Why structure Functions and Parton densities are Important?

Nucleon in a collider is a beam of partons. We use them to search for new physics. Take the LHC:

LHC is a “gluon collider”

Knowledge about the gluon content of proton at low x is essential, e.g. for Higgs, W/Z production

Ht

p

p( )2

2 2,x qg

( )2

1 1,x qg

g g H+ →

( )1

1

0

1 2

1

20( ) ( ) ( )g x g xpp H X dx dx gg Hσ σ→ + →∫ ∫∼

Good knowledge of PDF allow us to predict cross sections at coliders.

Deep Inelastic Scattering reveals internal structure of hadrons (mainly nucleon) and gives insight into the structure of other particles.

It shows that nucleon is composed of quarks (and gluons). It hasdetermined that quarks are spin ½ particles and attempts to determine their distributions, which turns out to be so important in search and discovery of new physics.

you may all have heard of all these. In these lectures I intend to shed some light on these topics and refresh your memory. In the process, I will also show some details.

Outline of the talk

Quick introduction to Form Factors: The nucleus

Elastic e-p Scattering and Nucleon form factors

Deep Inelastic Scattering (DIS) – mainly e-p, but polarization is also included

Parton Model, Scaling, scaling violation and QCD improved P.M.

Structure functions, parton distributions, and extraction

Polarized structure functions

DGLAP evolution equation

Looking inside a Hadron

Hadrons (mesons and baryons) are bound state of quarks. Like any composi system their spatial extend is complicated

:Leptons are the best probes: they don’t have structure

To “see” inside of them you need a probe. Light (photon) is a familiar tool to see things. But ordinary light is no good. Its wavelength is way too big. Need to make a magnifying glass with very high resolution.

Where to get it? Use electron, and neutrino

Begin with E&M Probes:γ ∗

Photon has a 4-momentum and an invariant massqμ22 2 0q q q qμ

μ= = − ≠q

i f= −q p pWe can change the invariant mass of photon by changing its energy, or scattering angle.

Acts as magnification of the magnifying glass1 λq ∼Increasing decreases the wavelength, e.g:q

I. Nucleus Form Factor

0

14p

ZeVπε

=r

( )3

0

14

V d rρ

πε′

=′∫r

r - r

These potentials are in position space. Lets go to the momentum space by a Fourier Transform, see what happens

Suppose 920KeV 10 cmλ −= ≈q ∼ , nucleous appears as a point chargeze Nuclear size: 1210 cm−

YOU (PHOTON) gets to see protons inside the nucleus

Nucleus appears as a charge distribution ( )ρ ′rIncrease to q 1220MeV 10 cmλ −= ≈q ∼

( ) ( ) ( )3 3 31 1;4 4

i ip

ZeV d r e V d r e d rr

ρπε πε

− ⋅ − ⋅ ⎛ ⎞′′= = ⎜ ⎟⎜ ⎟′⎝ ⎠

∫ ∫ ∫q r q r rq q

r - rI want to show that these two are related to each other by a measurable quantity, ( )2

NF q The Form Factor (FF)The Form Factor (FF)

( )

( ) ( ) ( )

3 3 2

1 cos2221 2

1 lim lim sin4 4 4

4 42 lim cos lim4 4 4

r ri i i

p

rr

Ze Ze e Ze eV d r e d r e r drd d er r r

Ze e Ze Zer dr e dr

μ μ

μ μ

μθ

μ μ

θ θ φπε πε πε

π ππ θπε πε πεμ

− −− ⋅ − ⋅ − ⋅

→ →

−∞ −

−→ →

= = =

= = =+

∫ ∫ ∫

∫ ∫

q r q r q r

q

q

qq

Let us evaluate these potentials:

( )V qAnd for we get:

qEverything is in terms of 3-vectorEverything is in terms of 3-vector

Not in C

ovariant

form either

Not in C

ovariant

form either

( ) ( )

( ) ( )

( )( )

