# Coordinate Representation of Divergence

## Theorem

Let $\struct {M, g}$ be a Riemannian manifold.

Let $U \subseteq M$ be an open set.

Let $\tuple {x^i}$ be local smooth coordinates.

Let $X$ be a smooth vector field on $M$.

Let $\operatorname {div}$ be the divergence operator.

Then:

$\map {\operatorname {div} } {X^i \dfrac \partial {\partial x^i}} = \dfrac 1 {\sqrt g} \dfrac \partial {\partial x^i} \paren {X^i \sqrt g}$

## Proof

Let $\omega$ be the Riemannian volume form on $M$.

Recall that the divergence of $X$ is defined by the relation:

$\map \d {\omega \rfloor X} = \map {\operatorname {div} } X \omega$, where $\rfloor$ denotes the interior product.

In local coordinates we have:

$\omega = \sqrt g \rd x_1 \wedge \dots \wedge \d x_n$

Thus:

$\omega \rfloor X = \paren {-1}^{i - 1} \sqrt g X^i \rd x_1 \wedge \cdots \wedge \widehat {\d x_i} \wedge \cdots \d x_n$

Thus:

$\d \omega = \map {\dfrac \partial {\partial x_i} } {\sqrt g X^i} \rd x_1 \wedge \cdots \wedge \d x_n$

Thus:

$\map {\operatorname {div} } X = \dfrac 1 {\sqrt g} \dfrac \partial {\partial x_i} \paren {X^i \sqrt g}$