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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.

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Connor Behan
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    This means that if in an inertial frame $:\rm S:$ the Lorentz force 3-vector is \begin{equation} \mathbf F= q\left(\mathbf E +\mathbf v\boldsymbol{\times}\mathbf B\right) \tag{01}\label{01} \end{equation} then in any other inertial frame $:\rm S':$ in uniform translational motion with respect to $:\rm S:$ the Lorentz force 3-vector is \begin{equation} \mathbf F'= q\left(\mathbf E' +\mathbf v'\boldsymbol{\times}\mathbf B'\right) \tag{02}\label{02} \end{equation} But be careful. The primed vectors are related with the unprimed via the Lorentz transformation. – Frobenius Dec 10 '21 at 00:24
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    Related : Are magnetic fields just modified relativistic electric fields?. – Frobenius Dec 10 '21 at 00:25

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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.

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mike stone
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