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Relativistic collision of two protons to produce a proton-antiproton

  1. May 3, 2014 #1
    1. The problem statement, all variables and given/known data

    What is the minimum proton energy needed in an accelerator to produce antiprotons by the

    [itex] P+P \rightarrow P+P+P+\bar{P} [/itex]

    The mass of both protons and antiprotons is [itex]m_p[/itex]. Assume first that the initial protons have equal
    energy (the lab frame is the center-of-momentum frame). What energy is required if protons are
    collided with a fixed target (one of the initial protons is at rest). This is one of the reasons
    modern particle accelerators collide two beams.

    2. Relevant equations

    [itex] P^{\mu} = \begin{pmatrix} \frac{E}{c}\\\vec{p}\end{pmatrix} [/itex]

    3. The attempt at a solution

    Setup: I think what I need to do is to look at the zeroth component of the momentum 4-vectors since it contains the relativistic energy of the particle. In the first case, since the protons have equal energy the center of momentum frame is already the lab frame so I don't need to boost to a different frame or anything like that. In the second case I will need to boost to a frame where the second particle at rest in the lab frame (K -frame) is moving at the same velocity [itex] v [/itex] as the first particle (K'-frame). Then I need to compare the zeroth component of the momentum 4-vectors in the K'-frame. Is my line of thinking correct, along the right direction, or flat out wrong? Thanks for your guidance guys
    Last edited: May 3, 2014
  2. jcsd
  3. May 3, 2014 #2
    Your logic is correct but that's not the easiest way to solve the problem. The easiest way to solve the problem is to write all the 4-momenta for all the particles and use energy and momentum conservation combined with the particles' energy-momentum relations. So for the incoming proton you have the 4-momenta Pin = (E/c,p), the target 4-momentum is Ptarget = (mc,0), and all the outgoing particles have identical 4-momenta Pout = (E'/c,p')
    Last edited: May 3, 2014
  4. May 3, 2014 #3
    So in the first case I would have the following:

    [itex]p^{\mu}_{1i}+p^{\mu}_{2i}= p^{\mu}_{1f}+p^{\mu}_{2f}+p^{\mu}+\bar{p}^{\mu} [/itex]

    [itex]\begin{pmatrix} \frac{E}{c}\\ \vec{p} \end{pmatrix} + \begin{pmatrix} m_pc \\ \vec{0} \end{pmatrix} = 4\begin{pmatrix} \frac{E'}{c}\\ \vec{p}' \end{pmatrix} [/itex]

    Which would give me 2 equations:

    Eqn 1 [itex] E+m_pc^2=4E' [/itex] and Eqn 2 [itex] \vec{p}=\vec{p}' [/itex]

    I'm not exactly sure what to do here, but if I use [itex] E'=\sqrt{(p'c)^2+(m_pc^2)^2}= \sqrt{(pc)^2+(m_pc^2)^2} [/itex] then I get:

    Eqn 1 [itex] E=4\sqrt{(pc)^2+(m_pc^2)^2}-m_pc^2 [/itex]
  5. May 3, 2014 #4
    Eqn 2 should be p = 4 p'
  6. May 3, 2014 #5
    Sorry to interrupt here - I've been using this method a lot to solve problems recently without understanding it too well.

    I was wondering what in this method actually constrains the energy required to be a minimum. Obviously all the products stick together which is necessary, but why does this method yield only the minimum energy solution and not infinitely many others for which say the incident proton has a larger energy/moves faster and thus the products have larger energies/move faster. I just can't see the step in my working (despite getting correct answers) that does this.

