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L18 add figures.

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1 parent dbd8776 commit 82b3116eb6fe8a9721c8fb3350f6b921effb7028 @peeterjoot committed Mar 26, 2013
@@ -19,7 +19,7 @@ \section{Disclaimer}
\paragraph{Review}
-Last time we found that the low temperature behaviour or the chemical potential was quadratic
+Last time we found that the low temperature behaviour or the chemical potential was quadratic as in \cref{fig:lecture18:lecture18Fig1}.
\begin{dmath}\label{eqn:basicStatMechLecture18:20}
\mu =
@@ -28,9 +28,7 @@ \section{Disclaimer}
%\kB T ?
\end{dmath}
-%\epsilon(k) = \frac{\hbar^2 k^2}{2m}
-
-F1
+\imageFigure{lecture18Fig1}{Fermi gas chemical potential}{fig:lecture18:lecture18Fig1}{0.2}
\paragraph{Specific heat}
@@ -62,13 +60,10 @@ \section{Disclaimer}
\lr{ n_{\mathrm{F}}(\epsilon, T) - n_{\mathrm{F}}(\epsilon, 0)}
\end{dmath}
-The only change in the distribution
-
-F2
+The only change in the distribution \cref{fig:lecture18:lecture18Fig2}, that is of interest is over the step portion of the distribution, and over this range of interest $N(\epsilon)$ is approximately constant as in \cref{fig:lecture18:lecture18Fig3}.
-that is of interest is over the step portion of the distribution, and over this range of interest $N(\epsilon)$ is approximately constant as in
-
-F3
+\imageFigure{lecture18Fig2}{Fermi distribution}{fig:lecture18:lecture18Fig2}{0.2}
+\imageFigure{lecture18Fig3}{Fermi gas density of states}{fig:lecture18:lecture18Fig3}{0.2}
\begin{subequations}
\begin{dmath}\label{eqn:basicStatMechLecture18:120}
@@ -144,7 +139,12 @@ \section{Disclaimer}
\int_{-\infty}^\infty \frac{ dy y^2 e^y }{ (e^y + 1)^2 \kB T^2} (\kB T)^3
=
N(\epsilon_{\mathrm{F}}) \kB^2 T
-\mathLabelBox{
+\mathLabelBox
+[
+ labelstyle={xshift=2cm},
+ linestyle={out=270,in=90, latex-}
+]
+{
\int_{-\infty}^\infty \frac{ dy y^2 e^y }{ (e^y + 1)^2 }
}{$\pi^2/3$}
=
@@ -161,7 +161,7 @@ \section{Disclaimer}
\epsilon_{\mathrm{F}} = \frac{\hbar^2 \kF^2}{2 m}
\end{dmath}
\begin{dmath}\label{eqn:basicStatMechLecture18:280}
-\kF = \lr{6 \pi \rho}
+\kF = \lr{6 \pi^2 \rho}
^{1/3},
\end{dmath}
\end{subequations}
@@ -181,22 +181,22 @@ \section{Disclaimer}
= \inv{4 \pi^2}
\lr{\frac{2m}{\hbar^2}}
^{3/2}
-\frac{\hbar }{\sqrt{2m}} \lr{6 \pi \rho}^{1/3}
+\frac{\hbar }{\sqrt{2m}} \lr{6 \pi^2 \rho}^{1/3}
= \inv{4 \pi^2}
\lr{\frac{2m}{\hbar^2}}
-\lr{6 \pi \frac{N}{V}}^{1/3}
+\lr{6 \pi^2 \frac{N}{V}}^{1/3}
\end{dmath}
-FIXME: ...
+FIXME: don't see how this leads to the board result?
\begin{dmath}\label{eqn:basicStatMechLecture18:300}
\frac{C}{N} =
\frac{\pi^2}{2} \kB \frac{ \kB T}{\epsilon_{\mathrm{F}}}.
\end{dmath}
-This is illustrated in
+This is illustrated in \cref{fig:lecture18:lecture18Fig4}.
-F4
+\imageFigure{lecture18Fig4}{Specific heat per Fermion}{fig:lecture18:lecture18Fig4}{0.25}
\paragraph{Relativisitic gas}
@@ -214,26 +214,31 @@ \section{Disclaimer}
\item massless Dirac Fermion
-F5
+%\cref{fig:lecture18:lecture18Fig5}.
+\imageFigure{lecture18Fig5}{Relativisitic gas energy distribution}{fig:lecture18:lecture18Fig5}{0.3}
+
+We can think of this state distribution in a condensed matter view, where we can have a hole to electron state transition by supplying energy to the system (i.e. shining light on the substrate). This can also be thought of in a relativisitic particle view where the same state transition can be thought of as a positron electron pair transition. A round trip transition will have to supply energy like $2 m_0 c^2$ as illustrated in \cref{fig:lecture18:lecture18Fig6}.
-We can think of this state distribution in a condensed matter view, where we can have a hole to electron state transition by supplying energy to the system (i.e. shining light on the substrate). This can also be thought of in a relativisitic particle view where the same state transition can be thought of as a positron electron pair transition.
+\imageFigure{lecture18Fig6}{Hole to electron round trip transition energy requirement}{fig:lecture18:lecture18Fig6}{0.2}
\end{itemize}
-Considering graphene, a 2D system, we want to determine the density of states $N(\epsilon)$,
+\paragraph{Graphene}
+
+Consider graphene, a 2D system. We want to determine the density of states $N(\epsilon)$,
\begin{dmath}\label{eqn:basicStatMechLecture18:380}
\int \frac{d^2 \Bk}{(2 \pi)^2} \rightarrow \int_{-\infty}^\infty d\epsilon N(\epsilon),
\end{dmath}
-We'll find
+We'll find a density of states distribution like \cref{fig:lecture18:lecture18Fig7}.
+
+\imageFigure{lecture18Fig7}{Density of states for 2D linear energy momentum distribution}{fig:lecture18:lecture18Fig7}{0.2}
\begin{dmath}\label{eqn:basicStatMechLecture18:400}
N(\epsilon) = \text{constant factor} \frac{\Abs{\epsilon}}{v},
\end{dmath}
-F7
-
\begin{dmath}\label{eqn:basicStatMechLecture18:420}
C \sim \frac{d}{dT} \int N(\epsilon) n_{\mathrm{F}}(\epsilon) \epsilon d\epsilon,
\end{dmath}
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