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 $\operatorname{div}_2$) ?
 A: 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
$$ i_v d\alpha = d\alpha(v) = v\alpha. $$
A: 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 : $$ \alpha \operatorname{div}_1(v) = \operatorname{div}_2(\alpha v) $$
