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What is the purpose of the electron neutrino?

  1. May 24, 2012 #1
    I'm still new to the world of physics... but I'm learning. I am reading a book that talks about the neutrino, and I understand the three neutrinos associated with the electron, muon, and tao particles.

    If the purpose of the neutrino is to conserve energy and mass when a particle decays, why does the electron have a neutrino? Isn't the electron supposed to be so fundamental that it does not decay into something else?

  2. jcsd
  3. May 24, 2012 #2


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    An electron neutrino is not the product of electron decay. It accompanies an electron when something else decays, such as a neutron decaying into a proton and an electron. The neutrino is needed for conservation of momentum and spin.
  4. May 24, 2012 #3
    Think of electrons and (electron) neutrinos as two different quantum (and mass) states of the same particle.

    Where electrons (or other charged leptons) take part in weak (charged current) interactions, the latter always converts a charged lepton into the corresponding neutrino, or vice versa, or the equivalent with their antiparticles. The Feynman rules also allow (eg) an antineutrino to be given off in lieu of an incoming neutrino being absorbed - permutations such as these are mathematically equivalent.

    For example the classic weak decay is

    [itex]n → p^+ + e^- + \bar{\nu}[/itex]​

    where the outgoing anti-neutrino is equivalent to an incoming neutrino getting 'converted' into the electron.
  5. May 24, 2012 #4
    If I'm not mistaken, back in the day people were looking at beta decay (like what Adrian the rock has posted), but it seemed like there was missing energy because nobody knew about neutrinos. The proton and electron didn't have enough, so it was either throw energy conservation in the bin, or postulate this very difficult-to-detect particle that could account for the missing energy.
    Also there is something call lepton flavour conservation. You've got three flavours of charged lepton - the electron, the muon and the tauon. In any interaction, lepton number and lepton flavour must be conserved. So, looking at [itex] n \rightarrow p^{+} + e^{-} + \overline{\nu_{e}} [/itex], there were no leptons on the left hand side of the arrow, but there is an electron appearing on the right. That means the lepton number has increased by +1. If lepton flavour is to be conserved, there has to be an anti-lepton (which would have a corresponding lepton number of -1) of the same flavour produced at the same time. Also, to conserve electric charge, it must be neutral. So, there is an electron anti-neutrino required. It's not coming out of the electron, it's being created the same time as the electron.
    Last edited: May 24, 2012
  6. May 25, 2012 #5
    Leptonic number was introduced later.
    The problem was the conservation of energy and mommenta that made Pauli say "there exist another particle" and for the charge reasons, it had to be neutral...
  7. May 25, 2012 #6
    It's a neutrino going backwards in time into the interaction, from whence it then comes out again, this time forward through time, as the electron!
  8. May 25, 2012 #7


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    welcome to pf!

    hello atxjoe512! welcome to pf! :smile:
    let's concentrate on creating an electron

    we can't create an electron on its own!!

    either we create it out of "nothing but energy", and then we have to create its antiparticle also (the particle with opposite charge)

    or we create it from neutron decay, and then we have to create its "weak partner" also (the particle with opposite isospin), ie the neutrino :wink:
  9. May 25, 2012 #8


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    Uh yeah, sure it is. :eek: How do you know it isn't the other way around?
  10. May 25, 2012 #9
    If it is a β+ decay then it is! :)
  11. May 26, 2012 #10
    Apart possibly from "dark energy", there is no such thing as "nothing but energy". The so-called "pure energy" states sometimes mentioned when talking about annihilation or pair production are actually states of particles such as photos, gluons or weak bosons.
    Actually, this is just a case of pair production.

    At fundamental particle level, neutron decay is actually a two-part interaction:

    [itex]d \rightarrow u + W^-;\ \ W^- \rightarrow e^- + \bar{\nu_e}[/itex]​

    But the Feynman diagram that describes the above also describes any of the following (amongst many others):

    [itex]d \rightarrow u + W^-;\ \ W^- + e^+ \rightarrow \bar{\nu_e}[/itex]​

    [itex]e^+ \rightarrow \bar{\nu_e} + W^+;\ \ d + W^+ \rightarrow u[/itex]​

    [itex]e^+ + \nu_e \rightarrow W^+;\ \ d + W^+ \rightarrow u[/itex]​
    These examples are all positron capture reactions, so don't normally happen in practice, but all are theoretically possible. In the first, the incoming positron is converted into its corresponding anti-neutrino. The second is similar except that the W boson propagates in the opposite direction (so has to become W+ instead of W-). In the third the positron annihilates the neutrino to produce the W+.

    All of these produce final states which have one more electron/one fewer positron, one fewer neutrino/one more anti-neutrino, one fewer down quark, and one more up quark, than their initial states. All are described by the same maths.
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