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I'm sure how fast a particle moves must have some relativistic effect, or maybe also classical ones too. Suppose you fixed the positions of two charged particles. Suppose you're in a lab frame and in a paradoxical but analytically useful way, you fix the position of a particle but change its velocity towards or away from you at varying speeds.

How does the charge the lab frame measure vary due to super-relativistic effects? If the particle moves away near the speed of light, is its charge higher than it would be classically or something like that?

Qmechanic
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Charge is independent of velocity. For example, the charge of a proton is $1.6\times 10^{-19}$ coulombs whether it is at rest or zooming around the Large Hadron Collider at 0.99999999 c. Charge is a Lorentz-invariant quantity.

Forces, whether electromagnetic or of some other kind, are not Lorentz-invariant quantities. The “three-force” familiar from Newtonian mechanics has a messy transformation law under Lorentz transformations which you can find in Wikipedia. In relativistic mechanics, physicists therefore prefer to talk about the “four-force” which has a nice Lorentz transformation that looks just like the transformation of spacetime differentials. Both are four-vectors.

The reason that charge can be invariant while four-force can transform like a four-vector is that the electromagnetic force on a charge depends not just on the charge but also on its velocity and the electromagnetic field produced by other charges. Both the velocity and the field transform under Lorentz transformations. The Lorentz force law looks like this in relativistic notation:

$$f^\mu=qF^\mu{}_\nu u^\nu.$$

Here $f^\mu$ is the four-vector describing the electromagnetic force, $F^\mu{}_\nu$ is the four-tensor describing the electromagnetic field, and $u^\mu$ is the four-vector describing the velocity of the charge.

If one understands this notation, it makes clear that forces, EM fields, and velocities have straightforward Lorentz transformations, but the charge $q$ must be Lorentz-invariant because otherwise the right side would fail to be a four-vector like the left side is.

G. Smith
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