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

Track momentum definition

  1. Jan 16, 2010 #1
    In experiments with nuclear emulsion (like CHORUS,OPERA...), the emulsion is scanned for tracks left by passing particles. In track analysis, there is a quantity called track momentum.

    I dont know what the track momentum is, so if anyone could help me, i would be grateful.
  2. jcsd
  3. Jan 16, 2010 #2
    In nuclear emulsion tracks, the developed blobs are due to the incident charged particle "hitting" atomic electrons and leaving ionization blobs in the emulsion, not to the charged particle hitting nuclei in the emulsion. There are 3 ways that the incident particle energy or momentum can be established;

    1) Multiple scattering: The small deviation of the track from the average trajectory is due to multiple Coulomb scattering of the incident particle on the nuclei in the emulsion. The mean square angular deviation of the incident particle is equal to
    θ2 = (21/βp)2XRAD, where β is the relativistic Lorentz factor, p is the particle momentum in MeV/c, and X is the radiation length of the emulsion.. This can be used to estimate the momentum of the incident particle.

    2) The density of blobs can be used to estimate the incident particle velocity at low kinetic energies (below about 1 m0c2). Review the Bethe-Bloch dE/dx energy loss equation.

    3) Track curvature in a magnetic field.
    Bob S

    [added] See correct projected angle multiple scattering equation at page 13 Eqn 27.12 in
    Last edited: Jan 16, 2010
  4. Jan 25, 2010 #3
    I want to thank you for your response. It will help a great deal in my understanding of particle momenta in emulsions.

    But, my problem is with definition of track, not particle momenta.
    Here is a excerpt from the text I was reading through:

    Each found track is fitted taking into account multiple scattering in the various materials
    between the hits. The track fit yields a χ2-probability P
    In this calculation,
    the hit resolution is assumed to be σ = 180 μm. Two effects cause the P(χ2)
    to deviate from the expected uniform distribution. First, the momentum of the track can
    be unknown
    . In this case, a momentum of 1GeV/c is assumed. Depending on the real
    momentum of the particle, this can over or under estimate the multiple scattering contribution
    to the hit resolution. Therefore, a cut on P(χ2)
    Introduces a bias on momentum,
    selecting preferably high momentum tracks. Second, the hit-residual distribution has
    tails with a rms width of about 750μm. These tails are probably due to cross-talk in
    the input window of the image intensifier or backscattering in the multi-channel plate.
    This leads to reconstructed tracks with very low χ2-probability

    If anyone is interested, full text can be found on:

    (excerpt is from page 70)
  5. Jan 25, 2010 #4
    The track momentum is closely related to the particle energy, kinetic energy, and rest mass:

    E2 = (pc)2 + (moc2)2

    For a given track, sometimes the mass of the particle is undetermined, for example differentiating between pions and muons. Multiple scattering lateral deviations and residual range (Bethe Bloch dE/dx equation) together can help identify the particle mass and the initial particle momenta (energy).

    The best nuclear emulsion work was probably done by the Barkas emulsion group (Barkas Birnbaum Smith) at Berkeley. See


    and the references therein. The Barkas group measurements on pions and muons determined their masses to around 0.1%, which were the reference mass values for many years.

    Bob S
    Last edited: Jan 25, 2010
  6. Jan 26, 2010 #5
    I now finally understand.

    Thank you so much for the time and effort it took you to clarify this for me.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook