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Help with the variation of the Ricci tensor to the metric

  1. Mar 26, 2015 #1
    I should calculate the variation of the Ricci scalar to the metric ##\delta R/\delta g^{\mu\nu}##. According to ##\delta R=R_{\mu\nu}\delta g^{\mu\nu}+g^{\mu\nu}\delta R_{\mu\nu}##, ##\delta R_{\mu\nu}## should be calculated. I have referred to the wiki page: http://en.wikipedia.org/wiki/Einstein–Hilbert_action which provided the deduction below
    $$g^{\mu\nu}\delta R_{\mu\nu}=\nabla_{\sigma}(g^{\mu\nu}\delta\Gamma^{\sigma}_{\nu\mu}-g^{\mu\sigma}\Gamma^{\rho}_{\rho\mu}),$$
    and this term can be converted to a total derivative when it's multiplied by ##\sqrt{-g}##:
    $$\sqrt{-g}\nabla_{\sigma}(g^{\mu\nu}\delta\Gamma^{\sigma}_{\nu\mu}-g^{\mu\sigma}\Gamma^{\rho}_{\rho\mu})=\partial_{\sigma}[\sqrt{-g}(g^{\mu\nu}\delta\Gamma^{\sigma}_{\nu\mu}-g^{\mu\sigma}\Gamma^{\rho}_{\rho\mu})].$$
    When we integrate it over the whole spacetime and assume that the variation of the metric ##\delta g^{\mu\nu}## vanishes at infinity, it will contribute nothing to the variation of the action ##\int dx^{4}\sqrt{-g}f(R)##.

    However i'm considering things in the spatially flat Friedmann universe ##ds^{2}=dt^{2}-a^2(t)\delta_{ik}dx^{i}dx^{k}.## The assumption above can't be true any more. To make my question specific, could you compute ##\delta R/\delta g^{\mu\nu}## in the spatially flat Friedmann universe? Thanks a lot.
     
  2. jcsd
  3. Mar 26, 2015 #2

    PeterDonis

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    Staff: Mentor

    Actually, the problem comes even before that: for a spacetime like the spatially flat FRW spacetime, the integral of the Lagrangian over all spacetime does not converge, so the action is not well-defined. But, as the wiki page says, "a modified definition where one integrates over arbitrarily large, relatively compact domains, still yields the Einstein equation". So you just have to stipulate that the variation of the metric vanishes on the boundary of the domain over which you are integrating. Since you can do this for any domain you wish, it effectively amounts to the EFE being valid everywhere.
     
  4. Mar 27, 2015 #3
    Thanks for reminding me this. I really ignored the point you mentioned above. However I also have some confusions. In cosmology, people always use the usual action to describe inflation, eg. $$S=\int dx^4\sqrt{-g}[-\frac{1}{16\pi}-\frac{1}{2}(\partial _{\mu }\phi )^{2}-V(\phi)]$$for the scalar inflaton field ##\phi##. A strict method to deduce the evolution of the universe should be like this: First, modify the action above, considering that it's not "well-defined" for a spatially flat Friedmann universe. Second, use variational method on the new action and obtain the evolution equation of the universe.

    However, people always process it like this: First, affirm the validity of the usual action and EFE. Second, calculate the energy-momentum tensor by ##T_{\alpha\beta}=\frac{2}{\sqrt{-g}}\frac{\delta S^m}{\delta g^{\alpha\beta}}##. Third, substitute ##T_{\alpha\beta}## to the right hand side of the EFE. Back to my original question, here is a paper http://arxiv.org/abs/0802.2068 which indirectly shows that $$g^{\alpha\gamma}g^{\mu\nu}\frac{\delta R_{\mu\nu}}{\delta g^{\gamma\beta}}A_{\lambda}A^{\lambda}=(\delta^{\alpha}_{\beta}\Box-\nabla^{\alpha}\nabla_{\beta})A_{\lambda}A^{\lambda},$$where ##A_{\mu}## is some vector field. ( I haven't get it proved yet.)

    You don't have to spend much time on this question unless you're very interested in it.:smile:
     
  5. Mar 27, 2015 #4

    PeterDonis

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    The same comment applies here as for the OP: we can still derive the EFE for this case by considering a large, relatively compact domain, and observing that the derivation works for any such domain. The RHS of the EFE would then just be the stress-energy tensor derived from the action for the inflaton field.
     
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