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Vxvx vyvy vzvz Classically, for free particles E = ½ mv 2 = ½ m(v x 2 + v y 2 + v z 2 ) Notice for any fixed E, m this defines a sphere of velocity points.

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Presentation on theme: "Vxvx vyvy vzvz Classically, for free particles E = ½ mv 2 = ½ m(v x 2 + v y 2 + v z 2 ) Notice for any fixed E, m this defines a sphere of velocity points."— Presentation transcript:

1 vxvx vyvy vzvz Classically, for free particles E = ½ mv 2 = ½ m(v x 2 + v y 2 + v z 2 ) Notice for any fixed E, m this defines a sphere of velocity points all which give the same kinetic energy. The number of “states” accessible by that energy are within the infinitesimal volume (a shell a thickness dv on that sphere). dV = 4  v 2 dv

2 Classically, for free particles E = ½ mv 2 = ½ m(v x 2 + v y 2 + v z 2 ) We just argued the number of accessible states (the “density of states”) is proportional to 4  v 2 dv dN dE  E 1/2

3 T  absolute zero The Maxwell-Boltzmann distribution describes not only how rms velocity increases with T but the spread about  in the distribution as well

4 With increasing energy (temperature) a wider range of velocities become probable. There are MORE combinations of molecules that can display this total E. That’s exactly what “density of states” describes. The absolute number is immaterial. It’s the slope with changing E, dN/dE that’s needed.

5 For a particle confined to a cubic box of volume L 3 the wave number vector is quantum mechanically constrained The continuum of a free particle’s momentum is realized by the limit L 

6 For a LARGE VOLUME these energy levels may be spaced very closely with many states corresponding to a small range in wave number The total number of states within a momentum range 

7 The total number of final states per unit volume or From E=p 2 /2m  dE=p dp/m  dE = v dp

8 number of final states per unit volume whose energies lie within the range from E to E+dE Giving (FINALLY) the transition rate of

9 Since transition rate = FLUX × cross-section / nv i Particle density of incoming beam but when dealing with one-on-one collisions n=1 If final states of target and projectile have non-zero spins or   rate/FLUX

10 Total cross sections for incident proton, antiproton, positive and negative pions, and positive and negative kaons on proton and neutron targets.

11 Breakdown of Rutherford Scattering formula When an incident  particle gets close enough to the target Pb nucleus so that they interact through the nuclear force (in addition to the Coulomb force that acts when they are further apart) the Rutherford formula no longer holds. The point at which this breakdown occurs gives a measure of the size of the nucleus. R.M.Eisberg and C.E.Porter Rev. Mod. Phys, 33, 190 (1961)

12 Total cross sections for  + p   + p and  + p   o  ++

13 The total  + p cross section

14 40 Ca 12 C 16 O Electron scattering off nuclei

15 Left: total cross sections for K + on protons (top) deuterons (bottom) Above: K-p total cross sections

16 Direct  Production in pp Interactions

17 Jet Production in pp and pp Interactions

18 Scattering Cross Sections ( light projectile off heavy target ) a + b  c + d counting number of final states counting number of initial states from  (E)=dN/dE “flux” Note: If a and b are unpolarized (randomly polarized) the experiment cannot distinguish between or separate out the contributions from different possibilities, but measures the scattered total of all possible spin combinations: averaging over all possible (equally probable) initial states

19 We will find that everything is in fact derivable from a comprehensive Lagrangian (better yet a L agrangian density, where. As a preview assume L describes some generic fields: f i, f j, f k wave functions…of matter fields …or interaction fields (potentials) Invariance principles (symmetries) will guarantee that schematically we will find 3 basic type of terms: 1mass terms: 2kinetic energy terms: 3interaction terms: mf i 2 or m 2 f i 2 FermionBoson  f i  f i or  f i   f i g fi fj fkg fi fj fk may also contain derivatives or >3 fields…or both

20 Cross sections or decay rates are theoretically computed/predicted from the Matrix Elements (transition probabilities) pick out the relevant terms in L can be expanded in a series of approximations [the coefficients of each term being powers of a coupling constant, g] Griffiths outlines the Feynmann rules that translate L terms into M factors mass & kinetic terms interaction terms propagator vertex g g

