Let $K=3(4\pi)^2$ and $L=\log(M^2/\mu^2)$. Running from $M$ down to $\mu$ gives
$$
\frac1{g_i^2(\mu)}=\frac1{g_i^2(M)}-\frac{b_i}{K}L.
$$
The assumed unified boundary condition implies
$$
\frac1{g_s^2(M)}=\frac1{g_w^2(M)}=\frac3{5g_Y^2(M)}.
$$
Subtracting the weak equation from the strong and normalized-hypercharge equations therefore gives
$$
\frac1{g_s^2(\mu)}-\frac1{g_w^2(\mu)}
=-\frac{b_s-b_w}{K}L,
$$
$$
\frac3{5g_Y^2(\mu)}-\frac1{g_w^2(\mu)}
=-\frac{\frac35b_Y-b_w}{K}L.
$$
Eliminating $L$ proves
$$
\boxed{
\frac1{g_s^2(\mu)}-\frac1{g_w^2(\mu)}
=\frac{b_s-b_w}{\frac35b_Y-b_w}
\left(\frac3{5g_Y^2(\mu)}-\frac1{g_w^2(\mu)}\right)}.
$$


Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-305/2/iv/solution