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Can you give an example of an inertial frame in presence of an electromagnetic field for all kinds of charged particles?

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closed as unclear what you're asking by Ben Crowell, Jon Custer, ZeroTheHero, John Rennie, Chair Feb 11 at 7:28

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  • $\begingroup$ Hi user341778. I removed your second subquestion. Phys.SE prefers 1 subquestion per post. $\endgroup$ – Qmechanic Feb 10 at 14:04
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    $\begingroup$ It is not at all clear what this would mean. Inertial frames are not typically defined "with respect to" anything, let alone an electromagnetic field. $\endgroup$ – Luke Pritchett Feb 10 at 14:28
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Presumably the OP is asking if there is any special inertial frame determined by a particular EM field, in which case the answer is often "yes". In the case where E is perpendicular to B and |E| > |B|, there is an inertial frame that is special: the one in which E or B vanishes.

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  • $\begingroup$ I think the correct statement is that if $\vec{E} \cdot \vec{B} = 0$ and $|E| > |B|$, then there is a frame in which $\vec{B} = 0$; and if $\vec{E} \cdot \vec{B} = 0$ and $|B| > |E|$, then there is a frame in which $\vec{E} = 0$. Your post conflates the two cases. $\endgroup$ – Michael Seifert Feb 10 at 15:05
  • $\begingroup$ I don't see how this can be an interpretation of the OP, since then "for all kinds of charged particles" doesn't make sense. $\endgroup$ – Ben Crowell Feb 10 at 16:56
  • $\begingroup$ @MichaelSeifert, saying E is perpendicular to B is of course equivalent to saying E dot B equals zero. $\endgroup$ – S. McGrew Feb 11 at 0:26
  • $\begingroup$ The inertial frame(s) determined by the vanishing of E or B are of course independent of the type of charged particle. When E and B are perpendicular and unequal, there is one and only one inertial frame in which E or B vanishes, leaving only a residual B or E, respectively. I haven't found or done the math to know if there is a special frame in the more general case, where E and B are not perpendicular. $\endgroup$ – S. McGrew Feb 11 at 0:40
  • $\begingroup$ My point was that your point only says "$|E| > |B|$", which might lead someone to believe that this doesn't hold when $|B| > |E|$. $\endgroup$ – Michael Seifert Feb 11 at 3:27

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