I'm working on adapting some of the formalism in this paper to a system I'm working with. The part I'm interested in amounts to convolving a density profile $\rho(r)$ with a smoothing kernel $S(r,\epsilon)$. The smoothing kernel used in this particular case is Plummer smoothing:

$$S(r;\epsilon) = \frac{3}{4\pi}\frac{\epsilon^2}{(r^2+\epsilon^2)^{5/2}}$$

and the density profile is left as a general (but spherically symmetric) function. The convolution (incidentally, I'm pretty sure this is a correct use of the term convolution, but would appreciate a confirmation/correction if I'm using it incorrectly) I want to evaluate is:

$$\rho(\vec{r};\epsilon) = \int d\vec{r}' \rho(\vec{r}') S(|\vec{r}-\vec{r}'|;\epsilon)$$

Following along with the paper, this is going to be a lot easier to evaluate in cylindrical coordinates with the vector $\vec{r}$ lying along the $z$-axis, and without too much difficulty I arrive at the first part of his equation (8), but get stuck trying to get the second part. Equation (8) reads:

$$\rho(r,\epsilon) = \frac{3\epsilon^2}{2}\int_{-\infty}^{\infty}dz\int_0^{\infty}dRR\frac{\rho\left(\sqrt{R^2+z^2}\right)}{(R^2+(z-r)^2+\epsilon^2)^{5/2}} \\= \frac{3\epsilon^2}{2}\int_{-\infty}^{\infty}dz\int_0^{\infty}dRR\frac{\rho\left(\sqrt{R^2+(z-r)^2}\right)}{(R^2+z^2+\epsilon^2)^{5/2}}.\tag{8}$$

He points out that "the equality holds because the outer integral is taken over the entire $z$ axis", but I don't see how this follows.

A couple details on the geometry. The cylindrical coordinates are $R$ and $z$ (and $\phi$ I guess, but that just gives a $2\pi$). $r$ is the distance from the origin of the spherically symmetric $\rho(r)$. It follows that:

$$|\vec{r}'| = \sqrt{R^2+z^2}$$


$$|\vec{r}'-\vec{r}|^2 = R^2 + (z-r)^2.$$

I thought the solution might be as simple as translating everything along the $z$-axis a bit, something like $\tilde{z}=z+r$, but that doesn't seem to work out. My next hunch is that the "trick" that I need to see has something to do with the behavior of $\lim_{r\to\infty}\rho(r)$: this had better be zero to impose a finite total mass (of course this does not guarantee finite mass, and indeed there are several commonly used density profiles that have unbounded total mass, in practice they get truncated at some $r$, and they always still limit to 0 as $r\to\infty$), but I'm still stuck. So, anyone see what makes that equality true?

  • $\begingroup$ I've hesitated a bit over which site to put this on. It would likely be on-topic on math.se or computational-science.se, but I have a hunch that the answer is based on intuition a physicist is more likely to have, based on the types of functions we deal with on a day-to-day basis. Plus I prefer a physically motivated answer to one that simply cites a theorem (though that works too). I'm open to discussion of migration if the community thinks that's appropriate, though. $\endgroup$
    – Kyle Oman
    Jan 30, 2014 at 23:17
  • $\begingroup$ Also had a hard time tagging this, so if anyone has ideas, by all means... $\endgroup$
    – Kyle Oman
    Jan 30, 2014 at 23:18
  • 2
    $\begingroup$ Ordinarily these sorts of questions go to Mathematics, but they are a little ambiguous. When you have reason to expect that a physically motivated argument will give you the solution whereas a purely mathematical one will not, I think it's fine here. $\endgroup$
    – David Z
    Jan 30, 2014 at 23:54

2 Answers 2


Hint: It seems the substitution $z \longrightarrow z^{\prime}=r-z$ will work to prove the second equality of eq. (8). (Geometrically, this corresponds to a reflection of the $z$-axis around $z=\frac{r}{2}$.)

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    $\begingroup$ Ack. Bit embarrassed that I missed that. Clearly I'd been working too hard for too long before I got to this point. $\endgroup$
    – Kyle Oman
    Jan 31, 2014 at 0:00

As @Qmechanic points out in his answer, making the substitution $z \rightarrow z - r$ (going from the first version to the second in the identity) works. That's the math answer.

The physical intuition is that, in order for the system to have perfect cylindrical symmetry, it must extend to $z = \pm \infty$. That corresponds to integrating your $z$ variable from $-\infty$ to $+\infty$, as you have. If you do that, then all values of $r$ must behave identically; $\rho(r,\epsilon)$ is independent of $r$. If there were any $r$ dependence left over, not all points along the $z$-axis would be the same, and you would not have cylindrical symmetry.

  • $\begingroup$ And right after posting this, I see the OP has accepted Qmechanics's answer. That will teach me not to refresh the page before writing. $\endgroup$ Jan 31, 2014 at 0:35
  • $\begingroup$ Well, I'll give you the upvote anyway. Just went ahead and accepted a one-liner because I saw the rest of the implications from there. This is still great if anyone else is reading, and if anyone up-votes Qmechanic's answer and not this one, I question their judgement ;) $\endgroup$
    – Kyle Oman
    Jan 31, 2014 at 1:42
  • $\begingroup$ That's why I didn't delete it as soon as I saw you accepted Qmechanic's answer. I hope it's a useful explanation of the physical reasoning behind that mathematical operation. $\endgroup$ Jan 31, 2014 at 2:49

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