How to Derive Christoffel Symbols for a Given Metric in General Relativity?

[Astrofiend]

Homework Statement

This is a problem in General Relativity, where I am trying to find the Christoffel symbols that correspond to a given metric. Any help would be greatly appreciated!

OK. I have been given the metric

$$<br /> ds^2 = (1+gx)^2 dt^2 - dx^2 - dy^2 - dz^2<br />$$

and have been told to show that the non-vanishing Christoffel symbols are given by

$$<br /> \Gamma^1 _0 _0 = g(1+gx)<br />$$

and

$$<br /> \Gamma^0 _1 _0 = \Gamma^0 _0 _1 = \frac{g}{(1+gx)}<br />$$

I can get the first out OK but I'm not getting how to get the second.

The Attempt at a Solution

So I wrote the Lagrangian out for the metric:

$$<br /> L\left(\frac{dx^\alpha}{d\sigma},x^\alpha\right) = - (1+gx)^2\frac{dt}{d\sigma}^2 +\frac{dx}{d\sigma}^2+\frac{dy}{d\sigma}^2+\frac{dz}{d\sigma}^2<br />$$

which, after a little substitution work getting taus instead of sigmas, basically gives us the following equation:

$$<br /> \frac{d^2x}{d\tau^2} = -\frac{1}{2} . 2(1+gx).g \frac{dt}{d\tau}^2<br />$$

...and comparing this to the definition of a C-symbol

$$\frac{d^2 x^\alpha}{d\tau^2} = -\Gamma ^\alpha _\beta _\gamma \frac{dx^\beta} {d\tau} \frac{dx^\gamma}{d\tau}$$

...it is pretty easy to see that

$$<br /> \Gamma^1 _0 _0 = g(1+gx)<br />$$

However, I can't understand how to get the $$\Gamma^0 _1 _0 = \frac{g}{(1+gx)}$$ one out.

$$\Gamma^0 _1 _0$$ would mean that we were looking for the C-symbols in equations of the form...

$$\frac{d^2 x^0}{d\tau^2} = -\Gamma^0 _1_0 \frac{dx^1} {d\tau} \frac{dx^0}{d\tau}$$

...yes?

I don't see any way of comparing that to the Lagrangian to find the C-symbol $$\Gamma^0 _1 _0 = \Gamma^0 _0 _1 = \frac{g}{(1+gx)}$$

Any help? Have I made a complete meal out of this?! Thanks for your time...

 

[CompuChip]

There is a well-known formula that allows you to calculate them directly:
$$\Gamma^i {}_{k\ell}=\frac{1}{2}g^{im} \left(\frac{\partial g_{mk}}{\partial x^\ell} + \frac{\partial g_{m\ell}}{\partial x^k} - \frac{\partial g_{k\ell}}{\partial x^m} \right)$$
where $g_{ij}$ can simply be read off and $g^{ij}$ are the coefficients of the inverse metric.

For the proof, you can refer to most GR or differential geometry texts.
Note that when you write it as
$$\Gamma^{\rho}_{\mu\nu} = \frac12 g^{\rho\sigma}\left( \partial_\nu g_{\sigma\mu} + \partial_\mu g_{\nu\sigma} - \partial_\sigma g_{\mu\nu} \right)$$
it's relatively easy to remember: there is only one upper index which goes in the inverse metric with a dummy variable, and then you cyclically permute the indices where the "canonical" order $\sigma\mu\nu$ has an opposite sign.

 

[Astrofiend]

Thanks very much for your help!

Much appreciated.

 

[George Jones]

I don't see any way of comparing that to the Lagrangian to find the C-symbol $$\Gamma^0 _1 _0 = \Gamma^0 _0 _1 = \frac{g}{(1+gx)}$$

For simple metrics like this, I quite like the Lagrangian method. You have used

$$\frac{\partial L}{\partial x} = \frac{d}{d \tau} \frac{\partial L}{\partial \frac{d x}{d \tau}}.$$

Now use

$$\frac{\partial L}{\partial t} = \frac{d}{d \tau} \frac{\partial L}{\partial \frac{d t}{d \tau}},$$

and differentiate with respect to $\tau$ the equation that you get.

 

[George Jones]

Oops, I didn't mean to say

and differentiate with respect to $\tau$ the equation that you get.

 

[Astrofiend]

Thanks a lot George - I appreciate the reply! I've now pushed through with both methods and they both work... funnily enough!

 

[Qyzren]

Why is it that when I try to solve it using the Lagrangian, my answer is off by a factor of 2 for the c-symbol $$\Gamma^0_{10}$$
Lagrange's equation for $$x^0$$ time component is
$$-\frac{d}{d\sigma}(\frac{\partial L}{\partial(\frac{\partial x^0}{\partial\sigma})})+\frac{\partial L}{\partial x^0}=0$$
where $$\frac{\partial L}{\partial x^0}=0$$ and $$\frac{\partial L}{\partial(\frac{\partial x^0}{\partial\sigma})}=\frac{1}{2}\frac{1}{L}(-2(1+gx^1)^2\frac{dx^0}{d\sigma}) =- \frac{1}{L}(1+gx^1)^2\frac{dx^0}{d\sigma}$$ as the Lagrangian $$L=\frac{d\tau}{d\sigma}$$, we can sub this in and also divide by L to convert the sigma's to tau's.
=>$$\frac{d}{d\tau}(\frac{d\sigma}{d\tau}(1+gx^1)^2\frac{dx^0}{d\sigma})=0$$
=>$$\frac{d}{d\tau}((1+gx^1)^2\frac{dx^0}{d\tau})=0$$
=>$$\frac{d}{d\tau}((1+gx^1)^2)\frac{dx^0}{d\tau}+(1+gx^1)^2\frac{d}{d\tau}(\frac{dx^0}{d\tau})=0$$
=>$$\frac{d^2x^0}{d\tau^2}=-\frac{2g}{1+gx^1}\frac{dx^1}{d\tau}\frac{dx^0}{d\tau}$$
so this shows $$\Gamma^0_{10}=\frac{2g}{1+gx^1}$$
So why do I have an extra factor of 2? T_T

 

[cristo]

Why is it that when I try to solve it using the Lagrangian, my answer is off by a factor of 2 for the c-symbol $$\Gamma^0_{10}$$

Because the EL equation is
$$0=\frac{d^2x^0}{d\tau^2}+\Gamma^0_{01}\frac{dx^1}{d\tau}\frac{dx^0}{d\tau}+\Gamma^0_{10}\frac{dx^0}{d\tau}\frac{dx^1}{d\tau}=\frac{d^2x^0}{d\tau^2}+2\Gamma^0_{01}\frac{dx^1}{d\tau}\frac{dx^0}{d\tau}$$

where the last equality is true since $$\Gamma^a_{bc}=\Gamma^a_{cb}$$

 

[Qyzren]

Thank you very much Cristo