In some papers [like https://arxiv.org/pdf/2204.06883.pdf, eq. (31) ], I see that the Electromagnetic tensor field, for a FRW metric (written in a conformal way) \begin{equation} ds^{2} = a^{2}(\tau) \Big( -d\tau^{2} + dx^{2} + dy^{2} + dz^{2} \Big) \end{equation} is
\begin{equation} F_{\mu \nu}=a^{2}(\eta)\left(\begin{array}{cccc} 0 & -E_{x} & -E_{y} & -E_{z} \\ E_{x} & 0 & B_{z} & -B_{y} \\ E_{y} & -B_{z} & 0 & B_{x} \\ E_{z} & B_{y} & -B_{x} & 0 \end{array}\right) \end{equation}
i.e. the Minkowski field tensor multiplied by $a^{2}(\tau)$. However, I'm having a hard time understanding where this formula comes from, even if it intuitively works.
I mean, in other papers [https://arxiv.org/pdf/1905.09968.pdf, pag.4] I read that if my metric is conformal to another metric, $g_{\mu\nu}=\Omega^{2}\tilde{g}_{\mu\nu} $, then $F_{\mu\nu}= \tilde{F}_{\mu\nu}$. In my case, $\tilde{g}_{\mu\nu}$ is the Minkowski metric and $\Omega^{2}=a^{2}$. So, I would expect that the e.m. tensor in the metric above is simply the Minkowski one, without the $a^{2}$ factor. Where does it come from? Or, in other words, given a generic metric, how can I get the e.m. tensor $F_{\mu\nu}$ ?
Thank you in advance!