# 9.4: How does divergence relate to the air parcel’s area change?

We see that divergence is positive when the parcel area grows and is negative when it shrinks. We call growth “divergence” and shrinking “convergence.” We wish to know whether air parcels come together (converge) or spread apart (diverge) or if the parcel area increases with time (divergence) or decreases with time (convergence).

Let’s see how divergence in the horizontal two dimensions is related to area change. We can do a similar analysis that relates divergence in three dimensions to a volume change, but we will stay with the two-dimensional case because it is easier to visualize and also has important applications. Consider a box with dimensions Δx and Δy. Different parts of the box are moving at different velocities (see figure below). A box that is moving at greater velocity for parts that are at greater x and greater y.

Credit: W. Brune

The box's area, A, is given by:

$$A=\Delta x \Delta y$$

\begin{aligned} \frac{d A}{d t}=& \frac{d(\Delta x \Delta y)}{d t}=\Delta x \frac{d(\Delta y)}{d t}+\Delta y \frac{d(\Delta x)}{d t}=\\ & \Delta x[v(y+\Delta y)-v(y)]+\Delta y[u(x+\Delta x)-u(x)] \end{aligned}

divide by $$A=\Delta x \Delta y$$

$\frac{1}{A} \frac{d A}{d t}=\frac{v(y+\Delta y)-v(y)}{\Delta y}+\frac{u(x+\Delta x)-u(x)}{\Delta x}$

Let $$\Delta y \rightarrow 0, \Delta x \rightarrow 0$$

$\frac{1}{A} \frac{d A}{d t}=\frac{\partial u}{\partial x}+\frac{\partial v}{\partial v}=\vec{\nabla}_{H} \bullet \vec{U}_{H}$

So we see that the fractional change in the area is equal to the horizontal divergence. Note that the dimension of divergence is time–1 and the SI unit is s–1.

We can do this same analysis for motion in three dimensions to get the equation:

$\frac{1}{V} \frac{d V}{d t}=\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}=\vec{\nabla}_{H} \cdot \vec{U}_{H}+\frac{\partial w}{\partial z}=\vec{\nabla} \cdot \vec{U}$

where V is the parcel volume. Thus, the 3-D divergence is just the fractional rate of change of an air parcel’s volume.

Exercise

Suppose that an air parcel has an area of 10,000 km2 and it is growing by 1 km2 each second. What is its divergence?

$$\frac{\Delta A}{\Delta t}=1 \mathrm{km}^{2} \mathrm{s}^{-}$$, so

$$\left(\frac{1}{A}\right)\left(\frac{\Delta A}{\Delta t}\right)=\left(\frac{1}{10^{4}} \mathrm{km}^{2}\right)\left(1 \mathrm{km}^{2} \mathrm{s}^{-1}\right)=10^{-4} \mathrm{s}^{-1}$$

Suppose that an air parcel has a area of 10,000 km2 and has a divergence of –10–4 s–1. Is the air parcel growing or shrinking?

divergence $$=\delta=\left(\frac{1}{A}\right)\left(\frac{\Delta A}{A t}\right),$$ or

$$\frac{\Delta A}{A}=\delta \Delta t=\left(-10^{-4} \mathrm{s}^{-1}\right)(1 \mathrm{s})=-10^{-4}$$. The air parcel is shrinking.

Check out this video (1:33) for further explanation:

Divergence Area