Relationship between divergence operators defined with respect to two different volume forms. 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 $\operatorname{div}_2$) ?

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 : $$ \alpha \operatorname{div}_1(v) = \operatorname{div}_2(\alpha v) $$
