## What is the feynman diagram for electromagnetic force between an electron and proton?

 Quote by Vanadium 50 3. At leading order (one photon exchange) there is no difference between same-sign and opposite-sign. You have the same motion scattering off a hump as off a well. It's only with multiple photon exchange that it makes any difference.
This is absolutely false. You can take the non-relativistic limit of the lowest order electron-electron and electron-positron scattering processes—which are single photon exchanges—and show that, in the first case, you get a repulsive Coulomb potential and, in the second case, you get an attractive Coulomb potential. You certainly do not need higher order corrections for the sign difference to be manifest. Consult p.125 of Peskin & Schroeder if you need to refresh yourself on how the calculation goes.

Recognitions:
 Quote by Vanadium 50 There's a lot of brush in this thread that's worth sweeping away.

 1. These are not Feynman diagrams without arrows. Full stop.
This is trivially true, because you cannot calculate a diagram without defining an arrow CONVENTION.

 2. While a proton is a composite particle, there's absolutely nothing that prevents one from calculating this using a (properly arrowed) diagram.
This is plane wrong: Because the proton is not a point-like, you CANNOT have vertex at the proton's end, i.e., you cannot assign $\gamma^{ \mu }$ as the proton’s vertex factor. This is why we draw a blob (instead of a point vertex) representing two independent Lorentz-invariant FORM FACTORS and reflecting our ignorance about the proton structure.

Sam
 Mentor A Feynman diagram is a shorthand for a calculation. There are two different ideas that are getting entangled: is the calculation done correctly, and are the calculation's approximations (and since this is a truncated perturbation series, there will always be approximations) valid? For Mott scattering, when calculating using Feynman diagrams the target (in this case, the proton's) charge appears only as Z2. So at 1st order, electron-proton and positron-proton scattering is the same. There's a very nice QM treatment of this in Schiff (p.141) where he shows the difference between attraction and repulsion is proportional to 1/v - i.e. for one to distinguish attraction from repulsion, the projectile needs to spend a lot of time near the target. This agrees with our classical intuition, and it corresponds to the case in the Feynman perturbation series of many photon exchanges. When texts make the usual Mott approximation that m_e is zero, that of course is far away from the large 1/v situation where attraction and repulsion are distinguishable. For the proton's form factor, of course it has one, and in certain kinematic regions - large momentum transfer - it is important. In others, it is not: in particular, so long as the projectile wavelength is large compared to the target radius, the target appears more or less pointlike. In this kinematic range, there is nothing wrong with treating the proton as a point. (For a purely electric interaction - there's a subtlety with the gyromagnetic ratio for purely magnetic scattering) This is no worse than treating the proton as a point when applying Coloumb's Law at large distances - indeed, it is the exact same approximation.

 Quote by samalkhaiat This is why we draw a blob (instead of a point vertex) representing two independent Lorentz-invariant FORM FACTORS and reflecting our ignorance about the proton structure. Sam
Just saying, I thought the blob was Green function. I guess this is probably the matter of convention.

Can we simply use a single Green function to represent the whole electron-proton interaction instead?

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