Show that radial worldlines with u = const are outgoing null rays

[so_gr_lo]

Homework Statement

The transformation from Schwarzschild coordinates (t,r,θ,φ) to outgoing Eddington– Finkelstein coordinates (u, r′, θ′, φ′) is given by
u = t−r∗
r′ = r
θ′ = θ
φ′ = φ,
where r∗ satisfies dr∗/dr = (1−2M/r)−1. Use the rule for the coordinate transformation
of metric components to obtain the line element in the (u, r′, θ′, φ′) coordinates.

(b) Show that radial worldlines with u = const are outgoing null rays.

I have managed part a but am not sure how to show how the line element I have is null.

I have used the fact that u = t−r∗ suggests du=0 and $d{\Omega}^2$ = 0 to simplify the line element in (a) and give the result below. Does anyone know how I can show that the result is null?

Relevant Equations

u = t−r∗
r∗ satisfies dr∗/dr = (1−2M/r)−1

$$ ds^2 = -(1-\frac{2M}{r'})(\frac{\frac{r'}{2M} -1+2M}{\frac{r'}{2M}-1})(dr')^2+(1-\frac{2M}{r'})^-1 (dr')^2 $$

 

[ergospherical]

With $u = t-r_{\star}$ you have $$du = dt - \frac{dr}{(1-\tfrac{2M}{r})}$$You rearrange this for $dt$ and stick that into the line element in Schwarzshild coordinates. You end up with the outgoing Eddington Finkelstein line element,$$ds^2 = -(1-\tfrac{2M}{r}) du^2 - 2du dr + r^2 d\Omega^2$$Manifestly, constant $u$ ($du = 0$) and radial motion ($d\theta = d\phi = 0$) corresponds to $ds^2 = 0$.

Why outgoing? In terms of the Regge-Wheeler coordinate, the line element is $ds^2 = -(1-\tfrac{2M}{r})[-dt^2 + dr_{\star}^2] + r^2 d\Omega^2$. That means radial null geodesics correspond to $dt/dr_{\star} = \pm 1$, or equivalently $t \mp r_{\star} = \mathrm{const}$ along radial null geodesics. The minus-sign option is outgoing, i.e. $u := t-r_{\star} = \mathrm{const}$ along the outgoing radial null geodesics.