Consider the following integral over a 2D plane,

$$\iint \mathrm{d}^2\mathbf{k}\ e^{i\mathbf{k}\cdot\mathbf{r}} = 4\pi^2\delta^2(\mathbf{r})$$

This is a Fourier transform of a distribution which is rotationally symmetric around the origin, and thus the result should also be rotationally symmetric around the origin. So it should be expressible in terms of the magnitude of $\mathbf{r}$ only, i.e. in terms of $r \equiv \lVert\mathbf{r}\rVert$.

$$\iint \mathrm{d}^2\mathbf{k}\ e^{i\mathbf{k}\cdot\mathbf{r}} = 4\pi^2 f(r)$$

What is the proper mathematical expression for $f(r)$, if it exists?

Clearly $f(r) = \delta(r)$ is unsuitable because it doesn't have the right units. Dimensional analysis suggests that it might be something like $f(r) = \frac{1}{r}\delta(r)$, but I'd like to have some sort of mathematical justification rather than just a guess.

I've looked at Delta function in curvilinear coordinates but that question is somewhat more abstract, plus it doesn't seem to eliminate the dependence on $\theta$ as I would like to do.

  • $\begingroup$ What's your definition of a rotationally symmetric distribution? Whatever it is, the delta function concentrated at the origin should satisfy it. $\endgroup$ – Jeff Nov 30 '11 at 16:38
  • $\begingroup$ Rotationally symmetric means it is independent of the polar coordinate $\theta$. And $\delta(r)$ is unsuitable because, for one thing, it has the wrong units. On dimensional grounds, $\frac{1}{r}\delta(r)$ seems like a possibility, but I'd like to have some sort of mathematical justification for it. (I edited that into the question) $\endgroup$ – David Z Nov 30 '11 at 17:03
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    $\begingroup$ Distributions are not functions, so it doesn't really make sense to say a distribution is independent of the polar coordinate $\theta$. You would have to make a definition based on how the distribution acts on test functions (maybe it assigns the same value to $\phi$ and any rotation of $\phi$?). $\endgroup$ – Jeff Nov 30 '11 at 18:11
  • $\begingroup$ Well, what I meant was "can be written in a form independent of $\theta$." I'm coming from physics, in which we tend not to be all that precise about the terminology ;-) If I understand your definition correctly to mean that $\iint f(r)g(r,\theta)\mathrm{d}^2\mathbf{r} = \iint f(r)g(r,\theta + \alpha)\mathrm{d}^2\mathbf{r}$ for all test functions $g$ and any $\alpha\in\mathbb{R}$, then that is an accurate statement of what I mean by "rotationally symmetric." But I'm really asking about an expression for the distribution, not about the distribution itself. $\endgroup$ – David Z Nov 30 '11 at 18:17

Expressions like $\delta(r)/r$ do not generally define a distribution. But if we define a distribution $T$ by setting

$$T(\phi)=\int_0^{2\pi}\int_0^{\infty} \phi(r\cos \theta,r\sin\theta) \delta(r) \mathrm dr \mathrm d\theta=\int_0^{2\pi}\int_0^{\infty} \phi(r\cos \theta,r\sin\theta) (\delta(r)/r) r \mathrm dr \mathrm d\theta$$

for test functions $\phi$, then $T=2\pi \delta(\mathbf x)$. So, informally, $\delta(\mathbf x)=\delta(r)/2\pi r$.

  • $\begingroup$ Thanks for the answer; I'll have to think about it. (Would it help if I made explicit in the question that when I refer to $f(r)$ as a distribution I really mean $f: \phi\to\iint \phi(\mathbf{r})f(r)\mathrm{d}^2\mathbf{r}$?) $\endgroup$ – David Z Nov 30 '11 at 18:25
  • $\begingroup$ My answer is sort of spelling out exactly that correspondence. Note that it is possible to write down an expression that does not give a distribution under that correspondence. For example, 1/x can't be integrated against 1-D test functions. $\endgroup$ – Colin McQuillan Nov 30 '11 at 18:45
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    $\begingroup$ You're glossing over the subtleties raised when one of the integration limits is $0$. If you treat $\delta$ as a distribution, such an integral isn't well-defined since it corresponds to an integral with a non-smooth characteristic function as a test function. Different ways of making sense of such an integral result in values of $0$, $1/2$ or $1$ times the test function value at $0$. See e.g. here (p. 7) and here. $\endgroup$ – joriki Dec 6 '11 at 7:17

Depending on taste, some pitfalls in the behavior near $0$ can be avoided as follows. Given a rotation-invariant (maybe tempered) distribution $u$ on $\mathbb R^n$, for test function $f$, letting $K=SO(n)$ be the rotation group, $$ u(f) = {1\over |K|} \int_K u(k\cdot f)\;dk = {1\over |K|} u\Big(\int_K k\cdot f\;dk\Big) $$ where $dk$ refers to an invariant ("Haar") measure on $K$, $|K|$ is the total measure of $K$, and the interchange of integral with application of $u$ can be justified at various levels of sophistication. (The key point is that $K$ is compact, and that rotations are nicely-continuous maps on functions... My own blanket preference for such justifications refers to Gelfand-Pettis' "weak" integrals, but there are many alternatives.)

The point I'm aiming at here is that the rotationally-averaged $f$, as a function of radius, is an even test function on $\mathbb R$. Conversely, an even test function on $\mathbb R$ gives a rotationally invariant test function on $\mathbb R^n$. Continuous linear functionals on rotationally-invariant test functions on $\mathbb R^n$ are thus naturally in bijection with continuous linear functionals on even test functions on $\mathbb R$. That is, this avoids talking about an "edge" at "r=0".


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