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Can anyone explain why $T_{\mu \nu} = \frac{2}{\sqrt{-g}} \frac{\delta \mathcal{L}_M}{\delta g^{\mu \nu}} $, other than justifying it from the einstein field equations?

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FYI That definition applies much more broadly than scalar field theory. –  Michael Brown Feb 23 '13 at 14:23
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up vote 2 down vote accepted

You surely need to consider Einstein's field equations at some point because the stress-energy tensor in general relativity is defined as the object that is set equal to the Einstein curvature tensor (with the appropriate coefficient).

So if you vary the whole action $$ \int \left(\frac{R}{16\pi g} +{\mathcal L}_{\rm matter} \right) \sqrt{-g}\,d^d x $$ with respect to the metric, you will simply obtained Einstein's equations that has a multiple of the Einstein tensor and a multiple of the stress-energy tensor defined in the way you mentioned.

What you could have asked is why this definition of the stress-energy tensor is equivalent to other definitions you may know – i.e. non-gravitational ones. To answer this question, you would have to mention which other definition you mean. Well, you would probably mean the stress-energy tensor that is locally covariantly conserved, $\nabla_\mu T^{\mu\nu}=0$ in the context of general relativity. You could derive this stress-energy tensor by the Noether procedure by considering spacetime translations in non-gravitational theory, and then by covariantizing the result that you get in some right way.

Why is the conserved stress-energy tensor the same thing as the stress-energy tensor obtained by varying the metric? They have to be the same because Einstein's equations set the stress-energy tensor equal to a multiple of the Einstein tensor and the latter is covariantly conserved as a matter of identity. The equation $$\nabla_\mu G^{\mu \nu}=0$$ holds identically, for any metric tensor field configuration, and because the Einstein tensor is set proportional to the stress-energy tensor, the same condition (vanishing of the covariant divergence) has to hold for the stress-energy tensor, too. This is no accident and this fact may be formulated in other ways. For example, we say that gravity has to use the diffeomorphism symmetry (it's doubly important as the gauge symmetry in the quantum mechanical framework to get rid of the unphysical, negative-norm polarizations of the gravitons). Such a local symmetry has to be coupled to a "conserved current" (conserved stress-energy tensor) for consistency.

You will therefore typically end up with the same formula for both stress-energy tensors. However, they may end up differ by certain terms whose covariant derivative is zero identically as well. For example, the non-gravitational conserved stress-energy tensor isn't always quite canonical and it doesn't even have to be a symmetric tensor. The GR definition of the tensor is automatically symmetric.

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