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How does a neutral particle interact with a photon?

  1. May 16, 2012 #1
    What happens if a photon is collided with a neutral particle? Can it also absorb the photon’s energy like a charged particle, can the photon be reflected, or does the photon just pass through the neutral particle?
  2. jcsd
  3. May 16, 2012 #2
    The only neutral elementary particles are the neutrinos from the fermions, and the Z0 vector boson and the gluons from the bosons. None interact with the photon directly (at tree level)

    All other neutral particles (e.g. neutrons, pions, atoms, etc.) have charged sub-particles in their composition, and, thus interact with photons.

    We have to mention that, at elementary particle level, all the interactions are scattering. Namely, an elementary particle cannot absorb a photon, and remain the same. Composite particles that have excited bound states, may absorb a photon, and transit to an excited state. As they de-excite, they re-emit an incoherent photon.
  4. May 16, 2012 #3
    So what exactly happens when a photon is collided with a neutral elementary particle?
  5. May 16, 2012 #4


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    To first approximation, the photon just passes through. Nothing happens. (I'm taking Dickfore's word that there are no tree level interactions between photons and neutral elementary particles)
  6. May 16, 2012 #5
    Do you mean by "incoherent photon" that there is no correlation between the photon's momentum vector (or spin) and the parent particle's momentum vector (or spin)?
  7. May 16, 2012 #6


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    Not to forget the photon on that list.
    In fact maybe the best studied interaction of a photon with a neutral particle is the interaction with another photon to produce e.g. electron positron pairs.
  8. May 16, 2012 #7
    Right. This is a tree level process, but a second order in the coupling [itex]e[/itex], whose transition amplitude is proportional to the fine-structure constant [itex]\alpha = \frac{e^2}{4 \pi}[/itex] (in natural units, and Lorentz-Heaviside units of electrodynamics), and the cross section is proportional to [itex]\alpha^2[/itex].

    In fact, Srednicki's QFT section 59 has the transition amplitude for the reverse process:
    e^{+} \, e^{-} \rightarrow \gamma \, \gamma
    One needs to multiply by the Lorentz Invariant phase-space (LIPS) volume of the final states (See section 11 of the said book) to obtain a differential cross section.

    Notice that this is quite different from a single photon pair production near a nucleus. Namely, this latter process is already accounted for in what I called scattering of a photon from a composite object (nucleus).
  9. May 16, 2012 #8
    So if two photons with enough energy collide, then they can be converted into the electron-positron pair. Can the electron-positron pair be used as a substitute for a hydrogen atom, since the positron has just as much charge as a proton? And can radiation from outer space be captured and individual pairs of photons collided to produce plenty of electron-positron pairs that will then be used as substitute hydrogen atoms?
  10. May 16, 2012 #9


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    Bound states of electrons and positrons are called positroniums and behave somewhat similar to hydrogen atoms although they decay within some ms again into two gamma quants. So it is difficult to store them.
  11. May 16, 2012 #10
    Note that there is a threshold energy of this reaction:
    \sqrt{s} \ge 2 m_e c^2 = 1.02 \, \mathrm{MeV}
    Thus, the two photons should be γ-rays.
  12. May 16, 2012 #11


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    Bound electron-positron pairs are called positronium, and have been extensively studied.


    I've never heard of positronium being produced by pair production from two photons. It seems to me that this would be an extremely delicate operation, because if the photons have too much energy, the e+ and e- will simply fly apart. The difference in energy between ground-state and "ionized" positronium is only 6.8 eV, compared to the 1.02 MeV (1020000 eV) that is necessary simply to produce the masses of the two particles.
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