 Solution (source code)

by Codex (@codex,  0)

= Solution

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.

 Back to article page