# Galilean Transformation of the EM fields

I was going through the proof that Maxwell's equations are not invariant under Galilean Transformations. If we consider two inertial frames (S and S' moving with velocity $$\vec u$$ with respect to the first) then the forces on a charge must be the same, which allows us to say that
$$\frac{\vec F}{q} = \vec E + \vec v\times\vec B = \vec E'+\vec v'\times\vec B' = \frac{\vec F'}{q}.$$
Since $$\vec v = \vec u + \vec v'$$ we can see that $$\vec E = \vec E'+\vec v\times(\vec B-\vec B')-\vec u\times\vec B'$$
$$\vec E' = \vec E + \vec v'\times(\vec B-\vec B')+\vec u\times\vec B$$
However, at this point, it is usually claimed that for these two expressions to be always satisfied, the only possible solution is that $$\vec E = \vec E'-\vec u \times\vec B'$$ $$\vec B = \vec B'$$ I wondered if someone could explain why we must assume that the magnetic fields are the same or what conditions would have to be imposed over the velocities if we did not make this assumption. Thanks

• Can you give a link or reference to that proof? May 2, 2023 at 11:22
• – JS30
May 2, 2023 at 12:20

If we consider particle velocity collinear with magnetic field in frame S, according to the Lorentz force formula per unit charge

$$\mathbf F = \mathbf E + \mathbf v \times \mathbf B,$$

magnetic force in that frame vanishes and thus net force in that frame is $$\mathbf E$$.

Assuming the Lorentz formula is valid also in frame S', and also that force is the same in both frames (which is the case in Newtonian mechanics), we have

$$\mathbf E = \mathbf E' + \mathbf v' \times \mathbf B'.$$

Using the Galilean relation between velocities $$\mathbf v = \mathbf u + \mathbf v'$$, we obtain $$\mathbf E = \mathbf E' + \mathbf v \times \mathbf B' - \mathbf u \times \mathbf B'.$$

Notice from the left-hand side that this force does not depend on velocity $$\mathbf v$$. Thus also the right-hand side must not depend on $$\mathbf v$$.

There are three terms on the right-hand side; the first and the last one obviously do not depend on $$\mathbf v$$ (because they are functions of fields and $$\mathbf u$$). The remaining term $$\mathbf v\times \mathbf B'$$ looks like it depends on velocity, but it must not; the only way this is possible is if the term is zero. Thus $$\mathbf v$$ is collinear with $$\mathbf B'$$ too.

This means $$\mathbf B'$$ must be collinear with $$\mathbf B$$, and thus the two vectors are linearly dependent: $$\mathbf B' = f\mathbf B$$ where $$f$$ is some real number quantity.

This relation holds generally, even if $$\mathbf v$$ is not collinear with $$\mathbf B$$.

Applying this result to the equation

$$\mathbf E + \mathbf v \times \mathbf B = \mathbf E' + \mathbf v' \times \mathbf B',$$ we obtain $$\mathbf E + \mathbf v \times \mathbf B = \mathbf E' - f \mathbf u \times \mathbf B + f\mathbf v \times \mathbf B.$$

This equation must be valid for all possible vectors $$\mathbf v$$. For $$\mathbf v = 0$$ the equation simplifies into

$$\mathbf E' = \mathbf E + f\mathbf u\times \mathbf B,~~~\tag{1}$$ which gives us a general relation between electric fields in the two frames.

Applying this result to the previous equation, we get

$$\mathbf v\times \mathbf B = f\mathbf v \times \mathbf B$$ which implies $$f=1$$.