Christoffel Symbols problem in GR

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  • #1
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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...
 

Answers and Replies

  • #2
CompuChip
Science Advisor
Homework Helper
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There is a well-known formula that allows you to calculate them directly:
[tex]\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)[/tex]
where [itex]g_{ij}[/itex] can simply be read off and [itex]g^{ij}[/itex] 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
[tex]\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)[/tex]
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 [itex]\sigma\mu\nu[/itex] has an opposite sign.
 
  • #3
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Thanks very much for your help!

Much appreciated.
 
  • #4
George Jones
Staff Emeritus
Science Advisor
Gold Member
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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]

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

[tex]
\frac{\partial L}{\partial x} = \frac{d}{d \tau} \frac{\partial L}{\partial \frac{d x}{d \tau}}.
[/tex]

Now use

[tex]
\frac{\partial L}{\partial t} = \frac{d}{d \tau} \frac{\partial L}{\partial \frac{d t}{d \tau}},
[/tex]

and differentiate with respect to [itex]\tau[/itex] the equation that you get.
 
  • #5
George Jones
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Oops, I didn't mean to say
and differentiate with respect to [itex]\tau[/itex] the equation that you get.
 
  • #6
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Thanks a lot George - I appreciate the reply! I've now pushed through with both methods and they both work... funnily enough!
 
  • #7
44
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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 [tex]\Gamma^0_{10}[/tex]
Lagrange's equation for [tex]x^0[/tex] time component is
[tex]-\frac{d}{d\sigma}(\frac{\partial L}{\partial(\frac{\partial x^0}{\partial\sigma})})+\frac{\partial L}{\partial x^0}=0[/tex]
where [tex]\frac{\partial L}{\partial x^0}=0[/tex] and [tex]\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}[/tex] as the Lagrangian [tex]L=\frac{d\tau}{d\sigma}[/tex], we can sub this in and also divide by L to convert the sigma's to tau's.
=>[tex]\frac{d}{d\tau}(\frac{d\sigma}{d\tau}(1+gx^1)^2\frac{dx^0}{d\sigma})=0[/tex]
=>[tex]\frac{d}{d\tau}((1+gx^1)^2\frac{dx^0}{d\tau})=0[/tex]
=>[tex]\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[/tex]
=>[tex]\frac{d^2x^0}{d\tau^2}=-\frac{2g}{1+gx^1}\frac{dx^1}{d\tau}\frac{dx^0}{d\tau}[/tex]
so this shows [tex]\Gamma^0_{10}=\frac{2g}{1+gx^1}[/tex]
So why do I have an extra factor of 2? T_T
 
  • #8
cristo
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Science Advisor
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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 [tex]\Gamma^0_{10}[/tex]

Because the EL equation is
[tex]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}[/tex]

where the last equality is true since [tex]\Gamma^a_{bc}=\Gamma^a_{cb}[/tex]
 
  • #9
44
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Thank you very much Cristo
 

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