Relationship between divergence operators defined with respect to two different volume forms.

Question

Let us assume that you have a volume form $\mu$ defined on a manifold $\mathcal{M}$. Then you can define the divergence operator with respect to this metric, such that the following relationship holds for all $\mathcal{U}\subset \mathcal{M}$ and $v \in T\mathcal{M}$ : $$ \int_{\mathcal{U}}{\operatorname{div} v \ \mu} = \int_{\partial \mathcal{U}}{ \mathbf{i}_v \mu }$$

Now suppose that you have two distinct non degenerate volume forms $\mu_1$ and $\mu_2$. Then there exists a scalar field $\alpha$ such as $\mu_1 = \alpha \mu_2$.

How are related (in terms of $\alpha$) the corresponding divergence operators (let's call them $\operatorname{div}_1$ and $\operatorname{div}_2$) ?

Answer 1

I've actually worked out the answer to the question from the following definition of the divergence : $\operatorname{div}(v)\mu = \operatorname{d}(\mathbf{i}_v \mu)$.

Short-circuiting a few computation lines, the result is : $$ \alpha \operatorname{div}_1(v) = \operatorname{div}_2(\alpha v) $$

Answer 2

If $\mu_1 = \alpha\mu_2$ for some positive scalar function $\alpha$, then for any vector field $v$, $$ \operatorname{div}_1 v = \operatorname{div}_2v + v(\log\alpha). $$ Here's a proof. Note that $(\operatorname{div}_j v)\mu_j = d(i_v\mu_j)$ for $j=1,2$. Thus \begin{align*} (\operatorname{div}_1 v)\mu_1 &= d(i_v\mu_1)\\ &= d(i_v(\alpha\mu_2))\\ &= d(\alpha i_v\mu_2)\\ &= d\alpha \wedge i_v\mu_2 + \alpha d(i_v\mu_2)\\ &= (v\alpha) \mu_2 + \alpha(\operatorname{div}_2 v)\mu_2\tag{$*$}\\ &= \left(\frac{v\alpha}{\alpha} + \operatorname{div}_2 v\right)\mu_1. \end{align*} The first term in ($*$) follows from the fact that interior multiplication by $v$ is an antiderivation. Since $d\alpha\wedge\mu_2$ is an $(n+1)$-form on an $n$-manifold, it is zero, so $$ 0 = i_v(d\alpha\wedge\mu_2) = (i_v d\alpha)\wedge \mu_2 - d\alpha\wedge (i_v\mu_2), $$ and $$ i_v d\alpha = d\alpha(v) = v\alpha. $$