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Angles in momentum four vectors

  1. Jul 25, 2012 #1
    1. The problem statement, all variables and given/known data

    A particle decays to two photons. In the rest frame the two photons are emitted on the x-y plane, in opposite directions along a line that forms an angle alpha with the the x axis. Derive the momentum four vector of the two photons in the lab frame.


    2. Relevant equations



    3. The attempt at a solution

    Photon 1 = (E1 sinθ1, E1 cos θ1 , 0 , E1)

    Photon 2 = (-E2 sinθ2 , -E2 cos θ2 , 0 , E2)

    I'm confused to what notations I should be using for the angles, can I use θ1 and θ2 for the angles of the two photons and then assume that alpha = (θ1+θ2) or is that the wrong way to go about it. Should alpha be in the vector equations?
    Also I'm not sure if the signs are correct as the photons are moving in opposite direction in the x - y plane, could someone check my answer and point me in the right direction please :)
     
  2. jcsd
  3. Jul 25, 2012 #2

    tiny-tim

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    hi twinklestar28! :smile:
    you're over-complicating it :redface:

    in the rest frame, α is θ1 (and θ2 - 180°) :wink:

    (but what is the relation between the rest frame and the lab frame? :confused:)
     
  4. Aug 26, 2012 #3
    ok so if i didn't sub in the values my vectors would be in the rest frame:

    Photon 1 = (E1 sina, E1 cos a , 0 , E1)

    Photon 2 = (-E2 sinθ , -E2 cos θ , 0 , E2)

    a and θ is what they've asked for in the question, so to get the lab frame i use the lorentz transformation but i then have to show that the cosine of the angle between the two trajectories as a function of a is:

    cos θ = β^2-1+^2sin^2a / 1-β^2cos^2a

    how do i do this? i tried p'x^2 + p'y^2 + p'z^2 -(E/c)^2 = p^2 but i get complicated answers. Do i equate the momentum vector i found to the initial momentum?
     
  5. Aug 26, 2012 #4
    sorry the angle is meant to be

    cos θ = β^2-1+β^2sin^2a / 1-β^2cos^2a

    I forgot the β in the first part of the fraction..
     
  6. Aug 26, 2012 #5

    vela

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    You missed tiny tim's point. In the rest frame, the photons have to have equal and opposite momenta because the total momentum has to be 0. So you should be able to write both four-momenta using one angle and one energy.
     
  7. Aug 30, 2012 #6
    oo ok can u check if my new vectors are correct please?

    photon 1= (E1 sina, E1 cos a , 0 , E1)
    Photon 2= (-E2 sinθ2 , -E2 cos θ2 , 0 , E2) = (0,E1,0,-E1)

    using E1+E2=0 and since they are photons E=p

    are these right, especially the signs?
     
  8. Aug 30, 2012 #7

    vela

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    They don't look right to me. Don't just post your answers. Explain how you came up with them.
     
  9. Aug 30, 2012 #8
    they are my initial vectors that i came up with, but i changed theta 2 into 180 because it is in the rest frame and also expressed E2 in E1 since in the rest frame they are equal but opposite, am i completely off the track here? or am i getting close... :S
     
  10. Aug 30, 2012 #9

    vela

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    The problem statement gives you a coordinate system to use. So let's say E is the energy of photon 1. As you noted, for a photon, E=p, so you know the magnitude of the photon's momentum. Now tell me, in terms of p and ##\alpha##, what are the x and y components of the photon's three-momentum?
     
  11. Aug 30, 2012 #10
    x component = (psina)^2
    y component=(pcosa)^2
     
  12. Aug 30, 2012 #11

    vela

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    No, that's not correct. Why are you squaring them for one thing? Also, the problem statement says, "the two photons are emitted on the x-y plane, in opposite directions along a line that forms an angle alpha with the the x axis." Try again.
     
  13. Aug 30, 2012 #12
    I was finding the magnitude, sorry got confused,
    would it simply be

    pcosa for x
    -psina for y
     
    Last edited: Aug 30, 2012
  14. Aug 30, 2012 #13

    vela

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    Why the negative sign in the y component?
     
  15. Aug 31, 2012 #14
    I put them both positive first and then i changed it, i think im thinking in the one dimension so im seeing the second photon in negative y quadrant, but looking in the x,y,z plane i think it would be positive, the signs r what i am most unsure of, is the rest correct?
     
  16. Aug 31, 2012 #15

    vela

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    I'm not sure what you mean by the negative y quadrant. There are two quadrants where y<0.

    I've attached a picture of what the problem describes. What are the components of the momentum of the upper right photon?
     

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