Past exam of the particle physics course of the University of Oxford / 2016 / 8 / 4 / Solution

For an electron-positron collider of fixed bending radius $R$, synchrotron radiation rapidly becomes the principal obstacle. Multiplying the supplied radiated power by the revolution period $2\pi R/c$ givesA bending radius of $3\mathrm{km}$ gives losses of about $2.95\mathrm{GeV}$ per turn at $100\mathrm{GeV}$ per beam, but $115\mathrm{GeV}$ per turn at $250\mathrm{GeV}$. At $100\mathrm{GeV}$, a $5\mathrm{mA}$ beam already radiates about $15\mathrm{MW}$. The radiofrequency system must replace this energy, and the vacuum chamber must absorb the radiation. The centre-of-mass energy is twice the beam energy for equal opposite beams.

For a collider with two proton beams, the same energy gives a much smaller Lorentz factor. At fixed energy and radius the radiated power scales as $m^{-4}$, so the suppression relative to electrons is $(m_e/m_p{)}^4\simeq 8.8\times 10^{-14}$. The main practical limit is the dipole magnetic field through magnetic rigidity:With $R\simeq 2.8\mathrm{km}$ and $B\simeq 8.3\mathrm{T}$, the beam energy is about $7\mathrm{TeV}$, or $14\mathrm{TeV}$ in the centre of mass. Fields near $16\mathrm{T}$ would give approximately $13.4\mathrm{TeV}$ per beam at that radius. These are realistic superconducting-magnet scales; a $27\mathrm{km}$ circumference includes straight sections, so its bending radius is less than $C/(2\pi )$. The proton radiation loss near $7\mathrm{TeV}$ is only a few keV per turn, though high beam currents still create substantial cryogenic heat loads. Magnet technology, aperture, protection and stored beam energy therefore dominate before the electron-like radiation limit.