Divergence of a Radial Vector Field

[cepheid]

Something we did in electrostatics that's a source of confusion for me:

We learned to use caution when taking the divergence of the (all important) radial vector field:

\vec{v} = \frac{1}{r^2} \hat{r}

Applying the formula in spherical coords gave zero...a perplexing result. The problem, it was revealed, was that the function increases without bound as the origin is approached (I don't know, I tried to phrase it properly. My physics prof just said...it "blows up").

So we were reminded of the Dirac Delta function, which saved the day:

\nabla \cdot \vec{v} = 4\pi\delta^3(r)

Fine. I tried to understand why there was a singularity at the centre. In other words, 1. The divergence tended to infinity at the origin, and 2. it was zero everywhere else. My reasoning was:

1. The divergence in this particular example is the rate at which the radial component of the vector field changes as r changes. So...from the centre, not only is the magnitude of the field "infinitely large" but, it immediately changes to having some finite value at some finite distance r from the origin. So the rate of change is also infinitely large...there is "infinite divergence" from the centre.

2. The divergence has a magnitude of zero everywhere else but the origin, because at any of these other points, the vector field is not "diverging away" from that point. It is confined to pass through it in only one direction (the radial direction).

There are two problems that I subsequently discovered with #1 and #2. My reasoning as it has been stated should apply to any vector field in the radial direction whose magnitude was inversely proportional to radial distance from the centre. This was clearly not true!

\nabla \cdot \left(\frac{1}{r}\hat{r}\right) = \frac{1}{r^2}

and in general:

\nabla \cdot \left(r^n \hat{r}\right) = n+2(r^{n-1})
(except for n = -2)

So my reasoning is shot to pieces! I don't understand this...there is no singularity at the centre for other radial vector fields of this form with n < 0
(even though they still "blow up" at the origin!). Not only that, but suddenly we have a non-zero divergence everywhere else. The field is still radial only, so from what point in space other than the origin could the field possibly be "diverging"? It still passes through these points in only one direction! Admittedly, its magnitude is diminishing as it does so, but that is true for 1/r^2 as well!

The math says what it says...so what is wrong with #1 and #2?

 

[arildno]

1)The point source vector field (\vec{F}=\frac{1}{r^{2}}\vec{i}_{r}), is NOT DEFINED at the origin.
2) Hence, neither is its divergence there.
3) The following surface integral is, however, a perfectly legitimate mathematical object:
\int_{S}\vec{F}\cdot\vec{n}dS
where S is a surface enclosing the origin.
It is legitimate, since the vector field \vec{F} is only required to be defined at the points on the surface S.
4) Precisely because we have a singularity within S, we cannot naively use the divergence theorem to rewrite the surface integral in terms of a volume integral (which requires the existence of \vec{F} at ALL interior points.
5) This is where Dirac's delta "function" comes in and saves the day, in that by the introduction of the delta function formalism, we are able to RETAIN the divergence theorem in the case of the point source field.

6)We can show, that in the case of the point source field,
\int_{S}\vec{F}\cdot\vec{n}dS=4\pi
that is, a quantity independent of our choice of S!
This LEGITIMATES the introduction of Dirac's delta function, so that we have:
\int_{S}\vec{F}\cdot\vec{n}dS=\int_{V}\nabla\cdot\vec{F}dV
PROVIDED that we DEFINE:
\nabla\cdot\vec{F}\equiv{4\pi}\delta^{3}(r)

7) The point source field is the only LOCALLY DIVERGENCE-FREE radial vector field in 3-D at the points where it is defined.
As you've found, the divergence of the general radial vectorfield \vec{F}=r^{n}\vec{i}_{r} equals
\nabla\cdot{F}=(n+2)r^{n-1}
(at the points where it is defined)
8) To get a feel with what is meant with "local divergence-freeness" (, I should say "solenoidal" instead), let us consider a region about a point \vec{x}_{0}\neq\vec{0}
Consider the cone with vertex at the origin, subtending the solid angle \gamma and consider the region V about \vec{x}_{0} bounded by the cone surfaces, and the (parts of the) spherical surfaces S_{0}, S_{1} (which, measured from the origin, are characterized by radii r_{0},r_{1}).
The area of S1 equals A_{1}=\gamma{r}_{1}^{2}; a similar expression holds for S0.

Clearly, net fluxes of a radial field \vec{F} only appear on S_{0},S_{1}, since \vec{F} is tangential to the cone surface.

When \vec{F}=r^{n}\vec{i}_{r} the quantity
\vec{F}\cdot\vec{n}=r^{n} is constant on both S_{0},S_{1} and the net flux through V is therefore given by:
(r_{1}^{n}\gamma{r}_{1}^{2}-r_{0}^{n}\gamma{r}_{0}^{2})=\gamma(r_{1}^{n+2}-r_{0}^{n+2})
This is seen to be zero (given non-zero solid angle and distinct radii) only if n=-2; i.e, by choosing the point-source field.

9) In the case of n=-2, we see that for arbitrary, non-zero choices of radii, the net flux through the boundaries of V is always zero, which corresponds to a local divergence-free behaviour within V.

 

[StatusX]

Another way to think of it is that the divergence at a point is a measure of the number of field lines originating at that point. The flux across a surface is the number of field lines crossing that surface. Since in this case, the flux over any sphere centered at the origin is the same(since the area of the spheres rises as r², exactly balancing how the field falls), all the field lines must originate at the origin. For the general case of radial fields falling as 1/rⁿ, the flux across these spheres changes as r increases. If n>2, it falls as r increases, and so there must be sinks at points r>0, and hence the divergence is negative here. For n<2, there must be more sources for r>0, and the divergence is positive.