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Trying to prove a consequence of harmonic gauge in GR

  1. Aug 10, 2015 #1
    So, I am following the PI lecture series by Neil Turok. He starts with the following description of harmonic gauge condition

    $$g^{\mu \nu}\Gamma^{\lambda}_{\mu \nu}=0$$
    He then claims that for linearized gravity (weak field) i.e.
    $$g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu} $$ with $$ |h_{\mu \nu}| <<1 $$ that harmonic gauge is equivalent to the condition that $$h^{\lambda}_{\nu , \lambda} - \frac{1}{2} h^{\lambda}_{\lambda , \nu} = 0$$

    My problem is in proving this is how do I prove this, Ive been trying for a few days now with no luck, I really need some pointers. I tried using the definition of the Christoffel symbol to work it out and I think I'm going nowhere. Any help would be appreciated.

  2. jcsd
  3. Aug 10, 2015 #2


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    If you don't show any of your work, how are we supposed to help? We don't know where exactly you got stuck. For example, did you make sure to work to only first order in the perturbation when computing the connection coefficients?
  4. Aug 10, 2015 #3


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    Use [tex]\nabla_{\mu}(\sqrt{-g}g^{\mu\nu}) = 0,[/tex] and rewrite the Harmonic condition in the form [tex]\partial_{\mu}(\sqrt{-g}g^{\mu\nu}) = 0 .[/tex] Expand and contract with [itex]g_{\nu\tau}[/itex] to obtain [tex]\frac{1}{\sqrt{-g}}\partial_{\tau}\sqrt{-g} + g_{\nu\tau}\partial_{\mu}g^{\mu\nu} = 0 .[/tex] This is the same as [tex]\frac{1}{2} g^{\rho\sigma}\partial_{\tau}g_{\rho\sigma} + g_{\nu\tau}\partial_{\mu}g^{\mu\nu} = 0 .[/tex] Now use [itex]g_{\mu\nu} = \eta_{\mu\nu} + \epsilon \ h_{\mu\nu}[/itex] and ignore the [itex]\mathcal{O}(\epsilon^{2})[/itex] terms.
  5. Aug 11, 2015 #4
    So what I have so far is $$0 = g^{\mu \nu} \Gamma^{\lambda}_{\mu \nu} = \frac{1}{2}g^{\mu \nu}g^{\tau \lambda}(g_{\mu \tau ,\nu} + g_{\nu \tau , \mu} -g_{\mu \nu , \tau}) = \frac{1}{2} g^{\tau \lambda}(g_{\mu \tau} \, ^{, \mu} + g_{\nu \tau} \, ^{, \nu}) - \frac{1}{2}g^{\mu \nu}g_{\mu \nu} \, ^{, \lambda} = g^{\tau \lambda} g_{\mu \tau} \, ^{, \mu} - \frac{1}{2}g^{\mu \nu}g_{\mu \nu} \, ^{, \lambda}$$

    Now I do the following
    $$g_{\mu \tau} \, ^{,\mu} = \partial_\mu g_{\mu \tau} = \partial_\mu( \eta_{\mu \tau} + h_{\mu \tau}) = \partial_\mu h_{\mu \tau} = h_{\mu \tau} \, ^{,\mu}$$ doing this for both terms gives me $$g^{\tau \lambda} h_{\mu \tau} \, ^{, \mu} - \frac{1}{2}g^{\mu \nu} h_{\mu \nu} \, ^{, \lambda} = 0$$

    This is what I got so far.

    Edit: What it looks like I should do is use the metric tensor to raise the indices on the derivatives of h in the final line, but I don't think I can do that because the derivatives of h aren't tensors so it would be like trying to raise indices on the christoffel symbol, am I wrong?
    Last edited: Aug 11, 2015
  6. Aug 13, 2015 #5
  7. Aug 13, 2015 #6


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    Why? Didn’t I show you exactly how to solve your problem? Do you even know why the de Donder condition ,[itex]g^{\mu\nu}\Gamma_{\mu\nu}^{\rho}=0[/itex], is called harmonic coordinate condition?
  8. Aug 13, 2015 #7
    First of all, thanks for answering.

    As I understand it, the reason why its called harmonic gauge is because under this condition the coordinate functions are harmonic i.e. $$\Box x^\mu = 0$$

    the bump was in response to this:

    who asked me to show the work I had done, but then didnt respond, and I wanted to know if I had gone in the totally wrong direction or if I was almost there.
  9. Aug 13, 2015 #8
    Of course. The de Dongle condition. Everybody knows it..
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