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Producing antiprotons

  1. Apr 28, 2008 #1
    1. The problem statement, all variables and given/known data
    What is the minimum proton energy needed in an accelerator to produce antiprotons [itex]\bar{p}[/itex] by the reaction

    [tex]p+p \to p+p + (p+\bar{p})[/tex]

    The mass of a proton and antiproton is m_p.

    2. Relevant equations

    3. The attempt at a solution
    I said 2 m_p c^2 because that is the sum of the rest energies of a proton and an antiproton. Apparently that is wrong.
  2. jcsd
  3. Apr 28, 2008 #2
    The velocities of the reaction products can't all be 0. Use conservation of momentum.
  4. Apr 28, 2008 #3
    Hmm. So, the momentum four-vector of the system has to conserved, right? Is it true that the total energy of the system has to be conserved, right? I was always confused about that.

    The total energy is relativistic kinetic energy + rest energy =[tex] \frac{1}{\sqrt{1-\frac{u^2}{c^2}}}mc^2[/tex].

    OK. I see now. The four-momentum contains the total energy as one of its components. And if the four-momentum is conserved then all of its components are conserved. But please confirm that four-momentum is conserved here.

    I am not really sure how to implement that though. I will probably need

    [tex]E^2 = p^2 c^2 +m^2 c^4[/tex]

    What is the first thing I should do?
    Last edited: Apr 28, 2008
  5. Apr 28, 2008 #4
    I think you may try to use four-momentum vector approach.
    This will never wrong~

    and i think the answer is something like [tex]7m_p c^2 [/tex]
    Last edited: Apr 28, 2008
  6. Apr 28, 2008 #5


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    Four momentum is always conserved. You can make the calculation much easier by noting that the square of the energy-momentum four-vector is Lorentz invariant (i.e. doesn't change with a Lorentz boost).
  7. Apr 28, 2008 #6


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    Does one assume that the two initial protons collide head on? That would produce the minimal energy. If one proton is at rest then there is recoil to consider.

    Also, when a particles kinetic energy is equal to rest mass, then one must consider relativistic effects. Total energy and momentum are conserved.
  8. Apr 28, 2008 #7


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    That would be the case if one of the protons were stationary, but that isn't the most efficient method of creating a particle-anti-particle pair :wink:

    Note also that your value contains the rest mass of the proton.

    Edit: It seems Astro beat me to it :smile:
  9. Apr 28, 2008 #8
    OK good so four-momentum is conserved. I am still trying to figure out what the first thing I should do is! So, if we label the six particles in the order from left to right that they appear in the reaction we get

    [tex]\sum_{i=1}^2 \mathbf{p}_i - \sum_{i=3}^6 \mathbf{p}_i = 0 [/tex]

    by conservation of four-momentum. What should I do with that? That is 4 equations containing 24 different quantities. How do I deal with that?
    Last edited: Apr 28, 2008
  10. Apr 28, 2008 #9


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    There is no way to answer the question before you say in what frame you want the energy to be calculated!! If the reaction is in the center of mass frame, then the energy of each colliding proton must be 2 mc^2 but you said that was wrong?
  11. Apr 28, 2008 #10
    The back of my book says the answer is

    Honestly, I cannot tell you what frame I want the energy calculated. I posted the question in its entirety! Can you tell me what frame to calculate the energy now that I gave you the answer?
  12. Apr 28, 2008 #11


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    1.The center of mass coordinate system will give the easiest approach to the problem--why?

    2.Simply apply the appropriate conservation rules -- how many?-- to relate before and after -- everything will work out just fine. (Think about vectors, four vectors, and/or components thereof, and which might be simplest descriptive tool available to you.

    Good luck,
    Reilly Atkinson

    (This is a very standard problem; an important one as well.)
  13. Apr 28, 2008 #12


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    What they are doing is considering a frame in which one proton is smashing against a second proton initially at rest.

    Call P_1 the four-momentum of the moving proton, p_2, the four-momentum of the proton at rest.

