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modified: ../../bin/latexRegex.pl

modified:   basicStatMechLecture17.tex
new file:   lecture17Fig1.png
new file:   lecture17Fig2.png
new file:   lecture17Fig3.png
new file:   lecture17Fig4.png
new file:   lecture17Fig5.png
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commit 453e5120d219c8c2e943715f87c04ee9409ee800 1 parent e0b745e
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2  bin/latexRegex.pl
@@ -40,7 +40,7 @@
#s/_ph/_{\\mathrm{ph}}/g ;
#s/_quantum/_{\\mathrm{quantum}}/g ;
#s/_subsystem/_{\\text{subsystem}}/g ;
-#s/_T/_{\\mathrm{T}}/g ;
+s/_T/_{\\mathrm{T}}/g ;
#s/_total/_{\\mathrm{total}}/g ;
#s/_V/_{\\mathrm{V}}/g ;
#s/{QM}/{\\mathrm{QM}}/g ;
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255 notes/blogit/basicStatMechLecture17.tex
@@ -13,100 +13,101 @@
%\chapter{Fermi gas thermodynamics}
\label{chap:basicStatMechLecture17}
-%\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
+\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{equation}\label{eqn:basicStatMechLecture17:20}
+\frac{E}{N} = \frac{3}{5} \epsilon_{\mathrm{F}}\qquad \mbox{where $T = 0$}.
+\end{equation}
+
+\item Pressure was found to have the form \cref{fig:lecture17:lecture17Fig3}
+
+\imageFigure{lecture17Fig3}{Pressure in Fermi gas}{fig:lecture17:lecture17Fig3}{0.3}
+
+\item The chemical potential was found to have the form \cref{fig:lecture17:lecture17Fig2}.
+
+\imageFigure{lecture17Fig2}{Chemical potential in Fermi gas}{fig:lecture17:lecture17Fig2}{0.3}
+
+We found that
+
+\begin{subequations}
+\begin{dmath}\label{eqn:basicStatMechLecture17:40}
+e^{\beta \mu} = \rho \lambda_{\mathrm{T}}^3
+\end{dmath}
+\begin{dmath}\label{eqn:basicStatMechLecture17:60}
+\lambda_{\mathrm{T}} = \frac{h}{\sqrt{ 2 \pi m \kB T}},
+\end{dmath}
+\end{subequations}
+
+so that the zero crossing is approximately when
+
+\begin{dmath}\label{eqn:basicStatMechLecture17:420}
+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?
+
+\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
+
\begin{dmath}\label{eqn:basicStatMechLecture17:140}
-\sum_\Bk f(\epsilon_\Bk) =
+\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
+=
+\sum_\Bk
+\int_{-\infty}^\infty d\epsilon
\delta(\epsilon - \epsilon_\Bk)
f(\epsilon_\Bk)
-=
-\int_{-\infty}^\infty d\epsilon
+=
+\int_{-\infty}^\infty d\epsilon
f(\epsilon_\Bk)
\lr{
-\sum_\Bk
+\sum_\Bk
\delta(\epsilon - \epsilon_\Bk)
-}
+}.
\end{dmath}
+This sum, evaluated using a continuum approximation, is
+
\begin{dmath}\label{eqn:basicStatMechLecture17:160}
N(\epsilon) \equiv
-\sum_\Bk
+\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}}
@@ -114,15 +115,18 @@
\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}
+=
+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.
+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 $\epsilon^{3/2} /\epsilon \sim \sqrt{\epsilon}$.
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} } \sim V,
\end{dmath}
and in \underline{1D}
@@ -131,27 +135,24 @@
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}
+\paragraph{What happens when we have linear energy momentum relationships?}
+
+Suppose that we have a linear energy momentum relationship like
\begin{dmath}\label{eqn:basicStatMechLecture17:220}
-\epsilon_\Bk = v \Abs{\Bk}
+\epsilon_\Bk = v \Abs{\Bk}.
\end{dmath}
-At high velocities energy and momentum of particles are also of this form
+An example of such a relationship is the high velocity relation between the energy and momentum of a particle
\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
+Another example is graphene, a carbon structure of the form \cref{fig:lecture17:lecture17Fig1}. The energy and momentum for such a structure is related in roughly as shown in \cref{fig:lecture17:lecture17Fig4}, where
-F3
-
-where the energy momentum is related in roughly the following form
-
-F4
-
-where
+\imageFigure{lecture17Fig1}{Graphene bond structure}{fig:lecture17:lecture17Fig1}{0.3}
+\imageFigure{lecture17Fig4}{Graphene energy momentum dependence}{fig:lecture17:lecture17Fig4}{0.3}
\begin{dmath}\label{eqn:basicStatMechLecture17:260}
\epsilon_\Bk = \pm v_{\mathrm{F}} \Abs{\Bk}.
