Rotating Ideal Gas: Homework Eqns & Boltzmann Dist

[Brewer]

Homework Statement

An ideal gas of N molecules of mass m is contained in a cylinder of length
L and radius R. The cylindrical container is rotating about its axis at an
angular velocity $$\omega$$, and is at equilibrium with temperature T.

Write down the energy for single particle states, and the Boltzmann distribution for the gas in the rotating cylinder given it experiences a centrifugal potential energy V(r)

Homework Equations

the energy is just E = $$\frac{p^2}{2m} + V(r)$$

I'm not entirely sure about the Boltzmann distribution. In my notes it states that $$Z_1 = \int \frac{d^3 rd^3p}{h^3} e^{-\beta E}$$

However here I get confused. I was pretty sure that Z is the partition function of the gas (which it has been so far), but a few lines further in the notes it calls this the Boltzmann distribution. As far as I was aware these aren't interchangeable names, and are completely separate things within the frame work of statistics.

The question goes onto to ask me to split the Boltzmann distribution into a translational and a interation part, which I know can be done with the above equation (another reason I'm hedging my bets on using this equation for the Boltzmann distribution). Whilst this is fine in theory I'm a little concerned about doing this in practice. I know one of the parts is an integral over 3d space with the kinetic part of the energy and the other bit is the integral over what seems to be 3d momentum with the potential part (I think I may have got the combinations back to front, but I'll check that - that's not the bit I'm confused with). My problem with this is the 3d momentum integral. I have never seen how to do this (or even seen anything like it before!) so could you suggest how I'd go about solving this bit. Is it anything like the the 3d integral over space in which you have various factors to add to it, or does it work completely differently?

 

[olgranpappy]

boltzman distribution--or is it called "gibbs distribution--is the integrand (suitably normalized)
$$e^{-\beta E}$$

 

[olgranpappy]

to do the momentum integral is the easy part. use
$$\int_{-\infty}^{\infty}dx e^{-\alpha x^2}=\sqrt{\frac{\pi}{\alpha}}$$