7
$\begingroup$

I'm taking a cosmology course this semester, and I don't really understand the concept of freeze out. Here is a short paragraph from our lecture.

During the initial phase of the Universe, particles are continuously subjected to interactions of the type: $A+B\leftrightarrow C+D$ and $D\leftrightarrow P+Q$. As long as $k_{B}T>m_{0}c^{2}$, particles of rest mass $ m_{0}$ were created. As the Universe evolves to a moment that $k_{B}T<m_{A}c^{2}$, particle $A$ vanishes completely due to $A+B\rightarrow C+D$ and $A+\overline{A}\rightarrow 2\gamma $. As the temperature lowering, the corresponding particle has thermally decoupled from the cosmic soup. The process of the decoupling particle is called freeze out.

Here's the thing I don't understand: What does the "corresponding particle" refer to? Is it $C$ and $D$ or $A$? Also, it is stated that "particle $A$ vanishes completely". But for example, in the process: $ \nu + \bar{\nu }\leftrightarrow e^{+} + e^{-}$, if particle $A$ is the neutrino, it should vanish completely in a few seconds after the big bang. How can neutrinos exist today? Also, I don't understand the difference between "decouple" and "freeze out". Are they the same thing?

$\endgroup$
  • $\begingroup$ The cosmic excitations of the fields vanish, the fields themselves are still there. We can still make neutrinos and plenty of natural processes do, they are just not in thermodynamic equilibrium with the other fields. This is no different from chemical species in a hot gas. Heat air to sufficient temperature and you get nitrogen-oxides. The concentration of those depend on the temperature. The hotter, the more NOx is detectable. Go much higher in temperature and NOx will disappear, again, and be replaced with free electrons and ions in a plasma. $\endgroup$ – CuriousOne Jun 14 '16 at 18:35
2
$\begingroup$

All these statements are meant to be understood in terms of large populations of particles. For example, if the mean energy per particle in the Universe is smaller than the mass of electron, it is impossible to produce many of them and they in general will annihilate or transform into other species. In thermal equilibrium this can be understood just by looking at the statistical weight of electrons — it is proportional to $e^{-m/T}$ which makes their population negligible. Electrons still exist, it is just there are never cosmologically many of them.

The other thing is that in cosmology not everything can achieve equilibrium. Equilibration requires time and the expansion of the Universe actively counteracts it by, basically, reducing the rates of interactions. Freeze out represents the case when interaction rate of the particle is smaller than the Hubble rate. In this case, reactions effectively cease and population of these particles as a whole changes only due to expansion of the space.

I recommend to read Kolb, Turner "The Early Universe" to find out more.


I think I should use an example in my lecture note to avoid those questions that are too complicated. Let me see if I get this right. In the neutrino decoupling case, it should be $ν+\overline{ν}↔e^++e^−$ when it is thermally equilibrium. After the temperature drops below the energy of electron, there is no sufficient energy for $ν+\overline{ν}→e^++e^−$ to happen. So only $e^++e^-→ν+\overline{ν}$ in the universe. So neutrinos propagates freely after since. Is the frozen out particle neutrino?

No, you are wrong. What makes you think that the reactions you show are the only ones happening in the plasma? What about $\nu+\nu \leftrightarrow \nu+\nu$ and many others?

You should to think about this in the following way.

In the past, the Universe was dense and hot and particles could reach both dynamical and chemical equilibrium (look them up, the difference is important!)

We are for some reason interested in temperatures around $1 MeV \approx 2 m_e$. Are electrons in equilibrium then? In the expanding Universe they are if

$$ H \ll \Gamma_e = \langle n \sigma v \rangle $$

where $\Gamma_e$ is rate of some reaction involving electrons. If all reactions are much faster than Hubble rate, electron is definitely in equilibrium. If none — than electron decouples. In the case when some are and some are not — it is hard to tell in general, but sometimes we can.

What reactions do we have for electrons? Hundreds. There are at least to prominent groups — weak reactions (e.g. $\nu+e \to \nu +e$) and electromagnetic reactions ($e + \gamma \to e + \gamma$). Luckily, the rate of electromagnetic reactions is really huge so electrons are definitely in equilibrium.

Cool, then what will happen with them? Well, they will follow the Boltzmann distribution $\propto e^{-\frac{m}{T}}$ and quickly disappear with the falling temperature.

And that's it? Actually, no. There is a small asymmetry in electrons related to baryon asymmetry and electro-neutrality of the Universe, so about 1 in a billion electrons will survive, because it won't find a positron to annihilate with.

Right. And what about neutrinos? They are not charged, they don't interact electromagnetically. Actually, a dimensional estimate of the weak rate ($\Gamma \propto G_f^2 T^5$) gives us that weak reactions rate becomes smaller than the Hubble rate at temperatures about $3 MeV$. So neutrinos will freeze-out.

And neutrons? Not charged and interact weakly as well. The same thing, neutrons freeze-out a bit later at $\approx 1 MeV$ (which is related to their compound nature) and propagate freely. But they still decay, so if nothing happens, they would completely turn into protons in an hour or so.

Does something happen to prevent this? Why, yes — it is Big Bang Nucleosynthesis. But this is a story for other time. (see here and here)


Adding any more details is waste of time as I am already giving you a synopsis of every book on cosmology and particle physics.

$\endgroup$
  • $\begingroup$ So, does it mean that some of them vanish and others just propagate freely after since. Because I looked up all the books I can find and they all say propagate freely after since, but our teacher said that they will all vanish. My understanding now is that both take place. Is this correct? $\endgroup$ – LY3000 Jun 17 '16 at 13:50
  • $\begingroup$ The thing you cite is vague and sounds incorrect. Particular examples: electrons vanish without freeze out while neutrinos freeze out and propagate. Neutrons is a different case: they freeze out and propagate, but neutron decay does not require interaction with other particles, so freeze out does not influence it. Neutrons decay and disappear after some time. (This does not happen in reality because of BBN, but I wanted to give you a qualitative understanding with a toy model. In reality a combination of coincidences in the much bigger system play role) $\endgroup$ – Andrii Magalich Jun 17 '16 at 13:58
  • $\begingroup$ I think you are looking for a simple answer, but there is none. You need to consider each system individually, estimate the interaction rates to find out if there can be equilibrium, then continue correspondingly. But the problem is that it can be really hard to assess equilibrium - some reactions are fast, some are slow and only numerical kinetic treatment will give you the answer in the complicated system $\endgroup$ – Andrii Magalich Jun 17 '16 at 14:11
  • $\begingroup$ I think I should use an example in my lecture note to avoid those questions that are too complicated. Let me see if I get this right. In the neutrino decoupling case, it should be $\nu +\overline{\nu }\leftrightarrow e^{+}+e^{-}$ when it is thermally equilibrium. After the temperature drops below the energy of electron, there is no sufficient energy for $\nu +\overline{\nu }\rightarrow e^{+}+e^{-}$ to happen. So only $ e^{+}+e^{-}\rightarrow \nu +\overline{\nu }$ in the universe. So neutrinos propagates freely after since. Is the frozen out particle neutrino? $\endgroup$ – LY3000 Jun 17 '16 at 16:37
  • $\begingroup$ See updated answer $\endgroup$ – Andrii Magalich Jun 17 '16 at 20:06

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.