Theory

Divergence as a derivative operator

Start with a vector field $\mathbf{F}$ with vector components $\mathbf{F}=(F_1,F_2)$, where $F_1(x,y)$ and $F_2(x,y)$ are functions of points in space.

From this $\mathbf{F}$ we create a scalar field called the divergence of $\mathbf{F}$, written $\nabla \cdot \mathbf{F}$. This field is calculated with the formula:

$$\nabla \cdot \mathbf{F}=\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}.$$

One can remember this formula using the “vector” $\nabla =\left(\frac{\partial }{\partial x},\frac{\partial }{\partial y}\right)$; the formula expresses a dot product of $\nabla$ and $\mathbf{F}$.

This formula naturally extends to higher dimensions. For example, in 3D we have:

$$\nabla \cdot \mathbf{F}=\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}+\frac{\partial F_3}{\partial z}.$$

Divergence as a derivative operator (part 2)

In physics it is common to use polar (or cylindrical or spherical) coordinates because many problems have rotational symmetry. The derivative operators of vector calculus can be expressed in these coordinate systems.

Divergence as a derivative operator (part 3)

Polar divergence looks different

Given a vector field with polar coordinates $F_r$ and $F_{\theta }$, its divergence is given by:

$$\nabla \cdot \mathbf{F}=\frac{1}{r}\frac{\partial }{\partial r}(r{F}_r)+\frac{1}{r}\frac{\partial }{\partial \theta }{F}_{\theta }$$

Divergence as measure of expansion

The divergence can be explained as a measure of expansion of the vector field at each point in space. (Negative divergence means contraction.) If the vector field corresponds to the velocity of a fluid flow, then divergence measures the rate at which fluid flows out of (or into) each point. We can verify this interpretation by using a vector line integral to measure the flow out of a circular loop, and then shrink the loop down to an infinitesimal loop.

Suppose $\mathbf{F}=(F_1,F_2)$ is given. Let ${\mathcal{C}}_r$ be a circle of radius $r$ centered at $\mathbf{p}=(a,b)$, oriented counter-clockwise. We will compute the vector line integral across ${\mathcal{C}}_r$ in the limit as $r\to 0$.

$${\int }_{{\mathcal{C}}_r}\mathbf{F}\cdot {\mathbf{e}}_{\mathbf{n}}ds.$$

Calculate this integral using linear approximations for $F_1$ and $F_2$. Let $\mathbf{r}(t)=(a+r\cos (t),b+r\sin (t))$, and write $q_1=F_1(\mathbf{p})$ and $q_2=F_2(\mathbf{p})$.

We have:

• $F_1(\mathbf{r}(t))\approx q_1+\frac{\partial F_1}{\partial x}r\cos (t)+\frac{\partial F_1}{\partial y}r\sin (t)$,
• $F_2(\mathbf{r}(t))\approx q_2+\frac{\partial F_2}{\partial x}r\cos (t)+\frac{\partial F_2}{\partial y}r\sin (t),$
  where the partial derivatives are evaluated at $\mathbf{p}$.

Now ${\mathbf{r}}^{\prime }(t)=(-r\sin (t),r\cos (t))$ and $\mathbf{n}(t)=(r\cos (t),r\sin (t))$, so the line integral becomes:

$$\begin{array}{rl}{\int }_{{\mathcal{C}}_r}\mathbf{F}\cdot {\mathbf{e}}_{\mathbf{n}}ds& \approx {\int }_0^{2\pi }\left(q_1+\frac{\partial F_1}{\partial x}r\cos (t)+\frac{\partial F_1}{\partial y}r\sin (t)\right)r\cos (t)\\ & +\left(q_2+\frac{\partial F_2}{\partial x}r\cos (t)+\frac{\partial F_2}{\partial y}r\sin (t)\right)r\sin (t)dt\\ & =\left(\frac{F_1}{\partial x}+\frac{F_2}{\partial y}\right)\pi r^2=(\nabla \cdot \mathbf{F})\pi r^2.\end{array}$$

In the limit as $r\to 0$, this becomes $\nabla \cdot \mathbf{F}dA$, where $dA$ is the infinitesimal area of the circle.