Equilibrium Statistical Mechanics

Abstract

We present here Boltzmann’s ideas concerning the microscopic basis of equilibrium statistical mechanics. It is based on the distinction between the microstates and the macrostates of a system and the fact that the number of microstates corresponding to a given value of a macrostate depends very much on that value. We also give the formula for the microscopic or Boltzmann’s entropy and then give the corresponding microscopic formulas for the free energy and the grand potential. We define the various ensembles or probability distributions on sets of microstates and show that they are equivalent in the limit of large systems.

Appendices

Appendix

6.A Asymptotics

6.1.1 6.A.1 Laplace’s Method

Consider integrals of the form

$$\begin{aligned} \int _0^\infty \exp (\lambda f(x))dx \end{aligned}$$

(6.A.1)

for $\lambda$ large and positive. Assume that $f(x)\rightarrow -\infty$ as $x\rightarrow \infty$, sufficiently last so that the integral converges for all $\lambda >0$. Assume also that the function f has a unique maximum at $x=x_0$, $0<x_0<\infty$. Write that integral as:

$$\begin{aligned}&\exp (\lambda f(x_0)) \\&\Big ( \int _{|x-x_0|\le \epsilon } \exp (\lambda (f(x)-f(x_0))dx +\int _{|x-x_0|> \epsilon } \exp (\lambda (f(x)-f(x_0))dx \Big ), \end{aligned}$$

for some $\epsilon >0$. Since, $\forall x, |x-x_0|> \epsilon$, $f(x)-f(x_0)< \delta$ for some $\delta >0$ (since $x_0$ is the unique maximum of f), and the integral in (6.A.1) converges, we have that the second integral in (6.A.1) is bounded by $\exp (-c\lambda )$, for some $c>0$.

When $|x-x_0| \le \epsilon$, we approximate $f(x)-f(x_0)\approx -\frac{a(x-x_0)^2}{2}$, with $-a= f''(x_0)$; since $x_0$ is a strict maximum of f, we have $f'(x_0)=0$ and $a>0$. Changing variables $y=\sqrt{\lambda a}(x-x_0)$ in $\int _{|x-x_0|\le \epsilon } \exp (-\lambda \frac{a(x-x_0)^2}{2})dx$, we approximate that latter integral by $\frac{1}{\sqrt{\lambda a}}\int _\mathbb R\exp (-\frac{y^2}{2})dy= \sqrt{\frac{2\pi }{\lambda a}}$ (we replace $\pm \lambda \epsilon$ by $\pm \infty$ in the limits of integration).

Altogether, we get the approximation:

$$\begin{aligned} \int _0^\infty \exp (\lambda f(x))dx \approx \exp (\lambda f(x_0)) \sqrt{\frac{2\pi }{\lambda a}}= \exp (\lambda f(x_0)) \sqrt{\frac{2\pi }{\lambda |f''(x_0)| }}, \end{aligned}$$

(6.A.2)

as $\lambda \rightarrow \infty$. With more work, one can obtain much more detailed estimates than (6.A.2), see e.g. Erdelyi [123].

6.1.2 6.A.2 Stirling’s Formula

Laplace’s method can be used to justify the approximation to the factorial function $N!\approx N^N e^{-N}$ as $N \rightarrow \infty$. First write:

$$ N!= \int _0^\infty x^N \exp (-x) dx \nonumber $$

which can be proven recursively from $\int _0^\infty x^n \exp (-x) dx=n \int _0^\infty x^{n-1} \exp (-x) dx$ (which follows from integration by parts) and $\int _0^\infty \exp (-x) dx=1$. Let $x=Ny$; we get:

$$ \int _0^\infty x^N \exp (-x) dx= N^{N+1} \int _0^\infty y^N \exp (-Ny) dy= N^{N+1} \int _0^\infty \exp (N(\ln y-y)) dy \nonumber $$

The function $f(y)=\ln y-y$ has a unique maximum at $y=1$ with $f''(1)=-1$, so that applying (6.A.2) gives:

$$\begin{aligned} N! \approx N^{N+1} \exp (-N) \sqrt{\frac{2\pi }{N }}= N^{N} \exp (-N) \sqrt{2\pi N } \end{aligned}$$

(6.A.3)

One can check that the same approximation holds for the Gamma function (6.4.4).

6.B “Derivation” of Formula (6.5.4)

In order to apply Laplace’s method to the integral (6.5.2), we need to verify that the function $f(e)= -\beta e+ \frac{s(e, v)}{k}$ has a unique maximum, which follows from the concavity of the function s(e, v), and check that the integral converges. That follows from the fact that $f(e)\approx -\beta e$, as $e\rightarrow \infty$, since $s(e, v)\approx \ln e$ as $e\rightarrow \infty$, see (6.4.7) (the latter computation is valid only for an ideal gas but the logarithmic growth of s(e, v) as a function of e when $e\rightarrow \infty$ holds more generally).

6.C An Intuitive Formula for the Entropy

An intuitive expression for the variation of entropy is most easily expressed in the situation of discrete variables.

Consider, as we did after (6.2.11), a system of N particles, each of which can be in L possible states, but with an “energy” variable $e_i (\lambda )$, $i=1, \ldots , L$, associated to each such state and depending on an external parameter $\lambda$, with the sum of the energies being fixed: $E(\mathbf{x}, \lambda )=\sum _{i= 1}^N e(x_{i}, \lambda )$.

