9
$\begingroup$

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)

|cite|improve this question
$\endgroup$
  • $\begingroup$ 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
  • 3
    $\begingroup$ @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
9
$\begingroup$

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.

|cite|improve this answer
$\endgroup$
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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