I'm trying to understand the modern description of high-energy scattering processes involving hadrons in the initial states. The phenomenological parton distributions functions play a central role, and as I understand it at the moment, if we are e.g. talking about gluons, the function $G(x, Q^2)$ is the probability of finding a gluon with momentum fraction $x$ inside the hadron if the transmitted four-momentum is $Q^2$.

When these functions are plotted, I often encounter plots showing $x G(x, Q^2)$ instead of simply $G(x, Q^2)$. Why is this so? Is this just because the plots look a lot nicer if plotted this way? Or is there some deeper reason behind this that I haven't figured out yet?

As an example, take a look at the plot used on Wikipedia. http://en.wikipedia.org/wiki/File:CTEQ6_parton_distribution_functions.png

(Picture from: http://en.wikipedia.org/wiki/File:CTEQ6_parton_distribution_functions.png)

  • $\begingroup$ Honestly I really don't know, but just a guess: to make it better visible how fast $G\to\infty$ for $x\to0$? Though an ordinary logarithmic scale would probably do that job better. $\endgroup$ – leftaroundabout Sep 12 '11 at 19:15
  • $\begingroup$ $G$ cannot approach infinity... it is a probability density function. $\endgroup$ – AdamRedwine Sep 12 '11 at 19:28
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    $\begingroup$ @Adam: there's no reason it can't approach infinity as $x\to 0$, as at least the gluon PDF seems to do, as long as the proper normalization conditions on the function as a whole are satisfied. $\endgroup$ – David Z Sep 12 '11 at 19:36
  • $\begingroup$ @David: Ahhh... right. I had trouble imagining this, but thinking about the Dirac delta function it makes sense. Sorry, I haven't been in school for a year and I'm a bit rusty. :) $\endgroup$ – AdamRedwine Sep 12 '11 at 19:52
  • $\begingroup$ Fair enough, I forget little things like that all the time :-) $\endgroup$ – David Z Sep 12 '11 at 20:00

As you say, "$G(x,Q^2)$ is the probability of finding a gluon with momentum fraction $x$ inside the hadron if the transmitted four-momentum is $Q^2$." In other words, $G(x,Q^2)$ is a probability density function. As you can see from the article, in this case the expectation value of the variable is

$E[X] = \int_{0} ^{1} x\cdot G(x,Q^2) dx$

The plot of $x\cdot G(x,Q^2)$ then gives an intuitive sense of the "contribution" to the expectation value of the probability density function.

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    $\begingroup$ the integral should be from 0 to 1. An additional interpretation is that the area under the curve is the total momentum fraction of this parton type. $\endgroup$ – luksen Sep 12 '11 at 19:34
  • $\begingroup$ Thanks for the correction on the integral. I'm not actually an HEP guy, I just recognized the statistics of the situation. $\endgroup$ – AdamRedwine Sep 12 '11 at 19:36

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