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Einstein assumed in the derivation of his field equations that the covariant derivative (defined through the metric) of the energy momentum tensor of matter is zero, i.e. $$\nabla_{\mu}T^{\mu\nu}_{\rm matter}=0,$$

no matter how distorted the spacetime is.

It seems such a conclusion cannot be reached at via Nöther's theorem (since the metric is itself changing over spacetime). So my question is how did he have this insight then, and how would one justify the correctness of this assumption in retrospect from a Lagrangian point of view now?

If it's based on the existence of infinitesimal coordinate transformations that fix the boundary of some volume in spacetime, then would this also imply that the energy momentum tensor of any field will always have zero covariant derivative with or without gravity?


Edit: Previous form of the question had an abuse of language, viz., that I referred to the "vanishing of the covariant derivative of the energy-momentum tensor" as "energy momentum tensor is covariantly conserved". This was rightly pointed out by Ben Crowell in his answer and, I've edited the question and made it mathematically precise for the sake of clarity.

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    $\begingroup$ I don't know how to explain this, but, I can hear Dr. Wheeler yelling from the grave "the boundary of a boundary is zero!" See MTW Chapter 15: The Bianchi Identities and the Boundary of a Boundary. $\endgroup$ Commented Oct 24, 2019 at 19:32
  • $\begingroup$ @StevenThomasHatton: I think that would be more like the reason why the Einstein tensor has zero divergence. It's a geometric fact about curvature, which doesn't really relate to the OP's question except indirectly because it helps to explain why the field equations make sense. $\endgroup$
    – user4552
    Commented Oct 24, 2019 at 19:42
  • $\begingroup$ @BenCrowell The answer to the first part of the question is "intuition". It seemed like the right thing to do, so Einstein tried it. For the rest of the question, I really think an understanding of the content of Chapter 15 of MTW will be required. The first paragraph talks about the conservation of sources of curvature. That is the conservation of the stress energy tensor. $\endgroup$ Commented Oct 24, 2019 at 19:58

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This isn't a fact about general relativity, it's a fact about special relativity. Based on hundreds of years worth of experiments, we know that the energy-momentum four-vector is conserved locally. This means that the stress-energy tensor has zero divergence. (The stress-energy tensor isn't conserved. What's conserved is the energy-momentum four-vector.)

If you want a deeper theoretical justification for conservation of the energy-momentum in special relativity, then you can appeal to Noether's theorem for translations. It doesn't have anything to do with more general coordinate transformations.

Once an equation like $\partial_\mu T^{\mu\nu}=0$ is established in special relativity, then carrying it over into general relativity is normally a trivial process. We just replace partial derivatives with covariant derivatives. Basically this is just the fact that GR is the same as SR, locally.

What is nontrivial is carrying over global things, like an equation stating the conservation of energy-momentum in an asymptotically flat spacetime. This kind of thing generally just doesn't work, the problem being that parallel transport is path-dependent, so we can't do things like summing momenta over a region of space.

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    $\begingroup$ If you add the equivalence principle to your answer, I believe it will be more complete. That is, Einstein's assumption is that spacetime is everywhere locally Minkowskian. Some say Lorentzean spacetime, but the recognition of 4-D spacetime really is due to Minkowski. Put differently we can say the tangent plane at every event is a Minkowski spacetime. $\endgroup$ Commented Oct 24, 2019 at 19:39
  • $\begingroup$ @StevenThomasHatton this may be beside the point but, wouldn't it be more appropriate to call it Lorentzian? Since we are talking about something of local character I would think that the most important space is the tangent space, i.e., the Lorentz vector space. On the other hand the Minkowski spacetime is instead a (flat) manifold, or a coordinate space if you will. $\endgroup$
    – user137661
    Commented Oct 24, 2019 at 20:21
  • $\begingroup$ @BenCrowell Thanks very much for your answer. Unfortunately when I searched for the tag "energy-momentum tensor", stack exchange threw up the other tag (whose existence doesn't make much sense to me). $\endgroup$
    – Vivek
    Commented Oct 24, 2019 at 20:28
  • $\begingroup$ @BenCrowell Regarding the question: I do appreciate how the energy-momentum tensor of matter is conserved in special relativity and how it is equivalent to the EOM of a classical field in flat spacetime. But this question is about how Einstein concluded that the covariant derivative (defined through the metric of spacetime) of the EM tensor of matter should be zero in arbitrary spacetimes. At the moment I do not think this answer addresses that - feel free to correct me if I am wrong. $\endgroup$
    – Vivek
    Commented Oct 24, 2019 at 20:35
  • $\begingroup$ @Vivek: Calling these things "energy-momentum" or "stress-energy" is a side issue. The distinction is that one is a four-vector (rank-1 tensor) while the other is a rank-2 tensor. Regardless of which English words we use, it's $p^\mu$ that's conserved, not $T^{\mu\nu}$. $\endgroup$
    – user4552
    Commented Oct 24, 2019 at 20:43
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  1. Diffeomorphism invariance implies (via Noether's 2nd theorem) that $$\nabla_{\mu} T^{\mu\nu}~\approx~0.\tag{1}$$ Perhaps surprisingly, there are no conserved quantities associated with the identity (1) per se for generic spacetime metric $g_{\mu\nu}$.

  2. However, if the metric $g_{\mu\nu}$ has a Killing vector field $K^{\mu}$, then the $4$-current $$J^{\mu}~:=~T^{\mu\nu} K_{\nu} \tag{2}$$ satisfies a continuity equation $$ \nabla_{\mu} J^{\mu}~\approx~0,\tag{3}$$ which leads to a conserved quantity in the standard fashion.

  3. See e.g. my related Phys.SE answer here for more details.

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  • $\begingroup$ #1 sounds wrong to me. My understanding is that Noether's theorem doesn't produce anything useful when applied to diffeomorphism invariance. Some discussion here: physicsforums.com/threads/… . #2 doesn't seem very relevant because most spacetimes don't have a killing vector. And in general this whole answer seems off to me, since zero divergence of the stress-energy has nothing to do with curvature or general relativity. $\endgroup$
    – user4552
    Commented Oct 24, 2019 at 20:18
  • $\begingroup$ Re #3, that seems to be about something other than what the OP is asking about. They simply want to know why the stress-energy tensor has zero divergence. $\endgroup$
    – user4552
    Commented Oct 24, 2019 at 20:58
  • $\begingroup$ @Qmechanic Thanks a lot - you read my mind, esp. the reference to Wald. #1 alone answers the question. The fact that I was not using was that on shell the contribution coming from the variation of matter fields during these infinitesimal coordinate transformations (diffeomorphisms) would be zero. Is it obvious that such diffeomorphisms always exist? I think that would complete the answer. Thanks, again! $\endgroup$
    – Vivek
    Commented Oct 24, 2019 at 21:34

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