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Maximum available energy for proton accelerators

  1. Oct 23, 2007 #1
    1. The problem statement, all variables and given/known data
    Suppose an accelerator has been constructed to provide 1 TeV proton beam. It must be decided how to use it, There are four possibilities:
    a) hit a fix target;

    b) collide with a 50 GeV electron to study ep collisions;

    c) collide with another proton beam also accelerated up to 1 TeV;

    d)collide with an antiproton beam, made from a source of antiprotons (more difficult to obtain).

    What is the maximum available energy for the production of new particles in each case? What other consideration might be important in making the decision?

    2. Relevant equations

    None given.

    3. The attempt at a solution

    Is the maximum available energy the same thing as the center of mass energy? In which case I thought the answer for part (d) was:

    [tex]\sqrt{s}=2E
    s=4E^{2}
    s=4TeV^{2}
    [/tex]


    and I found a formula for part (c) but cant work out how to apply it:
    [tex]\sqrt{s}=\sqrt{(x_{a}E)(x_{b}E)}
    [/tex]


    i'm still a bit stuck on the other parts...
     
  2. jcsd
  3. Oct 23, 2007 #2

    Astronuc

    User Avatar

    Staff: Mentor

    Look at the relativistic momentum and energy.

    Two beams of equal energy with equal and opposite momenta will utilize all the KE in the interaction (assuming a dead on center collision).
     
  4. Oct 23, 2007 #3
    So for part (a), the target hitting the wall,
    all energy is lost and so the total energy of each proton is available for production?
    Meaning [tex]E = \sqrt{p^{2}c^{2}+m^{2}_{0}c^{2}}

    E = \sqrt{p^{2}c^{2}+m^{2}_{0}c^{2}}

    proton rest mass = 938MeV

    E = \sqrt{p^{2}c^{2}+938^{2}}
    [/tex]
    but how do you get the velocity to workout the momentum?

    For part (b), an electron-proton collision, you are saying all the kinetic energy is used up meaning just the rest mass energy is left??

    for part (c) as above?

    and was part (d) incorrect?
     
  5. Oct 24, 2007 #4
    I found another formula that describes the total energy available for making particles:
    [tex]E=\sqrt{1+\frac{K}{2mc^{2}}}[/tex]
    Where K is the kinetic energy of the system beforehand...
    so for part (a)
    K=1TeV
    m=0.000938 TeV/c^2
    So E = 23 TeV

    but energy can't just be generated, so where have I gone wrong? And how do I expand for the other forumla?

    Thanks!
     
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