 Solution

Relativistic beaming concentrates the emission into a forward cone about the instantaneous velocity. The aberration of light maps a rest-frame ray at ${\theta }^{\prime }=\pi /2$ to $\cos \theta =\beta$, so the characteristic cone half-angle is $\theta \simeq {\gamma }^{-1}$. Thus its conventional full width is $2/\gamma =2\times 10^{-6}\mathrm{rad}$.

The isotropic-emission assumption has no beam nulls. For example, its received bolometric angular power is proportional to $D^4$, where the Doppler factor is $D=[\gamma (1-\beta \cos \theta ){]}^{-1}>0$. In particular it is nonzero at $\theta =1/\gamma$, disproving a literal null there. A true dipole pattern for an accelerated electron is not isotropic in its rest frame and needs a specified polarization and viewing plane. The width just quoted is a characteristic beaming angle, not a null-to-null width under the stated premise.

For a circular orbit seen in its orbital plane, let the velocity direction sweep past the observer through $-1/\gamma <\psi <1/\gamma$. The emission interval is $\Delta t_{\mathrm{em}}\simeq 2/(\gamma {\omega }_g)=1.14\times 10^{-2}\mathrm{s}$. The observed interval must also include light-travel compression:Its ratio to the gyromotion period is $\Delta t_{\mathrm{obs}}/T_g\simeq 2/(3\pi {\gamma }^3)=2.1\times 10^{-19}$. Since the boundary of a smooth beam is conventional, only the scale ${\omega }_g^{-1}{\gamma }^{-3}$ is invariant under reasonable pulse-width definitions. For helical motion the viewing geometry and pitch angle are additional inputs; the numerical estimate assumes transverse motion and a sightline intercepted by the cone.^[ref]