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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= |