Help with differential geometry: Hilbert's stress energy tensor

JustinLevy
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When varying the Hilbert action, we define the stress-energy tensor as:
T_{\mu\nu}:= \frac{-2}{\sqrt{-g}}\frac{\delta (\sqrt{-g} \mathcal{L}_\mathrm{matter})}{\delta g^{\mu\nu}} <br /> = -2 \frac{\delta \mathcal{L}_\mathrm{matter}}{\delta g^{\mu\nu}} + g_{\mu\nu} \mathcal{L}_\mathrm{matter}

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 g_{\mu\nu} \mathcal{L}_\mathrm{Matter}[/tex] term) That doesn&#039;t makes sense to me. Especially since a lagrangian is usually Kinentic energy - Potential energy ... and so doesn&#039;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&#039;s a more concrete one to help me get my feet wet.<br /> Is it possible to show (with appropriate assumptions about Lmatter) that the divergence of this stress-energy tensor is necessarily zero? I&#039;ve seen it worked out where they prove, from Einstein&#039;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.
 
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Maybe take a look at Eqn 144 of http://relativity.livingreviews.org/Articles/lrr-2007-1/ , also the commentary in sections 3 and 6.2
 
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