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Homework Help: Fixed Target & CoM Energy of colliding protons

  1. May 2, 2013 #1

    My question is with part (c) of the below:


    I wanted to check that I had done this write as I struggle to determine the correct invariants for each frame... despite their invariance. So for the lab frame I had

    [tex]\frac{\ E_{lab}^{2}}{c^{2}}-p^{2}=(\frac{E}{c}+m_{p}c)^{2}-p^{2}=2Em_{p}+m_{p}^{2}c^{2}=m^{2}c^{2}[/tex]

    (Where E is the energy of the cosmic ray proton)

    Where the last step is done by using the relativistic energy-momentum relation and the very last term is the invariant mass. The Centre of mass frame has:


    Finally we attain:

    [tex]2Em_{p}+m_{p}^{2}c^{2}=\frac{E_{CoM}^{2}}{c^{2}}\: ,\: E=\frac{\frac{E_{CoM}^{2}}{c^{2}}-m_{p}^{2}c^{2}}{2m_{p}}[/tex]

    I think this is correct but I used 7TeV for the centre of mass energy, should I use 14TeV? It says the 7TeV is the centre of mass energy per collision, so I'm guessing I use E(CoM)=7TeV and find E(lab) which was about 2.5*10^4 TeV.

    Thanks if anyone can tell me if this is correct!
  2. jcsd
  3. May 2, 2013 #2


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    Where is the difference between E and E_lab?
    What happened to E^2-p^2 at the second "="? It should give an additional mc^2.
    Why should this be equal to the total energy in the center of mass frame?
    There is no new particle produced, you don't have an invariant mass (apart from the masses of the protons).

    I wonder how you used momentum conservation here.

    Use 7 TeV as center of mass energy.

    E(lab) = 2.5*10^4 TeV has the right order of magnitude.
  4. May 3, 2013 #3

    Yeah I forgot the mc^2 term - there should be one in that second =, thanks! I thought the invariant for the lab frame has to be the same in any frame and for the CoM frame there's no momentum term due to equal and opposite momenta's of the particles. I guess I mean the mass of the protons then instead of invariant mass!

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