# 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?

• The problem is that the Lorentz force law breaks (the original version of) Newton's third law. In this scenario, Newton's laws are not fulfilled to begin with. Therefore, this does not show that Newton's laws are not Galilean invariant. I don't have the time at the moment, but I think one can prove that if your force law satisfies Newton's three laws, then your system is Galilean invariant. In your case, the Lorentz force law doesn't satisfy the proposition hypothesis, so it's not surprising you have something like velocity-dependence in this context. Dec 10, 2021 at 5:08
• Electromagnetism is inherently a relativistic theory, as you probably know, but when we teach it, we usually use a weird mix of some of Newton's laws and some of EM to talk about it (this is okay if you're careful with your assumptions... make sure your particles are not traveling close to the speed of light, etc, etc, etc). Dec 10, 2021 at 5:11
• @MaximalIdeal Actually, I was asking about the invariance of the second law, not the third. Are you saying that the invariance of the second law depends on the invariance of the third law under GT? Dec 10, 2021 at 5:15
• Yes. Newton's second law by itself does not lead to Galilean invariant systems (I think the form the equation $F=ma$ stays the same but the system being described might not be invariant under GT). This leads to an interesting question: why is the form of the equation $F=ma$ invariant under GT but the system is not invariant unless it also complies with the third law? Dec 10, 2021 at 5:21
• Please allow a small side remark. As @MaximalIdeal wrote, EM is inherently relativistic so Lorentz transformations are to be used. Some discussion and reference how the Lorentz force law (describing a charged particle in an EM field) should be formulated for any speed is found here. That includes a relativistic definition of force also. Dec 10, 2021 at 7:11

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: $${\bf F'}=q \left( {\bf E'} + {\bf v'}\times{\bf B'} \right)=q \left( {\bf V}\times{\bf B} + ({\bf v} - {\bf V}) \times{\bf B} \right)=q \left( {\bf v}\times{\bf B} \right)={\bf F}$$
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.