Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Bill_K

    User Avatar
    Science Advisor

    Or equivalently,

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

    Bill_K

    User Avatar
    Science Advisor

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Varying determinant of a metric
Loading...