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Conservation of momentum and HUP in PET

  1. Sep 8, 2005 #1
    "When a positron emitted in beta+ decay annihilates with an electron, the combined mass of the two particles, m = 2me, is converted into an amount of energy 2mec^2 in the form of two 511 keV gamma rays. Momentum conservation requires that the two gamma ray photons are emitted in opposite directions from the point of annihilation. Positron emission tomography exploits the fact that the detection of two oppositely-directed gamma rays defines a line on which the point of annihilation occurs."

    I haven't put this in college homework because this was just a side note in my textbook and is not something I imagine I will get asked about. As such, I imagine there are some details skimmed over. What these are will hopefully get cleared up with answers to the following questions. I'm an undergrad, and we've only studied QM to a relatively low level, so please don't baffle me with stuff there's no chance I'll understand. :cry:

    1. If the total momentum of the photons MUST be zero, then the total momentum of the electron-positron pair must also be zero. Why do such pairs only annihilate when they have equal and opposite momenta? What would happen if an electron and positron met with unequal momenta, either in magnitude or direction?

    2. If the total momentum of the pair must be equal and opposite, and we can determine the position of annihilation from the trajectories of the photons, where is the uncertainty?

    3. Presupposing an answer to the last question, I assume that proximity, and so precise position, is not an issue for pair annihilation, and that maybe the pair will annihilate if they are in the same quantum state. If so, what is the meaning of the point of annihilation?

    Thanks to all in advance.
  2. jcsd
  3. Sep 8, 2005 #2


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    No, this simply means that you should (for your own easiness) work in the center of gravity of the the electron-positron pair. You can always find such a center of gravity reference frame, and all that is written applies in THAT COG.

    Your detector is not necessarily in that COG frame ! The uncertainty resides in the uncertainty in what is the COG frame. You could only work it backwards from your photon detections if you knew perfectly the momentum AND impact positions ; but you can't because of the HUP, so you have an intrinsic uncertainty in this calculating back of the COG frame.

    See answer to 2.

  4. Sep 8, 2005 #3
    Hi Patrick. Thanks for the response.

    If I follow you correctly (doubtful :smile: ), then what you're saying is that in a reference frame chosen such that the centre of gravity of the electron-positron pair is at rest, the momenta of each will be equal and opposite. That, I dig.

    But momentum is conserved within a frame of reference. Surely then if the apparatus is set up such that the COG is not at rest in its frame (and this is by far the most likely situation), then the total momenta of the pair will not be zero. As such, the total momenta of the photon pair should not be zero.

    The above looks like an argument but it's really a question. I can follow how you can choose your frame such that the COG is at rest, but the only frame of any import I see is the frame of the apparatus.
  5. Sep 8, 2005 #4


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    That's correct. And they will thus not be back-to-back either.

    Exactly, and in that frame of the apparatus (the lab frame, say) the momenta are not exactly back-to-back.
  6. Sep 8, 2005 #5
    Ahhh, I see. So what you're saying is the momenta of the photon pair are not likely to be exactly equal and opposite, but close enough to pin the point of annihilation down with some uncertainty?
  7. Sep 8, 2005 #6
    It looks like you've read a little too deeply into the question and inserted things that aren't there.

    Total momentum doesn't need to be 0...it just needs to be conserved. Positrons carry some amount of kinetic energy when emitted. Usually by the time a positron interacts with an electron, much of it's initial energy has been lost through collisions and interactions and it ends up with 0 or very little energy. The annihilation photons get emitted 180° to each other.

    When the positron annihilates with an electron before losing a substantial fraction of it's energy, then conservation of momentum still requires the annihilation photons be emitted in opposite directions, but the angle becomes less than 180° Whatever momentum the positron had prior to annihilation is converted to the photons' momentum.
  8. Sep 8, 2005 #7
    imabug, ur2l8. Patrick had it covered, thanks, and with less dubious presumptions. My suspicion that the text was oversimplifying for succintness was correct, but the issue was confused by a poster on an earlier thread I've been on stating that the total momentum of an annihilating pair had to be zero. I thought that was wrong, but when I read the quoted text I started doubting my sanity.

    Anyway, I'm done here. Thanks again for the help. :smile:
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