Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 344 / 1 / b

Solution

ϕ (r) = A cos (q_{0} z)

, take

A \geq 0

without loss of generality. Spatial averaging gives

⟨ ϕ^{2} ⟩ = \frac{A ^{2}}{2}, ⟨(\nabla ϕ)^{2} ⟩ = \frac{q _{0}^{2} A ^{2}}{2}, ⟨(\nabla^{2} ϕ)^{2} ⟩ = \frac{q _{0}^{4} A ^{2}}{2},

and

⟨ ∣ ϕ ∣^{3} ⟩ = \frac{A ^{3}}{2 π} \int_{0}^{2 π} ∣ cos u ∣^{3} d u = \frac{4 A ^{3}}{3 π} .

Using

G (q_{0}) = a - a_{c}

\frac{F}{V} = \frac{a - a _{c}}{4} A^{2} + \frac{4 g}{3 π} A^{3} .

For

a \geq a_{c}

, its minimum is

A = 0

. For

a < a_{c}

, the nonzero stationary point is

A_{*} = \frac{π ( a _{c} - a )}{8 g},

and

\frac{F ( A _{*} )}{V} = - \frac{π ^{2} ( a _{c} - a ) ^{3}}{768 g ^{2}} < 0.

Thus the modulation amplitude grows linearly and continuously from zero below

a_{c}

. In the language of an order-parameter critical exponent, this nonanalytic

∣ ϕ ∣^{3}

theory has the mean-field value

β = 1

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