Put $L=(2\pi)^{-1}\log(M_{\rm GUT}/m_Z)$. At the <gauge coupling unification> scale, write the common normalized coupling as $\alpha_U=\alpha_2(M_{\rm GUT})=\alpha_3(M_{\rm GUT})=(5/3)\alpha_1(M_{\rm GUT})$. Running downward gives
$$
\alpha_3^{-1}(m_Z)=\alpha_U^{-1}-\beta_3L,
\quad
\alpha_2^{-1}(m_Z)=\alpha_U^{-1}-\beta_2L,
\quad
\frac35\alpha_1^{-1}(m_Z)=\alpha_U^{-1}-\frac35\beta_1L.
$$
Subtracting the second equation from the third determines
$$
L=\frac{\frac35\alpha_1^{-1}(m_Z)-\alpha_2^{-1}(m_Z)}{\beta_2-\frac35\beta_1}.
$$
Eliminating $L$ from the first two equations yields
$$
\boxed{
\alpha_3^{-1}(m_Z)=\alpha_2^{-1}(m_Z)
+\frac{\beta_3-\beta_2}{\frac35\beta_1-\beta_2}
\left[\frac35\alpha_1^{-1}(m_Z)-\alpha_2^{-1}(m_Z)\right].
}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 305 / 4 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-305/4/d/solution