21 g Vertices get hooked together with propagators, with each vertex contributing one power of coupling to the calculation of the matrix element for the process. g

22 You know from Quantum Mechanics: Amplitudes are, in general, COMPLEX NUMBERS In e + e  collisions, can’t distinguish   e+e+ e+e+ e+e+ e+e+ e-e- e-e- e-e- e-e- All diagrams with the same initial and final states must be added, then squared. The cross terms introduced by squaring describe interference between the diagrams (sometimes suppressing rates!) +

23 d  = [flux] × | M | 2 ×  (E) ×  4 initial state properties statistical factor counting the number of ways final state produced will be a Lorentz invariant phase space dp 3 2E kinematic constraints on 4-momentum Matrix elements get squared Basic vertex of any interaction introduces a coupling factor, g g 2 is the minimum factor associated with any process…usually the expansion parameter (coefficient) of any series approximation for the matrix element e electromagnetic e 2  1 137 g weak g 2  G F  10 -5 GeV -2  W±W± gsgs strong g s 2   s  0.1 g

24 Remember that when we are dealing with “free” (unbounded) particles q q The matrix element M is the fourier transform of the potential! pipi pipi q = k i  k f =(p i -p f )/ħ

25 Proton-proton (strong interaction) cross sections The strong force has a very short effective range (unlike the coulomb force) If assume a simple “black disk” model with fixed geometric cross section: typical hadron size r p ~1fm = 10 -13 cm (1 barn=10 -24 cm 2 ) This ignores the dependence on E or resonances, but from 1 to several 1000 GeV of beam energy its approximately correct!

26 35-40 mb pp collisions Note: Elastic scattering 1/E dependence

27 Letting one cat out of the bag: Protons, anti-protons, neutrons are each composed of 3 quarks The (lighter) mesons (  +,  0,  -, K +, K 0, K -, …) … 2 quarks Might predict: ~38mb ~42mb and   ± p ~ 25mb  K ± p ~ 20mb

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29 n  p + e   e    e   e +  Ne *  Ne +  N  C + e   e Pu  U +  20 10 20 10 13 7 13 6 236 94 232 92 Fundamental particle decays Nuclear decays Some observed decays The transition rate, W (the “Golden Rule”) of initial  final is also invoked to understand a  b+c (+  ) decays How do you calculate an “overlap” between ???

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33 It almost seems a self-evident statement: Any decay that’s possible will happen! What makes it possible? What sort of conditions must be satisfied? Total charge q conserved. J conserved.

34 probability of surviving to time t mean lifetime  = 1/  For any free particle (separation of space-time components) Such an expression CANNOT describe an unstable particle since Instead mathematically introduce the exponential factor:

35 then a decaying probability of surviving Note:  =  ħ Also notice: effectively introduces an imaginary part to E

36 Applying a Fourier transform: still complex! What’s this represent? E distribution of the unstable state

37 Breit-Wigner Resonance Curve Expect some constant

38 EoEo EE 1.0 0.5   MAX  = FWHM When SPIN of the resonant state is included:

39 130-eV neutron resonances scattering from 59 Co Transmission  -ray yield for neutron radiative capture

40  + p elastic scattering cross-section in the region of the Δ ++ resonance. The central mass is 1232 MeV with a width  =120 MeV

41 Cross-section for the reaction e + e   anything near the Z 0 resonance plotted against cms energy

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43 In general: cross sections for free body decays (not resonances) are built exactly the same way as scattering cross sections. DECAYS (2-body example) (2-body) SCATTERING except for how the “flux” factor has to be defined in C.O.M. in Lab frame: enforces conservation of energy/momentum when integrating over final states Now the relativistic invariant phase space of both recoiling target and scattered projectile

44 Number scattered per unit time = (FLUX) × N ×  total (a rate) /cm 2 ·sec A concentration focused into a small spot and small time interval density of targets size of each target Notice:  is a function of flux!


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