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278 notes/blogit/basicStatMechLecture17.tex
@@ -9,15 +9,279 @@
\input{../peeter_prologue_print2.tex}
\beginArtNoToc
-\generatetitle{PHY452H1S Basic Statistical Mechanics. Lecture 17: XXX. Taught by Prof.\ Arun Paramekanti}
-%\chapter{XXX}
+\generatetitle{PHY452H1S Basic Statistical Mechanics. Lecture 17: Fermi gas thermodynamics. Taught by Prof.\ Arun Paramekanti}
+%\chapter{Fermi gas thermodynamics}
\label{chap:basicStatMechLecture17}
-\section{Disclaimer}
+%\section{Disclaimer}
+%
+%Peeter's lecture notes from class. May not be entirely coherent.
+%
+%\section{Fermi gas thermodynamics}
+%
+%\begin{enumerate}
+%\item Energy was found to be
+%
+%\begin{dmath}\label{eqn:basicStatMechLecture17:20}
+%\frac{E}{N} = \frac{3}{5} \epsilon_{\mathrm{F}}\qquad T = 0.
+%\end{dmath}
+%
+%\item Pressure was found to have the form
+%
+%F1
+%
+%\item The chemical potential was found to have the form
+%
+%F2
+%
+%We found that
+%
+%\begin{subequations}
+%\begin{dmath}\label{eqn:basicStatMechLecture17:40}
+%e^{\beta \mu} = \rho \lambda_T^3,
+%\end{dmath}
+%\begin{dmath}\label{eqn:basicStatMechLecture17:60}
+%\lambda_T = \frac{h}{\sqrt{ 2 \pi m \kB T}}
+%\end{dmath}
+%\end{subequations}
+%
+%so that the zero crossing is approximately when
+%
+%\begin{subequations}
+%e^{\beta \times 0} = 1 = \rho \lambda_T^3,
+%\end{subequations}
+%
+%which gives $T \sim T_{\mathrm{F}}$.
+%
+%\end{enumerate}
+%
+%\paragraph{How about at other temperatures?}
+%
+%\begin{enumerate}
+%\item $\mu(T) = ?$
+%\item $E(T) = ?$
+%\item $\CV(T) = ?$
+%\end{enumerate}
+%
+%We had
+%\begin{dmath}\label{eqn:basicStatMechLecture17:80}
+%N = \sum_k \inv{e^{\beta (\epsilon_k - \mu)} + 1 = \sum_{\Bk} n_{\mathrm{F}}(\epsilon_\Bk).
+%\end{dmath}
+%\begin{dmath}\label{eqn:basicStatMechLecture17:100}
+%E(T) =
+%\sum_k \epsilon_\Bk n_{\mathrm{F}}(\epsilon_\Bk).
+%\end{dmath}
+%
+%We can define a density of states
+%
+%\begin{dmath}\label{eqn:basicStatMechLecture17:120}
+%\sum_\Bk
+%= \sum_\Bk \int_{-\infty}^\infty d\epsilon \delta(\epsilon - \epsilon_\Bk)
+%=
+%\int_{-\infty}^\infty d\epsilon
+%\sum_\Bk
+%\delta(\epsilon - \epsilon_\Bk),
+%\end{dmath}
+%
+%where the liberty to informally switch the order of differentiation and integration has been used. This construction allows us to write a more general sum
+%
+%XX
+\begin{dmath}\label{eqn:basicStatMechLecture17:140}
+\sum_\Bk f(\epsilon_\Bk) =
+= \sum_\Bk \int_{-\infty}^\infty d\epsilon \delta(\epsilon - \epsilon_\Bk) i
+f(\epsilon_\Bk)
+=
+\sum_\Bk
+\int_{-\infty}^\infty d\epsilon
+\delta(\epsilon - \epsilon_\Bk)
+f(\epsilon_\Bk)
+=
+\int_{-\infty}^\infty d\epsilon
+f(\epsilon_\Bk)
+\lr{
+\sum_\Bk
+\delta(\epsilon - \epsilon_\Bk)
+}
+\end{dmath}
+
+\begin{dmath}\label{eqn:basicStatMechLecture17:160}
+N(\epsilon) \equiv
+\sum_\Bk
+\delta(\epsilon - \epsilon_\Bk)
+=
+\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}}
+=
+\cdots
+V \lr{\frac{2 m}{\hbar^2}}^{3/2} \inv{4 \pi^2} \sqrt{\epsilon}
+\end{dmath}
+
+To make that last evaluation a change of vars and $k \sim \sqrt{\epsilon}$ so that $\epsilon^{3/2} /\epsilon \sim \sqrt{\epsilon}$. FIXME: chug through and understand this.
+
+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
+\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}}.
+\end{dmath}
+
+\paragraph{What happens when we have linear energy momentum relationships}
+
+\begin{dmath}\label{eqn:basicStatMechLecture17:220}
+\epsilon_\Bk = v \Abs{\Bk}
+\end{dmath}
+
+At high velocities energy and momentum of particles are also of this form
+
+\begin{dmath}\label{eqn:basicStatMechLecture17:240}
+\epsilon_\Bk = \sqrt{ m_0^2 c^4 + p^2 c^2 } \sim \Abs{\Bp} c.
