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Combining $d(\gamma m\mathbf v)/dt=q(\mathbf E+\mathbf v\times\mathbf B)$ with $mc^2\,d\gamma/dt=q\mathbf E\cdot\mathbf v$ gives
$$
\frac{d\mathbf v}{dt}
=\frac q{m\gamma}
\left[
\mathbf E+\mathbf v\times\mathbf B
-\frac{\mathbf v(\mathbf v\cdot\mathbf E)}{c^2}
\right].
$$


Coordinate acceleration of a relativistic charged particle

Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 1 / 37A / d / Solution

Coordinate acceleration of a relativistic charged particle

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Coordinate acceleration of a relativistic charged particle

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