Momentum space in spherical coordinate In statistical mechanics we calculate volume element in momentum space as  $4\pi p^2 dp$  to calculate microstate in phase space , but I don't understand why we write it in spherical polar coordinate ? Is it completely suitable to write it in spherical polar coordinate ? 
I read somewhere that we can write it because momentum in every direction is equally likely to occur  and  $\langle p_x ^2\rangle =\langle p_y^2\rangle = \langle p_z^2\rangle  $
Or something like this . But I don't understand it completely . So please explain this .
 A: This is valid as long as whatever function we're integrating only depends on $p$ and not on its direction, by the same argument used when integrating functions that only depend on $r$ in spherical coordinates.
When calculating integrals over momentum we have something like
$$I = \int dp_x dp_y dp_z \, f(\mathbf{p})$$
But if $f(\mathbf{p}) = f(|\mathbf{p}|)$, we can switch to spherical coordinates, and since the integrand doesn't depend on the angular variables $\theta$ and $\varphi$, we can just do those integrals to get the volume element:
$$I = \int_0^{2\pi} d\varphi\, \int_0^\pi d\theta\, \sin \theta \int_0^\infty dp\, p^2 f(p) = 4\pi \int_0^\infty dp\, p^2 f(p)$$
A: It's not a good idea to think of this in general terms, i.e. "in statistical mechanics, this is what we do". We do it when it's a useful mathematical trick to do it. That's it.
In your specific case, I suppose you're talking about a Hamiltonian that depends only on $|p|^2$, and not on $p$'s components, due to isotropy of the system.
So, if you're dealing with an integral of the form
$$ \int e^{-H(|p|^2)}d^3p$$
it's smarter to turn it into 
$$ \int e^{-H(|p|^2)}4\pi |p|^2d|p|.$$
This is purely maths, no physics here. If this doesn't answer your question, then we may need to know more about the specific calculation you're trying to solve. But yes, your intuition is correct, this is due to the fact that your system doesn't differentiate between $x, y, z$ directions.
