# A Classical statistical physics -- Number of microstates

1. May 21, 2017

### LagrangeEuler

Phase volume is it the same as the number of total microstates in some physical system? Phase volume= volume of phase space. Or there is some difference?

2. May 22, 2017

### grquanti

In the case of systems with a continuum of states (e.g. a classical gas) the concept of "number of states" is not well defined I think: let A be a state and B a second state, identical to A, but with this difference:

$v_i^A = (v_x,v_y,v_z) → v_i^B = (v_x + ε, v_y,v_z)$

where $i$ is the label of a generic (tipically identical to some other) particle. Well, states A and B are different and there is no way to count the minimum number of states for the system to go from A to B. It's a consequence of the continuity of phase space (continuity of energy, if you prefer).

In the case of classical systems with discrete states (e.g. Ising model) the volume of phase space is no more considered (you cannot do an integral over that space, as far as I know). It is usually considered a space of configuration (e.g. in the one dimensional Ising model with only spin up or down and $N$ sites, it is a space with all the possible $N$-dimensional vector with component $±1$). In this case the number of states is the number of element in this set (the cardinality of the set). As far as I know, this number is usually calculated "indirectly", in the sense that you do considerations over the number of degrees of freedom and simmetries (in principle you can calculate it summing $1$ for each state but you should know them all, i.e. you should know their number.)

I think that's right, but try to check what I told you.

3. May 23, 2017

### vanhees71

The problem is that classical statistical mechanics is plagued with conceptual problems which are easily solved only with quantum theory. To count microstates the most simple way is to introduce a finite volume, periodic boundary conditions for the Schrödinger wave function and then calculate the states contained in a momentum volume $\mathrm{d}^3 \vec{p}$, which turns out to be $V \mathrm{d}^3 \vec{p}/(2 \pi \hbar)^3$ in the large-volumen/thermodynamical limit (for bosons the limit is not that trivial due to the formation of a Bose-Einstein condensate at low temperatures, but that's far from the realm where the classical (Boltzmann) limit is valid).

4. May 24, 2017

### hilbert2

Yeah, one example of a problem of the type vanhees71 probably meant is the "paradox" that the entropy of any fluid should be increasing all the time without bound because its mixing with itself by diffusion isn't really different from the mixing of two different fluids (in classical mechanical reasoning).