${\cal E}_{\rm out}$ between the light ray and the hole boundary. *In the particular case of a spatially Euclidean FL solution* ($K=0$), the Cartesian coordinates $X^ i{_{\text{out}}}$ of ${\cal E}_{\rm out}$ satisfy the simple system of equations $$\left\{
\begin{aligned}
& \delta_{ij}\left(X^i_{\rm out} - X_{\rm h}^i\right)\left(X^j_{\rm out} - X_{\rm h}^j\right)=\rho_{\rm h}^2\\
& X^i_{\rm out} = X^i_0 +(\eta_0-\eta_{\rm out})\,d^i
\end{aligned}
\right.
,$$ where $X_{\rm h}^i$ and $X^i_0$ are the respective Cartesian coordinates of the hole and the observer, while $d^i$ is the spatial direction of observation. Although conceptually similar, the determination of $\mathcal{E}{_{\text{out}}}$ for a FL solution with arbitrary spatial curvature is technically harder.

In general, we deduce from Eq.  that the wave vector at $\mathcal{E}{_{\text{out}}}$ is $k^\mu{_{\text{out}}} = (a_0/a{_{\text{out}}})^ 2 k^\mu_0$, where $a{_{\text{out}}}\equiv a(\eta{_{\text{out}}})$.

### Wronski matrix

In the FL region, the Sachs basis $(s_1,s_2)$ is defined with respect to the fundamental observers, comoving with four-velocity $u=\partial_T$. The explicit form of this basis does not need to be specified here.

The Sachs equation can be solved analytically by means of a conformal transformation to the static metric $$\label{e.conf}
{{\rm d}}\tilde s^2 = a_0^2\left[-{{\rm d}}\eta^2 + f_K^2(\chi){{\rm d}}\Omega^2\right]\equiv\tilde g_{\mu\nu}{{\rm d}}x^\mu{{\rm d}}x^\nu.$$ Because the geometries associated with $g_{\mu\nu}$ and $\tilde g_{\mu\nu}$ are conformal, 