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Centre of mass energy in the p+p collision

  1. Nov 7, 2013 #1
    Hi, experts
    I got a very naive question.
    I read the following sentence in a paper: "New results on the production of charged pions in p+p interactions are presented. The data come from a sample of 4.8 million inelastic events obtained with the NA49 detector at the CERN SPS at 158 GeV/c beam momentum". (see the abstract of DOI:10.1140/epjc/s2005-02391-9)
    Then it says that the the centre of mass energy is 17.3 GeV. It is really hard for me to get this conclusion. Could sb. gives me some hint?
    Best wishes
  2. jcsd
  3. Nov 7, 2013 #2


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    Staff: Mentor

    I don't know your background in special relativity... the easiest exact way is to set up the 4-vectors of the resting and the moving proton and then to calculate the invariant mass of the sum of them.
    The quickest, approximate way for high energies
  4. Nov 7, 2013 #3


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    For a collision between two particles of equal mass M, the relationship is s = Ecm2 = 2M2 + 2M Elab where s is the usual Lorentz invariant.

    So M = 1 GeV and Elab = 158 GeV gives you Ecm = √318 = 17.8 GeV
  5. Nov 7, 2013 #4
    That's a beam target collision I suppose. Try using 4-vectos. I get the formula [itex]E_{CM}=\sqrt{2m(E_L+m)},[/itex] where ECM is the center of mass energy, EL is the Energy of the incoming proton in the LAB reference frame, and m is the mass of the proton.
  6. Nov 7, 2013 #5


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    I couldn't have said it better myself. :wink:
  7. Nov 7, 2013 #6
    Thanks for the replies. I think I got the answer. Say the four vector of the incoming proton in the lab frame is p1=(E_lab,0,0, p_z), the four vector for the rest proton is p2=(M,0,0,0), then s should be equal to (E_lab+M)^2-(p_z)^2=2M^2+2E_lab*M. Thanks a lot.
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