81 lines
2.4 KiB
TeX
81 lines
2.4 KiB
TeX
\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}
|