Deriving Lorentz force from relativistic force

Question

By EQ.$72$ of this, Richard Haskell proves Lorentz force law in subsequent sections. During the proof he says that,

Suppose a conductor of charge density $\rho'$ be at rest inside a frame of reference $S'$ which is in uniform motion relative to another frame of reference $S$. Then he states that on a test charge(which is invariant) $q$, the electric field as viewed from $S$ frame would be $\vec{F}=q\vec{E}$.

My question:-

If that conductor (source charge) is moving along with $S'$, then how can we say $\vec{F}=q\vec{E}$, why no other forces?

Answer 1

I think you have misunderstood author's argument. He is intentionally discussing how general (any) force transforms when changing frames in special relativistic world (formula 72). Then in section 17., he applies the general transformation to a special case where only electric force acts in frame $S'$, and derives how both electric and magnetic force may be present in the other frame $S$. Hence $\mathbf F' = q\mathbf E'$ is assumed in $S'$ as one possible case, and consequences in the frame $S$ are derived. In this, there is no requirement that all forces in all possible frames $S'$ have to be expressible as $q\mathbf E'$.

Also you seem to misquote the author, in the memoir there is no such formulation as "Suppose a conductor of charge density" and the subsequent text you wrote as far as I can see. When quoting other authors, it is best to do so verbatim, or at most with ellipsis (...) when omitting uninteresting parts.