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Determinant of metric tensor

  1. Apr 12, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that
    [tex]\Gamma^\mu_{\mu\lambda}=\frac{1}{\sqrt{-g}}\partial_\lambda(\sqrt{-g})[/tex]

    where g is the determinant of the metric, and [itex]\Gamma[/itex] are the Christoffel connection coefficients.

    3. The attempt at a solution
    From the general definition of the coefficients I got:
    [tex]\Gamma^\mu_{\mu\lambda}=(1/2)g^{\mu\rho}\partial_\lambda g_{\rho\mu}[/tex]

    But I have no idea how to work with the determinant of the metric. I'm not sure if I'm allowed to use this:

    det(g)=exp[Tr ln G]

    And if I did, would I have to use the GR definition of the trace?

    [tex] Tr R = R^\mu_\mu [/tex]

    I cleaned it up a little bit with the chain rule:

    [tex]\frac{1}{\sqrt{-g}}\partial_\lambda(\sqrt{-g})[/tex]
    [tex]=\frac{1}{2g}\partial_\lambda(g)[/tex]
     
  2. jcsd
  3. Apr 12, 2009 #2

    dextercioby

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    The last part you wrote can't be right.

    I'd start with

    [tex] \delta |g| = \delta \[\exp \(\mbox{Tr} \ln |g|\)\] [/tex]

    and then see what the 'delta' of the RHS brings me.
     
  4. Apr 12, 2009 #3

    Hurkyl

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    Isn't there an index-notation expression for the determinant?
     
  5. Apr 12, 2009 #4
    Thanks for the tips. Turns out it's a bit simpler:

    1. From Jacobi's formula we get:
    [tex] dg=A^{ik}dg_{ik}[/tex]
    where [itex]A^{ik}[/itex] are elements of the adjugate of the metric.

    2. Standard differential expansion:
    [tex]dg=\frac{\partial g}{\partial g_{ik}}dg_{ik}[/tex]
    [tex]\Rightarrow A^{ik}=\frac{\partial g}{\partial g_{ik}}[/tex]

    3. From Laplace's formula for the determinant we get:
    [tex]g^{ik}=(1/g)A^{ik}[/tex]


    Putting this all together we have:
    [tex]g^{ik}=\frac{1}{g}\frac{\partial g}{\partial g_{ik}}[/tex]

    Substitute this into my expression for [itex]\Gamma^\mu_{\mu\lambda}[/itex] above and the result follows.

    I'm still working on the proofs to steps 1 and 3 above, but essentially the problem is solved. Feels good to brush up on my linear algebra, haven't used it for years.
     
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