Interpreting divergence of velocity field

Question

The wikipedia article on divergence describes one interpretation of divergence: "The velocity of the air at each point defines a vector field. While air is heated in a region, it expands in all directions, and thus the velocity field points outward from that region."

If we have a vector field which represents a force, I interpret the divergence as representing the strength of the field at whatever point it's taken at. However I'm confused on how to interpret the divergence of a velocity field. Clearly if the divergence is positive gas is expanding outward and if it's negative it's contracting, however what quantity is actually represented?

If I have a velocity field with $m/s$ units, then the div presumably has $m/s^2$ units. Is the quantity we get the actual acceleration of gas away from that point?

Answer 1

I believe the derivative in the divergence operator is with respect to space, not time, therefore the unit is expected to be 1/s, not m/s^2. So it is not the acceleration, but frequency. I am not sure about the physical interpretation of that though.

Answer 2

The interpretation of the divergence of the velocity field as frequency can lead to confusion and not intuitive understanding of the concept. When analyzing the mass conservation equation, you can verify mathematically the role of the divergent velocity in the (Navier-Stokes) conservation equations, but for an intuitive understanding, imagine an expanding solid body.

$\displaystyle \vec{v} =\frac{d\vec{u}}{dt}$ (where $\displaystyle \vec{v}$ is the velocity and $\displaystyle \vec{u}$ is the displacement)

$\displaystyle \nabla \cdotp \vec{v} =\nabla \cdotp \frac{\partial \vec{u}}{\partial t} =\frac{\partial }{\partial t}( \nabla \cdotp \vec{u})$

And here $\displaystyle \nabla \cdot \vec{u}$ plays the role of the volumetric deformation $\displaystyle \varepsilon_{v}$, and therefore

$\displaystyle \nabla \cdot \vec{v} = \frac{\partial \varepsilon_{v}}{\partial t}$

$\displaystyle \varepsilon_{v}$ has no unit, but represents the volume deformation $(\displaystyle \frac{\Delta V}{V_{0}})$, so we see that although the $\displaystyle \nabla \cdotp \vec{v}$ unit is 1/s, its meaning is volumetric deformation per unit time

Answer 3

In solid mechanics, the divergence of the velocity field can be seen as the volumetric strain rate of the solid, which is an invariant of the strain rate tensor. $$ div(v_i) = \frac{\partial v_i}{\partial x_i} = \dot{\varepsilon}_{ii} $$ where $\dot{\varepsilon}_{ij}$ is the strain rate tensor. P.S.: summation convention holds.