This is a follow up to a recent question on SE asking about the apparent suppression of radiation shortward of the (red-shifted) Ly-$\alpha$ line of a quasar at redshift $z=6.53$. enter image description here

The general explanation for this is the Gunn-Peterson effect according to which beyond a certain redshift the universe has not re-ionized yet and thus the hydrogen is neutral throughout and can absorb the Ly-$\alpha$ line at any redshift and not just at redshifts corresponding to the localized clouds of neutral hydrogen (which result in the Ly-$\alpha$ forest (note that the spectra have been scaled back to the quasar reference in that reference)).

Now according to cosmological models it is thought that re-ionization has been completed at a redshift $z=6$, so only quasars with a redshift larger than this show complete absorption of the Ly-$\alpha$ line at wavelengths smaller than its peak wavelength ($9156 \mathring{A}$ in this case). However, if we take for instance the wavelength of $7000 \mathring{A}$ here, hydrogen absorbing locally at $1216 \mathring{A}$ (the Ly-$\alpha$ rest wavelength) must have been at a redshift

$$z_a(7000 \mathring{A})=\frac{7000}{1216} -1 = 4.76$$

($z_a$ should indicate that this is the (wavelength specific) redshift of the Ly-$\alpha$ absorbing hydrogen gas located between us and the quasar , not the redshift of the quasar (which has a fixed value of 6.53 in this case). )

This value of $z_a$ is way beyond the point where re-ionization is thought to have occurred, so hydrogen should not be completely neutral, and thus some of the quasar emission should still reach us at this wavelength. Yet the spectrum shows effectively zero intensity at this and even smaller wavelengths.

How can this paradox be resolved? Did I make some mistake here in my argument?

  • $\begingroup$ What spectrum? The only spectrum shown is a quasar at $z=6.53$. $\endgroup$
    – ProfRob
    Commented Feb 6, 2022 at 20:10
  • $\begingroup$ @ProfRob No radiation of the quasar with a wavelength less than 9156 A is present in the shown spectrum. It is 'absorbed' by Ly-$\alpha$ scattering on its way to us that is at smaller redshifts than 6.53. The radiation at 7000 A for instance is absorbed at z=4.76, but re-ionization would have happened by then quite a while ago, so the spectrum should look more like a Ly-$\alpha$ forest in this frequency region, not have zero intensity. $\endgroup$
    – Thomas
    Commented Feb 6, 2022 at 20:43
  • $\begingroup$ Ok, I understand. $\endgroup$
    – ProfRob
    Commented Feb 6, 2022 at 21:34
  • $\begingroup$ Hi Thomas, I just saw your edit. Just to make clear, I realize that the redshift "4.76" is just a random example (although chosen to be far from the redshift of the quasar). In my answer, the fact that one of the quasars in my example has z = 4.76 is a pure coincidence. I'm still referring to the differences in regions of the forest at various redshifts blueward of the quasar Lyα emission. $\endgroup$
    – pela
    Commented Feb 7, 2022 at 8:52

1 Answer 1


You're right that it should be possible to see the evolution of the ionization fraction of the intergalactic medium (IGM) in a single, high-redshift quasar spectrum, as you scan through the Lyman $\alpha$ forest toward progressively shorter wavelengths. But even though we usually say that reionization ended at $z\sim6$, residual neutral hydrogen was still able to scatter Ly$\alpha$ photons out of the line of sight for a very long time after.

A reionized Universe may still be opaque to Lyman α

The HI cross section to Ly$\alpha$ scattering is $\sigma=6\times10^{-14}\,\mathrm{cm}^2$, which is rather large compared to most other transitions. So even though the Universe is transparent to most light, Ly$\alpha$ is easily scattered out of the line of sight. In fact you can show that a neutral fraction of as little as $\sim10^{-4}$ is enough to render the IGM optically thick to Ly$\alpha$. In addition to the increased volume of the Universe, the full calculation depends on things like the background photoionization rate and the density field of the IGM (see e.g. this review by Choudhury & Ferrara).

Transmission as a function of redshift

At $z=4.76$, the IGM still only transmits a few tens of percent of the continuum blueward of the Ly$\alpha$ line, as seen in this figure from Songaila (2004) (with my own annotations):


Here the transmission is calculated as ratio between the total measured flux and an unabsorbed quasar template, in the interval 1080 Å to 1185 Å.

Your requested effect is there

Because of this, a spectrum has to have quite long wavelength range in order for the effect that you're looking for to be noticeable. But if you look at several spectra at various redshifts, you will see it. A classic comparison at high redshift is the one by Fan et al. (2006), but I found this figure, from a talk by Ross McLure (again with my own annotations), which shows the effect better (although it's not too good resolution):

Looking at the $z=6.29$ quasar spectrum, you do in fact see that the absorption is a little less at the bluemost wavelengths around 7600–7800 Å than it is just next to the quasar Ly$\alpha$ line around 8600–8700 Å. And for the $z=4.76$ quasar, the absorption also seems just a little larger at ~6800 Å than at ~6000 Å.

Note how in the latter example, the Ly$\beta$ line at $1026\,\mathrm{Å}\times(1+4.76)=5910\,\mathrm{Å}$ is easily seen.


  • $\begingroup$ Yes, the value of z=4.76 in this plot is a strange coincidence, but in this case it is the quasar redshift, whereas in my consideration it is the redshift of the hydrogen gas that would produce the absorption at 7000 A for the z=6.53 quasar. You can see though that the maximum peak amplitudes in the spectra are virtually constant both with z and wavelength, only the gaps are getting bigger with increasing z. So based on your very plot, one should expect some signal also at 7000 A for the z=6.53 quasar (which corresponds to hydrogen at z=4.76). Btw, your Ly$\beta$ figure should be 5910 A. $\endgroup$
    – Thomas
    Commented Feb 7, 2022 at 19:07
  • $\begingroup$ @Thomas Yes, I understand. I think though that I can in fact convince myself there your z = 6.53 does show more signal around 7000 Å than around 8500 Å. The problem is that your spectrum is quite noisy, so it's not easy to see. Anyway, thanks for catching my typo. $\endgroup$
    – pela
    Commented Feb 8, 2022 at 12:10
  • $\begingroup$ Yes, the spectrum for z=6.53 is quite noisy, but the noise is practically symmetrical around zero, which means that it is just that, noise. There is no systematic continuum intensity anymore at all, even not any obvious Ly$\beta$ emission (which should be at 7726 A). Note also that the differences in your highlighted regions for the z=4.76 and z=6.29 spectra are quite evidently due to Ly$\beta$ emission, not different continuum absorption. Do you know where to get the data for these kind of high z-spectra from? I would be interested in trying to analyze them in more detail in this respect. $\endgroup$
    – Thomas
    Commented Feb 8, 2022 at 18:52
  • $\begingroup$ By "less absorption" I don't mean the amplitude of the continuum, but the density of absorption lines, so it's not the Lyβ line (which in the z=6.29 spectrum is outside the range). Your spectrum is indeed practically symmetric around zero, but I don't think it is exactly zero at the blue end. You could try smoothing it, but I can't seem to find the data of that spectrum. But you can find other spectra e.g. here or here (hi-z) or here (lo-z). $\endgroup$
    – pela
    Commented Feb 9, 2022 at 9:47
  • $\begingroup$ I was thinking of the data in numerical form, so that I can do my own processing of it. Are there any publicly accessible, or would I have to contact the authors of the relevant papers? There have been quite a few systematic studies how the absorption changes with quasar redshift, but I have not found any how it changes within a given spectrum. $\endgroup$
    – Thomas
    Commented Feb 10, 2022 at 8:18

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .