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How Do They Create a Neutrino Beam?

  1. Nov 25, 2009 #1
    I've been reading about the T2K experiment where a "neutrino beam" is being sent from the J-PARC accelertor in Takio to the Super-Kamiokande detector in Japan.

    So if neutrinos only interact weakly, how do they create a beam of them?

    At J-PARC they create them by firing high-energy protons at a graphite target. You can't focus them magnetically or electrostatically, so how do they focus them into a beam?

    I've searched around and there seems to be something to do with target geometry in the production, but I haven't seen anything on how this works. Anyone know or have an ideas where I might look?

    I can see how they could create a cone because of conservation of momentum, but a beam?
     
    Last edited: Nov 25, 2009
  2. jcsd
  3. Nov 25, 2009 #2

    mgb_phys

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    Momentum is still conserved, if you fire protons into a target most of the neutrinos are going to come out of the other side in the same direction.
     
  4. Nov 25, 2009 #3
    Neutrino beams are produced by blasting a target with high energy protons. Neutrinos are not created in strong interactions, but only by the decay of short-lived strongly interacting particles; pions and kaons. The secondary charged pions and kaons are focused with a strong sign-selecting horn (solenoid), letting the pions and kaons decay in a long evacuated pipe (Lorentz transformed decay length is β γ c τ), and dumping all remaining particles in a thick target at end.( It takes hundreds of meters to stop muons). Solenoid pre-selects neutrinos or antineutrinos by selecting positive or negative pions and kaons..
    Bob S
     
  5. Nov 25, 2009 #4
    Thanks for the answers Bob_S and mgb_phys.

    - Curtis
     
  6. Nov 26, 2009 #5
    Here are some numbers regarding producing a muon neutrino beam from a 10-GeV pion decay. The mass of the charged pion is about 139.58 MeV. The muon mass is about 105 MeV. The difference, about 34.7 MeV in the center of mass, goes into the neutrino and muon kinetic energies. The decay in the rest rame is isotropic. In the lab frame, the decay products are in a forward cone of half-angle about 1/γ.

    If we use 10 GeV for the pion energy,
    γ = 71.6
    β =~1
    c = 3 x 108 m/s
    τ =~25 nsec pion decay lifetime

    So the pion decay length in the evacuated pipe is

    β γ c τ = 537 meters

    Bob S
     
  7. Nov 27, 2009 #6
    Thanks Bob S,

    I had rather thought it would be a cone rather than a narrow beam because there would have to be some component of the energy from decay that is not aligned with the original pion momentum vector.

    - Curtis
     
  8. Nov 27, 2009 #7

    jtbell

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    Yes, there is some transverse momentum imparted to the muon and the neutrino in the decay, but this remains constant (on the average) as the momentum of the pion increases.

    The easiest way to see this is to consider the pion decay first in the pion's rest frame. There are two outgoing particles, so the magnitudes of the momentum of both muon and neutrino are fixed by energy and momentum conservation. However, the directions are random.

    Now, transform the situation into the lab frame so that the pion initially has the desired (very large) momentum. The momentum components of the neutrino and the muon along the direction of the pion momentum are Lorentz-boosted, but the transverse components are not.
     
  9. Nov 27, 2009 #8
    You are correct. The pions, kaons, etc. emanating from the production target are in a forward cone of half-angle roughly 1/γ of the particle momentum. There is a magnetic cone-shaped solenoid with pulsed current just downstream of the production target to preferentially bend one sign (of charge) of particle into a parallel beam (and deflect the other sign). This roughly aligns the charged particles with the vacuum pipe, which is ~1 meter diameter. As pointed out by jtbell, the forward momentum of the decaying particles is much larger than the transverse momentum of the decay products, so the decay products are also in a forward cone..
    Bob S
     
  10. Nov 27, 2009 #9
    So for the 10 GeV pion example given earlier ([tex]\gamma \approx 70[/tex]), the half-angle of the cone is a little under one degree. Wikipedia http://en.wikipedia.org/wiki/T2K#Off-axis" (without specific citation, of course) the interesting tidbit that Super-K is more than two degrees off the axis of this cone. So (assuming that's accurate), the neutrinos that arrive there will come from pions with lower energy than 10 GeV (smaller γ, wider angle).
     
    Last edited by a moderator: Apr 24, 2017
  11. Nov 27, 2009 #10
    Here is the reason for misalignment (from Wiki):
    The neutrino beam at J-PARC is designed so that it is directed 2 to 3 degrees away from the Super-K detector. This lowers the flux of neutrinos reaching the detector but provides a more desirable neutrino energy spectrum. Off-axis the peak energy is low with higher energies being suppressed. At the distance between Tokai and Kamioka, maximal neutrino oscillation is expected to occur at energies lower than 1 GeV.
    Bob S
     
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