 Solution

ID: past-exam-of-the-optics-and-electrodynamics-course-of-the-university-of-cambridge/2013/paper-1/2/a/solution

Let

R

run from the charge's present position to the observation point, and let

β = v / c

be constant. Lorentz-transforming the Coulomb field gives

E (R, t) = \frac{q}{4 π ϵ _{0}} \frac{( 1 - β ^{2} ) R}{[ R ^{2} - ∣ β \times R ∣ ^{2} ] ^{3/2}}, B = \frac{v \times E}{c ^{2}} .

At fixed R, the radial magnitude is proportional to

(1 - β^{2}) / (1 - β^{2} sin^{2} θ)^{3/2}

. It is reduced by

γ^{- 2}

along the motion and enhanced by

γ

perpendicular to it. The field is therefore concentrated around the transverse plane at high speed. For negative charge all electric-field arrows reverse.

Although the fields depend causally on retarded motion, for constant velocity the electric-field direction points along the separation from the present position. This does not imply instantaneous action: the velocity-dependent part of the retarded field removes a naive directional lag. Both fields fall as

R^{- 2}

; uniform straight-line motion produces no acceleration-radiation term proportional to $R^{- 1}$ .

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