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First, the electromagnetic frequency spectrum, (EFS) is just a definition. It represents the change in intensity of the electric field and the magnetic field and they are defined as being at right angles to each other. The frequency is defined as the inverse of the average time from one null to the next or one peak to the next peak in field intensity. Don’t ...

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Briefly, no. Charged particle interaction is fundamentally a non-instantaneous photon exchange. The interaction can be written or expanded as the ''naive'' instantaneous interaction ($e^2/r$) plus a photon exchange portion which contains exactly a $-e^2/r$ term and a non-instantaneous (retarded in time) term. The author calls the first term the Coulomb ...

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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. This is a simple description of ...

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

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Since the field of each charged particle extends to infinity, the fields of two particles are "in contact" with each other (no "communication" is necessary). When the charges are not equal (+ & -), the fields "cancel" each other along the line connecting their centers. This causes the attraction of the particles. When the charges are the same (+ & + ...

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I just realized that the answer is just stupid. In the case of two outgoing photons when computing $M^*$ you get the epsilons (not complex conjugated) that in the full $|M|^2$ will allow you to use $\sum\epsilon_{\mu}\epsilon^*_{\nu}\to-\eta_{\mu\nu}$

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The unrenormalized vertex diagram is the dimensionally regularized Feynman integral corresponding to the following single Feynman diagram: It is called ‘unrenormalized’ because it is not accompanied by a diagram in which the momentum loop is replaced by a counterterm vertex generated by the QED Lagrangian. Adding a second Feynman diagram with the ...

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Now per QED, electrical charges interactions are effected by photons. Suppose you are one of the two charges. How do you know to attract or repel the other charge? You want something that does not exist - intuitive picture of physical process within a theory which is a demonstration of how far can one go with mathematisation of experience and ignoring ...

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The attraction of unlike charges and the repulsion of like charges is an experimental observation that has to be included in any model of electromagnetic reactions When talking of photons one is in the quantum mechanical regime. As in classical electrodynamics the sign of the charge defines the potential, attractive or repulsive, between the two charges, ...

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There are a couple of reasons why your question isn't answerable. Firstly if you're sitting outside the horizon trying to measure the colour of the particle then you'll never see it reach the horizon, let alone cross it, so you would just measure a white particle as usual. If you were sitting alongside the particle as it fell then there would be no horizon ...

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