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CERN SPS, some questions

  1. Feb 23, 2015 #1
    hello,
    I open this thread to ask some questions about
    SPS, it seems it is mounted 33 TeV, but how
    he has been able to reach him such energy?
    Is not it supposed to go up "only" 450 Gev Env?
    Good week
     
  2. jcsd
  3. Feb 23, 2015 #2

    Orodruin

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    Please provide references to where you have found this figure when asking questions like this. Otherwise it is virtually impossible for us to check your sources.
     
  4. Feb 23, 2015 #3
  5. Feb 23, 2015 #4

    Orodruin

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    My French is not very good, but the 33 TeV seems to refer to the energy of accelerated lead ions smashed together in order to create a QGP, not to the energy of accelerated protons.
     
  6. Feb 23, 2015 #5
    Re hello,
    thank you for your reply, I begin to think
    wikipedia that it was mistaken in citing this energy!

    I think I got the answer to my question! , Thank you;)
     
  7. Feb 23, 2015 #6

    Orodruin

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    Here is a CERN link about lead acceleration up to 33 TeV using the SPS: http://na49info.web.cern.ch/na49info/Public/Press/general.html
    Note that it is easier to accelerate lead to 33 TeV than it would be to accelerate protons to 33 TeV (just as it is easier to accelerate protons to high energies in a circular collider than it is to do it with electrons). The higher mass of the lead ions means you can get to higher energies with less losses in the form of synchrotron radiation.
     
  8. Feb 23, 2015 #7
    Just to state that the important difference between the lead ions and the protons here is the charge, not the mass. A lead ion has a charge of 82 times the proton charge, which means you can accelerate it to 82 times higher momentum with the same magnetic field. Unlike electrons, synchrotron radiation is already not the limiting factor for proton and ions, but the curvature of the motion in the magnetic field is.
     
  9. Feb 23, 2015 #8

    Orodruin

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    This is what I guess for just dealing with elementary particles I guess. :rolleyes:
    But of course this is true, the gyroradius is given by
    $$
    r_g = \frac{p}{qB}
    $$
    and for highly relativistic particles ##p \simeq E## so for fixed ##r_g## and ##B##, the energy would be directly proportional to the charge (surprise, surprise, ##82\times 450\mbox{ GeV} \simeq 37\mbox{ TeV}##).
     
  10. Feb 23, 2015 #9

    mfb

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    Protons at design energy lose about 10 keV per turn in the LHC, at the SPS the energy is lower by a factor of ~15 and radiated power scales with ##\gamma^4##, so even with the smaller curvature radius the energy loss should be in the range of tens of electronvolts per turn - completely negligible.
    Lead ions lose a factor of ##82^2## more energy at the same gamma-factor, but that is still no problem to compensate.

    Proton and ion synchrotrons are always limited by the dipole magnets for the curves, so the maximal energy is proportional to the charge of the accelerated particles.
     
  11. Feb 23, 2015 #10

    Vanadium 50

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    To avoid exactly this confusion, physicists usually use GeV per nucleon instead. So at the SPS, the beam energy is 177 GeV per nucleon.
     
  12. Feb 23, 2015 #11
    hi
    Thank you for all the answers
    good all week
     
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