# How does the force of charge between particles change with velocity?

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?

• Does this answer your question? How can we prove charge invariance under Lorentz Transformation? – DavidH Jun 27 '20 at 18:36
• So force is Lorentz invariant? – CheeseMongoose Jun 27 '20 at 18:52
• Your question asks two different things: How charge varies, and how force varies. – G. Smith Jun 27 '20 at 19:03
• So the force a charged particle exerts on another charged particle is in no way dependent on the charged particle's charge? Because if you say force isn't invariant yet force depends on charge which is invariant, it seems like a contradiction. – CheeseMongoose Jun 27 '20 at 19:05
• Force does depend on charge, but not only on charge. – G. Smith Jun 27 '20 at 19:08

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