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Energy, Mass and Frames

  1. Jul 14, 2009 #1
    So, here is the situation:
    One starts with 2 protons, they collide and one ends up with 3 protons and an antiproton.
    How much energy will you need to do this?

    Upon inspection, it seems you will just need 2mc[tex]^{2}[/tex] in kinetic energy to create 2 new proton-massed particles. This is wrong I guess.

    The way Griffiths does it is he starts in the lab frame. One proton has energy E, and the other is at rest. So the 4-vector looks like P=(E+m,[tex]\vec{p}[/tex]). (I will just set c=1, for simplicity.) The invariant mass of this is PP = E[tex]^{2}[/tex]+m[tex]^{2}[/tex]+2mE - p[tex]^{2}[/tex].
    One can eliminate p[tex]^{2}[/tex] with the formula E[tex]^{2}[/tex]=m[tex]^{2}[/tex]+p[tex]^{2}[/tex].
    This results in PP=2mE+2m[tex]^{2}[/tex].

    Since invariant mass is the same in any inertial frame, Griffiths then finds the invariant mass after the reaction in the Center-of-Mass frame. P = (4m,0).
    PP=16m[tex]^{2}[/tex]
    2mE+2m[tex]^{2}[/tex]=16m[tex]^{2}[/tex]
    E=7m

    This is confusing, but the reasoning is sound I guess. So I then tried to do the problem with PP before and after taken in the center-of-mass frame. I got a different answer.

    I started with 2 protons moving towards one another in the CM frame. So 3-d total momentum is zero. P1=(E,p) P2=(E,-p) P=P1+P2=(2E,0). So E denotes how much energy each proton should have.
    Initial PP = 4E[tex]^{2}[/tex]
    It still ends with PP = 16m[tex]^{2}[/tex]
    4E[tex]^{2}[/tex]=16m[tex]^{2}[/tex]
    E=2m

    I am not sure where the problem is. Does energy needed for a reaction depend on the frame? That is really confusing.
     
  2. jcsd
  3. Jul 15, 2009 #2

    George Jones

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    Yes, energy is frame-dependent.

    Let [itex]v[/itex] be the speed of proton 1 in the lab, with proton 2 at rest. The energy of proton 1 is

    [tex]E = \frac{m}{\sqrt{1 - v^2}}.[/tex]

    In the centre-of-momentum frame, both protons have the same speed, say [itex]v'[/itex]. The energy of proton 1 in this frame is

    [tex]E' = \frac{m}{\sqrt{1 - v'^2}}.[/tex]

    Since [itex]v \neq v'[/itex], [itex]E \neq E'[/itex]. You have set [itex]E = E'[/itex].
     
  4. Jul 15, 2009 #3
    Yes. That makes sense. I solved for velocities and got v = 3[tex]^{1/2}[/tex]4/7 in the lab frame. Then v = 3[tex]^{1/2}[/tex]/2 in the CM frame. With the velocity addition rule, u' = (u+v)/(1+vu) this makes sense.

    One question though. Do accelerators have to use more power if they are shooting one proton at a stationary one as opposed to doing it in the CM frame? If so, is this why accelerators try to do the experiment such that the lab frame and CM frame are one in the same? (It saves them money on their electric bill so to speak?)
     
  5. Jul 16, 2009 #4

    Ich

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    Yes. You achieve much higher CM energy this way. Calculate it yourself for the LHC (7 TEV per proton in the lab frame).
     
  6. Jul 16, 2009 #5
    I did calculate it. In the CM frame, I get E=2m per proton. (A total of 4m in the reaction).
    If I do it in the lab frame, I get E=7m for one proton, and E=m for the other, since it's stationary. (A total of 8m in the reaction).

    I don't see how you get 7TEV, if a proton is only 0.000983 TEV.
     
    Last edited: Jul 16, 2009
  7. Jul 16, 2009 #6

    Ich

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    These protons have a kinetic energy of 7 TeV in the lab frame. That means 14 TeV CM energy if two of them collide head-on.
    What is the CM energy if you shoot one proton against another one that is at rest in the lab frame?
     
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