Phase Transitions Exist Only In the Thermodynamic Limit Or

maverick_starstrider
Messages
1,118
Reaction score
7
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

[tex]M(H)=\frac{\partial F(H)}{\partial H}=\frac{\partial F(-H)}{\partial(- H)}=-\frac{\partial F(-H)}{\partial H}=-M(H)[/tex]

Thefore, [tex]M(H)=-M(H)[/tex] therefore at [tex]M=0[/tex] we have

[tex]M(0)=-M(0)[/tex] therefore M(0)=0, thus no spontaneous magnetization. However, in the thermodynamic limit [tex]f(H)[/tex] (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 [tex]F(H)=N f(H)[/tex] 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?
 
Physics news on Phys.org
Let [itex]f_N(H) = F(H)/N[/itex]. Let

[tex]f_\infty(H) = \lim_{N\rightarrow \infty} f_N(H)[/tex].

[itex]f_\infty(H)[/itex] and [itex]f_N(H)[/itex] 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 [itex]F(H) = -kT\log(\lambda_1(H,T)^N+\lambda_2(H,T)^N)[/itex], where the lamba's are eigenvalues of a certain matrix, and [itex]\lambda_1 > \lambda_2[/itex] (except when T = 0). You can check that under these conditions it is only when [itex]N \rightarrow \infty[/itex] that F(H) is asymptotically proportional to N. Hence, in general, [itex]f_\infty(H)[/itex] has a discontinuity at H = 0, as you have said, but [itex]f_N(H)[/itex] does not.

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

Similar threads

  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
4
Views
2K
  • · Replies 41 ·
2
Replies
41
Views
7K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
11
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K