Is Newton's law really invariant under Galilean transformation (for velocity-dependent Lorentz force)? 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?
 A: 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:
$$
{\bf F'}=q \left( {\bf E'} +  {\bf v'}\times{\bf B'}  \right)=q \left( {\bf V}\times{\bf B} +  ({\bf v} - {\bf V}) \times{\bf B}  \right)=q \left( {\bf v}\times{\bf B}  \right)={\bf F}
$$
A: Newton's second law $\vec F=m\vec a$ will only be invariant under Galilean transformation provided the force between the interacting objects depends on the separation vector between the interacting objects, $(\vec{r}_1-\vec{r}_2)$ and is directed along $(\vec{r}_1-\vec{r}_2)$. Since the separation vector $(\vec{r}_1-\vec{r}_2)$ remains unchanged under Galilean transformation, $\vec{r}'={\vec r}-\vec Vt$ , i.e. $$(\vec{r}_1-\vec{r}_2)=(\vec{r}'_1-\vec{r}'_2),$$ the forces will be the same in both the frames i.e. $\vec F=\vec F'$. Gravitational, electrostatic forces satisfy the above criterion but not the magnetic Lorentz force.
