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What is varing in the variational principle of GR

  1. Jan 10, 2012 #1
    Consider the variational principle used to obtain that in the vacuum the Einstein tensor vanish.
    So we set the lagrangian density as [itex]L(g,\partial g)=R[/itex]
    and asks for the condition
    [itex]0 = \delta S =\delta\int{d^4 x \sqrt{-g}L}[/itex]

    proceeding with the calculus I finally have to vary R such that
    [itex]\delta R = \delta{g^{\mu\nu}R_{\mu\nu}}[/itex]
    but what is [itex]\delta g[/itex]??

    i think (but not sure) [itex]\delta{g^{\mu\nu}}=g^{\mu\nu}-{g'}^{\mu\nu}[/itex]
    since g transform as a tensor
    [itex]\delta{g^{\mu\nu}}=g^{\mu\nu}-J_{\rho}^{\mu} J_{\sigma}^{\nu} {g}^{\rho\sigma}[/itex]

    where J is the jacobian of the transformation..am I right?
  2. jcsd
  3. Jan 10, 2012 #2


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    Science Advisor

    Usually the independent variables to be varied are taken as gμν. The equations you need are:

    δgμν = - gμσ gντ δgστ

    δ√-g = ½√-g gμν δgμν

    δR = gμν δRμν - Rμν δgμν

    δRμν = ½ gστ(gμν;στ + gστ;μν - gμσ;ντ - gντ;μσ)
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