\documentclass{article}

% set up telugu
\usepackage{fontspec}
\newfontfamily\telugufont{Potti Sreeramulu}[Script = Telugu]
\usepackage{polyglossia}
\setdefaultlanguage{english}
\setotherlanguage{telugu}

%other packages
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{physics}
\usepackage{siunitx}
\usepackage{todonotes}
\usepackage{luacode}
\usepackage{titling}
\usepackage{enumerate}

% custom deepak packages
\usepackage{luatrivially}
\usepackage{subtitling}

\usepackage{cleveref}

\begin{luacode*}
	math.randomseed(31415926)
\end{luacode*}

\newcommand{\kb}{k_{\mathrm{B}}}

\title{Problem 3.8}
\subtitle{Microcanonical energy fluctuations}
\author{\begin{telugu}హృదయ్ దీపక్ మల్లుభొట్ల\end{telugu}}
% want empty date
\predate{}
\date{}
\postdate{}

% !TeX spellcheck = en_GB
\begin{document}
	\maketitle
	
	This problem talks about about weakly coupling subsystems and looking at fluctuations.
	An earlier result is that
	\begin{equation}
		\sigma_{E_1}^2 = - \flatfrac{\kb}{\left(\pdv[2]{S_1}{E_1} + \pdv[2]{S_2}{E_2}\right)} \label{eq:1}
	\end{equation}
	
	\begin{enumerate}[(a)]
		\item Show that
		\begin{equation}
			\frac{1}{\kb} \pdv[2]{S}{E} = - \frac{1}{\kb T} \frac{1}{N c_v T}, \label{eq:2}
		\end{equation}
		where $c_v$ is the inverse of the total specific heat at constant volume.
		The specific heat $c_v$ is the energy needed per particle to change the temperature by one unit:
		\begin{equation}
			N c_v = \left.\left(\pdv{E}{T}\right)\right|_{V, N}
		\end{equation}
		\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
		\begin{equation}
			\frac{1}{c_v^{(1)}} + \frac{1}{c_v^{(2)}} = \frac{N \kb T^2}{\sigma_{E_1}^2} \label{eq:3}
		\end{equation}
		\item Using the equipartition theorem, write the temperature in terms of the mean kinetic energy $K = \left<E_1\right>$ (with standard deviation $\sigma_K$).
		Show that $c_v^{(1)} = \flatfrac{3\kb}{2}$ for the momentum degrees of freedom.
		In terms of $K$ and $\sigma_K$, solve for the total specific heat of the molecular dynamics simulation (configurational plus kinetic).
	\end{enumerate}

	\section{Solution} \label{sec:solution}
	\subsection*{(a)}
	We start with the general case, keeping volume and particle number constant.
	To start, we use $\frac{1}{T} = \left. \pdv{S}{E} \right|_{V, N}$, so 
	\begin{align}
		\frac{1}{\kb} \left.\pdv[2]{S}{E} \right|_{V, N} &= \frac{1}{\kb} \left. \pdv{}{E} \frac{1}{T} \right|_{V, N} \\
		&= \frac{1}{\kb} \left. \pdv{}{E} \frac{1}{T} \right|_{V, N} \\
		&= - \frac{1}{\kb} \left. \frac{1}{T^2} \pdv{T}{E} \right|_{V, N},
	\end{align}
	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
	\begin{align}
		\frac{1}{\kb} \left.\pdv[2]{S}{E} \right|_{V, N} &= - \frac{1}{\kb} \frac{1}{T^2} \frac{1}{N c_v} \\
		\frac{1}{\kb} \left.\pdv[2]{S}{E} \right|_{V, N} &= - \frac{1}{\kb T} \frac{1}{N c_v T},
	\end{align}
	as desired.
	\subsection*{(b)}
	We want to show
	\begin{equation}
		\frac{1}{c_v^{(1)}} + \frac{1}{c_v^{(2)}} = \frac{N \kb T^2}{\sigma_{E_1}^2}
	\end{equation}
	Start with the $\sigma_{E_1}^2$ term from \cref{eq:1}, and use our result from the first part \cref{eq:2}:
	\begin{align}
		\sigma_{E_1}^2 &= - \flatfrac{\kb}{\left(\pdv[2]{S_1}{E_1} + \pdv[2]{S_2}{E_2}\right)} \\
		\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)} \\
		- \frac{1}{\sigma_{E_1}^2} &= \frac{1}{\kb} \pdv[2]{S_1}{E_1} + \frac{1}{\kb}\pdv[2]{S_2}{E_2} \\
		- \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)\\
		\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} \\
		\frac{N \kb T^2}{\sigma_{E_1}^2} &= \frac{1}{c_v^{(1)}} + \frac{1}{c_v^{(2)}},
	\end{align}
	giving us our desired result.
	\subsection*{(c)}
	Sethna keeps these assumptions implicit, but this problem is for a monatomic ideal gas in a 3D space.
	I think that assuming 3D space is something you should explicitly state, but that probably says more about me than Sethna.
	In that case, there are three degrees of freedom for the kinetic energy.
	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).
	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.
	So our system $1$ will include the three linear momenta, and system $2$ includes everything else.
	What Sethna calls $K$ is then the kinetic energy from those three linear momenta, ignoring any rotational kinetic energy etc.

