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Christoffel Symbols problem in GR

  1. Aug 20, 2009 #1
    1. The problem statement, all variables and given/known data

    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.


    3. 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...
     
  2. jcsd
  3. Aug 20, 2009 #2

    CompuChip

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    Science Advisor
    Homework Helper

    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.
     
  4. Aug 20, 2009 #3
    Thanks very much for your help!

    Much appreciated.
     
  5. Aug 20, 2009 #4

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
     
  6. Aug 20, 2009 #5

    George Jones

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    Oops, I didn't mean to say
     
  7. Aug 21, 2009 #6
    Thanks a lot George - I appreciate the reply! I've now pushed through with both methods and they both work... funnily enough!
     
  8. Aug 21, 2009 #7
    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
     
  9. Aug 21, 2009 #8

    cristo

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    Staff Emeritus
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    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]
     
  10. Aug 21, 2009 #9
    Thank you very much Cristo
     
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