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Help with differential geometry: Hilbert's stress energy tensor

  1. Apr 20, 2010 #1
    When varying the Hilbert action, we define the stress-energy tensor as:
    [tex]T_{\mu\nu}:= \frac{-2}{\sqrt{-g}}\frac{\delta (\sqrt{-g} \mathcal{L}_\mathrm{matter})}{\delta g^{\mu\nu}}
    = -2 \frac{\delta \mathcal{L}_\mathrm{matter}}{\delta g^{\mu\nu}} + g_{\mu\nu} \mathcal{L}_\mathrm{matter}[/tex]

    I am still struggling to learn how to read physical meaning into differential geometry equations.
    Can someone help me "see the light" for how this makes sense as a stress-energy tensor?

    For T_uv to transform like a tensor, Lmatter must be a scalar (which makes sense) but should be interpretted as the energy density / pressure of a perfect fluid!? (the [itex]g_{\mu\nu} \mathcal{L}_\mathrm{Matter}[/tex] term) That doesn't makes sense to me. Especially since a lagrangian is usually Kinentic energy - Potential energy ... and so doesn't make sense to interpret as a total energy (as opposed to the Hamiltonian). And why must it have pressure?

    In case my questions are too vague, here's a more concrete one to help me get my feet wet.
    Is it possible to show (with appropriate assumptions about Lmatter) that the divergence of this stress-energy tensor is necessarily zero? I've seen it worked out where they prove, from Einstein's field equations, that the left hand side has zero divergence ... and therefore any solution demands that the right hand side, the stress-energy tensor, has no divergence either. But it would be more satisfying to see that this definition is reasonable on its own. I would like to see what assumptions about Lmatter are necessary to ensure that definition yields a stress-energy tensor that is divergenceless.
  2. jcsd
  3. Apr 20, 2010 #2


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    Maybe take a look at Eqn 144 of http://relativity.livingreviews.org/Articles/lrr-2007-1/ [Broken] , also the commentary in sections 3 and 6.2
    Last edited by a moderator: May 4, 2017
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