# Delta function from divergence

1. Jun 6, 2007

### smallphi

We know that

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

Why is there no generalized function (distribution) for

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

Last edited: Jun 6, 2007
2. Jun 8, 2007

### jostpuur

Isn't it

$$\nabla\cdot \frac{x}{|x|^3} = \nabla\cdot\frac{\hat{x}}{|x|^2} = 4\pi\delta^3(x)$$

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

$$\nabla\cdot\frac{x}{|x|^4}$$

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?

3. Jun 8, 2007

### 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

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

Last edited: Jun 8, 2007
4. Jun 8, 2007

### arildno

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

5. Jun 8, 2007

### smallphi

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

$$\frac{d}{dx} \left ( \frac{1}{x} \right ) = 2 \delta (x)$$

Last edited: Jun 8, 2007
6. Jun 8, 2007

### Hurkyl

Staff Emeritus
$\hat{r} / r^2$ 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

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

which is clearly convergent. On the other hand,

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

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

7. Jun 8, 2007

### Hurkyl

Staff Emeritus
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.

8. Jun 8, 2007

### 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.

Last edited: Jun 8, 2007
9. Jun 8, 2007

### Hurkyl

Staff Emeritus
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

$$\iiint (\nabla \cdot \vec{f}) g \, dV := -\iiint \vec{f} \cdot (\nabla g) \, dV$$

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.

10. Jun 8, 2007

### smallphi

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

11. Jun 8, 2007

### Hurkyl

Staff Emeritus
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:

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

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

Last edited: Jun 8, 2007
12. Jun 8, 2007

### 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 cant be a distribution.

Last edited: Jun 8, 2007
13. Jun 8, 2007

### Hurkyl

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

14. Jun 8, 2007

### 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

$$\frac{\hat{r}}{r^3 + \epsilon}$$

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:

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

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.

Last edited: Jun 8, 2007
15. Jun 8, 2007

### smallphi

Now lets change dimensions.

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

$$\frac{d}{dx} \, P \left( \frac{1}{x} \right) = \, ??$$

What would be the two dimensional analogue of

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

Last edited: Jun 8, 2007