Integrating a delta function with a spherical volume integral

[counterpoint1]

Homework Statement

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.

Homework Equations

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.

 

[Dick]

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.

 

[counterpoint1]

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.

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.

 

[Dick]

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.

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.

 

[Nikitin]

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?

 

[vela]

$\delta(x)$ 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.

 

[Nikitin]

Ok, so we just define it to be that way because the $\delta^3$ is different from the plain old $\delta$? Fine by me. Thanks!

BTW, could you help me here too? https://www.physicsforums.com/threads/what-does-the-sign-mean-in-this-text.784081/