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Electron/positron annihilation/momentum

  1. Jun 7, 2009 #1
    if an orbiting electron positron pair is moving through space with net momentum p and the 2 particles then annihilate each other producing 2 photons then where did the momentum go?
     
  2. jcsd
  3. Jun 8, 2009 #2
    Into the photons right? They have energy E=pc.
     
  4. Jun 8, 2009 #3
    well, theres nowhere else for it to go. but I cant really say that I understand it.
     
  5. Jun 8, 2009 #4
    What specifically do you not understand?
     
  6. Jun 8, 2009 #5
    Photons have momentum just as massive particles do. Their momentum however is not given by mv as usual (which would be zero), but rather by hf/c.
     
  7. Jun 8, 2009 #6
    how light can carry so much momentum. I always understood that light carried energy but no significant momentum. its often used as a reason for why certain transitions are forbidden.
     
    Last edited: Jun 8, 2009
  8. Jun 8, 2009 #7
    What makes you think photons carry much momentum? Simply looking at the equation I gave you should give you some idea of the magnitude of the momentum:
    [tex]p = \frac{hf}{c} = \frac{h}{\lambda}[/tex]

    Since h is of the order of [itex]10^{-34}[/itex] the wavelength [itex]\lambda[/itex] is of the order of [itex]10^{-7}[/itex] (for visible light), the momentum is of the order of [itex]10^{-27}[/itex] kg m / s, tiny!
     
  9. Jun 8, 2009 #8
    then where does the net momentum of the pair go? (see post #1)
     
  10. Jun 8, 2009 #9

    Vanadium 50

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    Be careful equating "photons" with "light". These photons are - at minimum - a quarter of a billion times as energetic as optical photons.

    As pointed out before, the photons carry away energy and momentum.
     
  11. Jun 8, 2009 #10

    Nabeshin

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    If you're confused, just look at it in the zero-momentum frame (where the orbiting pair is stationary). When the annihilate, momentum will be carried off by the photons produced. In a non-zero-momentum frame, the same is true, but in a way so that momentum is conserved in this frame also.
     
  12. Jun 8, 2009 #11
  13. Jun 8, 2009 #12

    Vanadium 50

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    That thread is about a completely different process.
     
  14. Jun 8, 2009 #13

    The electron and positron have very tiny masses. Since p=mv, we can say (9.109 x 10^-31) x velocity (let's just say 200,000,000 m/s) = the momentum of an electron. So an electron and a positron pair should have a net momentum of around 2 x (the above stated) = 3.6436 x 10^-22 kg m/s in this example. This is a very small momentum. It should be easy for the photons emitted to carry this momentum.
     
  15. Jun 8, 2009 #14
    small is a relative term.
     
    Last edited: Jun 8, 2009
  16. Jun 8, 2009 #15
    But I gave you an exact number to compensate for that.
     
  17. Jun 8, 2009 #16
    I dont know what you are referring to but I was referring to a pair moving through space with an arbitrary net momentum p (which could be as large as you want)
     
  18. Jun 8, 2009 #17
    yes. its about the reverse process
     
  19. Jun 8, 2009 #18
    So was I. You can't just pick a number like 3 and say it is the net momentum of the particle pair. Momentum depends upon mass, and the elecron and positron have very very small masses (9.109 x 10^-31). And the only way to compensate is from the velocity, which won't be enough to raise the momentum to any significant value. If an electron hits you, do you ever even feel it? No, because the momentum is not significant enough.
     
  20. Jun 8, 2009 #19
    What exactly is bothering you about this?

    For the reverse process, momentum is still conserved. A single photon cannot spontaneously errupt into an e-p pair; it has to do something else. I had no clue what, but in the thread you linked to it was said that it has to interact with a nucleus. So conservation of momentum still applies, in this process, and in the reverse (and every other possible process).



    I have been thinking myself and I have a question too (I'm probably over-thinking this):
    If you have a 'simple' annihilation, one electron moving to the right, and one positron to the left, annihilating in the center, then the net momentum is zero. For that reason, it is impossible that less than two photons are created, as only one photon would not be able to have zero momentum and thus conserve momentum.

    However, assuming the electron and positron already have momentum (like in the OP's question, they could be orbiting around each other, and their center of mass could be moving relative to some observer, no?), then in my reasoning it is perfectly allowed to have only one photon be created, since that can conserve momentum.

    What if we have this process, but then decide to co-move with the e-p pair? Net momentum, relative to us, would be zero, but only one photon would be emitted. What happens here? Where does relativity save us? :p
     
  21. Jun 8, 2009 #20
    Relativistic momentum is defined by

    [tex] p = \gamma m_0 v [/tex] where p = momentum, [tex] \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} [/tex] [tex] m_0 = [/tex] rest mass and v = velocity.

    Stick in any v you want < c, because particles with mass cannot travel at the speed of light. Let's do 0.99999999999c since you're seemingly hooked on superlatives.

    This would then mean that [tex] p = \frac{m_e * 0.99999999999c}{\sqrt{1-\frac{(0.99999999999c)^2}{c^2}}} = 6.10651447*10^{-17} kg m/s = \frac{h}{\lambda}[/tex]

    So a photon produced in this annihilation for that v, would have a wavelength = 1.085x10-8 nm which is extremely high gamma radiation. In fact this photon has energy of 114.26 GeV

    Satisfied?
     
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