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modified: basicStatMechProblemSet6Problem1.tex

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1 parent 295e8e2 commit 1ca6259eda8cceeeb7c5fc0fca2e949c01dc5c19 @peeterjoot committed Mar 31, 2013
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143 notes/blogit/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.

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