16.451 Lecture 5: Electron Scattering, continued... 18/9/2003

Details, Part III: Kinematics

e op

proton

e

p

q

Electrons are relativistic kinetic energy K ≠ p2/2m ...

Einstein mass-energy relations:

(total energy E, rest mass m)

KmcE

cpmcE

2)2

2222)1 )()(

Problem: units are awkward, too many factors of c ...

Notice that if c=1 then (E, m, p, K) all have the same units!

1

High Energy Units:

If we set c = 1 in Einstein’s mass-energy relations, then in order to “get the answerright”, the factor c has to be absorbed in the units of p and m:

KmE

pmE

)2

222)1

Let the symbol “[ ]” mean “the units of”, and then it follows that:

GeV/c2GeV/cGeV ][,][,][ pmE

(Frequently, physicists set c = 1 and quote mass and/or momentum in “GeV” units, asin the graph of the proton electric form factor, lecture 4. This is just a form ofshorthand – they really mean GeV/c for momentum and GeV/c2 for mass.... numericallythese have the same value because the value of c is in the unit – we don’t divide by the numerical value 3.00x108 m/s or the answer would be ridiculously small (wrong!)

2

From lecture 4: proton electric form factor data

Ref: Arnold et al., Phys. Rev. Lett. 57, 174 (1986)

4 – momentum transfer: Q2

(Inverse Fourier transform gives the electric charge density (r))

2a)

More units....

When we want to describe a scattering problem in quantum mechanics, we have towrite down wave functions to describe the initial and final states....

For example, the incoming electron is

a free particle of momentum:

The electron wave function is: where V is a normalization volume.

e op

proton

oo kp

rkieV

r o

1

)(

If we set ħ = 1, then momentum p and wave number k have the same units, e.g. fm-1;

to convert, use the factor: MeV.fm197c

Example:

An electron beam with total energy E = 5 GeV has momentum p = 5 GeV/c (m << E) ... the same momentum is equivalent to 5 GeV/(0.197 GeV.fm) or p = 25 fm-1.

So, p = 5 GeV, 5 GeV/c, and 25 fm-1 all refer to the same momentum!

3

Analysis: Kinematics of electron scattering

Note: Elastic scattering is the relevant case for our purposes here. This means that the beam interacts with the target proton with no internal energy transfer.

eoo pE

,

proton

e

pE ,'

qW,

• Specify total energy and momentum for the incoming and outgoing particles as shown.• Electron mass m << Eo. Proton mass is M.

Conserve total energy and momentum: qppWEME oo

','

Next steps: find the scattered electron momentum p’ in terms of the incidentmomentum and the scattering angle. Also, find the momentum transfer q2 as afunction of scattering angle, because q2 will turn out to be an important variablethat our analysis of the scattering depends on ...

4

Details ...

• conserve momentum:

cos'2)'( 222)1 ppppq oo

eoo pE

,

proton

e

pE ,'

qW,

• now use conservation of energy with W = M + K for the proton; E = p for electrons:

)'(' ppKKMEME oo

22222 2 KMKMqMW

• use kinematic relations to substitute for K = (po – p’) and q2:

)cos1(1')2

M

opop

pNote: these solutions 1), 2) are perfectlygeneral as long as the electron isrelativistic. The target can be anything!

)'( ppq o

5

Example: 5 GeV electron beam, proton target

e op

proton

e

p

q

(deg)

)(' GeV/cp

GeV/c5op

)cos1(1'

M

opop

p

Limits:

0º: p’ = po

180º: p’ po/(1+2po)

6

Relativistic 4 - momentum:

It is often convenient to use 4 – vector quantities to work out reaction kinematics.

There are several conventions in this business, all of them giving the same answerbut via a slightly different calculation. We will follow the treatment outlined inPerkins, `Introduction to High Energy Physics’, Addison-Wesley (3rd Ed., 1987).

