# peeterjoot/physicsplay

modified: basicStatMechProblemSet6Problem1.tex

 @@ -149,7 +149,148 @@ \makeSubAnswer{}{basicStatMech:problemSet6:1b} -TODO. +For the action'' that we want to minimize, let's write + +\begin{dmath}\label{eqn:basicStatMechProblemSet6Problem2:380} +f = S - \alpha \sum_j p_j - \beta \sum_j E_j p_j, +\end{dmath} + +for which we seek $\alpha$, $\beta$ such that + +\begin{dmath}\label{eqn:basicStatMechProblemSet6Problem2:400} +0 += \PD{p_i}{f} += -\PD{p_i}{} +\sum_j +p_j +\lr{ +\kB \ln p_j + \alpha + \beta E_j +} += +-\kB (\ln p_i + 1) - \alpha - \beta E_i, +\end{dmath} + +or + +\begin{dmath}\label{eqn:basicStatMechProblemSet6Problem2:420} +p_i = +\exp\lr{ +- \lr{\alpha - \beta E_i}/\kB - 1 +}. +\end{dmath} + +Our probability constraint is +\begin{dmath}\label{eqn:basicStatMechProblemSet6Problem2:440} +1 += \sum_j +\exp\lr{ +- \lr{\alpha - \beta E_j}/\kB - 1 +} += +\exp\lr{ +- \alpha/\kB - 1 +} +\sum_j +\exp\lr{ +- \beta E_j/\kB +}, +\end{dmath} + +or + +\begin{dmath}\label{eqn:basicStatMechProblemSet6Problem2:460} +\exp\lr{ +\alpha/\kB + 1 +} += +\sum_j +\exp\lr{ +- \beta E_j/\kB +}. +\end{dmath} + +Taking logs we have + +\begin{dmath}\label{eqn:basicStatMechProblemSet6Problem2:480} +\alpha/\kB + 1 = +\ln \sum_j +\exp\lr{ +- \beta E_j/\kB +}. +\end{dmath} + +We could continue to solve for $\alpha$ explicitly but don't care any more than this. Plugging back into the probability \eqnref{eqn:basicStatMechProblemSet6Problem2:420} obtained from the unconstrained minimization we have + +\begin{dmath}\label{eqn:basicStatMechProblemSet6Problem2:500} +p_i = +\exp\lr{ +-\ln \sum_j +\exp\lr{ +- \beta E_j/\kB +} +} +\exp\lr{ +- \beta E_i/\kB +}, +\end{dmath} + +or + +\begin{dmath}\label{eqn:basicStatMechProblemSet6Problem2:520} +\myBoxed{ +p_i = +\frac{ + \exp\lr{ + - \beta E_i/\kB + } +} +{ + \sum_j + \exp\lr{ + - \beta E_j/\kB + } +}. +} +\end{dmath} + +To determine $\beta$ we must look implicitly to the energy constraint, which is + +\begin{dmath}\label{eqn:basicStatMechProblemSet6Problem2:540} +\expectation{E} += \sum_i E_i p_i += +\sum_i +E_i +\lr{ + \frac{ + \exp\lr{ + - \beta E_i/\kB + } + } + { + \sum_j + \exp\lr{ + - \beta E_j/\kB + } + } +}, +\end{dmath} + +or +\begin{dmath}\label{eqn:basicStatMechProblemSet6Problem2:560} +\myBoxed{ +\expectation{E} = +\frac{ + \sum_i E_i \exp\lr{ - \beta E_i/\kB } +} +{ + \sum_j \exp\lr{ - \beta E_j/\kB } +}. +} +\end{dmath} + +The constraint $\beta$ ($=1/T$) is given implicitly by this energy constraint. + \makeSubAnswer{}{basicStatMech:problemSet6:1c} TODO.