@@ -160,7 +161,7 @@
Continuing with the \underline{3D} case we have
\begin{dmath}\label{eqn:basicStatMechLecture17:280}
-N = V \int_0^\infty
+N = V \int_0^\infty
\mathLabelBox{
n_{\mathrm{F}}(\epsilon)
}{$1/(e^{\beta (\epsilon - \mu)} + 1)$}
@@ -181,22 +182,22 @@
\frac{N}{V}
=
\lr{ \frac{2m}{\hbar^2 } }
-^{3/2} \inv{ 4 \pi^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}
+^{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
+where $z = e^{\beta \mu}$ as usual, and we write $x = \beta \epsilon$. For the low temperature aysmtotic behaviour see \citep{pathriastatistical} appendix \S E. 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}
+\frac{2}{3}
\lr{\ln z}
^{3/2}
\lr{
@@ -209,10 +210,10 @@
\begin{dmath}\label{eqn:basicStatMechLecture17:340}
\rho \approx
\lr{ \frac{2m}{\hbar^2 } }
-^{3/2} \inv{ 4 \pi^2}
+^{3/2} \inv{ 4 \pi^2}
\lr{\kB T}
^{3/2}
-\frac{2}{3}
+\frac{2}{3}
\lr{\ln z}
^{3/2}
\lr{
@@ -220,7 +221,7 @@
}
=
\lr{ \frac{2m}{\hbar^2 } }
-^{3/2} \inv{ 4 \pi^2}
+^{3/2} \inv{ 4 \pi^2}
\frac{2}{3}
\mu^{3/2}
\lr{
@@ -228,60 +229,62 @@
}
=
\lr{ \frac{2m}{\hbar^2 } }
-^{3/2} \inv{ 4 \pi^2}
+^{3/2} \inv{ 4 \pi^2}
\frac{2}{3}
\mu^{3/2}
\lr{
-1 + \frac{\pi^2}{8}
+1 + \frac{\pi^2}{8}
\lr{ \frac{\kB T}{\mu}}^2
}
-= \rho_{T = 0}
-\lr{ \frac{\mu}{ \ee_F } }
+= \rho_{T = 0}
+\lr{ \frac{\mu}{ \epsilon_{\mathrm{F}} } }
^{3/2}
\lr{
-1 + \frac{\pi^2}{8}
+1 + \frac{\pi^2}{8}
\lr{ \frac{\kB T}{\mu}}^2
}
\end{dmath}
-Assuming a quadratic form for the chemical potential at low temperature
+Assuming a quadratic form for the chemical potential at low temperature as in \cref{fig:lecture17:lecture17Fig5}, we have
+
+\imageFigure{lecture17Fig5}{Assumed quadratic form for low temperature chemical potential}{fig:lecture17:lecture17Fig5}{0.3}
-\begin{dmath}\label{eqn:basicStatMechLecture17:n}
-1 =
-\lr{ \frac{\mu}{ \ee_F } }
+\begin{dmath}\label{eqn:basicStatMechLecture17:360}
+1 =
+\lr{ \frac{\mu}{ \epsilon_{\mathrm{F}} } }
^{3/2}
\lr{
-1 + \frac{\pi^2}{8}
+1 + \frac{\pi^2}{8}
\lr{ \frac{\kB T}{\mu}}^2
}
=
-\lr{ \frac{\ee_F - a T^2}{ \ee_F } }
+\lr{ \frac{\epsilon_{\mathrm{F}} - a T^2}{ \epsilon_{\mathrm{F}} } }
^{3/2}
\lr{
-1 + \frac{\pi^2}{8}
-\lr{ \frac{\kB T}{\ee_F - a T^2}}^2
+1 + \frac{\pi^2}{8}
+\lr{ \frac{\kB T}{\epsilon_{\mathrm{F}} - a T^2}}^2
}
\approx
\lr{
-1 - \frac{3}{2} a \frac{T^2}{\ee_F}
+1 - \frac{3}{2} a \frac{T^2}{\epsilon_{\mathrm{F}}}
}
\lr{
-1 + \frac{\pi^2}{8} \frac{(\kB T)^2}{\ee_F^2}
+1 + \frac{\pi^2}{8} \frac{(\kB T)^2}{\epsilon_{\mathrm{F}}^2}
}
-=
-1 - \frac{3}{2} a \frac{T^2}{\ee_F} + \frac{\pi^2}{8} \frac{(\kB T)^2}{\ee_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}
\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}
+\begin{dmath}\label{eqn:basicStatMechLecture17:380}
+a = \frac{\pi^2}{12} \frac{\kB^2}{\epsilon_{\mathrm{F}}},
\end{dmath}
-we have
+we have
-\begin{dmath}\label{eqn:basicStatMechLecture17:n}
-\mu = \ee_F - \frac{\pi^2}{12} \frac{(\kB T)^2}{\ee_F}
+\begin{dmath}\label{eqn:basicStatMechLecture17:400}
+\mu = \epsilon_{\mathrm{F}} - \frac{\pi^2}{12} \frac{(\kB T)^2}{\epsilon_{\mathrm{F}}}.
\end{dmath}
\EndArticle
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