# peeterjoot/physicsplay

lecture 16 notes converted to sort of coherent form.

 @@ -47,7 +47,7 @@ \section{Fermi gas} gives \begin{dmath}\label{eqn:basicStatMechLecture16:60} -k_{\mathrm{F}} = (6 \pi \rho)^{1/3} +k_{\mathrm{F}} = (6 \pi^2 \rho)^{1/3} \end{dmath} \begin{dmath}\label{eqn:basicStatMechLecture16:80} @@ -58,7 +58,7 @@ \section{Fermi gas} \Bk = \frac{2\pi}{L}(n_x, n_y, n_z) \end{dmath} -This is for periodic boundary conditions \footnote{I filled in details in the last lecture using a particle in a box, whereas this periodic condition was intented. We see that both achieve the same result}, where +This is for periodic boundary conditions \footnote{I filled in details in the last lecture using a particle in a box, whereas this periodic condition was intended. We see that both achieve the same result}, where \begin{dmath}\label{eqn:basicStatMechLecture16:120} \Psi(x + L) = \Psi(x) @@ -104,10 +104,10 @@ \section{Fermi gas} Again \begin{dmath}\label{eqn:basicStatMechLecture16:240} -k_{\mathrm{F}} = (6 \pi \rho)^{1/3} +k_{\mathrm{F}} = (6 \pi^2 \rho)^{1/3} \end{dmath} -\paragraph{Spin considerations} +\makeexample{Spin considerations}{example:basicStatMechLecture16:1}{ %FIXME: %\begin{itemize} @@ -130,14 +130,15 @@ \section{Fermi gas} \begin{dmath}\label{eqn:basicStatMechLecture16:300} k_{\mathrm{F}} = -\lr{\frac{ 6 \pi \rho }{2 S + 1}}^{1/3} +\lr{\frac{ 6 \pi^2 \rho }{2 S + 1}}^{1/3} \end{dmath} and again \begin{dmath}\label{eqn:basicStatMechLecture16:320} \epsilon_{\mathrm{F}} = \frac{\hbar^2 \kF^2}{2m} \end{dmath} +} \paragraph{High Temperatures} @@ -165,22 +166,46 @@ \section{Fermi gas} e^{\beta \mu} \int \frac{d^3 \Bk}{(2 \pi)^3} e^{-\beta \epsilon_k} -\equiv -e^{\beta \mu} \inv{\lambda^3} += +e^{\beta \mu} +\int dk \frac{4 \pi k^2}{(2 \pi)^3} +e^{-\beta \hbar^2 k^2/2m}. \end{dmath} -where +Mathematica (or integration by parts) tells us that -\begin{dmath}\label{eqn:basicStatMechLecture16:400} -\lambda = \frac{h}{\sqrt{2 \pi m \kB T}}. +\begin{dmath}\label{eqn:basicStatMechLecture16:680} +\inv{(2 \pi)^3} +\int 4 \pi^2 k^2 dk +e^{-a k^2} = \inv{(4 \pi a )^{3/2}}, \end{dmath} -The density as a function of temperature is plotted in \cref{fig:lecture16:lecture16Fig6} for a few arbitrarily chosen values of the chemical potential $\mu$. +so we have -\imageFigure{lecture16Fig6}{Density as a function of temperature}{fig:lecture16:lecture16Fig6}{0.3} +\begin{dmath}\label{eqn:basicStatMechLecture16:700} +\rho += +e^{\beta \mu} \lr{ \frac{2m}{ 4 \pi \beta \hbar^2} }^{3/2} += +e^{\beta \mu} \lr{ \frac{2 m \kB T 4 \pi^2 }{ 4 \pi h^2} }^{3/2} += +e^{\beta \mu} \lr{ \frac{2 m \kB T \pi }{ h^2} }^{3/2} +\end{dmath} -Here $\lambda$ is the \underlineAndIndex{thermal de Broglie wavelength}, $\lambda^3 \sim T^{-3/2}$. +Introducing $\lambda$ for the \underlineAndIndex{thermal de Broglie wavelength}, $\lambda^3 \sim T^{-3/2}$ +\begin{dmath}\label{eqn:basicStatMechLecture16:400} +\lambda \equiv \frac{h}{\sqrt{2 \pi m \kB T}}, +\end{dmath} + +we have +\begin{dmath}\label{eqn:basicStatMechLecture16:720} +\rho = e^{\beta \mu} \inv{\lambda^3}. +\end{dmath} + +Does it make any sense to have density as a function of temperature? An inappropriately extended to low temperatures plot of the density is found in \cref{fig:lecture16:lecture16Fig6} for a few arbitrarily chosen numerical values of the chemical potential $\mu$, where we see that it drops to zero with temperature. I suppose that makes sense if we are not holding volume constant. + +\imageFigure{lecture16Fig6}{Density as a function of temperature}{fig:lecture16:lecture16Fig6}{0.3} We can write @@ -194,13 +219,13 @@ \section{Fermi gas} \frac{\mu}{\kB T} = \ln \lr{ \rho \lambda^3 } \sim -\frac{3}{2} \ln T \end{dmath} -or +or (taking $\rho$ (and/or volume?) as a constant) we have for large temperatures \begin{dmath}\label{eqn:basicStatMechLecture16:460} \mu \propto -T \ln T \end{dmath} -The chemical potential is plotted in \cref{fig:lecture16:lecture16Fig3}, whereas this $- \kB T \ln \kB T$ function is plotted in \cref{fig:lecture16TLogTPlot:lecture16TLogTPlotFig7}. The contributions to $\mu$ from the $\kB T \ln (\rho h^3 (2 \pi m)^{-3/2})$ term are dropped for the high temperature approximation. It's not entirely clear to me how we justify dropping the $\ln \rho$, since we see that $\rho = \rho(\mu, T)$ gets very small in the high temperature limit, but if all we are about is that $\mu$ is large and negative, then this isn't inconsisent. +The chemical potential is plotted in \cref{fig:lecture16:lecture16Fig3}, whereas this $- \kB T \ln \kB T$ function is plotted in \cref{fig:lecture16TLogTPlot:lecture16TLogTPlotFig7}. The contributions to $\mu$ from the $\kB T \ln (\rho h^3 (2 \pi m)^{-3/2})$ term are dropped for the high temperature approximation. %It's not entirely clear to me how we justify dropping the $\ln \rho$, since we see that $\rho = \rho(\mu, T)$ gets very small in the high temperature limit, but if all we are about is that $\mu$ is large and negative, then this isn't inconsisent. \imageFigure{lecture16Fig3}{Chemical potential over degenerate to classical range}{fig:lecture16:lecture16Fig3}{0.3} \imageFigure{lecture16TLogTPlotFig7}{High temp approximation of chemical potential, extended back to $T = 0$}{fig:lecture16TLogTPlot:lecture16TLogTPlotFig7}{0.3} @@ -220,10 +245,13 @@ \section{Fermi gas} \end{dmath} For a Fermi gas at $T = 0$ we have +%FIXME: \eqnref{eqn:basicStatMechLecture15:400} \begin{dmath}\label{eqn:basicStatMechLecture16:560} E = \sum_\Bk \epsilon_k n_k += \sum_\Bk \epsilon_k \Theta(\mu_0 - \epsilon_k) += \frac{V}{(2\pi)^3} \int_{\epsilon_k < \mu_0} \frac{\hbar^2 \Bk^2}{2 m} d^3 \Bk = \frac{V}{(2\pi)^3} \int_0^{\kF} \frac{\hbar^2 \Bk^2}{2 m} d^3 \Bk = \frac{V}{(2\pi)^3} \frac{\hbar^2}{2 m}
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