0
$\begingroup$

Context

From what I understand about the Cosmic Microwave Background Radiation is that it was from the big bang, and since space has streched and become bigger since then the wavelength of the CMBR has increased over time, which means that when the universe stopped being opaque it wasn't microwaves but gamma rays, that must mean that it went from gamma rays to x-rays to ultra violet to visibile light (blue then all the way to red) then infra red and now microwave.

The CMB was emitted at an energy of $E_{em}=13.6\text{ eV}$, which is the binding energy of hydrogen. This corresponds to a wavelength of $$ \lambda_{em} = \frac{hc}{E_{em}} \approx 9.12\times 10^{-8}\text{ m}$$

Question

So the question is simply, if the CMB was emitted at the ionization energy of hydrogen, why is it a blackbody spectrum instead of a line spectrum?

$\endgroup$
7
$\begingroup$

At the epoch of (re)combination, the radiation field of the universe was a blackbody at a temperature of roughly 3000 K.

I am not sure where you get the idea that the temperature is directly given by the binding energy of a hydrogen atom - that is not the case. A blackbody at 3000 K has an average photon energy of 0.7 eV, but any hotter and there are enough photons with energies above 13.6 eV to ionise any H atoms.

Prior to recombination, the universe consisted of free electrons, protons, helium atoms and a very small number of hydrogen atoms and alpha particles in thermodynamic equilibrium.

Since the plasma is in thermodynamic equilibrium all emission processes are balanced by absorption processes and the source function is given by the Planck blackbody function. In addition, since the mean free path of a photon is much smaller than the speed of light multiplied by the age of the universe, mainly due to the opacity provide by free electrons, the universe is effectively optically thick. In such circumstances, the radiative transfer equation tells us that the radiation field approximates to the source function, which in this case is the Planck blackbody function.

The recombination occurs at 3000 K because at that temperature and below, there are insufficient photons with high enough energy to ionise lots of hydrogen. You can calculate the ionisation fraction using the Saha equation - this shows that the ionisation fraction decreases abruptly at $\sim 3000$ K (see here for some details). As the hydrogen atoms recombine, the opacity falls by orders of magnitude due to the disappearance of free electrons and the radiation field that exists at that time is then free to propagate throughout the universe.

Your real question might be why can a gas consisting of electrons protons and atoms produce a continuous spectrum? The answer to this is that there is enough opacity at all wavelengths to make the universe optically thick at those wavelengths and that absorption and emission processes must be in detailed balance. The relevant processes that absorb photons over a continuum of wavelengths are electron scattering, inverse bremsstrahlung, and photoelectric absorption, with inverse continuum emission processes of inverse Compton scattering, bremsstrahlung and recombination radiation.

$\endgroup$
  • $\begingroup$ making this the correct answer for a more detailed, comprehensive answer. I thought the temperature is related to the binding energy of hydrogen because that's the moment when hydrogen could form (not enough energetic photons), this is when the universe became transparent. Am I mistaken in my understanding? $\endgroup$ – Dheeraj Bhaskar Jun 20 '17 at 15:30
  • 1
    $\begingroup$ @DheerajBhaskar The temperature at recombination is approximately 3000K = 0.26 eV. A blackbody spectrum with a temperature any hotter than this has sufficient photons with energy above 13.6eV to ionise any hydrogen atoms that form. $\endgroup$ – Rob Jeffries Jun 20 '17 at 21:02
  • 1
    $\begingroup$ @alex The average energy of a photon in a blackbody spectrum is 2.7kT. In this case, recombination takes place at 3000K, when the average photon energy is 0.7 eV. $\endgroup$ – Rob Jeffries Jun 20 '17 at 21:05
  • $\begingroup$ "but any hotter and there are enough photons", how would one define enough here? Is there a number to it? $\endgroup$ – Dheeraj Bhaskar Jul 13 '17 at 10:34
  • 1
    $\begingroup$ @DheerajBhaskar The radiation field is given by the Planck formula; this gives the energy density of the radiation field that can be translated into photons per cubic metre as a function of frequency if you wish. The only free parameter is temperature - which is 3000K. The critical temperature is found from the Saha equation, which also depends on the density. $\endgroup$ – Rob Jeffries Jul 13 '17 at 11:42
4
$\begingroup$

It would have been a line spectrum if it was solely comprised of photons emitted by the transition of electrons from a free state to the ground state of the hydrogen atom during recombination. But most of the radiation was just photons scattering of charged electrons and protons in a 'plasma'. Because of this scattering any light could not escape and therefore plasma was opaque to light of all wavelengths ( which is effectively the definition of a black body) Since the photons and the plasma were in equilibrium (this is the reason why the CMB is uniform across the sky, because it reached thermodynamic equilibrium) and had the same average energy they had a definite temperature. This is the reason the plasma behaved like a blackbody in equilibrium at a fixed temperature. Once recombination occurred the plasma turned almost transparent and let the black body radiation pass through which is what we receive today.

EDIT: As Rob Jeffries pointed out, the temperature of the photons I initially mentioned was wrong The correct temperature of the photons is Given by the Saha equation, as he has written in his answer.

$\endgroup$
  • $\begingroup$ on second thoughts, making this the correct answer for an easier to understand answer. Both alex's and @RobJeffries's answers are correct $\endgroup$ – Dheeraj Bhaskar Jun 20 '17 at 15:33
  • 1
    $\begingroup$ This is incorrect in the sense that the temperature at recombination is not 13.6 eV$/k_B$. $\endgroup$ – Rob Jeffries Jun 20 '17 at 20:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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