= Solution
Interpret $3$ GeV as the <electron>'s total <energy>; the alternative kinetic-energy reading changes the estimate only slightly at this scale. The <Lorentz factor> and orbital acceleration are
$$
\gamma=\frac{3000}{0.510999}=5.871\times10^3,\qquad
a=\frac{v^2}{R},\qquad a_0=\gamma^2\frac{v^2}{R}.
$$
Using the supplied <Larmor formula>, $\mu_0=1/(\epsilon_0c^2)$ and the revolution time $2\pi R/v$,
$$
P=\frac{e^2\gamma^4v^4}{6\pi\epsilon_0c^3R^2},\qquad
\boxed{U_{\rm turn}=\frac{e^2}{3\epsilon_0R}\gamma^4\beta^3.}
$$
For $R=561$ m this gives $P=1.7405\times10^{-10}$ W per <electron> and $T_{\rm turn}=1.17577\times10^{-5}$ s. Their product is
$$
\boxed{U_{\rm turn}=2.0464\times10^{-15}\ \mathrm J
=1.2773\times10^4\ \mathrm{eV}\simeq12.8\ \mathrm{keV}.}
$$
This is the <energy> loss of one <electron> on a full circular orbit, not its <energy> loss per metre or the total loss of the whole bunch. The fractional loss is only about $4.26\times10^{-6}$ of the $3$ GeV <energy> each revolution. Maintaining a steady stored <energy> requires replacing that <energy> through the accelerator's rf system.
Back to article page
