The delta function in spherical coordinates is given by:


(The ordering of the coordinates inside the $\delta$'s isn't important). If I have a particular location in mind, say $(r_0,\theta_0,\phi_0)=(r_0,0,0)$, is there a neat way to "re-normalize" the following:


since $\cot\theta_0$ is also singular at $\theta_0\in \pi \mathbb{Z}$.

To be clear, I'm trying to see if there is a way to cleverly absorb this cotangent function into the delta function so that it remains relatively unchanged. It's also possible this question is entirely unfounded, and I apologize if so.

  • $\begingroup$ Do you mean $\cot\theta$ outside the parentheses? $\endgroup$ Feb 7, 2018 at 23:21
  • $\begingroup$ @probably_someone - no, I didn't. But would you have an answer if it was just $\cot\theta$? If so, let me know. $\endgroup$
    – Karl
    Feb 7, 2018 at 23:24
  • $\begingroup$ Well, the question would make a bit more sense from the perspective of delta functions if it did. As it is now, when you integrate the $\cos\theta$ delta function, you're replacing all instances of $\cos\theta$ with $\cos\theta_0$, but no such instances exist, so it doesn't do anything. $\endgroup$ Feb 7, 2018 at 23:27
  • $\begingroup$ I've voted to move to math, since this seems to me to be about properties of dirac delta functions and how to rewrite $g(x) \delta(f(x, y))$ as $\delta(f'(x,y))$ $\endgroup$
    – innisfree
    Feb 8, 2018 at 4:58

2 Answers 2


Using the identity $$ \delta(ax) = \frac{1}{|a|}\delta(x) , $$ one can do the following $$ \cot\theta_0\delta(\cos\theta_0-\cos\theta) =\delta\left(\frac{\cos\theta_0}{\cot\theta_0}-\frac{\cos\theta}{\cot\theta_0}\right) =\delta\left(\sin\theta_0-\frac{\cos\theta\sin\theta_0}{\cos\theta_0}\right) , $$ assuming $\cot\theta_0>0$. (If not, the whole thing must get a minus sign.)

Is this what you wanted to know?


It seems OP wants to avoid having poles inside & outside the Dirac delta distribution. Using a hopefully obvious notation $$s~\equiv~ \sin\theta,\qquad c~\equiv~ \cos\theta,$$ one possibility is to replace

$$ \frac{c}{s} \delta(c-c_0)~=~ \frac{sc}{|f^{\prime}(c)|} \delta(c-c_0)~=~sc~\delta(f(c)-f(c_0)), $$ where $$f(c)~:=~c(1-\frac{c^2}{3}),\qquad f^{\prime}(c) ~=~1-c^2, \qquad c~\in~[-1,1].$$ Of course such rewritings are essentially just cosmetics. The singular nature of the construction does not disappear but manifests itself in other ways.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .