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Homework Help: Variation of the metric tensor determinant

  1. Dec 12, 2012 #1
    1. The problem statement, all variables and given/known data

    This is not homework but more like self-study - thought I'd post it here anyway.

    I'm taking the variation of the determinant of the metric tensor:


    2. Relevant equations

    The answer is

    [itex]\delta(det[g\mu\nu])[/itex] =det[g\mu\nu] g[itex]\mu\nu[/itex] [itex]\delta[/itex]g[itex]\mu\nu[/itex]

    Here, g[itex]\mu\nu[/itex] is the metric tensor, [g[itex]\mu\nu[/itex]] is the matrix of the components of the metric tensor, and [itex]\delta[/itex] is a variation.

    3. The attempt at a solution

    I have managed to get close to the answer, I hope, with
    =det[g\mu\nu] tr([g\mu\nu]-1[itex]\delta[/itex]([g\mu\nu]))

    The problem, in my view, is the trace. I cannot see how to remove it.

    Also, if someone could kindly describe how to tidy the LaTeX up, I will do that.

    Thank you!
  2. jcsd
  3. Dec 13, 2012 #2
    You can't use the tags in latex. Instead, write g_{\mu \nu}, g^{\mu \nu} etc.

    This is a little bit tricky, as you dont really keep track of how your matrices are multiplied together. If you do it with more care, you should find something like
    [tex] \delta \det(g) = \det(g) \mathrm{Tr}( g \cdot \delta(g) ), [/tex]
    with [itex] (g \cdot \delta(g))_{\mu \nu} = g_{\mu \lambda} \delta ({g^\lambda}_\nu) [/itex]
    Now it's easy to see what the trace is:
    [tex] \mathrm{Tr}( g \cdot \delta(g) ) = (g \cdot {\delta(g))_\mu}^{\mu} = g_{\mu \lambda} \delta (g^{\lambda \mu}) [/tex]
  4. Dec 13, 2012 #3
    now when you apply δ on g you can use the second of the above to get whole result.(i am not going to do it in full,because of the requirement of homework section)
  5. Dec 14, 2012 #4
    Hmm,may be tough.
    so use Tr (ln M)=ln(det M)
    variation of it yields,
    Tr(M-1 δM)=(1/det M)δ(det M)
    JUST PUT M=gμv
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