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P+p -> p + (anti)p

  1. Oct 16, 2012 #1
    p+p --> p + (anti)p

    Hi everyone,

    I'm looking at the following problem from C. Bertulani's Ch.1, Problem 6. The problem statement is:
    "Using relativistic expressions for momentum and energy conservation, show that a proton must have energy greater than 5.6 GeV to produce a proton-antiproton pair in a collision with another proton at rest."

    My first question is why does this reaction occur? It seems to violate charge conservation the way it stands.

    My second question is, if I use energy conservation, and let initial p of the moving proton be my unknown, and assume the minimum energy required of this incoming proton is when both final momenta are zero, I get that

    [tex] p=0 [/tex]... which is setting my unknown to zero which I don't want. So there must be a mistake in my reasoning. Here is how I get this result:

    1. Requiring energy conservation yields:

    [tex] E_1 + E_2 = E'_1 + E'_2 [/tex]

    2. For the minimum incoming momentum, p, we would require that both final particles have zero momentum, so the above reduces to:

    [tex] m_p + sqrt(m_p^2+ p^2) = m_p + m_\bar p [/tex]

    3. And if we just rearrange the above we get: [tex] p=0 [/tex] since the mass of a particle and its antiparticle are equal.

    I appreciate any clarifications on this problem, thanks!

    iman
     
  2. jcsd
  3. Oct 16, 2012 #2

    Nugatory

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    Re: p+p --> p + (anti)p

    This one may belong in the homework section... But chances are that you're being asked to analyze the process in which the collision produces a new proton/antiproton pair:

    [itex]p + p \Rightarrow p + p + p + \overline{p}[/itex]
     
  4. Oct 16, 2012 #3
    Re: p+p --> p + (anti)p

    Hi Nugatory!

    Thanks for suggesting this reaction! It seems to conserve what it has to conserve (charge, baryon number).

    When I use the same calculation steps as I did originally, and assume the minimum required momentum for the incoming particle is when all outgoing particles have zero momentum, I get that the incoming particle should now have [tex] p = 3.250 GeV [/tex]... which isn't what's required. Maybe the reasoning to my calculation is wrong?

    This time I use:
    [tex]
    sqrt(m_p^2 + p^2) + sqrt(m_p^2 + p^2)= 4 m_p
    [/tex]
    and solve for p. I am working in the initial system's centre of mass frame. I than multiply this momentum, p, by two to write the answer in the rest frame of one of the two initial protons.

    Maybe where I'm going wrong is in working in the CM frame? This assumes that the product protons and antiproton are stationary in the CM frame, but I want the momentum for which they are stationary in the lab frame?

    Nugatory: I haven't put this question in the homework section because it wasn't on my homework problems.. but if you think this would be more useful for others I can look for ways to transfer the post.
     
  5. Oct 16, 2012 #4

    Bill_K

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    Re: p+p --> p + (anti)p

    Working in the CM frame is a good idea, but I think your answer of 3.25 GeV is still in the CM frame also, and you'll need to transform this to the lab frame, in which one of the protons is at rest.
     
  6. Oct 16, 2012 #5

    Nugatory

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    Re: p+p --> p + (anti)p

    What did you get for the center of mass energy?
     
  7. Oct 16, 2012 #6

    PAllen

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    Re: p+p --> p + (anti)p

    " I than multiply this momentum, p, by two to write the answer in the rest frame of one of the two initial protons."

    This is not correct (all else seems correct, as far as your formulas go - haven't checked the arithmetic). Once you have solved for p for one of the protons in COM, you must compute velocity, then a apply a boost for the velocity to the 4-momentum. That will not be just multiplying by 2.
     
  8. Oct 16, 2012 #7
    Re: p+p --> p + (anti)p

    PAllen: Thanks for the reminder! So I can find the boost using the ratio of energy and z-momentum, and supposing [tex] y [/tex] is my boost, the energy in the lab frame would be:

    [tex] E' = E cosh(y) + p^z sinh y [/tex]

    Is this correct?

    Nugatory, Bill_K, PAllen: Thanks for all of your input. I've thought of another way I can (possibly) do this. If I consider the four vector [tex] P^\mu [/tex] for each particle, this should be an invariant for the reaction (i.e. the same in all frames). And the nice thing about using the four-vector squared (an invariant) is that it is indepdent of frame.. so I can work in the proton frame for the initial system and the CM frame for the final system and set

    [tex] (\sum P_i^\mu)^2 = (\sum P_f^\mu)^2 [/tex] , where ths sum is over particles initially and finally, respectively, and each side can be calculated in a different frame. Please call this Equation (1).

    Using this invariant, I can than write:

    [tex] \sum P_i^\mu = (m_p +\sqrt{m_p^2+p^2}, 0, 0, p) [/tex] , and

    [tex] \sum P_f^\mu = (4m_p, 0, 0, 0) [/tex]

    Plugging these into Equation 1 and remembering the metric requires a negative sign for the time component (or space component, depending on your convention) and solving for the momentum [tex] p [/tex], I get

    [tex] p = 6.5 GeV [/tex]

    ... but again this is not quite the value they wanted. Any hints why this may be wrong? I'll try to do this by calculating the boost.
     
