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I have seen a few posts here on stack exchange that suggest that energy is not conserved in general relativity on a large scale.

I am a pretty confused with this. I don’t understand how such a conclusion is reached.

$\nabla_\mu T^{\mu\nu}$ $=0$

Doesn’t the above equation signify that energy momentum is conserved in GR.

Please note there are other questions already on such issues. But they seem to suggest that energy is not conserved in GR.

See for eg: If the energy of the photon is conserved along a geodesic why is it redshifted

But I am asking that when we have $\nabla_\mu T^{\mu\nu}$ $=0$ why will energy not be conserved in the first place itself

Edit after the comment:

I am asking why $\nabla_\mu T^{\mu\nu}$ $=0$ does not imply energy conservation

Shashaank
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  • I am not familiar with the mathematics of GR, but is the above equation true between reference frames or only in one reference frame? – jng224 Feb 08 '21 at 12:59
  • @Jonas It is an equation so it’s valid in an reference frame. If I had a partial derivative instead of a covariant derivative then it would have been valid in just an inertial frame. – Shashaank Feb 08 '21 at 13:02
  • This question is already answered in the link you provide. This question should be closed. – Oбжорoв Feb 08 '21 at 13:03
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    @Oбжорoв Read the question again. I know that a conserved quantity can be constructed from a killing vector. I am asking why the vanishing of covariant derivative of energy momentum tensor not imply energy conservation. – Shashaank Feb 08 '21 at 13:05
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    Also possible duplicates: https://physics.stackexchange.com/q/2597, https://physics.stackexchange.com/q/109532 – Nihar Karve Feb 08 '21 at 13:08
  • Possible duplicates: https://physics.stackexchange.com/q/306838/2451 and links therein. – Qmechanic Feb 08 '21 at 13:39
  • A proper conservation law means $\partial_{a} T^{ab}=0$. The covariant conservation law isn't really a conservation law in the sense you're talking about because it's implicitly related to the gravitational field. In other words, $\nabla_{a} T^{ab}=0$ also describes the transfer of energy between the gravitational field (not included in $T^{ab}$) and the energy of $T^{ab}$. – Eletie Feb 08 '21 at 13:45
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    @Eletie I vaguely understand what you are saying in your comment. I guess things would be clearer if the explanation is expanded a bit. If you would like to put and answer, I will be glad to accept it. – Shashaank Feb 08 '21 at 13:50

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