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Calculate the particle energy in a collision

  1. Aug 1, 2017 #1
    1. The problem statement, all variables and given/known data
    The particle J/Ψ can be produced in both proton-proton collisions and electron-positron collisions.
    a) Consider a proton beam incident upon a fixed hydrogen target. Calculate the energy of the proton beam in the reaction
    p1 + p2 → p + p + J/Ψ

    b) Consider two counter propagating beams of electrons and positrons respectively. The two beams have identical energy. Calculate the value of the electron energy in the reaction
    e+ + e− → J/Ψ

    2. Equations:
    (a)
    E + mc2 = 2E' + EJ
    P = 2P' + PJ
    (Proton energy before and after are denoted as E and E')
    (b)
    2Ee = mJc2
    Pe+ + Pe- = 0

    3. Attempt:

    I applied the law of energy and momentum conservation.
    (a) I tend to use E'2 = (p'c)2 + (mc2)2 to get E in terms of P and PJ, but then I would have problem as I don't know about PJ.
    (b) The positron and electron have equal and opposite momenta so the meson must only have rest mass energy mJc2. So, I can try to solve for Ee.

    Can anyone please tell if my approach is correct?
     
    Last edited: Aug 1, 2017
  2. jcsd
  3. Aug 1, 2017 #2

    mfb

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    For (a), what can you say about the relative motion of the produced particles at the minimal energy? Higher energies can lead to the same process, but I guess the question asks about the minimal energy.

    (b): Right (and that "solving or Ee" is very easy).
     
  4. Aug 1, 2017 #3
    For (a), I think the question asks the minimum energy required to produce those output particles (p + p + J/Ψ).

    For (b), yes, Ee is just the rest mass energy of the meson divided by 2.
     
  5. Aug 1, 2017 #4

    mfb

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    Yes, that is the typical question.
    What do you know about the relative motion of the particles at this energy?
     
  6. Aug 1, 2017 #5
    They share the momentum of the particles at the initial state, I think? By the conservation of 4 momentum:

    P1 + P2 = ΣNi=1Pi

    In this case, the initial state is a laboratory system, one of the protons (target) has zero momentum.

    Would it be easier to solve it with 4-momenta?

    So, if P1 is the incoming proton, it has the 4-momentum P1 = (E/c,p), and
    P2 is the target, its 4-momentum is P2 = (mc,0), and the output particles have 4-momenta 2(E'/c,p') and (E'J/c,p') for the two output protons and the J/psi meson respectively.
     
    Last edited: Aug 2, 2017
  7. Aug 2, 2017 #6

    mfb

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    That is correct, but not what I asked.
    That is very similar.

    Let's ask differently: Do you expect the three particles to fly apart at high speed relative to each other after the collision? If yes, why, if not, why not?
     
  8. Aug 2, 2017 #7
    No, it's a laboratory system. After the collision, the two output protons remain fixed in the target while the J/psi meson gets emitted.
     
  9. Aug 2, 2017 #8
    From #7:

    I now have:

    upload_2017-8-2_18-1-10.png

    That give two equations:
    E/c + mpc = 2mpc + EJ/c
    P = PJ
     

    Attached Files:

  10. Aug 2, 2017 #9

    mfb

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    That's not what will happen.
     
  11. Aug 2, 2017 #10
    After collision, it's more suitable to use the CoM system, in which the two colliding protons will have equal and opposite momentum, resulting zero momentum. And by conservation of momentum, the only particle which has a momentum is the J/psi meson and its momentum is the same as that of the incoming proton at the initial state.

    The other possibility is that these three output particles will have the same momentum P'=1/3*P.
     
    Last edited: Aug 2, 2017
  12. Aug 2, 2017 #11

    mfb

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    In the center of mass system, what is the minimal energy you can have? This translates to the minimal energy in every frame.
    All three particles are at rest in the CoM system.
    What does that mean for the relative velocity of the particles?
     
  13. Aug 2, 2017 #12
    Rest mass energy. (But the law of momentum conservation should still apply, otherwise, all three particles would have zero momentum, and hence zero velocity.)
     
  14. Aug 2, 2017 #13

    mfb

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    They have zero velocity in the center of mass frame.
    In the lab frame they all have the same velocity.

    That is the missing equation you need to solve the system.
     
  15. Aug 2, 2017 #14
    I will try to get that equation when I get to a nearby library in the next half hour or so.

    Nevertheless, I think the initial states I got for the particles are correct. It's only that in the center of mass frame (final states) their total momentum should be zero.
     
  16. Aug 2, 2017 #15
    E/c + (m_p)c = E'/c + E'/c +(E_j)/c

    Use E' = [(p'c)^2 + (m_p)^2c^4]^1/2

    Take p'=0

    And solve for E.
     
  17. Aug 2, 2017 #16

    mfb

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    That is not correct.
    Your first equation for conservation of energy was in the lab frame. In the the lab frame, p' won't be zero.

    You can either work in the lab frame or in the CoM frame, but keep it consistent. The latter is a bit shorter but both are possible.
     
  18. Aug 3, 2017 #17
    If I work in the lab frame:
    E/c + (m_p)c = E'/c + E'/c +(E_j)/c

    p = p' + p' + p_j = 2(m_p)v + (m_j)v

    as the output particles all have the same velocity.
     
  19. Aug 3, 2017 #18

    mfb

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    You should use the relativistic momentum. Apart from that: right.
     
  20. Aug 3, 2017 #19
    But I cannot solve for the proton energy E.

    E2 = (pc)2 + (mpc2)2

    p = ϒ(2mpv + mjv)

    Substitute for p, I have:

    E2 = ϒ2v2c2(4mp2 + 4mpmj + mj2) + mp2c4

    Because ϒ2 = c2/(c2-v2)

    E2 = [v2c4/(c2-v2)]*(4mp2 + 4mpmj + mj2) + mp2c4

    It still left me with the ‘v’ which I can't get rid of.
     
  21. Aug 3, 2017 #20

    mfb

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    That is the energy of the incoming proton I guess? It is not the total energy.

    You didn't consider the total energy after the collision in your last post.
     
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