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Clarifications to L16 and L17. Still more to go.

modified:   ../../bin/latexRegex.pl
modified:   ../METADATA
modified:   basicStatMechLecture16.tex
modified:   basicStatMechLecture17.tex
new file:   lecture16Fig6.pdf
new file:   lecture16Fig6p.png
new file:   lecture16TLogTPlotFig7.pdf
new file:   lecture16TLogTPlotFig7p.png
modified:   renumber
modified:   ../phy452/basicStatMechLecture15.tex
new file:   ../phy452/mathematica/lecture16DensityPlot.nb
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1 parent 84d8307 commit 85df4caf6e1fa5fdede2cb8eddecf22f4c9a80b0 @peeterjoot committed Mar 26, 2013
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@@ -26,7 +26,6 @@
#s/_div/_{\\mathrm{div}}/g ;
#s/_E/_{\\mathrm{E}}/g ;
#s/_eff/_{\\mathrm{eff}}/g ;
-s/_F/_{\\mathrm{F}}/g ;
#s/_final/_{\\mathrm{final}}/g ;
#s/_full/_{\\text{full}}/g ;
#s/_G/_{\\mathrm{G}}/g ;
@@ -40,7 +39,6 @@
#s/_ph/_{\\mathrm{ph}}/g ;
#s/_quantum/_{\\mathrm{quantum}}/g ;
#s/_subsystem/_{\\text{subsystem}}/g ;
-s/_T/_{\\mathrm{T}}/g ;
#s/_total/_{\\mathrm{total}}/g ;
#s/_V/_{\\mathrm{V}}/g ;
#s/{QM}/{\\mathrm{QM}}/g ;
@@ -59,4 +57,6 @@
#s/xialpha\b/x_{i_\\alpha}/g ;
#s/xidotalpha/\\dot{x}_{i_\\alpha}/g ;
s/\\ee/\\epsilon/g ;
-s/\\pp(.)/{($1)}/g ;
+#s/\\pp(.)/{($1)}/g ;
+#s/_F/_{\\mathrm{F}}/g ;
+#s/_T/_{\\mathrm{T}}/g ;
View
@@ -4709,7 +4709,12 @@ Generate figures for continuum mechanics problem set II figure 1. Using Show an
{
DATE => 'March 21, 2013',
path => 'phy452/mathematica/kittelZipper.nb',
- WHAT => qq(Kittel zipper problem and plot. Found how to use Placed PlotLegends to get the legend into the figure for use in Save As.),
+ WHAT => qq(Kittel zipper problem and plot. Found how to use Placed PlotLegends with {Left, Top} instead of Left, to get the legend into the figure for use in Save As.),
+},
+{
+ DATE => 'March 25, 2013',
+ path => 'phy452/mathematica/lecture16DensityPlot.nb',
+ WHAT => qq(Plot the period boundary conditions density. Used Mathematica Map, pure functions, Placed PlotLegends, ToString, Text, text concatonation operator),
},
# not all of these are committed to the repo. Some are, but are not described here.
#blogit/imageProcessingExperimentation.cdf
@@ -58,7 +58,7 @@ \section{Fermi gas}
\Bk = \frac{2\pi}{L}(n_x, n_y, n_z)
\end{dmath}
-This is for periodic boundary conditions (not particle in a box), where
+This is for periodic boundary conditions \footnote{I filled in details in the last lecture using a particle in a box, whereas this periodic condition was intented. We see that both achieve the same result}, where
\begin{dmath}\label{eqn:basicStatMechLecture16:120}
\Psi(x + L) = \Psi(x)
@@ -152,7 +152,7 @@ \section{Fermi gas}
\mu(T \rightarrow \infty) \rightarrow - \infty
\end{dmath}
-so that for large $T$ we haave
+so that for large $T$ we have
\begin{dmath}\label{eqn:basicStatMechLecture16:380}
\inv{e^{\beta(\epsilon_k - \mu)} + 1} \rightarrow e^{-\beta(\epsilon_k - \mu)}
@@ -172,10 +172,15 @@ \section{Fermi gas}
where
\begin{dmath}\label{eqn:basicStatMechLecture16:400}
-\lambda = \frac{h}{\sqrt{2 \pi m \kB T}}
+\lambda = \frac{h}{\sqrt{2 \pi m \kB T}}.
\end{dmath}
-where $\lambda$ is the \underlineAndIndex{thermal de Broglie wavelength}, $\lambda^3 \sim T^{-3/2}$.
+The density as a function of temperature is plotted in \cref{fig:lecture16:lecture16Fig6} for a few arbitrarily chosen values of the chemical potential $\mu$.
+
+\imageFigure{lecture16Fig6}{Density as a function of temperature}{fig:lecture16:lecture16Fig6}{0.3}
+
+Here $\lambda$ is the \underlineAndIndex{thermal de Broglie wavelength}, $\lambda^3 \sim T^{-3/2}$.
+
We can write
@@ -195,9 +200,10 @@ \section{Fermi gas}
\mu \propto -T \ln T
\end{dmath}
-Plotting this in \cref{fig:lecture16:lecture16Fig3}.