1 3 34

1 1 13

2

3 224

iV d r e d r

i

F q

e iV d R d r e F qN

N

ρπε

ρπε ε

⎛ ⎞′− ⋅ ′= ⇒⎜ ⎟∫ ∫⎜ ⎟′⎝ ⎠− ′− ⋅′ ′⇒ = =∫ ∫

≡

rq rqr - r

q.R q rq rR q

( ) ( )22 3 iNF q d r e ρ′− ⋅′ ′= − ≡ ∫ q rq r

Is the Fourier Transform of charged proton distribution, in nucleus( )ρ ′r

Obviously, and that is the Normalization we take.( )0NF Ze=

( ) ( ) ( )( )

( )2 22 2

2N N

PN

F q F qV q V q

F q= = −⇒

( )22p

ZeV qqε

= −Compare With FF is equal to the Fourier transform of potential due to charge Ze

2 2 211 cos2

ie i r θ′− ⋅ ′= − ⋅ − +q r q r qUse the expansion

And integrate over we get3d r′

( ) ( ) ( ) ( )2 22 2 3 21 12 61 cos 1NF i r r d r F rθ ρ ⎡ ⎤′ ′ ′= − ⋅ − + = − +⎣ ⎦∫q q r q q

( ) ( )2

22 2 3

2

0

6 N

q

dF qr r r d r

dqρ

=

≡ =∫Integration of linear term vanishes, by spatial symmetry

The derivative of FF gives the mean size of the nucleus.The greater the object is, the faster its form factor decreases.For Point objects, FF is INDEPENDENT of 2q• FF can be measured in Rutherford or Mott scattering.

( )( )

2 2 2 22

2 4

1 422

ifi

d m m z mV d V edσ α

ν ππ⋅= = =

Ω ∫ q rp r rq For Point charge

( ) ( ) ( ) ( )V V d V ρ′ ′ ′ ′⇒ → = −∫r r r r r rIf Charge is spread:

And the matrix element:

23 23

12

iNd qe F qρ

π

⎛ ⎞⎜ ⎟⎛ ⎞

⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎜ ⎟⎛ ⎞ ⎝ ⎠⎜ ⎟⎝ ⎠

+ ⋅ ′= =−′ ∫ qrr q

• Inverse Fourier Transform of FF gives the charge (proton) distribution of The nucleus:

( ) ( ) ( ) ( ) ( ) 2( )ii ifi fi fiV V d V e d V e d e V Fρ− ′⎡ ⎤

⎢ ⎥⎣ ⎦→ = = − − × =′ ′ ′ ′ ′ ′∫ ∫ ∫

q. r rq.r q.rr r r r r r r r q

So, the cross- section for an extended object is:

( ) ( )2

2ptd d Fσ θ σ= q

So , lets make it covariant, using Lorentz invarianceSo , lets make it covariant, using Lorentz invariance22 2q q= − q

In Fourier transform of static coulomb potential, energy does not get transformed , only momentum is transformed , so and we can replace In Fourier transform of static coulomb potential, energy does not get transformed , only momentum is transformed , so and we can replace

22q = − q 2 2q→ −q

So , we get for the point charge potentialSo , we get for the point charge potential

( )22p

ZeV qqε

= −

♣ The point is : NONRELATIVISTIC LIMIT OF is the propagator of massless boson Interchanged between electron and the nucleus.

Exchange of a meson with non-relativistic propagator is the source of interaction potential between the two particles.

♣ The point is : NONRELATIVISTIC LIMIT OF is the propagator of massless boson Interchanged between electron and the nucleus.

Exchange of a meson with non-relativistic propagator is the source of interaction potential between the two particles.

21 q 12q

−

Not in Covariant formNot in Covariant form1 1 2

2V F qNε⎛ ⎞

⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎜ ⎟⎝ ⎠

=qq

Static, No recoil

II. The NucleonWe can also use the concept of FF for the nucleon and learn that nucleon is an extended object, with structure like proton.

