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?


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?"


1 Answer 1


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 $\nabla_\nu A^{\mu\nu}=0$.

  • 3
    $\begingroup$ Covariant derivatives do not necessarily commute. $\endgroup$
    – jacob1729
    Commented May 29, 2020 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
    Commented May 29, 2020 at 16:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.