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Momentum increased in electron-positron annihilation?

  1. Mar 24, 2013 #1
    1. The problem statement, all variables and given/known data

    An electron and a positron is on a collision course. Both have a speed of 1,80*10^8 m/s.
    What are the frequencies of the two photons created after the electron and the positron has annihilated?

    Electron mass=positron mass= 9,11*10^-31 kg

    2. Relevant equations


    3. The attempt at a solution

    When I first tried solving this problem using the fact that momentum is conserved before and after the collision, I got the wrong answer:

    Pbefore=γmv=4,10*10^-22 kgm/s


    Pafter=2*(hf/c) => f=(c*Pbefore)/(2h)=9,27*10^19 Hz.
    This is wrong according to my textbook. But when I do the same calculation, using that energy is conserved, I get that the frequency of each photon is 1,55*10^20, which is correct according to the textbook.

    This is weird because my textbook says that momentum and the total energy of an isolated system is always conserved.

    When I calculate the momentum of the system after the collision, using that each photon has a frequency of 1,55*10^20 Hz, I get:

    Pafter= 2(hf)/c = 6,85*10^-22 kgm/s. Doesn't that mean that the momentum of the system has INCREASED? What am I doing wrong?
  2. jcsd
  3. Mar 24, 2013 #2


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    The momentum of the system depends on the direction of electron and positron. If they collide head-on (and I think you are supposed to assume this here), the total momentum is zero.
    The photons will be produced back to back with the same frequency, so the total momentum afterwards is zero, too.
  4. Mar 25, 2013 #3
    Yes they do collide head-on. And the Photons afterwards have opposite directions. But still the absolute value of the momentum has increased after the collision. Isn't this against the laws of physics?
  5. Mar 25, 2013 #4


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    The momentum is 0, and the absolute value of 0 is 0 as well.
    You sum the absolute momenta of individual particles - but that is not a conserved quantity in physics.
  6. Mar 25, 2013 #5


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    No, it isn't. Only the total momentum of the system is conserved, not the individual momenta.
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