Editing Landau theory (section)

===First-order transitions===
Landau theory can also be used to study [[Phase_transition#Modern_classifications|first-order transitions]]. There are two different formulations, depending on whether or not the system is symmetric under a change in sign of the order parameter.

====I. Symmetric Case====
Here we consider the case where the system has a symmetry and the energy is invariant when the order parameter changes sign.
A first-order transition will arise if the quartic term in <math>F</math> is negative. To ensure that the free energy remains positive at large <math>\eta</math>, one must carry the free-energy expansion to sixth-order,<ref>{{cite book|title=The Landau Theory of Phase Transitions|chapter=Chapter 5: First-Order Transitions|last1=Tolédano|first1=J.C.|last2=Tolédano|first2=P.|publisher=World Scientific Publishing Company|date=1987|url=https://books.google.com/books?id=6gU8DQAAQBAJ&q=landau+first+order|isbn=9813103949}}</ref><ref>{{cite book|last1=Stoof|first1=H.T.C.|last2=Gubbels|first2=K.B.|last3=Dickerscheid|first3=D.B.M.|title=Ultracold Quantum Fields|publisher=Springer|year=2009|isbn=978-1-4020-8763-9}}</ref>

:<math>F(T,\eta) = A(T) \eta^2 - B_0 \eta^4 + C_0 \eta^6,</math>

where <math>A(T)=A_0(T-T_0)</math>, and <math>T_0</math> is some temperature at which <math>A(T)</math> changes sign. We denote this  temperature by <math>T_0</math> and not <math>T_c</math>, since it will emerge below that it is not the temperature of the first-order transition, and since there is no critical point, the notion of a "critical temperature" is misleading to begin with. <math>A_0, B_0,</math> and <math>C_0</math> are positive coefficients.

We analyze this free energy functional as follows: (i) For <math> T > T_0 </math>, the <math>\eta^2</math> and <math>\eta^6</math> terms are concave upward for all <math>\eta</math>, while the <math>\eta^4</math> term is concave downward. Thus for sufficiently high temperatures <math>F</math> is concave upward for all <math>\eta</math>, and the equilibrium solution is <math>\eta = 0</math>. (ii) For <math> T < T_0 </math>, both the <math>\eta^2</math> and <math>\eta^4</math> terms are negative, so <math>\eta = 0</math> is a local maximum, and the minimum of <math>F</math> is at some non-zero value <math>\pm\eta_0(T)</math>, with
<math> F(T_0,\eta_0(T_0)) < 0</math>. (iii) For <math> T </math> just above <math> T_0 </math>, <math>\eta = 0</math> turns into a local minimum, but the minimum at <math>\eta_0(T)</math> continues to be the global minimum since it has a lower free energy. It follows that as the temperature is raised above <math>T_0</math>, the global minimum cannot continuously evolve from
<math>\eta_0(T)</math> to 0. Rather, at some intermediate temperature <math>T_*</math>, the minima at <math>\eta_0(T_*)</math> and <math>\eta = 0</math> must become degenerate. For <math>T > T_*</math>, the global minimum will jump discontinuously from <math>\eta_0(T_*)</math> to 0.

To find <math>T_*</math>, we demand that free energy be zero at <math>\eta = \eta_0(T_*)</math> (just like the <math>\eta=0</math> solution), and furthermore that this point should be a local minimum. These two conditions yield two equations,
:<math>0=A(T) \eta^2 - B_0 \eta^4 + C_0 \eta^6,</math>
:<math>0=2A(T) \eta - 4 B_0 \eta^3 + 6 C_0 \eta^5,</math>

[[File:LandauFirstOrderTransition.svg|thumb|First-order phase transition demonstrated in the discontinuity of the order parameter as a function of temperature]]
which are satisfied when <math>\eta^2(T_*) = {B_0}/{2C_0}</math>. The same equations also imply that <math>A(T_*) = A_0(T_*-T_0) = B_0^2/4C_0</math>. That is,
:<math> T_* = T_0 + \frac{B_0^2}{4 A_0 C_0}.</math>

From this analysis both points made above can be seen explicitly. First, the order parameter suffers a discontinuous jump from <math>(B_0/2C_0)^{1/2}</math> to 0. Second, the transition temperature <math>T_*</math> is not the same as the temperature <math>T_0</math> where <math>A(T)</math> vanishes. 

At temperatures below the transition temperature, <math>T<T_*</math>, the order parameter is given by
:<math>\eta_0^2 = \frac{B_0}{3C_0} \left[ 1 + \sqrt{1 - \frac{3A(T) C_0}{B_0^2}} \right]</math>
which is plotted to the right. This shows the clear discontinuity associated with the order parameter as a function of the temperature. To further demonstrate that the transition is first-order, one can show that the free energy for this order parameter is continuous at the transition temperature <math>T_*</math>, but its first derivative (the entropy) suffers from a discontinuity, reflecting the existence of a non-zero latent heat.

====II. Nonsymmetric Case====
Next we consider the case where the system does not have a symmetry. In this case there is no reason to keep only even powers of <math>\eta</math> in the expansion of <math>F</math>, and a cubic term must be allowed (The linear term can always be eliminated by a shift <math> \eta \to \eta</math> + constant.) We thus consider a free energy functional

:<math>F(T,\eta) = A(T) \eta^2 - C_0 \eta^3 + B_0 \eta^4 + \cdots.</math>

Once again <math>A(T)=A_0(T-T_0)</math>, and <math>A_0, B_0, C_0</math> are all positive. The sign of the cubic term can always be chosen to be negative as we have done by reversing the sign of <math>\eta</math> if necessary.

We analyze this free energy functional as follows: (i) For <math> T < T_0 </math>, we have a local maximum at <math>\eta = 0</math>, and since the free energy is bounded below, there must be two local minima at nonzero values <math>\eta_-(T) < 0</math> and <math>\eta_+(T) > 0</math>. The cubic term ensures that <math>\eta_+</math> is the global minimum since it is deeper. (ii) For <math>T</math> just above <math>T_0</math>, the minimum at <math>\eta_-</math> disappears, the maximum at <math>\eta = 0</math> turns into a local minimum, but the minimum at <math>\eta_+</math> persists and continues to be the global minimum. As the temperature is further raised, <math> F(T,\eta_+(T)) </math> rises until it equals zero at some temperature <math>T_*</math>. At <math>T_*</math> we get a discontinuous jump in the global minimum from <math>\eta_+(T_*)</math> to 0. (The minima cannot coalesce for that would require the first three derivatives of <math>F</math> to vanish at <math>\eta = 0</math>.)

To find <math>T_*</math>, we demand that free energy be zero at <math>\eta = \eta_+(T_*)</math> (just like the <math>\eta=0</math> solution), and furthermore that this point should be a local minimum. These two conditions yield two equations,
:<math>0=A(T) \eta^2 - C_0 \eta^3 + B_0 \eta^4,</math>
:<math>0=2A(T) \eta - 3 C_0 \eta^2 + 4 B_0 \eta^3,</math>
which are satisfied when <math>\eta(T_*) = {C_0}/{2B_0}</math>. The same equations also imply that <math>A(T_*) = A_0(T_*-T_0) = C_0^2/4B_0</math>. That is,
:<math> T_* = T_0 + \frac{C_0^2}{4 A_0 B_0}.</math>
As in the symmetric case the order parameter suffers a discontinuous jump from <math>(C_0/2B_0)</math> to 0. Second, the transition temperature <math>T_*</math> is not the same as the temperature <math>T_0</math> where <math>A(T)</math> vanishes.