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