Define the relativistic 4-momentum: 4...1),,( iEpp

The length of any 4 – vector is the same in all reference frames:

2222 mEpp

`length’ squared

For completeness, a Lorentz boostcorresponding to a relative velocity along the x-axis is accomplished by the 4x4 “rotation” matrix , with:

00

0100

0010

00

i

i

2/12 )1(,/v c

7

Analysis via 4 – momentum:

e

),(, ooo iEpP

proton

e

)','(' iEpP

),(, iWqPR

Define: 4 - momentum transfer Q between incoming and outgoing electrons:

)'(,)'(,')'( EEiqEEippPPQ oooo

Since Q is a 4 – vector, the square of its length is invariant:

222 )'()'( EEppQ oo

Expand, simplify, remembering to use p2 – E2 = - m2 and m << E,p ...

)cos1('22 ppQ o

8

Four-momentum versus 3-momentum, or Q2 versus q2 for e-p scattering

4–momentum transfer squared is the variable used to plot high energy electron scattering data:

)cos1('22 ppQ o

For a nonrelativistic quantum treatment of the scattering process (next topic!) the form factor is expressed in terms of the 3-momentum transfer (squared):

cos'2)'( 222 ppppq oo

(deg)p’

Q2

q2

GeV or GeV2

5 GeV beam

huge difference for the proton!

significant variationwith angle.

9

What happens if the target is a nucleus? (Krane, ch. 3)

The difference between numerical values of Q2 and q2 decreases as the mass of thetarget increases we can “get away” with a 3-momentum description (easier) toderive the cross section for scattering from a nucleus. Note also the simplificationthat p’ po and becomes essentially independent of as the mass of the targetincreases. (Why? as M ∞, the electron beam just ‘reflects’ off the target –just like an elastic collision of a ping pong ball with the floor – 16.105!!)

M = 16 M = 100

5 GeV electron beam in both cases, as before, but the target mass increasedfrom a proton to a nucleus (e.g., 16O, 100Ru )

p’ p’

q2

q2

Q2

Q2

10

Details, Part IV: electron scattering cross section and form factor

e op

proton

e

p

q

detector

Before After

Recall from lecture 4:

22 )()( qFd

d

d

d

o

point charge result (known) “Form factor” gives the Fouriertransform of the extended targetcharge distribution. Strictly correct for heavy nuclei: same idea but slightly more complicated expression for the proton...

Experimenters detect elastically scattered electrons and measure the cross section:

Last job: we want to work out an expression for the scattering crosssection to see how it relates to thestructure of the target object.

11

Scattering formalism: (nonrel. Q.M.)

Basic idea:

The scattering process involves a transition between an initial quantum state: |i = incoming e-, target p and a final state |f = scattered e-, recoil p .

The transition rate if can be calculated from “Fermi’s Golden Rule”, a basic prescription in quantum mechanics: (ch. 2)

Units: s-1

fifif M 22

where the `matrix element Mif’ is given by: rdrVM ifif3* )(

The potential V(r) represents the interaction responsible for the transition,in this case electromagnetism (Coulomb’s law!),

and the `density of states’ f is a measure of the number of equivalent final states per unit energy interval – the more states available at the same energy, the faster the transition occurs.

ff dEdn /

This formalism applies equally well to scattering and decay processes! We will also use it to analyze and decay later in the course...

e

e

12

Relation of transition rate if to d/d:

Recall:A beam particle will scatter from the target particle into solid angle d at (, ) if it approaches within the corresponding area d = (d/d) d centered on the target.

A

dtc

e

• Electron (speed c) is in a plane wave state normalized in volume V as shown.

• Probability of scattering at angle is given by the ratio of areas:

• Transition rate = (electrons/Volume) x (Volume/time) x P()

A

d

d

dP

)(

)(

d

d

d

V

cd

A

dd

dt

dtcA

Vif /1

Next time: we will put this all together and calculate the scattering cross section...

V = A c dt

13