feat: Adds 3.11

tex/3.11.tex

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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{luacode}
+\usepackage{titling}
+\usepackage{enumitem}
+\usepackage{mathtools}
+
+% custom deepak packages
+\usepackage{luatrivially}
+\usepackage{subtitling}
+
+\usepackage{cleveref}
+
+\begin{luacode*}
+	math.randomseed(31415926)
+\end{luacode*}
+
+\newcommand{\kb}{k_{\mathrm{B}}}
+\newcommand{\defeq}{\vcentcolon=}
+%\newcommand{\defeq}{\equiv}
+
+\title{Problem 3.11}
+\subtitle{Maxwell Relations}
+\author{\begin{telugu}హృదయ్ దీపక్ మల్లుభొట్ల\end{telugu}}
+% want empty date
+\predate{}
+\date{}
+\postdate{}
+
+% !TeX spellcheck = en_GB
+\begin{document}
+	\maketitle
+	
+	We want to derive our Maxwell relations based on our thermodynamic potentials.
+	For convenience work in units where $\kb = 1$, (if you want to put it back, just put it in front of the temperatures).
+	Start these with our \emph{definitions}:
+	\begin{align}
+		\frac{1}{T} &\defeq \left. \pdv{S}{E} \right|_{V, N} \label{eq:defT} \\
+		\frac{P}{T} &\defeq \left. \pdv{S}{V} \right|_{E, N} \label{eq:defP} \\
+		\frac{\mu}{T} &\defeq \left. \pdv{S}{N} \right|_{E, V} \label{eq:defMu}
+	\end{align}
+	
+	Show the following:
+	\begin{enumerate}[label=(\alph*)]
+		\item \begin{equation}
+			\left. \pdv{T}{V} \right|_{S, N} = - \left. \pdv{P}{S} \right|_{V, N}
+		\end{equation}
+		\item \begin{equation}
+			\left. \pdv{C_P}{P} \right|_{T} = - T \left. \pdv[2]{V}{T} \right|_{P}
+		\end{equation}
+		\item \begin{equation}
+			\left. \pdv{E}{T} \right|_{P} = - \frac{T}{C_P} \left. \pdv{S}{P} \right|_{T}
+		\end{equation}
+	\end{enumerate}
+	
+	\section{Solution} \label{sec:solution}
+	
+	These can either be manipulated by thinking in terms of manipulating expressions like $\dd{E} = T \dd{S} - P \dd{V} + \mu \dd{N}$ casually, using the `physicist's identities' strategy, or by doing it more formally.
+	Being too informal runs the risk of accidentally working off the wrong intuition, like the minus signs if we write a chain rule too casually:
+	\begin{equation}
+		\pdv{P}{V} \neq \pdv{P}{T} \pdv{T}{V},
+	\end{equation}
+	but instead 
+	\begin{equation}
+		\pdv{P}{V} = - \pdv{P}{T} \pdv{T}{V}!
+	\end{equation}
+	Why?
+	Because these aren't actually compatible chain rule derivatives, we're ignoring what we're holding constant.
+	Which is why it's useful to write the labels for what's not changing for these guys.
+	It's actually
+	\begin{equation}
+		\left. \pdv{P}{V} \right|_{T} = - \left. \pdv{P}{T} \right|_{V} \left. \pdv{T}{V} \right|_{P}.
+	\end{equation}
+
+	
+	Our two main operational identities are the cyclic triple product relation:
+	\begin{equation}
+		\left. \pdv{x}{y} \right|_{z} \left. \pdv{y}{z} \right|_{x} \left. \pdv{z}{x} \right|_{y} = - 1 \label{eq:trip}
+	\end{equation}
+	and the inverse derivative relation:
+	\begin{equation}
+		\left. \pdv{x}{y} \right|_{z} = \flatfrac{1}{\left( \left. \pdv{y}{x} \right|_{z} \right)}. \label{eq:inv}
+	\end{equation}
+	Those basically just let us write things down.
+
+	\subsection{(a)}
+
+	Want to show
+	\item \begin{equation}
+		\left. \pdv{T}{V} \right|_{S, N} = - \left. \pdv{P}{S} \right|_{V, N}
+	\end{equation}
+	
+	Start with the expression
+	\begin{equation}
+		\pdv[2]{E}{V_{S, N}}{S_{V, N}},
+	\end{equation}
+	where the subscript in the denominator tells us what we're keeping constant.
+	If $E$ is sufficiently well-behaved (fun question, what're the necessary and sufficient conditions here?), then these derivatives commute:
+	\begin{align}
+		\pdv[2]{E}{V_{S, N}}{S_{V, N}} &= \pdv[2]{E}{S_{V, N}}{V_{S, N}}
+	\end{align}
+	The lhs starts with the reciprocal (using \cref{eq:inv}) and our temperature definition \cref{eq:defT}, and the $V$ derivative in the rhs comes out of the triple product rule \cref{eq:trip,eq:defP}.
+
+	\begin{align}
+		\pdv{T}{V_{S, N}} &= - \pdv{P}{S_{V, N}}
+	\end{align}
+
+	\subection{(b)}
+	
+	We want:
+	\begin{equation}
+		\left. \pdv{C_P}{P} \right|_{T} = - T \left. \pdv[2]{V}{T} \right|_{P}
+	\end{equation}
+
+	\newpage
+	\listoftodos
+
+\end{document}