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New Monte Carlo method to compute the free energy of arbitrary solids. Application to the fcc and hcp phases of hard spheres
1.2K
Citations
13
References
1984
Year
Free EnergyAbsolute Free EnergySolid PhaseEngineeringMaterial SimulationComputational ChemistryComputational MechanicsMolecular DynamicsArbitrary SolidsMolecular ThermodynamicsNumerical SimulationThermophysicsThermodynamicsSolidificationCrystal FormationHard SpheresMaterials SciencePhysicsCrystal MaterialMonte CarloSolid MechanicsQuantum ChemistryCrystallographyCondensed Matter TheoryPhase EquilibriumNatural SciencesMonte Carlo MethodCondensed Matter PhysicsApplied PhysicsMaterial ModelingContinuum Modeling
The major source of error in solid‑phase free‑energy calculations is the extrapolation to the thermodynamic limit. The authors introduce a Monte Carlo method to compute the absolute free energy of arbitrary solid phases. The method constructs a reversible path from the target solid to an Einstein crystal of the same crystallographic structure, enabling free‑energy evaluation via Monte Carlo simulation. It reproduces the fcc hard‑sphere free energy at melting in agreement with Hoover and Ree, and finds the hcp–fcc free‑energy difference near melting to be essentially zero (−0.001 < ΔF < 0.002).
We present a new method to compute the absolute free energy of arbitrary solid phases by Monte Carlo simulation. The method is based on the construction of a reversible path from the solid phase under consideration to an Einstein crystal with the same crystallographic structure. As an application of the method we have recomputed the free energy of the fcc hard-sphere solid at melting. Our results agree well with the single occupancy cell results of Hoover and Ree. The major source of error is the nature of the extrapolation procedure to the thermodynamic limit. We have also computed the free energy difference between hcp and fcc hard-sphere solids at densities close to melting. We find that this free energy difference is not significantly different from zero: −0.001&lt;ΔF&lt;0.002.
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