3.1: Modeling the Approach to Equilibrium

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Equilibrium

A thermodynamic system typically consists of an enormously large number of constituent particles, a typical ‘large number’ being Avogadro’s number, $\NA=6.02\times 10^{23}$. Nevertheless, in equilibrium, such a system is characterized by a relatively small number of thermodynamic state variables. Thus, while a complete description of a (classical) system would require us to account for $\CO\big(10^{23}\big)$ evolving degrees of freedom, with respect to the physical quantities in which we are interested, the details of the initial conditions are effectively forgotten over some microscopic time scale $\tau$, called the collision time, and over some microscopic distance scale, $\ell$, called the mean free path¹. The equilibrium state is time-independent.

The Master Equation

Relaxation to equilibrium is often modeled with something called the master equation. Let $P\ns_i(t)$ be the probability that the system is in a quantum or classical state $i$ at time $t$. Then write

$${dP\ns_i\over dt}=\sum_j\big(W\ns_{ij}\,P\ns_j- W\ns_{ji}\,P\ns_i\big)\ . \label{MEQN}$$

Here, $W\ns_{ij}$ is the rate at which $j$ makes a transition to $i$. Note that we can write this equation as

$${dP\ns_i\over dt}=-\sum_j\Gamma\ns_{ij}\,P\ns_j\ ,$$

where

$$\Gamma\ns_{ij}=\begin{cases} -W\ns_{ij} & {if}\ i\ne j\\ \sum'_k W\ns_{kj} & {if}\ i=j\ , \end{cases}$$

where the prime on the sum indicates that $k=j$ is to be excluded. The constraints on the $W\ns_{ij}$ are that $W\ns_{ij}\ge 0$ for all $i,j$, and we may take $W\ns_{ii}\equiv 0$ (no sum on $i$). Fermi’s Golden Rule of quantum mechanics says that

$$W\ns_{ij}={2\pi\over\hbar}\,\big|\sexpect{i}{\hat V}{j}\big|^2\,\rho(E\ns_j)\ ,$$

where $\HH\ns_0\,\ket{i}=E\ns_i\,\ket{i}$, $\hat V$ is an additional potential which leads to transitions, and $\rho(E\ns_i)$ is the density of final states at energy $E\ns_i$. The fact that $W\ns_{ij}\ge 0$ means that if each $P\ns_i(t=0)\ge 0$, then $P\ns_i(t)\ge 0$ for all $t\ge 0$. To see this, suppose that at some time $t>0$ one of the probabilities $P\ns_i$ is crossing zero and about to become negative. But then Equation $\ref{MEQN}$ says that $\DP\ns_i(t)=\sum_j W\ns_{ij} P\ns_j(t) \ge 0$. So $P\ns_i(t)$ can never become negative.

Equilibrium distribution and detailed balance

If the transition rates $W\ns_{ij}$ are themselves time-independent, then we may formally write

$$P\ns_i(t)=\big(e^{-\Gamma t}\big)\ns_{ij}\,P\ns_j(0)\ .$$

Here we have used the Einstein ‘summation convention’ in which repeated indices are summed over (in this case, the $j$ index). Note that

$$\sum_i\Gamma\ns_{ij}=0\ ,$$

which says that the total probability $\sum_i P\ns_i$ is conserved:

$${d\over dt}\sum_i P\ns_i=-\sum_{i,j}\Gamma\ns_{ij}\,P\ns_j =-\sum_j \bigg(P\ns_j\> \sum_i\Gamma\ns_{ij}\bigg)=0\ .$$

We conclude that ${\vec\phi}=(1,1,\ldots,1)$ is a left eigenvector of $\Gamma$ with eigenvalue $\lambda=0$. The corresponding right eigenvector, which we write as $P^{eq}_i$, satisfies $\Gamma\ns_{ij} P^{eq}_j=0$, and is a stationary ( time independent) solution to the master equation. Generally, there is only one right/left eigenvector pair corresponding to $\lambda=0$, in which case any initial probability distribution $P\ns_i(0)$ converges to $P^{eq}_i$ as $t\to\infty$, as shown in Appendix I (§7).

