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I'm reading Special Relativity by Woodhouse at the moment, and he shows the inconsistency of electromagnetism with Galilean Relativity the following way: Two observers $O, O'$ are measuring the force acting on a test charge moving with velocity v. $O'$ is moving at velocity v relative to $O$. Then the forces measured by the two observers have to be equal. $O$ is measuring $\textbf F=q(\textbf E+\textbf v \times \textbf B)$ and $O'$ $\textbf F=q\textbf E'$. Hence $\textbf E'=\textbf E+\textbf v \times \textbf B$. Now the author assumes says that if their roles are exchanged $\textbf E=\textbf E'-\textbf v\times \textbf B'$. How is that? Why does the observer $O'$ measure a magnetic force acting on the particle if the particle is stationary in his frame of reference? What am I missing?

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When explaining the experiment you say "$O'$ is moving with velocity ${\bf v}$ relative to $O$". Turns out that this statement can be written as "$O$ is moving with velocity $-{\bf v}$ relative to $O'$".

This means that if there's a particle of charge $q$ moving with velocity $\color{red}{-}{\bf v}$ with respect to $O'$, then the observer $O$ will see it at rest, that is ${\bf F}' = q({\bf E}' - {\bf v}\times{\bf B}')$ whereas ${\bf F} = q {\bf E}$, which leads to

$$ {\bf E} = {\bf E}' - {\bf v}\times{\bf B}' $$

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  • $\begingroup$ Oh i see these are two different particles. Thanks for explaining it to me! $\endgroup$ – Jannik Pitt May 3 '17 at 21:38

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