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I am happy with the concept of electrons interacting with each other through the emission and absorption of photons, but what I don't understand is how the negative charge on an electron is expressed through the medium of these photons.

How does a photon produced by a negatively charged particle differ in terms of frequency to that produced by a positively charged one?

Magpie
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puco
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  • this is a duplicate of http://physics.stackexchange.com/q/148794/ – anna v Nov 30 '14 at 14:52
  • @annav this is a very specific question about the frequency of the emitted photon. – Magpie Nov 30 '14 at 14:56
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    Why do you assume there should be a difference in terms of frequency? – Jim Nov 30 '14 at 15:00
  • @Magpie virtual photons have a spectrum of energies but because they are virtual they are off mass shell, not really photons. I have answered the content of this question in the linked duplicate – anna v Nov 30 '14 at 16:15
  • @annav your answer may also answer this question but that does not make the questions themselves the same. This question is very specific and imho it is not asking the same thing that the question you linked to was asking. – Magpie Nov 30 '14 at 16:17
  • I have never seen a nice explanation of an answer to this question. The closet I have come across is http://math.ucr.edu/home/baez/physics/Quantum/virtual_particles.html – DJBunk Nov 30 '14 at 17:13
  • I join this question too. I got explanation only about what is NOT, but not about what is YES. – Sofia Nov 30 '14 at 17:29

2 Answers2

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The frequencies are the same but they are 180 degrees out of phase with each other.

Guill
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  • what frequencies !!! – anna v Dec 04 '14 at 05:30
  • I'm not so sure about that, phase is a matter of timing implying that the charged particles were in cahoots before any photons were exchanged. To rephrase my question for all you quantum physicists out there: picture a hypothetical charged particle( ill leave the other properties to your imagination ), write down an expression for it , now the component of the equation(s) that represent the negative charge;what fundamentals does it contain, how is it validated, where did you get it from. – puco Dec 04 '14 at 07:17
  • You are asking for a course in quantum mechanics and quantum electrodynamics in one answer to a question in this site? it requires two books to really get into if you have not had any courses. – anna v Dec 04 '14 at 08:32
  • Puco, the particles are "in cahoots" as you say, because their fields extend to infinity, so they are "aware" of each other, before the photons are exchanged. – Guill Dec 12 '14 at 00:38
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The concept of photon exchange comes from Feynman diagrams and the exchange is of a virtual photon, a four vector carrying the quantum numbers of the photon ( or whatever the off mass shell particle is) but the mass is not on the mass shell, and has a different value according to the calculation of the process under study.

repulsive virtual

This is a simple description of heuristically understanding how one can write with Feynman diagrams electron electron and electron positron interactions. For the diagram above it "understands" repulsion by the momentum carried by the virtual photon four vector.

Clearly the recoil of the first electron and the impact of the second electron with the photon drive the electrons away from each other.

The attractive force between electron and positron pulls in the Heisenberg uncertainty principle for an intuition ( it is the boomerang analogy turned to Feynman diagrams for those who know it from elsewhere):

attractive force

The uncertainty principle makes this possible . The attraction between an electron and a positron may be described as follows: the electron emits a photon with momentum directed away from the positron and thus recoils towards the positron. This entails a degree of definiteness in the momentum of the photon. There must be a corresponding uncertainty in its position - it could be on the other side of the positron so that it can hit it and knock it towards the electron .In the diagram above the wavy line represents the SAME photon.

The real reason that attractive and repulsive forces appear in quantum mechanics is that the solutions of the Schrodinger equation have different potentials and thus the wave functions differ . In this link there is reference to an example that can be solved simply, of attractive and repulsive potentials.

Now on frequencies and virtual photons. The simple exchange diagrams give a formula of how to create the integrals that have to be evaluated so that a calculation of the scattering crossection, i.e. probability of interaction, can be done. If you read the link, you will see that in equation 23 on the four momentum vector dot product, q^2 the four momentum transfer squared, of the exchanged virtual photon enters in the denominator of an integral over q. As for a real photon q^2 is zero ( its mass) it is evident that this photon for most of the integration is off mass shell, i.e. its energy^2 minus its momentum^2 can be negative or positive ( depending on the diagram kinematics), not a definite value so as to be substituted into E=h*nu and give a frequency.

If it is negative, more so, as for the positive one might use the de Broglie wavelength for the effective mass. Positive q^2 can happen for electron positron scattering generating particles, as at LEP, or annihilating into two gammas. Your question though, asking about an electron vaguely interacting with the charges in the environment is in an undefined regime of interactions and thus the term frequency cannot be applied.

anna v
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