 Solution

ID: past-exam-of-the-thermal-and-statistical-physics-course-of-the-university-of-cambridge/2013/paper-1/4/iii/solution

For

a > 0

and

H > 0

, the field tilts the single well toward positive M. Since

F^{''} = a + 3 b M^{2} > 0

everywhere, there is exactly one minimum, at the positive solution

M_{H}

of the equation of state. At that solution,

⟨ M ⟩ = M_{H} > 0, F_{m i n} = - \frac{a}{2} M_{H}^{2} - \frac{3 b}{4} M_{H}^{4} .

This follows by eliminating

μ_{0} H = a M_{H} + b M_{H}^{3}

from F. At weak field,

M_{H} ≃ μ_{0} H / a

and

F_{m i n} ≃ - μ_{0}^{2} H^{2} / (2 a)

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