Simulation of the interaction of macro- particles with the LHC proton beam Zhao Yang, EPFL

Equation of motion Parameters Introduction The purpose of simulation Study the influence of magnetic field/mass of particle/number of protons in the beam on the particle trajectory/beam loss rate/loss duration/total lost protons. Mass of macro-particle A=10 12, 10 13, 10 14, 10 15, (unit: M p ) Initial position of in x-axisX 0 = m Mass of photonM p =1.6726* kg Magnetic fieldB=8.33 T Radius of the vacuum chamber b=0.02 m Circumference of the storage ring C= m Total number of protons in the beam (protons/beam) N p = 0.9*10 11 *25(low, early commissioning) N p =1.15*10 11 *2808(high, nominal LHC) rms beam sizeσ= m The classical proton radiusr p =1.5* m Properties of aluminum A atom =27; Z atom =13; ρ=2700 kg/m 3 ; Nuclear interaction cross section=0.420* m 2 Properties of copper A atom =64; Z atom =29; ρ=9000 kg/m 3 ; Nuclear interaction cross section=0.553* m 2 -g-g

Aluminum Macro-particle Influence of magnetic field on particle trajectory. Np=0.9*10 11 *25 A equals to ~10 15, B equals to 0 T, 8.33 T and 80 T (Blue, red and green) The X-Y trajectory gets close to the Y-axis as the mass of particle decreases and the magnetic field intensity is increased. The strong magnetic field may result in a cyclotron oscillation of the particle. The period of this oscillation is reduced for increasing magnetic field and smaller particle mass. A=10 12 A=10 13 A=10 14 A=10 15

Aluminous Macro-particle Influence of magnetic field on beam loss rate. Np=0.9*1011*25 A equals to ~10 15, B equals to 0 T, 8.33 T and 80 T (Blue, red and green) The maximum of loss rate of the 1 st crossing is independent of magnetic field; it increases with increasing mass A. The maximum of loss rate of the 2 nd crossing increases with increasing magnetic field strength. It decreases with increasing mass at normal magnetic field strength. A=10 12 A=10 13 A=10 14 A=10 15 The reason for a strange behavior of the loss rate in the case A=10 14 B=80 T is unclear

Aluminous Macro-particle Particle trajectory of A=10 12, B=8.33 T Np=1.15*10 11 *2808 At t= s, the macro-particle gets close to the beam and there comes the 1 st crossing, it is repelled upwards and hits the wall at t= s. Then it falls down and approaches to the beam. During this process it hits the wall a 2 nd time at t= s. Finally it falls down at s. The time between two crossing is s s s s s

Aluminous Macro-particle Plot the figures of beam loss rate for a varying total number of protons in the beam. We determine the maximum of each beam loss rate curve, and fit the curve around this maximum. Then we calculate the total number of lost protons. The maximum of each beam loss rate curve is fitted by the function log 10 y=a+bx+cx2 Original plotsFitting curves N p =1.15*10 11 *2808N p =1.15*10 11 *1600 N p =1.15*10 11 *400

Aluminous Macro-particle 1.15*10 11 * *10 11 * *10 11 * *10 11* *10 11 * *10 11 * *10 11 * The number of lost protons calculated with sigma= m and B=8.33T

Aluminous Macro-particle Plots of lost protons versus Np with different particle masses The solid lines are the original plots, the dotted line are the fitting ones. The number of lost protons is fitted by the function y=1/(a+bx+cx 2 ) when the mass is relatively small. A=10 12 A=10 13 A=10 14

Aluminous Macro-particle B=4.17 T is the present value of magnetic field in LHC at 3.5 TeV beam energy. We use sigma= m, B=4.17 T and N p =1.15*10 11 *2808 to simulate the beam loss rate and calculate the total number of lost protons *10 11 * The results are exactly the same as the results with B=8.33 T the magnetic field in LHC will not affect the total number of lost protons. We increase the value of transverse rms beam size by a factor 5, to sigma= m, and keep the other parameters constant *10 11 *2 808 N/A *10 11 * The increase of the transverse beam size reduces the total number of lost protons.

Copper Macro-particle We change the material of macro-particle. Compared with aluminum, copper has larger density, cross section and atomic mass number. A=10 12 A=10 13 A=10 14 sigma= m, x 0 = m, N p =0.9*10 11 *25 The X-Y trajectories of Aluminum are closer to the Y-axis than the X-Y trajectories of Copper. The maximum of beam loss rate with aluminum macro- particle is larger than that with a copper macro- particle.

Copper Macro-particle We plot the curves of beam loss rate versus time, then calculate the loss duration and the total number of lost protons for the 1 st crossings Total lost protons obtained for varying value of mass A and total protons intensity N p (B=8.33 T). x 0 = m, B=8.33 T, A=10 12, and N p =1.15*10 11 *2808N p =1.15*10 11 *1600N p =1.15*10 11 * *10 11 * *10 11 * *10 11 * *10 11 * *10 11 * *10 11 * Loss duration obtained for varying mass A and total proton intensity N p in unit of second (B=8.33 T). The loss duration is almost independent of the material of the macro-particle. The total number of lost protons for a copper particle is smaller than that for an aluminum particle.

Final conclusions The time separation between first and second crossing is consistent with some beam observations of multiple successive events. However, the losses at the second crossing are always much lower than for the first crossing, which is different from actual observations. The computed peak losses should be compared with the quench threshold which corresponds to about 10 7 protons / second. Effect of magnetic force on the macro-particle motion is weak and can be neglected, even for a field of 8.33 T. The loss duration and the number of lost protons decrease with higher total beam intensity; the losses roughly in inverse proportion. Increasing the beam size by a factor of 5 reduces the total proton loss by about a factor of 3. This might be part of the explanation why events have not been important at LHC injection. Future work may extend this discussion to some other materials and to other macro-particle shapes, e.g. to needle-like objects.

The end Thank you