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Suppose I have two electrons alone in space. They will repel each other and after some time they will have exchanged some momentum.

That momentum has been transfered by a number of fundamental interactions, with one photon being exchanged in each interaction.

That means we can calculate the average frequency for the interaction to happen that should, assuming the electrons are at rest, only depend on the distance.

I'm asking if that's sensible and then what the frequency is for a given distance.

I'm also curious what the respective frequency one gets for strong interactions between quarks in a proton or a neutron. In that case, the distance is already given by the distance of the quarks, which I assume is known.

John
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  • What do you mean with "after some time they will have exchanged some energy in momentum." both gain kinetic energy and loose potential energy – trula Sep 14 '21 at 13:54
  • @trula You're right, I've removed talk about energy here. – John Sep 14 '21 at 14:45
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    "That momentum has been transfered by a number of fundamental interactions, with one photon being exchanged in each interaction." [citation needed] – ACuriousMind Sep 14 '21 at 15:04
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    See https://physics.stackexchange.com/q/142159/50583 and its linked questions (in particular https://physics.stackexchange.com/q/2244/50583) for other discussions of how the classical Coulomb force is actually related to the quantum field theoretical picture – ACuriousMind Sep 14 '21 at 15:05
  • Thanks, but let me rephrase: I'm interested in the average time that a Schroedinger equation for the decribed setups has to it's interaction. Clearly some points in time and space are more likely than others, and there will be a "center of mass" in space and time (which might itself be an unlikely point of interaction). What's the time part? – John Sep 14 '21 at 16:52
  • You have to understand, by reading the links given by ACuriousMind, that the solutions of Schrodinger equation for two electrons is complicated , see https://www.sciencedirect.com/science/article/pii/0021999173900302 . Furthermore your " with one photon being exchanged in each interaction." you are already talking in terms of a quantum field theory depiction of the interaction of two particles. QFT developed exactly because the solutions of the equation were verh complicated , not as in the opposite charge bound states, and the description of particles scattering off eachother needed a new – anna v Sep 15 '21 at 05:10
  • tool . QFT allows to describe scatterings, as it will be when one electron is repulsed by the other with the coulomb force. If you study QFT you will understand that the exchange of virtual particles is given by continuous mathematical functions, so there is no " frequency of interaction", but a continuum. see http://hyperphysics.phy-astr.gsu.edu/hbase/Particles/expar.html third page for e e scattering – anna v Sep 15 '21 at 05:13

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