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EXAM I : Physics 926 February 14, 2004

EXAM I : Physics 926 February 14, 2004. some useful information. dr 3 = dx dy dz = ( r sin  d )( r d ) dr = r 2 sin  d d dr. 1 Å = 10 -10 m. 1 barn = 10 -28 m 2. m e = 0.511003 MeV/c 2 = 9.10953 10 -31 kg m  = 105.6583 MeV/c 2 = 1.88355 10 -28 kg. 1 0 0 0

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EXAM I : Physics 926 February 14, 2004

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  1. EXAM I : Physics 926 February 14, 2004 some useful information dr3 = dx dy dz = (r sin d )(r d ) dr = r2sin d d dr 1 Å = 10-10 m 1 barn = 10-28 m2 me = 0.511003 MeV/c2 = 9.1095310-31 kg m = 105.6583 MeV/c2 = 1.8835510-28 kg

  2. 1 0 0 0 0 1 0 0 0 0 -1 0 00 0 -1 g0 = 0 si -si 0 I 0 0-I g0 = gi = 0 0 0 1 0 0 1 0 0 -1 0 0 -1 0 0 0 g1 = The block diagonal form suggests it may sometimes be simpler to work with the “reduced” notation of A 0 0 0 -i 0 0 +i 0 0 +i 0 0 -i 0 0 0  g2 = = B 1 3 0 0 1 0 0 0 0 -1 -10 0 0 0 1 0 0 A B g3 = = = where 2 4

  3. 1 2 3 4 1 2 3 4 1 0 0 0 0 1 0 0 0 0 -1 0 00 0 -1 0 0 0 1 0 0 1 0 0 -1 0 0 -1 0 0 0 1 2 3 4 1 2 3 4 1 2 3 4 0 0 0 -i 0 0 +i 0 0 +i 0 0 -i 0 0 0 0 0 1 0 0 0 0 -1 -10 0 0 0 1 0 0 A A A 0 si -si 0 I 0 0-I B B B

  4. i ħ E c i ħ i ħ i ħ Recall that a “free particle” has, in general, a PLANE WAVE solution: e-Ete-p·r evolution of the time-dependent part solution to the space-dependent part e-[(ct)-pjrj ] = e-pmxm

  5. Assume a form (free particle of4-momentum pm) u a 4-component “Dirac spinor” carrying any needed normalization factors If we can find u’s that satisfy this, then the  above will be a solution to the Dirac equation

  6. Now, note that So our equation looks like: a two component vector (of 2 component vectors)

  7. and

  8. So returning to: we must have and ? which, notice together give:

  9. and We can start picking uA’s and solve for uB’s and/or uB’s and solve for uA’s We need 4 linearly independent solutions, right? An obvious starting point: What’s wrong with the simpler basis:

  10. What do these components mean? Let’s look at them in the limit where p  0

  11. in the limit where p  0 for Emc2 for E-mc2

  12. These ARE eigenvectors of with “spin” +1/2 -1/2 +1/2 -1/2

  13. 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?

  14. 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 ?????????

  15. 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!

  16. g

  17. 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)

  18. 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

  19. 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 16m2c4 = 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

  20. 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:0.78c51nsec 1955 - Chamberlain, Segre, Wiegrand, Ypsilantis

  21. 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

  22. 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.

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

  24. 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

  25. 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

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