Showing Rigid Rotor Microstates with Angular Momentum ≤ M

[halley00]

Homework Statement

By evaluating the "volume" of the relevant region of its phase space, show that the number of microstates available to a rigid rotor with angular momentum less or equal to M is (M/ħ)².

Homework Equations

Consider the motion in the variables θ and φ, with M^{2}=(p_{θ})²+{p_{φ}/{sin(θ)}².

The Attempt at a Solution

I just integrated the "volume": (1/h²)∫₀^πdθ∫₀^2πdφ∫₀^Mdp_θ∫₀^Msinθdp_φ,

For some reason I'm missing a π. Why? I'm getting a "volume" 4π(M/h)²
Maybe I'm making a mistake in the choice of the limits of integration.

Sorry, I don't know how to write the equations in clean way.

 

[TSny]

Hello. Welcome to PF!

Note that while integrating over $p_\phi$, the variable $p_{\theta}$ is fixed at some value. Using the relation $M^2 = p_{\theta}^2 +\frac{ p_{\phi}^2}{\sin^2 \theta}$, what are the limits for the integration over $p_{\phi}$?

(Can the angular momenta components $p_\theta$ and $p_\phi$ be negative?)

 

[Met119]

No angular momentum can't be negative so those components can't be negative either. But I don't know what the limits for p_phi are supposed to be. (I'm also having trouble with this problem.)

 

[TSny]

No angular momentum can't be negative so those components can't be negative either. But I don't know what the limits for p_phi are supposed to be. (I'm also having trouble with this problem.)

The components of angular momentum can be negative as well as positive.

For the limits of integration, it might help to consider a simpler, but similar, problem. Suppose you have a single particle moving in two dimensions. In Cartesian coordinates the magnitude of linear momentum, p, is related to the components of momentum, p_x and p_y, by

p² = p_x² + p_y².

Suppose you want to integrate over the momentum portion of phase space to get the "volume" corresponding to all momentum values 0 < p < P₀. How would you choose the limits of integration in ∫dp_y ∫dp_x?