1 Introduction. 2 Relativistic Kinematics. 2.1 Particle Decay

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1 1 Introduction Relativistic Kinematics.1 Particle Decay Due to time dilation, the decay-time (i.e. lifetime) of the particle in its restframe is related to the decay-time in the lab frame via the following equation T γτ τ 1 v where v is the particle s velocity, as measured in the lab frame. Thus we get the following relationship:. So it s clear that near the speed of light, the decay-time of the particle approaches infinity (it will never decay). The decaying particle s 4-momentum in its own rest frame is given by p (M, 0, 0, 0). After the decay, we assume there are two resultant particles with 4-momentum q and k respectively. By momentum conservation we have p q + k. p q Mq 0 and p k Mk 0 ] (q + k) q + q k + k. Thus q k 1 [(q + k) q k. But [ q + k p, so that q k 1 M m q m ] k. 1

2 Thus, q 0 p q M 1 m q M 1 m k M. (q+k) q M q +k q M mq + 1 [ M m q m ] k M 1 M + By symmetry of argument we also have k 0 1 M 1 m q M + 1 m k M. We can also calcualte the magnitude of the 3-momentum q ( q 0) [ ] m 1 q M + 1 m q M 1 mq m k M 1 4M [M 4 M ( m q + m ) k + ( m q m ) ] k. Of course q + k 0 and so k q. The direction of the 3-momentum, however, remains unknown.. Two-Particle to Two-Particle Scattering 1. Incoming particles are labeled by 1 and.. Outgoing particles are labeled by 3 and DEF: elastic scattering iff m 1 m 3 and m m Conservation of 4-momentum dictates that p 1 + p p 3 + p 4. s : (p 1 + p ) 5. Define the Mandelstam variables: t : (p 1 p 3 ). u : (p 1 p 4 ) 6. Claim: s + t + u 4 i1 m i. 7. The zero-total-3-momentum-frame is defined by p 1 + p The Breit-frame is defined by p 1 + p 3 0. (used in deep inelastic scattering) 9. Define the Källen function as λ (a, b, c) : a + b + c ab ac bc 10. Work in the zero-total-3-momentum-frame: ( ) (a) p 1 + p (p 1 ) 0 + (p ) 0, 0 (( )) ( (b) s (p 1 + p ) (p 1 ) 0 + (p ) 0, 0 (p 1 ) 0 + (p ) 0) ( ) 0 p1 + ( ) 0 ( ) ( p + 0 p1 p ) 0 (c) (p 1 + p ) p 1 + p 1 p + p so that p 1 p (p1+p) p 1 p (p 1+p ) p 1 p s m1 m

3 (p 1 ) 0 [ 1 s s + m1 m ] (p ) 0 [ 1 (d) We can calculate that s s m1 + m ] (p 3 ) 0 [ 1 s s + m3 m ] as well as 4 (p 4 ) 0 [ 1 s s m3 + m ] 4 p 1 1 4s λ ( m 1, m, s ) p 1 that 4s λ ( m 1, m, s ) p 3 1 4s λ ( m 3, m 4, s ) p 4 1 4s λ ( m 3, m 4, s )..1 Scattering Angle ( 1. Define cos (θ) : p1 p3 s(t u)+ m1 p 1 p 3. Then we can prove that cos (θ) m )( m 3 m ) 4. λ(s, m1, m ) λ(s, m 3, m 4 ).. Elastic Scattering When the scattering is elastic the equations simplify... (surprise!)..3 Angular Distribution π d[cos(θ)] dt πs λ(s, m1, m ) λ(s, m 3, m 4 )..4 Relative Velocity πs s p 1 p 3 π p 1 p First note that v p E in general.. Now define the relative velocity: v 1 : v 1 v p1 E 1 p E p 1E p E 1 E 1E p 1 E E 1E p p E 1 1. Now recall that in the zero-3-momentum-frame, p p 1 p and so: v 1 1 p E E 1E + E 1 1 E 1E s s E E1 1E m 1 s 1 E 1E 4s (s + m 1 m ) m 1 1 E 1E (m 1 + p 1 p ) (m 1 + m + p 1 p ) m 1 1 E 1E (p 1 p ) m m 1 3. Define E 1 E v 1 as the Moller flux factor. From the above relation it is clear that E 1 E v 1 is Lorentz invariant...5 Center of mass and laboratory Systems In the laboratory system we are usually interested in the situation in which p 0 (fixed target). 1. s in lab frame (p 1L + p L ) p 1L + p L + p 1L p L m 1 + m + (p 1L ) 0 (p L ) 0 p 1L p }{{} L m 1 + m + m (p 1L ) 0 0 3

