Phase Transitions Exist Only In the Thermodynamic Limit Or

[maverick_starstrider]

It is often said that phase transitions only exist in the thermodynamic limit based on some proof like:

-A system has time-reversal symmetry thus its TOTAL free energy F(H)=F(-H) (H is the field) therefore the magnetization is

M(H)=\frac{\partial F(H)}{\partial H}=\frac{\partial F(-H)}{\partial(- H)}=-\frac{\partial F(-H)}{\partial H}=-M(H)

Thefore, M(H)=-M(H) therefore at M=0 we have

M(0)=-M(0) therefore M(0)=0, thus no spontaneous magnetization. However, in the thermodynamic limit f(H) (the free energy per site) may have a discontinuity at H=0 and therefore the above is not necessarily true, and so on thus true phase transitions only exist in the thermodynamics limit.

Here's where I get confused. Can we not just say that the total free energy F(H) for a finite system of N spins is really F(H)=N f(H) and the same argument applies. Therefore, isn't it really just the case the the thermodynamic limit EXIST and not necessarily that the system be at it?

 

[Mute]

Let f_N(H) = F(H)/N. Let

f_\infty(H) = \lim_{N\rightarrow \infty} f_N(H).

f_\infty(H) and f_N(H) are not the same function. The free energy F(H) is only proportional to N in the thermodynamic limit. For example, the free energy of a 1d Ising ring is F(H) = -kT\log(\lambda_1(H,T)^N+\lambda_2(H,T)^N), where the lamba's are eigenvalues of a certain matrix, and \lambda_1 > \lambda_2 (except when T = 0). You can check that under these conditions it is only when N \rightarrow \infty that F(H) is asymptotically proportional to N. Hence, in general, f_\infty(H) has a discontinuity at H = 0, as you have said, but f_N(H) does not.

The operations of taking N \rightarrow \infty and the derivation of the free energy do not commute, so you cannot establish the phase transition without the thermodynamic limit.

 

[genneth]

See: http://link.aps.org/doi/10.1103/PhysRev.87.404 : [1] C. N. Yang and T. D. Lee, Physical Review 87, 404-409 (1952).