Let \((M,g)\) be a Riemannian manifold . Then the local expression of the divergence of a smooth vector field \(X^i\partial_i\) is given by:
\begin{equation*} \frac{1}{\sqrt{\det g}} \frac{\partial }{\partial x^i}\bigl(\sqrt{\det g} X^i\bigr), \end{equation*}where \(\det g := \det (g_{ij})\) is the determinant of the component matrix of \(g\) in these coordinates.
Proof (Sketch) Link to heading
To prove this identity First calculate \(X\lrcorner dV_g\) by determining the basis components. Afterwards apply the defining sequentially operations on the resulting object.
Remarks
- For the euclidean metric the divergence coincide with the common definition: \begin{equation*} \div X = \sum_{i=1}^{n} \frac{\partial X^i}{\partial x^i}. \end{equation*}
- Using (0x66deed94) (trace representation of divergence) and the local representation of the total covariant derivative we may simplify the local expression for the divergence by \(\div X = {X^i}_{;i}\).