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Uncertainty principle and particle accelerators

  1. Feb 23, 2012 #1
    Fairly simple question but it's been bugging me for a while:

    Particle accelerators such as the LHC publish some impressive images of the tracks of particles in their detectors. Can someone explain why that is possible considering the uncertainty principle?
  2. jcsd
  3. Feb 23, 2012 #2

    Vanadium 50

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    Work out the uncertainty relation for a track and you can see for yourself.
  4. Feb 23, 2012 #3
    Great question! I asked this question here before also, and got no conclusive answer.

    Why high momentum when Heisenberg's relation only talks about uncertainty of momentum? Why should high momentum decrease uncertainty in position?

    Here I asked these kind of questions, especially starting from post 6 onwards.
  5. Feb 23, 2012 #4


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    You're thinking the appearance of the track depends somehow on the width of the particle's wavepacket, but it doesn't. The track is marked by the aftermath of a series of atomic collisions, which in the old days produced bubbles, or developed the granules on a photographic plate. Nowadays the innermost detectors in CMS and ATLAS are silicon chips about 100 microns in diameter. The resolution is the same regardless of the particle's energy.

    High energy is relevant in the collision itself but not the track.
  6. Feb 23, 2012 #5
    Just for clarification: the need for the high energies in particle accelerators has nothing to do with HUP?
  7. Feb 23, 2012 #6


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    The question in this thread, Lapidus, was about the track. I said that high energy is relevant in the collision itself but not the track.

    As an illustration of your own question, consider the electron scattering experiments that determine the size ("charge radius") of a nucleus. At low energy the differential scattering cross section σ(θ) doesn't change much as you vary the energy, it's known as the Mott cross section σM(θ). However as you raise the energy, to the point where the electron's deBroglie wavelength q approaches the size of the nucleus, σ(θ) begins to decrease: σ(θ) = σM(θ) |F(q)|2 where F(q) is the Fourier transform of the nuclear charge density, it's known as the form factor. Approximately, F(q) = 1 - (qa)2/6 where a is the charge radius. So you can see the effect of the finite size of the nuclear charge distribution, but only if the electron's deBroglie wavelength is small enough, i.e. a comparable size.

    The need for the much higher energy used in LHC collisions is simply to provide enough energy to create the particles.
  8. Feb 24, 2012 #7
    Thanks for this long and clear answer! Very much appreciated.
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