#### Transcript lecture 6

Topics in Contemporary Physics Basic concepts 4 Luis Roberto Flores Castillo Chinese University of Hong Kong Hong Kong SAR January 23, 2015 PART 1 • Brief history • Basic concepts • Colliders & detectors 5σ • From Collisions to papers S ATLA (*) 15 GeV d Selecte s=7 2000 1800 TeV, s=8 ò 5.9 fb òLdt = -1 TeV, 1600 1400 1200 1000 800 600 10 ATLAS 400 5 150 0 GeV) sample 2 126.5 and 201 fit (m H = 2011 usive Data Bkg incl -1 Sig + nomial 4.8 fb er poly Ldt = 4th ord n diphoto 2400 2200 / Events -1 fb t = 4.8 V: òLd -1 5.8 fb Ldt = TeV: ò Te s=7 s=8 ®4l 100 nary Prelimi 200 250 0 ] [GeV 200 m100 4l - Bkg Even • The Higgs discovery c. Un Syst. 20 H®ZZ Data V ts/5 Ge (*) Data ZZ round s, tt Backg Z+jet round V) Backg 25 Ge l (m H=1 Signa 25 150 140 -100 100 160 V] mg g [Ge 130 0 120 110 • BSM • MVA Techniques • The future L. R. Flores Castillo CUHK January 23, 2015 2 … last time: Basic concepts 3 • Relativistic kinematics – Lorentz transformations – Consequences – A few examples – Four-vector notation • Four-vectors • Lorentz transformation in matrix form • The metric, the scalar product, invariants • Energy and momentum L. R. Flores Castillo CUHK January 23, 2015 3 Reminder: interactions QED: QCD: Weak: W/Z: L. R. Flores Castillo W/Z/γ: CUHK January 23, 2015 4 Reminder: the CKM matrix • The W boson can ‘jump across generations’ • In the SM, the weak force “sees” slightly rotated versions of the down quarks: Cabibbo-Kobayashi-Maskawa matrix L. R. Flores Castillo CUHK January 23, 2015 5 Reminder: Relativistic kinematics Maxwell equations æ v ö t ' = g ç t - 2 x÷ è c ø x' = g (x - vt) y' = y z' = z Lorentz transformations c for all observers x0 ' = g (x0 - b x1 ) Four-vector xm ' = Lnm xn x1 ' = g (x1 - b x0 ) x2 ' = x 2 x0 = ct, x = x, 1 x3 ' = x3 x2 = y, x =z 3 bº time-position: proper velocity: energy-momentum: v c E = g mc2 = mc2 + 12 mv2 + 83 m vc4 +... 4 contravariant I º (x0 )2 - (x1 )2 - (x2 )2 - (x3 )2 I = xm x a× b º am bm covariant Energy-momentum Useful: p = g mv pm pm = (mhm )(mh m ) = m2 c2 E2 pm p = 2 - p2 c m L. R. Flores Castillo Scalar product: m is Lorentz-invariant CUHK E = g mc 2 E = m c +p c 2 xμ = (ct, x, y, z) ημ=dxμ/dτ = γ(c, vx, vy, vz) pμ = mημ = (E/c, px, py, pz) 2 4 2 2 January 23, 2015 p / E = v / c2 v = pc2 / E 6 Relativistic Kinematics II 7 Energy and momentum What if m=0? Would the energy/momentum necessarily be 0? mv p= 2 E= 2 1- v / c 2 mc 2 2 1- v / c not if v=c !! The equation can still hold. Hence: E =mc +p c 2 v = c, 2 4 2 2 E = |p| c And L. R. Flores Castillo CUHK January 23, 2015 E = hv 8 Collisions • Classical – Mass is conserved: mA+mB=mC+mD – Momentum is conserved: pA+pB=pC+pD – Kinetic energy may or may not be conserved: • Sticky (kinetic energy decreases): TA+TB > TC+TD • Explosive (kinetic energy increases): TA+TB < TC+TD • Elastic (kinetic energy decreases): TA+TB = TC+TD • Relativistic – Energy is conserved: EA+EB=EC+ED m m m m p + p = p + p A B C D – Momentum is conserved: pA+pB=pC+pD – Kinetic energy may or may not be conserved • Sticky (T decreases): rest energy and mass increase • Explosive (T increases): rest energy and mass decrease • Elastic (T is conserved): rest energy and mass are conserved L. R. Flores Castillo CUHK January 23, 2015 9 Collisions • In general, mass is not conserved: π0 γ + γ • Elastic: same particles come