EXAM I : Physics 926February 14, 2004
1 Å = 10-10 m
33
)(2
1)( drerfkF rki
)()()(1
)( ooo2o rrr
rr
)...()2(2)2(2)2()()(4
M 4144
43
43
33
33
2221
221
22
ppE
pcd
E
pcd
cmmpp
Sd
!21
!
)1(lim
)()()(
2
0
xx
n
xe
eN
x
x
rfxrfrrf
n
nx
xNN
ii
some useful information
dr3 = dx dy dz = (r sin d )(r d ) dr = r2 sin d d dr
1 barn = 10-28 m2
me = 0.511003 MeV/c2 = 9.1095310-31 kg
m = 105.6583 MeV/c2 = 1.8835510-28 kg
f
f
iNi
v
pEVE
v 3
22
2
4||
12
2
2
2
2
42
1
2ln
4/
I
mv
A
Z
mv
ezNdxdE A
0 =I 0
0 -Ii =
0 i
-i 0
0 =
1 0 0 00 1 0 00 0 -1 00 0 0 -1
1 =
0 0 0 1 0 0 1 0 0 -1 0 0-1 0 0 0
2 =
0 0 0 -i 0 0 +i 0 0 +i 0 0-i 0 0 0
3 =
0 0 1 0 0 0 0 -1-1 0 0 0 0 1 0 0
The block diagonal form suggests it may sometimes be simpler to
work with the “reduced” notation of
=A B
where A =1 2
B =3 4
0)( mci
1 0 0 00 1 0 00 0 -1 00 0 0 -1
itc 1
1
2
3
4
i 0 0 0 1 0 0 1 0 0 -1 0 0-1 0 0 0 x
1
2
3
4
iy
1
2
3
4
0 0 0 -i 0 0 +i 0 0 +i 0 0-i 0 0 0
iy
1
2
3
4
0 0 1 0 0 0 0 -1-1 0 0 0 0 1 0 0
0 mc
itc 1I 0
0 -I B A
i ix 0 i
-i 0 B A
1
2
3
4
0 mcB A
Recall that a “free particle” has, in general, a PLANE WAVE solution:
e Et e p·r iħ
iħ
evolution of the time-dependent part
solution to the space-dependent part
e [ (ct) pjrj ]
iħ
Ec = e px
iħ
0 mcp 0)(
mci
Assume a form (free particle of 4-momentum p)
pxi
ex
)( ua 4-component “Dirac spinor”
carrying any needed normalization factors
0//
umceueipixpix
0)( umcp
If we can find u’s that satisfy this, then the above will be a solution to the Dirac equation
Now, note thatppp
00
0
0
0
0
p
I
I
c
E
cEp
pcE
So our equation looks like:
u)( cmp I 0
B
A
uu
mccEp
pmccE
0)(
)(
BcE
A
BcE
A
umcpu
pumcu
a two component vector(of 2 component vectors)
10
01
0
0
01
10zyx p
i
ippp
zyx
yxz
pipp
ippp
2)( p
and
222
2222
0
0
zyx
yxz
pipp
pipp
I22
2
0
0p
p
p
0
0
)(
)(
BcE
A
BcE
A
umcpu
pumcu
So returning to:
we must have
BA umcE
pcu
2)(
AAu
cmE
pcu 422
22 )(
AB umcE
pcu
2)(
and
which, notice together give:
422
22
1cmE
pc
22422 cpcmE
BA umcE
pcu
2)(
AB umcE
pcu
2)(
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?
What’s wrong with the simpler basis:
1
0
0
0
,
0
1
0
0
,
0
0
1
0
,
0
0
0
1
An obvious starting point:
1
0,
0
1,
1
0
,0
1
zyx
yxz
pipp
ipppp
BA umcE
pcu
2)(
AB umcE
pcu
2)(
1
0,
0
1,
1
0
,0
1
2
2
2
2
2
2
2
2
)(
)()(
)(
mcE
cpmcE
ippc
mcE
ippcmcE
cp
mcE
cpmcE
ippc
mcE
ippcmcE
cpz
yx
yx
z
z
yx
yx
z
What do these components mean?
Let’s look at them in the limit where p 0
1
0
0
0
,
0
1
0
0
,
0
0
1
0
,
0
0
0
1
1
0,
0
1,
1
0
,0
1
2
2
2
2
2
2
2
2
)(
)()(
)(
mcE
cpmcE
ippc
mcE
ippcmcE
cp
mcE
cpmcE
ippc
mcE
ippcmcE
cpz
yx
yx
z
z
yx
yx
z
in the limit where p 0
for Emc2 for E-mc2
1
0
0
0
,
0
1
0
0
,
0
0
1
0
,
0
0
0
1
These ARE eigenvectors of
3
3
0
0
with “spin”
+1/2 1/2 +1/2 1/2
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
16m2c4 = 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: 0.78c 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
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
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