Understanding why Einstein found Maxwell's electrodynamics not relativistic

GregAshmore

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.

The B field is the map of the strength of the magnetic field in space. (That is, the map at any instant. Variations in the magnetic field over time are accounted for in the second term of the equation.) That being so, the subject of your statement (the B field) is the same as the subject of my counter (the strength of the magnetic field). I don't see how your statement and my counter can both be true.

I wanted to understand the particular example Einstein gives in the introduction to his 1905 paper. The biggest part of my difficulty was not understanding that "energy in itself" is a technical term, as you have explained. It sounded to me as though Einstein was asserting that Maxwell's theory does not assign a cause to the rise of the emf in the moving conductor.

I also had some difficulty in understanding why the Maxwell equation for emf is not relativistic, in the sense of depending only on the relative velocity of the magnet and conductor. The equation is relativistic in that sense, so long as the "stationary" body is at rest relative to the ether. But of course, that stipulation opens the whole can of relativistic worms, doesn't it?

DaleSpam

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.

DaleSpam

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:
[itex]E' = E + u \times B[/itex]
[itex]B' = B[/itex]
[itex]f'=q(E' + v \times B') = q(E + u \times B + (v + u) \times B) \neq f[/itex]

Electric limit:
[itex]E' = E [/itex]
[itex]B' = B - u \times E[/itex]
[itex]f'=q(E' + v \times B') = q(E + (v + u) \times (B - u \times E)) \neq f[/itex]

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.