150 lines
7.6 KiB
TeX
150 lines
7.6 KiB
TeX
\documentclass{article}
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% set up telugu
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\usepackage{fontspec}
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\newfontfamily\telugufont{Potti Sreeramulu}[Script = Telugu]
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\usepackage{polyglossia}
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\setdefaultlanguage{english}
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\setotherlanguage{telugu}
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%other packages
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\usepackage{amsmath}
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\usepackage{amssymb}
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\usepackage{physics}
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\usepackage{siunitx}
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\usepackage{todonotes}
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\usepackage{luacode}
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\usepackage{titling}
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\usepackage{enumerate}
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% custom deepak packages
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\usepackage{luatrivially}
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\usepackage{subtitling}
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\usepackage{cleveref}
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\begin{luacode*}
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math.randomseed(31415926)
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\end{luacode*}
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\newcommand{\kb}{k_{\mathrm{B}}}
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\title{Problem 3.8}
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\subtitle{Microcanonical energy fluctuations}
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\author{\begin{telugu}హృదయ్ దీపక్ మల్లుభొట్ల\end{telugu}}
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% want empty date
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\predate{}
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\date{}
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\postdate{}
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% !TeX spellcheck = en_GB
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\begin{document}
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\maketitle
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This problem talks about about weakly coupling subsystems and looking at fluctuations.
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An earlier result is that
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\begin{equation}
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\sigma_{E_1}^2 = - \flatfrac{\kb}{\left(\pdv[2]{S_1}{E_1} + \pdv[2]{S_2}{E_2}\right)} \label{eq:1}
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\end{equation}
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\begin{enumerate}[(a)]
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\item Show that
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\begin{equation}
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\frac{1}{\kb} \pdv[2]{S}{E} = - \frac{1}{\kb T} \frac{1}{N c_v T}, \label{eq:2}
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\end{equation}
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where $c_v$ is the inverse of the total specific heat at constant volume.
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The specific heat $c_v$ is the energy needed per particle to change the temperature by one unit:
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\begin{equation}
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N c_v = \left.\left(\pdv{E}{T}\right)\right|_{V, N}
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\end{equation}
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\item If $c_v^{(1)}$ and $c_v^{(2)}$ are the specific heats per particle for two subsystems of $N$ particles each, show using \cref{eq:1,eq:2} that
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\begin{equation}
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\frac{1}{c_v^{(1)}} + \frac{1}{c_v^{(2)}} = \frac{N \kb T^2}{\sigma_{E_1}^2} \label{eq:3}
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\end{equation}
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\item Using the equipartition theorem, write the temperature in terms of the mean kinetic energy $K = \left<E_1\right>$ (with standard deviation $\sigma_K$).
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Show that $c_v^{(1)} = \flatfrac{3\kb}{2}$ for the momentum degrees of freedom.
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In terms of $K$ and $\sigma_K$, solve for the total specific heat of the molecular dynamics simulation (configurational plus kinetic).
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\end{enumerate}
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\section{Solution} \label{sec:solution}
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\subsection*{(a)}
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We start with the general case, keeping volume and particle number constant.
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To start, we use $\frac{1}{T} = \left. \pdv{S}{E} \right|_{V, N}$, so
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\begin{align}
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\frac{1}{\kb} \left.\pdv[2]{S}{E} \right|_{V, N} &= \frac{1}{\kb} \left. \pdv{}{E} \frac{1}{T} \right|_{V, N} \\
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&= \frac{1}{\kb} \left. \pdv{}{E} \frac{1}{T} \right|_{V, N} \\
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&= - \frac{1}{\kb} \left. \frac{1}{T^2} \pdv{T}{E} \right|_{V, N},
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\end{align}
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and we use the Physicist's Theorem to invert the derivative, giving us $\flatfrac{1}{N c_v} = \left.\left(\pdv{T}{E}\right)\right|_{V, N}$, so
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\begin{align}
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\frac{1}{\kb} \left.\pdv[2]{S}{E} \right|_{V, N} &= - \frac{1}{\kb} \frac{1}{T^2} \frac{1}{N c_v} \\
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\frac{1}{\kb} \left.\pdv[2]{S}{E} \right|_{V, N} &= - \frac{1}{\kb T} \frac{1}{N c_v T},
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\end{align}
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as desired.
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\subsection*{(b)}
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We want to show
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\begin{equation}
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\frac{1}{c_v^{(1)}} + \frac{1}{c_v^{(2)}} = \frac{N \kb T^2}{\sigma_{E_1}^2}
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\end{equation}
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Start with the $\sigma_{E_1}^2$ term from \cref{eq:1}, and use our result from the first part \cref{eq:2}:
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\begin{align}
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\sigma_{E_1}^2 &= - \flatfrac{\kb}{\left(\pdv[2]{S_1}{E_1} + \pdv[2]{S_2}{E_2}\right)} \\
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\sigma_{E_1}^2 &= - \flatfrac{1}{\left(\frac{1}{\kb} \pdv[2]{S_1}{E_1} + \frac{1}{\kb}\pdv[2]{S_2}{E_2}\right)} \\
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- \frac{1}{\sigma_{E_1}^2} &= \frac{1}{\kb} \pdv[2]{S_1}{E_1} + \frac{1}{\kb}\pdv[2]{S_2}{E_2} \\
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- \frac{1}{\sigma_{E_1}^2} &= \left(- \frac{1}{\kb T} \frac{1}{N c_v^{(1)} T}\right) + \left(- \frac{1}{\kb T} \frac{1}{N c_v^{(2)} T} \right)\\
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\frac{1}{\sigma_{E_1}^2} &= \frac{1}{\kb T} \frac{1}{N c_v^{(1)} T} + \frac{1}{\kb T} \frac{1}{N c_v^{(2)} T} \\
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\frac{N \kb T^2}{\sigma_{E_1}^2} &= \frac{1}{c_v^{(1)}} + \frac{1}{c_v^{(2)}},
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\end{align}
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giving us our desired result.
