# Multiplicity and Inverse of probability distribution, what do they mean?

1. Oct 10, 2012

### silverwhale

Hello Everybody.

I have a rather simple question, which still kept me thinking for two hours without any result.

If we want to determine the multiplicity in the microcanonical ensemble we just devide the volume of the shell containing the accessible microstates over the volume of one microstate; let us write the multplicity as:
$$\Gamma = \frac{\omega }{\omega_{0}}.$$
Now, we can write the RHS as:
$$\frac{\omega }{\omega_{0}} = \frac{\int' d^{3N}q d^{3N}p}{d^{3N}q d^{3N}p}.$$
Next multiplying by $$\rho_{0},$$ where $$\rho_0$$ is the constant density distridubtiion of the ensemble members in a microcanonical ensemble. we get:
$$\frac{\omega }{\omega_{0}} = \frac{\int' d^{3N}q d^{3N}p \, \rho_0}{d^{3N}q d^{3N}p \, \rho_0};$$
Now this would mean that the mutliplicity $$\Gamma$$ is given by:
$$\Gamma = \frac{N}{n_0};$$
where N is the total number of ensemble members and n_0 is the number of ensemble elements in a given microstate.

First question, is this correct? Is this ok to write?

Next, let's suppose everything is fine. What would this expression then mean?

$$\frac{\int d^{3N}q d^{3N}p \, \rho(q,p)}{d^{3N}q d^{3N}p \, \rho(q,p)} = \frac{N}{n(q,p)};$$

where $$\rho(q,p)$$ is the density distribution again but this time it is not constant anymore, it varies from microstate to microstate. We would get something like $$\Gamma (q,p).$$ What does it mean? Can anyone help? I can't firugre out what this "multiplcity", that is (q,p) dependent, means. Incidently its inverse is similar to the probability distribution of the canonical ensemble and grand canonical ensemble.

Any help would be greatly appreciated.