5
$\begingroup$

I am just beginning to learn about the ideas of the early universe, so this is probably a beginner question.

I understand that protons and neutrons (which are baryons, which are hadrons) are made out of quarks, and quarks are held together by gluons (at a high level). So those are all the ingredients listed in the title "quark-gluon plasma".

But how do electrons and photons come into the picture? Could you describe how this works?

In the wikipedia article Chronology of the Universe, the quark-gluon plasma cools to form hadrons. That makes sense, because hadrons are made out of quarks and gluons.

Then it quickly glosses over a bunch of stuff, basically saying "hadrons and anti-hadrons annihilate to form leptons and anti-leptons, which annihilate... and now the universe is largely dominated by photons interacting frequently with free protons and electrons".

What exactly is happening there? Where did these anti-hadrons come from, and leptons and anti-leptons, and electrons? That's the first time electrons were mentioned.

How did it go from quark-gluon plasma to the formation of atomic nuclei, photons, and electrons? From that point I can imagine the rest and there are lots of resources on how that works, but before that there is very little explanation of the reasoning from what I've seen so far (just wikipedia and arxiv.org mainly).

$\endgroup$

4 Answers 4

8
$\begingroup$

Gluons and quark-antiquark pairs, like photons and lepton-antilepton pairs, are excitations of the vacuum. Any volume of space with enough energy density to contain a quark-gluon plasma also has, by definition, enough energy density to contain a gas of photons and electron-positron pairs. The difference is that the quark-gluon plasma is governed by the dynamics of the strong interaction, which isn't very well understood, so that's what people are interested in when they talk about it.

Even if you could produce a "purely" hadronic plasma, it would contaminate with leptons and photons as it cooled. The same-charge quark-antiquark pairs in the plasma could annihilate either by producing gluon pairs, thanks to their color charge, or photon pairs, thanks to their electric charge. (Think of the decay $\pi^0\to\gamma\gamma$.) These photons would be available to find each other in the plasma and produce $e^\pm$ pairs. And electroweak transitions between quark members of isospin doublets would also produce charged and neutral leptons directly (think $\pi^+\to\mu^+\nu_\mu$).

$\endgroup$
4
  • 1
    $\begingroup$ fine , but we all are glossing over the CP problem, i.e. the "why we end up with baryon galaxies", but it is a basic research problem in particle physics, as the standard model does not have enough CP violation . $\endgroup$
    – anna v
    Commented Sep 14, 2014 at 8:04
  • $\begingroup$ @annav The missing CP violation is certainly an important problem, must mostly irrelevant to the question of lepton/antilepton density in a QGP. Remember that the baryon asymmetry, baryon number density relative to CMB photon number density, is only about $10^{-9}$ — most questions about evolution of a QGP can be answered to eight decimal places assuming that CP is exact. There are, however, some models that put the missing CP violation in the lepton sector. $\endgroup$
    – rob
    Commented Sep 14, 2014 at 14:34
  • $\begingroup$ Thank you this was helpful. Yeah what am really hoping to see is a summary of all the decay pathways like $\pi^0\to\gamma\gamma$ that led from the quark-gluon plasma to the sea of electrons, photons, and baryons. $\endgroup$ Commented Sep 14, 2014 at 17:11
  • 1
    $\begingroup$ @LancePollard Well, keep in mind that the difference between cold nuclear matter and a QGP is that "bound" states like the pion sort of go away. However if you'd like a comprehensive list of decay modes you should check out the meson summary tables at pdg.lbl.gov $\endgroup$
    – rob
    Commented Sep 14, 2014 at 19:45
2
$\begingroup$

The chronology in Wikipedia is emphasizing what is changing in each epoch. Particles that are lighter than the ambient temperature are presumed to be in thermal equilibrium, but not doing much of interest at that time. In particular, the light leptons and photons are created and annihilated all the time, so there is a sea of electrons, positrons, and photons around at all times before the temperature drops below 0.5 MeV. The individual ones come and go, but the bulk of them doesn't change much. They get emphasized in the lepton epoch as the temperature drops enough for the hadrons to annihilate, but they were there all along.

$\endgroup$
2
$\begingroup$

In most quantum processes you cannot get just particles (without antiparticles) as products. That would violate some conservation laws (charge conservation mainly). So the quark-gluon plasma was a mixture of quarks and antiquarks. As a consequence, after the QCD cooled, you get both hadrons and antihadrons. These annihilated, but there was certain assymetry (more hadrons than antihadrons) and thus some hadrons remained after the annihilation.

The leptons (neutrinos, electrons, antielectrons, etc.) and photons were there in the quark-gluon plasma, all the time. They just were not doing anything important at that time, that's why they are not mentioned. More leptons (+antileptons) and photons are created as the QCD cools down (annihilation of hadrins and antihadrons, decay of unstable hadrons). Then the leptons annihilate with antileptons.

$\endgroup$
0
$\begingroup$

First clearing up a misconception.

I understand that protons and neutrons (which are baryons, which are hadrons) are made out of quarks, and quarks are held together by gluons (at a high level). So those are all the ingredients listed in the title "quark-gluon plasma".

Nope. There are no hadrons in the QGP phase (regarded as a pure phase, I'm discounting the mixed phase - that's mostly a crossover region and isn't a ''phase'', so to speak). In fact, if you look at the QCD Phase Diagram, you can see clearly that the hadrons don't exist at temperatures typical of QGP phase. This fact is also known from lattice QCD simulations, which progressively keep refining the temperature where such a cross-over from hadronic to QGP Phase takes place. (A state-of-the-art example is this one.) So, as pure phases, they are mutually exclusive.

Now, as QGP cools, you go lower down in that graph and get to the hadronic phase - so quarks aren't free anymore, they are confined within hadrons, and these are all you'll find in this ''hadronic regime''. As you probably know, hadrons themselves have decay modes, some of them are the so-called leptonic or semileptonic ones, in which leptons (e.g. electrons) are emitted. Of course, it isn't rare to have photons emitted in hadronic decays too, e.g. see here for examples of radiative decay modes of baryons.

I think this much answers your question in the title. There can be accompanying sidetracks such as cosmic asymmetry, which aren't included in my answer, but then, they were not explicitly a part of the question either.

Hope that helps :)

$\endgroup$
4
  • 1
    $\begingroup$ While there are radiative hadronic decays (the recently-observed $n\to pe^-\nu\gamma$ comes to mind) it's worth noting that there are only putative decay modes known for the proton; there is zero evidence for any proton decay. $\endgroup$
    – rob
    Commented Sep 14, 2014 at 5:33
  • $\begingroup$ @rob - Oops did I link to $p$ instead of $n$?!! Yes, you are right. I was sprinting to complete it, and the idea was to post an example of a decay modes of a nucleon and picked the wrong example. I'll edit my answer and link to neutron now, that's known I'm sure - mean life of 880 s. Thanks anyways. :) $\endgroup$
    – 299792458
    Commented Sep 14, 2014 at 6:25
  • $\begingroup$ No but again, a collision time scale is much shorter. Oh I chose the wrong example again. Editing again. $\endgroup$
    – 299792458
    Commented Sep 14, 2014 at 6:27
  • $\begingroup$ @rob - Thanks for pointing out the choice of a wrong example. $\endgroup$
    – 299792458
    Commented Sep 14, 2014 at 6:30

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.