16.451 Lecture 7: Understanding the cross-section results 25/9/2003 1

From last class, the cross section for electron scattering from nuclear charge Z:

2)2(

2

4

2)2(

2

4

4

22

4

2

222

22

22

2

)(

)(

4

)()(

4)(

qFo

e

qFo

e

q

cp

c

Z

q

p

c

Z

d

d

f

f

point charge cross-section:most notably, falls off as q-4

(units should be fm2)

form factor squared(dimensionless)

Check units: -1

fmMeVMeV.fm ][;][;]4[][ 2 qcpec o

24-

22

4fm

)()(

)(

1

fm

MeVMeV.fm

MeV.fm

d

d

2)2(22

)2(2

4)()(

)(

4)(4

22

4

2

qFZqF

o

e

o

f

d

d

q

cp

c

Z

d

d

What does this function look like?

First calculate the point charge cross section for Z = 1:

294

22

4)(1045.544.1

)5.197(

4

)(

4MeV.fm

42

o

e

c

2

point charge

constants:

For the kinematic factors, we need our results from lecture 5 for the momenta,being of course very careful with the units:

e

e

ip

fp

q

)cos1(1

M

pi

if

pp

cos2)()( 222fifi ppppq

Point charge cross section, continued...

First, let’s consider a heavy nuclear target, so that pf pi and work out Z2d/d)o

Choose the example of 248 MeV/c electrons scattering from 208Pb as in Krane, fig. 3.2

(deg)208Pb, pi = 248 MeV/c

2or

(GeV/c) GeV/c

pf pi

(ħq)2

2fm44.6

constant if pp

)cos1(2)( 22 i

pq

Kinematic relations become:

...)2/(sin2cos1

~

2

4

2

q

p

d

d f

)2/(sin

1~

.44

q

const

d

d

o

3

Point charge calculation for 248 MeV/c electrons scattering from 208Pb, Z = 82...

)2/(sin

0567.0.44

2 /sr2fm

q

const

d

dZ

o

(deg)

/sr2fm2

od

dZ

Log scale! The pointcharge cross section dropslike a stone!!!

NB, the graph blows up at = 0.In this limit, the practical cutoffis given by -2, a scale set by the size of the atom....

4

How are we doing? Compare to data from Krane:

= 10°, d/d = 105 b/sr = 10 fm2/sr ...

Point charge calculation:

d(10°)/d = 982 fm2/sr

(F(q2))2 at 10° =

10/982 = 0.01 !!!

222 )(qFd

dZ

d

d

o

Also, the graphs are not smooth like 1/q4 -- evidence that the target has finite size!

5

Summary so far:

We can predict the cross-section exactly for a pointlike target with nonrelativistic quantum mechanics.

(This approach is correct for a target particle that has charge but no magnetic moment, i.e. intrinsic angular momentum of zero. We can’t use this for the proton without adding some refinements, so along the way we are stopping to look at the charge distributions of nuclei. Nuclei with (Z, N) both even, such as 208Pb, have angular momentum zero, so our theory is perfect for this case!)

2

)2(sectioncrosschargepoint)(

qF

d

d

Recall our basic result from lecture 6:

Where F(q2) is the Fourier transform of the target charge density:

rdrrqieqF 32 )()(

Measuring d/d and dividing by the point charge result yields a value for F(q2)

6

How to find the charge distribution?

rdrrqieqF 32 )()(

form factor:

qdqFrqier 323

)()()2(

1

inverse Fourier transform:

In principle, one could measure the form factor, and numerically integrate toinvert the Fourier transform and find (r).

However, this doesn’t work in practice, because the integral has to be done overa complete range of q from 0 to ∞, and no experiment can ever span an infiniterange of momentum transfer!

(It is bad enough trying to acquire data at large momentum transfer because the basic cross-section drops like q-4 the rate of scattered particles into a detector gets too small – see slide 14, lecture 4)

What to do? ...

7

Experimental data are fitted to a functional form for F(q2); parameters extracted from the fit are used to invert the transform and deduce (r)....

A practical solution:

Example: elastic electron scatteringfrom gold (A=197, Z = 79). Best fit to data is given by solid curves.

(Ref: R. Hofstadter, Electron Scattering& Nuclear Structure, 1963)

Discontinuities are evidenceof diffraction-like behavior,characteristic of a Fouriertransform, but the edges arefuzzy!

8

Illustration of “Diffraction” behavior in F(q2): Uniform sphere

diqredrrrdrdrrqieqF

2

0 0 0

232 cossin)()()(

Rr

Rro

,0

;,

qRxx

xxxqF

32 )cos(sin3)(

Density:

Zeroes never quite get tozero on a log scale!

22 )(qF

In contrast, the minima are not as sharp for nuclei...

9

Nuclear charge distributions from experiment:

fm5.0

fm3/11.1

/)(1

:

a

AR

aRre

formFunctional

o

diffuse surface excludes sharp diffraction minimaApprox. constant

central density

10

Contrast to the proton (recall lecture 4):

2

GeV71.01)(

2

22

QQG p

E

Proton Form factor data:

fm80.012

fm33.4

)exp()(

2/12

1

Mr

M

rMer o

3fm

e

Electric charge distribution:

11