1. In the rest frame of the spin-½ particle: spin up electron spin down electron ?? Is the E=  mc 2 unphysical? Meaningless? Can we enforce  B always be zero?

2. 1932 Carl Anderson publisher’s this cloud chamber photograph. Droplet density (thickness) of track identifies it as that of an electron ????????? Curvature of track confirms the charge to mass ratio (q/m) is that of an electron ?????????

3. B-field into page Direction of curvature clearly indicates it is POSITVELY charged! The particle’s slowing in its passage through lead foil establishes its direction ( UP! )

4. 

6. Additional comments on Matter/Antimatter Production e+ee+e Particles are created in pairse+e+ ee and annihilate in pairs Conserves CHARGE, SPIN (and other quantum numbers yet to be discussed)

7. p+p  p+p+p+p Lab frame (fixed target) Center of Momentum frame a b a c d b a b at threshold of production final state total energy E a lab E b lab =mc 2 p a lab p b lab =0 So conservation of energy argues: E a COM +E b COM =4mc 2 = 4m proton c 2

8. By conservation of energy: E a COM + E b COM =4mc 2 and by the invariance of the inner produce of the 4-vector p  p  (E a COM +E b COM ) 2  (p a COM + p b COM ) 2 c 2 =(E a lab +E b lab ) 2  (p a lab + p b lab ) 2 c 2 p a COM + p b COM = 0 mc2mc2 0 ( 4 mc 2 ) 2 = m 2 c 4 + 2E a lab mc 2 + m 2 c 4 16mc 2 = 2m 2 c 4 + 2E a lab mc 2 E a lab = 7mc 2 = 6.5679 GeV (using m p =938.27231 MeV/c 2 ) (E a COM +E b COM ) 2 = (E a lab + mc 2 ) 2  (p a lab ) 2 c 2 = (E a lab ) 2  2E a lab mc 2 + m 2 c 4  (p a lab c) 2 = {m 2 c 4 +(p a lab c) 2  2E a lab mc 2 + m 2 c 4  (p a lab c) 2

9. Bevatron Beam Carbon Target M1 Q1 Shielding S1 Q2 M2 C1 C2 C3 S2 S3 1955 - Chamberlain, Segre, Wiegrand, Ypsilantis Berkeley BEVATRON accelerating protons up to 6.3 GeV/c 10 ft magnetic steering selects 1.19 GeV/c momentum negatively charged particles Čerenkov counters thresholds distinguish  > 0.75  > 0.79 scintillation counters measure particle “time of flight” 1.19 GeV/c  s: 0.99c40nsec K s: 0.93c43nsec p s: 0.99c51nsec

10. 0.5 1.0 Ratio: m/m proton Selecting events with TOF: 40  1 nsec and 0.79<  0.148=m  /m p Selecting events with TOF: 51  1 nsec and 0.75<  <0.79

11. 2345678 GeV Anti-proton production rate (per 10 5   ) vs beam energy 2.0 1.0 The Fermi energy of the confined target protons smears the turn-on curve.

12. 4.05.06.07.0 Anti-protons per 10 5  - s proton kinetic energy GeV The Fermi energy of the confined target protons smears the turn-on curve.

13. We factored the Klein-Gordon equation into then found solutions for:

14. Free particle solution to Dirac’s equation  (x) = u e -ix  p  /h u(p)u(p) 1 01 0 cp z E+mc 2 c(p x +ip y ) E+mc 2 0 10 1 c(p x  ip y ) E+mc 2  cp z E  mc 2 1 01 0 cp z E  mc 2 c(p x +ip y ) E  mc 2 1 01 0 c(p x  ip y ) E  mc 2  cp z E  mc 2

15. What if we tried to solve: We would find 4 nearly identical Dirac spinors with the u A, u B (matter/antimatter entries) interchanged: E+mc 2  E  mc 2

16. In general, any ROTATION or LORENTZ Transformation mixes vector components: space-time coordinates not the spinor components! a  = sin , cos , 1, 0 for R = , , 1, 0 for  If we want to preserve “lengths” and “distances” 

17. Now watch this: The transformation matrices must be ORTHOGONAL! Somust mean

18. Somust mean

19. chain rule (4 terms!) or Finally