# Divergence of vector field

## Main Question or Discussion Point

Hello I am trying to get my head around what the divergence actually represents physically.

If you have some vector field v, and the components of v, vx, vy, vz have dimensions of kg/s ("flow" - mass of material per second) the divergence will have units of kg/(s*m) (mass per time distance)

Say the divergence of v is constant in some region R with volume a.

div(v)*a has units (kg*m^2)/s (mass area per time) - this is the flux of v through area(R)

(div(v)*a)/area(R) has units (mass per time) - the net mass flowing out of R in some time

So what exactly is divergence - kg/(s*m)

Would it be accurate to think of the divergence as "Flux per volume" in general?

its where the flux lines end

Ben Niehoff
Gold Member
Would it be accurate to think of the divergence as "Flux per volume" in general?
Exactly.

Take some small volume V. Let $\Phi$ be defined as the flux of a vector field u out of the volume V. That is, if S is the closed surface bounding V, then $\Phi$ is the flux across S, in the outward direction. Then the divergence is given by the limit

$$\mathrm{div} \vec u = \lim_{V \rightarrow 0} \frac{\Phi}{V}$$

exactly. Or to be more explicit on the above result,
$$\mathrm{div}\vec{u} = \frac{\oint_{S} \vec{u}\cdot \vec{dS}}{V}$$.
In fact we can derive this using the mean value theorem on integrals on the divergence theorem.

Divergence is simply flux density, flux per volume. In different vector fields the flux density can vary at different points in space. If you add all of the fluxes/volume up and multiply by volume then you will get the total flux through the solid.

What is the usefulness in knowing the Divergence of a vector field? I mean I realize it is important with regards to stuff like Maxwells equations. But I only learned those in Integral Form, and not differential form.

Also can someone conceptually explain what the Curl represents?

Ben Niehoff