	The equipartition theorem tells us that each degree of freedom in the kinetic energy should satisfy $\frac12 N \kb T$.
	So, using the three degrees of freedom as described above, we can just write that $K = \left< E_1 \right> = \frac32 N \kb T$.
	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.
	
	Now lets use \cref{eq:3} and solve for $c_v^{(2)}$.
	\begin{align}
		\frac{1}{c_v^{(1)}} + \frac{1}{c_v^{(2)}} &= \frac{N \kb T^2}{\sigma_{E_1}^2} \\
		\frac{1}{\frac32 \kb} + \frac{1}{c_v^{(2)}} &= \frac{N \kb T^2}{\sigma_K^2} \\
		\frac{1}{c_v^{(2)}} &= \frac{N \kb T^2}{\sigma_K^2} - \frac{1}{\frac32 \kb} \\
		\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} \\
		\frac{1}{c_v^{(2)}} &= \frac{\frac32 N \kb^2 T^2 - \sigma_K^2}{\frac32 \kb \sigma_K^2}.
	\end{align}
	Solve for $K$ rather than $T$:
	\begin{align}
		K &= \frac32 N \kb T \\
		T &= \frac23 \frac{1}{N \kb} K.
	\end{align}
	Plug this back in:
	\begin{align}
		\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} \\
		\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 \\
		\frac{\frac32 \kb \sigma_K^2}{c_v^{(2)}} &= \frac23 N \kb^2 \frac{1}{N^2 \kb^2} K^2 - \sigma_K^2 \\
		\frac{\frac32 \kb \sigma_K^2}{c_v^{(2)}} &= \frac23 \frac{1}{N} K^2 - \sigma_K^2 \\ 
		\frac{c_v^{(2)}}{\frac32 \kb \sigma_K^2} &= \flatfrac{1}{\left(\frac23 \frac{1}{N} K^2 - \sigma_K^2\right)} \\ 
		c_v^{(2)} &= \flatfrac{\frac32 \kb \sigma_K^2}{\left(\frac23 \frac{1}{N} K^2 - \sigma_K^2\right)} \\ 
		c_v^{(2)} &= \frac{9 N \kb \sigma_K^2}{4 K^2 - 6 N \sigma_K^2}
	\end{align}
	That's not a pretty equation.
	Our total specific heat is $c_v^{(1)} + c_v^{(2)}$.
	\begin{align}
		c_v^{(1)} + c_v^{(2)} &= \frac32 \kb + \frac{9 N \kb \sigma_K^2}{4 K^2 - 6 N \sigma_K^2} \\
		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}  \\
		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}  \\
		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}  \\
		c_v^{(1)} + c_v^{(2)} &= \frac{6 \kb K^2}{4 K^2 - 6 N \sigma_K^2}  \\
		c_v^{(1)} + c_v^{(2)} &= \frac{3 \kb K^2}{2 K^2 - 3 N \sigma_K^2} 
	\end{align}

	\newpage
	\listoftodos

\end{document}
