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In LHC collisions are protons aligned one-into-one?

  1. Aug 13, 2012 #1
    My understanding, please correct me if I am wrong, is that in the LHC proton-to-proton collision experiments it is technically not feasible to align protons directly into each other. Why is that and where can I look this up?
  2. jcsd
  3. Aug 13, 2012 #2
    The protons are small (around 10-15m diameter) so it is nigh impossible logistically to make one proton directly hit another one. Instead they make a beam of about 1011 protons with a radius of approximately 10-6 m. When two of these beams hit each other, the chances are that some protons will collide.

    Here's a good place to look for more information http://www.lhc-closer.es/php/index.php?i=1&s=3&p=9&e=0
  4. Aug 14, 2012 #3
    Rooted is correct.

    However, maybe you were referring to a different issue.

    The two colliding proton beams are also not aligned intentionally, not because of technical limitations.

    This is because if they were aligned there would be collisions at many points across the beam axis (every point proton bunches cross), and there would be a "mess"
    Therefore, the beams have a small relative angle which makes sure they only cross at one point
    Which is the collision point
  5. Aug 14, 2012 #4
    Thanks for the link. I saw that page before and found it most informative. I was wondering if there was an article about this impossibility (direct hit alignment) somewhere?
  6. Aug 14, 2012 #5
    Rooted's answer was what I was referring to. Tricky stuff, it is amazing what they can and cannot do.
  7. Aug 15, 2012 #6
    Magnetic 'lenses' of a sort (casually called 'optics', a handy term to use for research) are used to control the protons around the ring, to bend them into a circle, focus and defocus the beam, and accelerate them. The protons are continually repelling each other, so has a tendency to widen the beam. You probably will have seen the term 'luminosity' in your research? This is an important concept. Integrated luminosity can give you information about how many 'events' you are likely to see. The luminosity is directly related to properties of the beam. Perhaps this will be useful to you? http://pdg.web.cern.ch/pdg/2011/reviews/rpp2011-rev-accel-phys-colliders.pdf [Broken]

    Just thinking about it a little bit, if you were actively trying to collide one proton directly against another there would be a significant Heisenberg Uncertainty in the position of the proton as the velocities involved are so relativistic. This page does a calculation that shows the uncertainty in position is ten times greater than the radius of the proton. http://www.relativitycalculator.com/Heisenberg_Uncertainty_Principle.shtml This effect by itself would be enough to make a head-on collision very challenging.
    Last edited by a moderator: May 6, 2017
  8. Aug 15, 2012 #7


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    "1-into-2 or more protons." does not happen with any reasonable probability.

    Concerning selection and so on:

    CMS and ATLAS see a bunch crossing each 50ns (design: 25ns), with ~15-25 collisions per crossing (design: 25). Here, a collision is anything hard enough to give tracks in the detector. Most events are "boring" - simple electromagnetic deflection of the proton, or just some pions or other light particles are produced. Some events are more high-energetic - especially all events with heavy particles (top, Higgs).
    It is impossible to store all this data. Therefore, both detectors use trigger systems: Events without high-energetic particles are dropped, the remaining events are reconstructed and again events without something interesting are dropped. Those triggers reduce the input rate of 20MHz (bunch crossing rate) to ~100-200 Hz. Just those rare events (one out of 100.000) get stored. In the analysis afterwards, another filter step is applied, reducing the effective rate even further.

    The idea behind all this filtering is to remove as much background as possible, while keeping as much signal as possible at the same time. The triggers remove some Higgs events, indeed. But this fraction is well-known from simulations and has been taken into account.
  9. Aug 16, 2012 #8

    Vanadium 50

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    This thread has been set adrift on the Sea of Crackpottery. Since the OP's original question is answered, the thread is closed.
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