Intro to Nuclear and Particle Physics (5110)

Transcription

1 Intro to Nulear and Partile Physis (5110) Marh 7, 009 Relativisti Kinematis 3/7/009 1

2 Relativisti Kinematis Review! Wherever you studied this before, look at it again, e.g. Tipler (Modern Physis), Hyperphysis Maxwell EM theory, Mihelson-Morley experiment Lorentz/FitzGerald+Poinaré+Einstein Speial Theory of Relativity. Inertial-referene-frame independene of the laws of physis in general and the speed of light in partiular give... Lorentz Transformation S S vx x' = γ ( x vt) y' = y z' = z t' = γ t x= γ ( x' + vt' ) y = y' z = z' t = γ t' + r r v r ( ) 1 β = β = βxˆ γ = 1 β vx ' 3/7/009

3 Lorentz 4-Vetor4 Coordinates of event in 4D spae-time. Energy-momentum vetor or 4-momentum. x µ = x x x x = t x y z = r t x ( 0 1 3,,, ) (,,, ) (, ) µ E E r p = ( p, p, p, p ) =, px, py, pz=, p Components of a 4-vetor depend on the hoie of RF. Laws of physis and ertain physial quantities derived from 4-vetors do not (Lorentz invariant). x =Λx Implied summation (Einstein) 3 µ µ ν µ ν = Λ ν =Λν ν = 0 x x x Λ γ βγ 0 0 βγ γ = 3/7/009 3

4 Lorentz Transformations r r v r β = β = βxˆ γ = 1 β ( ) 1 t γ βγ 0 0 t x βγ γ 0 0 x = y y z z ( t, x, y, z) ( t, x, y, z ) 3/7/009 4

5 Lorentz Transformations r r v r β = β = βxˆ γ = 1 β ( ) 1 t γ βγ 0 0 t x βγ γ 0 0 x = y y z z ( t, x, y, z ) ( t, x, y, z) 3/7/009 5

6 Lorentz Transformations r r v r β = β = βzˆ γ = 1 β ( ) 1 t γ 0 0 βγ t x x = y y z βγ 0 0 γ z ( t, x, y, z) ( t, x, y, z ) 3/7/009 6

7 Lorentz Transformations r r v r β = β = β ˆ ˆ xx+ βyy+ βzzˆ γ = 1 β ( ) 1 ( t, x, y, z) ( t, x, y, z ) 3/7/009 7

8 Lorentz Transformations r r v r β = β = βxˆ γ = 1 β ( ) 1 E γ βγ 0 0 E βγ γ 0 0 p x px = p p y y p p z z E E, px, py, pz, p x, p y, p z 3/7/009 8

9 Lorentz Invariant Just as a vetor s length is invariant under a oordinate transformation in lassial physis, there are quantities invariant under Lorentz transformation in 4D spae-time. Salar (or inner or dot) produt of two 4-vetors: r r x y= xy xy xy xy = xy x y E.g., the length of the spae-time oordinate 4-vetor: Invariant Spaetime Interval s = x x= t x y z E.g., the magnitude of the 4-momentum: Energy- Momentum Invariant s s s > 0 = 0 < 0 E p p p p p p M r 4 E + p = M = = x y z = timelike lightlike spaelike 3/7/009 9

10 1 ( ) v 1 γ = 1 β β = β = 1 γ r In the rest frame (CM frame) of a partile: p = 0 r ECM Rest Energy p p = = M E CM = M Nuts and Bolts Total Energy E = M + T =γ M Rest Mass When T << M we an safely use nonrelativisti kinematis Almost always for nulear phenomena; rarely for partile physis 5.3 MeV α (He nuleus) from Thorium deay T = = 5.3 MeV M 3.76 GeV << T 5.3 MeV γ = 1+ = 1+ = M 3.76 GeV 1 7 m β = 1 = v= β= γ s 1 TNR 7 m TNR = MvNR vnr = = M s 50 MeV eletron produed in the deay at rest of the osmi ray muon that stops in matter T < = = 50 MeV not M MeV T γ = 1+ = 98.8 M 1 β = 1 = γ 1 3/7/009 10

11 Lorentz Trans. Energy Energy-Momentum S' S r In : (,0) E r S p= M In S : p =, p Consider a partile at rest in S. What are its momentum and energy in S'? r r v r β = β = β xˆ γ = 1 β ( ) 1 E 0 0 E γ βγ p βγ γ 0 0 x p x = p p y y p p z z E E, p x, p y, p z, p x, p y, p z E E = γ + βγ ( 0 ) = γm E = γ M E p' = βγ + γ ( 0) = βγ M p = γm β = γmv ( ) p = γ M β E = γ M E p = γ M γ M β 4 4 = γ M ( 1 β ) 4 4 E p = M 3/7/009 11

12 Lab CM Time Dilation and Partile Deay CM Unstable partile reated at and deays at t, x. t 1, x τ = t t 1 Proper Lifetime Lab Unstable partile reated at and deays at t, x. In the lab, the partile travels ( )( ) ( ) 1 1 x= β t t = βγ t t = βγτ 1 1 β t1 = γ t1 + x 3/7/009 1 t, x β t = γ t + x If we know the speed (mass, momentum/energy), we an determine the lifetime by fitting the distribution of path lengths to the expeted deay distribution (exponential).

