Started better discussion of the equilibrium solution.

1 files changed, 13 insertions, 3 deletions

diff --git a/frsb_kac_new.tex b/frsb_kac_new.tex
index f5e9422..d8f7f87 100644
--- a/frsb_kac_new.tex
+++ b/frsb_kac_new.tex
@@ -118,11 +118,21 @@ minima are exponentially subdominant with respect to saddles, because a saddle i
 \section{Equilibrium}
 
 Here we review the equilibrium solution. \cite{Crisanti_1992_The, Crisanti_1993_The, Crisanti_2004_Spherical, Crisanti_2006_Spherical}
-
+The free energy is well known to take the form
 \begin{equation}
-  \beta F=-\frac12\lim_{n\to0}\frac1n\left(\beta^2\sum_{ab}f(Q_{ab})+\log\det Q\right)-1-\log2\pi
+  \beta F=-\frac12\lim_{n\to0}\frac1n\left(\beta^2\sum_{ab}^nf(Q_{ab})+\log\det Q\right)-1-\log2\pi
+\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
+\begin{equation}
+  \chi(q)=\int_1^qdq'\,\int_0^{q'}dq''\,P(q'')
 \end{equation}
-We insert a  $k$ step RSB ansatz (the steps are standard, and are reviewed in Appendix A) , and obtain 
+Since it is the double integral of a probability distribution, $\chi$ must be convex, monotonically decreasing and have $\chi(1)=0$, $\chi'(1)=1$. The free energy can be written as a functional over $\chi$ as
+\begin{equation}
+  \beta F=-\frac12\int dq\,\left(\beta^2f''(q)\chi(q)+\frac1{\chi(q)}\right)-1-\log2\pi
+\end{equation}
+
+
+We are especially interested We insert a  $k$ step RSB ansatz (the steps are standard, and are reviewed in Appendix A) , and obtain 
 $q_0=0$
 \begin{align*}
   \beta F=