+\end{dmath}
+
+Another example is graphene, a carbon structure of the form
+
+F3
+
+where the energy momentum is related in roughly the following form
+
+F4
+
+where
+
+\begin{dmath}\label{eqn:basicStatMechLecture17:260}
+\epsilon_\Bk = \pm v_{\mathrm{F}} \Abs{\Bk}.
+\end{dmath}
+
+Continuing with the \underline{3D} case we have
+
+\begin{dmath}\label{eqn:basicStatMechLecture17:280}
+N = V \int_0^\infty
+\mathLabelBox{
+n_{\mathrm{F}}(\epsilon)
+}{$1/(e^{\beta (\epsilon - \mu)} + 1)$}
+\mathLabelBox
+[
+ labelstyle={below of=m\themathLableNode, below of=m\themathLableNode}
+]
+{
+N(\epsilon)
+}{
+$\epsilon^{1/2}$
+}
+\end{dmath}
+
+\begin{dmath}\label{eqn:basicStatMechLecture17:300}
+\rho
+=
+\frac{N}{V}
+=
+\lr{ \frac{2m}{\hbar^2 } }
+^{3/2} \inv{ 4 \pi^2}
+\int_0^\infty d\epsilon \frac{\epsilon^{1/2}}{z^{-1} e^{\beta \epsilon} + 1}
+=
+\lr{ \frac{2m}{\hbar^2 } }
+^{3/2} \inv{ 4 \pi^2}
+\lr{\kB T}
+^{3/2}
+\int_0^\infty dx \frac{x^{1/2}}{z^{-1} e^{x} + 1}
+\end{dmath}
+
+where $z = e^{\beta \mu}$ as usual, and we write $x = \beta \epsilon$. For the low temperature aysmtotic behaviour see one of the appendices (FIXME:lookup) of \citep{pathriastatistical}. For $z$ large it can be shown that this is
+
+\begin{dmath}\label{eqn:basicStatMechLecture17:320}
+\int_0^\infty dx \frac{x^{1/2}}{z^{-1} e^{x} + 1}
+\approx
+\frac{2}{3}
+\lr{\ln z}
+^{3/2}
+\lr{
+1 + \frac{\pi^2}{8} \inv{(\ln z)^2}
+},
+\end{dmath}
+
+so that
+
+\begin{dmath}\label{eqn:basicStatMechLecture17:340}
+\rho \approx
+\lr{ \frac{2m}{\hbar^2 } }
+^{3/2} \inv{ 4 \pi^2}
+\lr{\kB T}
+^{3/2}
+\frac{2}{3}
+\lr{\ln z}
+^{3/2}
+\lr{
+1 + \frac{\pi^2}{8} \inv{(\ln z)^2}
+}
+=
+\lr{ \frac{2m}{\hbar^2 } }
+^{3/2} \inv{ 4 \pi^2}
+\frac{2}{3}
+\mu^{3/2}
+\lr{
+1 + \frac{\pi^2}{8} \inv{(\beta \mu)^2}
+}
+=
+\lr{ \frac{2m}{\hbar^2 } }
+^{3/2} \inv{ 4 \pi^2}
+\frac{2}{3}
+\mu^{3/2}
+\lr{
+1 + \frac{\pi^2}{8}
+\lr{ \frac{\kB T}{\mu}}^2
+}
+= \rho_{T = 0}
+\lr{ \frac{\mu}{ \ee_F } }
+^{3/2}
+\lr{
+1 + \frac{\pi^2}{8}
+\lr{ \frac{\kB T}{\mu}}^2
+}
+\end{dmath}
+
+Assuming a quadratic form for the chemical potential at low temperature
+
+\begin{dmath}\label{eqn:basicStatMechLecture17:n}
+1 =
+\lr{ \frac{\mu}{ \ee_F } }
+^{3/2}
+\lr{
+1 + \frac{\pi^2}{8}
+\lr{ \frac{\kB T}{\mu}}^2
+}
+=
+\lr{ \frac{\ee_F - a T^2}{ \ee_F } }
+^{3/2}
+\lr{
+1 + \frac{\pi^2}{8}
+\lr{ \frac{\kB T}{\ee_F - a T^2}}^2
+}
+\approx
+\lr{
+1 - \frac{3}{2} a \frac{T^2}{\ee_F}
+}
+\lr{
+1 + \frac{\pi^2}{8} \frac{(\kB T)^2}{\ee_F^2}
+}
+=
+1 - \frac{3}{2} a \frac{T^2}{\ee_F} + \frac{\pi^2}{8} \frac{(\kB T)^2}{\ee_F^2}
+\end{dmath}
+
+We have used a Taylor expansion $(1 + x)^n \approx 1 + n x$ for small $x$. With
+
+\begin{dmath}\label{eqn:basicStatMechLecture17:n}
+a = \frac{\pi^2}{12} \frac{\kB^2}{\ee_F}
+\end{dmath}
-Peeter's lecture notes from class. May not be entirely coherent.
+we have
-\section{XXX}
+\begin{dmath}\label{eqn:basicStatMechLecture17:n}
+\mu = \ee_F - \frac{\pi^2}{12} \frac{(\kB T)^2}{\ee_F}
+\end{dmath}
-%\EndArticle
-\EndNoBibArticle
+\EndArticle

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