Past exam of the optics and electrodynamics course of the University of Cambridge / 2016 / Paper 1 / 1

Interpreting the cone as Cherenkov radiation, its angle gives the speed, not directly the mass:Combining this with the relativistic kinetic energy $T=(\gamma -1)m{c}^2$ gives the Cherenkov mass inferenceThe material must be transparent at the observed frequency, $n$ is its phase refractive index there, and $n\cos \theta >1$. For $T>0$ this requires $0<\theta <\arccos (1/n)$ for a massive particle above threshold. Radiation of an unspecified other origin would not determine the mass from these data alone.

The transverse spatial coherence width is much larger than the longitudinal coherence length. The longitudinal scale is controlled by spectral bandwidth:For broad, unfiltered visible sunlight it is of order a wavelength to a few wavelengths, roughly $10^{-6}\mathrm{m}$. This is an order-of-magnitude scale, not a uniquely defined solar constant; the precise value depends on spectral response and the coherence-width convention.

The Sun subtends angular diameter $\delta \simeq 1919^{\prime \prime }\simeq 9.3\times 10^{-3}\mathrm{rad}$ at one astronomical unit.^[ref] Consequently the sunlight coherence comparison gives a transverse scaleusing $\lambda =550\mathrm{nm}$ only as an illustrative visible wavelength. The uniform-disk spatial coherence model has first coherence zero at $1.22\lambda /\delta$, about $72\mu \mathrm{m}$ at that wavelength. The small angular size permits coherence across tens of micrometres, while the broad spectrum destroys coherence over much shorter path differences. Spectral filtering increases ${\ell }_c$; spatial filtering changes the transverse scale.^[ref]

Write $C=e^2/(4\pi {ϵ}_0)=m{c}^2{r}_0$. For a slowly evolving circular Coulomb orbit, force balance and the mechanical energy giveThe supplied Larmor formula therefore yieldsSet $dE/dt=-P$. Since $dE/dt=C\dot{r}/(2r^2)$,Thus the classical radiation collapse of a Coulomb orbit obeysThe decreasing mechanical energy means a more tightly bound orbit, not a smaller speed. The formal collapse time is $a_0^3/(4r_0^{2}c)$. The nonrelativistic, small-loss-per-orbit approximation eventually fails; the result demonstrates the classical instability rather than describing an actual hydrogen atom.