Anomalous Magnetic Moment of nucleon is another implication that nucleon has an internal structure. (let us see this)

Take a point like fermion current , interacting with electromagnetic field A μ( ) ( )u p u pμγ′

( ) ( ) ( ) ( ) ( ) ( )e u p p pu p i p p u p Ap Au mν μ

μνμμ

μ σγ ⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦

= + + −′ ′ ′′ 2 Gordon Decomposition

Non-relativistic limit: second term reduces to:

( ) ( ) ( )e m u p u p⋅′ σ B2: External Field2:proton's Spin

Bσ

Proof: Non-Relativistic Limit means

( ) 2 , iij jp p A m i q A i q Aμ μ

μν μμφ σ σ′⇒ + → → ⋅∇× = ⋅σ A σ B∼

( ),p p Mμ μ′= = 0

This shows that the magnetic moment of a point particle of charge e, mass m, is one Bohr Magneton: 2B e mμ ⎛ ⎞⎜ ⎟

⎝ ⎠≡

Experimentally: Proton and neutron Magnetic moments are: 2.79 , 1.91p B n Bμ μ μ μ= = −

electron 0.0016μΔ =CONCLUSION: Nucleon is not point fermion, they have internal structure

Large magnetic moment Quarks!

II. The Nucleon: Electromagnetic form factore

Can’t use the same argument to tackle the Nucleon FF as we did for Nucleus. Because

1. It is not static charge distribution. Upon collision with electron, the nucleon will recoil.

2. In addition to electric charge, it also posses magnetic moment which contributes to electromagnetic interaction.

22 2 2322 4 11

lab

cos 2 sin 22cos 2 sin 2 lab labplab

E qdE md E

σ α π θ θθ θ

⎛ ⎞⎜ ⎟ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

= −

If Proton had no structure and no recoil, i.e. it could be considered as a heavy point particle. we know the e- p cross section: Mott scattering.

If proton is considered as a point particle of charge e and magnetic moment

still we know the cross- section: replace muon mass with proton mass in

2 pe m

e eμ μ→

Mag. Moment contribution( )1

23

1

21 sin 2p

EE m θ

−⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

= +Effect of recoil

If muon was spinless (no mag. Moment) this term would not be there

This is an important formula for our story

1kμγ

2k

eJ μ

ap bp

pJ μ

42

1e pfi qT i j J d xμ

μ⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

= − −∫Transition amplitude

2 12 1

i k k xej eu k u k eμμ γ

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

− ⋅=−

But Cannot use for proton vertexYet, initial and final proton can be treated ad Dirac particle

a bq p p= −But No It is not independentSo, only ( ) ( )and, a ab bp p p pμγ − +

Since proton is not point charge, we need to introduce a Form factor for charge distribution, and one more to describe the distribution of the magnetic moment.

Two Form factors. What are they? Or what is the most general form of pJ μ

Well, pJ μ is a Lorentz 4- vector, and can use the general considerations to construct itfrom 4-vectors at the proton vertex: ap , ,bp q And Dirac γ Matrices

22 2 2322 4 11

lab

cos 2 sin 22cos 2 sin 2 lab lab

plab

E qdE md E

σ α π θ θθ θ

⎛ ⎞⎜ ⎟ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎛ ⎞ ⎛ ⎞ ⎝ ⎠ ⎝ ⎠⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎜ ⎟⎝ ⎠⎝ ⎠

= −

Due to coulomb scattering Due to magnetic moment

Recall: for scatt. of electron from a point charge:

μγ

( )51 2 3 4p a a a a ab b b b bJ eu p k i p p k i p p k p p k p p k u pμ μμ μ μν μν

ν νγ σ σ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦= + − + + + − + +

The most general 4-vector

The most general 4-vector Use Gordon decomposition and simplify itμγWrite out abp p

μ⎛ ⎞⎜ ⎟⎝ ⎠

+ andIn terms of abp pμνν

σ ⎛ ⎞⎜ ⎟⎝ ⎠

−

( ) ( ) ( )1 12 2a a a a ab b b b bu p p p u p u p u p u p i p p u pm m

μ μ μνν

γ σ⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦+ = − −

And plug into

( )( ) ( )