In equilibrium, the net rate of transitions into a state $\sket{i}$ is equal to the rate of transitions out of $\sket{i}$. If, for each state $\sket{j}$ the transition rate from $\sket{i}$ to $\sket{j}$ is equal to the transition rate from $\sket{j}$ to $\sket{i}$, we say that the rates satisfy the condition of detailed balance. In other words,

$$W\ns_{ij} \, P^{eq}_j= W\ns_{ji} \, P^{eq}_i .$$

Assuming $W\ns_{ij}\ne 0$ and $P^{eq}_j\ne 0$, we can divide to obtain

$${W\ns_{ji}\over W\ns_{ij}}={P^{eq}_j\over P^{eq}_i}\ .$$

Note that detailed balance is a stronger condition than that required for a stationary solution to the master equation.

If $\Gamma=\Gamma^\Rt$ is symmetric, then the right eigenvectors and left eigenvectors are transposes of each other, hence $P^{eq}=1/N$, where $N$ is the dimension of $\Gamma$. The system then satisfies the conditions of detailed balance. See Appendix II (§8) for an example of this formalism applied to a model of radioactive decay.

Boltzmann’s $\SH$-theorem

Suppose for the moment that $\Gamma$ is a symmetric matrix, $\Gamma\ns_{ij}=\Gamma\ns_{ji}$. Then construct the function

$$\SH(t)=\sum_i P\ns_i(t)\,\ln P\ns_i(t)\ .$$

Then

$$\begin{split} {d\SH\over dt}&=\sum_i{dP\ns_i\over dt}\,\big(1+\ln P\ns_i) = \sum_i{dP\ns_i\over dt}\,\ln P\ns_i\\ &=-\sum_{i,j}\Gamma\ns_{ij}\,P\ns_j\ln P\ns_i\\ &=\sum_{i,j}\Gamma\ns_{ij}\,P\ns_j\big(\!\ln P\ns_j-\ln P\ns_i\big)\ , \end{split}$$

where we have used $\sum_i\Gamma\ns_{ij}=0$. Now switch $i\leftrightarrow j$ in the above sum and add the terms to get

$${d\SH\over dt}={1\over 2}\sum_{i,j}\Gamma\ns_{ij} \big(P\ns_i-P\ns_j\big)\,\big(\!\ln P\ns_i-\ln P\ns_j\big)\ .$$

Note that the $i=j$ term does not contribute to the sum. For $i\ne j$ we have $\Gamma\ns_{ij}=-W\ns_{ij}\le 0$, and using the result

$$(x-y)\,(\ln x - \ln y)\ge 0\ ,$$

we conclude

$${d\SH\over dt}\le 0\ .$$

In equilibrium, $P^{eq}_i$ is a constant, independent of $i$. We write

$$P^{eq}_i={1\over\ROmega}\quad,\quad\ROmega=\sum_i 1 \quad\Longrightarrow\quad \SH=-\ln\ROmega\ .$$

If $\Gamma\ns_{ij}\ne\Gamma\ns_{ji}$, we can still prove a version of the $\SH$-theorem. Define a new symmetric matrix

$${\overline W}\ns_{ij}\equiv W\ns_{ij} \, P_j^{eq} = W\ns_{ji} \, P^{eq}_i = {\overline W}\ns_{ji}\ ,$$

and the generalized $\SH$-function,

$$\SH(t)\equiv \sum_i P\ns_i(t)\,\ln\!\bigg({P\ns_i(t)\over P_i^{eq}}\bigg)\ .$$

Then

$${d\SH\over dt}=-{1\over 2}\> \sum_{i,j} {\overline W}\ns_{ij}\,\bigg({P\ns_i\over P^{eq}_i}-{P\ns_j\over P^{eq}_j}\bigg) \Bigg[\ln\!\bigg({P\ns_i\over P^{eq}_i}\bigg)-\ln\!\bigg({P\ns_j\over P^{eq}_j}\bigg)\Bigg]\le 0\ .$$