Assumptions of Landau theory

  • #1
Assume that we can expand the Helmholtz potential about [tex]T=T_c[/tex], [tex]M=0[/tex] in a standard Taylor series form of functions of the variables,
[tex]A(T,M)=\sum^{\infty}_{j=0}L_j(T)M^j=L_0(T)+L_2(T)M^2+L_4(T)M^4+...[/tex]

Why [tex]A(T,M)[/tex] must be even function of [tex]M[/tex]?

Coefficients can be expanded about [tex]T=T_c[/tex]

[tex]L_j(T)=\sum^{\infty}_{k=0}l_{jk}(T-T_c)^k=l_{j0}+l_{j1}(T-T_c)+...[/tex]

How I could no that coefficients are analytic functions od [tex]T[/tex].
 

Answers and Replies

  • #2
25
1
Consider a system of spins, where M is the coarse-grained magnetization of a certain “block.” In that block, if I flip every spin then there is no change in the energy in the system. Flipping of all the spins would result in M → -M. That’s why A(T,M) is even in M. I didn't quite understand your second question about analyticity
 
Last edited:
  • #3
Mute
Homework Helper
1,388
10
Assume that we can expand the Helmholtz potential about [tex]T=T_c[/tex], [tex]M=0[/tex] in a standard Taylor series form of functions of the variables,
[tex]A(T,M)=\sum^{\infty}_{j=0}L_j(T)M^j=L_0(T)+L_2(T)M^2+L_4(T)M^4+...[/tex]

Why [tex]A(T,M)[/tex] must be even function of [tex]M[/tex]?
In the absence of an external magnetic field, you can change the sign of every spin and the total energy stays the same (as you can see by looking at the Ising model Hamiltonian), as tejas777 said. Flipping the sign of every spin flips the sign of the magnetization, so if the energy doesn't change, then changing the sign of M cannot matter, so the free energy must only depend on even powers of M.


Coefficients can be expanded about [tex]T=T_c[/tex]

[tex]L_j(T)=\sum^{\infty}_{k=0}l_{jk}(T-T_c)^k=l_{j0}+l_{j1}(T-T_c)+...[/tex]

How I could know[\b] that coefficients are analytic functions of [tex]T[/tex].


As far as Landau theory is concerned, t's an assumption, the same with how you assume that the free energy is analytic in M and can be expanded in integer powers of M (otherwise terms like |M| could be present, which are not even powers but still obey the symmetry that the free energy doesn't change if you take M -> -M).

You'll find out eventually that it's actually a bad assumption.
 
  • #4
I think that this is very hard problem, and not precise theory.

Helmholtz potential is convex function of magnetisation.

[tex]A(T,M)=\sum^{\infty}_{j=0}L_j(T)M^j[/tex]

that must put some demands on coefficients in series, and if I say

[tex]L_j(T)=l_{j0}+l_{j1}(T-T_c)+...[/tex]

I have a problem with different behavior below [tex]T_c[/tex] and for [tex]T>T_c[/tex].

So is there coefficients [tex]l_j[/tex] positive or negative?
 
  • #5
member 11137
Assume that we can expand the Helmholtz potential about [tex]T=T_c[/tex], [tex]M=0[/tex] in a standard Taylor series form of functions of the variables,
[tex]A(T,M)=\sum^{\infty}_{j=0}L_j(T)M^j=L_0(T)+L_2(T)M^2+L_4(T)M^4+...[/tex]

Why [tex]A(T,M)[/tex] must be even function of [tex]M[/tex]?

Coefficients can be expanded about [tex]T=T_c[/tex]

[tex]L_j(T)=\sum^{\infty}_{k=0}l_{jk}(T-T_c)^k=l_{j0}+l_{j1}(T-T_c)+...[/tex]

How I could no that coefficients are analytic functions od [tex]T[/tex].
The Mac Laurin Taylor development that you have writen is the one depending on the M2 variable. This is the reason why you get only even coefficients. The M variable of the Landau theory is more often written as [itex]\Psi[/itex] and represent the bosonic field amplitude. The square of it, in extenso [itex]\Psi[/itex]2, is the bosonic density.
 

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