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Pair production and another body?

  1. Oct 30, 2006 #1
    Why does pair production cannot occur except in the presence of another body? The explanation is so that linear momentum is conserved but why does a photon must first collide with something? I can see how linear momentum is conserved without a collision first. i.e. the two particles move with equal magnitude of momentum in y axis but in oppsite directions. Their horizontal momentum is in same direction and add up to the photon's original momentum and it is all conserved.
  2. jcsd
  3. Oct 30, 2006 #2


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    Consider the pair-production process in the inertial reference frame in which the outgoing electron and positron have zero total momentum (i.e. they come out in opposite directions with equal magnitudes of momentum). It is always possible to find such an IRF.

    If no other body participates in the pair-production process, then by conservation of total momentum, the original photon must also have zero momentum, in that IRF. But there's no such thing as a photon with zero momentum (and therefore zero energy). Therefore some other body must participate in the process.
  4. Oct 30, 2006 #3

    This is clearly explained there:

    Pair Production in Empty Space?

    - assume now photon -> electron + positron -

    Without a third body, the maximum momentum tranferred to the produced pair cannot impart enough energy to absorb all the energy available from the photon.

    However, if a third body could by an interation absorb this excess of energy, then the process becomes possible. It is funny that the needed interaction could only take place between the procuced pair and a third charged particle. This mean: after the pair is created. This must again be some magic of QM ?

  5. Oct 30, 2006 #4
    So the basic idea is that if linear momentum is conserved than it is impossible to transfer all of it to the two particles (when considering the most efficient transfer and even allowing the particles to travel at c as shown on the slides) in a vaccum.
  6. Oct 30, 2006 #5
    ...from/to a single photon.
  7. Oct 30, 2006 #6
    from a single photon is true.

    to a single photon would mean that initially there must be more than the positron and electron being ready to turn into a photon. There must be some other matter with momentum in the same direction (this is to make up for the extra momentum needed for conservation unlike the first case where some initial momentum needs to be given away).

    But if the two particles annilate to form two photons than the two photons could cancel some momentum off so as to reduce the two final momentum to a level of initial total momentum.
    Last edited: Oct 30, 2006
  8. Oct 30, 2006 #7
    Said in the pair's center of mass ref. frame: the pair momentum is zero and the total momentum of the photons is zero as well.
  9. Nov 2, 2006 #8
    How can the total momentum of two photons be zero? In any frame, the photons will still be photons, meaning they travel at c. So given that the ref frame of the particles is not travelling at c, for the photons to have zero momentum in that frame, one of the photons must have negative energy which implies negative momentum. How can that be?

    Or is it the case that one of the photons is travelling in the negative x direction so one of the photons can have negative momentum. In that case, the formula for momentum for a photon is E/c. Looks like c in here has a direction as well so can be negative.
    Last edited: Nov 2, 2006
  10. Nov 2, 2006 #9
    I have an alternative answer to the original question in this thread.

    Suppose a photon could turn into two particles by itself. Lets jump into a frame where the momentum of the two particles is 0. This frame will travel at v<c. Now let the pair production begin in this frame. But in this frame, the photon won't have 0 momentum since it can only have 0 momentum in a frame that is traveling at c. So momentum is not conserved. Therefore a photon cannot turn into two particles by itself.
  11. Nov 2, 2006 #10


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    Momentum is a vector. Energy is a scalar.

    Two photons with equal energy have equal magnitude of momentum. If they also travel in opposite directions, the momentum vectors are in opposite directions. Therefore the total (vector) momentum is zero, whereas the total energy is twice the energy of either of them.
  12. Nov 2, 2006 #11


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    Which is basically the same thing I said in post #2. :biggrin:

    Now that I think of it, after the other post I just made, I can sharpen this argument a bit.

    Suppose a photon could produce a pair of massive particles by itself. Switch to the reference frame in which the outgoing particles have a total momentum of zero. in that frame, the original photon must have a momentum of zero.

    However, the original photon must also have an energy equal to the total energy of the two outgoing particles, which must be nonzero. For a photon E = pc, so the momentum must be nonzero also. This contradicts the conclusion of the preceding paragraph.
  13. Nov 2, 2006 #12


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    Dearly Missed

    I've got to say that these "motivational" explanations bring the impossibility home to me much more strongly than an analytical proof does.
  14. Nov 2, 2006 #13
    I didn't read you first post carefully. But the proof I suggested and this one is almost identical. Except I used a reference argument, namely "[the photon] can only have 0 momentum in a frame that is traveling at c" because it will be stationary in that frame and any stationary object will have 0 momenum. It raises the interesting question of what would be like to be in a ref frame travelling at c. In that frame, the photon itself would have 0 energy (since momentum=0) so it cannot exist in that frame. Does that also imply that no other 'thing' can exist in that frame since matter and radiation are intimately linked. No radiation => no matter. Moreoever, observations cannot be made in that frame because of the impossibility of the existence of photons. It could also raise the possibility that the frame itself is the photon and so there is 'nothing' inside the photon.
    Last edited: Nov 3, 2006
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