Why Is There No Generalized Function for div (r̂ / r²)?

[smallphi]

We know that

div \; (\hat{r} / r ) = 4 \pi \delta (r)

Why is there no generalized function (distribution) for

div \; (\hat{r} / r^2) = ??

 

[jostpuur]

Isn't it

<br /> \nabla\cdot \frac{x}{|x|^3} = \nabla\cdot\frac{\hat{x}}{|x|^2} = 4\pi\delta^3(x)<br />

I checked something quickly on my notes, if the mistake was yours and not mine, then you probably just ask what is

<br /> \nabla\cdot\frac{x}{|x|^4}<br />

next? I don't know about that yet...

EDIT: Oh, I didn't stop to think about what dimension you are in. I guess you were in three, because of the 4\pi constant. Was I correct?

 

[smallphi]

You are right about the correction, I work in 3D. The first relation simply expresses the divergence of the electric field of a point charge which gives the charge density (from one of the Maxwell's equations). The question should have been:

We know that

div \; (\hat{r} / |r|^2 ) = 4 \pi \delta^3 (r)

Why is there no generalized function (distribution) for

div \; (\hat{r} / |r|^3) = ??

 

[arildno]

Because that expression cannot be rigorously reformulated in terms of distributions.

 

[smallphi]

Which is the corresponding distribution in 1 dimension? Is it something like

<br /> \frac{d}{dx} \left ( \frac{1}{x} \right ) = 2 \delta (x)<br />

 

[Hurkyl]

\hat{r} / r^2 is well-behaved everywhere, and so it represents a tempered distribution. Thus, the distributional divergence operator can be applied to it.

\hat{r} / r^3, on the other hand, is ill-behaved at the origin, and thus does not represent a tempered distribution. Thus, the distributional divergence operator cannot be applied to it.


In particular, if we try to convolve \hat{r} / r^2 with a Schwartz function (i.e. test function), we get

\iiint f(\vec{r}) \frac{\hat{r}}{r^2} dV<br /> = \int \int \int f(\vec{r}) \hat{r} \sin \varphi \, d\rho d\varphi d\theta

which is clearly convergent. On the other hand,

\iiint f(\vec{r}) \frac{\hat{r}}{r^3} dV<br /> = \int \int \int f(\vec{r}) \frac{\hat{r}}{\rho} \sin \varphi \, d\rho d\varphi d\theta

which is usually a divergent integral, due to the bad behavior at the origin.

 

[Hurkyl]

Which is the corresponding distribution in 1 dimension? Is it something like

<br /> \frac{d}{dx} \left ( \frac{1}{x} \right ) = 2 \delta (x)<br />

Not quite:

\frac{d}{dx} \left( \frac{x}{|x|} \right) = 2 \delta(x)

x/|x|, of course, is simply the sign function. And, of course, d/dx here means the distributional derivative.

 

[smallphi]

Well (something) and div(something) can have very different behaviors at origin r =0 so one has to consider div(something) directly to decide if that could be a distribution or not.

I see the point that

\frac{\hat{r}}{r^3}

is not tempered even before application of the div operator which will make it even 'less tempered' after div.

 

[Hurkyl]

Does the 'distributional divergence operator' applied on tempered distribution, guarantee you will get tempered distribution as a result?

Yep. This works because the derivative of a test function is also a test function. (so, this property is true of any kind of distribution)

The distributional divergence is defined by, for a vector distribution f and test function g

<br /> \iiint (\nabla \cdot \vec{f}) g \, dV := -\iiint \vec{f} \cdot (\nabla g) \, dV<br />

This integral always exists (because \vec{f} is a distribution and \nabla g is a test function), so \nabla \cdot \vec{f} is a scalar distribution.

 

[smallphi]

Is it possible that (something) is not tempered but div(something) is tempered distribution?

 

[Hurkyl]

Well (something) and div(something) can have very different behaviors at origin r =0 so one has to consider div(something) directly to decide if that could be a distribution or not.

Fair enough. \hat{r} / r^3 is not a function on all of R^3, so is not in the domain of the divergence operator you learned in calculus. It's not a tempered distribution, so you cannot apply the divergence operator for tempered distributions.

If you choose another definition of the divergence operator, then whether its domain includes \hat{r} / r^3 is yet another question.



For example, if you define distributions on a space of test functions have the property that they are all zero at the origin, then \hat{r} / r^3 ought to be a distribution and have a divergence. But such a choice of test functions means you are effectively working only on \mathbb{R}^3 - \{\, (0, 0, 0)\, \}.



Ostensibly, if you could define a divergence for \hat{r} / r^3, you would want the product rule to hold, so:

\nabla \cdot \left( \frac{\hat{r}}{r^3} \right) =<br /> \nabla \left(\frac{1}{r}\right) \cdot \frac{\hat{r}}{r^2} + \frac{1}{r} \nabla \cdot \left( \frac{\hat{r}}{r^2} \right)<br /> = <br /> -\frac{1}{r^4} + \frac{4\pi}{r} \delta^3(\vec{r})

I suppose you could cross your fingers and try to use this expression in a formal manner. Such optimism works now and then!

 

[smallphi]

I can think of a series of functions well behaving at the origin and having r_hat/r^3 as limit. I can apply the normal calculus divergence to them and see if the new series converges in the distributional sense (i.e. taking integrals with test functions) to a certain distribution.

That's why I think, just because r_hat/r^3 blows up worse than r_hat/r^2 at origin, doesn't mean its divergence can't be a distribution.

 

[Hurkyl]

(p.s. I added more to my previous post)

 

[smallphi]

I think I understand why it doesn't work for r_hat/r^3 - Hurkyl you are right. I formed a sequence of functions

<br /> \frac{\hat{r}}{r^3 + \epsilon}<br />

that converge to r_hat/r^3 when epsilon goes to zero. Then I formed a seguence of their divergences, which are good behaved at the origin, and applied that sequence to a test function f(r). Integrating by parts and dropping the boundary term:

<br /> \iiint \nabla \cdot \left( \frac{\hat{r}}{r^3+\epsilon}\right)f(r) \, d^3 r = -\iiint \frac{\hat{r}}{r^3+\epsilon} \cdot \nabla f(r) \, d^3 r<br />

There is no way the right hand side go to finite number when epsilon goes to zero for every test function because the differential volume can't cancel the power of r^3 around the origin.

 

[smallphi]

Now let's change dimensions.

In one dimension, I know that there is a distribution called principal value P(1/x). What is

<br /> \frac{d}{dx} \, P \left( \frac{1}{x} \right) = \, ??<br />

What would be the two dimensional analogue of

div \; (\hat{r} / |r|^2 ) = 4 \pi \delta^3 (r)