# peeterjoot/physicsplay

1 parent dbd8776 commit 82b3116eb6fe8a9721c8fb3350f6b921effb7028 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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