divergence

In Riemannian geometry, the **divergence** of a vector field $X$ over a Riemannian manifold $(M,g)$ is the real valued smooth function $div(X)$ defined by

$div(X) = \star_g^{-1} d_{dR} \star_g g(X)
,$

where $\star_g$ is the Hodge star operator of $(M,g)$,

$\star_g\colon \Omega^i(M;\mathbb{R}) \to \Omega^{dim M-i}(M;\mathbb{R})
,$

and $d_{dR}$ is the de Rham differential.

Alternatively, the divergence of a vector field $\vec\mathcal{A}$ in some point $x\in M$ is calculated (or alternatively defined) by the integral formula

$div \vec\mathcal{A} = \lim_{vol D\to 0} \frac{1}{vol D} \oint_{\partial D} \vec{n}\cdot \vec\mathcal{A} d S$

where $D$ runs over the open submanifold?s containing point $x$ and with smooth boundary $\partial D$ and $\vec{n}$ is the unit vector of outer normal to the hypersurface? $S$. The formula does not depend on the shape of boundaries taken in limiting process, so one can typically take a coordinate chart and balls with decreasing radius in this particular coordinate chart.

Although an orientation is required for the usual notion of Hodge star as given above, we may take it as valued in pseudoforms to show that the orientation (or even orientability) of $M$ is irrelevant (since the Hodge star is applied twice, returning us to untwisted forms, and since a bounding hypersurface has a natural ‘outwards’ pseudoorientation). However, the metric, which is hidden in the volume form and in the “dot product”, is relevant.

If $(M,g)$ is the Cartesian space $\mathbb{R}^n$ endowed with the canonical Euclidean metric, then the divergence of a vector field $X^i \partial_i$ is

$div(X) = \sum_{i=1}^n\frac{\partial X^i}{\partial x^i}
.$

The divergence was first developed in quaternion analysis, where its opposite appeared most naturally, called the *convergence* $con(X) = - div(X)$. In many applications of the divergence to the successor field, classical vector analysis?, the metric is irrelevant and we may use differential forms instead: we translate a vector field $X$ into the $(n-1)$-form $\star_g g(X)$ and a scalar field $f$ into the $n$-form $\star_g f$, so that the divergence *is* simply the de Rham differential, and simply use the differential forms from the start.

Revised on June 11, 2013 02:09:54
by Toby Bartels
(64.89.53.249)