e -

1. Two important BASIC CONCEPTS • The “coupling” of a fermion • (fundamental constituent of matter) • to a vector boson • (the carrier or intermediary of interactions) e- • Recognized symmetries • are intimately related to CONSERVED quantities in nature • which fix the QUANTUM numbers describing quantum states • and help us characterize the basic, fundamental interactions • between particles

2. Should the selected orientation of the x-axis matter? As far as the form of the equations of motion? (all derivable from a Lagrangian) As far as the predictions those equations make? Any calculable quantities/outcome/results? Should the selected position of the coordinate origin matter? If it “doesn’t matter” then we have a symmetry: the x-axis can be rotated through any direction of 3-dimensional space or slid around to any arbitrary location and the basic form of the equations…and, more importantly, all the predictions of those equations are unaffected.

3. If a coordinate axis’ orientation or origin’s exact location “doesn’t matter” then it shouldn’t appear explicitly in the Lagrangian! EXAMPLE:TRANSLATION Moving every position (vector) in space by a fixed a (equivalent to “dropping the origin back” –a) a original description of position r r' or new description of position under the newly shifted basis qi

4. For a system of particles: function of separation acted on only by CENTAL FORCES: no forces external to the system generalized momentum (for a system of particles, this is just the ordinary momentum) for a system of particles T may depend on q or r but never explicitly on qi or ri =

5. For a system of particles acted on only by CENTAL FORCES: ^ ai -Fi net force on a system experiencing only internal forces guaranteed by the 3rd Law to be 0 Momentum must be conserved along any direction the Lagrangian is invariant to translations in.

6. Particle properties/characteristics specifically their interactions are often interpreted in terms of CROSS SECTIONS.

7. For elastically scattered projectiles: The recoiling particles are identical to the incoming particles but are in different quantum states Ef , pf Ei , pi  EN , pN The initial conditions may be precisely knowable only classically! The simple 2-body kinematics of scattering fixes the energy of particles scattered through .

8. Nuclear Reactions Besides his famous scattering of  particles off gold and lead foil, Rutherford observed the transmutation: or, if you prefer Whenever energetic particles (from a nuclear reactor or an accelerator) irradiate matter there is the possibility of a nuclear reaction

9. Classification of Nuclear Reactions • inelastic scattering • individual collisions between the incoming projectile and a single target nucleon; the incident particle emerges with reduced energy • pickup reactions • incident projectile collects additional nucleons from the target • O + d  O + H (d, 3H) • Ca + He Ca +  (3He,) 16 8 3 1 15 8 41 20 3 2 40 20 • stripping reactions • incident projectile leaves one or more nucleons behind in the target Zr + d  Zr + p 90 40 (d,p) (3He,d) 91 40 Na + He Mg + d 23 11 3 2 24 12

10. [ Ne]* 20 10

11. The cross section is defined by the ratio rate particles are scattered out of beam rate of particles focused onto target material/unit area a “counting” experiment notice it yields a measure, in units of area number of scattered particles/sec incident particles/(unit area  sec)  target site density how tightly focused or intense the beam is density of nuclear targets With a detector fixed to record data from a particular location ,  we measure the “differential” cross section: d/d.

12. scattered particles Incident mono-energetic beam v Dt A dW N = number density in beam (particles per unit volume) Solid angle dWrepresents detector counting the dN particles per unit time that scatter through qinto dW Nnumber of scattering centers in target intercepted by beamspot FLUX = # of particles crossing through unit cross section per sec = NvDt A / Dt A = Nv Notice: qNv we call current, I, measured in Coulombs. dN NF dW dN = s(q)NF dW dN =NFds -

13. dN = FNs(q)dWNFds(q) the “differential” cross section R  R R R R

14. R R the differential solid angle d for integration is sin d d Rsind Rd Rsind Rd Rsin

15. Symmetry arguments allow us to immediately integrate  out and consider rings defined by  alone R Rsind  R R R Nscattered= NFsTOTAL Integrated over all solid angles

16. Nscattered= NFsTOTAL The scattering rate per unit time Particles IN (per unit time) = FArea(ofbeamspot) Particles scattered OUT (per unit time) = F NsTOTAL

17. Earth Moon

18. Earth Moon

19. for some sense of spacing consider the ratio orbital diameters central body diameter ~ 10s for moons/planets ~100s for planets orbiting sun • In a solid • interatomic spacing:1-5 Å (1-5  10-10 m) • nuclear radii: 1.5 -5 f(1.5-5  10-15 m) the ratio orbital diameters central body diameter ~ 66,666 for atomic electron orbitals to their own nucleus Carbon 6C Oxygen 8O Aluminum 13Al Iron 26Fe Copper 29Cu Lead 82Pb

20. A solid sheet of lead offers how much of a (cross sectional) physical target (and how much empty space) to a subatomic projectile? 82Pb207 w Number density,n: number of individual atoms (or scattering centers!) per unit volume n= rNA / A where NA = Avogadro’s Number A = atomic weight (g) r = density (g/cc) n= (11.3 g/cc)(6.021023/mole)/(207.2 g/mole) = 3.28  1022/cm3

21. 82Pb207 w For a thin enough layer n(Volume)  (atomic cross section) = n(surface areaw)(pr2) as a fraction of the target’s area: = n(w)p(5  10-13cm)2 For 1 mm sheet of lead: 0.00257 1 cm sheet of lead: 0.0257

22. Actually a projectile “sees” nw nuclei per unit area but Znw electrons per unit area!

24. that general description of cross section let’s augmented with the specific example of Coulomb scattering

25. BOTH target and projectile will move in response to the forces between them. q1   q2  Recoil of target But here we are interested only in the scattered projectile q1

27. impact parameter, b

28.  b q2 d A beam of N incident particles strike a (thin foil) target. The beam spot (cross section of the beam) illuminates n scattering centers. If dN counts the average number of particles scattered between and d dN/N = n d d = 2 b db using becomes:

29.  b q2 d and so

30.  b q2 d