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?
Asked
Active
Viewed 88 times
1 Answers
4
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
- 27,235