4
$\begingroup$

My understanding of the microscopic "mechanism" of blackbody radiation* in a gas is as follows: interactions between fluctuating charge distributions in particles of a gas create a microscopic random electromagnetic field. The particles interact with this field and with each other directly (for example, in collisions), producing electromagnetic radiation in a continuous spectrum that is consistent with the blackbody spectrum.

This means that only gases which actually couple to the electromagnetic field are allowed to produce blackbody radiation. This is fine for almost all normal matter** in the universe, as nearly all matter is either overtly electrically charged, or is a composite particle that has charged constituents, and therefore has a fluctuating charge distribution. (In the case of a photon gas, the constituents themselves are not charged, but the blackbody radiation and the blackbody spectrum are equivalent, because they essentially are blackbody radiation.) This is not the case for neutrinos. They are long-lived, pointlike particles with no electrical charge, and as such, they should not be able to interact with the electromagnetic field. So a gas of neutrinos would not produce blackbody radiation.

Is this believed to be true? The only "neutrino gas" in thermal equilibrium that I'm aware of is the ensemble of relic neutrinos created along with the CMB, and the difficulty of measurement of low-energy neutrinos means that we probably won't be measuring this any time soon. Nevertheless, are there observations that support this line of reasoning?

*I realize that blackbody radiation is a statistical, macroscopic phenomenon, so any semblance of a microscopic "mechanism" is stretching the truth a bit. This is why I don't endeavor to go into specifics in this explanation.

**Except dark matter, obviously. Which, by mass, should by all rights be considered the "normal matter" in this universe, but that's beside the point.

Improve this question
$\endgroup$
2
  • $\begingroup$ I would assume they produce a very faint blackbody radiation of $Z$ bosons via the neutral current interactions, although due to its high mass, that would only occur at the very high end of the scale. $\endgroup$
    – Slereah
    May 22, 2018 at 13:45
  • $\begingroup$ @Slereah But $Z$ boson production can't be called "radiation" in any real sense, they decay much too quickly. Then you have a blackbody $Z$ spectrum convolved with the $Z$ decay spectrum, which is problematic because there's a nontrivial probability of $Z\to\nu\bar{\nu}$ taking place, so I'm pretty sure that wouldn't look like a blackbody either. $\endgroup$
    – probably_someone
    May 22, 2018 at 13:50

1 Answer 1

Reset to default
4
$\begingroup$

There is a difference between local and complete thermodynamic equilibrium. The relic cosmic neutrinos are an example of the former. Their distribution of kinetic energies can be described by a single temperature, but they are pointedly not, and have not been for a long time, in equilibrium with a radiation field at the same temperature.

In order to be in complete thermodynamic equilibrium and emit blackbody radiation, the neutrinos would need to be optically thick to the radiation in the universe. That is not the case now, but it effectively was before the neutrinos "decoupled" from baryonic matter about 1s after the big bang. Prior to this, the neutrino gas could be said to be in CTE with the radiation field because of weak interactions that kept them in equilibrium with charged leptons, protons and neutrons.

So note that the key here is that the neutrinos can in principle be optically thick, but that is because they can be in equilibrium with charged baryons/leptons through reactions like $$ e^+ + e^- \leftrightarrow \nu_e + \bar{\nu_e}$$ $$ p + e \rightarrow n + \nu_e$$ $$ n \rightarrow p + e + \bar{\nu_e}$$

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ similar to the case of the decoupled photons of CMB, after all. $\endgroup$
    – anna v
    May 22, 2018 at 18:40
  • $\begingroup$ This makes sense for the cosmic neutrino background, for which cosmological processes are keeping it out of equilibrium. But if I were able to hypothetically just put a bunch of neutrinos in some sort of confinement and wait a while, are you saying the neutrino gas would eventually thermalize and emit blackbody radiation? If so, then what sort of microscopic processes would we take as the "mechanism" for blackbody radiation emission? $\endgroup$
    – probably_someone
    May 22, 2018 at 18:48
  • $\begingroup$ @probably_someone No, I am not saying that. Not unless you collect enough neutrinos together to be optically thick (which a blackbody needs to be). $\endgroup$
    – ProfRob
    May 22, 2018 at 18:51

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.