Vol 2 - Landau, Lifshitz - The classical theory of fields (4th, 1994) (1) · Problem 2

The Electromagnetic Field Equations

Physics — Landau & Lifshitz · relativistic-motion-perpendicular-equal-em-fields

Problem

Determine the relativistic motion of a charge in electric and magnetic fields which are mutually perpendicular and equal in magnitude.\textsuperscript{†} Hint: Choosing the

z

axis along

\mathbf{H}

and the

y

axis along

\mathbf{E}

and setting

E = H

, write the equations of motion:

\frac{dp_x}{dt} = \frac{e}{c}Ev_y, \quad \frac{dp_y}{dt} = eE\left(1 - \frac{v_x}{c}\right), \quad \frac{dp_z}{dt} = 0

and, as a consequence of them,

\frac{d\mathscr{E}_{\mathrm{kin}}}{dt} = eEv_y.

From these equations:

p_z = \mathrm{const}

\mathscr{E}_{\mathrm{kin}} - cp_x = \mathrm{const} \equiv \alpha

. The motion is determined in parametric form (parameter

p_y

) by

2eEt = \left(1 + \frac{\varepsilon^2}{\alpha^2}\right)p_y + \frac{c^2}{3\alpha^2}p_y^3,

x = \frac{c}{2eE}\left(-1 + \frac{\varepsilon^2}{\alpha^2}\right)p_y + \frac{c^3}{6\alpha^2 eE}p_y^3, \quad y = \frac{c^2}{2\alpha eE}p_y^2, \quad z = \frac{p_z c^2}{eE\alpha}p_y,

where

\varepsilon^2 = m^2c^4 + c^2p_z^2 = \mathrm{const}

. The velocity increases most rapidly in the direction perpendicular to

\mathbf{E}

and

\mathbf{H}

(the

x

axis). \textsuperscript{†} The problem of motion in mutually perpendicular fields

\mathbf{E}

and

\mathbf{H}

which are not equal in magnitude can, by a suitable transformation of the reference system, be reduced to the problem of motion in a pure electric or a pure magnetic field (see §25).

Related theory

Electromagnetic Fields Forces And Fields

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