Varying $\Phi$ and integrating by parts gives
$$
\boxed{\nabla_a\nabla^a\Phi
-\alpha e^{\alpha\Phi}F_{ab}F^{ab}=0}.
$$
Since $\delta F_{ab}=2\nabla_{[a}\delta A_{b]}$, variation of the potential and another integration by parts give
$$
\boxed{\nabla_a(e^{\alpha\Phi}F^{ab})=0}.
$$
These are the scalar and Maxwell equations of <Einstein-Maxwell-dilaton theory>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 309 / 4 / b / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-309/4/b/i/solution