Replica method
Free energy of fluid in a matrix of configuration in the Canonical () ensemble is given by:
- Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle -\beta F_{1}(q^{N_{0}})=\log Z_{1}(q^{N_{0}})=\log \left({\frac {1}{N_{1}!}}\int \exp[-\beta (H_{01}(r^{N_{1}},q^{N_{0}})+H_{11}(r^{N_{1}},q^{N_{0}}))]~d\{r\}^{N_{1}}\right)}
where is the fluid partition function, and is the Hamiltonian of the matrix. Taking an average over matrix configurations gives
\cite{JPFMP_1975_05_0965,JPAMG_1976_09_01595} Important mathematical trick to get rid of the logarithm inside of the integral:
one arrives at
The Hamiltonian written in this form describes a completely equilibrated system of Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle s+1} components; the matrix and identical non-interacting copies (replicas) of the fluid. Thus the relation between the free energy of the non-equilibrium partially frozen and the replica (equilibrium) system is given by
- Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle -\beta {\overline {F}}_{1}=\lim _{s\rightarrow 0}{\frac {\rm {d}}{{\rm {d}}s}}[-\beta F^{\rm {rep}}(s)]}