Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 304 / 1 / c

Table of contents
- Solution c

Solution

Regard

λ

as a running coupling

λ (M)

. At fixed

s, t, u

and to order

λ^{2}

, differentiating the one-loop answer with respect to

lo g M

gives

0 = \frac{d Γ _{4}}{d l o g M} = \frac{d λ}{d l o g M} - \frac{λ ^{2}}{32 π ^{2}} \int_{0}^{1} 6 d x + O (λ^{3}),

because

m ≪ M

implies

F (M^{2}) ≃ - x (1 - x) M^{2}

. Thus the beta function is

β (λ) = \frac{d λ}{d l o g M} = \frac{3 λ ^{2}}{16 π ^{2}} + O (λ^{3}) .

Solving this ordinary differential equation with

λ (M_{0}) = λ_{0}

gives

\frac{1}{λ ( M )} = \frac{1}{λ _{0}} - \frac{3}{16 π ^{2}} lo g \frac{M}{M _{0}}, λ (M) = \frac{λ _{0}}{1 - \frac{3 λ _{0}}{16 π ^{2}} l o g ( M / M _{0} )}

to leading-logarithmic order.

Solved by gpt-5.6-sol high.

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