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Ricci tensor proportional to divergence of Christoffel symbol?

  1. May 15, 2014 #1
    I'm reading an old article published by Kaluza "On the Unity Problem of Physics" where i encounter an expression for the Ricci tensor given by

    $$R_{\mu \nu} = \Gamma^\rho_{\ \mu \nu, \rho}$$

    where he has used the weak field approximation ##g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}## where ##h## is a small deviation tensor. I have never seen this expression before. Is it true in some gauge in the weak field?
     
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  3. May 15, 2014 #2

    Bill_K

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    Working backwards, the Riemann tensor is

    Rμνστ = Γμντ,σ - Γμνσ,τ + (quadratic terms)

    Dropping the quadratic terms, the Ricci tensor is

    Rντ = Γμντ,μ - Γμμν,τ

    The first term is what you want - how do we get rid of the second?

    Γμμν = (1/2) gμλgμλ,ν = (1/2) g

    To first order, g = 1 + hμμ, so I'd say a sufficient gauge condition is that h is traceless.
     
  4. May 15, 2014 #3
    I agree Bill K. Now I am a little bit familiar with the transverse-traceless gauge, which is a combination of the harmonic gauge ##\partial_\mu h^\mu_\nu = 0## and the trace condition ##h=\eta^{\mu \nu}h_{\mu \nu}= 0##, but as I've understood it, this gauge is only allowable in a vacuum. Is there a gauge with ##h=0## without the harmonic gauge condition for which we do not need a vacuum?
     
  5. May 15, 2014 #4

    Bill_K

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    That's probably true if you want the field equation to be ◻hμν = Tμν, since then it would work only if Tμν was traceless also.

    However in general, it seems to me you can always pick the coordinates to be "isovolume", that is, det(gμν) = const. (Although I haven't a clue how you would actually do that!)
     
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