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Problems with intuition for scattering / x-sections

by hagi
Tags: intuition, scattering, xsections
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hagi
#1
Jun15-14, 06:44 AM
P: 11
Dear PF,

although I've gone through many particle phyics lectures and text books, I still have problems with wrap my mind around the whole scattering theory and cross section topics.

1. Is there a deep reason why cross sections for charged, point-like particles decrease with the center-of-mass energy (the Mandelstam s) as [itex]\frac{1}{s}[/itex]? I read that it can be explained with the Heisenberg uncertainty principle, but I don't really see the connection.

2. Moving on to the differential cross section with respect to momentum transfer, I guess that the factor [itex]q^{-2}[/itex] is due to the Fourier transform of the Coulomb potential. In the space picture I understand that it's less likely for a particle to interact with a Coulomb potential if it's further away from it. What is the corresponding intuition in momentum space?

3. Fermi's Golden rule tells us that the interaction rate scales with the phase space, i.e., with the number of possible final states. How can I understand this intuitively? How does an electron know before annihilating with a positron to a photon how many possibilities of decay the photon will have? Similarly, why should the electron care about the fact that quarks have colors so that the cross section for the [itex]q\bar{q}[/itex] final state is three times higher than for muons?

Thanks so much in advance!
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Bill_K
#2
Jun15-14, 07:31 AM
Sci Advisor
Thanks
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P: 4,160
Quote Quote by hagi View Post
1. Is there a deep reason why cross sections for charged, point-like particles decrease with the center-of-mass energy (the Mandelstam s) as [itex]s^{-2}[/itex]? I read that it can be explained with the Heisenberg uncertainty principle, but I don't really see the connection.
Don't you mean s-1? For example, σ(e+e- → μ+μ-) = 4πα2/3s. It has to be s-1 from dimensionality considerations, since s is the only variable, and s-1 is the only way you can form an area. Intuitively, I guess, Heisenberg tells us that the target particle looks smaller at higher energy.

Quote Quote by hagi View Post
3. Fermi's Golden rule tells us that the interaction rate scales with the phase space, i.e., with the number of possible final states. How can I understand this intuitively? How does an electron know before annihilating with a positron to a photon how many possibilities of decay the photon will have? Similarly, why should the electron care about the fact that quarks have colors so that the cross section for the [itex]q\bar{q}[/itex] final state is three times higher than for muons?
This one's easy: "average over initial, sum over final." The experimenter can't distinguish the color of the quarks involved, and they all happen, so he has to sum over the various possibilities.
Jilang
#3
Jun15-14, 07:39 AM
P: 517
It's worth remembering too that the electron and positron don't annihilate with 100% certainty.

hagi
#4
Jun15-14, 07:43 AM
P: 11
Problems with intuition for scattering / x-sections

Quote Quote by Bill_K View Post
Don't you mean s-1?
Yes, of course I meant that, sorry.

I still don't understand. So if my projectile has low momentum and the spatial resolution is low, the particle doesn't look point-like but like a disc and I have to fold the potential with this disk? If I increase the momentum, the particle will approach its point-like nature and the potential is folded with a delta function?

This one's easy: "average over initial, sum over final." The experimenter can't disitnguish the color of the quarks involved, and they all happen, so he has to sum over the various possibilities.
It seems to be so obvious, but my problem is that I don't understand how we can sum the possibilities. If there were infinitely many colors the cross section and hence the interaction rate would be infinite as well?

In my intuition, the interaction between two particles determines how often they interact and just after this interaction I need to care about how I distribute this interaction rate among the possible final states. I'm sure I'm missing something...
Jilang
#5
Jun15-14, 08:20 AM
P: 517
Yes, you are missing the idea that the number of possible final states actually determines the interaction rate in the first place.
hagi
#6
Jun15-14, 09:21 AM
P: 11
Quote Quote by Jilang View Post
Yes, you are missing the idea that the number of possible final states actually determines the interaction rate in the first place.
Yes, and how can I understand that (apart from lengthy calculations)?
The_Duck
#7
Jun15-14, 09:53 AM
P: 855
Quote Quote by hagi View Post
Yes, and how can I understand that (apart from lengthy calculations)?
It's just regular probability theory: if you want to know the probability of any of a number of different things happening, you have to sum up the probabilities of all the possibilities.

Quote Quote by hagi View Post
If there were infinitely many colors the cross section and hence the interaction rate would be infinite as well?
Yeah, I think if you want a sensible limit when you take the number of colors ##N_c## to infinity you have to scale the coupling ##g## like ##1/\sqrt{N_c}## to compensate for this.
kurros
#8
Jun15-14, 05:56 PM
P: 364
Quote Quote by hagi View Post
Yes, and how can I understand that (apart from lengthy calculations)?
As The_Duck says, this is "normal" :). If I take a die and colour 5 sides red and one side blue, then you would not be surprised that it is 5 times more probable that the die lands on a red side than on a blue side.

As for this bit:

Quote Quote by hagi
It seems to be so obvious, but my problem is that I don't understand how we can sum the possibilities. If there were infinitely many colors the cross section and hence the interaction rate would be infinite as well?
Consider if the die had infinitely many sides. Probability is conserved, so the individual probabilities you are summing would each be infinitesimal. QFT also conserves probabilities so the cross-section isn't going to explode unless there is some bad problem occurring.


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