Why are the coefficients of the Landau theory Taylor expansion even?

In summary: This is the reason why the coefficients of the Taylor development are even in this case.In summary, the conversation discusses the expansion of the Helmholtz potential as a Taylor series about T=T_c and M=0. It is pointed out that A(T,M) must be an even function of M due to the fact that in a system of spins, flipping every spin results in no change in energy. The conversation also mentions the expansion of coefficients about T=T_c and the assumption that they are analytic functions of T. However, it is noted that this may not always be a valid assumption. Finally, the discussion touches on the behavior of coefficients for different temperatures and the role of the bosonic field amplitude in the Landau theory.
  • #1
matematikuvol
192
0
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].
 
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  • #2
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
 
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  • #3
matematikuvol said:
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
matematikuvol said:
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 written 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.
 

Related to Why are the coefficients of the Landau theory Taylor expansion even?

1. What is the basic premise of Landau theory?

Landau theory is a theoretical framework used to describe the behavior of a system undergoing a phase transition, such as the transition from liquid to solid. It assumes that the free energy of the system can be described by a polynomial expansion in the order parameter, which is a quantity that characterizes the transition.

2. How do assumptions about symmetry play a role in Landau theory?

Landau theory assumes that the symmetry of the system is broken at the transition point, meaning that the system has a lower symmetry in one phase compared to another. This broken symmetry is captured by the order parameter, which reflects the differences between the two phases.

3. Can Landau theory be applied to all types of phase transitions?

No, Landau theory is only applicable to second-order phase transitions, which are characterized by continuous changes in the order parameter as a function of temperature or other external parameters. It does not apply to first-order transitions, which involve a discontinuous change in the order parameter.

4. How do external factors, such as pressure or magnetic field, affect Landau theory?

Landau theory can be extended to include the effects of external factors, such as pressure or magnetic field, on the phase transition. This is done by adding additional terms to the polynomial expansion of the free energy, which account for the influence of these factors on the behavior of the system.

5. Are there any limitations to Landau theory?

While Landau theory is a powerful tool for understanding phase transitions, it has some limitations. It assumes that the system is in thermal equilibrium at all times, and it does not take into account fluctuations or critical phenomena near the transition point. In some cases, more advanced theories, such as renormalization group theory, may be required to fully describe a phase transition.

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