- #1
Astrofiend
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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
[tex]
ds^2 = (1+gx)^2 dt^2 - dx^2 - dy^2 - dz^2
[/tex]
and have been told to show that the non-vanishing Christoffel symbols are given by
[tex]
\Gamma^1 _0 _0 = g(1+gx)
[/tex]
and
[tex]
\Gamma^0 _1 _0 = \Gamma^0 _0 _1 = \frac{g}{(1+gx)}
[/tex]
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:
[tex]
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
[/tex]
which, after a little substitution work getting taus instead of sigmas, basically gives us the following equation:
[tex]
\frac{d^2x}{d\tau^2} = -\frac{1}{2} . 2(1+gx).g \frac{dt}{d\tau}^2
[/tex]
...and comparing this to the definition of a C-symbol
[tex]
\frac{d^2 x^\alpha}{d\tau^2} = -\Gamma ^\alpha _\beta _\gamma \frac{dx^\beta} {d\tau} \frac{dx^\gamma}{d\tau}
[/tex]
...it is pretty easy to see that
[tex]
\Gamma^1 _0 _0 = g(1+gx)
[/tex]
However, I can't understand how to get the [tex]\Gamma^0 _1 _0 = \frac{g}{(1+gx)}[/tex] one out.
[tex]\Gamma^0 _1 _0[/tex] would mean that we were looking for the C-symbols in equations of the form...
[tex]
\frac{d^2 x^0}{d\tau^2} = -\Gamma^0 _1_0 \frac{dx^1} {d\tau} \frac{dx^0}{d\tau}
[/tex]
...yes?
I don't see any way of comparing that to the Lagrangian to find the C-symbol [tex]\Gamma^0 _1 _0 = \Gamma^0 _0 _1 = \frac{g}{(1+gx)}[/tex]
Any help? Have I made a complete meal out of this?! Thanks for your time...