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\subsection*{(c)}
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Sethna keeps these assumptions implicit, but this problem is for a monatomic ideal gas in a 3D space.
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I think that assuming 3D space is something you should explicitly state, but that probably says more about me than Sethna.
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In that case, there are three degrees of freedom for the kinetic energy.
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The problem implicitly sets up a division of the degrees of freedom in the problem as the kinetic and configurational, where configuration refers to functions of position (including the background potential and the interactions).
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If there are rotational degrees of freedom, you could group the rotational kinetic energy in with the ``configurational'' ones and recover these results anyway, but then it's not quite configurational so the terminology doesn't make a ton of sense.
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So our system $1$ will include the three linear momenta, and system $2$ includes everything else.
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What Sethna calls $K$ is then the kinetic energy from those three linear momenta, ignoring any rotational kinetic energy etc.
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The equipartition theorem tells us that each degree of freedom in the kinetic energy should satisfy $\frac12 N \kb T$.
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So, using the three degrees of freedom as described above, we can just write that $K = \left< E_1 \right> = \frac32 N \kb T$.
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Our specific heat per particle satisfies $N c_v^{(1)} = \left.\left( \pdv{E_1}{T} \right)\right|_{V, N}$, which is clearly $c_v^{(1)} = \frac32 \kb$, as Sethna wants us to find.
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Now lets use \cref{eq:3} and solve for $c_v^{(2)}$.
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\begin{align}
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\frac{1}{c_v^{(1)}} + \frac{1}{c_v^{(2)}} &= \frac{N \kb T^2}{\sigma_{E_1}^2} \\
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\frac{1}{\frac32 \kb} + \frac{1}{c_v^{(2)}} &= \frac{N \kb T^2}{\sigma_K^2} \\
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\frac{1}{c_v^{(2)}} &= \frac{N \kb T^2}{\sigma_K^2} - \frac{1}{\frac32 \kb} \\
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\frac{1}{c_v^{(2)}} &= \frac{\frac32 N \kb^2 T^2}{\frac32 \kb \sigma_K^2} - \frac{\sigma_K^2}{\frac32 \kb \sigma_K^2} \\
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\frac{1}{c_v^{(2)}} &= \frac{\frac32 N \kb^2 T^2 - \sigma_K^2}{\frac32 \kb \sigma_K^2}.
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\end{align}
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Solve for $K$ rather than $T$:
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\begin{align}
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K &= \frac32 N \kb T \\
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T &= \frac23 \frac{1}{N \kb} K.
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\end{align}
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Plug this back in:
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\begin{align}
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\frac{1}{c_v^{(2)}} &= \frac{\frac32 N \kb^2 \left(\frac23 \frac{1}{N \kb} K \right)^2 - \sigma_K^2}{\frac32 \kb \sigma_K^2} \\
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\frac{\frac32 \kb \sigma_K^2}{c_v^{(2)}} &= \frac32 N \kb^2 \left(\frac23 \frac{1}{N \kb} K \right)^2 - \sigma_K^2 \\
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\frac{\frac32 \kb \sigma_K^2}{c_v^{(2)}} &= \frac23 N \kb^2 \frac{1}{N^2 \kb^2} K^2 - \sigma_K^2 \\
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\frac{\frac32 \kb \sigma_K^2}{c_v^{(2)}} &= \frac23 \frac{1}{N} K^2 - \sigma_K^2 \\
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\frac{c_v^{(2)}}{\frac32 \kb \sigma_K^2} &= \flatfrac{1}{\left(\frac23 \frac{1}{N} K^2 - \sigma_K^2\right)} \\
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c_v^{(2)} &= \flatfrac{\frac32 \kb \sigma_K^2}{\left(\frac23 \frac{1}{N} K^2 - \sigma_K^2\right)} \\
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c_v^{(2)} &= \frac{9 N \kb \sigma_K^2}{4 K^2 - 6 N \sigma_K^2}
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\end{align}
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That's not a pretty equation.
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Our total specific heat is $c_v^{(1)} + c_v^{(2)}$.
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\begin{align}
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c_v^{(1)} + c_v^{(2)} &= \frac32 \kb + \frac{9 N \kb \sigma_K^2}{4 K^2 - 6 N \sigma_K^2} \\
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c_v^{(1)} + c_v^{(2)} &= \frac{3 \kb}{2}\frac{2 K^2 - 3 N \sigma_K^2}{2 K^2 - 3 N \sigma_K^2} + \frac{9 N \kb \sigma_K^2}{4 K^2 - 6 N \sigma_K^2} \\
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c_v^{(1)} + c_v^{(2)} &= \frac{6 \kb K^2 - 9 \kb N \sigma_K^2}{4 K^2 - 6 N \sigma_K^2} + \frac{9 N \kb \sigma_K^2}{4 K^2 - 6 N \sigma_K^2} \\
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c_v^{(1)} + c_v^{(2)} &= \frac{6 \kb K^2 - 9 \kb N \sigma_K^2 + 9 N \kb \sigma_K^2}{4 K^2 - 6 N \sigma_K^2} \\
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c_v^{(1)} + c_v^{(2)} &= \frac{6 \kb K^2}{4 K^2 - 6 N \sigma_K^2} \\
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c_v^{(1)} + c_v^{(2)} &= \frac{3 \kb K^2}{2 K^2 - 3 N \sigma_K^2}
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\end{align}
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\newpage
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\listoftodos
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\end{document}
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