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Physics LibreTexts

3.1: Modeling the Approach to Equilibrium

  • Page ID
    18555
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    \)
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    \)
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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

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