My question is

How to simplify $$f(x,y)=\delta(\cos x\cos y-k)\delta(\cos x\sin y)$$ ($0<k<1$) in terms of Dirac combs?

Background: I was working on a physics problem, in which I used $$g(\theta,\phi)=\lambda\cdot\delta(r\cos \theta\cos \phi-a)\delta(r\cos \theta\sin \phi)$$ (in spherical coordinates, $\lambda$ being the linear charge density) to represent a line of charge at $x=a,y=0$ in 3D Cartesian coordinates.

For some mathematical reasons, I need to find its Fourier transform, i.e. $$\int^\infty_{-\infty}\int^\infty_{-\infty}d\theta d\phi \cdot e^{-ik_1\theta-ik_2\phi}\cdot g(\theta,\phi)$$

I tried to work it out by brute force, and obtained $$\frac{4\lambda C(k_1)C(k_2)}{a\sqrt{r^2-a^2}}\cos\left(k_1\cos^{-1}\frac ar\right)$$ where $C(\cdot)$ is the $1$-periodic Dirac comb.

However, when I checked its correctness by performing an inverse Fourier transform on it, the original expression is not recovered.

Therefore, I would like to first simplify $g$ to reduce the chance of making errors in the second attempt. My ansatz is something proportional to $$C\left(\frac{\theta+\cos^{-1}\frac ar}{\pi}\right)C\left(\frac\phi\pi\right)$$ but I cannot proceed due to my lack of understanding in distribution theory.

Thanks in advance for any help.

  • $\begingroup$ The formula for $\delta(f_1(x, y)) \, \delta(f_2(x, y))$ is given here. There are four intersection points in $[0, 2 \pi) \times [0, 2 \pi)$, the norms of the gradients and the angle at which the curves intersect are the same for each point. $\endgroup$
    – Maxim
    May 15, 2020 at 12:46

1 Answer 1


My result is that the expression just becomes zero.

First there is this formula for the composition of $\delta$ with some smooth function $f$: $$ \delta(f(x,y)) = \sum_{(x_0,y_0)\in f^{-1}(0)} \frac{1}{|\nabla f(x_0,y_0)|}\delta(x-x_0,y-y_0). $$ Here $f^{-1}(0)$ is the set of zeroes of $f$.

Applying this to the second factor gives $$ \delta(\cos x\sin y) = \sum_{(x_0,y_0) \text{ s.t. } \cos x_0 \sin y_0=0} \frac{1}{\sqrt{\sin^2 x_0 + \cos^2 y_0}} \delta(x-x_0,y-_0) \\ = \sum_{m\in\mathbb Z}\sum_{n\in\mathbb Z} \frac{1}{\sqrt{\sin^2 (\frac{\pi}{2}+m\pi) + \cos^2 (n\pi)}} \delta(x-(\frac{\pi}{2}+m\pi),y-n\pi) \\ = \sum_{m\in\mathbb Z}\sum_{n\in\mathbb Z} \frac{1}{\sqrt{2}} \delta(x-(\frac{\pi}{2}+m\pi),y-n\pi) . $$

Now we include the first factor and simplify: $$ \delta(\cos x\cos y-k)\delta(\cos x\sin y) = \delta(\cos x\cos y-k) \sum_{m\in\mathbb Z}\sum_{n\in\mathbb Z} \frac{1}{\sqrt{2}} \delta(x-(\frac{\pi}{2}+m\pi),y-n\pi) \\ = \frac{1}{\sqrt{2}} \sum_{m\in\mathbb Z}\sum_{n\in\mathbb Z} \delta(\cos x\cos y-k) \delta(x-(\frac{\pi}{2}+m\pi),y-n\pi) \\ = \frac{1}{\sqrt{2}} \sum_{m\in\mathbb Z}\sum_{n\in\mathbb Z} \delta(\cos (\frac{\pi}{2}+m\pi)\cos n\pi-k) \delta(x-(\frac{\pi}{2}+m\pi),y-n\pi) \\ = \frac{1}{\sqrt{2}} \sum_{m\in\mathbb Z}\sum_{n\in\mathbb Z} \delta(-k) \delta(x-(\frac{\pi}{2}+m\pi),y-n\pi) \\ = \frac{1}{\sqrt{2}} \delta(k) \sum_{m\in\mathbb Z}\sum_{n\in\mathbb Z} \delta(x-(\frac{\pi}{2}+m\pi),y-n\pi) . $$

But since you have $k \neq 0$ the factor $\delta(k)$ will just be $0$ so our final result will also be $0$.


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