Why does the free energy depend on the configuration, in the Peierls argument?
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The origin of my doubt lays on the Peierls argument to show that there is no phase transition in the 1D Ising Model.
I understand that the Helmholtz Potential is the partial Lengendre transform of UU, and therefore can be written as
F=U−TSF=U−TS
Now, making use of the previous expression we can easily compute FF for an unidimensional Ising system with its all NN spins pointing in the same direction:
F1=−NJ−kBTln2F1=−NJ−kBTln2
where JJ is the interaction factor and I calculated the entropy as kBlnΩkBlnΩ being ΩΩ the number of possible microstates with U=−NJU=−NJ.
If we invert a strip of spins of undefined lenght the new Helmholtz Free Energy is the following:
F2=−NJ+4J−kBTln[N(N−1)]F2=−NJ+4J−kBTln[N(N−1)]
where 4J4J is the contribution due to the walls and again I calculated the entropy counting the number of possible states.
I understand that the Helmholtz Potential is the partial Lengendre transform of UU, and therefore can be written as
F=U−TSF=U−TS
Now, making use of the previous expression we can easily compute FF for an unidimensional Ising system with its all NN spins pointing in the same direction:
F1=−NJ−kBTln2F1=−NJ−kBTln2
where JJ is the interaction factor and I calculated the entropy as kBlnΩkBlnΩ being ΩΩ the number of possible microstates with U=−NJU=−NJ.
If we invert a strip of spins of undefined lenght the new Helmholtz Free Energy is the following:
F2=−NJ+4J−kBTln[N(N−1)]F2=−NJ+4J−kBTln[N(N−1)]
where 4J4J is the contribution due to the walls and again I calculated the entropy counting the number of possible states.
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