Distribution Theory – Dirac Delta in Polar Coordinates

Question

Given

$$x=r\,\cos\theta\\y=r\,\sin\theta$$ and
$$x'=r'\,\cos\theta'\\y'=r'\,\sin\theta'$$

how can I express

$$\delta(x'-x)\delta(y'-y)$$ in terms of the polar coordinates?

And the more general case:

$$\delta(x'-x-a)\delta(y'-y-b)$$

Answer 1

If the transformation between coordinates ${\bf x}$ and ${\boldsymbol \xi}$ is not singular then $$\delta({\bf x}-{\bf x_0}) = \frac{1}{|J|}\delta({\boldsymbol \xi}-{\boldsymbol \xi}_{0}),$$ where $J$ is the Jacobian of the transformation. This is analogous to $\delta(f(x)) = \delta(x-x_0)/|f'(x_0)|$, for $x$ near an isolated zero $x_0$ of $f$.

The Jacobian is $r$ so, assuming $r'\ne 0$, $$\delta(x-x')\delta(y-y') = \frac{1}{r}\delta(r-r')\delta(\theta-\theta').$$ (We take $\theta'\in[0,2\pi)$.) Notice that $$\int_0^\infty r dr\int_0^{2\pi}d\theta \ \frac{1}{r}\delta(r-r')\delta(\theta-\theta') = 1$$ as required.

If $r'=0$ we must integrate out the ignorable coordinate $\theta$, $J\to \int_0^{2\pi}d\theta \ J = 2\pi r$. Thus $$\delta(x)\delta(y) = \frac{1}{2\pi r}\delta(r).$$ Again, notice $$\int_0^\infty r dr\int_0^{2\pi}d\theta \ \frac{1}{2\pi r}\delta(r) = 1.$$