Replica method
Free energy of fluid in a matrix of configuration in the Canonical () ensemble is given by:
where is the fluid partition function, and is the Hamiltonian of the matrix. Taking an average over matrix configurations gives
- Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle -\beta {\overline {F}}_{1}={\frac {1}{N_{0}!Z_{0}}}\int \exp[-\beta _{0}H_{00}(q^{N_{0}})]~\log Z_{1}(q^{N_{0}})~d\{q\}^{N_{0}}}
\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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s} 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 (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle - \beta \overline{F}_1 = \lim_{s \rightarrow 0} \frac{{\rm d}}{{\rm d}s} [- \beta F^{\rm rep} (s) ] }