\dot{t}^{2}

as \theta is a constant.

How do I proceed to show that this indeed gives us a time-like geodesic.

Could someone also tell me if I have computed the Christoffel symbol components correctly? My result is

\Gamma^{t}_{\theta \theta} = 0-sinh(t) \times cosh(t)

\Gamma^{\theta}_{t \theta} = tanh(t)

and all other components vanish.

Cheers!

P.S. How do I type minus sign? It doesn't seem to work if I have left the 0 out at above.

 

[NanakiXIII]

Maybe I'm not understanding the problem correctly, but if you're holding \theta constant, couldn't you just use d\theta = 0?

 

[diazona]

P.S. How do I type minus sign? It doesn't seem to work if I have left the 0 out at above.

\Gamma^{t}_{\theta \theta} = -sinh(t) \times cosh(t)
Works fine for me
Anyway, I get the same thing as you for the Christoffel symbols except without the minus sign. I could have made a mistake but you might want to recheck your calculations.

 

[gabbagabbahey]

I don't see the need to compute the Christoffel coefficients at all. The solution to the Euler-Lagrange equations will be a geodesic, so if R and \theta are constants, you want to solve

\frac{\partial L}{\partial t}-\frac{d}{d\tau}\frac{\partial L}{\partial \dot{t}}=0

(Where I'm using \tau to represent your affine parameter...i.e. \dot{t}=\frac{dt}{d\tau} )