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Photon-photon collision - pair creation

  1. Jun 2, 2017 #1
    1. The problem statement, all variables and given/known data
    Two photon (wavelengths 1pm, 2pm) collide, decaying into a pair positron-electron find the velocities of these particles
    Everything happens in the x axis


    2. Relevant equations
    The system
    hc(1*1012+0.5*1012)=E++E-
    h(1*1012-0.5*1012)=P+-P-
    E=sqrt(c2P2+m2c4)


    3. The attempt at a solution
    Well this yields a 2 page attempt at solving this system so i wont bother pasting it here. I have a feeling that like for finding the compton wave length, there is a specific path i should take that would make everything go smoother. I just cant find it
     
    Last edited: Jun 2, 2017
  2. jcsd
  3. Jun 2, 2017 #2

    vela

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    Your relevant equations are wrong. It usually helps to refrain from plugging numbers in right from the start.

    Try solving the problem in the center-of-mass frame and then transforming back to the lab frame.
     
    Last edited: Jun 2, 2017
  4. Jun 2, 2017 #3
    Ok so i understand the momentum part might be wrong because i dont know if the pair goes in opposite directions, I haven't plugged in any numbers... But how do i use a center of mass frame with photons?
     
  5. Jun 2, 2017 #4

    vela

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    Where did the 1*10^12 and 2*10^12 come from then?

    In the case of photons, you're looking for the case where the total momentum of the system is zero.
     
  6. Jun 2, 2017 #5
    Ah those are 1/λ the second one should be 0.5 and not 2.

    Im not sure if i covered CM collisions in class but il give it a shot.
    So i have to figure out the velocity frame in which the sum of momentums of the photons is 0 right? Then the velocities of the pair would be the same?
     
  7. Jun 2, 2017 #6

    vela

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    The electron and positron would move with the same speed but in opposite directions so that their momentum would sum to 0.
     
  8. Jun 2, 2017 #7
    So P1 is the momentum of photon 1, P2 photon 2, Pcm momentum of the CM; so
    (P1-Pcm)+(-P2-Pcm) =0?
    And then for the pair i substitute the new momentum of the photons in the equations and i get the energy of both photons and from the energy i get the velocity in the cm frame? Im confused
     
  9. Jun 2, 2017 #8
    Ok so here is what i did; i used the formula in my last post to find the momentum of the cm; then used that momentum to find the momentum of both photons in that frame, (same momentum opposite directions);
    And that momentum i transformed in energy so i get the same energy with same "signal" used the photon energy to find the energy of positron (electron is the same energy because of the CM frame) used that energy to calculate momentum;
    Subtracted/added momentum of the CM to find momentum of the lab frame;
    Used that momentum to find energy then velocity and got that both of them were at aprox. 0.65c
    Did i do this right? I can provide a pic of resolution if you want.
    Thank you!

    Edit: ok this seems right!! Momentum was conserved: original photon resulting momentum was 3.315*10^-22 and final pair was 3.2*10^-22 im checking energy conservation(disregard that velocity i wrote above i made an aproximation error im recalculating that also)

    Edit2: ok energy also conserved, inicial 2.98 *10^-13 j ; final 2.99*10^-13 j.
    Electron/positron velocities -0.9c and 0.66c

    Since everything was conserved im assuming my answer is correct. Thank you very much! Much easier this way!!!
     
    Last edited: Jun 2, 2017
  10. Jun 3, 2017 #9

    vela

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    The description of your calculations doesn't sound right, but your answers are pretty close to what I found, the differences perhaps due to rounding error.

    You shouldn't transform between reference frames by just adding and subtracting vectors, like you did with the momenta. You should use the Lorentz transformations. On the other hand, the speed of the center of mass is pretty low, only 1/3 c, so maybe it was a decent approximation.

    Recall Newtonian mechanics. The momentum of the center of mass is just the sum of the individual momenta of the particles making up the system. The same holds true in relativistic mechanics, so ##p_{cm} = p_1 - p_2##. It looks like you put in a factor of two.
     
  11. Jun 3, 2017 #10
    I found the momentum of the CM so that the momentums of the photons would be the same in the CM reference. I will redo the problem with that CM and compare results
    How would i use the lorentz transforms to find the momentum of the pair?
    What other parts of my resolution dont seem right to you?
     
  12. Jun 3, 2017 #11

    vela

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    You'd need to show your work. Verbal description are often too vague to clearly describe what you're doing. For example, you wrote, "I found the momentum of the CM so that the momentums of the photons would be the same in the CM reference." If you the calculation correctly, then great, but if you didn't, I don't know what you actually did.
     
  13. Jun 3, 2017 #12
    thank you for your patience, here is my work:
    for Pcm
    (P1-Pcm)+(-P2-Pcm)=0 ; this is "the momentum of the CM so that the momentums of the photons would be the same in the CM reference"
    then :
    P1+Pcm=P1cm; P2+Pcm=P2cm
    and so P1cm+P2cm=0.
    found the energies of the photons now E1=cP1cm and E2=E1 so:
    2E1=2Ep
    2Ep is the sum of the energy of the pair, since we are in the CM the velocities are equal.
    since E2=c2p2+m2c4
    found P of pair, and P+/-Pcm is the momentum of the pair turned this back into energy and used:
    E=(mc2)/(sqrt(1-(V2/C2)) to find each velocity
     
  14. Jun 3, 2017 #13

    vela

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    If a particle has energy E and momentum p in one frame, then in the other frame, its energy and momentum are
    \begin{align*}
    E' &= \gamma[E - (v/c)(pc)] \\
    p'c &= \gamma[pc - (v/c)E].
    \end{align*} where ##v## is the speed of the second frame relative to the first.
     
  15. Jun 4, 2017 #14
    And the speed of the second frame is Pcm/2me?
     
  16. Jun 4, 2017 #15
    doing this did not give correct results
    i got 0.9c and 0.72c this does not conserve energy...

    i got vcm =0.30c; i used the previously calculated energy and momentum, and used the formula you stated to calculate the new energy; γ[E−(v/c)pc]., with γ= 1/sqrt(1-v2/c2) E and p of the particle and v of the cm
    what am i doing wrong?
     
  17. Jun 4, 2017 #16

    vela

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    The speed's not right for one thing. Assume the photons have momentum ##p_1## and ##p_2## in the lab frame. Then calculate what their momenta would be in the CM frame, set those two expressions equal, and solve for ##v##.
     
  18. Jun 4, 2017 #17
    i dont get it
    photons have no mass how do i find "v" of the cm with photons? am i to use:
    γ(p1cmc−(vcm/c)E)= γ(p2cmc−(vcm/c)E) ?
     
  19. Jun 4, 2017 #18

    vela

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    I'm not sure what you mean by p1cm, etc.
     
  20. Jun 4, 2017 #19
    momentum of photon 1 in CM frame
     
  21. Jun 4, 2017 #20

    vela

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    In the CM frame, you want p1cm = p2cm, right?
     
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