( ) ( ) ( )

( ) ( ) ( ) ( )( ) ( )

5

1 2 3

5 54

5 51 2

2 4

2

2

2

12

ab

a ab b

p ab a ab b

p p k

abb

ab

k i p p k i m i p p k

J eu p u pp p k m k i p p k

k m k k k i p peu p u p

p p k I g k

μ

μ μν μν μ μνν ν

μμ μ μνν

μ μνν

μμν

γ σ σ γ σ

γ σ

γ σ

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

+

⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

+ − + − −

= + − + − −

+ + − −=

+ − +( )a

Note 12Iμνμνσ σ =

Note: no 5γ Parity conservation

they can be written in terms of 2qanda bp q p q⋅ ⋅Note: Are also scalars, but are not independent

Functions of scalar variables, and we got only one: 2q'sik

2 2 21 2 32p ab p

iJ eu p F q F q q q F q u pmμ μ μν μ

νκγ σ

⎡ ⎤⎢ ⎥⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

= + +

Collect and rearrange. We get

Use proton current conservation 0p pJ q Jμ μμ μ∂ = =

2 2 21 2 32p ab p

iq J eu p q F q F q q q q q F q u pmμ μ μν μ

μ μ μ μνκγ σ

⎡ ⎤⎢ ⎥⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

= + + =

=0, by Dirac Eq. =0, since μνσ is anti-symmetric

The third term 32 2

3 since 00 0 Fq F q m =⇒= ≠ ≠ =

We are left with two FFs

2 21 22p ab p

iJ eu p F q F q q u pmμ μ μν

νκγ σ

⎡ ⎤⎢ ⎥⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

= +

When 2 0q → That is photon wavelenght larger than a Fermi, it sees proton as a point charge, and its crrent also reduces to the current of a point charge. This requires that we set

( ) ( )2 21 20 0 1F q F q= = = =

22 1 1 2 22 2 p ai f b

e u k u k u p F F K m F u pq

μ μμγ γ

⎧ ⎫⎡ ⎤ ⎡ ⎤⎪ ⎪⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎢ ⎥ ⎢ ⎥ ⎛ ⎞⎪ ⎪⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥ ⎜ ⎟⎨ ⎬⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎪ ⎪⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥⎪ ⎪⎣ ⎦⎣ ⎦

⎩ ⎭

→ = + −′ ′

Now we can write the transition amplitude

,2 2F Fκ≡′Where, a bK p p≡ +

And calculate the cross section. Summing over final state spins and averaging over the initial state particles spin, redefining some symbols, we arrive at 22 2

2spins14

pei f i f

e Lq

μνμν

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

∗→ →= =∑

Evaluating Traces, using linear combinations of 1 2andF F

( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )

( )( )

22 2 2 2 2 2

1 2 1 22

2 2 2 222 2

1 22 2 2

,4

0 1, 0 0, 0 2.97, 0 1.91

44 1 4

E M

p n p nE E M M

E M

qG q F q F q G q F q F qM

G G G G

G q M GqF FM q M

= + = +

= = = = −

−− =

−

The Coefficient 2tan 2θ is due to magnetic moment and reflects the spin of the proton.If The target was spinless, like a pion, this term would not be there.

