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Neutrinos: why uninteractive?

  1. May 17, 2005 #1
    If i remember correctly from astrophysics. Neutrinos are said to be relatively uninteractive. Now is that statement an theory or a fact?
    whereby "fact" i mean that we see it pass through without interacting
    and "theory" we observe the effects of 1 neutrino going into a system
    and coming out of a system and we assume its the same one.
  2. jcsd
  3. May 17, 2005 #2

    I don't know if I understand your question very well, but here is a shot. Neutrinos only interact through the weak force. They have cross sections of the order 10^-30 cm^2 even at high energies (see http://pdg.lbl.gov/ for more info). This very small cross section indicates that they do not interact very easily with ordinary matter. This has been seen experimentally in supernova explosions in which the neutrinos hit the detectors hours before the photons did (there was a thread around here about this just the other week). There is nothing weird about that happening when you understand that neutrinos did not get scattered (very often) once they were produced but the photons which interact via electrodynamics, are rescattered many times in matter. So both experimentally and theoretically the weakly interacting neutrino is understood.
    hope this helps.
  4. May 17, 2005 #3
    I'm quoting from memory (so may be wrong) but about 250 000 neutrinos pass through every cm^2 of the face of the earth facing the Sun every second! These pass straight through and out the other side. After neutrinos were theorised as an entity (to help explain the apparent energy and momentum loss in Beta decay) scientists spent about 30 years trying to detect neutrinos before they managed to finally accomplish this - this was despite the huge number of neutrinos coming from the Sun.

    That gives some idea of how unreactive with matter they are. I read somewhere that the average neutrino could pass through several light years of solid lead without interacting!!

    Do a Google search and read about the discovery of the neutrino - it certainly took some doing!!
  5. May 17, 2005 #4


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    Negligible mass. No charge. No strong interactions.

    Neutrino detectors are typically buried very deep in the ground to filter out alternative interference. We know how many neutrios are produced in nuclear reactions from experiment. We can estimate how many of those reactions take place in the sun based on the aggregate observations of light from the sun. We can observe how many interactions there are between the heavy water in a neutrio dectector and neutrios because the interactions produce a burst of light which photomultipliers can catch. Interactivity is a function of observed detections and the number of neurtinos known to be headed our way from the sun.
  6. May 23, 2005 #5

    Meir Achuz

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    In the Glashow-Salam-Weinberg theory of electro-weak interaction, the "charge" for "weak" interaction of a neutrino is identical to the charge for EM interactions.
    The neutrino appears to interact weakly because its effective charge squared in beta decay or low energy scattering is alpha*(M_p/M_W)^2, which is quite small.
    In high (very high) energy scattering neutrinos interact as strongly as electrons.
  7. May 24, 2005 #6


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    Dear Meir Achuz,

    > alpha*(M_p/M_W)^2

    What's M_p in this formula? The proton mass? If so, then why does this come into the effective charge?

    Also, I thought there was a small difference between the magnitude of electric charges and weak charges, something like sqrt(3).

  8. May 24, 2005 #7

    Meir Achuz

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    M_p is the mass of the proton. I will have to go into a bit more detail.
    The difference in EM and Weak interactions (WI) is in the mass of the exchanged particle. For WI, it is the W or Z particle, whilch have masses of 80 and 90 GeV.
    This puts a factor like Q^2/(M_W^2+Q^2) into the interaction (in momentum space).
    Q^2 is the momentum transfer in a scattering process.
    For EM, the exchanged particle is the massless photon.
    The factor is of order Q'^2/Q^2~1. The two Q^2 are different, but of the same order of magnitiude. At low Q^2, the WI has the dependence Q^2/M^2.
    Fermi's theory of beta decay was developed 50 years before the W and Z were known, so he used Q^2/M_p^2 in defining what became known as the "Fermi coupling constant" of WI. This is the origin of the factor (M_p/M_W)^2.
    Factors like sqrt(3) or sometimes sqrt(4pi) come from varying conventions.
    They don't affect the order of magnitude. The absolulte equality of the fundamental dimensionless WI and EM couplings is important in the GSW theory, because it leads to a cancellation of some high energy divergences. This was one of the original motivations.
    Now, I have written a chapter.
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