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Heavy sterile neutrino decay into Z bosons. (Feynman Diagram)

  1. Dec 1, 2011 #1
    Hello there,

    I am looking at the decays of Heavy sterile neutrinos in the their mass states to Z bosons.

    Using the feynman rules, how would I go about calculating the decay amplitude. I have only ever seen cases where the Z boson is a propagator. Would my expression look something like this:

    [itex]
    M = \kappa\frac{g}{2\cos\theta_{W}}\overline{\nu_{1}}(2)\gamma^{\mu}P_{L}N_{m}(1)\epsilon_{\mu}^{*}(3)
    [/itex]

    where [itex]\kappa[/itex] is some constant. Essentially the Feynman diagram looks like a standard weak vertex between two neutrinos. The only difference is that one of them is Heavy (GeV-TeV) scale. However my question really is about whether I just take the standard vertex and include the [itex]\epsilon^{*\mu}[/itex] vector for the outgoing Z-boson.

    The standard vertex factor reads
    [itex]
    -i\frac{g}{2\cos\theta_{W}}\kappa\gamma^{\mu}P_{L}
    [/itex]

    Is it enough for me to just add [itex]\epsilon^{\mu *}[/itex] for the outgoing Z boson?

    Thank you to anyone who can help.
     
    Last edited: Dec 1, 2011
  2. jcsd
  3. Dec 1, 2011 #2
    i think the part of z (ϵμ∗ )boson is correct but there is a problem in vertex. Because sterile neutrino does not interact in weak interaction so it will not couple to z boson. so you need to do something different here. You can check Mahapotro's " Massive neutrinos in physics and astrophysics " . check out the chapter on Left right symmetric model. there check sections like FCNC and heavy neutrino decay. I think you can find your answer there. So check if possible.
     
  4. Dec 1, 2011 #3

    Vanadium 50

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    I think that's right - if it's sterile, the vertex is 0, right?

    More realistically, you will have to have to define your couplings and decide how sterile is sterile.
     
  5. Dec 3, 2011 #4
    Vanadium, in my experience, sterile neutrino models (well, interesting ones at least) generally have small mixings between the sterile and active neutrinos. I think that's what Hypnofunk's [itex]\kappa[/itex] is intended to quantify.
     
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