# peeterjoot/physicsplay

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
1 parent e0b745e commit 453e5120d219c8c2e943715f87c04ee9409ee800 committed Mar 21, 2013
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 ;
255 notes/blogit/basicStatMechLecture17.tex
 @@ -13,116 +13,120 @@ %\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}} = \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{ @@ -209,79 +210,81 @@ \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{ 1 + \frac{\pi^2}{8} \inv{(\ln z)^2} } = \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} \inv{(\beta \mu)^2} } = \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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