+The chemical potential is plotted in \cref{fig:lecture16:lecture16Fig3}, whereas this $- \kB T \ln \kB T$ function is plotted in \cref{fig:lecture16TLogTPlot:lecture16TLogTPlotFig7}. The contributions to $\mu$ from the $\kB T \ln (\rho h^3 (2 \pi m)^{-3/2})$ term are dropped for the high temperature approximation. It's not entirely clear to me how we justify dropping the $\ln \rho$, since we see that $\rho = \rho(\mu, T)$ gets very small in the high temperature limit, but if all we are about is that $\mu$ is large and negative, then this isn't inconsisent.
\imageFigure{lecture16Fig3}{Chemical potential over degenerate to classical range}{fig:lecture16:lecture16Fig3}{0.3}
+\imageFigure{lecture16TLogTPlotFig7}{High temp approximation of chemical potential, extended back to $T = 0$}{fig:lecture16TLogTPlot:lecture16TLogTPlotFig7}{0.3}
\paragraph{Pressure}
@@ -51,7 +51,7 @@ \section{Fermi gas thermodynamics}
e^{\beta \times 0} = 1 = \rho \lambda_{\mathrm{T}}^3.
\end{dmath}
-That last identification provides the relation $T \sim T_{\mathrm{F}}$. FIXME: how?
+That last identification provides the relation $T \sim T_{\mathrm{F}}$. FIXME: that bit wasn't clear to me.
\end{enumerate}
@@ -72,6 +72,8 @@ \section{Fermi gas thermodynamics}
\sum_k \epsilon_\Bk n_{\mathrm{F}}(\epsilon_\Bk).
\end{dmath}
+FIXME: review where these came from.
+
We can define a density of states
\begin{dmath}\label{eqn:basicStatMechLecture17:120}
@@ -87,7 +89,7 @@ \section{Fermi gas thermodynamics}
\begin{dmath}\label{eqn:basicStatMechLecture17:140}
\sum_\Bk f(\epsilon_\Bk)
-= \sum_\Bk \int_{-\infty}^\infty d\epsilon \delta(\epsilon - \epsilon_\Bk) i
+= \sum_\Bk \int_{-\infty}^\infty d\epsilon \delta(\epsilon - \epsilon_\Bk)
f(\epsilon_\Bk)
=
\sum_\Bk
@@ -113,26 +115,46 @@ \section{Fermi gas thermodynamics}
\frac{V}{(2 \pi)^3} \int d^3 \Bk \delta\lr{ \epsilon - \frac{\hbar^2 k^2}{2 m}}
=
\frac{V}{(2 \pi)^3} 4 \pi \int_0^\infty k^2 dk \delta\lr{ \epsilon - \frac{\hbar^2 k^2}{2 m}}
+\end{dmath}
+
+Using
+\begin{dmath}\label{eqn:basicStatMechLecture17:440}
+\delta(g(x)) = \sum_{x_0} \frac{\delta(x - x_0)}{\Abs{g'(x_0)}},
+\end{dmath}
+
+where the roots of $g(x)$ are $x_0$, we have
+\begin{dmath}\label{eqn:basicStatMechLecture17:460}
+N(\epsilon) =
+\frac{V}{(2 \pi)^3} 4 \pi \int_0^\infty k^2 dk \delta
+\lr{ k - \frac{\sqrt{2 m \epsilon}}{\hbar} }
+\frac{m \hbar }{ \hbar^2 \sqrt{2 m \epsilon}}
=
-\cdots
+\frac{V}{(2 \pi)^3} 2 \pi \frac{2 m \epsilon}{\hbar^2}
+\frac{2 m \hbar }{ \hbar^2 \sqrt{2 m \epsilon}}
=
V \lr{\frac{2 m}{\hbar^2}}^{3/2} \inv{4 \pi^2} \sqrt{\epsilon}.
\end{dmath}
-FIXME: chug through detail of $\cdots$ and understand this.
-
-To make that last evaluation a change of vars and $k \sim \sqrt{\epsilon}$ so that $\epsilon^{3/2} /\epsilon \sim \sqrt{\epsilon}$.
+%To make that last evaluation a change of vars and $k \sim \sqrt{\epsilon}$ so that $\epsilon^{3/2} /\epsilon \sim \sqrt{\epsilon}$.
In \underline{2D} this would be
\begin{dmath}\label{eqn:basicStatMechLecture17:180}
-N(\epsilon) \sim V \int dk k \delta \lr{ \epsilon - \frac{\hbar^2 k^2}{2m} } \sim V,
+N(\epsilon) \sim V \int dk k \delta \lr{ \epsilon - \frac{\hbar^2 k^2}{2m} }
+=
+V \frac{\sqrt{2 m \epsilon}}{\hbar} \frac{m \hbar}{\hbar^2 \sqrt{ 2 m \epsilon}}
+\sim
+V
\end{dmath}
and in \underline{1D}
\begin{dmath}\label{eqn:basicStatMechLecture17:200}
-N(\epsilon) \sim V \int dk k \delta \lr{ \epsilon - \frac{\hbar^2 k^2}{2m} } \sim \inv{\sqrt{\epsilon}}.
