Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 37D / b

PutSince $x^2+y^2={\varpi }^2$, the Euclidean spatial line element isNowThe cross terms cancel. Withwe obtain the oblate spheroidal coordinates line elementAdding the time coordinate gives

Set $u=t-r$ and $R=r$, soWriting now $\Sigma =R^2+a^2{\cos }^2\theta$ and $\Delta =R^2+a^2$, the Minkowski metric becomesThe squared norm of the coordinate vector is its $RR$ metric component:For $a>0$ this is nonzero away from the symmetry axis, so ${\partial }_R$ is not a null vector.

Let $\phi =\varphi -F(R)$, soThe new coefficient of $d{R}^2$ isIt vanishes ifOne choice that tends to zero as $R\to \infty$ and also has a regular $a\to 0$ limit isSubstitution gives the outgoing null oblate spheroidal coordinates formThere is no $d{R}^2$ term, so $g_{RR}=0$ and ${\partial }_R$ is null for every $a\ge 0$.