Editing Landau theory (section)

====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.