# peeterjoot/physicsplay

problem 1a done.

modified:   basicStatMechProblemSet6Problem1.tex
 @@ -44,7 +44,7 @@ where $E_i$ is the energy of microstate $i$. Use the method of Lagrange multipliers, doing an unconstrained minimization of the function \begin{dmath}\label{eqn:basicStatMechProblemSet6Problem1:100} -S - \alpha \sum_i p-I - \beta \sum_i E_p p_i, +S - \alpha \sum_i p_i - \beta \sum_i E_i p_i, \end{dmath} then fix $\alpha, \beta$ by demanding that the constraint be satisfied. What is the resulting $p_i$? @@ -80,7 +80,73 @@ \makeanswer{basicStatMech:problemSet6:1}{ \makeSubAnswer{}{basicStatMech:problemSet6:1a} -TODO. +Writing + +\begin{dmath}\label{eqn:basicStatMechProblemSet6Problem2:220} +f += S - \alpha \sum_{j = 1}^\Omega p_j, += -\sum_{j = 1}^\Omega p_j \lr{ \kB \ln p_j + \alpha }, +\end{dmath} + +our unconstrained minimization requires + +\begin{dmath}\label{eqn:basicStatMechProblemSet6Problem2:240} +0 += \PD{p_i}{f} += +-\lr{ +\kB \lr{ \ln p_i + 1 } + \alpha +}. +\end{dmath} + +Solving for $p_i$ we have + +\begin{dmath}\label{eqn:basicStatMechProblemSet6Problem2:260} +p_i = e^{-\alpha/\kB - 1}. +\end{dmath} + +The probabilities for each state are constant. To fix that constant we employ our constraint + +\begin{dmath}\label{eqn:basicStatMechProblemSet6Problem2:280} +1 += \sum_{j = 1}^\Omega p_j += \sum_{j = 1}^\Omega e^{-\alpha/\kB - 1} += \Omega e^{-\alpha/\kB - 1}, +\end{dmath} + +or + +\begin{dmath}\label{eqn:basicStatMechProblemSet6Problem2:300} +\alpha/\kB + 1 = \ln \Omega. +\end{dmath} + +Inserting \eqnref{eqn:basicStatMechProblemSet6Problem2:300} fixes the probability, giving us the first of the expected results + +\begin{equation}\label{eqn:basicStatMechProblemSet6Problem2:320} +\myBoxed{ +p_i = e^{-\ln \Omega} = \inv{\Omega}. +} +\end{equation} + +Using this we our Gibbs entropy can be summed easily + +\begin{dmath}\label{eqn:basicStatMechProblemSet6Problem2:340} +S += -\kB +\sum_{j = 1}^\Omega p_j \ln p_j += -\kB +\sum_{j = 1}^\Omega \inv{\Omega} \ln \inv{\Omega} += -\kB \frac{\Omega}{\Omega} \lr{ -\ln \Omega }, +\end{dmath} + +or + +\begin{dmath}\label{eqn:basicStatMechProblemSet6Problem2:360} +\myBoxed{ +S = \kB \ln \Omega. +} +\end{dmath} + \makeSubAnswer{}{basicStatMech:problemSet6:1b} TODO.
 @@ -338,9 +338,7 @@ \makeSubAnswer{}{basicStatMech:problemSet6:2c} -FIXME: why $z \le 1$. - -For the Boson case, we substitute $z = e^{-\alpha}$ to look at the $z \rightarrow 1$ case. +For the Boson case, we require $z \le 1$ so that the occupancy will be strictly positive. Given this restriction the substitution $z = e^{-\alpha}$ is convenient to look at the $z \rightarrow 1$ case. \begin{dmath}\label{eqn:basicStatMechProblemSet6Problem2:540} G_\nu(e^{-\alpha})