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feat: Adds 3.8 file and starts solution for part a

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Deepak Mallubhotla 2022-03-04 19:52:06 -06:00
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\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[plain]{fancyref}
\usepackage{luacode}
\usepackage{titling}
\usepackage{enumerate}
% custom deepak packages
\usepackage{luatrivially}
\usepackage{subtitling}
\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)}
\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},
\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}
\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.
\newpage
\listoftodos
\end{document}