From f2c13afb838a88577d4e189a5087ca65a9c02954 Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Fri, 8 Jul 2022 15:32:54 +0200
Subject: Some cleanup of the equilibrium section.

---
 frsb_kac-rice.tex | 21 ++++++++++++++-------
 1 file changed, 14 insertions(+), 7 deletions(-)

(limited to 'frsb_kac-rice.tex')

diff --git a/frsb_kac-rice.tex b/frsb_kac-rice.tex
index 880cca7..b1c25ab 100644
--- a/frsb_kac-rice.tex
+++ b/frsb_kac-rice.tex
@@ -115,22 +115,29 @@ Fyodorov \cite{Fyodorov_2007_Replica} for a detailed discussion).
 
 \section{Equilibrium}
 
-Here we review the equilibrium solution. \cite{Crisanti_1992_The, Crisanti_1993_The, Crisanti_2004_Spherical, Crisanti_2006_Spherical}
-The free energy averaged over disorder is 
+Here we review the equilibrium solution \cite{Crisanti_1992_The,
+Crisanti_1993_The, Crisanti_2004_Spherical, Crisanti_2006_Spherical}.  The free
+energy averaged over disorder is
 \begin{equation}
-\beta F = - \overline{\ln \int ds \; e^{-\beta H(s)}}
+  \beta F = - \overline{\ln \int d\mathbf s \;\delta(\|\mathbf s\|^2-N)\, e^{-\beta H(\mathbf s)}}
 \end{equation}
 Computing the logarithm as the limit of $n \rightarrow 0$ replicas is standard, and it take the form
 \begin{equation}
   \beta F=-1-\log2\pi-\frac12\lim_{n\to0}\frac1n\left(\beta^2\sum_{ab}^nf(Q_{ab})+\log\det Q\right)
 \end{equation}
-which must be extremized over the matrix $Q$. When the solution is a Parisi matrix, this can also be written in a functional form. If $P(q)$ is the probability distribution for elements $q$ in a row of the matrix, then
+which must be extremized over the $n\times n$ matrix $Q$, whose diagonal must
+be one. When the solution is a Parisi matrix, the free energy can also be
+written in a functional form. If $P(q)$ is the probability distribution for
+elements $q$ in a row of the matrix, then define $\chi(q)$ by
 \begin{equation}
-  \chi(q)=\int_1^qdq'\,\int_0^{q'}dq''\,P(q'')
+  \chi(q)=\int_q^1dq'\,\int_0^{q'}dq''\,P(q'')
 \end{equation}
-Since it is the double integral of a probability distribution, $\chi$ must be concave, monotonically decreasing and have $\chi(1)=0$, $\chi'(1)=1$. The free energy can be written as a functional over $\chi$ as
+Since it is the double integral of a probability distribution, $\chi$ must be
+concave, monotonically decreasing, and have $\chi(1)=0$ and $\chi'(1)=-1$.
+Using standar arguments, the free energy can be written as a functional over
+$\chi$ as
 \begin{equation}
-  \beta F=-1-\log2\pi-\frac12\int dq\,\left(\beta^2f''(q)\chi(q)+\frac1{\chi(q)}\right)
+  \beta F=-1-\log2\pi-\frac12\int_0^1dq\,\left(\beta^2f''(q)\chi(q)+\frac1{\chi(q)}\right)
 \end{equation}
 
 
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