For the one-mode <smectic phase> ansatz $\phi(\mathbf r)=A\cos(q_0z)$, take $A\geq0$ without loss of generality. Spatial averaging gives
$$
\langle\phi^2\rangle=\frac{A^2}{2},
\qquad
\langle(\nabla\phi)^2\rangle=\frac{q_0^2A^2}{2},
\qquad
\langle(\nabla^2\phi)^2\rangle=\frac{q_0^4A^2}{2},
$$
and
$$
\langle|\phi|^3\rangle
=\frac{A^3}{2\pi}\int_0^{2\pi}|\cos u|^3\,du
=\frac{4A^3}{3\pi}.
$$
Using $G(q_0)=a-a_c$, the <free-energy density> is
$$
\frac FV
=\frac{a-a_c}{4}A^2+\frac{4g}{3\pi}A^3.
$$
For $a\geq a_c$, its minimum is $A=0$. For $a<a_c$, the nonzero stationary point is
$$
A_*=\frac{\pi(a_c-a)}{8g},
$$
and
$$
\frac{F(A_*)}{V}
=-\frac{\pi^2(a_c-a)^3}{768g^2}<0.
$$
Thus the modulation amplitude grows linearly and continuously from zero below $a_c$. In the language of an <order-parameter critical exponent>, this nonanalytic $|\phi|^3$ theory has the mean-field value $\beta=1$.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-344/1/b/solution