any null geodesic for $g_{\mu\nu}$ affinely parametrized by $v$ is also a null geodesic for $\tilde g_{\mu\nu}$ affinely parametrized by $\tilde{v}$, with $a^2  {{\rm d}}\tilde{v} = a_0^2 {{\rm d}}v$. As ${{\rm d}}v = (a^2/a_0){{\rm d}}\eta$, it follows that $\tilde{v}=a_0\eta$.

For the static geometry, the optical tidal matrix reads $\widetilde{\boldsymbol{\mathcal R}}=-(K/a_0^2) \boldsymbol{1}$, so that the Sachs equation is simply $$\label{gdefl}
\frac{{{\rm d}}^2\tilde{\boldsymbol{\mathcal \xi}}}{{{\rm d}}\eta^2}= - K \,\tilde{\boldsymbol{\mathcal \xi}}.$$ We then easily obtain the Jacobi and scale matrices: $$\widetilde{\boldsymbol{\mathcal D}} = a_0 f_K(\eta-\eta{_{\text{init}}})\boldsymbol{1},\qquad
 \widetilde{\boldsymbol{\mathcal C}} = f'_K(\eta-\eta{_{\text{init}}})\boldsymbol{1}.$$

To go back to the original FL spacetime, we use that ${{\rm d}}v= a^2{{\rm d}}\eta$ and the fact that the screen projections of the separation vectors for both geometries are related by $a \tilde{\boldsymbol{\xi}}=a_0 \boldsymbol{\xi}$. The final result is $$\begin{aligned}
{\boldsymbol{\mathcal D}}{_{\text{FL}}} &= a{_{\text{init }}}\frac{a}{a_0} f_K(\eta-\eta{_{\text{init}}}) \boldsymbol{1},\\
{\boldsymbol{\mathcal C}}{_{\text{FL}}} &= \frac{a}{a{_{\text{init}}}} \Big[ f'_K(\eta-\eta{_{\text{init}}}) - \mathcal{H}{_{\text{init}}} f_K(\eta-\eta{_{\text{init}}}) \Big] \boldsymbol{1},\end{aligned}$$ where $\mathcal{H} \equiv a'(\eta)/a(\eta)$ is the conformal Hubble function. This completely determines ${\boldsymbol{\mathcal W}}{_{\text{FL}}}$.

Note that we can recover the standard expression of the angular distance by taking the initial condition at the observer. The relation (\[defDA\]) then implies $$D{_{\text{A}}} = \sqrt{\det\boldsymbol{\mathcal{D}{_{\text{FL}}}}} = \frac{a_0}{(1+z)} \, f_K(\eta{_{\text{source}}}),
\label{eq:FL_DA-z_relation}$$ where $z=a_0/a-1$ is the redshift of a photon that only 