For $V=r\partial_r$, <Cartan's magic formula> gives
$$
\mathcal L_Vdr=d(i_Vdr)+i_Vd(dr)=d(r)=dr,
$$
whereas contraction and exterior differentiation both vanish for $du,d\theta,d\phi$. Thus
$$
\mathcal L_Vdu=\mathcal L_Vd\theta=\mathcal L_Vd\phi=0.
$$
Applying the <Lie derivative> to
$g=-du^2-2,du,dr+r^2d\Omega^2$ yields
$$
\mathcal L_Vg=-2,du,dr+2r^2d\Omega^2.
$$
Here $n_a=-(du+dr)_a$ and $V_a=-r(du)_a$, so comparison with $2g$ gives
$$
\boxed{(\mathcal L_Vg)_{ab}
=2g_{ab}+\frac2r n_{(a}V_{b)}},
\qquad \boxed{\alpha=2}.
$$
Using tracelessness from part (b)(i),
$$
\nabla_a(T^{ab}V_b)
=\frac1rT^{ab}n_aV_b\geq0
$$
by the <dominant energy condition>, because $n$ is future timelike and $V$ is future null.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 309 / 2 / c / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-309/2/c/ii/solution