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Force due to a moving charge = qvB

Immagine a charge moving with some velocity on earth and I calculate the force due to its magnetic field with Earth as reference frame for me. An astronaut in space also calculates the force but space as reference frame.

For ease lets imagine that everything is happening in XY-plane then the force would be perpendicular to both of us, so our motion in XY plane would hove no effect on the force.

The force calculated by me and the astronaut would be different...right? If so, then how is this possible that the force produced is different when it should be same?

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  • $\begingroup$ Please consider rephrasing your question (especially the second paragraph) to make it as clear as possible. When you are asking a question involving relativity you don't want even a shred of uncertainty over what you are exactly asking. $\endgroup$ – Aritra Aug 16 '16 at 22:22
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Immagine a charge moving with some velocity on earth and I calculate the force due to its magnetic field with Earth as reference frame for me.

A charge that exists in the universe generates both an electric and a magnetic field, in general. An observer in the reference frame $S$ (whether at rest or integral with the charge) measures $\mathbf{E}, \mathbf{B}$ and a total force $\mathbf{F} = q(\mathbf{E} + \mathbf{v}\times\mathbf{B})$. An observer in $S'$ (whether at rest or integral with the charge) measures $\mathbf{E}', \mathbf{B}'$ and a total force $\mathbf{F}' = q(\mathbf{E}' + \mathbf{v}'\times\mathbf{B}')$.

The electric and magnetic field transform into each other as component of the electromagnetic tensor $F^{\mu\nu}$ under Lorentz transformations. Applying the just mentioned to the force expressions one obtains that the force transforms as the spatial components of a four-vector under change of reference frame.

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The magnetic field and velocity vectors are not Lorentz invariants, so yes, the resulting force is frame-dependent.

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