out as went in • Consistency? – Rest mass vs internal energy – Hadron mass L. R. Flores Castillo CUHK January 23, 2015 10 Example 1 Two lumps of clay, each of mass m, collide head-on at 3c/5; they stick together. What is the mass M of the final composite lump? Conservation of momentum: trivial. p1 + p2 = pM = 0 Conservation of Energy: 2mc2 5 Mc = 2Em = = (2mc2 ) 1- (3 / 5)2 4 2 5 M= m 2 Notice that M > m1 + m2 L. R. Flores Castillo CUHK January 23, 2015 11 Example 2 A particle of mass M, initially at rest, decays into two pieces, each of mass m. What is the speed of each piece as it flies off? Conservation of momentum: equal and opposite speeds Conservation of energy: Mc = 2g mc = 2 2 2mc2 1- v2 / c2 v = c 1- (2m/ M ) L. R. Flores Castillo CUHK 2 Only makes sense if M>2m For M=2m, v=0 For M>>2m, vc β depends only on m/M January 23, 2015 12 Example 3 A pion at rest decays into a muon plus a neutrino. What is the speed of the muon? (mμ=105.66 MeV/c2; mπ=139.57 MeV/c2; mν=0) Ep = Em + En pp = pm + pn pp = 0 ® pm = -pn mp c2 = pm c+ c mm2c2 + pm2 pm = mp2 - mm2 2mp c Subst. in E2=m2c4+p2c2: E2 = m2c4 + p2c2 : Ep = mp c2 En = pn c = pm c Em = c mm2 c2 + pm2 L. R. Flores Castillo CUHK Em = mp2 + mm2 2mp c2 Subst. in v = p c2 / E : mp2 - mm2 vm = 2 c = 0.27c 2 mp + mm January 23, 2015 13 Example 3. Second solution. A pion at rest decays into a muon plus a neutrino. What is the speed of the muon? (mμ=105.66 MeV/c2; mπ=139.57 MeV/c2; mν=0) pp = pm + pn Scalar product w/itself: In all frames: pn = pp - pm pn2 = pp2 + pm2 - 2 pp × pm pn2 = 0; pp2 = mp2c2 ; pm2 = mm2 c2 With the pion at rest: pp × pm = Similarly, from : 2 2 Em = Ep Em Ep = 2 Em = mp Em c c c mp2 + mm2 2mp c2 pm = pp - pn mm c = mp2c2 - 2mp En Similarly, from En = pn c = pm c and v = p c2 / E: L. R. Flores Castillo 0 = mp2c2 + mm2 c2 - 2mp Em CUHK mp2 - mm2 vm = 2 c = 0.27c 2 mp + mm January 23, 2015 14 Example 4 The Bevatron at Berkeley aimed at producing antiprotons via p+ p® p+ p+ p+ p What is the minimum energy of the incident proton that allows it? Lab frame: CM frame: L. R. Flores Castillo CUHK January 23, 2015 15 Example 4 The Bevatron at Berkeley aimed at producing antiprotons via p+ p® p+ p+ p+ p What is the minimum energy of the incident proton that allows it? Minimum = all particles at rest in the CM frame. m æ E + mc2 ö =ç , p , 0, 0 ÷ è c ø Before the collision: pTOT After the collision: m pTOT ' = ( 4mc, 0, 0, 0) m m pTOT pm TOT = pTOT ' pm TOT ' æ E + mc2 ö 2 2 ç ÷ - p = (4mc) è c ø 2 Eliminating p w/ E2=m2c4+p2c2: E = 7mc 2 i.e., the incident proton should have a kinetic energy above 6TeV (and, indeed, the antiproton was discovered with E ~ 6000MeV) L. R. Flores Castillo CUHK January 23, 2015 16 Example 5 Two identical particles, each with mass m and kinetic energy T, collide head-on. What is their relative kinetic energy, T’? (the kinetic energy of one on the rest system of the other) Fixed target Colliding beams m pTOT ( p ) = ( p ') m 2 TOT æ E'+ mc2 ö pTOT ' = ç , p'÷ è c ø æ 2E ö = ç , 0÷ è c ø m TOT m 2 æ 2E ö æ E'+ mc2 ö 2 ç ÷ =ç ÷ - p' è c ø è c ø 2 2 : ( Eliminating p w/ E2=m2c4+p2c2: 2E2 = mc2 E'+ mc2 In terms of L. R. Flores Castillo T=E-mc2 and CUHK T’=E’-mc2: ) æ T ö T ' = 4T ç1+ ÷ è 2mc2 ø January 23, 2015 Classical result: only T’=4T Reduces to it when T<<mc2 In B’s frame, A has twice the speed, hence 4 times the E. 17