# Coordinate and tensor divergence theorems

The expression previously obtained for the divergence theorem was

\displaystyle \begin{aligned}\int_{V}\mathrm{div}(u)\mathrm{d}V & =\int_{\partial V}i_{u}\mathrm{d}V\\ & =\int_{\partial V}\left\langle u,\hat{n}\right\rangle \mathrm{d}S, \end{aligned}

where $${V}$$ is an $${n}$$-dimensional compact submanifold of $${M^{n}}$$, $${\hat{n}}$$ is the unit normal vector to $${\partial V}$$, and $${\mathrm{d}S\equiv i_{\hat{n}}\mathrm{d}V}$$ is the induced volume element (“surface element”) for $${\partial V}$$. If we choose an orthonormal frame with $${e_{1}=\hat{n}}$$ on $${\partial V}$$, the divergence theorem can be written

\begin{aligned}\int_{V}\mathrm{div}(u)\mathrm{d}V & =\int_{\partial V}u^{1}\mathrm{d}S,\end{aligned}

and if we can choose coordinates with $${x^{1}}$$ constant on $${\partial V}$$ and normal to it, the divergence theorem can be written

\begin{aligned}\int_{V}\partial_{\lambda}\left(\sqrt{g}u^{\lambda}\right)\mathrm{d}^{n}x & =\int_{\partial V}\sqrt{g}\mathrm{d}x^{1}\left(u\right)\mathrm{d}^{n-1}x\\
& =\int_{\partial V}u^{1}\sqrt{g}\mathrm{d}^{n-1}x,
\end{aligned}

where $${\mathrm{d}^{n}x\equiv\mathrm{d}x^{1}\wedge\cdots\wedge\mathrm{d}x^{n}}$$ and $${\mathrm{d}^{n-1}x\equiv\mathrm{d}x^{2}\wedge\cdots\wedge\mathrm{d}x^{n}}$$.

Since the “divergence” of a tensor $${T}$$ with order greater than 1 is tensor-valued, and the parallel transport of tensors is path-dependent, we cannot in general integrate to get a divergence theorem for tensors. In the case of a flat metric and zero torsion however, we can choose coordinates whose coordinate frame is orthonormal, so that the frame is its own parallel transport, i.e. $${\nabla_{v}\left(\beta^{\mu}\right)=0}$$. For e.g. a tensor $${T^{ab}}$$, we can then define a coordinate-dependent vector $${J^{\mu}}$$ for each index $${\mu}$$

\begin{aligned}\require{cancel}J^{\mu}&\equiv T\left(\beta^{\mu},\quad\right)\\\Rightarrow\left(J^{\mu}\right)^{b}&=T^{\mu b}\\\Rightarrow\overline{\nabla}_{v}J^{\mu}&\overset{\cancel{R}}{=}\beta^{\mu}\overline{\nabla}_{v}T\\\Rightarrow\overline{\nabla}_{b}\left(J^{\mu}\right)^{b}&\overset{\cancel{R}}{=}\overline{\nabla}_{b}T^{\mu b}\\\Rightarrow\int_{V}\overline{\nabla}_{b}T^{\mu b}\mathrm{d}V&\overset{\cancel{R}}{=}\int_{V}\overline{\nabla}_{b}\left(J^{\mu}\right)^{b}\mathrm{d}V\\&=\int_{\partial V}T^{\mu}{}_{b}\hat{n}^{b}\mathrm{d}S.
\end{aligned}

For arbitrary coordinates, the components of the coordinate frame are by definition constant, i.e. $${\partial_{v}\left(\mathrm{d}x^{\mu}\right)=0}$$; we can therefore write

This relation remains true in the presence of both curvature and torsion, however it is important to note that $${\partial_{\nu}\left(\sqrt{g}T^{\mu\nu}\right)}$$ is not a “divergence” and $${T^{\mu b}=\left(J^{\mu}\right)^{b}}$$ is coordinate-dependent. In the special case of an anti-symmetric tensor under zero torsion, we can write