13 Example: CLEO- Experiment ee D + DD K π S 0 KS π + π π 0 K S CM π + M = τ = GeV s p = GeV E = GeV E 1 γ= =1.997 β= βγ=1.78 = M γ p Shortut : βγ = = 1.78 M 0 Flight path of "typial" from 0 -body deay is D K S βγτ = m. CDF Experiment pp TeV of partiles, inluding energeti K 0 S 's. To illiustrate, take p = 0 GeV E = GeV γ = 40 β = Flight path of "typial" 0 - is 0 GeV KS βγτ = 1.1 m. 3/7/009 13

14 Build a Partile The sum of two 4-vetors is a 4-vetor. Consider the 4-momentum of a system of two partiles, for example the daughters of a deaying parent partile. E1 r E r p1 =, p1 p =, p E1+ E r r p1+ p =, p1+ p Compute length : ( ) p1+ p = p1 + p + p1 p 1 1 EE r r = m + m + p 1 p EE 1 r r = m 1 + m + p 1 p osθ 4 = m p + p = m ( ) θ p r p r 1 m 1 is the effetive mass of the twopartile system 3/7/009 14

15 Strategies for RK Problems Use full Lorentz transformation for boosts of 4-momenta, but rely on invariants as muh as possible to simplify omputations. Speifially, for threshold alulations : What is the minimum energy of to? Relate the initial state in one frame to the final state in another with invariants and 4-momentum onservation. Initial Final 4 -momentum = 4 -momentum in Lab in CM Frame Frame 3/7/009 15

16 O. Chamberlain E. Segré Berkeley Bevatron 1955 Disovery of the Antiproton Stationary Target Can t get this p+ p p+ p+ p+ p Beam Partile Min. KE? without this Lab CM i Before After r p = E+ m p (, ) p f = r ( 4 m,0) p i = r p ( E+ m ) p = ( 4m ) f r E + Em + m p = 16m 4 4 Em = 14m 4 E = m T = m 7 or GeV 3/7/009 16

17 Fixed Target vs. Colliding Beams The CM frame is natural for threshold problems. Why not forget the lab and its energy-wasting overall motion and ollide a speeding partile with a speeding partile? E E r r p + p = +, p + p E1 E r = +,0 ( ) For equal energies E 1 =E =E beam, the available energy for partile reation is E beam. E r r p1+ p = + M, p1+ 0 E1 r = + M, p1 ( ) 1 For beam energy E 1 =E beam and target mass M, the available energy for partile reation is M Ebeam. Almost all new major HEP aelerators built sine the 1970 s have been olliding beam mahines or have had a olliding beam option. 3/7/009 17

18 Natural Units Take h = 1and = 1by onvention. is the fundamental unit of veloity Feynman : "Keeping trak of h and is a omplete waste of time." and h is the fundamental unit of angular momentum. Your unit system is then fully speified with one more quantity, usually the unit of energy. Choose for onveniene: ev or kev or MeV or GeV or TeV or m e or m p or amu Thus your basi units are E E h Mass : Momentum : Length : Time : E E h 3/7/009 18

19 Example: the energymomentum invariant 4 E = p + m in natural units beomes E = p + m And a quantity of dimensionality p q r M LT p q r E in natural units is equivalent to After using NU, onvert to real-world units for alulation Restore missing fators by dimensional analysis Use standard onversions: m = h = s h = MeV fm MeV s e 1 α = = 4πε h fm = m 3/7/009 19

20 Example: Thomson Sattering γ + e - γ+ e - with E γ << m e σ = 8π α 3 me a b σ Find and that give the dimensions of area. 8πα a b σ = h 3 m a m 1 kg m m kg s s [ σ ] = = See that a = + gets rid of the kg. b 4 m= = b 1 kg m m m m b kg s s s s b = -gets rid of s, leaves m. e in natural units b 8πα σ = h m σ = What s the probability? What s the ross setion? Compute it 3 e παh = 4 3 m e ( MeV fm) ( ) 8π MeV σ = 66.5 fm or m 9 1barn = 10 m = 10 m 8 4 σ = barn or 665 mb 3/7/009 0

21 Mandelstam Variables Three independent relativisti invariants that together give a omplete desription of the kinematis of this reation. (Now in natural units!) A + B C+ D A D C B s = p + p ( ) A t = p p B ( ) A C u = p p ( ) A D (CM energy) (4-momentum transfer) (something else) Nifty feature: s+ t+ u = m + m + m + m A B C D 3/7/009 1