in the rest frame of the spin-½ particle:
Post on 20-Jan-2016
26 Views
Preview:
DESCRIPTION
TRANSCRIPT
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
top related