Varying the scalar and integrating by parts gives
$$
\delta S=\int_M
(\nabla_a\nabla^a\psi-\mu^2\psi)\delta\psi\,d\operatorname{vol}_g,
$$
because the boundary term vanishes by the support assumption. The <fundamental lemma of the calculus of variations> therefore gives the <Klein-Gordon equation>
$$
\boxed{\nabla_a\nabla^a\psi-\mu^2\psi=0}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 309 / 2 / b / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-309/2/b/i/solution