feat: Adds 3.19

tex/3.19.tex

@@ -62,9 +62,57 @@
 	(Hint: as $N$ grows, the probability density $P(N \epsilon)$ decreases exponentially, which the total number of states $2^N$ grows exponentially.
 	Which one wins?)
 
+	\subsubsection*{(b) Entropy}
+	What does an exponentially growing number of states mean?
+	Let the entropy per particle be $s(\epsilon) = \flatfrac{S(N\epsilon)}{N}$.
+	Then (setting $\kb = 1$) $\Omega(E) = \exp(S(E)) = \exp(N s(\epsilon))$ grows exponentially whenever the entropy per particle is positive.
+	
+	How do we calculate the entropy per particle $s(\epsilon)$ of a typical REM?
+	Can we just use the annealed average
+	\begin{equation}
+		s_{\mathrm{annealed}}(\epsilon) = \lim_{N \rightarrow \infty} \left(\frac{1}{N}\right) \log \left<\Omega(N \epsilon) \right>_{\mathrm{REM}}?
+	\end{equation}
+	
+	Show that $s_{\mathrm{annealed}} = \log2 - \epsilon^2$.
+
+	\subsubsection*{Not a part just an answer to that question}
+	Sethna says no, because that limit doesn't work in the glassy state below $\epsilon_\ast$.
+
 	\section{Solution} \label{sec:solution}
 	
+	\subsection*{(a)}
+	We want $\left< \Omega(N\epsilon) \right>_{REM}$.
 
+	Pretty simply, $\Omega(N \epsilon)$ is the total number of states with a given energy, and the average of that over possible configurations is just the number of possible states $M$ times the probability of any individual state having energy $N \epsilon$, $P(N \epsilon)$:
+	\begin{align}
+		\left< \Omega(N\epsilon) \right>_{REM} &= M P(N \epsilon) \\
+		&= 2^N \frac{1}{\sqrt{\pi N}} e^{\flatfrac{-(N \epsilon)^2}{N}} \\
+		&= \frac{1}{\sqrt{\pi N}} e^{N \log 2} e^{\flatfrac{-(N \epsilon)^2}{N}} \\
+		&= \frac{1}{\sqrt{\pi N}} e^{N \log 2 - \flatfrac{(N \epsilon)^2}{N}}
+	\end{align}
+	This grows exponentially if the argument of the exponential is positive, so
+	\begin{align}
+		N \log 2 &> \frac{(N \epsilon)^2}{N} \\
+		N^2 \log 2 &> N^2 \epsilon^2 \\
+		\sqrt{\log 2} &> \abs{\epsilon}
+	\end{align}
+	This is the condition we were asked to show.
+
+	\subsection*{(b)}
+	We can just jump straight in.
+	
+	\begin{align}
+		s_{\mathrm{annealed}}(\epsilon) &= \lim_{N \rightarrow \infty} \left(\frac{1}{N}\right) \log \left<\Omega(N \epsilon) \right>_{\mathrm{REM}} \\
+	&= \lim_{N \rightarrow \infty} \left(\frac{1}{N}\right) \log(\frac{1}{\sqrt{\pi N}} e^{N \log 2 - \flatfrac{(N \epsilon)^2}{N}}) \\
+	&= \lim_{N \rightarrow \infty} \left(\frac{1}{N}\right) \left( \log \frac{1}{\sqrt{\pi N}} + N \log 2 - \flatfrac{(N \epsilon)^2}{N} \right) \\
+	&= \lim_{N \rightarrow \infty} \frac{1}{N} \log \frac{1}{\sqrt{\pi N}} + \log 2 - \epsilon^2 \\
+	\end{align}
+
+	The first term clearly goes to $0$ for $N \rightarrow \infty$, because it ends up proportional to $\flatfrac{\log{N}}{N}$.
+	So that just leaves the last two terms which are $N$-independent, giving us
+	\begin{equation}
+		s_{\mathrm{annealed}} = \log2 - \epsilon^2.
+	\end{equation}
 	\newpage
 	\listoftodos