Re: attractive force via particle exchange - how?


shevek4@xxxxxxxxx wrote:
> Every now and then somebody expalains electromagnetic forces as
> "exchanges of virtual photons". My question is simple: how can this
> be, for the case of attractive forces?
>
> For example, let's take an electron and a positron. If the electron
> emits a photon in the direction away from the positron, this will give
> it momentum toward the positron. However, this is not an exchange of
> particles but a net emission of them. If the emitted particles go
> towards the other particle (an exchange), this will be a net repulsion.
>
>
> Some have answered that the exchanged particles have a "negative
> momentum" - supposedly a momentum directed oppositely to the velocity
> vector! This sounds completely crazy. Others have answered that the
> analogy of force-carrying particles is a mathematical one and the
> Feynman diagram reasoning involved in my question is simply misplaced.
>
>
> Is there a better answer?
>
> Thanks!

I think the error you make has to do with associating too classical a
picture to the Feynman diagram. The diagram is a mnemonic for writing
terms that contribute to an integral. The integration is over all
possible internal values of momentum and energy *including apparently
non-physical ones*.

However, this does point to something that has always bothered me.
Changing the sign of one of the charges of the scattering particle
changes the sign of one factor in the leading terms (say, t-channel
photon exchange between electron-electron or between electron
positron). Summing the lowest-order terms and squaring to get the
cross-sections produces *identical* integrals for electron-electron
scattering and electron-positron scattering. In a way, this makes
sense, because a deflection away from a charge and a pulling around a
charge at small impact parameters should produce the canonical
Rutherford scattering formula, either way. One the other hand, I would
think that so fundamental difference between electrostatic attraction
and repulsion should show up somewhere in the lowest-order calculation.

Have I missed something?

PD

.

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