This is a homework question and it goes like this:

"In spherical coordinates the Delta function is written in the form


Show that this is identical to $\delta(x-x_o)\delta(y-y_o)\delta(z-z_o)$ "

Now I think he has a mistake because I've never seen the delta function written this way in spherical coordinates. In fact I see it this way:$$\frac{\delta(r-r_o)\delta(\theta-\theta_o)\delta(\phi-\phi_o)}{r^2 \sin\theta}$$

I could prove that the above one is equal to the delta function for cartesian coordinates.

I have no idea how to prove the equation I'm given by me professor is equal to the one of cartesian coordinates. Any ideas?



1 Answer 1


Use the identity $\delta (f(x)) =\delta (x-a)/|f'(x)|$, where $a$ is a root of $f$, and apply it to the case for $f(x) =\cos \theta-\cos \theta_0$.

To show that this works, observe the integral

$$\int_0^{\pi} \Phi(\theta) \delta(\cos \theta-\cos \theta_0)\sin \theta d\theta$$

where $\Phi$ is any test function and make the change of variable $u=\cos \theta -\cos \theta_0$. Then,

$$\int_{-1-\cos \theta_0}^{1-\cos \theta_0} \Phi(\arccos(u+\cos \theta_0)) \delta(u)du=\Phi(\theta_0)$$

which holds for all test functions $\Phi$. This is precisely the same result as one obtains when writing

$$\int_0^{\pi} \Phi(\theta) \delta(\cos \theta-\cos \theta_0)\sin \theta d\theta=\int_0^{\pi} \Phi(\theta) \frac{\delta(\theta- \theta_0)}{|\sin \theta|}\sin \theta d\theta$$

  • $\begingroup$ Nice I think this works. I was looking on google for a proof of that identity and can't find it. Any ideas? Thanks! $\endgroup$
    – DLV
    Apr 12, 2015 at 17:35
  • $\begingroup$ It makes no sense to say that the resulting expression is "equal to" the one for cartesian ones right? It just that they're the same operator or something along those lines? $\endgroup$
    – DLV
    Apr 12, 2015 at 17:38
  • $\begingroup$ The absolute value of the sine function using the theorem you've shown won't lead me to the exact expression in spherical coordinates $\endgroup$
    – DLV
    Apr 12, 2015 at 17:42
  • $\begingroup$ I've added some more detail. See if this works better for you. $\endgroup$
    – Mark Viola
    Apr 12, 2015 at 17:47
  • 1
    $\begingroup$ @David You're welcome. My pleasure. Note that here $0\le \theta \le \pi$, and thus $|\sin \theta|=\sin \theta$. So, here the absolute value is superfluous. Also, the general case might actually involve multiple roots in the domain of integration. So, one actually needs to account for all of the roots. Here, we have only one root for the "$f(x)$" in the domain of interest. $\endgroup$
    – Mark Viola
    Apr 12, 2015 at 17:59

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