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I was looking at a plot of the parton distribution functions today and had a question. On the y axis, it seems like the value of x f(x) for gluons is greater than one at small x. I was under the impression that parton distribution functions are probability densities and cannot be greater than one. Also, x is a fraction of momentum and can also not be greater than one. Does anyone know why this is?

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Thanks!

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  • $\begingroup$ "Also, x is a fraction of momentum and can also not be greater than one. Does anyone know why this is?" Bjorken x can be greater than one, thought this is rare. It implies the rest of the mass of the compound object moving momentarily backwards in the lab frame. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Aug 16, 2012 at 18:27
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    $\begingroup$ I was under the impression that parton distribution functions are probability densities and cannot be greater than one. here is the wrong statement. $\endgroup$
    – Grisha Kirilin
    Commented Aug 16, 2012 at 18:47
  • $\begingroup$ More about x>1. See the two experiment from JLAB's Hall C with "x>1" in their titles. I also understand that there is another instance planned once the 12 GeV upgrade is complete. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Aug 16, 2012 at 19:13

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A probability cannot be greater than 1, but a probability density can be. The parton distribution represents basically the probability per unit momentum fraction, so it can be large over a small region of $x$ without contributing much to the actual probability, $\int f(x)\mathrm{d}x$.

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  • $\begingroup$ Nor is the integral required to be one, as there are multiple up valence quarks in a proton and an indefinite number of sea quarks and gluons. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Aug 16, 2012 at 18:54
  • $\begingroup$ Actually, I think the integral $\int_0^1 f(x)\mathrm{d}x$ can be interpreted as the average value of the particle number operator, and that average is going to be 1 by definition. (I had to think about it for a while to reconvince myself of this) $\endgroup$
    – David Z
    Commented Aug 17, 2012 at 4:30

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