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Energy momentum of linearized gravity.

  1. Jan 19, 2012 #1

    I am doing a problem of GR dealing with weak field limit equations. After decoupling the static and dynamical part the usual way ([itex]g^{ab}=\eta^{ab}-h^{ab}[/itex]) I arrived at those field equations:

    \frac12 \eta^{ab}\partial_a \partial_b h_{\mu\nu} - \frac12 h^{ab}(\partial_a\partial_b h_{\mu\nu} -\partial_a \partial_{\nu} h_{\mu b} - \partial_{\mu}\partial_b h_{a\nu} + \partial_{\nu}\partial_{\mu}h_{ab}) -\frac14 (\partial_a h_{b\nu} +\partial_{\nu}h_{ab} -\partial_b h_{a\nu})(\partial_{\mu}h^{ab} + \partial^a h^{b}_{\mu} - \partial^bh_{\mu}^a) + \mathcal{O}(h^3) = 0

    The first term being the operator on the field, the rest should be the source (stress energy tensor of the spin-2 field because we are in vaccum) but I can't derive this stress energy from the naive Lagrangian [itex]\cal{L} = \frac14 (\partial^ah_{\mu\nu})^2[/itex]. Plus one question is: "What are the conditions to interpret that as a stress-energy tensor?"

    Some help/tips will be more than welcomed.
    Last edited: Jan 19, 2012
  2. jcsd
  3. Jan 19, 2012 #2
    You say "equations" but there are no equal signs...
  4. Jan 19, 2012 #3
    Yeah, fixed it. By the way, I think I "found my Lagrangian", [itex]\cal{L}=\frac14 g^{ab}\partial_a h_{\mu\nu} \partial_b h^{\mu\nu}[/itex]. I should have seen from beginning that I needed terms cubic in the Lagrangian to get quadratic in the field equations.
  5. Jan 20, 2012 #4


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    Usually there is no stress-energy-tensor for the gravitational field; the r.h.s. of the Einstein field equations is T expressed in terms of non-gravitational fields; in vacuum you have T=0. The weak field limit for gravitational waves in vacuum is nothing else but an approximation for the vacuum Einstein equations with T=0.
  6. Jan 20, 2012 #5


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    You are right; however, in the linearized regime, the second order perturbations in the metric can "act like" an effective stress-energy tensor, but with some caveats. The stress energy tensor constructed from the second order perturbations are not gauge invariant.

    To get this result, you don't need to go quartic or cubic in the field equations. I refer to Wald pg. 84 to 86 for a more detailed discussion.
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