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Varying determinant of a metric

  1. Feb 6, 2013 #1
    Hi does anyone know how to calculate:

    [itex]\delta (det|g_{\mu\nu}|) or simply \delta g[/itex]
  2. jcsd
  3. Feb 6, 2013 #2
  4. Feb 6, 2013 #3


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    Or equivalently,

    δg = g gμν δgμν
  5. Feb 7, 2013 #4


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    Actually I believe this answer has a sign error, and does not correspond to what I wrote, since δgμν = - gμα gνβ δgαβ.

    A quick way to check the sign is to consider a special case, gμν = λ ημν under the variation δλ. For this case,

    gμν = λ-1 ημν and g = -λ4, so δgμν = ημν δλ and δg = -4 λ3 δλ.

    The minus sign agrees with the expression I gave, namely g gμν δgμν = (-λ4)(λ-1ημν)(ημν δλ).

    However the other expression g gμν δ(gμν) = (-λ4)(λ ημν)(-λ-2) (ημν δλ) comes out positive, which is incorrect.
  6. Feb 7, 2013 #5
    You're right. Thanks for the correction.
  7. Feb 8, 2013 #6
    yeah thanks i knew the answer but had absolutely no idea how to get it, I end up with [itex]g^{\nu\rho}g_{p\mu}\delta g^{\mu p}=-g^{\nu\rho}g^{\mu p} \delta g_{p\mu}[/itex] and am not sure how to proceed from here, there is clearly relabelling but im still a bit stuck.
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