From 929067069100dbdeb78f3982fd3d1a5735183a3e Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Wed, 8 Jun 2022 20:11:56 +0200
Subject: Tweaked equilibrium section.

---
 frsb_kac_new.tex | 11 ++++++++---
 1 file changed, 8 insertions(+), 3 deletions(-)

diff --git a/frsb_kac_new.tex b/frsb_kac_new.tex
index 34ca3c9..514fda9 100644
--- a/frsb_kac_new.tex
+++ b/frsb_kac_new.tex
@@ -162,7 +162,7 @@ carefully treating the $k$th term in each sum separately from the rest, we get
     \right\}
   \end{aligned}
 \end{equation}
-$F$ is a $k-1$ RSB ansatz with all eigenvalues scaled by $\tilde\beta z^{-1}$ and shifted by
+This is a $k-1$ RSB ansatz with all eigenvalues scaled by $\tilde\beta z^{-1}$ and shifted by
 1, with effective temperature $\tilde\beta$, and an extra term. This can be seen
 more clearly by rewriting the result in terms of the matrix $\tilde Q$, a
 $(k-1)$-RSB Parisi matrix built from the $\tilde x_1,\ldots,\tilde x_{k-1}$ and
@@ -174,8 +174,13 @@ $q_1,\ldots,q_{k-1}$, which gives
   \right)
 \end{equation}
 {\em We have lost one level of RSB because at zero temperature the states become points.}
-
-{\bf Jaron: should'nt we put here the continuum solution?}
+In the continuum case, this is
+\begin{equation} \label{eq:ground.state.free.energy}
+  \lim_{\beta\to\infty}\tilde\beta F
+  =-\frac12z\tilde\beta f'(1)-\frac12\int dq\left(
+    \tilde\beta^2f''(q)\tilde\chi(q)+\frac1{\tilde\chi(q)+\tilde\beta z^{-1}}
+  \right)
+\end{equation}
 
 
 
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