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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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  • $\begingroup$ Related: Does a single moving charge produce magnetic field in an empty universe $\endgroup$
    – Ghoster
    Commented Mar 27 at 20:08
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    $\begingroup$ Interesting… two accounts ask the same question on the same day. $\endgroup$
    – Ghoster
    Commented Mar 27 at 20:12
  • $\begingroup$ Can I get a proper answer? quiet new to this site professor said I could ask for help here. $\endgroup$ Commented Mar 27 at 20:14
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    $\begingroup$ @AdonayGandolffi Please slow down and take your time when writing questions. The questions will likely be downvoted and closed if you do not ask the question clearly. Please also use the search bar at the top of the website to first search for similar question. See: physics.stackexchange.com/help/how-to-ask $\endgroup$
    – hft
    Commented Mar 27 at 20:15
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    $\begingroup$ A moving charge produces a magnetic field. Why won’t your professor answer your questions? $\endgroup$
    – Ghoster
    Commented Mar 27 at 20:23

1 Answer 1

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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 $\vec v$.

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    $\begingroup$ 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 physics.stackexchange.com/questions/412924/… $\endgroup$
    – Rich
    Commented Mar 27 at 21:19

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