4 . We could also use the s cm formula in a fixed target, but then we would need to convert the given beam energy from the lab frame (in which it is given) to the cm frame (in which we want to do the calculation)...6 Compton s Scattering 1. For Compton s scattering we can show that the energy of the photon after the scattering is related to the energy of the photon before the scattering via E after E before where θ is the angle with which the photon 1+ Ebefore me [1 cos(θ)] is scatterred relative to the axis of incidence..3 Crossing Symmetry Definition: X-channel reaction is a reaction in which only the X Mandelstam variable is positive. An amplitude for an s-channel process is given by M s (s, t, u). For this process we know that s > 0, t 0 and u 0. What if we were to exchange p and p 3 as follows: M s (s, t, u) new M s (s, t, u) old p p 3, p 3 p. Then s new ( p 1 p 3 old ) t old, t new ( p1 ( p old )) s old, and u new (p 1 p 4 ) u old. Thus the result of this switch is that we merely exchange s t. We interpret a particle with p new p old as p new describing the antiparticle of the particle p old was describing. This is become the emission of a particle with energy E is equivalent to the absorption of a particle with energy E. 3 Lorentz Invariant Scattering Cross Section and Phase Space 3.1 Ass-Operator The Ass-Operator is the quantum-mechanical time-evolution operator between two states. It gives us the amplitude for a transition, and thus the probability for transition from i to f is given by f Ŝ i, or in matrix-component notation, S fi f Ŝ i. 1. Write S fi δ fi + i (π) 4 δ (4) (p f p i ) M fi }{{} T fi or otherwise as Ŝ 1 + i ˆT. 4

5 . Assuming an off-diagonal element (f i), we have then f Ŝ i 3. Fermi s Golden Rule 1. Define the cross-section as σ (π) 8 δ (4) (p f p i ) δ (4) (p f p i ) M fi (π) 4 d 4 xe i(p f p i)x δ (4) (p f p i ) M fi (π) 4 d 4 xe i0x δ (4) (p f p i ) M fi (π) 4 T V δ (4) (p f p i ) M fi number of particles scattered number of particles incident N velocity of particles in beam number density of beam time duration N Φ T ime where Φ is the flux.. Φ 1 V v if there is only one particle incident at a fixed target with velocity v. However, in the CM frame, we would have instead Φ v1 v V. 3. In quantum mechanics we speak of probabilities instead of numbers, so that we have σ P robability ΦT ime. f Ŝ i (π) 3 4. We can calculate the (differential) probability as dp f f i i dπ where Π is a volume in momentum space to be integrated over. It is V given by the state-density in momentum space times an infinitesimal momentum volume element d 3 p: dπ V d 3 p. We have such a (π) 3 factor for each final particle: dπ N f V f1 d 3 p (π) 3 f. 5. Observe that dπ 1 because dp π L. 6. Observe also that f f 1! That would violate Lorentz invariance. Instead we have p i p i E i (π) 3 δ (3) ( p i p i ). But δ (3) (0) V so (π) 3 that p i p i E i V. 5

