# Is particle number a problem for formulating statistical physics in a mathematically rigorous manner?

Quantities like the chemical potential can be expressed as something like

$$\mu=-T\left(\tfrac{\partial S}{\partial N}\right)_{E,V}.$$

Now the entropy is the log some volume, which depends on the particle number $N$. As in this definition, we sum over natural numbers of particles, is there any good way of actually evaluating the derivative?

What one practically does, i.e. when dealing with an ideal gas, is computing the quantity $\Gamma_N$, which might turn out to be $\frac{\pi^{N/2}}{(n/2)!}$, and then one will get an expression $S(E,N,V)$ which can of course be treated as if it was a function over $\mathbb{R}^3$. Even if that assumes that one has a closed expression which is a function $N$. In principle I'd be fine with that - if one has a given function (Or at least the bunch of values for all $N$) over a grid and a procedure to introduce more and more grid points to get a finer mash, then there is a notion of convergence to a derivative. But here the N's are clearly always at least 1 value apart - no matter how many N s there are (thermodynamics limit), the mesh doesn't get finer between any two given points.

You might define the derivative as computing the average rate of change between two partcle numbers $n$ and $n+d$ and say $\tfrac{\partial S}{\partial N}$ evaluated at $N'$ gives the same value for all $N'$ in one of the $d$-Intervals, but then you would have to postulate how to come up with $d$ in every new situation. This might be overcome in very specific situations in coming up with a "reasonable" fraction of Avogadros number, but this is not quite mathematical and the values of different finite difference approximation schemes are always different.

In full generality, I feel there is no categorical understanding of what the fractional dimensional space (phase space in this case) has to be and so the procedure of evaluation of the derivative should be explicitly postulate.

-

If you work with instead with the canonical or microcanonical ensemble, the formula is valid only in the thermodynamic limit. In this limit, the thermodynamic variable that makes sense in the limit is not $N$ but $\overline N = N\overline V/V$, where $\overline V$ is a finite reference volume that in the thermodynamic formulas takes the place of $V$.Again, $\overline N$ takes in the limit a continuum of values.
 Okay, so in the second paragraph you say that the variable that one shold use rigorously is not really $N$, right? Regarding the first paragraph, to compute the grand canonical phase space density function you already need the change of the phase space volume. How do you get at that then, as you need that object to get at the expectation values? – Nick Kidman Jul 27 '12 at 11:57 Whenever a thermodynamic limit is involved, $N$ goes to infinity.The actual material is regarded as a finite piece (of the infinitely extended materinal with infinite $N$) of volume %\overline V\$. This explains my formula. - The grand canonical ensemble works already at finite volume; a limit need not be taken (except for actually computing a partition function) See, e.g., my book for the derivation of the thermodynamical formulas from the grand canonical ensemble. A change of volume means integrating over a different phase space region. – Arnold Neumaier Jul 27 '12 at 14:15