Magnetic field around a current carrying wire and Special Theory of Relativity

Question

I understand how the magnetic field and electric field (around a current-carrying conductor) swap their roles depending on the frame of reference, due to the effect of length contraction (in the special theory of relativity); however, this question has been troubling me.

Consider two charges separated by a vertical distance and both moving horizontally with equal velocities (with respect to the ground frame). In the ground frame, we can observe both a magnetic effect (each moving charge produces a magnetic field, in which the other charge is moving) and an electric effect of each upon the other. But in the charges' frame, there is only the electrostatic force (of the same value as that of observed in the ground frame because the charge and the vertical length are invariant). How is this possible? What is the actual value of force between the charges?

Answer 1

In a frame with two equal charges (a source charge and a test charge) separated by $y$:

The electric field from the source at the test charges is:

$$ \vec E = k_e \frac q {y^2} \hat y$$

and the force on the test charge is:

$$ \vec F = q\vec E = k_e \frac {q^2} {y^2} \hat y$$

If we boost this by $\vec v = -v\hat x$ so that the two charges appear to move in the $+x$ direction, then the electric field at the test becomes:

$$ \vec E' = \gamma(\vec E + \vec v \times \vec B)-(\gamma-1)(\vec E\cdot\hat v)\hat v = \gamma \vec E$$

There is also a magnetic field:

$$ \vec B' = \gamma(\vec B - \frac{\vec v \times \vec E}{c^2})-(\gamma-1)(\vec B\cdot\hat v)\hat v = \gamma\frac v {c^2} ||E|| \hat z$$

The Lorentz force law

$$ \vec F' = q(\vec E' + \vec v' \times \vec B) $$

gives:

$$ \vec F'=q(\gamma \vec E - \frac{v^2}{c^2}\vec E) = \vec F/\gamma $$

So the 3-force is not a Lorentz invariant. Note that when the charges appear to be moving, the electric field attraction is stronger, but it is mitigated by an opposing magnetic force.