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by greysphere
660 days ago
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In a classical view of 2 particles accelerating towards each other v x r will always be 0 so B will always be 0 even if the particles are accelerating towards each other. I believe all this holds under QFT [1]. Looking further a redefinition E is necessary when including the \beta factor [2]. So that was a mistake on my part - relativity does change the rhs of Coulomb's law. Admittedly the problem as stated (two particles falling towards each other) constrains things in such a way that there is no off-axis contribution. Or to put it another way, 1d electromagnetism doesn't have magnetism. [1] https://physics.stackexchange.com/questions/142159/deriving-... [2] https://en.wikipedia.org/wiki/Coulomb%27s_law#In_relativity |
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It doesn't even hold in ordinary QM (because there are no point charges anymore, only charge densities) and it fails completely in QFT for interacting fields. What you linked is an intro example from many textbooks that shows how the tree level diagram in the nonrelativistic limit can indeed yield the Coulomb potential. At the tree level you will often see such "classical" behaviour. But if you consider higher order corrections, the picture changes rapidly. See the Uhling potential for a practical example, but for higher loops this gets analytically intractable very quickly. The world starts to work very differently once you reach these length scales.