Although I've forgotten the proof (and cannot find it in, say, Carroll's book), the following formula holds for the covariant divergence in general relativity:

$$\nabla_{\mu} A^{\mu} = \frac{1}{\sqrt{|g|}} \partial_{\mu} \left( \sqrt{|g|} A^{\mu}\right),$$

where $g = \det(g_{\alpha\beta})$. I was wondering if this formula holds if $A^{\mu}$ is replaced with a general rank $(n,m)$ tensor

$$T^{\mu \mu_1\mu_2 \cdots \mu_{n-1}}_{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\nu_1\cdots \nu_m}?$$

If not, could you point me to any references that have divergence formulas for higher rank tensors?

  • 1
    $\begingroup$ It does not. It holds when $A^\mu$ is replaced with a general $p$-form (which is a totally antisymmetric $(p,0)$ tensor. $\endgroup$ – Prahar Mitra Oct 1 '18 at 14:40

No, this does not hold in general for higher-rank tensors. The general equation for the divergence of a completely contravariant tensor in terms of a coordinate derivative operator $\partial_\mu$ is $$ \nabla_\mu T^{\mu \nu_1 \dots \nu_n} = \partial_\mu T^{\mu \nu_1 \dots \nu_n} + \Gamma^\mu {}_{\mu \rho} T^{\rho \nu_1 \dots \nu_n} + \sum_{i = 1}^n \Gamma^{\nu_i} {}_{\mu \rho} T^{\mu \nu_1 \dots \rho \dots \nu_n}. $$ We also have the fact that $$ \Gamma^\mu {}_{\mu \rho} = \frac{1}{\sqrt{|g|}} \partial_\mu \sqrt{|g|}. $$ Thus, $$ \nabla_\mu T^{\mu \nu_1 \dots \nu_n} = \frac{1}{\sqrt{|g|}} \partial_\mu \left( \sqrt{|g|} T^{\mu \nu_1 \dots \nu_n} \right) + \sum_{i = 1}^n \Gamma^{\nu_i} {}_{\mu \rho} T^{\mu \nu_1 \dots \rho \dots \nu_n}. $$ This last sum will not vanish for a general tensor. However, some or all of the terms may vanish for tensors with a particular symmetry structure. In particular, if $T^{\mu \nu_1 \dots \nu_n}$ is antisymmetric in all of its indices, then any contraction of two of its indices with the symmetric indices of the Christoffel symbols automatically vanishes; and thus the entire sum goes away.

For references as to how to take the covariant derivative of a general tensor, see Chapter 5 of Schutz's A First Course in General Relativity for a coordinate-based approach, or Chapter 3 of Wald's General Relativity for a more general approach.

  • $\begingroup$ I believe that for general Christoffel symbols and non-zero $T$, total antisymmetry is the only case where the extra sum vanishes. $\endgroup$ – Void Oct 1 '18 at 15:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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