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Proton-proton collisions

by Quarlep
Tags: collisions, protonproton
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Quarlep
#1
Aug14-14, 09:29 AM
P: 78
I learned that Pauli principle says two fermions cannot be same place than how we can collide two protons actually its also impossible (for me) to show this in feymann diagram.I want to know how its possible.

Thanks
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mfb
#2
Aug14-14, 12:38 PM
Mentor
P: 12,023
I learned that Pauli principle says two fermions cannot be same place
It does not. They cannot be identical in all quantum numbers. The protons are moving relative to each other, so this is not an issue.

What do you want to show in a Feynman diagram?
DrDanny
#3
Aug14-14, 04:05 PM
P: 5
Generally, particle interactions like collisions should not be thought off as two billiard balls colliding, this is two classical a picture.

In particle physics, interactions between protons are usually mediated by photons. Two protons colliding means: they get near enough each other that their mutual electric fields change their directions. On a Feynman diagram, you would show this as a photon being exchanged by the two protons.

ChrisVer
#4
Aug14-14, 07:50 PM
P: 1,008
Proton-proton collisions

additionally to DrDanny or a pions.... For weak interactions (beta decay) you can only work with protons at a 4point Fermi interaction approximation. I am not sure if in general you can use the vector bosons for proton to neutron or vice versa coupling.
exception: for if you work at high energies and so the interactions are hard scattering processes (interactions between the partons), so you would need to take into account the quarks and gluons.
Quarlep
#5
Aug15-14, 02:44 AM
P: 78
thanks
UVCatastrophe
#6
Aug17-14, 06:22 PM
P: 3
Not sure why you're using field theory to compute this process, unless you're at energies much higher than the QCD scale, in which case you'd be using perturbative QCD. In this case, the issue is irrelevant because the degrees of freedom there are quarks and gluons.

Two nucleons can interact via a contact interaction at low energies ala this paper. But don't take such a contact interaction literally; Feynman diagrams are equivalent to propagators which actually describe amplitudes in terms of wave mechanics. It is certainly possible for two identical fermions to have overlapping wave functions.

I think another source of your confusion is stemming from thinking about bound systems, whose spectra are discrete. A bound state can be described by [itex]\left\lfloor n \, (L\, S) J \, M_J ... \right\rangle[/itex], where the "..." denotes the possibility of internal degrees of freedom. In this case, you definitely can't put two fermions in a state where all the quantum numbers are the same. But in scattering states, a continuum of states is possible: [itex]\left\lfloor n \, (L\, S) J \, M_J ... \right\rangle \rightarrow \left\lfloor E \, (L\, S) J \, M_J .. \right\rangle.[/itex] Asymptotically, the overlap between two protons in a scattering experiment is zero, and their indistinguishability can be neglected.


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