Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

The GLAP equation describes the evolution of parton distribution under successive branchings. When branching takes place at a finite angle, the outgoing parton carries some finite transverse momentum. However, this is ignored in deep inelastic scattering calculations, where the parton momentum is always assumed to be longitudinal. I have trouble understanding why the angle should be ignored. The GLAP equation effectively comes from resummation of collinear branching to all orders in perturbation theory. So you can imagine 100 branchings, each at angle of $\pi/200$, resulting in an outgoing parton at an angle of $\pi/2$, i.e. perpendicular, to the velocity of the incoming nucleon. Why is this not a problem?

share|improve this question
1  
If we assume the branching is a memoryless process, then the central limit theorem indicates that the contributions from wide angles are superexponentially damped. That is, angles have a strong tendency to roughly cancel. (This is a mathematical statement, and I don't know enough about the physics to say anything substantial there.) –  Scott Carnahan Jul 18 '11 at 3:43
add comment

1 Answer

up vote 1 down vote accepted

A couple of reasons: first, the whole framework of parton distributions is developed to apply to high-energy hadrons, as they appear in scattering experiments. These hadrons move with a very large longitudinal momentum, typically hundreds or thousands of GeV, and except at very small $x$, the partons will also have large longitudinal momenta. If a parton is going to diverge from the hadron path by any significant angle, it'll have to have a similarly large transverse momentum, and outside of hard scattering events, there just isn't enough energy around to do that.

At small $x$, however, it's conceivable that a parton could have an amount of transverse momentum comparable to its longitudinal momentum. The reason we don't care about this is that the cross section for parton splitting is roughly inversely related to the transverse momentum transfer ($\hat\sigma \sim 1/p_T^2$ I think), so it exhibits a sharp peak at $p_T^2 = 0$. This means we can consider the only non-negligible contribution to come from $\theta = 0$, i.e. true collinear branching, and any significant deviations of partons from the original longitudinal path are exponentially suppressed.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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