In the rest frame of the spin-½ particle: spin up electron spin down electron ?? Is the E= mc 2...

1

0

0

0

,

0

1

0

0

,

0

0

1

0

,

0

0

0

1

In the rest frame of the spin-½ particle:

spin upelectron

spindown

electron

? ?

Is the E=mc2 unphysical? Meaningless?

Can we enforce B always be zero?

ue rpEti )(

2242 cpcmE )()( tEtEtE

1932 Carl Andersonpublisher’s thiscloud chamberphotograph.

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?????????

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! )

Additional comments on Matter/Antimatter Production

e+eParticles are created in pairs e+

e

and annihilate in pairs

Conserves CHARGE, SPIN(and other quantum numbers

yet to be discussed)

p+pp+p+p+p

Lab frame (fixed target) Center of Momentum frame

a b a c db

a b

at thresholdof production final state

total energyEalab Eb

lab=mc2

palab pb

lab=0

So conservation of energy argues: EaCOM+Eb

COM=4mc2

= 4mprotonc2

By conservation of energy: EaCOM+Eb

COM=4mc2

and

by the invariance of the inner produce of the 4-vector pp

(EaCOM+Eb

COM)2 (paCOM + pb

COM)2c2

=(Ealab+Eb

lab)2 (palab + pb

lab)2c2

paCOM + pb

COM = 0 mc2 0

( 4 mc2 )2 = m2c4 + 2Ealabmc2+ m2c4

16mc2 = 2m2c4 + 2Ealabmc2

Ealab = 7mc2 = 6.5679 GeV

(using mp=938.27231 MeV/c2)

(EaCOM+Eb

COM)2 = (Ealab+ mc2)2 (pa

lab)2c2

= (Ealab )22Ea

labmc2+ m2c4(palabc)2

= {m2c4+(palabc)22Ea

labmc2+ m2c4(palabc)2

BevatronBeam

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 steeringselects

1.19 GeV/c momentum negatively

charged particles

Čerenkov counters

thresholdsdistinguish > 0.75 > 0.79

scintillation countersmeasure particle“time of flight”

1.19 GeV/c s: 0.99c 40nsec Ks: 0.93c 43nsec

ps: 0.99c 51nsec

0.5 1.0

Ratio: m/mproton

Selecting events with TOF: 401 nsec

and 0.79<

0.148=m/mp

Selecting events with TOF: 511 nsecand 0.75<<0.79

2 3 4 5 6 7 8 GeV

Anti-proton production rate(per 105 ) vs beam energy

2.0

1.0

The Fermi energyof the confinedtarget protons

smears the turn-on curve.

4.0 5.0 6.0 7.0

Ant

i-pro

tons

per

105

- s

proton kinetic energy GeV

The Fermi energy of the confinedtarget protons smears the

turn-on curve.

0))((

mcimci

We factored the Klein-Gordon equation into

then found solutions for:

0)( mci

Free particle solution to Dirac’s equation

(x) = ue-ixp/h

u(p)

1

0

cpz

E+mc2

c(px+ipy)E+mc2

0

1

c(pxipy)E+mc2

cpz

Emc2

1

0

cpz

Emc2

c(px+ipy)

Emc2

1

0

c(pxipy)Emc2

cpz

Emc2

0)( mci

What if we tried to solve:

We would find 4 nearly identical Dirac spinors with the uA, uB (matter/antimatter entries) interchanged:

E+mc2 Emc2

0))((

mcimci

xax In general, any ROTATION or LORENTZ Transformation mixes vector components:

33221100 xaxaxaxaxa

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”

33221100 xxxxxxxxxxxx

xxaaxaxa ))(( aa

aa

Now watch this:

)(

1

1

aa

aa

aa

taa 1 The transformation matrices must be ORTHOGONAL!

axx ' 'xax tSo must mean

axx ' 'xax tSo must mean

xxaa

xaaxa

xaaxa

xax

'

)(')(

'

1

11

xxa '

xxa '

xax'

a

xx

x

xx

''chain rule (4 terms!)

xa

x

' x

ax

'

or

Finally