Understanding why Einstein found Maxwell's electrodynamics not relativistic

[Dale]

But I'm not sure why I'm not contradicting you. You said, "Although the energy does come from the B field, the force and the B field are not the same thing and are in fact not even proportional." I countered, "On inspection, it appears as though the force is proportional to the strength of the magnetic field." I expected you to tell me I am wrong. Instead you agreed with me.

A vector has a magnitude (strength) and a direction. So one vector (force) can be proportional to the strength (your claim) of another vector (B field) while not being proportional (my claim) to the other vector (B field) if they point in different directions. A closer inspection would show that they don't point in the same direction.

However, this is a minor point. The more important point is the one about work over closed paths.

 

[Dale]

This is easy enough to check. The Lorentz force is given by $f=q(E+v \times B)$. When we do a Galilean boost by a velocity u then q is unchanged as are E, and B, but v transforms to v+u. So in the boosted frame $f=q(E+(v+u) \times B)$ which is not the same as in the original frame. So the electromotive force is definitely not invariant wrt a Galilean transform.

So, if we return to this post and use the correct pair of Galilean transforms we get the following (in units where c=1):

Magnetic limit:
$E' = E + u \times B$
$B' = B$
$f'=q(E' + v \times B') = q(E + u \times B + (v + u) \times B) \neq f$

Electric limit:
$E' = E$
$B' = B - u \times E$
$f'=q(E' + v \times B') = q(E + (v + u) \times (B - u \times E)) \neq f$

So the Lorentz force is not Galilean invariant under either limit. For Einstein's scenario, clearly the magnetic limit is the appropriate one. For the special case of u=-v which he described f'=f even though it is not, in general, invariant.