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- Homework Statement
- calculating the probability density of a particle being in a radius $r=r_0$ in a rotating cylinder. ($\omega, R, L$ are the frequency, radius, and height of the cylinder)

- Relevant Equations
- $$Z=\sum_i{exp(-\beta*H(q,p))}$$

First of all, I've calculated the partition function:

Z=1h3∫e−βH(q,p)d3pd3q=1h3∫e−β(p22m−12mrω2)d3prdrdθdz=2πL(2mπh2β)3/2e12βmω2R2−1ω2mβ

The probability of being of one particle in radius $r_0$ is:

p(r=r0)=1Z∫e−βHd3pd3q=∫1Z2πL(2mπh2β)3/2eβmrω22rdr

So I've thought that because, by definition, the probability is the integral over the probability density:

p(r=r0)=∫f(r)dr

The probability density will the integrand above:

f(r)=1Z2πL(2mπh2β)3/2eβmrω22r0

Am I right here?

Plus, If I am right, and I want to know the number of the particles $n(r)$ in radius $r_0$, am I need just to integrate to find the probability, and then multiply it by the number of total molecules in the system?

n(r)=N⋅∫f(r)dr

Z=1h3∫e−βH(q,p)d3pd3q=1h3∫e−β(p22m−12mrω2)d3prdrdθdz=2πL(2mπh2β)3/2e12βmω2R2−1ω2mβ

The probability of being of one particle in radius $r_0$ is:

p(r=r0)=1Z∫e−βHd3pd3q=∫1Z2πL(2mπh2β)3/2eβmrω22rdr

So I've thought that because, by definition, the probability is the integral over the probability density:

p(r=r0)=∫f(r)dr

The probability density will the integrand above:

f(r)=1Z2πL(2mπh2β)3/2eβmrω22r0

Am I right here?

Plus, If I am right, and I want to know the number of the particles $n(r)$ in radius $r_0$, am I need just to integrate to find the probability, and then multiply it by the number of total molecules in the system?

n(r)=N⋅∫f(r)dr

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