Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 305 / 4 / a / i

Table of contents
- Solution i

Solution

Separating variables gives

\frac{d}{d l o g μ} \frac{1}{α _{3}} = \frac{b}{2 π} .

Therefore

\frac{1}{α _{3} ( μ )} = \frac{1}{α _{3} ( μ _{0} )} + \frac{b}{2 π} lo g \frac{μ}{μ _{0}} .

For

b > 0

, define the strong-coupling scale

Λ_{QCD} = μ exp [- \frac{2 π}{b α _{3} ( μ )}] .

Then

α_{3} (μ) = \frac{2 π}{b l o g ( μ / Λ _{QCD} )} .

It decreases logarithmically toward zero in the ultraviolet and grows toward the infrared, becoming nonperturbative when

μ

approaches

Λ_{QCD}

.

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