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Help computing Christoffel coefficient

  1. Apr 13, 2007 #1
    Not so much a homework problem, more needing help understanding where something comes from. I've attached a jpg file with what I need help on.

    I've done a general relativity course at uni but can't seem to work out what should be a very simple problem.
     

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    Last edited: Apr 13, 2007
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  3. Apr 13, 2007 #2

    Mentz114

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    I get this,

    [tex]\Gamma^i_{0j} = \frac{1}{2}g^{i\nu}\partial_0(g_{\nu j}) [/tex] with [tex]\nu[/tex] running over 1,2,3.

    Differentiating
    [tex]\Gamma^i_{0j} = \frac{1}{2}g^{i\nu}(2aa'\eta_{\nu j} + 2aa'h_{\nu j} + a^2\partial_0 h_{\nu j})[/tex]

    which gives
    [tex]\Gamma^i_{0j} = (\eta^{i\nu}-h^{i\nu})(\frac{a'}{a}\eta_{\nu j} + \frac{a'}{a}h_{\nu j} + \frac{1}{2}\partial_0 h_{\nu j})[/tex]


    and so
    [tex]\Gamma^i_{0j} = \frac{a'}{a}g^{i\nu}g_{\nu j} + \frac{1}{2}g^{i\nu}\partial_0 h_{\nu j})[/tex]

    summing we get,
    [tex]\Gamma^i_{0j} = \frac{a'}{a}\delta^i_j + \frac{1}{2}h'^i_j[/tex]


    Which is not quite the right answer !
     
    Last edited: Apr 13, 2007
  4. Apr 15, 2007 #3
    That's a bit closer than I can get. Does [tex] \delta^i_j = \delta_{ij}[/tex]?
     
  5. Apr 15, 2007 #4

    Mentz114

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    [tex] \delta^i_j = g^{ik}\delta_{kj}[/tex]

    which requires a summation. I hope you don't mind me asking, but do you know how to raise and lower indexes ? I notice in your equ 5 you've lost the dummy index.
     
  6. Apr 15, 2007 #5
    I'm starting to think that the answer I've been given has a typo, and should read [tex] \delta^i_j [/tex] instead of [tex] \delta_{ij} [/tex].

    I've lost the dummy indices as I've summed over [tex] v [/tex]. When [tex] v=0 [/tex] everything equalled zero, so the next index was j (where j = 1,2,3). Is this not the way to proceed?

    Also, I understand that [tex] g_{iv}g^{vj}=\delta^i_j [/tex], but [tex] (\eta_{iv}+h_{iv})*(\eta^{vj}-h^{vj})=0 [/tex]?
     
    Last edited: Apr 15, 2007
  7. Apr 15, 2007 #6

    Mentz114

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    Aha - yes, I made an error, so my answer should have [tex] \delta_{ij}[/tex].
    Don't know why I wrote a mixed index there. That still leaves the sign and the prime on h.
     
  8. Apr 15, 2007 #7
    there should be a prime on h, that was a typo on my part.
     
  9. Apr 15, 2007 #8

    Mentz114

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    Well, if you can track down the why the sign disagrees, it's mission accomplished.

    I'm busy right now but I'll check it a little later.
     
  10. Apr 15, 2007 #9

    Mentz114

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    There's another typo in your doc. The expression you quote as the answer is wrong. The first delta must have one upper and one lower index like the other terms.

    So I stand my earlier result because

    [tex] g^{ik}g_{kj} = g_j^i = \delta_j^i[/tex]

    I haven't tracked down the sign yet.
     
  11. Apr 16, 2007 #10
    aha! got it.

    [tex]\Gamma^i_{0j} = \frac{a'}{a}\delta^{ij}+\frac{1}{2}g^{i\nu}\partial_0(h_{\nu j}) [/tex].

    If I expand the last term out,

    [tex] \frac{1}{2}(\eta^{i\nu}-h^{i\nu})\partial_0(h_{\nu j}) [/tex].

    Ignore higher order terms (ie h^2), using the metric signiture (+,-,-,-) coupled with the fact that only the spatial part is non zero, then,

    [tex] \eta^{i\nu} = \eta^{ij}=-\delta^{ij} [/tex]

    Thanks for the help :smile:
     
    Last edited: Apr 16, 2007
  12. Apr 16, 2007 #11

    Mentz114

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    I'm glad you're content. I don't follow your last step, but I suppose it's time to move on ...
     
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