6 7. Thus we have dp dσ Φ T ime (π) 4 T V δ (4) (p f p i) M fi Ni i1 EiV N f j1 E f V V 3..1 Total Decay Rate v 1 v V T ime Nf f1 V (π) 3 d 3 p f v 1 v N i i1 E iv M fi (π) 4 δ (4) (p f p i ) N f f1 d 3 p f (π) 3 E f 1. The decay rate is defined as dγ 1 T dp, that is, the probability for a decay per unit time. We can think of a decay as a scattering with only one incident particle and thus we get dγ 1 f Ŝ i T f f i i dπ N 1 f M fi (π) 4 δ (4) (p f p i ) E a f1 d 3 p f (π) 3 E f. If there are more there is more than one set of resultant particles then the total decay rate is a sum on the sets. 3.. Scattering of Two Particles 1. Because in most practical cases it is only possible to collide two particles, we have N i and so: dσ N 1 f M fi (π) 4 δ (4) (p f p i ) v 1 v E 1 E f1 1 M fi (π) 4 δ (4) (p f p i ) 4 (p 1 p ) m m 1. Thus σ 1 3. Observe that d3 p f E f 4 (p 1 p ) m m 1 d 3 p f (π) 3 E f N f f1 M fi (π) 4 δ (4) (p f p i ) N f f1 is a Lorentz-invariant quantity: d 3 p f E f 0 de f δ ( p f m f ) d 3 p f d 4 p f δ ( p f m f ) θ (E f ) where we note that the fact that E f > 0 is Lorentz invariant. d 3 p f (π) 3 E f d 3 p f (π) 3 E f 6

7 3.3 Scattering Cross Section 1. Try to compute (π) 4 δ (4) (p f p i ) f1 d 3 p f (π) 3 (π) 4 E f δ (4) d 3 p 1 d 3 p (p 1 + p p a p b ) (π) 6 E 1 E. Now use the fact that δ ( p m ) 1 p 1 p δ ( p E m ) 3. Thus we have (π) 4 δ (4) (p f p i ) f1 (π) 4 d 4 p 1 d 4 p (π) 6 δ (4) (p 1 + p p a p b ) δ ( p 1 m ) 1 θ (E 1 ) δ (π) 4 d 4 p 1 (π) 6 δ ( p 1 m ) 1 θ (E 1 ) δ ( (p a + p b p 1 ) m ) δ ( p E m ) + δ ( p + ) E m }{{} 0 as p >0 d 3 p f (π) 3 (π) 4 d 4 p 1 1 ( E f (π) 6 p 1 δ p 1 ) E 1 m 1 θ (E 1 ) δ ( (p a + p b (π) de 1 d p 1 p 1 dω 1 ( R 0 p 1 δ p 1 E 1 m 1 Ea+E (π) b E1 m 1 de 1 dω δ ( (p a + p b p 1 ) m 4. Note that (p a + p b p 1 ) m (p a + p b ) (p a + p b ) p 1 + m 1 m s (p a + p b ) p 1 + m 1 m 5. Assume we are in the center of mass frame, so that p a + p b 0. Thus (p a + p b ) (E a + E b ) s. 6. Thus we get (π) 4 δ (4) (p f p i ) f1 0 d 3 p f (π) 3 (π) s E1 m 1 de 1 dω δ ( s se 1 + m 1 m E f 0 1 (π) (m s 1 m + s) 4sm 1 dω 4 s 1 (π) s λ (m1, m, s) dω 4 s (π) p 1 4 dω s (π) λ (m1, m, s) dω 8s 7

