Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2015 / S19 / 2

For an ultrarelativistic electron of Lorentz factor $\gamma =E/(m_e{c}^2)=3000/0.511=5870.84$, the synchrotron critical frequency gives ${\epsilon }_c=\hslash {\omega }_c=3\hslash c{\gamma }^3/(2\rho )$. Thus, using the supplied constants,The angle-integrated spectral power per unit photon energy is proportional to $x{\int }_x^{\infty }{K}_{5/3}(u)du$, with $x=\epsilon /{\epsilon }_c$. It peaks at $x\simeq 0.286$, soThis is a power spectrum per linear energy interval; power per logarithmic interval and photon-count spectra have different peaks. The magnetic rigidity condition is $p=|e|B\rho$, with $pc\simeq E$, givingThe critical energy sets a broad spectral scale, not a maximum allowed photon energy.^[ref]

For a ring whose total bending angle $2\pi$ is supplied by dipoles with the computed radius, the single-electron energy loss per turn isA stored current $I$ corresponds to $I/|e|$ electron turns per second: the number of stored electrons times the revolution rate. ThereforeThis is the whole beam's radiated power, not the power accepted by one beamline. If the specified radius describes only one dipole and the other bends are different, the radiation integral is needed instead:The numerical result uses the standard equal-radius-bend assumption; the ring circumference, including straight sections, is not required. The RF system must replenish this loss to maintain stored beam energy.

Near the critical photon energy, the characteristic vertical divergence of synchrotron radiation is ${\sigma }_{\theta }\simeq 1/\gamma =1.70\times 10^{-4}\mathrm{rad}$ in the approximation requested. Ignoring electron source size and divergence, the rms vertical width $10\mathrm{m}$ downstream isThe full aperture width is $10\mathrm{mm}$, so its half-width is $5\mathrm{mm}=2.94{\sigma }_y$. It therefore passes almost all the central $8\mathrm{keV}$ photon beam. A Gaussian angular approximation would giveThe exact synchrotron angular distribution is not Gaussian, so this number is an illustrative containment estimate, not an exact throughput calculation. A finite aperture cannot transmit every photon in a distribution with tails. The assumed infinite horizontal aperture removes horizontal fan clipping.

Below the critical photon energy, radiation is emitted over a wider angular cone. The usual low-frequency characteristic-angle estimate isup to an order-one angular-width convention. At $\epsilon =0.012\mathrm{eV}$ this givesThe aperture should therefore be enlarged vertically. A full width of order $0.30\mathrm{m}$ covers $\pm 1$ characteristic width; approximately $0.90\mathrm{m}$ covers $\pm 3$ and gives containment comparable with the preceding X-ray estimate in the same Gaussian model. The original $10\mathrm{mm}$ width would pass only roughly $\mathrm{erf}[0.005/(\sqrt{2}0.149)]\simeq 2.7\%$ in that model. Equivalent optics could instead reduce the spot at a relocated aperture, at the expense of angular acceptance. These estimates concern geometric acceptance; THz coherence and bunch structure require further beam data.