Generalized divergence of tensor in GR

Question

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?

Answer 1

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.