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In short:

For a stress-energy tensor $T^{\mu\nu}$, what are possible additions that will leave the tensor equations of motion $\nabla_\nu T^{\mu\nu} = 0$ unchanged?

Context:

Any modification, $T^{\mu\nu} \rightarrow T^{\mu\nu} + A^{\mu\nu}$ where $\nabla_\nu A^{\mu\nu} = 0$ as an identity, would work.

In various publications and text books, it appears that people say one can add something that looks like

$A^{\mu\nu} = \nabla_\lambda B^{\mu\nu\lambda}$, where $B^{\mu\nu\lambda}=-B^{\mu\lambda\nu}.$

See for instance, here for a discussion in the context of curved space-time (beginning of Section 2).

However, I calculate (using the antisymmetry of $B$ and Eq. 3.68 from Carroll's Lecture Notes on GR)

$\nabla_\nu A^{\mu\nu} = \nabla_\nu \nabla_\lambda B^{\mu\nu\lambda} = \frac{1}{2}R^\mu_{\lambda\nu\alpha}B^{\lambda\nu\alpha}$ ($R$ is the curvature tensor)

so it wouldn't effect the equations of motion in flat space-time, but in curved space-time, it does affect them.

  1. Am I missing something here? Some identity that makes the expression above zero even in curved space-time? The derivations only a few lines, and I don't believe I made a mistake.

    1. If this is not zero in curved space-time, is there a general form that describes the family of tensors one can add to the stress-energy tensor without changing the equations of motion?

PS: This is not a question about how to derive the stress-energy tensor in GR (varying the Lagrangian with respect to the metric), or its uniqueness. In GR, it seems to me the stress-energy tensor is uniquely defined by varying Lagrangian with respect to the metric. Any non-zero addition would add to the curvature of space-time, which implicitly changes the equations of motion by changing the covariant derivative. This is just a question about portions of the stress-energy tensor that contribute to the curvature, but not the equations of motion explicitly; I could rephrase as: "Given a stress-energy tensor, what portions of the tensor do not explicitly contribute to the tensor equations of motion for matter?"

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Applying the covariant derivative $\nabla_\nu$ to $A^{\mu \nu}$ you have

$$\nabla_\nu \nabla_\lambda B^{\mu \nu \lambda}.$$

You then have a contraction between two tensors in the $\nu, \lambda$ indices. The first tensor $\nabla_\nu \nabla_\lambda$ is obviously symmetric in these two indices since covariant derivatives commute. Whereas $B$ is anti-symmetric in these two indices. The contraction of a symmetric and anti-symmetric tensor vanishes, hence $\nabla_\nu A^{\mu\nu}=0$.

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    $\begingroup$ Covariant derivatives do not necessarily commute. $\endgroup$ – jacob1729 May 29 '20 at 11:48
  • $\begingroup$ If you take a look at Eq. 3.68 from Carroll's Lecture Notes on GR, you'll see the commutator of the covariant derivative is related to the curvature, which is how I came up with the expression in my question. $\endgroup$ – juacala May 29 '20 at 16:42

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