Is Newton's third compatible with retarded Lorentz force?

Question

In Griffiths Introduction to electrodynamics it is said that Newton's third law is not valid in electrodynamics, but, in the example given, the it does not consider the retarded values for the fields and looks at the contradiction only on the magnetic force.

Just putting the books example: Two positive charges are moving along axes x and y towards the origin. The electric forces satisfy Newtons 3rd law, but the magnetic don't.

Below an image with the original example and a 'diagram' showing where my doubt comes from:

Question: Does Newton's third law hold for Lorenz force with appropriate retarded fields?

Answer 1

Your images do not explain very well what is your doubt.

If the charges moved uniformly for long enough in the past time, the electric forces are along the joining line of the particles, even if retardation is taken into account. The magnetic fields are perpendicular to velocities, so there are arrangements in which the momentum of the particles is not conserved; perhaps there is some special where momentum is conserved - can you find one?

If the particles accelerate, not even the electric fields are along the joining line, and the forces may not be even of the same magnitude. In such case, in general the momentum of the particles will no be conserved.

If electromagnetic momentum is defined based on Maxwell equations and the Lorentz force formula, and if the total momentum includes it, the total momentum is still conserved, but the electromagnetic momentum will not be as "visible", since it is not ordinary momentum of any particle, but is distributed in the whole space, most of it near the particles.

Answer 2

For the first figure shown, and assuming equal velocity magnitudes for both charges, I get that Newton's third law does not hold because:

1) The electric force of charge 2 on charge 1 is equal in magnitude and opposite in direction to that of charge 1 on charge. Both have x and y components.

2) The magnetic force of charge 2 on charge 1 is equal in magnitude to that of charge 1 on charge 2. The former has only y-component and the latter only x-component.

3) The total force of charge 2 on charge 1 is equal in magnitude to that of charge 1 on charge 2. However, the former and the latter will only be in approximately opposite directions for charge velocities small in comparison to c.

4) Bear in mind that, at a given field point (x,t) or (y,t), the electric and magnetic fields do not depend on the attributes of the charge passing through such point, i.e. its velocity or equation of motion.