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problem 1a done.

modified:   basicStatMechProblemSet6Problem1.tex
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commit 295e8e245dbde2249794df82c3bc9c967f73532b 1 parent f2884a8
@peeterjoot authored
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4 notes/blogit/basicStatMechProblemSet6.tex
@@ -18,8 +18,8 @@
%
%This is an ungraded set of answers to the problems posed.
-%\input{basicStatMechProblemSet6Problem1.tex}
-\input{basicStatMechProblemSet6Problem2.tex}
+\input{basicStatMechProblemSet6Problem1.tex}
+%\input{basicStatMechProblemSet6Problem2.tex}
%\input{basicStatMechProblemSet6Problem3.tex}
%\input{basicStatMechProblemSet6Problem4.tex}
%\input{basicStatMechProblemSet6Problem5.tex}
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70 notes/blogit/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.
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4 notes/blogit/basicStatMechProblemSet6Problem2.tex
@@ -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})
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4 notes/blogit/renumber
@@ -1,4 +1,3 @@
-#~/bin/lgrep basicStatMechProblemSet6Problem1.tex | tee o ; . ./o
#~/bin/lgrep basicStatMechProblemSet6Problem3.tex | tee o ; . ./o
#~/bin/lgrep basicStatMechProblemSet6Problem4.tex | tee o ; . ./o
#~/bin/lgrep basicStatMechProblemSet6Problem5.tex | tee o ; . ./o
@@ -11,4 +10,5 @@
#~/bin/lgrep pipeFlowConstPressureGradient.tex | tee o ; . ./o
#~/bin/lgrep t.tex | tee o ; . ./o
#perl -p -i ./p energyProbabilityPathriaQuestion.tex
-~/bin/lgrep basicStatMechProblemSet6Problem2.tex | tee o ; . ./o
+#~/bin/lgrep basicStatMechProblemSet6Problem2.tex | tee o ; . ./o
+~/bin/lgrep basicStatMechProblemSet6Problem1.tex | tee o ; . ./o
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