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

Collider luminosity is the proportionality factor between a process cross section and its interaction rate: $dN/dt=\mathcal{L}\sigma$ before efficiency corrections. It has units of inverse area per time. Integrated collider luminosity is ${\mathcal{L}}_{\mathrm{int}}=\int \mathcal{L}(t)dt$, with inverse-area units, and gives $N=\sigma {\mathcal{L}}_{\mathrm{int}}$ for a constant cross section and unit efficiency.

At an $e^{+}{e}^{-}$ collider, count small-angle Bhabha scattering in a dedicated forward detector. Its calculable quantum electrodynamics cross section in a precisely defined angular acceptance provides $\mathcal{L}=N/(ϵ\sigma \Delta t)$, after background and radiative corrections. The inner angular boundary is especially important because the rate is sharply forward-peaked.

At a collider with two proton beams, a van der Meer scan calibrates a visible interaction monitor by separating the beams horizontally and vertically while recording the overlap-dependent rate. Known bunch populations and the measured overlap give an absolute collider luminosity and hence the monitor's visible cross section. Its rate then tracks collider luminosity during normal running. Correlations between transverse coordinates require a nonfactorizing scan analysis rather than blindly multiplying two one-dimensional profiles.

For two identical, head-on beams with Gaussian distributions, integrating the product of their normalized transverse densities gives an overlap $1/(4\pi {\sigma }_x{\sigma }_y)$. Each of the $n_b$ colliding bunch pairs crosses once per revolution, with $f_{\mathrm{rev}}\simeq c/C$. ThusThereforeThe factor $4\pi$ accounts for the convolution of two beams, rather than the $2\pi$ normalization of just one beam. The result assumes all stated bunch pairs collide at the chosen interaction point, speeds approximately $c$, zero crossing angle and negligible hourglass reduction. Additional interaction points do not multiply the local collider luminosity at a single detector.

The geometric rms beam emittance measures transverse phase space area. In one plane,The betatron amplitude function ${\beta }_x(s)$ describes how the beam envelope varies around the lattice. At a waist, $⟨x{x}^{\prime }⟩=0$ and $⟨x^{\prime 2}⟩={ϵ}_x/{\beta }_x$. This betatron amplitude function is unrelated to either the relativistic velocity ratio $v/c$ or a renormalization-group beta function.

With the beam emittances and bunch populations held fixed, reducing each interaction-point amplitude function to one third makes both transverse rms widths smaller by $\sqrt{3}$. Hence their product is smaller by three, andThe change is not a factor of nine: widths are proportional to the square root of the amplitude function. The transverse angular spread increases by $\sqrt{3}$. In a real machine, stronger final focusing, finite bunch length, crossing angle, aperture and beam-beam limits can reduce the nominal gain; the question's ideal symmetric head-on approximation neglects these effects.

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.