Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2012 / S19 / 2

Relativistic radiation is concentrated into an angle of order $1/\gamma$ around the instantaneous velocity. The emission direction sweeps past a tangent observer over an orbital interval of order $\rho /(\gamma c)$. Since the charge moves toward the observer, arrival times are compressed by an additional factor of order $1/{\gamma }^2$, giving the spectral scale $\omega \sim c{\gamma }^3/\rho$.

The conventional numerical factor follows from the radiation Fourier phase. Near the tangent direction, circular motion givesWith $u=c\gamma t/\rho$, this becomes $[\omega \rho /(2c{\gamma }^3)](u+u^3/3)$. The resulting Airy-type integrals are expressible through modified Bessel functions with tangent-plane argument $\xi =\omega \rho /(3c{\gamma }^3)$. The standard angle-integrated synchrotron radiation spectrum has shapeIt therefore identifiesThe beaming argument supplies the physical scaling; the spectral integral fixes the conventional factor $3/2$. Critical energy is a broad spectral scale, not a hard photon-energy cutoff.^[ref]

For a $5$ GeV electron, ${\gamma }_e=E/(m_e{c}^2)\simeq 9785$. With $\rho =10$ m, the stated synchrotron critical frequency formula givesAt the same total beam energy and bend radius, a proton has ${\gamma }_p=5/0.9383\simeq 5.329$, and the same supplied formula givesThe ratio is $({\gamma }_p/{\gamma }_e{)}^3\simeq (m_e/m_p{)}^3$, explaining the strong suppression for protons. The critical-energy formula is an ultrarelativistic approximation, excellent for the electron and only approximate for this moderately relativistic proton. If “5 GeV proton beam” instead denotes kinetic energy, then ${\gamma }_p=1+5/0.9383$ and the estimate is about $7.50\times 10^{-6}$ eV; the two beam-energy conventions should not be mixed.

Substituting the electron Lorentz factor in the supplied energy-loss formula givesWithout replenishment, the electron's momentum decreases. The reference dipoles then bend it more tightly, producing a displaced orbit. In linear beam optics, $x=D_x\delta$ and $y=D_y\delta$, where $\delta =\Delta p/p\simeq -U_0/E\simeq -1.11\times 10^{-3}$ per uncompensated turn. In an ideal planar ring the horizontal dispersion is nonzero and the orbit shifts horizontally, inward where $D_x>0$. There is no corresponding vertical closed-orbit shift when $D_y=0$. Real coupling or vertical dispersion can change that idealization.

With current $I=0.500$ A, the number of circulating particles is $N=I{T}_{\mathrm{rev}}/e$, so the radiated power isThe $2\mu \mathrm{s}$ revolution time cancels when using beam current. Radio-frequency cavities restore the lost energy longitudinally, with synchronous gain $e{V}_{\mathrm{RF}}\sin {\phi }_s=U_0$ in the sine-phase convention. Their RF power system must supply at least this beam power, in addition to its own losses. Radiation damping and quantum excitation determine beam emittances; they should not be confused with the dispersive closed-orbit shift from a momentum change.