The uniqueness of Kerr-Schild form

Question

The Kerr-Schild form is defined as $$g_{ab}=\eta_{ab}+Sk_ak_b,$$ where $k_a$ is the null vector measured by both $g_{ab}$ and $\eta_{ab}$. The Kerr-Schild form of the Schwarzschild metric is always written as $$ d s^{2}= -dt'^{2}-d r^2+r^{2}\left(d \theta^{2}+\sin ^{2} \theta d \phi^{2}\right) +\frac{2M}{r} (dt'^2-2dt'dr+dr^2).$$ However, it could be seen that the Schwarzschild metric in the Eddington-Finkelstein coordinate also exhibits the Kerr-Schild structure: $$d s^{2}= -du^{2}-2 d ud r+r^{2}\left(d \theta^{2}+\sin ^{2} \theta d \phi^{2}\right) +\frac{2M}{r} du^{2}.$$ The $u$ and $t'$ are linked by $u=t'+r$.

Moreover, as a coordinate will not mix the $\eta_{ab}$ and $k_ak_b$ term, it is possible to write the metric in Kerr-Schild form but with a different coordinate.

My question is, is there any particular reason most people use the first coordinate? Or the Kerr-Schild form in different coordinates are totally equivalent?

Answer 1

They are equivalent. In the first form the ingoing and outgoing light paths are horizontal and vertical lines, and if you transform to the second form the ingoing or outgoing ones will have ±45°.

With Schwarzschild you could just give your local set of observers the squared escape velocity like you give them the regular one in raindrop coordinates in order to get the {ť, r} metric.

But in general it's easier to first eliminate the g_rr term to get the horizontal and vertical worldlines in {u, r} and then take ť=u±r to get the ±45° photon worldlines on the {ť, r} spacetime diagram.