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Relativistic Lorentz force

Null crossed electromagnetic field

Choose
$$
\mathbf B=B\,\mathbf e_z,\qquad
\mathbf E=cB\,\mathbf e_x,
\qquad
\Omega=\frac{qB}{m}.
$$
A particle released from rest at the origin has the proper-time trajectory
$$
t(\tau)=\tau+\frac{\Omega^2\tau^3}{6},
\qquad
x(\tau)=\frac{c\Omega\tau^2}{2},
\qquad
y(\tau)=-\frac{c\Omega^2\tau^3}{6},
\qquad z(\tau)=0.
$$
Eliminating <proper time> gives the semicubical parabola
$$
\boxed{y^2=\frac{2qB}{9mc}x^3.}
$$
This signed coefficient assumes that the positive $x$ direction is chosen along $\mathbf E$; reversing that axis changes the signs of both $x^3$ and its coefficient.

Relativistic trajectory in a constant null crossed field

Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 1 / 36D / Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 1 / 37C / d / Solution

Relativistic trajectory in a constant null crossed field

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Relativistic trajectory in a constant null crossed field

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