Consider the motion of a charged particle of charge $q$ and mass $m$ from two different inertial frames $S$ and $S'$ connected by Galilean transformation equation ${\vec r}'={\vec r}-{\vec V}t$. This readily implies that ${\vec a}^\prime=\vec{a}$. Since $m'=m$, for the invariance of of ${\vec F}=m{\vec a}$ we need that ${\vec F}'={\vec F}$. However, a magnetic field is velocity-dependent, and also a pure magnetic field in one frame becomes a combination of an electric and a magnetic field.
Let me just show the non-invariance of Newton's second law. Let $\frac{d\vec r}{dt}={\vec v}$ and $\frac{d\vec r^\prime}{dt}={\vec v}^\prime$. Then Galilean transformation implies $${\vec v}^\prime={\vec v}-{\vec V}.$$ Newton's second law for the charged particle from $S$ is $${\vec F}=m{\vec a}=q(\vec v\times\vec B)\tag{1}$$ and from
from $S'$ is $${\vec F}^\prime=m{\vec a}^\prime=q(\vec v^\prime\times\vec B^\prime).$$ Using that the magnetic field transforms under GT as (the $c\to\infty$ limit of Lorentz transformation) $$\vec B_{||}=\vec B_{||},~{\rm and}~ {\vec B_\perp}^\prime={\vec B_\perp}^\prime\Rightarrow {\vec B}'=\vec B$$ we see that $$\vec F^\prime=m\vec a^\prime=q(\vec v -\vec V)\times \vec B.\tag{2}$$
Since $\vec a'=\vec a$, we a contradiction between (1) and (2). Does this not mean that Newton's law is not always invariant under Galilean transformation?
Best Answer
Newton's law is invariant under Galilean transformation, provided the proper non-relativistic limit of the Lorentz transformation of the electromagnetic field is taken into account.
As recalled in the question, in the non-relativistic limit, the magnetic field in the $S'$ reference frame is the same as in the $S$ frame. However, even if in $S$ there is no electric field, in the $S'$ frame, there will be an electric field $$ {\bf E'}={\bf V}\times{\bf B}. $$ Therefore, we have the equality of the force in the two reference frames: $$ {\bf F'}=q \left( {\bf E'} + {\bf v'}\times{\bf B'} \right)=q \left( {\bf V}\times{\bf B} + ({\bf v} - {\bf V}) \times{\bf B} \right)=q \left( {\bf v}\times{\bf B} \right)={\bf F} $$