Require
$$
(\partial_\mu+A'_\mu)(g\phi)=g(\partial_\mu+A_\mu)\phi
$$
for every $\phi$. Expanding and cancelling $g\partial_\mu\phi$ gives
$$
(\partial_\mu g)+A'_\mu g=gA_\mu.
$$
Right multiplication by $g^{-1}$ yields the <gauge-field transformation law>
$$
A'_\mu=gA_\mu g^{-1}-(\partial_\mu g)g^{-1}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 302 / 4 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-302/4/a/solution