Question

1) Consider an accelerator that collides beams of electrons and protons. The beams actually consist of bunches of particles, that are traveling essentially at the speed of light. For both the electron and proton beams, the bunche:s are 2 cm long (in the direction of motion) and they are circulating in rings of 300 m circumference. Each ring contains six bunches of protons (or electrons). Each bunch contains 1013 particles. Assume that the particles in the bunches are distributed uniformly over a cross sectional area of 0.2mm and the beams overlap completely (so the cross-sectional area, over which the collisions occur is also 0.2mm. In order to solve this problem you may need to assume that the beams are traveling near the speed of light. Whether it is 0.99c or 0.999c doesn't make that much difference to the luminosity. You can just take this to be c and you will only be (1% or 0.1%) off in luminosity. 2 (a) Determine the luminosity of this collider. (b) If the cross-section for collisions is 10 μb, determine the number of scattering events that would be ob- served in an experiment that completely surrounds the interaction region. NB. the units for cross-section (barns or b) are actually 10-24 cm2. There is an urban legend that the actual name comes from either the saying "Couldn't hit the broad side of a barn" which refers to someone with very poor aim. Clearly mak ing things collide when they have cross-section of 10-24 or even 10-30 cm2 is going to be challenging Alternatively, some say it comes from the first common usage in the 1970s when experiments were first being carried out at Fermilab, which was built on reclaimed farm land west of Chicago and many of the lab service buildings were (and some still are today!) recycled barns (c) Find the average flux of electrons (d) Consider the situation where, instead of colliding with the proton beam, the beam of electrons is extracted from the machine, and scatters from a stationary target of liquid hydrogen (density 0.1 g/cm3) which is 2 metres long. In this case find the number of scattering events. Compare this with your answer in (b). To do this calculation you should assume that the cross section for ep collisions is independent of the centre-of mass energy. This is not actually true as the higher energy collisions (in collider mode) are actually more probable, but not by more than an order of magnitude. The change is cross-section is not the dominant effect here

Answer 1

(a) Luminosity L is given by

$L=N/tS=6*10^{13}/(2*30000/3.0*10^{10})*0.2=6*10^{13}/0.4*10^{6}=15*10^7~~ cm^{-2} s^{-1}$

where N is the number of proton, t is the time taken by proton to cross the accelerator and S is the cross sectional area.

(b) The number of scattering event is given by

$\frac{dR}{dt}=L\sigma=15*10^7*10*10^{-30}=15*10^{-22}$

where $\sigma$ is the cross section.

(c) The average flux of electrons

$\phi=L/l=15*10^7/2=7.5 * 10^7$

where l is the length of the proton beam.

(d) number of scattering events

$=L/\phi \rho=15 * 10^7 /7.5 *10^7*0.1=20$