Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2018 / S19 / 3

Let the bunch tail have the higher energy so that it catches the head. For an ultrarelativistic proton, $\delta v/v\simeq (\delta E/E)/({\beta }^2{\gamma }^2)$. A drift of length $S$ then changes the bunch length by approximately $S\delta v/v$. Requiring that to match the initial length givesTaking $\gamma \simeq 62.5/0.938\simeq 66.6$, $\beta \simeq 1$, and the given lengths yields $\Delta E/E\simeq 0.023$, about $2.3\%$. Treating $62.5$ GeV as kinetic rather than total energy changes that by only a few percent.

This is the linear ballistic compression requirement; a real finite-emittance bunch, nonlinear velocity-energy relation, uncorrelated energy spread and collective fields prevent exact zero length. The chirp is a correlated full head-to-tail spread, not automatically an rms energy spread.

The target collision energy requires $175$ GeV per lepton beam. Its initial energy after the half-linac is $62.5$ GeV, so each lepton must gain $112.5$ GeV. Define the transformer ratio as accelerating field experienced by the witness divided by decelerating field experienced by the driver. If the useful stage length exhausts the driver's $62.5$ GeV while delivering the required gain, thenThis matches depletion to acceleration and avoids leaving unused driver energy at the witness target. It is an ideal energy-transfer estimate, assuming suitable loading and longitudinal field profiles; a field ratio alone does not specify a practical wake or its efficiency.

For ideal transfer of the proton bunch's kinetic energy to the witness bunch,ThusThis is a minimum driver population for unit efficiency and full depletion under the stated energy budget. If only a fraction $\eta$ is usefully transferred, the required ratio becomes $1.8/\eta$. The equality to the transformer ratio uses the same ideal matching assumption as part b.

The mean witness gradient must be $112.5\mathrm{GeV}/1.1\mathrm{m}\simeq 102\mathrm{GV}{\mathrm{m}}^{-1}$. To turn this into a density estimate, assume an accelerating wake amplitude of order the cold plasma wave-breaking scale,Eliminating ${\omega }_p$ givesThe associated plasma wavelength is about $31$ micrometres, illustrating why a short driver structure is required. The given length and energy gain do not uniquely fix density without an assumption about wake amplitude relative to the plasma scale. Driver shape, beam loading, relativistic wave-breaking corrections and matching to a Positron witness are substantive additional issues; the estimate is a scale calculation, not a completed accelerator design.

If the driver can be fully depleted with transformer ratio six, the largest ideal witness gain from a $62.5$ GeV proton is $375$ GeV. Each lepton then hasFor equal opposing electron and Positron beam energies,The associated ideal population ratio would be $N_p/N_e=6$. This energy-budget ceiling ignores wake-generation and transfer losses, staging limits, emittance preservation and the difficulty of accelerating both charge signs with suitable focusing. It is the consequence of the stipulated transformer-ratio bound, not a prediction of an achievable upgrade.