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Scattering Experiment Monoenergetic particle beam Beam impinges on a target Particles are scattered by target Final state particles are observed by detector.

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Presentation on theme: "Scattering Experiment Monoenergetic particle beam Beam impinges on a target Particles are scattered by target Final state particles are observed by detector."— Presentation transcript:

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2 Scattering Experiment Monoenergetic particle beam Beam impinges on a target Particles are scattered by target Final state particles are observed by detector at . 2

3 Flux : Number of particles/ unit area / unit time Area: perpendicular to beam For a uniform beam: particle density Number of particles / unit volume Consider box in Figure. Box has cross sectional area a. Particles move at speed v with respect to target. Make length of box a particle entering left face just manages to cross right face in time  t. Volume of Box: So Flux 3

4 How many targets are illuminated by the beam? Multiple nuclear targets within area a Target Density, Number of targets per kg: Recall: 1 mol of a nuclear species A will weigh A grams. i.e. the atomic mass unit and Avogadro’s number are inverses: (N A x u) = 1 g/mol So: 4 L Area a Density 

5 5 L Area a Density 

6 For a detector subtending solid angle d  If the detector is at an angle  from the beam, with the origin at the target location: 6

7 Compute: Fraction of particles that are scattered Area a contains N t scattering centers Total number of incident particles (per unit time) N i =Fa Total number of scattered particles (per unit time) N s =F N t  tot So Fraction of particles scattered is: N s /N i =F N t  tot / (F a) = N t  tot / a Cross section: effective area of scattering Lorentz invariant: it is the same in CM or Lab. For colliders, Luminosity: Rate: 7

8 Quantum Mechanics: Fermi’s (2 nd ) Golden Rule Calculation of transition rates In simplest form: QM perturbation theory Golden Rule: particles from an initial state  scatter to a final state  due to an interaction Hamiltonian H int with a rate given by: 8

9 Quantum theory of interaction between nucleons 1949 Nobel Prize Limit m → ∞. Treat scattering of particle as interaction with static potential. Interaction is spin dependent First, simple case: spin-0 boson exchange Klein-Gordon Equation Static case (time-independent): 9

10 Steps to calculating and observable: Amplitude: f = Probability ~ |f| 2. Example: Non-Relativistic quantum mechanics Assume  is small. Perturbative expansion in powers of . Problem: Find the amplitude for a particle in state with momentum q i to be scattered to final state with momentum q f by a potential H int (x)=V(x). 10

11 q = momentum transfer q = q i - q f 11

12 QFT case, recover similar form of propagator! Applies to single particle exchange Lowest order in perturbation theory. Additional orders: additional powers of . Numerator: product of the couplings at each vertex. g 2, or . Denominator: Mass of exchanged particle. Momentum transfer squared: q 2. In relativistic case: 4-momentum transfer squared q  q  =q 2. Plug into Fermi’s 2 nd Golden Rule: Obtain cross sections 12

13 Nuclear forces are short range Range for Yukawa Potential R~1/M x Exchanged particles are pions: R~1/(140 MeV)~1.4 fm Nuclei interact when their edges are ~ 1fm apart 0 th Order: Hard sphere Bradt & Peters formula b decreases with increasing A min J.P. Vary’s formula: Last term: curvature effects on nuclear surfaces 13

14 So: Bevalac Data Fixed Target Beam: ~few Gev/A AGS, SPS: works too Bonus question: Intercept: 7mb ½ What is r 0 ? Hints: 1 b = 100 fm 2, √0.1=0.316, √  =1.772 14

15 Vernier Scan Invented by S. van der Meer Sweep the beams across each other, monitor the counting rate Obtain a Gaussian curve, peak at smallest displacement Doing horizontal and vertical sweeps: zero-in on maximum rate at zero displacement Luminosity for two beams with Gaussian profile 1,2 : blue, yellow beam N i : number of particles per bunch Assumes all bunches have equal intensity Exponential: Applies when beams are displaced by d 15

16 van der Meer Scan. A. Drees et al., Conf.Proc. C030512 (2003) 1688 Cross Section: STAR: 16

17 World Data on pp total and elastic cross section PDG: http://pdg.arsip.lipi.go.id/2009/hadronic-xsections/hadron.html http://pdg.arsip.lipi.go.id/2009/hadronic-xsections/hadron.html RHIC, 200 GeV tot~50 mb el~8 mb nsd=42 mb LHC, 7 TeV tot=98.3±2.8 mb el=24.8±1.2 mb nsd=73.5 +1.8 – 1.3 mb (TOTEM, Europhys.Lett. 96 (2011) 21002 ) 17 CERN-HERA Parameterization

18 Froissart Bound, Phys. Rev. 123, 1053–1057 (1961) Marcel Froissart: Unitarity, Analiticity require the strong interaction cross sections to grow at most as for Particles and Antiparticles Cross sections converge for Simple relation between pion-nucleon and nucleon- nucleon cross sections 18

19 19

20 Charge densities: similar to a hard sphere. Edge is “fuzzy”. 20

21 Woods-Saxon density: R = 1.07 fm * A 1/3 a =0.54 fm A = 208 nucleons Probability : 21

22 Each nucleon is distributed with: Angular probabilities: Flat in , flat in cos(  ). 22

23 Like hitting a target: Rings have more area Area of ring of radius b, thickness db: Area proportional to probability 23

24 2 Nuclei colliding Red: nucleons from nucleus A Blue: nucleons from nucleus B 24 M.L.Miller, et al. Annu. Rev. Nucl. Part. Sci. 2007.57:205-243

25 After 100,000 events Beyond b~2R Nuclei miss each other Note fuzzy edge Largest probability: Collision at b~12-14 fm Head on collisions: b~0: Small probability 25

26 If two nucleons get closer than d<  they collide. Each colliding nucleon is a “participant” (Dark colors) Count number of binary collisions. Count number of participants 26

27 Nuclear Collisions 27

28 Multiplicty Distributions in STAR 28 MCBS, Ph.D Thesis Phys.Rev.Lett. 87 (2001) 112303

29 Each nucleon-nucleon collision produces particles. Particle production: negative binomial distribution. Particles can be measured: tracks, energy in a detector. CMS: Energy deposited by Hadrons in “Forward” region 29

30 From CMS MC Glauber model. CMS: HIN-10-001, JHEP 08 (2011) 141 30


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