Is it possible to observe Lyman-continuum emission from extragalactic objects?

Question

Lyman-continuum (LyC) emission is everything blueward of 912 Angstroms (so it includes extreme UV photons, X-rays and gamma rays). There are many low-redshift astrophysical plasmas such as the diffuse halo gas around galaxies that are hot enough ($\sim10^6$ K) that we expect them to produce copious UV/X-ray emission (either continuum or metal line emission). On the other hand, neutral hydrogen can absorb any LyC photon (no matter how energetic, as long as the wavelength is $<912$ Angstroms) in order to become ionized.

Since there is tons of neutral hydrogen in the interstellar medium between Earth and other galaxies (and even between us and gas in our own Milky Way's hot corona), I'm struggling to understand how we can ever detect LyC emission from extragalactic objects at $z\sim0$. Assume that we are perfectly sensitive to the diffuse, faint X-ray-emitting gas at low density and further assume we use space-based telescopes to avoid extinction by the Earth's atmosphere -- under what conditions would we then expect to detect extragalactic LyC emission? The only thing I can think of is that there might be patches of the sky that are relatively underdense in neutral atomic hydrogen in our ISM (maybe it's ionized or molecular hydrogen, or no hydrogen at all), so those parts of the sky might be more "transparent" to extragalactic LyC photons.

There is also the question of high-redshift sources of LyC emission. Given that their light will cumulatively encounter much more intergalactic gas on the way to us, I also find it hard to imagine how we can ever measure LyC emission from them. I guess at $z<6$ the Universe is reionized, so there may not be much neutral atomic hydrogen in the IGM to cause significant absorption/scattering. But it's confusing because the IGM was reionized by LyC photons yet we cannot measure them directly...?

Answer 1

I think the short answer is "You can't observe $z=0$ LyC leakers, but you can observe $z\sim0$ LyC leakers."

No z = 0 leakers…

Measuring the Lyman continuum (LyC) — i.e. photons capable of ionizing neutral hydrogen (HI) — is indeed very hard. The HI cross section to LyC is $$ \sigma_\mathrm{LyC} = \sigma_\mathrm{LL} \times \left( \frac{\lambda}{912\,\mathrm{Å}} \right)^3, \qquad(\mathrm{for\,}\lambda<912\,\mathrm{Å}) $$ where $\sigma_\mathrm{LL}=6.3\times10^{-18}\mathrm{cm}^2$ is the cross section at the $912\,\mathrm{Å}$ Lyman limit (Osterbrock 1989).

With an escape fraction $$ f_\mathrm{esc} = e^{-N_\mathrm{HI}\sigma_\mathrm{LyC}}, $$ where $N_\mathrm{HI}$ is the HI column density, it doesn't take more than $$ N_\mathrm{HI} \simeq \frac{1}{\sigma_\mathrm{LL}} \sim 10^{17}\,\mathrm{cm}^{-2} $$ to stop this radiation (which is why gas clouds with column densities larger than this value are known as "Lyman limit systems").

Indeed, most galaxies are not LyC leakers, and those that are tend to have very clumpy interstellar media, and still have escape fractions $f_\mathrm{esc}\lesssim 10\%$ (see e.g. Izoutov et al. 2016; Rutkowski et al 2017; Gazagnes et al. 2018; Guseva et al. 2020).

But if you look at a $N_\mathrm{HI}$ map of the Milky Way, such as this one from HI4PI Collaboration et al. (2017):

you'll see that, even if you look directly away from the plane of the Milky Way, the column density barely goes below $N_\mathrm{HI} = 10^{20}\,\mathrm{cm}^{-2}$. In fact, one of the lowest-column density directions is the so-called "Lockman Hole", which has $N_\mathrm{HI} \simeq 5.7^{19}\,\mathrm{cm}^{-2}$, still high above the threshold for transmitting LyC.

For this reason, $z=0$ LyC leakers aren't directly observable (you also can't see UV through Earths' atmosphere, but that is circumvented by going to space, e.g. with Hubble's COS spectrograph).

…but some z ~ 0 leakers

However, if you relax your $z=0$ criterion a little, we do have some "local" LyC leakers, most notably Haro 11 at $z=0.02$ and Tol 1247 at $z=0.05$. From the equation above, you see that already at $z=0.02$, the Lyman limit has been redshifted to $930\,\mathrm{Å}$, meaning that all LyC in the range $[912\text{–}930]\,\mathrm{Å}$ is transmitted.

The intergalactic medium

As you say, the intergalactic medium (IGM) is virtually fully ionized, and on top of that is also very dilute. Starting roughly at $z\sim15$ or so, the LyC that did escape galaxies started reionizing the Universe. Reionization was complete enough by $z=6$ that we "easily" see galaxies here, but there was still enough residual neutral hydrogen until $z\sim5$ that the part of the spectrum blueward of Lyman $\alpha$ at $1216\,\mathrm{Å}$ was erased, resulting in the so-called Gunn-Peterson trough.