feat: finishes 3.8

tex/3.8.tex

@@ -13,7 +13,6 @@
 \usepackage{physics}
 \usepackage{siunitx}
 \usepackage{todonotes}
-\usepackage[plain]{fancyref}
 \usepackage{luacode}
 \usepackage{titling}
 \usepackage{enumerate}
@@ -22,6 +21,8 @@
 \usepackage{luatrivially}
 \usepackage{subtitling}
 
+\usepackage{cleveref}
+
 \begin{luacode*}
 	math.randomseed(31415926)
 \end{luacode*}
@@ -43,19 +44,26 @@
 	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)}
+		\sigma_{E_1}^2 = - \flatfrac{\kb}{\left(\pdv[2]{S_1}{E_1} + \pdv[2]{S_2}{E_2}\right)} \label{eq:1}
 	\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},
+			\frac{1}{\kb} \pdv[2]{S}{E} = - \frac{1}{\kb T} \frac{1}{N c_v T}, \label{eq:2}
 		\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}
+		\item If $c_v^{(1)}$ and $c_v^{(2)}$ are the specific heats per particle for two subsystems of $N$ particles each, show using \cref{eq:1,eq:2} that
+		\begin{equation}
+			\frac{1}{c_v^{(1)}} + \frac{1}{c_v^{(2)}} = \frac{N \kb T^2}{\sigma_{E_1}^2} \label{eq:3}
+		\end{equation}
+		\item Using the equipartition theorem, write the temperature in terms of the mean kinetic energy $K = \left<E_1\right>$ (with standard deviation $\sigma_K$).
+		Show that $c_v^{(1)} = \flatfrac{3\kb}{2}$ for the momentum degrees of freedom.
+		In terms of $K$ and $\sigma_K$, solve for the total specific heat of the molecular dynamics simulation (configurational plus kinetic).
 	\end{enumerate}
 