8 7. If there is spherical symmetry for the integrand we can replace dω 4π and so we have 1 σ M fi (π) s p1 4π v 1 v E 1 E 8s 1 p a s M fi (π) s p1 π 8s 1 p 1 16πs p a M fi 8. There must be some factor of 4π missing because the usual result is 1 p 1 64π s p M a fi 3.4 Unitarity of the S-Operator Ŝ must be unitary to preserve probabilities The Optical Theorem 1. Because S is unitary, we must have. Thus T i ( T T ) M f i M i f i (π) 4 n δ (4) (p n p i ) M f n M i n 3. Take f i to obtain M i i M i i i (π) 4 n δ (4) (p n p i ) M i n Im {M i i } 1 (π)4 n δ (4) (p n p i ) M i n 4. Use the relation above to get Im {M i i } 1 1 T V i S n n 5. Actually we should not have written M i i but rather i M i i i i i. 6. Assuming we have only two incident particles, a and b, we would then have (for elastic scattering!) ab M ab E av E b V E av E b V. 8

9 7. Thus we have Im { ab M ab } 16E a E b V T V i S n n 8. We know that σ P robability T ime F lux and so σ total i S n n i i n n T Φ. So that we can write: Im { ab M ab } 16E a E b V i S n T V n... 1 i S n T V n n n V E a E b n V E a E b n E a E b v a v b n E a E b v a v b n E a E b v a v b σ tot i S n E a V E b V n n T probability i n T λ (s, m a, m b )σ tot probability i n T va v b V probability i n time flux 4 Accelerators and Collider Experiments 4.1 Particle Accelerators: Motivations λ h p By using a fixed target, one can furthermore produce a beam of secondary particles that may be stable, unstable, charged or neutral Center of Mass Energy In a fixed target experiment, s m + me inc. However, for two beams colliding with each other, s E inc. 4. Acceleration Methods 4..1 Cyclotron mv For a cyclotron, R qvb and so mv qbr, and so the maximum momentum attainable with a cyclotron is p qbr. 9

10 Maximal energy is of the order 0MeV. The period is given by T πr v and so f 1 T v πr p/m πr qbr/m πr qb πm : f qb πm. Isochronous Cyclotron Can calculate relativistic effects by γmv R qvb. Can then make the B field non-uniform in order to make the particle s speed synchorinized with the AC voltage. Alternatively the AC current can be adjusted to compensate for these relativistic effects. 4.. Synchrotron The trajectory radius is kept constant. In a synchrotron, the adaptation for relativistic effects is done by variation of the magnetic field strength in time, rather than in space. 4.3 Particle Physics Experiments Cross Section quantum probability σ time incoming flux where flux velocity of particles in beam number density of beam 1 v volume 4.3. Luminosity L number of events per unit time σ number of events cross section integrated luminosity Particle Detectors Want to measure: 1. Spatial coordinates and timing of final state. Momentum 3. Energy 4. Type of particle 4.4 Kinematics and Data Analysis Methods Pseudorapidity [ ( )] 1 η log tan θ cm 10

11 4.4. Claim The pseudorapidity is invariant under longitudinal boosts The Missing Mass Method Sometimes some particles are not detected (especially neutral particles). Missing Mass (E in E out ) ( p in p out ) We use the missing mass to identify them experimentally The Invariant Mass Method ( N ) The invariant mass is defined as M : p i i1 where there are N particles in the system. It allows us to take two or more particles together and identify them as a short-lived intermediate particle by looking at a peak in a histogram plot. Example: Using the π 0 γ, we can find the mass of the π 0. 5 Elements of QED 5.1 Quantum Mechanical EoM Brining E m + p into quantum mechanics we get the Klein-Gordon equation: (i t ) m + i ) ( and so µ µ m 0. This is not good because it leads to negative energy solutions (when interpreted as operating on a single particle wave function). Thus postulate an equation with only first-order derivatives: (iγ µ µ m) 0. This has to be compatible with p m, thus, (γ µ p µ )! p. This can only be fulfilled if the γ µ objects are not numbers but rather matrices. Thus we have the Dirac equation: (iγ µ µ m) ψ 0. There is also the adjoint Dirac equation: i µ ψγ µ + m ψ 0, [ where ψ ] ψ γ 0, and {γ [ µ, γ ν } ] 0 η µν. One possible representation: γ σ and γ 1 0 i i σ i 0 [ ] [ ] [ ] i 1 0 where σ 1, σ 1 0 and σ i Then define S µν : i 4 [γµ, γ ν ] and under a Lorentz transformation, ψ e i ωµνsµν ψ. There is a conserved current j µ ψγ µ ψ. 11

