# peeterjoot/physicsplay

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

modified:   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