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From the covariant formulation of electromagnetism we know that the fields transform as:

$$\vec{E}'=\gamma \vec{E}-\frac{(\gamma-1)}{u^2}(\vec{u}\cdot \vec{E})\vec{u}+\frac{\gamma}{c}[\vec{u}\times\vec{H}]$$ $$\vec{H}'=\gamma \vec{H}-\frac{(\gamma-1)}{u^2}(\vec{u}\cdot \vec{H})\vec{u}-\frac{\gamma}{c}[\vec{u}\times\vec{E}]$$ Here I am using Gaussian units, and $\vec{u}$ is the relative velocity between the frames. From the Lorentz force law we can derive the "Galilean" transformations on the fields (assuming of course the Newtonian notion of $F'=F$): $$\vec{E}'+\frac{1}{c}[\vec{v}'\times\vec{H}']=\vec{E}+\frac{1}{c}[\vec{v}\times\vec{H}]$$ The primed velocity $\vec{v}'$is $\vec{v}-\vec{u}$, therefore: $$\vec{E}'+\frac{1}{c}[(\vec{v}-\vec{u})\times\vec{H}']=\vec{E}+\frac{1}{c}[\vec{v}\times\vec{H}]$$ After some manipulation we can get the following Galilean transformations: $$\vec{E}'=\vec{E}+\frac{1}{c}[\vec{u}\times\vec{H}]$$ $$\vec{H}'=\vec{H}$$

The general "principle" (not really a principle per se, but hear me out) of the Lorentz transformations is that in the low velocity limit they can look like the Galilean transformations, but here we have a problem. The low velocity limit of the first two equations give: $$\vec{E}' \approx \vec{E}+\frac{1}{c}[\vec{u}\times\vec{H}]$$ $$\vec{H}' \approx \vec{H}-\frac{1}{c}[\vec{u}\times\vec{E}]$$ Which are not the equations that were derived from the Lorentz force. Why is this the case?

P.S.: If we use SI units, this problem seems to disappear, because in the low velocity approximations we would have no $\frac{1}{c}$ in the electric transf. and in the magnetic transf. we would have $\frac{1}{c^2}$ instead of $\frac{1}{c}$.

Please help me, at least by some nudge in the right direction, so that I can understand the problem and then maybe fix it.

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    $\begingroup$ Could you clarify where $\vec{H}' = \vec{H}$ comes from? $\endgroup$
    – ProfRob
    Commented Oct 12 at 17:04
  • $\begingroup$ @ProfRob From the Lorentz force expression we get $\vec{E}'=\vec{E}+\frac{1}{c}[\vec{v}\times(\vec{H}-\vec{H}')]+\frac{1}{c}[\vec{u}\times\vec{H}']$ from that we assume $\vec{H}=\vec{H}'$. $\endgroup$
    – Athanasius
    Commented Oct 12 at 17:11

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