Divergence

Introduction

In physics, it is often necessary to be able to get a sense of how much a vector field tends to point towards or away from a particular point. This is precisely what divergence tells us.

Divergence

Imagine a vector field $\mathbf{F}$ and some point $\mathbf{p}$ in ${\mathbb{R}}^3$. To quantify the extent to which $\mathbf{F}$ points towards or away from $\mathbf{p}$, we can construct a sphere $S_r$ of radius $r$ which is centred at $\mathbf{p}$ and then see how much of the field enters and exits through $S_r$. This can be done by using the integral of $\mathbf{F}$ over $S_r$, which is equal to ${\iint }_{S_r}\mathbf{F}\cdot \mathbf{n}\mathrm{d}{S}_r$, where $\mathbf{n}$ is the unit surface normal.

Recall that the dot product quantifies the extent to which two vectors are oriented in a similar direction. Essentially, at each point of the sphere $S_r$, the dot product $\mathbf{F}\cdot \mathbf{n}$ tells us how aligned $\mathbf{F}$ is with $\mathbf{n}$ there. A positive value means that $\mathbf{F}$ points in a direction which is generally aligned with $\mathbf{n}$, i.e. away from the sphere. Conversely, a negative value means that $\mathbf{F}$ points towards the sphere. The integral is just the sum of these values. If it is positive, then there are more points on the sphere where $\mathbf{F}$ tends to point outwards. If it is negative, then there are more points on the sphere where $\mathbf{F}$ tends to point inwards.

Moreover, as we make the radius $r$ smaller and smaller, the sphere closes in on $\mathbf{p}$. In the limit of $r\to 0$, the part of the field which enters the sphere is essentially the part of the field which points towards $\mathbf{p}$. Similarly, the part of the field which exits the sphere is the part of the field which points away from $\mathbf{p}$.

Theorem: Existence of Divergence

Let $\mathbf{F}:\mathcal{D}\subseteq {\mathbb{R}}^3\to {\mathbb{R}}^3$ be a real vector field which is differentiable at $\mathbf{p}$ and let $S_r$ be a parametric surface whose image is a sphere of radius $r$ centered at $\mathbf{p}$ such that its normal vector is always oriented outwards.

If there exists some neighbourhood of $\mathbf{p}$ where $\mathbf{F}$ is continuously differentiable, then the limit

$$\underset{r\to 0^{+}}{\lim }\frac{1}{V_r}{\iint }_{S_r}\mathbf{F}\cdot \mathrm{d}{\mathbf{S}}_r,$$

where ${\iint }_{S_r}\mathbf{F}\cdot \mathrm{d}{\mathbf{S}}_r$ is the surface integral of $\mathbf{F}$ over $S_r$ and $V_r$ is the volume of $S_r$, exists.

PROOF

TODO

Definition: Divergence

The divergence of $\mathbf{F}$ at $\mathbf{p}$ is defined as precisely

$$\underset{r\to 0^{+}}{\lim }\frac{1}{V_r}{\iint }_{S_r}\mathbf{F}\cdot \mathrm{d}{\mathbf{S}}_r,$$

NOTATION

$$\nabla \cdot \mathbf{F}(\mathbf{p})\mathrm{div}\mathbf{F}(\mathbf{p})$$

Note: Divergence as a Function

When there is no specific $\mathbf{p}\in \mathcal{D}$ mentioned, the term “divergence” is used to refer to the scalar field $\mathrm{div}\mathbf{F}:\mathcal{D}\to \mathbb{R}$ which maps each $\mathbf{x}$ to its divergence $\nabla \cdot \mathbf{v}(\mathbf{x})$.

Theorem: Divergence in Cartesian Coordinates

Let $\mathbf{v}:\mathcal{D}\subseteq {\mathbb{R}}^3\to {\mathbb{R}}^3$ be a real vector field with component functions $v_1,v_2,v_3$ and let $\mathbf{p}\in {\mathbb{R}}^3$.

If $\mathbf{v}$ is continuously differentiable at $\mathbf{p}$, then its divergence there can be calculated using the partial derivatives of $v_1,v_2,v_3$ with respect to Cartesian coordinates:

$$\nabla \cdot \mathbf{v}(\mathbf{p})=\frac{\partial v_1}{\partial x}(\mathbf{p})+\frac{\partial v_2}{\partial y}(\mathbf{p})+\frac{\partial v_3}{\partial z}(\mathbf{p})$$

PROOF

TODO

Properties of Divergence

Theorem: Linearity of the Divergence

The divergence is linear - for all $\lambda ,\mu \in \mathbb{R}$ and all vector fields $\mathbf{u},\mathbf{v}$ which are differentiable at $\mathbf{p}\in {\mathbb{R}}^3$ we have

$$\mathrm{div}(\lambda \mathbf{u}+\mu \mathbf{v})(\mathbf{p})=\lambda \mathrm{div}\mathbf{u}(\mathbf{p})+\mu \mathrm{div}\mathbf{v}(\mathbf{p})$$

PROOF

TODO