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| -rw-r--r-- | figs/316_complexity_detail.pdf | bin | 0 -> 18185 bytes | |||
| -rw-r--r-- | frsb_kac-rice.tex | 21 |
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diff --git a/figs/316_complexity_detail.pdf b/figs/316_complexity_detail.pdf Binary files differnew file mode 100644 index 0000000..5cf3ef2 --- /dev/null +++ b/figs/316_complexity_detail.pdf 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} |