= Solution
An accelerated charge radiates. For ultrarelativistic circular motion, the emitted radiation is concentrated into a cone of angular width of order $\gamma^{-1}$ about the instantaneous <velocity>. As the cone sweeps past a distant observer, propagation-time compression produces a short pulse of width of order $\rho/(\gamma^3c)$, where $\rho$ is orbit curvature radius. Its Fourier spectrum extends to high harmonics, with conventional critical <angular frequency>
$$
\boxed{\omega_c=\frac{3\gamma^3c}{2\rho}.}
$$
This is a spectral scale, not a sharp upper cutoff. For transverse <acceleration> and $v\simeq c$, the classical radiated <power> is
$$
P=\frac{q^2\gamma^4a^2}{6\pi\epsilon_0c^3}
\simeq\frac{q^2c\gamma^4}{6\pi\epsilon_0\rho^2}.
$$
The radiation is linearly polarized for an observer in the orbit plane, with <elliptical polarization> away from it. The two sides of the orbit plane have opposite handedness. The strong forward preference and high-frequency spectrum follow from the relativistic radiation calculation.<On the Classical Radiation of Accelerated Electrons by Julian Schwinger>[1949, abstract: angular distribution, forward preference and high-frequency radiation]{ref}
\Image[/past-exam-of-the-optics-and-electrodynamics-course-of-the-university-of-cambridge/2013/paper-1-oe2013-synchrotron-cone.png]
{title=Original tangent-beaming geometry and normalized illustrative pulse showing the origin of a broad synchrotron spectrum; the pulse curve is schematic rather than measured}
Magnetic bending fields steer the charge but do no work, so maintaining a storage-ring beam requires <energy> replacement by accelerating fields. Bright <electron> beams provide useful radiation from <infrared> to <X-rays>; stronger curvature or higher <electron> <energy> shifts the spectral scale. The classical formulas assume negligible quantum recoil.
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