# Lorentz force expression and its implications

I found this information about Lorentz force in my textbook (as an extra point):

But I couldn't understand the meaning of the last statement: "Lorentz force expression does not imply a universal preferred frame of reference in nature." Can someone please elaborate on it.

• This means that if in an inertial frame $\:\rm S\:$ the Lorentz force 3-vector is $$\mathbf F= q\left(\mathbf E +\mathbf v\boldsymbol{\times}\mathbf B\right) \tag{01}\label{01}$$ then in any other inertial frame $\:\rm S'\:$ in uniform translational motion with respect to $\:\rm S\:$ the Lorentz force 3-vector is $$\mathbf F'= q\left(\mathbf E' +\mathbf v'\boldsymbol{\times}\mathbf B'\right) \tag{02}\label{02}$$ But be careful. The primed vectors are related with the unprimed via the Lorentz transformation. Dec 10, 2021 at 0:24
• Dec 10, 2021 at 0:25

They mean that the appearance of $${\bf v}$$ in the formula $${\bf F}= q({\bf E}+{\bf v}\times {\bf B})$$ appears to require that the velocity of the charged particle has to be measured with respect to specific "rest frame." This is not the case. If you are in a frame moving with velocity $${\bf V}$$ then, to you, the charged particle moves with velocity $${\bf v}-{\bf V}$$ and

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

This is the same force $${\bf F}$$ but $${\bf E}'={\bf E}+{\bf V}\times{\bf B}$$, so the electric field has changed in your new frame.

• Mike, your answer is relativitically totally wrong. Especially as concerns to the velocity and the electromagnetic field in the new frame. Related : Are magnetic fields just modified relativistic electric fields?. Dec 10, 2021 at 0:01
• @Frobenius. Of course!. I am just using the Galilean limit. I doubt that the OP has got of the point of studying SR, and the book is using Galilean language. Dec 10, 2021 at 1:14
• Ok, Mike, accepted. Dec 10, 2021 at 10:57