The set of microstates is $\Omega = \{1, \ldots , L\}^N$, and we can take as macrostates the fractions of particles in each state: $\mathbf{n}= (n_1, \ldots , n_L)$, where $n_i= \frac{N_i}{N}$ with $N_i$ being the number of particles in state i. In terms of the macrostate, the total energy equals: $E(\mathbf{x}, \lambda )= \sum _{i= 1}^L N_i e_i(\lambda )$ and $\frac{E(\mathbf{x}, \lambda )}{N}= \sum _{i= 1}^L n_i e_i(\lambda )$.

In the canonical ensemble, we get:

$$ E =\sum _\mathbf{x} E(\mathbf{x}, \lambda ) \frac{\exp (-\beta E (\mathbf{x}, \lambda ))}{Z}=N\sum _{i =1}^L e_i (\lambda ) n_i (\beta ,\lambda ) \nonumber $$

with $E= \sum _{i =1}^N e(x_i, \lambda )$ and $n_i (\beta ,\lambda )$ the fraction of states with energy $e(x_i, \lambda )=e_i (\lambda )$, given by $n_i (\beta ,\lambda )= \frac{\exp (-\beta e_i (\lambda ))}{Z}$. We get:

$$\begin{aligned} dE =N \Big ( \sum _{i =1}^L d e_i (\lambda ) n_i (\beta ,\lambda ) + \sum _{i =1}^L e_i (\lambda ) d n_i (\beta ,\lambda )\Big ). \end{aligned}$$

(6.C.1)

In the sum, the first term equals, using (6.7.2), (6.7.3),

$$ N\sum _{i =1}^L \frac{\partial e_i}{\partial \lambda } n_i (\beta ,\lambda ) d \lambda = \sum _\mathbf{x} \frac{\partial E(\mathbf{x}, \lambda )}{\partial \lambda } \frac{\exp (-\beta E (\mathbf{x}, \lambda ))}{Z} d \lambda =-dW. \nonumber $$

Therefore, we get from (6.7.1):

$$\begin{aligned} Tds = \frac{T}{N}dS = \sum _{i =1}^L e_i (\lambda ) d n_i (\beta ,\lambda ). \end{aligned}$$

(6.C.2)

So we see that the variation of entropy is given (up to the factor T) by the variation of the occupation numbers of the different energy levels, while the work is given by the variation of the individual energies. If no work is done ($\lambda$ is constant) then we can still have a variation of the total energy through heat transfer and that translates microscopically into a variation of the occupation numbers $n_i$’s: the occupation numbers of higher energy levels can increase or decrease (depending on the direction of the heat transfer) and that leads to a change of entropy.

One can also obtain directly (6.C.2) from the formula (6.7.7)

$$Tds =\frac{1}{N}\left( kT\frac{\partial \ln Z}{\partial \lambda } -\frac{\partial ^2 \ln Z}{\partial \beta \partial \lambda }\right) d\lambda - \frac{\partial ^2 \ln Z}{\partial ^2 \beta } d\beta $$

and $n_i (\beta ,\lambda )= \frac{\exp (-\beta e_i (\lambda ))}{Z}$:

We have:

$$\frac{1}{N}kT\frac{\partial \ln Z}{\partial \lambda }= -<\frac{\partial e (\lambda )}{\partial \lambda }> $$

and

$$ \frac{1}{N}\frac{\partial ^2 \ln Z}{\partial \beta \partial \lambda }= -<\frac{\partial e (\lambda )}{\partial \lambda }> + \beta \left(<\frac{\partial e (\lambda )}{\partial \lambda } e (\lambda )> -<\frac{\partial e (\lambda )}{\partial \lambda }> <e (\lambda )>\right) . \nonumber $$

Also,

$$ \frac{1}{N}\frac{\partial ^2 \ln Z}{\partial ^2 \beta } =<e^2(\lambda )>- <e(\lambda )>^2. $$

So,

$$\begin{aligned} \begin{aligned}&Tds = -<\frac{\partial e (\lambda )}{\partial \lambda }> d\lambda +<\frac{\partial e (\lambda )}{\partial \lambda }> d\lambda \\&- \beta \left(<\frac{\partial e (\lambda )}{\partial \lambda } e (\lambda )> -<\frac{\partial e (\lambda )}{\partial \lambda }> <e (\lambda )>\right) d\lambda \\&- (<e(\lambda )^2>-<e(\lambda )>^2) d\beta \\&=-\beta \left(<\frac{\partial e (\lambda )}{\partial \lambda } e (\lambda )> -<\frac{\partial e (\lambda )}{\partial \lambda }> <e (\lambda )>\right) d\lambda \\&- (<e(\lambda )^2>- <e(\lambda )>^2) d\beta \\ \end{aligned}\end{aligned}$$

(6.C.3)

Let us compute the right hand side of (6.C.2)

$$\begin{aligned} \sum _{i =1}^L e_i (\lambda ) d n_i (\beta ,\lambda )= \sum _{i =1}^L e_i (\lambda ) \frac{\partial n_i (\beta ,\lambda )}{\partial \beta } d\beta + \sum _{i =1}^L e_i (\lambda ) \frac{\partial n_i (\beta ,\lambda )}{\partial \lambda } d\lambda \end{aligned}$$

(6.C.4)

Computing each term in (6.C.4), we have:

$$ \sum _{i =1}^L e_i (\lambda ) \frac{\partial n_i (\beta ,\lambda )}{\partial \beta }= - (<e(\lambda )^2>- <e(\lambda )>^2),$$

and

$$ \sum _{i =1}^L e_i (\lambda ) \frac{\partial n_i (\beta ,\lambda )}{\partial \lambda }= - \beta \left(<\frac{\partial e (\lambda )}{\partial \lambda } e (\lambda )> -<\frac{\partial e (\lambda )}{\partial \lambda }> <e (\lambda )>\right) . $$

Inserting this in (6.C.4) and comparing with (6.C.3) proves (6.C.2).