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. $$\dot{t}=\frac{dt}{d\tau}$$ )

 

[paranormal]

To show that the metric has time-like geodesics, you can use the geodesic equation:

\frac{d^2x^{\mu}}{d\lambda^2} + \Gamma^{\mu}_{\alpha\beta}\frac{dx^{\alpha}}{d\lambda}\frac{dx^{\beta}}{d\lambda} = 0

where \lambda is an affine parameter. For a time-like geodesic, we have v^{\mu}v_{\mu} = -1, so we can set v^t = \frac{dt}{d\lambda} = \sqrt{-\frac{1}{R^2}} and v^{\theta} = \frac{d\theta}{d\lambda} = \frac{cosh(t)}{R}. Plugging these into the geodesic equation, we get:

\frac{d^2t}{d\lambda^2} + \Gamma^{t}_{\theta\theta}\frac{dt}{d\lambda}\frac{d\theta}{d\lambda} + \Gamma^{t}_{tt}\left(\frac{dt}{d\lambda}\right)^2 = 0

\frac{d^2\theta}{d\lambda^2} + \Gamma^{\theta}_{t\theta}\frac{dt}{d\lambda}\frac{d\theta}{d\lambda} + \Gamma^{\theta}_{\theta\theta}\left(\frac{d\theta}{d\lambda}\right)^2 = 0

Plugging in the values for the Christoffel symbols that you calculated, we get:

\frac{d^2t}{d\lambda^2} + \frac{1}{R}\frac{dt}{d\lambda}\frac{cosh(t)}{R} = 0

\frac{d^2\theta}{d\lambda^2} + \frac{1}{R}\frac{dt}{d\lambda}\frac{dt}{d\lambda}\frac{cosh(t)}{R} = 0

Simplifying, we get:

\frac{d^2t}{d\lambda^2} + \frac{1}{R^2}cosh(t) = 0

\frac{d^2\theta}{d\lambda^2} + \frac{1}{