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Topics in Contemporary Physics Basic concepts 3 Luis Roberto Flores Castillo Chinese University of Hong Kong Hong Kong SAR January 16, 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 TeV, ò Ldt = -1 5.9 fb 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 16, 2015 2 … last time: Basic concepts 2 • Review of units & particle dynamics – Feynman diagram examples • Decays and conservation laws • A word on Unification L. R. Flores Castillo CUHK January 16, 2015 3 Reminder: interactions QED: QCD: Weak: W/Z: L. R. Flores Castillo W/Z/γ: CUHK January 16, 2015 4 Reminder: adding possible histories +… L. R. Flores Castillo CUHK January 16, 2015 5 A few key concepts • Formally, the W boson can only link ‘up-type’ quarks (u,c,t) into the corresponding ‘down-type’ (d,s,b). • However, experimentally, some times it mixes generations • Solution: the weak force “sees” slightly rotated versions of the down quarks: Cabibbo-Kobayashi-Maskawa matrix L. R. Flores Castillo CUHK January 16, 2015 6 About unification • • • • Electricity + Magnetism Glashow, Weinberg and Salam: EM + Weak = EW Chromodynamics + EW ? The “running” of the coupling constants hints at it ? L. R. Flores Castillo CUHK January 16, 2015 7 Today’s outline • Relativistic Kinematics • Symmetries L. R. Flores Castillo CUHK January 16, 2015 8 Relativistic Kinematics 9 Lorentz transformations • S’ moves with velocity v wrt S • Along the common x/x’ axis • t = t’ = 0 when x = x’ = 0. L. R. Flores Castillo CUHK January 16, 2015 10 Lorentz transformations An event occurs at (x,y,z) and time t in S; coordinates in S’: x ' = g (x - vt) x = g (x '+ vt) y' = y y = y' z' = z æ v t ' = g çt - 2 è c z = z' æ v t = g ç t '+ 2 è c ö x÷ ø g= L. R. Flores Castillo CUHK ö x÷ ø 1 1- v 2 / c 2 Inverse transformation: v-v January 16, 2015 11 Consequences • Relativity of simultaneity: events A and B, at the same time in S, but in different locations, do not occur at the same time in S’: t ' A = t 'B + gv c 2 (xB - x A ) • Lorentz contraction: a stick at rest in S’, from x’=0 to L’; size in S? – its ends at the same time in S (t=0): x=0, x = L’/γ (from LT) – Length = L’/ γ – γ > 1, so moving objects are shortened by a factor γ L. R. Flores Castillo CUHK January 16, 2015 12 Consequences • Time dilation: – Clock at x’=0 ticks at t’=0, t’=T’; interval in S? from t=0 to (t’=T’, x’=0), so t=γT’ – T = γT’ ; clocks in S mark a longer interval – “Moving clocks run slow” – All processes are affected – Including the “internal clock” of elementary particles – Moving particles lasts longer by a factor of γ – Examples: • Muons from the upper atmosphere • B mesons in our HEP detectors L. R. Flores Castillo CUHK January 16, 2015 13 A few quick examples v 130 km/h beta gamma 14,000 km/h 0.5 c 0.99 c ? L. R. Flores Castillo CUHK 6,500 January 16, 2015 14 Four-vector notation Position-time four-vector: x = ct, x = x, x = y, x = z 0 1 2 3 Lorentz transformations take a nicely symmetrical form: æ v ö t ' = g çt - 2 x ÷ è c ø x ' = g (x - vt) y' = y x 0 ' = g (x 0 - b x1 ) z' = z x3 ' = x3 L. R. Flores Castillo CUHK x ' = g (x - b x ) 1 1 0 x2 ' = x2 January 16, 2015 v bº c 15 Four-vector notation This can be written as: 3 x ' = åL x m m n n (m = 0,1, 2, 3) v=0 with Einstein’s ‘summation convention’: m m n n x '=L x L. R. Flores Castillo CUHK January 16, 2015 16 Four-vector notation Very concise, and allows generalizations. • It works also for transformations not along the x axis From S to S’, coordinate values change, but the following combination remains the same: I º (x ) - (x ) - (x ) - (x ) 0 2 1 2 2 2 3 2 = (x ') - (x ') - (x ') - (x ') 0 2 1 2 2 2 3 2 3 To express it as sum m m x å x m =0 we need to change the sign of the spatial components. L. R. Flores