+N(\epsilon) \sim V \int dk \delta \lr{ \epsilon - \frac{\hbar^2 k^2}{2m} }
+=
+V
+\frac{m \hbar}{\hbar^2 \sqrt{ 2 m \epsilon}}
+\sim \inv{\sqrt{\epsilon}}.
\end{dmath}
\paragraph{What happens when we have linear energy momentum relationships?}
@@ -158,7 +180,9 @@ \section{Fermi gas thermodynamics}
\epsilon_\Bk = \pm v_{\mathrm{F}} \Abs{\Bk}.
\end{dmath}
-Continuing with the \underline{3D} case we have
+\paragraph{Continuing with the \underline{3D} case we have}
+
+FIXME: Is this (or how is this) related to the linear energy momentum relationships for Graphene like substances?
\begin{dmath}\label{eqn:basicStatMechLecture17:280}
N = V \int_0^\infty
@@ -271,17 +295,16 @@ \section{Fermi gas thermodynamics}
\lr{
1 + \frac{\pi^2}{8} \frac{(\kB T)^2}{\epsilon_{\mathrm{F}}^2}
}
-=
-1 - \frac{3}{2} a \frac{T^2}{\epsilon_{\mathrm{F}}} + \frac{\pi^2}{8} \frac{(\kB T)^2}{\epsilon_{\mathrm{F}}^2}
+\approx
+1 - \frac{3}{2} a \frac{T^2}{\epsilon_{\mathrm{F}}} + \frac{\pi^2}{8} \frac{(\kB T)^2}{\epsilon_{\mathrm{F}}^2},
\end{dmath}
-We have used a Taylor expansion $(1 + x)^n \approx 1 + n x$ for small $x$. With
-
+or
\begin{dmath}\label{eqn:basicStatMechLecture17:380}
a = \frac{\pi^2}{12} \frac{\kB^2}{\epsilon_{\mathrm{F}}},
\end{dmath}
-we have
+We have used a Taylor expansion $(1 + x)^n \approx 1 + n x$ for small $x$, for an end result of
\begin{dmath}\label{eqn:basicStatMechLecture17:400}
\mu = \epsilon_{\mathrm{F}} - \frac{\pi^2}{12} \frac{(\kB T)^2}{\epsilon_{\mathrm{F}}}.
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@@ -1,18 +1,18 @@
+#~/bin/lgrep basicStatMechLecture16.tex | tee o ; . ./o
+#~/bin/lgrep basicStatMechProblemSet6Problem2.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
+#~/bin/lgrep basicStatMechProblemSet6Problem6.tex | tee o ; . ./o
+#~/bin/lgrep basicStatMechProblemSet6Problem7.tex | tee o ; . ./o
#~/bin/lgrep desaiCh24.tex | tee o ; . ./o
#~/bin/lgrep desaiCh24.tex | tee o ; . ./o
#~/bin/lgrep energyProbabilityPathriaQuestion.tex | tee o ; . ./o
#~/bin/lgrep landauSection11Problem2b.tex | tee o ; . ./o
#~/bin/lgrep pipeFlowConstPressureGradient.tex | tee o ; . ./o
#~/bin/lgrep t.tex | tee o ; . ./o
#perl -p -i ./p basicStatMechLecture16.tex
-#~/bin/lgrep basicStatMechLecture16.tex | tee o ; . ./o
#perl -p -i ./p energyProbabilityPathriaQuestion.tex
-#~/bin/lgrep basicStatMechLecture17.tex | tee o ; . ./o
-#perl -p -i ./p basicStatMechLecture17.tex
-~/bin/lgrep basicStatMechProblemSet6Problem1.tex | tee o ; . ./o
-#~/bin/lgrep basicStatMechProblemSet6Problem2.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
-#~/bin/lgrep basicStatMechProblemSet6Problem6.tex | tee o ; . ./o
-#~/bin/lgrep basicStatMechProblemSet6Problem7.tex | tee o ; . ./o
+#~/bin/lgrep basicStatMechProblemSet6Problem1.tex | tee o ; . ./o
+perl -p -i ./p basicStatMechLecture17.tex
+~/bin/lgrep basicStatMechLecture17.tex | tee o ; . ./o
@@ -350,8 +350,12 @@ \section{Fermions and Bosons}
We can also calculate the chemical potential at high temperatures. We'll find that this has the form
\begin{dmath}\label{eqn:basicStatMechLecture15:700}
-e^{\beta \mu} = \frac{4}{3} \rho \lambda_T^3,
+e^{\beta \mu} =
+%\frac{4}{3}
+\rho \lambda_T^3,
\end{dmath}
+%
+%FIXME: this isn't the result we found in lectures 16 and 17 (no $4/3$ factor).
where this quantity $\lambda_T$ is called the \underlineAndIndex{Thermal de Broglie wavelength}.
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