It is commonly said that after the universe cooled enough for ionized Hydrogen to settle down into neutral Hydrogen, i.e. recombination, the universe became transparent. A reason I have heard for this is that most photons don’t have the right energy to be absorbed by H atoms.

But the free electrons before recombination weren’t absorbing the photons either, they were scattering them. Doesn’t light still scatter off bound electrons? For instance, my understanding is that Compton’s original experiment on Compton scattering used graphite as the source of electrons. Certainly then, photons were scattering off the electrons bound in carbon atoms?

I suspect the answer has something to do with the scattering cross section of bound electrons in neutral Hydrogen being much less than that of free electrons, but then why is that the case?

  • $\begingroup$ Are you asking why neutral hydrogen is transparent to visible light? $\endgroup$
    – PM 2Ring
    Commented Feb 16, 2019 at 22:25
  • $\begingroup$ Yes, that is what I’m asking. $\endgroup$
    – WillG
    Commented Feb 16, 2019 at 22:39
  • $\begingroup$ Similar discussion- physics.stackexchange.com/questions/25447/… $\endgroup$
    – Judas503
    Commented Feb 17, 2019 at 7:11
  • $\begingroup$ Related, yes, but still doesn’t answer my question above. Most answers in that post simply state that neutral hydrogen is transparent without addressing the issue I raise above, about light still being able to scatter off bound electrons. $\endgroup$
    – WillG
    Commented Feb 20, 2019 at 22:43

2 Answers 2


The scattering cross-section for free electrons is known as the Thomson scattering cross-section $\sigma_T$.

It is relatively easy to show that once the electrons are bound, the cross-section can be treated like that of a damped harmonic oscillator. If the oscillator is driven at frequencies below the resonant frequency $\omega_r$, then the (Rayleigh) scattering cross-section goes as $\sigma_T (\omega/\omega_r)^4$.

After recombination most of the hydrogen atoms are in the ground state, so the lowest energy resonance corresponds to Lyman alpha absorption in the UV. But most photons have frequencies in the infrared, much lower than the resonant frequency, so $\omega \ll \omega_r$. Hence the scattering cross-section is orders of magnitude lower than $\sigma_T$.


When a photon with insufficient energy to move an electron from one energy level to a higher energy level (including to complete ionization) the photon may change direction, but not energy. During a period of time, say e.g. one second, about the same same number of photons move towards the observer, just as if there were no atoms in the way. That is, for each photon that was heading for the observer, and gets deflected to a different direction, there will (on the average) be another photon not heading towards the observer, but after hitting an electron it gets deflected to move exactly towards the observer.

Only a very small fraction of photons emitted from the last ionized atoms will move the energy of an electron high enough so that photons of a different energy are subsequently emitted.

I hope you find this answer to be acceptable.

  • $\begingroup$ I am very curious about why this answer was flagged with a -1. I would much appreciate the flagger to post a comment explaining this. $\endgroup$
    – Buzz
    Commented Nov 19, 2020 at 21:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.