$$x=r\,\cos\theta\\y=r\,\sin\theta$$ 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:



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 $$\int_0^\infty r dr\int_0^{2\pi}d\theta \ \frac{1}{2\pi r}\delta(r) = 1.$$

  • $\begingroup$ Thanks! And what about the more general case added to the question? $\endgroup$ – JFNJr May 22 '13 at 7:03
  • $\begingroup$ @JFNJr: Glad to help. If we have $\delta(x-x'-a)$ let $x''=x'+a = r''\cos\theta''$ and similarly for $y$. We simply add a couple of primes to the above answer. $\endgroup$ – user26872 May 22 '13 at 17:10
  • $\begingroup$ but then I don't have explicit dependency on $r'$ anymore!? $\endgroup$ – JFNJr May 23 '13 at 8:59
  • $\begingroup$ @JFNJr: Solve the system $r'\cos\theta'+a=r''\cos\theta''$, $r'\sin\theta'+b=r''\sin\theta''$ for $(r'',\theta'')$ if you want this explicit dependence. The shift from $(x',y')$ to $(x'+a,y'+b)$ doesn't really introduce anything new into the problem. $\endgroup$ – user26872 May 23 '13 at 16:13

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


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