# Deriving "Galilean Electromagnetism"

I'm taking an introductory E&M course, and we're currently covering what our text calls "Galilean Electromagnetism" (i.e. the transformation of electric and magnetic fields between non-relativistic inertial frames).

Our text "derives" the transformation(s) in the following way (given mostly verbatim):

Consider two inertial frames, call them $$A$$ and $$B$$.

Suppose that frame $$B$$ moves with velocity $$\vec{V_{BA}}$$ w.r.t. frame $$A$$. Suppose further that there exists a charge $$+q$$, call it $$C$$, moving with velocity velocity $$\vec{V_{CA}}$$ w.r.t frame $$A$$/$$\vec{V_{CB}} = 0$$ w.r.t frame $$B$$. Lastly, suppose that there exists a uniform magnetic field $$\vec{B_A}$$ in the region of the charged particle, as viewed from frame $$A$$.

The Lorentz Force law gives the force on $$C$$ due to $$B$$ as viewed by frame $$A$$ as $$q\vec{v_{CA}} \times \vec{B_A}$$. By the invariance of Newton's laws under the Galilean transformation, we may equate this to the force on $$C$$ as viewed in frame $$B$$.

"But there can be no magnetic force on the charged particle as viewed in frame $$B$$, as the particle has no velocity w.r.t. frame $$B$$. So, the force observed must be that of an electric field!"

We then have, by the Lorentz Force law, $$q\vec{v_{CA}} \times \vec{B_A} = q\vec{E_B}$$. Clearly, then, in general $$\vec{E_B} = \vec{E_A} + \vec{v_{CA}} \times \vec{B_A}$$.

Particularly coming from a mathematical background, this all seems very "hand-wavy" and, at times, arbitrary to me.

Why must it be an electric field that is measured in frame $$B$$?

Clearly this can be determined empirically, however I feel as though I'm missing some deeper fundamental understanding.

• You probably meant either "wishy-washy" or "hand-wavy"... Sep 15 '21 at 2:01

This is because, as you have supposed, $$\overrightarrow V_{CB}=0$$, so viewed from frame B, the charge is not moving, so there is no Lorentz force. Since the total force the charge experiences is the electric force plus the magnetic force (Lorentz force), and the magnetic force is $$0$$, we conclude that whatever force it experiences must be from the electric force.