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Energy of relativistic particle in LHC

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  1. Mar 5, 2013 #1
    1. The problem statement, all variables and given/known data
    (see attachment)
    The proton charge is ##1.6 \times 10^{-19} C## and the speed of light is ##3 \times 10^8 m/s##. The proton's mass is not necessary in this problem.


    2. Relevant equations



    3. The attempt at a solution
    The particle revolves in a circular path, hence
    [tex]\frac{mv^2}{R}=qvB[/tex]
    [tex]mv=eRB[/tex]
    (e is the charge of particle)
    Since p=mv (momentum)
    [tex]p=eRB[/tex]
    I don't understand how the mass of proton is not necessary for the problem. Is my expression for p correct?

    Any help is appreciated. Thanks!
     

    Attached Files:

    • lhc.png
      lhc.png
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  2. jcsd
  3. Mar 5, 2013 #2

    mfb

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    Staff: Mentor

    Your equations do not use the proton mass, how could it be relevant? You can use the approximation ##E=pc## to get the energy, as the mass of the proton is negligible relative to its momentum.

    Your formulas are good for nonrelativistic particles only, you cannot use this approximation for protons in the LHC.
    The result is the same with relativistic formulas, but that is just a coincidence here.

    The sketch has wrong positions for LHCb and ALICE :(.
     
  4. Mar 5, 2013 #3
    Thanks a lot mfb for the help! :smile:
     
  5. Mar 5, 2013 #4

    rude man

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    Yes it is, even relativistically, if you remember p is the relativistic momentum = mv = γm0v.
     
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