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Varying determinant of a metric

  1. Feb 23, 2013 #1
    Hi can anyone explain how to find [itex]\delta \sqrt{-g}[/itex] when varying with respect to the metric tensor [itex]g^{\mu\nu}[/itex]. i.e why is it equal to [itex]\delta \sqrt{-g} = -\frac{1}{2} \sqrt{-g}g_{\mu\nu} \delta g^{\mu \nu} [/itex]
     
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  3. Feb 23, 2013 #2

    Bill_K

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    Because... what else could it be? Seriously, that's a legitimate way to proceed. It has to be a scalar density linear in δgμν, and there is only one such. All that remains is the numerical factor in front, so prove it's -1/2 for a diagonal metric and you're done.
     
  4. Feb 23, 2013 #3
    δ(√g)= (1/2√g) δg

    g=εabcdg0ag1bg2cg3d
    δg=....
     
  5. Feb 23, 2013 #4
    Hi i have been lookig at the wiki articles and can anyone explain why [itex] tr(g^{\mu \nu} \delta g_{\mu \nu}) = g^{\mu \nu}\delta g_{\mu \nu} [/itex], is it just a property of metric tensors?
     
  6. Feb 23, 2013 #5

    Bill_K

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    Do you accept my answer to your original question?
     
  7. Feb 23, 2013 #6
    well i can prove [itex] \delta \sqrt{-g}=-0.5(\sqrt{-g})^{-1} \delta g [/itex] by creating a chain rule situation for the variation. and i see that [itex] \delta g=g\left( TRACE(g^{\mu \nu} \delta g_{\mu \nu}\right) [/itex], from the jacobi equation. I just don't see how to go from [itex]TRACE(g^{\mu \nu} \delta g_{\mu \nu}) = g^{\mu \nu} \delta g_{\mu \nu}) [/itex].
     
    Last edited: Feb 23, 2013
  8. Feb 23, 2013 #7

    Bill_K

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    You don't need Trace here, and in fact Trace can only be applied to a matrix. Whereas gμνδgμν is already summed over all of its indices, and is already a scalar.

    (The approach I suggested is much, much easier! :wink:)
     
  9. Feb 23, 2013 #8
    the weirdest thing is that if the metric in variation is with indices up it appears with a minus sign...
     
  10. Feb 24, 2013 #9
    yeah but i read this:

    http://en.wikipedia.org/wiki/Einstein–Hilbert_action

    and the formula it uses is a trace but, are you saying that a trace would effectively just be a metric tensor [itex] g^{\mu \nu} \delta g_{\mu \nu} [/itex] anyway so it is in effect negated, also that minus sign is weird?
     
  11. Feb 24, 2013 #10

    Fredrik

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    Do you know the definition of trace, and the definition of matrix multiplication?

    Edit: OK, since there are some notational issues, I'll just explain what I'm thinking. Suppose that A is a matrix, and that we use the notation ##A^{\mu\nu}## for the component on row ##\mu##, column ##\nu##. Suppose that B is a matrix, and that we use the notation ##B_{\mu\nu}## for the component on row ##\mu##, column ##\nu##. If we use the notation ##(AB)^\mu{}_\nu## for the component on row ##\mu##, column ##\nu## of AB, then by definition of trace and matrix multiplication, we have
    $$\operatorname{Tr}(AB) =(AB)^\mu{}_\mu =A^{\mu\nu}B_{\nu\mu}.$$ If instead A is a (2,0) tensor, and B is a (0,2) tensor (like g), then
    $$A^{\mu\nu}B_{\mu\nu} =A(e_\mu,e_\nu)B(e^\mu,e^\nu) =(A\otimes B)(e_\mu,e_\nu,e^\mu,e^\nu).$$ The right-hand side is a "contraction", and I suppose someone might want to use use the Tr notation for it.
     
    Last edited: Feb 24, 2013
  12. Feb 24, 2013 #11

    Bill_K

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    gμνgμν ≡ 4, so vary this and get (δgμν)gμν + gμνδgμν = 0, or (δgμν)gμν = - gμνδgμν
     
  13. Feb 25, 2013 #12
    i see and understand al except still struggling to get my head around the trace. So simply [itex] TRACE g^{\mu \nu} \delta g_{\mu \nu} = g^{\mu \nu} \delta g_{\mu \nu} [/itex] or is the trace just not there in the first place? Thanks for all your help up to know
     
  14. Feb 25, 2013 #13

    Fredrik

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    What sort of thing can you take the trace of? Matrices for sure. Maybe some sort of tensors and tensor fields, with an appropriate generalization of the definition of "trace". But ##g^{\mu\nu}\delta g_{\mu\nu}## is a scalar field. So the only way I can make sense of your left-hand side is to assume that it's a bizarre notation for a contraction of a tensor.
     
  15. Feb 25, 2013 #14
  16. Feb 25, 2013 #15

    Fredrik

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    Nothing like the notation ##\mathrm{TRACE}(g^{\mu \nu} \delta g_{\mu \nu})## appears in either of those articles. This is something you introduced. What you're doing is the equivalent of taking the equality I wrote as ##\operatorname{Tr}(AB) =(AB)^\mu{}_\mu =A^{\mu\nu}B_{\nu\mu}## and rewriting it in notation that doesn't make sense.

    As opposed to? The components of the metric can of course be viewed as the components of a matrix.
     
  17. Feb 25, 2013 #16
  18. Feb 25, 2013 #17

    Fredrik

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    It's a trace, but there are no indices on the thing to the right of the Tr. For any matrix, the definition of trace is ##\operatorname{Tr}A=\sum_i A_{ii}##. A notation like ##Tr\left(\sum_{i}A_{ii}\right)## would not make much sense. The only possible interpretation of it is that you're taking a trace of a 1×1 matrix. This will of course just give you the number inside the matrix.
     
  19. Feb 25, 2013 #18
    then how is [itex] \delta \det g_{\mu \nu} = g g^{\mu \nu} \delta g_{\mu \nu} [/itex]
     
  20. Feb 25, 2013 #19
  21. Feb 25, 2013 #20
    Thanks but shouldn't it be:


    [itex] \delta g = g Tr(g^{-1}. \delta g) [/itex] ?
     
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