 	\section{Solution} \label{sec:solution}
@@ -73,6 +81,67 @@
 		\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.
+	\subsection*{(b)}
+	We want to show
+	\begin{equation}
+		\frac{1}{c_v^{(1)}} + \frac{1}{c_v^{(2)}} = \frac{N \kb T^2}{\sigma_{E_1}^2}
+	\end{equation}
+	Start with the $\sigma_{E_1}^2$ term from \cref{eq:1}, and use our result from the first part \cref{eq:2}:
+	\begin{align}
+		\sigma_{E_1}^2 &= - \flatfrac{\kb}{\left(\pdv[2]{S_1}{E_1} + \pdv[2]{S_2}{E_2}\right)} \\
+		\sigma_{E_1}^2 &= - \flatfrac{1}{\left(\frac{1}{\kb} \pdv[2]{S_1}{E_1} + \frac{1}{\kb}\pdv[2]{S_2}{E_2}\right)} \\
+		- \frac{1}{\sigma_{E_1}^2} &= \frac{1}{\kb} \pdv[2]{S_1}{E_1} + \frac{1}{\kb}\pdv[2]{S_2}{E_2} \\
+		- \frac{1}{\sigma_{E_1}^2} &= \left(- \frac{1}{\kb T} \frac{1}{N c_v^{(1)} T}\right) + \left(- \frac{1}{\kb T} \frac{1}{N c_v^{(2)} T} \right)\\
+		\frac{1}{\sigma_{E_1}^2} &= \frac{1}{\kb T} \frac{1}{N c_v^{(1)} T} + \frac{1}{\kb T} \frac{1}{N c_v^{(2)} T} \\
+		\frac{N \kb T^2}{\sigma_{E_1}^2} &= \frac{1}{c_v^{(1)}} + \frac{1}{c_v^{(2)}},
+	\end{align}
+	giving us our desired result.
+	\subsection*{(c)}
+	Sethna keeps these assumptions implicit, but this problem is for a monatomic ideal gas in a 3D space.
+	I think that assuming 3D space is something you should explicitly state, but that probably says more about me than Sethna.
+	In that case, there are three degrees of freedom for the kinetic energy.
+	The problem implicitly sets up a division of the degrees of freedom in the problem as the kinetic and configurational, where configuration refers to functions of position (including the background potential and the interactions).
+	If there are rotational degrees of freedom, you could group the rotational kinetic energy in with the ``configurational'' ones and recover these results anyway, but then it's not quite configurational so the terminology doesn't make a ton of sense.
+	So our system $1$ will include the three linear momenta, and system $2$ includes everything else.
+	What Sethna calls $K$ is then the kinetic energy from those three linear momenta, ignoring any rotational kinetic energy etc.
+
+	The equipartition theorem tells us that each degree of freedom in the kinetic energy should satisfy $\frac12 N \kb T$.
+	So, using the three degrees of freedom as described above, we can just write that $K = \left< E_1 \right> = \frac32 N \kb T$.
+	Our specific heat per particle satisfies $\N c_v^{(1)}) \left.\left( \pdv{E_1}{T} \right)\right|_{V, N}$, which is clearly $c_v^{(1)} = \frac32 \kb$, as Sethna wants us to find.
+	
+	Now lets use \cref{eq:3} and solve for $c_v^{(2)}$.
+	\begin{align}
+		\frac{1}{c_v^{(1)}} + \frac{1}{c_v^{(2)}} &= \frac{N \kb T^2}{\sigma_{E_1}^2} \\
+		\frac{1}{\frac32 \kb} + \frac{1}{c_v^{(2)}} &= \frac{N \kb T^2}{\sigma_K^2} \\
+		\frac{1}{c_v^{(2)}} &= \frac{N \kb T^2}{\sigma_K^2} - \frac{1}{\frac32 \kb} \\
+		\frac{1}{c_v^{(2)}} &= \frac{\frac32 N \kb^2 T^2}{\frac32 \kb \sigma_K^2} - \frac{\sigma_K^2}{\frac32 \kb \sigma_K^2} \\
+		\frac{1}{c_v^{(2)}} &= \frac{\frac32 N \kb^2 T^2 - \sigma_K^2}{\frac32 \kb \sigma_K^2}.
+	\end{align}
+	Solve for $K$ rather than $T$:
+	\begin{align}
+		K &= \frac32 N \kb T \\
+		T &= \frac23 \frac{1}{N \kb} K.
+	\end{align}
+	Plug this back in:
+	\begin{align}
+		\frac{1}{c_v^{(2)}} &= \frac{\frac32 N \kb^2 \left(\frac23 \frac{1}{N \kb} K \right)^2 - \sigma_K^2}{\frac32 \kb \sigma_K^2} \\
+		\frac{\frac32 \kb \sigma_K^2}{c_v^{(2)}} &= \frac32 N \kb^2 \left(\frac23 \frac{1}{N \kb} K \right)^2 - \sigma_K^2 \\
+		\frac{\frac32 \kb \sigma_K^2}{c_v^{(2)}} &= \frac23 N \kb^2 \frac{1}{N^2 \kb^2} K^2 - \sigma_K^2 \\
+		\frac{\frac32 \kb \sigma_K^2}{c_v^{(2)}} &= \frac23 \frac{1}{N} K^2 - \sigma_K^2 \\ 
+		\frac{c_v^{(2)}}{\frac32 \kb \sigma_K^2} &= \flatfrac{1}{\left(\frac23 \frac{1}{N} K^2 - \sigma_K^2\right)} \\ 
+		c_v^{(2)} &= \flatfrac{\frac32 \kb \sigma_K^2}{\left(\frac23 \frac{1}{N} K^2 - \sigma_K^2\right)} \\ 
+		c_v^{(2)} &= \frac{9 N \kb \sigma_K^2}{4 K^2 - 6 N \sigma_K^2}
+	\end{align}
+	That's not a pretty equation.
+	Our total specific heat is $c_v^{(1)} + c_v^{(2)}$.
+	\begin{align}
+		c_v^{(1)} + c_v^{(2)} &= \frac32 \kb + \frac{9 N \kb \sigma_K^2}{4 K^2 - 6 N \sigma_K^2} \\
+		c_v^{(1)} + c_v^{(2)} &= \frac{3 \kb}{2}\frac{2 K^2 - 3 N \sigma_K^2}{2 K^2 - 3 N \sigma_K^2} + \frac{9 N \kb \sigma_K^2}{4 K^2 - 6 N \sigma_K^2}  \\
+		c_v^{(1)} + c_v^{(2)} &= \frac{6 \kb K^2 - 9 \kb N \sigma_K^2}{4 K^2 - 6 N \sigma_K^2} + \frac{9 N \kb \sigma_K^2}{4 K^2 - 6 N \sigma_K^2}  \\
+		c_v^{(1)} + c_v^{(2)} &= \frac{6 \kb K^2 - 9 \kb N \sigma_K^2 + 9 N \kb \sigma_K^2}{4 K^2 - 6 N \sigma_K^2}  \\
+		c_v^{(1)} + c_v^{(2)} &= \frac{6 \kb K^2}{4 K^2 - 6 N \sigma_K^2}  \\
+		c_v^{(1)} + c_v^{(2)} &= \frac{3 \kb K^2}{2 K^2 - 3 N \sigma_K^2} 
+	\end{align}
 
 	\newpage
 	\listoftodos