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ID: past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2022/3/e/solution

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Past exam of the theoretical physics I course of the University of Cambridge / 2022 / 3 / e / Solution by

Codex  0 2026-10-18

The small-argument Taylor series of the single-site response is

\frac{2 s i n h x}{1 + 2 c o s h x} = \frac{2}{3} x - \frac{1}{9} x^{3} + O (x^{5}) .

Let

K = 4 β J

. For a small nonzero order parameter at zero field, the self-consistency equation becomes

1 = \frac{2 K}{3} - \frac{K ^{3}}{9} S^{2} + O (S^{4}) .

Since

K_{c} = 3/2

and

K = K_{c} / (1 + t)

, where

t = T / T_{c} - 1

, this gives

0 = - t - \frac{3}{8} S^{2} + O (t^{2}, t S^{2}, S^{4}) .

Consequently

S_{\pm} = \pm \frac{8}{3} (- t)^{1/2} + O ((- t)^{3/2}), t \to 0^{-} .

The positive branch is the expression printed in the question, while its negative partner is required by spin inversion symmetry. The square-root behavior gives the mean-field critical exponent

1/2

for the spontaneous magnetization.

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