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I'm sorry for this question but, I just don't get it. According to the electromagnetic field theory, electrons repel each other by exchanging photons. How do protons attract electrons, by photon exchange?

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Try this first: How can [photons] be responsible for attractive forces? Also, look at… – Alfred Centauri Mar 24 '13 at 13:59
The idea that photon is a particle leaving and entering charged particle is no good. It would require a massive number of photons in each charged particle. In addition, since charged particles have mass this also means that using mass to move charged particles also leads to discharging of photons. In other words, gravitational interaction must supply charged particles with new photons. – user121963 Jun 25 at 17:10
maybe this answer of mine might help… – anna v Jun 25 at 18:57
@annav I have my own model. I think it is better than Feymann diagram. I just don't know how to refine a few things because many astro physics experiments seem so fraudulent. – user121963 Jun 25 at 21:20
@InstructedA look up "virtual particles" . The names, "photon" in this case , are just place holders with the quantum numbers but not the mass. – anna v Jun 26 at 3:08

ELECTRON-PROTON ATTRACTION: a simple, semi-classical analysis to avoid full scale QFT.

The exchange of the photon between the proton and the electron leads to attraction, only because the total energy of the electron is negative.

Let us consider the hydrogen atom for simplicity, and imagine the funnel-like shape of the electron energy. The total energy of the electron in the hydrogen atom at distance $r_0$ from the proton is

$E(r_0)=-\frac{e^2}{8\pi\epsilon_0 r_0}$.

If the distance $r_0$ is sufficiently short, then the electron will emit a photon which will be absorbed by the proton, and the amount of energy of the exchanged photon will be dictated by the uncertainty principle:

$\Delta E\Delta t=\hbar.$

But $\Delta t=\frac{r_0}{c}$ so that

$\Delta E r_0=\hbar c\rightarrow \Delta E= \frac{\hbar c}{r_0}$

So the new energy of the electron will be

$E_1=-\frac{e^2}{8\pi\epsilon_0 r_0}-\frac{\hbar c}{r_0}=-\frac{e^2+8\pi\epsilon_0\hbar c}{8\pi\epsilon_0r_0}$

or the equivalent amount of energy corresponding to some new position $r_1$

$-\frac{e^2}{8\pi\epsilon_0 r_1}=-\frac{e^2+8\pi\epsilon_0\hbar c}{8\pi\epsilon_0 r_0}$

from which we get $r_1$ in terms of $r_0$

$r_1=r_0\frac{e^2}{e^2+8\pi\epsilon_0\hbar c}<r_0$

Therefore the electron moves closer to the proton rather than farther from it (an attractive force.)

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In non-bound states, the total energy of the electron is positive. Does your analysis indicate that the EM interaction would be repulsive in that case? – Dan Mar 27 '13 at 19:14
According to your picture, it seems the energy lost by the electron is transferred to the proton so the proton now has more energy with which to climb out of the potential well. I must be missing something. – Dan Piponi Mar 27 '13 at 19:33
@Dan The simple analysis I presented does not rediscover the wheel. It simply shows consistency of the two pictures. You can do the analysis for the Proton-positron case (repulsive "force") in a similar, but not identical way. – JKL Mar 27 '13 at 20:57
@DanPiponi When two particles echange energy (photons) their energy and momentum do change. You need to take into account that the exchange of virtual photons lasts a very short time, for the electron or proton to climb or tunnel through the potential well. The proton does recoil, but this recoil is very small and pulls the electron with it anyway. This is how hydrogen atoms can move about without getting ionized by the slightest impact. – JKL Mar 27 '13 at 21:09

Photons aren't like billiard balls. A photon isn't a particle in the sense of a hard, massive bit of stuff, it's a "chunkyness" of the Electric and Magnetic fields. When we say protons and electrons are attractive because they exchange photons, we mean protons create EM fields, and electrons create EM fields that cancel out those created by the proton.

The fields themselves are exactly the same as the classical EM fields, the only exception is that they can only be made and cancelled out in chunks. The reason that photon exchange can be attractive is precisely the same reason classical electric fields can be attractive.

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protected by ACuriousMind Jun 25 at 17:02

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