    It's obvious when working in the CM frame because we constrain the products not to move. In the lab though the products can have any energy/momentum E' and p'.
  7. May 3, 2014 #6
    If the proton had higher energy, the particles wouldn't stick together. As you said it yourself, that fact is obvious when working in the CM frame.
  8. May 3, 2014 #7
    Okay I corrected the factor of 4: how does it look now?
  9. May 3, 2014 #8
    Its okay I'd like to learn why the method works as well :)
  10. May 3, 2014 #9
    So if the products all stick together, there is only one mathematical solution for this to happen, which by judging it from the CM frame has to be the minimum energy one?
  11. May 3, 2014 #10
    This might be a really dumb question, but why would the particles not stick together if the proton had a higher energy. Also is the in-elasticity of the collision required for the proton anti-proton production to occur?
  12. May 3, 2014 #11
    Now use the relation between E, p, and m along with the equation you found and solve for E
  13. May 3, 2014 #12
  14. May 3, 2014 #13
    If there was extra energy that energy would end up as kinetic energy of the products in the CM frame. That means the products would fly away from each other (they can't fly together because that would give them a total momentum different from zero, but that's impossible because the CM is defined as the referential where the total momentum is zero.
  15. May 3, 2014 #14
    Haven't I already solved for [itex] E [/itex]? I used [itex] E' = \sqrt{(\frac{1}{4}pc)^2+(m_pc^2)^2} [/itex] to get and equation for [itex] E [/itex] in terms of [itex] p[/itex] an [itex]m_p [/itex]
  16. May 3, 2014 #15
    You solved for E but p is unknown. Use E2 = (cp)2 + (mc2)2 to eliminate p

    EDIT: the equation had a typo. I fixed it
    Last edited: May 3, 2014
  17. May 3, 2014 #16
    ah ok I see thank you! I'll post my result when I finish the calculation(:
  18. May 3, 2014 #17
    Now I'll let [itex] E^2 = (pc)^2+(m_pc^2)^2 [/itex] then

    Eqn 1 [itex] (pc)^2+(m_pc^2)^2 = [4\sqrt{(\frac{1}{4}pc)^2+(m_pc^2)^2}-m_pc^2]^2 [/itex]

    [itex] (pc)^2+(m_pc^2)^2 = 16(\frac{1}{4}pc)^2+16(m_pc^2)^2-8\sqrt{(\frac{1}{4}pc)^2+(m_pc^2)^2}m_pc^2+(m_pc^2)^2 [/itex]

    [itex] (m_pc^2)^2=16(m_pc^2)^2-8\sqrt{(\frac{1}{4}pc)^2+(m_pc^2)^2}m_pc^2+(m_pc^2)^2 [/itex]

    [itex] 1= 16-\frac{8\sqrt{(\frac{1}{4}pc)^2+(m_pc^2)^2}}{m_pc^2}+1 [/itex]

    [itex] 2=\frac{\sqrt{(\frac{1}{4}pc)^2+(m_pc^2)^2}}{m_pc^2} [/itex]

    [itex] (2m_pc^2)^2-(m_pc^2)^2=(\frac{1}{4}pc)^2[/itex]

    [itex]\frac{4\sqrt{3(m_pc^2)^2}}{c}= p [/itex]

    [itex] p = 4\sqrt{3}m_pc [/itex] plugging back into Eqn 1 I get:

    [itex] E = 4\sqrt{(\frac{1}{4}[4\sqrt{3}m_pc]c)^2+(m_pc^2)^2}-m_pc^2 [/itex]

    [itex] E = [4\sqrt{13}-1]m_pc^2 [/itex]
  19. May 3, 2014 #18
    Okay now my next question is: is this the answer to part 1 or part 2 because it seems like this is the answer to part 2.
  20. May 3, 2014 #19
    Yeah, that's part 2. Part 1 is easier.
  21. May 3, 2014 #20
    Ah ok yea I see: for part 1 all I have to do is set all the energies on the RHS of the reaction equal to the rest energy for a proton and the 3-momentum equal to 0 such that

    [itex] 2E=4E' [/itex] and [itex] p'=0 [/itex] Using [itex] E'^2 = (m_pc^2)^2 [/itex] I get

    [itex] E=2m_pc^2 [/itex] since [itex] 2 < 4\sqrt{13}-1 [/itex] thats why they use two proton beams instead of one proton beam and a "stationary" proton.
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