Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Delta function from divergence

  1. Jun 6, 2007 #1
    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]
     
    Last edited: Jun 6, 2007
  2. jcsd
  3. Jun 8, 2007 #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?
     
  4. Jun 8, 2007 #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]
     
    Last edited: Jun 8, 2007
  5. Jun 8, 2007 #4

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Because that expression cannot be rigorously reformulated in terms of distributions.
     
  6. Jun 8, 2007 #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]
     
    Last edited: Jun 8, 2007
  7. Jun 8, 2007 #6

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    [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.
     
  8. Jun 8, 2007 #7

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

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

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
     
  11. Jun 8, 2007 #10
    Is it possible that (something) is not tempered but div(something) is tempered distribution?
     
  12. Jun 8, 2007 #11

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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!
     
    Last edited: Jun 8, 2007
  13. Jun 8, 2007 #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 cant be a distribution.
     
    Last edited: Jun 8, 2007
  14. Jun 8, 2007 #13

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    (p.s. I added more to my previous post)
     
  15. Jun 8, 2007 #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.
     
    Last edited: Jun 8, 2007
  16. Jun 8, 2007 #15
    Now lets 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]
     
    Last edited: Jun 8, 2007
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Delta function from divergence
  1. Delta function (Replies: 1)

  2. Delta function (Replies: 3)

Loading...