feat: adds 3.15

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}