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a single electron and a detector for magnetic fields. If the electron is moving at a constant velocity, will the detector ever register a magnetic field?

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If the electron is moving at a constant velocity, will the detector ever register a magnetic field?

If a point charge (e.g., an electron) is moving at a constant velocity, the electromagnetic field can be determined by boosting the field strength tensor of a static point charge.


In the rest frame of the charge, the electric field of a point charge of charge q is: $$ \vec E_{rest}=\frac{q \vec r}{4\pi \epsilon_0 r^3} \;. $$ and the magnetic field is: $$ \vec B_{rest}= 0 \;. $$

And the field strength tensor is thus: $$ F^{\mu\nu}_{rest}=\left(\begin{matrix} 0 & -\frac{q x}{4\pi c \epsilon_0 r^3} & -\frac{q y}{4\pi c \epsilon_0 r^3} & -\frac{q z}{4\pi c \epsilon_0 r^3} \\ \frac{q x}{4\pi c \epsilon_0 r^3} & 0 & 0 & 0 \\ \frac{q y}{4\pi c \epsilon_0 r^3} & 0 & 0 & 0 \\ \frac{q z}{4\pi c \epsilon_0 r^3} & 0 & 0 & 0 \end{matrix} \right) $$

In the boosted frame, where the electron is moving with a fixed velocity $\vec v$, the field strength tensor is: $$ F^{\mu\nu}(\vec v) = \Lambda(\vec v)^{\mu}_{\mu'}\Lambda(\vec v)^{\nu}_{\nu'} F^{\mu'\nu'}_{rest}\;, $$ where $\Lambda(\vec v)$ is the usual boost Lorentz transformation matrix for a boost of speed $v$ in the $\hat v$ direction.

If you find that any of the purely spatial ($i,j$) components of $F^{\mu\nu}(\vec v)$ are non-zero then you will have found that there is a magnetic field for the electron moving at velocity $\vec v$.

hft
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    If the detector is in the same inertial frame of reference as the charge (i.e moving at the same velocity), with no acceleration, it will not register a magnetic field, otherwise it will. This is one of the issues that led Einstein to write his 1905 paper on SR. A similar question is answered here https://physics.stackexchange.com/questions/412924/magnetic-field-due-to-a-single-moving-charge – Rich Mar 27 '24 at 21:19