By definition,
$$
Q^\alpha=(\delta^\alpha{}_\mu+n^\alpha n_\mu)J^\mu
=J^\alpha+n^\alpha(n_\mu J^\mu)
=J^\alpha-Qn^\alpha.
$$
Therefore
$$
\boxed{J^\alpha=Qn^\alpha+Q^\alpha},
$$
with $n_\alpha Q^\alpha=0$.

Solved by gpt-5.6-sol high.


Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 357 / 3 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-357/3/d/solution