Particle Physics Particle Physics Chris Parkes April/May 2003  Hydrogen atom Quantum numbers Electron intrinsic spin  Other atoms More electrons! Pauli Exclusion Principle Periodic Table  Collisions Fixed-Target Colliding beam q 2  Cross-sections Differential xsecs Transition proby Luminosity Luminosity Reaction Kinematics Atomic Structure 2 nd Handout Second Handout

Natural Units,  =c=1  Energy GeV  Momentum GeV/c (abbreviated to GeV)  Mass GeV/c 2  Length (GeV/  c) -1  c=0.197GeVfm=1 [1fm=1E-15m] Natural unit of length 1GeV -1 =0.197fm  Time (GeV/  ) -1  =6.6E-25GeVs Natural unit of time 1GeV -1 =6.6E-25s  Cross-section (GeV/  c) -2 1barn= m 2 Natural unit of xsec =1GeV -2 =0.389mb  Charge - ‘Heavyside-Lorenz units’ ε 0 =1  Use dimensionless ‘fine structure constant’

Fixed Target experiment e.g. NuTeV Scatter neutrinos off nucleons (iron target) Measure sin 2  W Why does this have to be fixed target? Interaction consider with four momenta (E a,p a ) etc.. Total CM energy, a frame invariant [show this] b at rest:E b =m b See Appendix A Martin&Shaw for

Colliding Beam LEP,Tevatron, LHC – synchotrons. SLC – 1990s e + e- 90GeV Linear Collider ILC – International Linear Collider, 500GeV e + e - ? [Now see Question 2.2] Symmetric beams – lab frame =CM frame Particle & anti-particle collision Four Momentum Transfer Defined as a c b d where ** Scattered through angle (in CM) ** When particles are not changed in the interaction i.e. a=c, b=d – elastic scattering process, magnitudes of momenta unchanged [Here * indicates CM frame] Hence q 2  0, when  *  0, forward scattering, otherwise negative [Q 2 =t=-q 2 ] For large momenta in CM, can neglect masses, all momenta same

Cross-Sections We perform an experiment: How many pions do we expect to see ?  Duration of expt(t)  Volume of target seen by beam (V)  Density of p in target (  )  Beam incident /sec/Unit area (I)  Solid angle of detector (  Ω)  Efficiency of experiment (trigger/analysis) (  )  (I t) (V  )  Ω   (1/Area)(N o )  Ω  Smashing beam into a target The constant of proportionality – the bit with the real physics in ! – is the differential cross-section NN Integration over 4  gives total cross-section Can divide total xsec into different reactions e.g. Xsec measured in barn, pb etc…

Luminosity For colliding beams no V (target volume) term. Require two narrow beams with complete overlap at collision point Typical beam sizes  m in xy and cm in z Interaction rate is n 1,n 2 are number of particles in a bunch f is the frequency of collisions e.g. rotation in circular collider, this can be high, LHC 40 MHz! a is the bunch area of overlap at collision point (100% overlap) jn s -1 is known as the luminosity LHC plans cm -2 s -1 Number of events = lumi x xsec x time Typically good machine running time is ~1/3 yr (1x10 7 s)

Accelerators Considerations Considerations for an accelerator. reaction to be produced Energy required Luminosity required Events expected Particles are accelerated by electric field cavities. Achievable Electric fields few MV/m Higher energy = longer machine Fixed target expt. – not energy efficient but sometimes unavoidable (e.g. neutrino expts) Particles are bent into circles by magnetic fields. Synchrotron radiation – photons radiated as particle travels in circle E lost increases with  4, so heavy particles or bigger ring LEP/LHC 27km ring, long-term future a VLHC of 700km??! Or straight line… Linac – repetition rate slower as beams are not circulating Synchrotron – beams can circulate for several hours

Deep Inelastic Scattering Quarks confined inside proton by potential [more about this later] Quarks have momentum distribution, each one carries a Varying fraction of the protons E,p call this fraction x At low q 2, scattering shows shape of proton At high q 2, small wavelength, scatter off quarks inside proton electron Proton At rest quark E,p xM E’,p’ m Consider scattered quark in proton v=E-E` q=p`-p Where q is 4-vector v,q Can tell momentum of quark by Looking only at electron! Find only ~½ momentum in quarks! Rest in gluons

Transition Probability reactions will have transition probability How likely that a particular initial state will transform to a specified final state e.g. decays Interactions We want to calculate the transition rate between initial state i and final state f, We Use Fermi’s golden rule This is what we calculate from our QFT, using Feynman graphs This tells us that  fi (transition rate) is proportional to the transition matrix element T fi squared (T fi 2 ) Transition rate  Prob y of decay/unit time  cross-section x incident flux  IV 