12 5. Solutions to the Dirac Equation 5..1 Free Particle at Rest 1. Plug in a Fourier transform of ψ, ψ (x) dpe ipx u (p), into the Dirac equation to get in momentum space: (γ µ p µ m) u (p) 0.. By Lorentz invariant we should be able to solve the equation in any frame and then boost to an arbitrary ( frame ) to get a general solution. Thus solve it in the frame where p m, 0, and so, ( ) γ u (p) 0. The x solutions to this equation are u (p) y x for any (x, y) C. y 5.. Free Particle Now we can boost these solutions and get: 5..3 Explicit form of u and v p σx u (p) p σy p σx were σµ ( 1, σ i) and σ µ ( 1, σ i). p σy Similarly we will also get an equation for v (p) (the Fourier transform of ψ) Operators on Spinor Spaces Hamiltonian If H i 0 then H mγ 0 iγ 0 γ by the Dirac equation. [ ] σ 0 Helicity Helicity is defined as h 1 p 0 σ p (the projection of the spin onto the momentum). Observe that [H, h] 0, and so we can diagonalize both simultaneously. Chirality Define γ 5 iγ 0 γ 1 γ γ 3 and so γ 5 is the chirality operator. It corresponds to an irreducible subspace of the 4-spinor representation space. 5.3 Field Operator of the Dirac Field a s p annihilates a particle, b s p annihilates an anti-particle. ψ and ψ [ E p e ipx a s (p) ū s (p) + e ipx b s (p) v s (p) ]. Then { a s p, a } r q d 4 p (π) 4 1 (π) 3 δ rs δ 3 ( p q) and similarly for the b s. All other combinations anti-commute. d 4 p 1 [ (π) 4 E p s1 e ipx a s p u s (p) + e ipx b s p v s ( 1

13 5.4 The Dirac Propagator S F (p) i(pµγµ +m) p m +iε 6 QED Tests 7 QCD 8 QCD in electron-positron Annihilations Examples of jet Algorithms The JADE Algorithm Compute all possible pairs of y ij where y is a chosen metric (for example, m ij E cm ). Search for the smallest y among all pairs, say it s y kl. If y kl < y cut for some pre-chosen y cut then combine particles k and l (combination via some pre-chosen method, for example, p (kl) new p k + p l ). Go back to step one until any y below y cut Event Shape Variables ({ Thrust: T : sup i pi v v S i pi is perfectly back-to-back whereas T 1 symmetric. 9 Questions from Exam Protocols 1. What is the S-matrix? How to calculate it? }). Thus T 1 means the event means the event is spherically. What is the interaction picture? (is this necessary with the path integral?) ( n 3. Wick s theorem: T ˆφ ) ( n j1 (x j ) N ˆφ ) j1 (x j ) +all possible contractions 4. Memorize all propagators. Also be able to explain the iε prescription. 5. How to prove experimentally that quarks have spin- 1? Obey the color statistics? Look at angular dependence of cross-section? (1 + cos (θ) for spin- 1 particles, sin (θ) for spin-0 particles). Also possible to look at the thrust parameter to decipher how the jets hadronize and get angular dependence. 13

