In a Riemannian manifold $\mathcal{S}$ with metric $\boldsymbol{g}$, given a chart $\{x^a\}$, it is fairly easy to prove that the divergence of a vector field $\boldsymbol{w} : \mathcal{S} \to T\mathcal{S}$ is given by $$ \mathrm{div}\boldsymbol{w} = w^a{}_{|a} = \frac{1}{\sqrt{\det[g_{mn}]}} \frac{\partial}{\partial x^a} \left[ \sqrt{\det[g_{mn}]} \, w^a \right], $$ where $[g_{mn}]$ is the matrix representation of $\boldsymbol{g}$ in the chart $\{x^a\}$, and the vertical bar in ${}_{|a}$ denotes the covariant derivative with respect to the Levi-Civita connection induced by $\boldsymbol{g}$.

I am wondering whether there is an equivalent expression for the divergence of a higher-order tensor, say, a second-order tensor $\boldsymbol{\sigma} : \mathcal{S} \to T\mathcal{S} \otimes T\mathcal{S}$, in which case the component form is $(\mathrm{div}\boldsymbol{\sigma})^a = \sigma^{ab}{}_{|b}$, with the divergence taken, e.g., on the last index.

PS: In case you are curious, yes, I am looking for an expression of the divergence of the (fully contravariant) Cauchy stress.

  • $\begingroup$ For a general $2$-tensor $\sigma^{ab}$, there are two different divergences, namely $\sigma^{ab}{}_{,a}$ and $\sigma^{ab}{}_{, b}$. For a symmetric $2$-tensor (like the Cauchy stress), however, these two contractions produce the same tensor. $\endgroup$ – Travis Jun 11 '15 at 2:47
  • $\begingroup$ I am assuming that the divergence is taken with respect to the last tensor leg (i.e., the last index), although you are correct that it makes no difference for a symmetric tensor. I have now specified that in the question. $\endgroup$ – S.F. Jun 11 '15 at 4:06

There is indeed such expression, you have written it yourself: $\sigma^{ab}|_{b}$. Now it is a matter of writing the Christoffel symbols in the form you want, on the one hand the covariant derivative acts on both indices $a,b$ so the answer is:

$$ \sigma^{ab}|_{b} = \frac{1}{\sqrt{\det g}}\partial_b (\sqrt{\det g} \sigma^{ab}) + \Gamma^{a}_{bc}\sigma^{cb} $$ but I doubt that the second term could be made into a total derivative because it invloves all the components of the Christoffel symbol.

  • $\begingroup$ I certainly agree with your expression, as the first term on the right-hand side reduced to $\sigma^{ab}{}_{,b} + \Gamma^b_{cb} \sigma^{ac}$, which, together with the second term $\Gamma^a_{bc} \sigma^{cb}$, completes the covariant derivative of $\boldsymbol{\sigma}$, but I was really hoping to find an expression giving $\sigma^{ab}{}_{|b}$ as the derivative of "something". I am afraid that you are right in that this cannot be achieved. $\endgroup$ – S.F. Jun 11 '15 at 5:26

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