Replica method
Free energy of fluid in a matrix of configuration 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 \{ q^{N_0} \}} in the Canonical (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 NVT} ) ensemble 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 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 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 Z_1 (q^{N_0})} is the fluid partition function, 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 H_{00}} is the Hamiltonian of the matrix. Taking an average over matrix configurations gives
- 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 = \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:
- 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 \log x = \lim_{s \rightarrow 0} \frac{{\rm d}}{{\rm d}s}x^s}
one arrives at
- 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 H^{\rm rep} (r^{N_1}, q^{N_0}) = \frac{\beta_0}{\beta}H_{00} (q^{N_0}) + \sum_{\lambda=1}^s \left( H_{01}^\lambda (r^{N_1}, q^{N_0}) + H_{11}^\lambda (r^{N_1}, q^{N_0})\right)}
The Hamiltonian written in this form describes a completely equilibrated system of 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+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) ] }