EXAM I : Physics 926 February 14, 2004
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Transcript of EXAM I : Physics 926 February 14, 2004
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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
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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
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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
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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
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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
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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)
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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
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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
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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
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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
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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
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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
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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
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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?????????
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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! )
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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)
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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
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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
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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
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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
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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.
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0))((
mcimci
We factored the Klein-Gordon equation into
then found solutions for:
0)( mci
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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
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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