Dirac delta in polar coordinates 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:
$$\delta(x'-x-a)\delta(y'-y-b)$$
 A: By defintion the dirac delta function should satisfy the following condition.
$\int\limits_{-\infty} ^\infty \delta(\bar x - \bar x_{0}) \bar dx = 1$ . Now in polar coordinates $\bar dx = rdr d\theta$ which makes our integral $\int\limits_{0} ^\infty \int\limits_{0} ^{2\pi} \delta(\bar x- \bar x_{0}) rdrd\theta = 1$ For this integral to satisfy the defintion: $\delta(\bar x-\bar x_{0}) = \frac{1}{r} \delta(r-r_{0})\delta(\theta -\theta_{0})$. fixed it now
A: 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 
$$\int_0^\infty r dr\int_0^{2\pi}d\theta \
\frac{1}{2\pi r}\delta(r) = 1.$$
