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conf(T) in lattice models. For this reason, we will not use the subscript "conf" to denote the configurational entropy in this work.
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33645056786
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It should be stressed at this point that the existence of a negative entropy in our exact calculation is caused by the use of the continuation and/ or of solutions that are not the equilibrium one, i.e., solutions that do not have the lowest possible free energy F̃. Recall that the exact calculation of the free energy through the use of the partition function only requires that the free energy has the correct convexity property, which in this case corresponds to having a non-negative specific heat. This remains true for all states in our calculation, equilibrium or not. The requirement that the entropy be non-negative is an additional postulate, justified by the obvious fact that the number of realizable configurations be at least one. The continuation need not satisfy the entropy requirement, the latter being completely independent of the convexity property of the free energy. There will never be a negative-entropy equilibrium solution in any exact calculation, as the equilibrium state must correspond to a number of realizable configurations. This also means that if there is no phase transition in a model, there will be no possibility of any continuation that can then violate the entropy requirement. In this case, there cannot exist any Kauzmann temperature.
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63
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33645071413
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private communication
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F. Semerianov (private communication).
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Semerianov, F.1
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