Matter v Antimatter III: Leptogenesis

By Mark Trodden | August 18, 2008 10:00 am

I’ve been blogging the last few weeks about the question of the baryon asymmetry of the universe – the measured excess of matter over antimatter in the universe. Having already discussed electroweak baryogenesis, I’d now like to turn to another possible way that this asymmetry may have come about – leptogenesis.

In order to explain this, I’ll switch gears for a moment to a seemingly unrelated issue in contemporary particle physics, that of neutrino masses. Neutrinos are electrically neutral particles that, until the last decade, were thought to be exactly massless (and indeed are precisely so in the standard model of particle physics). There exists one neutrino particle associated with each electron-like particle (one for the electron, one for the muon, and one for the tau lepton, going by the imaginative names electron neutrino, etc.).

However, careful experiments on neutrinos created both in the Sun and in the upper atmosphere have conclusively demonstrated that neutrinos have an extremely small but non-zero mass. Fortunately for me, and for you, I don’t need to go into my own explanation of why this is so because Heather Ray did a fantastic job of it in her guest post on the MiniBooNE results from April of last year.

Explanations for unusually small masses, like those that neutrinos seem to have, aren’t easy to come by, but one popular and seemingly natural way to achieve them in the so-called seesaw mechanism. This isn’t the place to go into that in detail, but the important point is that one needs to postulate a heavy right-handed neutrino, the mass of which then conspires with the electroweak scale to generate the unusually light scale characterizing the masses of the left-handed neutrinos.

Having told this little story about the possible origin of small neutrino masses, let’s return to the issue of the baryon asymmetry of the universe. As I’ve just described, there is a somewhat compelling argument for the existence of a very heavy (it turns out to be close to the GUT scale) right-handed neutrino. The interactions of this field are such that they violate lepton number, and so this opens up the possibility of generating an asymmetry of leptons over antileptons, as long as a couple of other crucial conditions are satisfied.

If our right-handed neutrino was light, say at the electroweak scale, then the expansion rate of the universe when it was decaying would be so slow that the decay products could easily find one another, and thus the reverse interactions would balance the decays and no net lepton asymmetry would result. This is a heuristic way of saying that the interactions of a sufficiently light neutrino would be in approximate thermal equilibrium, and hence no net evolution can occur. However, because the right-handed neutrino in question is heavy, it decays at early times, when the universe is expanding extremely rapidly, and because of this the decays occur out of equilibrium.

A final requirement is that the decays be CP-violating, but this is a generic feature of theories beyond the standard model and so we needn’t dwell on it here.

So, we’ve used the measured non-zero masses of neutrinos to infer the possible existence of a heavy right-handed neutrino decaying out of equilibrium in a lepton number and CP-violating way in the early universe. All the ingredients necessary to generate an asymmetry of leptons over their antiparticles.

But wait a minute. Weren’t we looking to explain the baryon asymmetry, not a lepton asymmetry? Fortunately, the standard model contains a handy way of partially converting a lepton asymmetry to a baryonic one – and we’ve already discussed it! My post on anomalous baryon number violation in the standard model described the way in which nonperturbative effects allow for processes in which more baryons than antibaryons are produced. One thing I neglected to say though, is that there is an identical anomaly in the lepton number symmetry and, remarkably, the combination of the two symmetries, B-L, has no anomaly at all! Another way to say this is that whenever there is an anomalous process that produces more baryons than antibaryons, it is guaranteed to be accompanied by the production of more leptons than antileptons (up to subtle numerical factors that need not concern us here, equal numbers are produced).

So if we have a way of efficiently generating a leptonic asymmetry in the early universe, anomalous electroweak processes can then equilibrate that asymmetry into one partly in leptons and partly in baryons – yielding our required baryon asymmetry! This whole process is known as leptogenesis.

Leptogenesis has always been a cute idea, but the discovery of neutrino masses has provided more support for it and it may well be that this is how the baryon asymmetry was generated, leading to the light elements, and ultimately us.

CATEGORIZED UNDER: Science
  • http://blog.chungyc.org/ Yoo

    Very nice exposition. I have a question, though. Could both electroweak baryogenesis and leptogenesis happened? Or would the B-L symmetry mean that if the only possibilities are these two, then only one of these could have occurred?

  • Aramael

    Wow.

    Neutrinos have mass, therefore heavy right handed neutrinos, therefore baryonic asymmetry, therefore us.

    Physics is so beautiful. I wish I was smarter, so I could follow it better!

    A

  • Sili

    If the baryon and lepton asymmetries are correlated, does that mean they’re sorta, kinda related through some other other conserved number? A new symmetry of sorts? (Or an old one, since there seems to be a lepton generation for every quark ditto?)

    I take it the GUT scale is way beyond experimental reach, so this is not something that’ll pop up in Geneva any time soon.

  • http://www.gregegan.net/ Greg Egan

    If neutrinos have mass, how is the distinction between left-handed and right-handed neutrinos maintained? Naively, I’d have thought that you could always (in principle) find a frame in which the neutrino was at rest; in that frame, how do you know the handedness of the neutrino?

  • http://golem.ph.utexas.edu/~distler/blog/ Jacques Distler

    If neutrinos have mass, how is the distinction between left-handed and right-handed neutrinos maintained? Naively, I’d have thought that you could always (in principle) find a frame in which the neutrino was at rest; in that frame, how do you know the handedness of the neutrino?

    That’s why these are bad names.

    The best way to disentangle what is being talked about is to refer to all the fermions as left-handed Weyl spinors (their CTP conjugates are right-handed Weyl spinors). A mass term couples a pair of left-handed Weyl spinors together to make a Lorentz-scalar.

    The neutrino (the guy who is usually called “left-handed”) is part of an SU(2) doublet, L, with hypercharge -1/2. SU(2)×U(1) gauge invariance forbids a mass term for him. Now, imagine there’s another Weyl fermion (the guy who is confusingly called “right-handed”), Ψ, which is an SU(2)×U(1) singlet. Gauge invariance allows a mass term, MΨΨ, for him. It also allows a Yukawa coupling λ &Psi L·h, where h is the Higgs doublet.

    When the Higgs gets a VEV, this gives us a 2×2 mass matrix, with off-diagonal entries m=λ⟨h⟩, and diagonal entries M and 0. When you diagonalize this mass matrix, you find one heavy fermion (whose mass ∼ M) and one light fermion (whose mass ∼ $latex m^2/M$). These mass eigenstates are, of course, linear combinations of the fermion in the SU(2) doublet and the singlet, Ψ.

  • http://golem.ph.utexas.edu/~distler/blog/ Jacques Distler

    Wow! Those equations look pretty damned awful. Curse the lack of a preview!

  • http://blogs.discovermagazine.com/cosmicvariance/mark/ Mark

    Jacques is, of course, correct (thanks for answering Greg, Jacques). Nevertheless, it is conventional to refer to these neutrinos in the way that I did in the post (although it is definitely confusing if you’re not used to it).

  • http://lablemminglounge.blogspot.com/ Lab Lemming

    I have a very basic (and probably stupid) question:
    If the sun is producing lots of neutrinos, and if reactors and radioactive decay is producing lots of anti-neutrinos, then can we see any sort of radiation produced when these particles collide and cancel each other out?

  • TimG

    This business of left-handed and right-handed has always confused me. Am I correct in understanding that there are two meanings of “left-handed”, and that being left-handed in terms of belonging to the left-handed doublet doesn’t necessarily mean you have left-handed helicity? What is the connection (if any) between these two definitions of “left-handed”?

    Also, could someone expand on this statement:

    The neutrino (the guy who is usually called “left-handed”) is part of an SU(2) doublet, L, with hypercharge -1/2. SU(2)×U(1) gauge invariance forbids a mass term for him.

    In particular, why is it that the left-handed electron, which is in the same SU(2) doublet, is allowed to have a mass, but the neutrino is not? What’s the difference?

  • http://golem.ph.utexas.edu/~distler/blog/ Jacques Distler

    In particular, why is it that the left-handed electron, which is in the same SU(2) doublet, is allowed to have a mass, but the neutrino is not? What’s the difference?

