3
$\begingroup$

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?

link to PDF image

Thanks!

$\endgroup$
3
  • $\begingroup$ "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. $\endgroup$ Aug 16, 2012 at 18:27
  • 2
    $\begingroup$ I was under the impression that parton distribution functions are probability densities and cannot be greater than one. here is the wrong statement. $\endgroup$ 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. $\endgroup$ Aug 16, 2012 at 19:13

1 Answer 1

2
$\begingroup$

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, $\int f(x)\mathrm{d}x$.

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$ 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
    Aug 17, 2012 at 4:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.