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Right handed neutrinos

  1. Dec 25, 2007 #1
    Does that fact that it has been shown that neutrinos have mass in any way imply that there must be right handed neutrinos?
  2. jcsd
  3. Dec 25, 2007 #2
    Follow up question. If there are right handed neutrinos is there an explanation as to why they haven't been observed?
  4. Dec 25, 2007 #3


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    not necessarily, because you can have a majorana mass term consisting of only one type of neutrino field: [tex](\nu_L)^c \nu_L \Delta[/tex] but to make this term (after spontaneous symmetry breaking) to be invariant under the unbroken gauge group (SU(3) color and U(1) electric charge), then you need [tex]\Delta[/tex] to be a triplet field under weak-SU(2). So you can do without the right handed neutrino [tex]\nu_R[/tex] but have to introduce a new scalar field [tex]\Delta[/tex] to the Standard Model instead to give neutrino a mass.

    having said that there are reasons to favor the existence of the right-handed neutrino to explain neutrino mass (eg. see-saw mechanism)

    One explanation (the typical one) is that they are too heavy to be seen at colliders at current operating energies. The see-saw mechanism where light neutrino masses are related to heavy right-handed neutrino masses via
    [tex]M_\text{light} \simeq \frac{\langle\phi\rangle^2}{M_\text{heavy}}[/tex]
    where [tex]\langle\phi\rangle[/tex] is the electroweak breaking VEV which is about 174GeV, tells us that if [tex]M_\text{light} \sim 0.1 \,\text{eV}[/tex] then that implies a [tex]M_\text{heavy}[/tex] of the order of [tex]10^{14}\,\text{GeV}[/tex] which is 100,000,000,000 times higher in energy than the current colliders can reach. :smile:
    Last edited: Dec 25, 2007
  5. Dec 25, 2007 #4


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    Actually, if you allow for non-renormalizable operators in the SM (coming from a GUT theory, for example) then you immediately get a Majorana mass without adding anything (no new scalars)! In fact, the UNIQUE(!) dimension-5 operator will do it:


    where L is the lepton doublet and H is the Higgs doublet, and c is some dimensionless coupling, and M is the UV scale where the SM breaks down (GUT scale, for instance). When you set H equal to its vev, then this will become a Majorana mass for the left handed neutrino, whose value is the same as what you'd expect from the see-saw mechanism. No RH neutrinos necessary. No new scalars necessary.
  6. Dec 26, 2007 #5


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    yes, perhaps it would be a good idea to point this out to the OP as well.
    hope I didn't confuse anyone.
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