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In the rest frame of the spin-½ particle:

In the rest frame of the spin-½ particle:. spin down electron. ?. ?. spin up electron. Is the E = - mc 2 unphysical? Meaningless?. Can we enforce  B always be zero?. 1932 Carl Anderson publisher’s this cloud chamber photograph.

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In the rest frame of the spin-½ particle:

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  1. In the rest frame of the spin-½ particle: spin down electron ? ? spin up electron Is the E=-mc2 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 The particle’s slowing in its passage through lead foil establishes its direction ( UP!) Direction of curvature clearly indicates it is POSITVELY charged!

  4. g

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

  6. p+pp+p+p+p Center of Momentum frame Lab frame (fixed target) a c d b a b a b at threshold of production final state total energy EalabEblab=mc2 palab pblab=0 = 4mprotonc2 Soconservationofenergyargues:EaCOM+EbCOM=4mc2

  7. Byconservationofenergy:EaCOM+EbCOM=4mc2 and by the invariance of the inner produce of the4-vectorpmpm (EaCOM+EbCOM)2- (paCOM + pbCOM)2c2 =(Ealab+Eblab)2- (palab +pblab)2c2 paCOM + pbCOM = 0 0 mc2 ( 4 mc2 )2 = m2c4 + 2Ealabmc2+ m2c4 16mc2 = 2m2c4 + 2Ealabmc2 Ealab = 7mc2 = 6.5679 GeV (usingmp=938.27231 MeV/c2) (EaCOM+EbCOM)2= (Ealab+ mc2)2- (palab)2c2 = (Ealab )2+2Ealabmc2+ m2c4-(palabc)2 = {m2c4+(palabc)2}+2Ealabmc2+ m2c4-(palabc)2

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

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

  10. Anti-proton production rate (per 105p-) vs beam energy 2.0 1.0 2 3 4 5 6 7 8 GeV The Fermi energy of the confined target protons smears the turn-on curve.

  11. Anti-protons per 105-s 4.0 5.0 6.0 7.0 proton kinetic energy GeV The Fermi energy of the confined target protons smears the turn-on curve.

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

  13. Free particle solution to Dirac’s equation (x) = ue-ixp/h u(p) cpz E-mc2 c(px+ipy) E-mc2 c(px-ipy) E-mc2 -cpz E-mc2 1 0 0 1 c(px-ipy) E+mc2 -cpz E+mc2 cpz E+mc2 c(px+ipy) E+mc2 1 0 1 0

  14. What if we tried to solve: We would find 4 nearly identical Dirac spinors with the uA, uB (matter/antimatter entries) interchanged: E+mc2 E-mc2

  15. In general, anyROTATION or LORENTZ Transformation mixes vector components: space-time coordinates not the spinor components! amn= sin, cos, 1, 0 forR = , , 1, 0 for If we want to preserve “lengths” and “distances”          

  16. Now watch this: The transformation matrices must be ORTHOGONAL! So must mean

  17. So must mean

  18. Finally chain rule (4 terms!) or

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