# peeterjoot/physicsplay

modified: ../../env/.aspell.en.pws

modified:   basicStatMechProblemSet6Problem1.tex
 @@ -149,7 +149,7 @@ \makeSubAnswer{}{basicStatMech:problemSet6:1b} -For the action'' that we want to minimize, let's write +For the action'' like quantity 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, @@ -293,5 +293,90 @@ \makeSubAnswer{}{basicStatMech:problemSet6:1c} -TODO. +Again write + +\begin{dmath}\label{eqn:basicStatMechProblemSet6Problem2:580} +f = S - \alpha \sum_j p_j - \beta \sum_j E_j p_j - \gamma \sum_j N_j p_j. +\end{dmath} + +The unconstrained minimization requires + +\begin{dmath}\label{eqn:basicStatMechProblemSet6Problem2:600} +0 += \PD{p_i}{f} += -\PD{p_i}{} +\lr{ +\kB (\ln p_i + 1) + \alpha + \beta E_i + \gamma N_i +}, +\end{dmath} + +or + +\begin{dmath}\label{eqn:basicStatMechProblemSet6Problem2:620} +p_i += +\exp\lr{ -\alpha/\kB - 1 } +\exp\lr{ -(\beta E_i + \gamma N_i)/\kB }. +\end{dmath} + +The unit probability constraint requires + +\begin{dmath}\label{eqn:basicStatMechProblemSet6Problem2:640} +1 += \sum_j p_j += +\exp\lr{ -\alpha/\kB - 1 } +\sum_j +\exp\lr{ -(\beta E_j + \gamma N_j)/\kB }, +\end{dmath} + +or +\begin{dmath}\label{eqn:basicStatMechProblemSet6Problem2:660} +\exp\lr{ -\alpha/\kB - 1 } = +\inv{ +\sum_j +\exp\lr{ -(\beta E_j + \gamma N_j)/\kB } +}. +\end{dmath} + +Our probability is then + +\begin{dmath}\label{eqn:basicStatMechProblemSet6Problem2:680} +\myBoxed{ +p_i = +\frac{ +\exp\lr{ -(\beta E_i + \gamma N_i)/\kB } +}{ +\sum_j +\exp\lr{ -(\beta E_j + \gamma N_j)/\kB } +}. +} +\end{dmath} + +The average energy $\expectation{E} = \sum_j p_j E_j$ and average number of particles $\expectation{N} = \sum_j p_j N_j$ are given by + +\begin{subequations} +\begin{dmath}\label{eqn:basicStatMechProblemSet6Problem2:700} +\expectation{E} = +\frac{ +E_i +\exp\lr{ -(\beta E_i + \gamma N_i)/\kB } +}{ +\sum_j +\exp\lr{ -(\beta E_j + \gamma N_j)/\kB } +} +\end{dmath} +\begin{dmath}\label{eqn:basicStatMechProblemSet6Problem2:720} +\expectation{N} = +\frac{ +N_i +\exp\lr{ -(\beta E_i + \gamma N_i)/\kB } +}{ +\sum_j +\exp\lr{ -(\beta E_j + \gamma N_j)/\kB } +}. +\end{dmath} +\end{subequations} + +The values $\beta$ and $\gamma$ are fixed implicitly by requiring simultaneous solutions of these equations. }