0
$\begingroup$

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.

enter image description here

$\endgroup$
2
  • 1
    $\begingroup$ 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. $\endgroup$
    – Frobenius
    Dec 10, 2021 at 0:24
  • 1
    $\begingroup$ Related : Are magnetic fields just modified relativistic electric fields?. $\endgroup$
    – Frobenius
    Dec 10, 2021 at 0:25

1 Answer 1

3
$\begingroup$

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.

$\endgroup$
3
  • 1
    $\begingroup$ 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?. $\endgroup$
    – Frobenius
    Dec 10, 2021 at 0:01
  • 1
    $\begingroup$ @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. $\endgroup$
    – mike stone
    Dec 10, 2021 at 1:14
  • $\begingroup$ Ok, Mike, accepted. $\endgroup$
    – Frobenius
    Dec 10, 2021 at 10:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.