14 6. How to study the e e + µ µ + process? (a) Silicon detectors (b) Drift chambers (c) Put everything in a magnetic field to measure the momentum and charge by measuring the curvature and bending direction. (d) Calorimeters measure the energy of the produced particles. 7. Why is the muon long-lived? Because it decays weakly! 8. Time dilation: T γt 0. This is how we observe short-lived particles. 9. How to experimentally verify that there are 3 colors in QCD? σ e + e µ + µ σ e + e q q N C flavor e flavor. 10. How does the top quark decay? (weakly, but it has a huge phase space integral which means it will decay very quickly!) 11. Phase space formula: R n d 3 p 1 d... 3 p n δ (4) (P (π) 3 E 1 (π) 3 E f P i ). n phase space integral is Lorentz invariant. The 1. The decay time or the decay width is not Lorentz invariant (think of time dilation). 13. How to compute the color factors of the various vertices? 14. Be able to list all particles in the standard model. 15. How to measure the non-linear nature of QCD? Look for 3-vertex interactions. 16. How to decipher quarks from jets? (the scheme with the metric) 17. OZI rule: any strongly occurring process will be suppressed if its Feynman diagram can be split in two by cutting only internal gluon lines. 18. Decay of the J/ψ meson. 19. How to get from M to σ? 0. What s the difference between the e and the µ as they go through matter? Be able to draw the ionization and bremstrahlung vs. energy graphs. Critical energy (below which ionization is dominant) is 10M ev for electrons and 1T ev for muons. 1. How to accelerate particles (explain a particular accelerator).. The π 0 decays into two photons (use invariant mass of two-photons to get the mass of the pion). 14

15 3. How to measure the mass of π 0 using the two photons? Invariant mass method. 4. The motivation behind Dirac s equation. 5. Clifford Algebra. 6. Free solutions to Dirac s equation. 7. How many generators are there in SU (N)? N How to know gluons are real? Three jet events. 9. How to detect a pion? It would be visible in the hadron calorimeter. 30. What are rapidity and transverse mass? 31. Pseudorapidity. 15

16 10 Particle Detectors 16

17 11 Particle Accelerators 11.1 Cyclotron 11. Electrostatic Accelerator (Cockroft and Walton) 11.3 Synchrotron 11.4 Linear Accelerator 1 Colliders 13 Energy Loss Through Materials 13.1 Photons Photoelectric Effect A photon ionizes an atom. Dominant in low-energy photons Compton Scattering A photon scatters off an electron, giving it energy Pair Production If E γ > 1MeV then the photon can create an electron-positron pair. 13. Muons Ionization Described by the Bethe-Bloch formula. Dominant for heavy particles like the muon until a few T ev s Electrons Ionization Dominates at low energies Moller Scattering (e e e e ) Merely takes place at low energies Bhabha scattering (e e + e e + ) Merely takes place at low energies. 17

18 e + e γ annihilation Merely takes place at low energies Bremsstrahlung Loss of energy by radiation of photons as a result of change of velocity. Above a few tens of MeV starts to dominate. Energy loss proportional to de dx E X 0 where X 0 is the radiation length (the length at which the particle loses 63% of its energy) Hadrons Nuclear Interactions 14 Cherenkov Radiation The result of a massive particle passing through a material faster than the speed of light in that material. cos (θ) 1 vn where v is the speed of the particle, n is the refractive index of the material, and θ is the angle of radiation with respect to the axis formed by the flight of the particle. Minimum energy for Cherenkov radiation is around 30M ev for an electron passing through Hydrogen gas EM 1. electron positron or photon with less than a few MeV will have mainly the photoelectric effect and Compton scattering. E c ). Above a few MeV, photons interact with matter primarily via pair production, whereas high energy electrons or positrons mainly emit photons (Bremsstrahlung). The characteristic length transversed until the energy falls below a few MeV is called the radiation ( length X 0, which is E characteristic for the material: X X 0 log 0 where E c is the critical energy (the energy in which the bremsstrahlung and ionization rates are equal). E c 800MeV (Z+1.) and E 0 is the initial energy of the particle. de dt E 0 b (bt)a 1 e bt Γ(a) where t X X 0, E 0 is the initial energy and a and b are parameters to be fitted with experimental data. 18