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An UHECR neutron (kinetic energy 10^19 eV) approaches a magnetar (B = 10^10 T)

  1. Oct 13, 2007 #1
    What happens to the neutron? :smile:
  2. jcsd
  3. Oct 14, 2007 #2


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    Never heard of UHECR neutrons, how can they be accelerated to those velocities?

    And that magnet, pretty strong ^^

    I Guess it would have its trajectory sligtly bended due to its (very)small magnetic moment.
  4. Oct 14, 2007 #3
    Well, just assume that a UHECR proton interacted with a neutrino giving rise to the neutron.

    If you picture the neutron naively as bag containing three charges, what will be the stress the bag will experience? How strong is the bag? What will happen if it breaks? :smile:
  5. Oct 14, 2007 #4


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    So the fundemental question you ask is you can separate the nucleons by making them bend in a magnetic field? I dont think that there exist a magnet (or we will be able to construct one) that is powerful enough to split nucleons. There is much easier just to bombard nucleons with other particles in accelerators, and that has been done centuries ago.
  6. Oct 14, 2007 #5


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    We certainly don't see B's of 1010 on earth. The best static field is about 45T, and one can produce a transient field ~1000 T but only for microseconds.

    Ref - http://solomon.as.utexas.edu/~duncan/magnetar.html [Broken]

    See also - http://www.magnet.fsu.edu/

    This might be of interest - Physics in Ultra-strong Magnetic Fields

    I haven't read it yet.
    Last edited by a moderator: May 3, 2017
  7. Oct 14, 2007 #6
  8. Oct 16, 2007 #7
    As the cited paper points out, the mechanism by which the neutron or proton is pulled apart will result in the traditional effect anyway; the process of separating the differently charged quarks leads to the production of quark-antiquark pairs inside the baryon, and hence the strain between quarks is alleviated by the formation of a resultant baryon and meson, with the meson and baryon having opposite charges and seperating in the magnetic field. This does not defeat confinement; it just shows that confinement is strong enough to prevent quarks from becoming completely free, even when the standard strength of bonds between color charges is overwhelmed by the strength of extreme electromagnetic forces acting on electric charges.

    What is interesting, though, about this idea is that it will definitely favor charged meson production over neutral meson production. For example, the proton contains two up quarks with Q = +2/3 and a down quark with Q = -1/3. Upon entering field T, the up quarks will clearly be pulled in the opposite direction from the down quark, with twice the force in fact. When the up quarks stretch away from the down quark, eventually the force of confinement will build up enough energy for spontaneous pair production, resulting in a quark-antiquark pair. With the production of ddbar and uubar equally probable, the occurance of ddbar production will actually NOT result in meson production; the two down quarks (negatively charged) cannot separate from the two up quarks and down antiquark (positively charged). Thus, only uubar production will result in meson emission;

    uud --> uud + uubar --> uuu + dubar

    which is the same as;

    p --> Delta++ pi-

    In order for this process to go forward, there must be sufficient energy contributed by the magnetic field to impart enough mass to support the formation of the spin-3/2 Delta++ baryon and an additional spin-0 negative pion. This means that 433 MeV in additional binding energy must be applied to the valence quarks of the proton for this process to proceed. For a neutron, the process is similar: the two down quarks (negatively charged) and the up quark (positively charged) will be joined by a ddbar pair in order to proceed to fruition. The uubar pair production will not allow the reaction to proceed. Thus;

    udd --> udd + ddbar --> ddd + udbar

    which is equivalent to;

    n --> Delta- pi+

    Similarly, about 433 MeV in extra binding energy is required. In either case, the meson emission will proceed at half the rate calculated for normal light quark flavor symmetry.

    Although... on second thought, uubar production in a neutron could result in a viable reaction just the same;

    udd --> udd + uubar --> uud + dubar

    which would describe;

    n --> Delta+ pi-

    The Delta+ pi- final state requires the original 433 MeV over and above the initial state. Since protons will be readily accelerated, while neutrons are not, we are likely to see protons in the initial state much more often than neutrons. The result is that the reaction p --> Delta++ pi- is produced the majority of the time. It will require an energy that can only be produced by the proton constituents in a 6x10^15 Tesla magnetic field. The neutron initial state can transition in a 9x10^15 Tesla magnetic field. Delta+ pi- and Delta- pi+ final states from initial neutrons should be equal in occurance due to light quark flavor symmetry, which is not broken in the case of an initial neutron. Neutron decay into Delta baryons by pion emission will be equal to the rate calculated for light quark symmetry, with 50% split between the Delta+ pi- and Delta- pi+ modes each.
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