What is divergence of a field?

Question

As far as I know, divergence can be grasp physically as the flux of the field passing through an infinitesimal small volume (correct me if I'm wrong).

Can I say now the following: $V_i$ (an infinitesimal volume) is firstly placed in field $F_1$ and secondly in field $F_2$ and after observation it comes out that the flux passing through $V_i$ in $F_1$ is lesser than in $F_2$. So can we say that $F_1$ is more diverge than $F_2$, because it causes less flux to pass through $V_i$?

Answer 1

Perhaps the example of the electric field can help visualize this concept:

A. Metaphor of the electric field Let’s picture an electric field that does not diverge:

1. Imagine a condenser : two conducting plates facing each other; one holding a given charge $Q$. The resulting field is homogeneous (ideally) and equal to $V = Q / C$ (where Q is the capacity of the condenser).

2. If you place a cubic box (order arbitrary size) between the plates (with face $F1$ and $F2$ aligned with the plates) and measure the flux going through $F1$ and $F2$, it will be the same on both faces (with opposite sign) and the sum therefore will be zero: all the electric line fields are parallel, they do not diverge or converge, the divergence is zero.

So, in the everyday sense, the field does not diverge, but if you imagine the box as being of arbitrarily small dimension then the divergence in the mathematical sense is also zero.

B. An electric field that diverges isotropically:

1. Imagine a particle of charge Q (in the middle of nothing), according to Maxwell’s laws the corresponding electric field spawned by said charge is:

$$ \nabla{E} = \rho / \epsilon_0 $$

Where $\rho$ is the density of the charge distribution.

In that case the field lines diverge from the spot where the charge is located in an isotropic way (like metaphorical sun rays emanating from the sun).

2. If you calculate the flux of said electric field lines through a sphere of arbitrary radius it will be constant:

$\iint E.ds = 1/\epsilon_0 \iiint \rho dv = Q/\epsilon_0 $

The electric field lines diverge, but the surface increases in the same proportion, so it cancels out.

Again, since you can imagine the sphere as being of an arbitrary small radius, for an infinitesimally small value of the radius, the divergence of the electric field lines will match the intuitive everyday definition of divergence.

C. Metaphor taken from optics

Another possible metaphor for the divergence of a field is the related concept of divergence / convergence in optics:

1. If you take a magnifying glass (a convergent lens) and you focus sun rays onto a piece of paper you can set fire to said piece of paper, because the rays get focused, concentrated: they converge onto the focal point achieving (ideally) an arbitrary large energy density.

The variation of $divergence$ of the beam as it is being focused by the lens is also in this case, a variation of flux of the light beams between an entry surface (the surface of the lens) and the exit surface: the small area on the piece of paper onto which the beams are redirected by the lens.

Conclusion:

The physical, intuitive, meaning of divergence is the measure of the way the flux of a field varies between the entrance and the exit of a volume traversed by said field.

Alternatively, equivalently: it measures the way a field diverges or converges or gets focused/unfocused, gets fanned out, etc. in space (think light beam, optics).