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Charge Conservation in Particle Collisions

  1. Apr 13, 2015 #1
    I did a quick search through the forums and didn't find the answers I was looking for, so I thought I'd ask. Does charge conservation still apply for collisions between elementary particles? I'm taking a second year foundations of physics course, and we were given a fairly simple looking reaction in lecture: (hopefully I can get tex to work this time)

    $$
    p \rightarrow n + \mu^{-} + \bar{\nu}_\mu
    $$

    and told that this decay is possible since baryon number and lepton number are conserved. Now as far as I can tell, this violates conservation of charge. Am I missing something, or did my prof goof when making the example?

    Thanks!
     
  2. jcsd
  3. Apr 13, 2015 #2

    e.bar.goum

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    Charge conservation absolutely applies to elementary particle collisions. I reckon your prof just made a mistake. If you change that muon to an antimuon, and the muon anti-neutrino to a muon neutrino, it'd be ok.
     
  4. Apr 14, 2015 #3
    Since neutrons are 0.14% heavier than protons, and muons have about 11% the mass of protons, this decay violates the conservation of energy, too
     
  5. Apr 14, 2015 #4

    vanhees71

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    The correct ##\beta##-decay formula is
    $$n \rightarrow p + e^- + \overline{\nu_e}.$$
    Proton decay is impossible due to to energy conservation, because the proton mass is a bit smaller than the neutron mass.

    Electric charge must be always conserved, because otherwise the so far very successful standard model of elementare particle physics wouldn't make any sense anymore, because charge-non conservation would violate the fundamental electromagnetic gauge symmetry.
     
  6. Apr 14, 2015 #5

    e.bar.goum

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    You're allowed to have ##\beta##-plus decay, inside a nucleus, of course.

    $$p \rightarrow n + e^+ + \nu_e$$

    (Though it's better to write ##^A_Z X \rightarrow ^{A} _{Z-1} X' + e^+ + \nu_e##)
     
  7. Apr 14, 2015 #6

    ChrisVer

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    or it's better to write:
    [itex]p (+N) \rightarrow n + e^+ +\nu_e (+N)[/itex]
    where [itex]N[/itex] stands for nucleon and tells you that this process is possible within an environment of nuclei.

    As for the energy violation, it doesn't apply for the case of a decaying proton to neutron, since you already consider this process in a nucleus environment. It wouldn't be the same for:
    [itex]n \rightarrow p + \mu^- + \bar{\nu}_\mu[/itex]
    for free-neutrons, since this would violate the energy conservation.
     
  8. Apr 14, 2015 #7
    I just want to say I'm pretty sure there doesn't exist a Z→Z±1 nuclear transition with enough energy to produce a muon (106 MeV).
     
  9. Apr 14, 2015 #8

    mfb

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    Every macroscopic conservation law (apart from statistical properties) has to apply on a fundamental level as well, otherwise there would be a way to violate it.
    Yes, charge is conserved in particle physics.
     
  10. Apr 14, 2015 #9

    e.bar.goum

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    I'm sure you're right.
     
  11. Apr 14, 2015 #10
    Charge conservation is obeyed in collisions, but while energy is conserved, the energy can be supplied by kinetic energy of colliding particles.

    What can come from proton-proton collisions?
    Reaction
    p+p->d+e++nue
    would produce energy, but has never been observed.
    Reaction
    p+p->p+p+hnu
    loses energy - yet it is standard braking radiation.
    But how likely are the following reactions:
    p+p->p+p+e-+e+?
    p+p->p+n+e++nue?
    p+p->p+n+mu++numu?
    p+p->d+mu++numu?
     
  12. Apr 14, 2015 #11

    ChrisVer

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    However you can still have:
    [itex] \nu_\mu p \rightarrow n \mu[/itex].
     
  13. Apr 14, 2015 #12

    mfb

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    Indirectly, the sun relies on that reaction and the neutrinos from the reaction have been detected recently.

    pp -> ppee looks like a reaction with reasonable probability, I'm sure there is some old paper discussing it. The other reactions all need weak production modes, I guess they are quite unlikely (and might be hard to detect as well).
     
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