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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

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