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Matter/Antimatter Collision Radius

  1. Dec 31, 2009 #1
    I am curious, if a matter particle and it's corresponding antimatter particle meet, they will annihilate to produce pure energy. Which is the minimal distance between the particles that would allow the reaction to occur and how is that determined? (I ask this because I know that the nice idea of thinking of particles as spheres with boundaries is false, so then what is the definition of "collide" if you know what I mean?)

    Asked another way; If there are no true boundaries to a particle (and if I am mistaken, let me know) how can two "collide"? (not so much in an electromagnetic repulsion, but more specifically regarding matter/antimatter collisions)
     
  2. jcsd
  3. Dec 31, 2009 #2
    For fast positrons annihilating with electrons at rest, the annihilation-in-flight probability plot is shown on page 274 and 385 of Heitler "Quantum Theory of Radiation" 3rd edition. The cross section is maximum near 0.5 MeV. The cross section, derived in section 27 (page 268), is originally attributed to P.A. M. Dirac (1930). According to Heitler, "... in most cases, a fast positron will first lose all its energy and is then annihiliated [at rest]".
    For antiprotons annihilating on protons, see the plots for p-bar p and pp total cross sections in Fig. 40.11 in
    http://pdg.lbl.gov/2009/reviews/rpp2009-rev-cross-section-plots.pdf
    The total cross sections for p-bar p (anti-proton annihilation) and pp are nearly equal above 10 GeV, and between 40 and 100 millibarns (slightly above geometric; ~pi R2). Below 10 GeV, the p-bar p cross section rises above 300 millibarns.
    Bob S
     
  4. Dec 31, 2009 #3

    bcrowell

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    Bob S, I don't think the OP was asking about cross sections. I think the simple answer to his question is zero. The Feynman diagram for this process has the world-lines of the particle and antiparticle intersecting at a point. The reason you get a finite cross-section, even though the range of the effect is zero, is the Heisenberg uncertainty principle; each particle is a wave-packet that covers a certain amount of space.
     
  5. Dec 31, 2009 #4
    Why is the p-bar p interaction probability larger than the pp interaction probability at GeV energies, if the finite cross section is due only to the Heisenberg uncertainty principle?
    Bob S
     
  6. Dec 31, 2009 #5
    I am a little confused, if each particle is a wave-packet that covers a certain amount of space, how can the minimum distance of a particle/antiparticle collision be zero? Also, if this distance were zero, how could these reactions ever occur? If they have to be at no distance with each other wouldn't such a reaction have zero probability of occurring?
     
  7. Dec 31, 2009 #6

    diazona

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    Here's a rough way of thinking about it: when you think about particles as wavepackets, they don't really "collide." Rather, there is some probability that the two particles will disappear and be replaced with one or more new particles. This probability is related to the "overlap" (a term that I am intentionally using vaguely) of the two wavepackets. When the wavepackets are closer together, the probability of an interaction becomes larger.
     
  8. Dec 31, 2009 #7
    Ooh i see, that is cool. Also i realized I meant to ask what is the MAXimum distance between them, but after your explanation I guess it would be infinity, except the probability would be very low. (Am I right?)

    Thanks
     
  9. Dec 31, 2009 #8

    bcrowell

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    I didn't say that the H.u.p. was the only factor in determining the cross section.
     
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