Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Centre of mass energy in the p+p collision

Tags:
  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
    W.
     
  2. jcsd
  3. Nov 7, 2013 #2

    mfb

    User Avatar
    2016 Award

    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

    Bill_K

    User Avatar
    Science Advisor

    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

    Bill_K

    User Avatar
    Science Advisor

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Centre of mass energy in the p+p collision
  1. P+P=Deuterium? how? (Replies: 2)

  2. Peskin p. 160 (Replies: 2)

Loading...