Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 305 / 2 / iv

Table of contents
- Solution iv

Solution

Let

K = 3 (4 π)^{2}

and

L = lo g (M^{2} / μ^{2})

. Running from

M

down to

μ

gives

\frac{1}{g _{i}^{2} ( μ )} = \frac{1}{g _{i}^{2} ( M )} - \frac{b _{i}}{K} L .

The assumed unified boundary condition implies

\frac{1}{g _{s}^{2} ( M )} = \frac{1}{g _{w}^{2} ( M )} = \frac{3}{5 g _{Y}^{2} ( M )} .

Subtracting the weak equation from the strong and normalized-hypercharge equations therefore gives

\frac{1}{g _{s}^{2} ( μ )} - \frac{1}{g _{w}^{2} ( μ )} = - \frac{b _{s} - b _{w}}{K} L,

\frac{3}{5 g _{Y}^{2} ( μ )} - \frac{1}{g _{w}^{2} ( μ )} = - \frac{\frac{3}{5} b _{Y} - b _{w}}{K} L .

Eliminating

L

proves

\frac{1}{g _{s}^{2} ( μ )} - \frac{1}{g _{w}^{2} ( μ )} = \frac{b _{s} - b _{w}}{\frac{3}{5} b _{Y} - b _{w}} (\frac{3}{5 g _{Y}^{2} ( μ )} - \frac{1}{g _{w}^{2} ( μ )}) .

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