diff --git a/tex/3.15.tex b/tex/3.15.tex new file mode 100644 index 0000000..7bb2fa7 --- /dev/null +++ b/tex/3.15.tex @@ -0,0 +1,60 @@ +\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 3.15} +\subtitle{Entropy maximum and temperature} +\author{\begin{telugu}హృదయ్ దీపక్ మల్లుభొట్ల\end{telugu}} +% want empty date +\predate{} +\date{} +\postdate{} + +% !TeX spellcheck = en_GB +\begin{document} + \maketitle + + Explain in words why, for two weakly coupled systems + \begin{equation} + \rho(E_1) = \flatfrac{\Omega_1(E_1) \Omega_2(E - E_1)}{\Omega(E)} + \end{equation} + is intuitive for a system where all states of energy $E$ have equal probability density. + Using $S = \kb \log(\Omega)$, show in one step that maximising the probability of $E_1$ makes the two temperatures $\frac{1}{T} = \pdv{S}{E}$ the same, and hence that maximising $\rho(E_1)$ maximises the total entropy. + + \section{Solution} \label{sec:solution} + + Basically just a probability calculation of independent stuff, so it's intuitive. + + Derivative is easy enough. + + \newpage + \listoftodos + +\end{document}