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Integrating a delta function with a spherical volume integral

  1. Oct 6, 2014 #1
    1. The problem statement, all variables and given/known data

    Integrate $$\int_V \delta^3(\vec r)~ d\tau$$ over all of space by using V as a sphere of radius r centered at the origin, by having r go to infinity.

    2. Relevant equations


    3. The attempt at a solution

    This integral actually came up in a homework problem for my E&M class and I'm confused about how to handle it. I understand that the delta function will make the volume integral come out to one, but when I express that integral as a triple integral over spherical coordinates, $$\int_0^{2\pi}\int_0^\pi\int_0^\infty \delta^3(\vec r)~ r^2 sin(\theta)~ dr~ d\theta~ d\phi$$ The integrating factor apparently makes the integral go to zero, since ##\delta^3(\vec r)~r^2=0##.

    In my textbook, the author uses the delta function to say that ##\delta^3(\vec r)~r=0## and I can understand that since ##\delta^3(\vec r)~r = \delta(x)\delta(y)\delta(z)~\sqrt{x^2+y^2+z^2} = 0##. But then, given that, I'm at a loss as to what went wrong when I set up the integral.
     
  2. jcsd
  3. Oct 6, 2014 #2

    Dick

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    The delta function is not a function. It's a distribution. If you want to do this rigorously then the actual definition of
    ##\delta^3(\vec r)## is the limit of real live functions that are not 0 when r is not zero. To say it's infinity at r=0 and zero otherwise is a gross simplification. I assume that's what you are basing your argument on.
     
  4. Oct 6, 2014 #3
    So ... the volume integral doesn't go to zero because the limiting process is different in spherical coordinates as opposed to Cartesian, and it matters whether I'm using a volume integral like ##\iiint dx~dy~dz## as opposed to ##\iiint dr~d\theta~d\phi## ?

    EDIT: That is, I understand that the delta "function" only has meaning inside an integral, and I'm vaguely aware that a distribution is defined as a limit of functions in an integral, so the reason why the volume integral I used doesn't just go to zero is because of the specifics of integrating in spherical coordinates? I'm not too familiar with that math, honestly.
     
    Last edited: Oct 6, 2014
  5. Oct 6, 2014 #4

    Dick

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    No, the limiting process is the same in cartesian and spherical. You can define a representation of the delta function as, for example, a sequence of functions that have value 0 outside of r=1/n, but inside have value n^3*3/(4*pi). The total integral of each representative is 1 and they vanish outside of radius 1/n.. And yes, if you integrate that against a function and take the limit as n->infinity, that's the delta function. The volume element doesn't have much to do with it. You could also use cubes. Notice the the functions in the sequence are not 0 when r is not zero.

     
  6. Nov 26, 2014 #5
    I have the exact same problem: from the definition ##\int f(x) \delta(x) dx = f(0)##, one'd think ##\int \delta^3(\vec r)~ r^2 dr d\Omega = 0##, but this is obviously wrong.

    However, I struggle with understanding Dick's math-talk explanation, so is there some way to explain it based on what's written in Griffiths text on electromagnetism?
     
  7. Nov 26, 2014 #6

    vela

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    ##\delta(x)## and ##\delta^3(\vec{r})## are different objects, so while you have ##x\delta(x) = 0##, it doesn't necessarily follow that ##r\delta^3(\vec{r}) = 0## (as opposed to ##r\delta(r)=0##). To write ##x\delta(x) = 0## is relatively safe because the expression will appear as part of a one-dimensional integral. You tack on a ##dx## and integrate. On the other hand, ##\delta^3(\vec{r})## only makes sense in a three-dimensional integral, so there's possible confusion arising from the volume element in different coordinate systems. You need to think carefully about what the expression ##r\delta^3(\vec{r})## means.

    One of the properties of the three-dimensional delta function is
    $$\int f(\vec{r}'-\vec{r})\delta^3(\vec{r}')\,d^3\vec{r}' = f(\vec{r}).$$ If you're avoiding distributions, this would, in fact, be one of the defining properties of the three-dimensional delta function. The factors that can cause that integral to vanish come from ##f##, not the volume element. So when the author says ##r \delta^3(\vec{r}) = 0##, he means that
    $$\int r\delta^3(\vec{r})\,d^3\vec{r} = \iiint r\delta^3(\vec{r})\,r^2\sin\theta\,dr\,d\theta\,d\phi=0.$$ The mistake in saying that
    $$\iiint \delta^3(\vec{r})\,r^2\sin\theta\,dr\,d\theta\,d\phi = \iiint [r\delta^3(\vec{r})]\,r\sin\theta\,dr\,d\theta\,d\phi = 0$$ is that the second integral is no longer an integral of the form ##\int [f(\vec{r})\delta^3(\vec{r})]\,d^3\vec{r}## so the meaning of the delta function is no longer clear. Specifically, in the context of that integral, you can't say that ##r\delta^3(\vec{r}) = 0## and deduce that the integral vanishes.
     
    Last edited: Nov 28, 2014
  8. Nov 27, 2014 #7
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