This is a homework question and it goes like this:
"In spherical coordinates the Delta function is written in the form
$\frac{1}{r^2}\delta(r-r_o)\delta(\cos\theta-\cos\theta_o)\delta(\phi-\phi_o)$
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?
Thanks.
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
$$\int_0^{\pi} \Phi(\theta) \delta(\cos \theta-\cos \theta_0)\sin \theta d\theta=\int_0^{\pi} \Phi(\theta) \frac{\delta(\theta- \theta_0)}{|\sin \theta|}\sin \theta d\theta$$