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In general relativity, the energy-momentum tensor is written as

$$ T^{\mu \nu } \sim \frac{1}{\sqrt{|g|}}\frac{\delta \mathcal L}{\delta g_{\mu\nu}},$$

while in field theory it is written as

$$ T^{\mu}{}_{ \nu } = \frac{\delta \mathcal L}{\delta \partial_{\mu} \phi} \partial_{\nu} \phi - \delta^{\mu}_{\nu} \phi .$$

I am having trouble reconciling these two expressions. Also since diffeomorphisms, and hence translational invariance, are a gauge symmetry in general relativity, how can there be a non-zero conserved current for translational invariance, i.e. the energy-momentum tensor?

JamalS
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Himanshu Khanchandani
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  • what do you mean by "since diffeomorphism is gauge symmetry in general relativity, how can there be a non zero conserved current for translational invariance"? In gauge theory, we still have the conserved current corresponding the global gauge symmetry which certainly is a true symmetry. The same is true for Lorentz symmetry. – Wein Eld Aug 20 '16 at 18:43
  • Yes, but in General relativity translational invariance is not necessarily a global gauge symmetry. I mean there may not be a global translational invariance. Also if somebody could tell me where I can find a derivation of the above formula for energy momentum tensor in General relativity using translational invariance, it would be helpful. In most books I saw, it is being derived calling it as density of energy-momentum etc. – Himanshu Khanchandani Aug 21 '16 at 03:09
  • I would recommend reading the chapters § 32 and § 94 "The energy-momentum tensor" of L. D. Landau and E.M. Lifshitz - 1971 - The Classical Theory Of Fields - 2nd ed if you are interessted in the derivation of the energy momentum tensor in GR. In principle it is similar to the one in field theory and done by varying the action. – N0va Aug 21 '16 at 12:51
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    Question already answered in detail here http://physics.stackexchange.com/q/119895 – TwoBs Aug 21 '16 at 21:53
  • Possible duplicates: http://physics.stackexchange.com/q/119838/2451 , http://physics.stackexchange.com/q/270877/2451 and links therein. – Qmechanic Aug 27 '16 at 23:40

1 Answers1

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You are talking about a very particular example. If you minimally couple a scalar field to gravity and use the first equation you mention you get indeed the tensor you quote in the second line "the QFT one". However this is not always the case, if you couple a spinor minimally to gravity and find the energy momentum tensor it will not match the one you get by applying Noether's theorem to the Dirac lagrangian.

One can think about General Relativity as a gauge theory with a very large gauge group, the group of diffeomorphisms as you mention. These should not be considered as symmetries giving rise to conserved charges but rather as redundancies in our mathematical description of the underlying physics.

However if your spacetime has "Killing vectors" there will be conserved charges, the tyical metrics one encounters have Killing vectors that lead to energy momentum conservation. This means there is an independent observer statement of the kind $\frac{d}{dt}Q=0$. Where $Q$ is the integral over a spatial slice of the conserved component.

If you do not have Killing vectors all you can say is $\nabla_{\mu} T^{\mu\nu}=0$. This is not a conservation law, is actually an assumption not even a mathematical fact, all this formula says is that all reasonable matter that we know about satisfies this equation.

user128505
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