  9. Oct 16, 2012 #8
    Re: p+p --> p + (anti)p

    Correction: Sorry, I meant [tex] (P^\mu)^2 [/tex] is an invariant..
     
  10. Oct 16, 2012 #9

    Vanadium 50

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    Re: p+p --> p + (anti)p

    5.6 GeV is wrong. Simple as that.
     
  11. Oct 16, 2012 #10
    Re: p+p --> p + (anti)p

    Vanadium: Thank you... I was actually starting to think the same.

    There is one little problem: when I calculate the boost and calculate the momentum in the proton rest frame using the boost I get something different, I get 15.869 GeV.

    Here are my equations:

    y is the boost,

    [tex] p'_z = p_z cosh(y) + E sinh(y) [/tex]

    I call the above eqation (2). I calculate the boost using

    [tex] y = atanh(p_z/E) [/tex]

    and I get y = 1.7595.

    Now, in order to use Equation (2), I use that [tex] p_z = 2.653 GeV [/tex] which I calculated using conservation of energy (my first attempt at this problem which can be found at the beginning of this post), and E is equal to [tex] E = \sqrt{m_p^2+p_z^2} [/tex].

    Then, using, Equation (2) I get

    [tex] p'_z = 15.869 GeV [/tex]

    Sorry for all the troubles, but if the answers agree than it would mean the answer is correct.. thanks for your input.
     
  12. Oct 16, 2012 #11

    Bill_K

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    Re: p+p --> p + (anti)p

    Is it? Funny, that's what I got.
     
  13. Oct 16, 2012 #12
    Re: p+p --> p + (anti)p

    Bill_K: Can you please let me know how you got to this answer? Thanks..
     
  14. Oct 16, 2012 #13

    PAllen

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    Re: p+p --> p + (anti)p

    I also get 5.6 Gev simply using OPs own formula for COM (which is correct), then getting v, then using velocity addition:

    2v/(1+v^2) // of course c=1

    then computing m(γ-1) for this.

    [To be unambiguous, OP problem should ask for KE, but I think that is reasonable to assume for the problem.]
     
  15. Oct 16, 2012 #14
    Re: p+p --> p + (anti)p

    Hi PAllen,
    I'm guessing there's a square root around the denominator? Do you get the velocity from relativisitc momentum?
    Why is [tex] m(\gamma -1) [/tex] needed?

    Are you using [tex] KE = E-mc^2 [/tex]?

    I'm studying for a test and it'd be nice to clarify this problem now :) Thanks for the input!
     
  16. Oct 16, 2012 #15

    PAllen

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    Re: p+p --> p + (anti)p

    Have you not seen the velocity addition formula: (u+v)/(1+uv)? If not, please review basics. If you have protons going v and -v in a frame, this tells you that in a frame where one is at rest, the other is 2v/(1+v^2). No sqrt needed.

    m(γ-1) is KE. Again, if this is news to you, you need to review very basic SR facts. Of course, it does follow immediately from E-mc^2.
     
  17. Oct 16, 2012 #16

    Vanadium 50

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    Re: p+p --> p + (anti)p

    Good heavens.

    You don't want to be messing with velocities, or rapidities or any of that. It will make a horrible mess.

    You have a proton of mass m and momentum p, collide with another proton. The invariant mass is equal to 4m.

    [tex]E^2 - p^2 = 16m^2[/tex]

    [tex](\sqrt{m^2+p^2}+m)^2 - p^2 = 16m^2[/tex]

    A little algebra and you get [itex]p = 4\sqrt{3}m[/itex]. Plug in the proton mass and you get 6.5 GeV for momentum, and 6.57 GeV for energy. It is true that 5.6 GeV is the kinetic energy of the proton, but that's not what they are asking for.
     
  18. Oct 16, 2012 #17

    PAllen

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    Re: p+p --> p + (anti)p

    Obviously, the author and two of us here thought KE is what they were asking for.

    As for extra complexity, I was interested in showing the proper way to get from COM momentum to KE in frame where one proton is at rest (i.e. what you do instead of doubling this momentum). Of course, your approach to the problem as a whole is much simpler.
     
  19. Oct 16, 2012 #18
    Re: p+p --> p + (anti)p

    Thanks a bunch everyone! This discussion was really useful for me, and I hope a few others.

    Vanadium: Thanks for confirming the invariant momentum approach.. I was getting worried! And I finally found the proper way of calculating the velocity between the two frames :) So things like KE become reachable.

    PAllen: To answer your question, I have seen the formula, but it got forgotten once we learned four vector notation.. thanks for your advice, I should find time to review the basics.

    Sincerely,
    iman
     
  20. Oct 17, 2012 #19

    Vanadium 50

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    Re: p+p --> p + (anti)p

    Just about the only time one talks about a proton's kinetic energy is when the energy is so low as to make it unambiguous: say 100 MeV. Otherwise, when they say energy, they mean energy. :smile:

    As a general rule, working with velocities or even rapidities in relativistic mechanics is a good way to make a mess. Deal with E, m and p, and you have nice Lorentz invariants to work with, and if you need velocities at the end, one of the following three equations is usually helpful: E/m = γ, p/m = βγ and p/E = β.
     
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