    Call P_3,P_4,P_5, P_6 the four-momenta of the produced particles. The minimum energy corresponds to the case when they have all the same four-momentum.

    we have [tex] P_1 = (E_1, \vec{p_1}), P_2 = (m_p c^2, \vec{0}), P_3=P_4=P_5=P_6 = (E', \vec{p'}) [/tex]

    Conservation of four-momentum gives

    [tex] P_1 + P_2 = 4 P' [/tex]

    squaring, you get

    [tex] P_1^2 +P_2^2 + 2 P_1 \cdot P_2 = 16 P'^2 [/tex]

    all the squares give P^2 = m_p^2 c^4.

    Finish the calculation and isolate E_1.

  14. Apr 28, 2008 #13
    OK. So we go to the CM coordinate system. In that system the 3-momentum is 0 before the collision, so it has to be zero after the collision. The energy in that system before the collision is given by

    [tex]E^2 = p_1^2+p_2^2 + 2m_p c^2[/tex]

    where p_1 and p_2 are the relativistic momentum of protons 1 and 2.

    E^2 must also be conserved.

    Is that right?

    EDIT: I didn't read the post above when I wrote this.
  15. Apr 28, 2008 #14
    Do you know that because you have done the problem before or because that is obvious or what? To prove that statement, should I first assume that the four-momenta are different and then use Lagrange multipliers or calculus or something?

    What do you mean by this? Should that be P'?
  16. Apr 28, 2008 #15


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    I know that because in the center of mass frame, it's obvious that the minimum energy corresponds to all the particles in th efinal state being at rest. If you boost to the lab frame, they all have the same four-momentum
    I meant that for *any* of the P you get that. So that P could be P' or P_1 or P_2 , etc.
  17. Apr 28, 2008 #16


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    That problem is generally given for the target proton at rest, which agrees with the answer in the book. Use the invariance of E^2-p^2, so (4m)^2=(E_lab + m)^2-(p_lab)^2.
    Solve for E_lab.
  18. Apr 28, 2008 #17
    See if i have done anything wrong or not~

    First do it in the frame where the target proton (marked 2 below) is stationary.
    [tex] P_1 = (E_1/c,p_1,0,0) \mbox{ ; } P_2 = (mc,0,0,0)[/tex]

    By conservation of four-momentum,
    [tex] P_1 + P_2 = P_3 + P_4 +P_5 +P_6 [/tex]

    [tex] (P_1 + P_2)^2 = (P_3 + P_4 +P_5 +P_6)^2 [/tex]

    [tex] (P_1 + P_2)^2 = P_1^2 + 2P_1 \cdot P_2 + P_2^2 = -2m^2c^2 -2mE_1 [/tex]

    [tex] (P_3 + P_4 +P_5 +P_6)^2 = -4m^2c^2[/tex] (evaluate in CM frame for minimum energy)

    [tex] -2m^2 c^2 -2mE_1 = -16 m^2c^2 [/tex]

    [tex] E_1 = 7mc^2[/tex]

    Last edited: Apr 28, 2008
  19. Apr 29, 2008 #18
    OK. I understand everything except how you proved that the 4-momenta of the four products of the reaction should be the same to obtain the minimum energy E_1.

    So nrqed and tnho say they go to the center-of-mass frame to do this. What do we know about the center-of-mass frame in special relativity? In the nonrelativistic case, we know that the sum of the linear momenta of the particles is 0 in the center-of-mass frame. In special relativity, is it true that the sum of the relativistic momenta is 0 in the center-of-mass frame?

    My book talks about the "center-of-momentum" frame although I am not really sure what that means. Is that the same as the center-of-mass frame here?
  20. Apr 29, 2008 #19
    Yes. I think so. The sum of linear momentum of the particles is zero in the centre of mass frame in SR.
  21. Apr 30, 2008 #20


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    For relatively simple scattering problems, elastic scattering for example, the lab frame - target at rest -- works nicely, and is the standard frame used in most particle physics.

    However, for more complex situations involving particle production, the Center of Mass/Momentum frame is preferred, because of the simplicity it affords. (In practice, the Center of Mass and Center of Momentum mean the same thing, here the frame in which the two protons have equal and opposite momentum.) But, to get back to the lab frame requires a Lorentz transformation.

    A hint for simplification: let the initial protons move along the z axis - can this be true in both the lab and CM frames?

    With this approach, the answer given by nqrd should be very apparent.

    Reilly Atkinson
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