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

In summary, the divergence of \hat{r} / r^3 cannot be expressed as a distribution because it is not well-behaved at the origin. This is in contrast to \hat{r} / r^2, which can be expressed as a tempered distribution and have the divergence operator applied to it. While \hat{r} / r^3 may have a limit as a sequence of functions, its divergence cannot be defined in a distributional sense.
  • #1
smallphi
441
2
We know that

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

Why is there no generalized function (distribution) for

[itex]div \; (\hat{r} / r^2) = ??[/itex]
 
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  • #2
Isn't it

[tex]
\nabla\cdot \frac{x}{|x|^3} = \nabla\cdot\frac{\hat{x}}{|x|^2} = 4\pi\delta^3(x)
[/tex]

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

[tex]
\nabla\cdot\frac{x}{|x|^4}
[/tex]

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 [tex]4\pi[/tex] constant. Was I correct?
 
  • #3
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

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

Why is there no generalized function (distribution) for

[itex]div \; (\hat{r} / |r|^3) = ??[/itex]
 
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  • #4
Because that expression cannot be rigorously reformulated in terms of distributions.
 
  • #5
Which is the corresponding distribution in 1 dimension? Is it something like

[tex]
\frac{d}{dx} \left ( \frac{1}{x} \right ) = 2 \delta (x)
[/tex]
 
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  • #6
[itex]\hat{r} / r^2[/itex] is well-behaved everywhere, and so it represents a tempered distribution. Thus, the distributional divergence operator can be applied to it.

[itex]\hat{r} / r^3[/itex], 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 [itex]\hat{r} / r^2[/itex] with a Schwartz function (i.e. test function), we get

[tex]\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[/tex]

which is clearly convergent. On the other hand,

[tex]\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[/tex]

which is usually a divergent integral, due to the bad behavior at the origin.
 
  • #7
smallphi said:
Which is the corresponding distribution in 1 dimension? Is it something like

[tex]
\frac{d}{dx} \left ( \frac{1}{x} \right ) = 2 \delta (x)
[/tex]

Not quite:

[tex]\frac{d}{dx} \left( \frac{x}{|x|} \right) = 2 \delta(x)[/tex]

x/|x|, of course, is simply the sign function. And, of course, d/dx here means the distributional derivative.
 
  • #8
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

[tex] \frac{\hat{r}}{r^3}[/tex]

is not tempered even before application of the div operator which will make it even 'less tempered' after div.
 
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  • #9
smallphi said:
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

[tex]
\iiint (\nabla \cdot \vec{f}) g \, dV := -\iiint \vec{f} \cdot (\nabla g) \, dV
[/tex]

This integral always exists (because [itex]\vec{f}[/itex] is a distribution and [itex]\nabla g[/itex] is a test function), so [itex]\nabla \cdot \vec{f}[/itex] is a scalar distribution.
 
  • #10
Is it possible that (something) is not tempered but div(something) is tempered distribution?
 
  • #11
smallphi said:
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. [itex]\hat{r} / r^3[/itex] 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 [itex]\hat{r} / r^3[/itex] 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 [itex]\hat{r} / r^3[/itex] ought to be a distribution and have a divergence. But such a choice of test functions means you are effectively working only on [itex]\mathbb{R}^3 - \{\, (0, 0, 0)\, \}[/itex].



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

[tex]\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})[/tex]

I suppose you could cross your fingers and try to use this expression in a formal manner. Such optimism works now and then!
 
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  • #12
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.
 
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  • #13
(p.s. I added more to my previous post)
 
  • #14
I think I understand why it doesn't work for r_hat/r^3 - Hurkyl you are right. I formed a sequence of functions

[tex]
\frac{\hat{r}}{r^3 + \epsilon}
[/tex]

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:

[tex]
\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
[/tex]

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.
 
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  • #15
Now let's change dimensions.

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

[tex]
\frac{d}{dx} \, P \left( \frac{1}{x} \right) = \, ??
[/tex]

What would be the two dimensional analogue of

[itex]div \; (\hat{r} / |r|^2 ) = 4 \pi \delta^3 (r)[/itex]
 
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FAQ: Why Is There No Generalized Function for div (r̂ / r²)?

What is the delta function from divergence?

The delta function from divergence is a mathematical concept used in vector calculus to solve for the divergence of a vector field. It is defined as the limit of a function as the radius approaches zero.

How is the delta function from divergence related to the Dirac delta function?

The delta function from divergence is related to the Dirac delta function in that they both have the property of being zero everywhere except at one point where they are infinite. However, the Dirac delta function is one-dimensional, while the delta function from divergence is three-dimensional.

What is the physical interpretation of the delta function from divergence?

The physical interpretation of the delta function from divergence is that it represents a point charge in the vector field. This means that the vector field diverges at that point, and the delta function from divergence allows us to calculate the magnitude and direction of this divergence.

What are the applications of the delta function from divergence?

The delta function from divergence has many applications in physics and engineering, particularly in the study of electric and magnetic fields. It is also used in fluid dynamics to model point sources or sinks in a fluid flow. Additionally, it is used in signal processing and image reconstruction algorithms.

Are there any limitations or special properties of the delta function from divergence?

One limitation of the delta function from divergence is that it is only defined in three dimensions. Additionally, it is a distribution and not a regular function, meaning that it cannot be evaluated at a specific point. It also has the property of being odd, which means that it is equal to its negative when integrated over all space.

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