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ID: past-exam-of-the-optics-and-electrodynamics-course-of-the-university-of-cambridge/2016/paper-1/1/c/solution

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

Codex  0 2026-10-20

Write

C = e^{2} / (4 π ϵ_{0}) = m c^{2} r_{0}

. For a slowly evolving circular Coulomb orbit, force balance and the mechanical energy give

v^{2} = \frac{C}{m r} = \frac{c ^{2} r _{0}}{r}, ∣ α ∣ = \frac{C}{m r ^{2}}, E = - \frac{C}{2 r} .

The supplied Larmor formula therefore yields

P = \frac{e ^{2} ∣ α ∣ ^{2}}{6 π ϵ _{0} c ^{3}} = \frac{2 m c ^{3} r _{0}^{3}}{3 r ^{4}} .

d E / d t = - P

. Since

d E / d t = C \overset{r}{˙} / (2 r^{2})

,

\overset{r}{˙} = - \frac{4 c r _{0}^{2}}{3 r ^{2}}, \frac{d ( r ^{3} )}{d t} = - 4 c r_{0}^{2} .

Thus the classical radiation collapse of a Coulomb orbit obeys

r (t)^{3} = a_{0}^{3} - 4 r_{0}^{2} c t .

The decreasing mechanical energy means a more tightly bound orbit, not a smaller speed. The formal collapse time is

a_{0}^{3} / (4 r_{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.

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