Consider a reference frame $S$ and which we observe some electric field $\mathbf{E}$ and magnetic field $\mathbf{B}$.

Let $S'$ be a reference frame moving at a constant velocity $\mathbf{u}$ with respect to $S$. In $S'$ we observe an electric field $\mathbf{E'}$ and magnetic field $\mathbf{B'}$.

The obvious question is how is the EM field in the frame $S'$ related to the EM field in the frame $S$. We know that special relativity gives us an exact answer, but **let's suppose we don't know about SR.**

First let's impose that if we observe some charged particle $q$ with velocity $\mathbf{v_0}$ experiencing the Lorentz force in the frame $S$, then it should experience the same force in all other frames.

Hence $$\mathbf{F}=q(\mathbf{E}+\mathbf{v} \times \mathbf{B})=\mathbf{F'}=q(\mathbf{E'}+\mathbf{v'} \times \mathbf{B'}) \tag{*}$$

Assuming the Galilean velocity transformation, we have $\mathbf{v'}+\mathbf{u}=\mathbf{v}$. Now, since (*) must hold for all velocities $\mathbf{v}$, we deduce that

$$\mathbf{E'}=\mathbf{E}+\mathbf{u}\times\mathbf{B} \tag{I}$$

$$\mathbf{B'}=\mathbf{B} \tag{II}$$

Now, this analysis leads to an obvious paradox. If we choose $S'$ to be the frame of reference of the moving charge $q$, then in $S'$ we have no magnetic field, which would impose by (II) that $\mathbf{B}=0$ in all frames $S$, a clear contradiction.

Now apparently there is something called the Galilean field transformation which retains (I) but replaces (II) by:

$$\mathbf{B'}=\mathbf{B}-\frac{1}{c^2}\mathbf{u} \times \mathbf{E} \tag{III}$$

Furthermore, apparently the above can be derived without assuming SR.

My question is how can we derive (III) without assuming SR? We obviously have to drop the assumption that the Lorentz force is invariant under a change of reference frame, which is physically unintuitive.

Also, does the Galilean transformation solve the paradox described above?

## Best Answer

In the charge's frame, the Lorentz force law becomes $q({\bf E} + {\bf 0} \times {\bf B}) = q {\bf E}$. The magnetic field doesn't apply a force on the particle, but that doesn't mean it has to be ${\bf 0}$. It just means that you can't use that particular particle to measure the ${\bf B}$ field in that particular frame.