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Parton distribution functions plot axes

  1. Jan 2, 2013 #1
    I am slightly confused by the labelling of the vertical axis on parton distribution function plots.

    Take the one here: http://www.hep.phy.cam.ac.uk/~wjs/partons2008nlo.jpg [Broken]
    as an example.

    The vertical axis is labelled as xf(x, Q^2), where f is the probability density of finding a particular parton with a given longitudinal momentum fraction x.

    I think it should simply be f(x, Q^2). Why the extra 'x' in front of f?
     
    Last edited by a moderator: May 6, 2017
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  3. Jan 2, 2013 #2

    Vanadium 50

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    f and xf are two different things. f is the fraction of partons carrying a momentum fraction f, and xf is the fraction of the proton's momentum carried by partons that each carry a momentum fraction x.
     
  4. Jan 2, 2013 #3
    Is the purpose of plotting the product xf instead of f to compress the functions along the vertical direction? That is to say, f goes up for low x quite a bit, but the product xf increases less as one decreases x.
     
  5. Jan 2, 2013 #4

    Vanadium 50

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    It does that, but the purpose is to represent something different than the fraction of partons carrying a momentum fraction x. It's to represent the fraction of the proton's momentum carried by partons that each carry a momentum fraction x.
     
  6. Jan 3, 2013 #5
    I am having difficulty parsing this sentence. "fraction of the proton's momentum" would imply the vertical axis wouldn't go above 1, but it does. Could you phrase it longer/differently?

    If I think of the proton as a sack of balls with the sack moving with a large momentum, then all the balls must be moving at the same speed or they wouldn't stay together. In the approximation that the differences in mass are negligible then all the balls will have the same momentum. That is to say, if a ball has momentum fraction x, then there must be 1/x balls in total in the sack each carrying fraction x of the sack's momentum.

    Being a quantum sack, the number of balls is not fixed and it is the task of the balls' distribution function to determine the number and kind of them (e.g 10 gluon balls, 2 up balls, 1 down ball).

    Is this the right way to think about it?
     
  7. Jan 3, 2013 #6

    Vanadium 50

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    Maybe examples will help:

    The integral of xf(x) must be 1. (The fraction of momentum carried by all of its constituents must sum to 1).

    The integral of f(x) can sum to anything.

    The integral of u(x)-ubar(x) must sum to 2. (2 up quarks)

    The integral of d(x)-dbar(x) must sum to 1. (1 down quark)
     
    Last edited: Jan 3, 2013
  8. Jan 3, 2013 #7

    Bill_K

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    The point of having a parton distribution function of course is that they do not all have the same momentum. Paraphrased from Halzen and Martin: fi(x) represents the probability that a particular parton carries a fraction x of the proton's total momentum. All the fractions have to add up to 1, therefore ∑i∫ xfi(x) dx = 1.

    And thus good reason for focusing on the quantity xfi(x), which as 50V said, is "the fraction of the proton's momentum carried by partons that each carry a momentum fraction x."
     
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