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A Antilepton excess of leptogenesis and sphalerons

  1. Oct 21, 2016 #1
    Hello

    In sphaleron interactions (B - L) is conserved.
    So we need an antilepton excess (provided by decay of heavy RH neutrinos in leptogenesis).
    This is then converted to a baryon excess, via sphalerons, to explain the baryon excess that we observe.

    But the conversion factor is ~1/3.
    For example in this paper https://arxiv.org/abs/hep-ph/0502082 (Equation 17, where a_sph = 1/3 )

    I first interpreted this as saying only 1/3 of antilepton doublets are converted, meaning we are left with 2/3.. and I was wondering what happens to this 'left over' excess.
    Now I am thinking about it again, I think that sounds a bit stupid.
    Does it actually just mean that it takes three antilepton doublets to convert to one baryon?
    (and so we convert all antilepton excess to baryons).

    I guess 1/3 is because baryons have three quarks? although, actually, it is not exactly 1/3 it is approximate, so maybe it is for another reason?

    Also... do we not need a mechanism to generate a lepton asymmetry of the universe?
    For example, I presume the universe has more electrons than positrons? How did this occur?

    Thanks
     
  2. jcsd
  3. Oct 21, 2016 #2

    Orodruin

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    No, it is a matter of where the lepton and baryons enter chemical equilibrium, ie, the spalerons go as much in one direction as the other.

    Yes.
     
  4. Oct 21, 2016 #3
    Ah ok. Right. Thanks for the reply. So I'll go back to my original way of thinking, only 1/3 of antileptons are converted.

    The sphaleron interactions reach chemical equilibrium as 1/3 of the antilepton excess is converted to baryons.
    So at the end of leptogenesis, you are left with an antilepton excess (2/3 of what was created by the heavy neutrino decays).
    And we need a mechanism that favours leptons to create a lepton excess.

    I know we have deltaCP, but people never talk about needing deltaCP for lepton asym, only for baryon asym (often giving people the wrong impression that delaCP generates the CP asym in all lepto models).

    I never hear about a lepton asymm 'problem', does this mean there is an obvious answer (to generate lepton excess) im missing?
     
  5. Oct 21, 2016 #4

    mfb

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    We know CP violation in the baryon system is small, in the neutrino sector it can be larger. In addition, we can see all baryons, while many leptons are in the cosmic neutrino background, which has not been observed yet. Charge conservation plus the baryon masses (no long-living negative baryons) are sufficient to explain why we have so many electrons around as dominant charged lepton.
     
  6. Oct 21, 2016 #5
    Thanks for the reply. Im being a bit slow here, but I still don't understand, could you please give me an example of an interaction that creates a lepton excess?
    Are you suggesting a negative baryon decays somehow to give an electron??


    I would have guessed a lepton asym would need the same requirements (sakharov conditions) as a baryon asym.
    As far as I know, we have no proof of CP violation in the lepton sector at present (strong hints, but no discovery yet).

    OK the neutrino background has not been observed, but what is the significance?
    Are you saying this could contain an antineutrino excess, or are you implying something else?


    On a related note, this makes me wonder, if neutrinos turn out to be majorana then does a simple nue + d -> e + u interaction violate L ?

    Sorry for bombarding you with questions!
     
  7. Oct 21, 2016 #6

    mfb

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    Not necessarily directly, but as stable result: sure. Plus an antineutrino to keep lepton number conserved. Lepton number is not changed in this process, but we still see electrons accumulating.
    Right. The point is the upper limit, or lack thereof. CP violation with neutrinos can be (relatively) large.
    It is possible. Orodruin should know more.
     
  8. Oct 22, 2016 #7
    Thank you very much for the replies.

    You mention deltaCP has no limit and that it could be large - sure, and for neutrino oscillation physics this is a big focus. Some experiments have started to rule out certain values with at least 2 sigma, and one of the big selling points of obtaining funding for the upcoming experiments is always deltaCP. What I don't understand though, is that deltaCP is always 'sold' as something we need to determine because we need CP violation for baryon asymm (notice how I dont say it is THE cp violation that generates baryon asym). In my opinion this is always presented in a purposely missleading way, and many experimentalists end up thinking deltaCP is the CP violation in leptogenesis ( I belive theories do exist that use deltaCP, but in general the most popular lepto models use the CP violation of the heavy neutrinos) . But from what you have just said, I dont understand why the need for a lepton asymmetry is not a selling point for deltaCP ? This is never mentioned when people try to sell new experiments.

    Are the two possible answers for a leptoin asym the following?
    1 - relic neutrino bkg contains antinu
    2 - deltaCP

    Neither of these are confirmed - is there something in the SM that explains lepton asymm? if not, I feel this question should get more attention that it does, or maybe im missing something...?!

    Or maybe this falls into Dirac V Majorana?
    If neutrinos are majorana then it naively looks to me like ( nue + neutron -> proton + electron ) violates L
     
  9. Oct 22, 2016 #8

    ohwilleke

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    As a basic point, I would note that we don't know with any great confidence if there is lepton asymmetry, or if there is, how much lepton asymmetry there is or in what direction it lies. There are vastly more neutrinos than there are quarks or charged leptons (both of which are almost entirely matter as opposed to antimatter), and we have no reliable estimate of the ratio of neutrinos to antineutrinos in the universe.
     
  10. Oct 22, 2016 #9
    ah ok. So we know the universe is neutral, therefore we know we have more charged leptons (negative) in the form of electrons than we do charged antileptons (positive). So we must have a charged lepton asymmetry.
    But since neutrinos are neutral, and we have not yet measured the relic neutrino background, we do not understand what direction the total lepton asymmetry is in.
    But the situation must be different for Majorana V Dirc....

    ** If neutrinos are dirac
    - then the lepton asym may go in either direction
    - the relic neutrinos may be dominated by RH antineutrinos

    ** If neutrinos are majorana
    - they cannot contribute to the asymmetry
    - so the lepton asym is due only to the charged lepton asym
    - and the charge of this asym must cancel the charge of the baryon asym, so we know we need a lepton asym in this case - so dont we need sakharov conditions?
    - (majorana) neutrino interactions with charged leptons would violate L
    - But we still need some CP violation, dont we ??
    -> So we hope for a non-zero CP violating phase?


    Thank you very much for the reply.
     
    Last edited: Oct 22, 2016
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