    Because there’s another left-handed Weyl fermion, $latex tilde{e}$, which is an SU(2) singlet, and has hypercharge +1. When the Higgs gets a VEV, the Yukawa coupling $latex tilde{e} L cdot overline{h}$ gives a mass to $latex tilde{e}$ and one component of the lepton doublet. For the other component of the lepton doublet, $latex Psi$ plays the same role that $latex tilde{e}$ did. Except that $latex Psi$ ican have a large gauge-invariant mass term all by itself. That leads to the “seesaw” mechanism discussed above.

  • TimG

    Ah, I get it now. Thanks for the explanation.

  • Ralph

    Still one more question about electrons v. neutrino masses: The neutrino singlet Phi has a large mass, but the electron singlet e~ does not. Is there any reason for that?

    Hmmm… IIUC the neutrino singlet has no weak isospin, zero electric charge and so zero weak hypercharge … is it this lack of charges that allows the neutrino singlets to have large masses?

  • TimG

    Ralph, if I’m understanding the discussion correctly the difference is that the right-handed neutrino $latex Psi$ can have a mass term that couples it to itself (a Majorana mass, if I’m remembering the terminology) whereas the right-handed electron only has a mass term that couples it to the left-handed electron (a Dirac mass, again if remember the terminology.)

    So the electron mass matrix looks like:
    0 m
    m 0

    which when you diagonalize to find the mass eigenstates gives you two equal masses. Whereas, the neutrino mass matrix looks like:
    0 m
    m M

    with M >> m, which gives you one large mass eigenstate (~ M) and one small one ($latex ~ m^2/M$)

    So in terms of the difference between electron masses and neutrino masses, I think the question is not just “Why does the neutrino have such a large Majorana mass?” but “Why does it have any Majorana mass at all?” I think you’re right in thinking the Majorana mass is allowed because of its lack of charges, but I could be wrong.

    I’m sure if any of the above isn’t right that Jacques, Mark or someone else will be along soon to correct it. :)

  • http://golem.ph.utexas.edu/~distler/blog/ Jacques Distler

    Ah, yes, Dirac versus Majorana masses. Another pointlessly confusing bit of nomenclature.

    If you have two left-handed Weyl fermions, $latex psi,tilde{psi}$, transforming in complex conjugate representations of the gauge group, you can write down a mass term

    $latex m psi tilde{psi}$

    That’s called a “Dirac mass” (because this pair of Weyl fermions can be assembled into a Dirac fermion). On the other hand, if $latex psi$ is in a real representation of the gauge group, you can write down a mass term for it, all by its lonesome

    $latex m psi psi$

    That’s called a “Majorana mass.”

    And, just to confuse you, if $latex psi$ is in a pseudoreal representation of the gauge group, then you still need a pair of such fermions to write down a mass term,

    $latex m psi_1 psi_2$

    even though both fermions transform in the same representation of the gauge group.

  • Pingback: Matter v Antimatter III: Leptogenesis : Sophoblog()

  • Mike

    “Another way to say this is that whenever there is an anomalous process that produces more baryons than antibaryons, it is guaranteed to be accompanied by the production of more leptons than antileptons”

    The anomalous process should occur above the electroweak scale, right? Not that this is a problem here.

  • http://blogs.discovermagazine.com/cosmicvariance/mark/ Mark

    Hi Mike. In practice this is true, just because such processes are mediated by instantons of high Euclidean action, and therefore will be fantastically rare at today’s temperatures. In principle though, one could happen today (there’s always a tiny chance).

  • Shantanu

    Mark, thanks for the nice post. On a different note I haven’t seen any
    blog on this year’s SLAC summer school (where from the agenda I see
    you gave a talk). Do you (or Joanne) know if the videos of the talks are
    going to be put up?
    Thanks

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Cosmic Variance

Random samplings from a universe of ideas.

About Mark Trodden

Mark Trodden holds the Fay R. and Eugene L. Langberg Endowed Chair in Physics and is co-director of the Center for Particle Cosmology at the University of Pennsylvania. He is a theoretical physicist working on particle physics and gravity— in particular on the roles they play in the evolution and structure of the universe. When asked for a short phrase to describe his research area, he says he is a particle cosmologist.

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