Castillo CUHK January 16, 2015 17 Four-vector notation We can do it by introducing: so that We can also define L. R. Flores Castillo CUHK January 16, 2015 18 Four-vector notation So that, now m m I º xm x = x xm This expresses the invariant in a very simple expression. Too much trouble just to include those minus signs? It generalizes to non-Cartesian coordinates and to GR. L. R. Flores Castillo CUHK January 16, 2015 19 Four-vector notation Other quantities also transform as the position-time fourvector xμ. m m n n a '=L a “contravariant” We define a four-vector as any object that transforms like this when we go from one inertial system to another. We can always define a corresponding vector with negative signs in the spatial components: n am = gmn a (which is easy to convert back: “covariant” m mn a = g an ) [ formally, gμν are the elements of g-1, but it’s its own inverse ] L. R. Flores Castillo CUHK January 16, 2015 20 Four-vector notation For any two four-vectors, their scalar product is m m a bm = am b = a b - a b - a b - a b 0 0 1 1 It is also invariant. a × b = a 0b0 - a × b a º a × a = (a ) - a 2 0 2 2 2 3 3 a × b º am b m 2 Four vectors are called timelike if a2>0 spacelike if a2<0 lightlike if a2=0 L. R. Flores Castillo CUHK January 16, 2015 21 Energy and momentum When two reference frames are considered, the proper time advances by a smaller amount dt = dt g This is also an invariant (all observers agree on the proper time of a given particle, satellite or space traveler). A related quantity, similar to velocity is the “proper velocity”: dx h= dt dx v= dt h = gv Useful because, under a LT, only the numerator transforms (and as a four-vector); this is indeed a four-vector too. L. R. Flores Castillo CUHK January 16, 2015 22 Energy and momentum Components: h = g (c, vx , vy, vz ) m In this case, the invariant ημημ is: hmh m = g 2 (c2 - vx2 - vy2 - vz2 ) = g 2c2 (1- v 2 / c2 ) = c2 How should we define momentum? mv or mη ? Better use η, because then, if momentum is conserved in one system, it will also be in all others. m m p º mh L. R. Flores Castillo CUHK January 16, 2015 23 Energy and momentum Proper velocity is a four-momentum, so also is pμ p º mh m m p0 = g mc p = g mv = mv 1- v 2 / c 2 We define the relativistic energy as E = g mc = 2 mc 2 1- v 2 / c 2 So, the 0-th component of the four-momentum is E/c. æE ö p = ç , px , py , pz ÷ èc ø m L. R. Flores Castillo CUHK January 16, 2015 24 Energy and momentum Computing the invariant pμpμ: From the definition: pm pm = (mhm )(mh m ) = m2hmh m = m2c 2 From its components: E pm p = 2 - p2 c 2 m E2 2 2 2 -p = m c 2 c E 2 = m2 c 4 + p2 c 2 L. R. Flores Castillo CUHK January 16, 2015 25 Energy and momentum Energy: æ 1 2 2ç E = g mc = mc ç 1- v22 è c ö ÷ = 1+ 1 v22 + 3 v44 +... 2 c 8 c ÷ ø = mc 2 + 12 mv 2 + 83 m vc4 +... 4 Rest energy Relativistic kinetic energy T º mc (g -1) 2 L. R. Flores Castillo CUHK January 16, 2015 26 Energy and momentum What if m=0? Would the energy/momentum necessarily be 0? mv p= 1- v / c 2 E= 2 mc 2 1- v / c 2 2 not if v=c !! The equation can still hold. Hence: E =m c +p c 2 v = c, 2 4 2 2 E = |p| c And L. R. Flores Castillo CUHK January 16, 2015 E = hv 27 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 16, 2015 28 Collisions • In general, mass is not conserved: π0 γ + γ • Elastic: same particles come out as went in L. R. Flores Castillo CUHK January 16, 2015 29 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: 2mc 2 5 2 Mc = 2Em = = (2mc 2 ) 1- (3 / 5)2 4 5 2 M = mc 2 Notice that M > m1 + m2 L. R. Flores Castillo CUHK January 16, 2015 30 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 2 = 2g mc 2 = 2m 1- v 2 / c 2 v = c 1- (2m / M ) L. R. Flores Castillo CUHK 2 Only if M>2m For M=2m, v=0 For M>>2m, vc January 16, 2015 31 32