2 2 2 22 2 2Mott

22lab

2 2 2 22 2 2

2 tan 211 sin 2

2 tan 21

E MME

M

E MM

lab NS

G q G qd dd G qd

G q G qd G qd

τσσ τ θτθ

τσ τ θτ

⎧ ⎫⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎪ ⎪⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎪ ⎪⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎛ ⎞⎪ ⎪⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎜ ⎟⎨ ⎬⎜ ⎟⎪ ⎪⎝ ⎠⎪ ⎪⎪ ⎪⎩ ⎭

⎧ ⎫⎛ ⎞ ⎛ ⎞⎪ ⎪⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎪ ⎪⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎛ ⎞⎪ ⎪⎝ ⎠ ⎝ ⎠⎜ ⎟ ⎜ ⎟⎨ ⎬⎜ ⎟ ⎜ ⎟⎪ ⎪⎝ ⎠⎜ ⎟⎜ ⎟ ⎪ ⎪⎝ ⎠ ⎪⎩ ⎭

+Ω= +

+Ω +

+= +

+Ω⎪

Rosenbluth Cross-section

Experimentally: for a number of 2q plot the measured values of cross section

as a function of 2tan 2θ Since it is linear, i.e ( ) ( )2 2 2tan 2d d A q B qσ θΩ∝ +

Determine The Form Factors

Exercise: electromagnetic FF of π ±cab be measured in e eπ π− ± − ±+ → +

Dcattering Amplitude is ( ) ( ),e q T k kμμε± ′ with ( ) ( ) ( ) ( )em, 0T k k k J kμ μπ π± ±≡′ ′

2,T k k k k F qμμ

π⎛ ⎞⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠= +′ ′Show that

Experiment shows that as ( ) ( )2 2,E MG q G q2q increases, The Form Factors Behave as

2 2 22

222

1 ; 00 0 0

1

p p nE M M n

Ep p nE M M q

G q G q G qG q

G G G

⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎛ ⎞⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎜ ⎟

⎜ ⎟⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎝ ⎠⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎜ ⎟⎜ ⎟⎝ ⎠

Λ

= = = =

−A Dipole Form

0.84 G eVΛ =

2

3 3 33

2 222 2 2 132

0

2

2

, 182

1

6 6 1 0.81 10 cm0.84

i r

Eq

d q er e d x r

qdr dG q dqdq

ρ ρππ

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎛ ⎞

⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎜ ⎟

⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎜ ⎟⎝ ⎠ ⎝ ⎠

⋅ −Λ

−−

=

Λ

Λ= = =

+

= = − ≈ ×

∫ ∫q x

q

Inverse Fourier Transform of this gives proton charge distribution (quarks) rρ ⎛ ⎞⎜ ⎟⎝ ⎠= x

and its mean square radius

Form factors fall off rapidly as increases. Remember this2q

II Deep Inelastic Scattering (DIS)

II DEEP INELASTIC SCATTERING (DIS)

Incoming beam of lepton with Energy E scatters off a

fixed hadron target. The Energy and the direction (scattering angle) of the scattered lepton is measured.

No Final state hadron (denoted by X ) is measured.

Lepton interacts with hadron through the exchange of a virtual photon, Z ( or W, if lepton is a neutrino). The target hadron absorbs the virtual photon, to produce the final state hadrons X .

,k E ,k E′ ′

q k k ′= −

p X

Basic diagram for DIS

Warning! various kinematic variables are used. I will chose z-axes along the incident lepton beam direction. The kinematic variables are:

X may be the hadron itself (Elastic Scatt.) or an excited state of it. If q is large the initial hadron breaks up

Kinematic Variables

M: Mass of target

k: momentum of initial lepton

if lepton mass neglected.

the solid angle into which final lepton scattered.

Momentum of scattered lepton

( ,0,0, )k E E=

Ω

k ′

( , sin cos , sin sin , cos )k E E E Eθ φ θ φ θ′ ′ ′ ′ ′=p: Momentum of target. For fixed target, ( ,0,0,0)p M=

q: momentum transfer , i.e the momentum of virtual photon q k k ′= −

ν The energy loss of the lepton E E q p Mν ′= − = ⋅

:y The fractional energy loss of lepton p qE p ky ν ⋅

⋅= =( )2 2 2

22 1 cos 4 sinQ q EE EE θθ′ ′≡ − = − =2 2 2

2 2 21

Q Q QxM p q MEy

x

ν

ω

= = =⋅

=

,k E ,k E′ ′

q k k ′= −

p X

Basic diagram for DIS

x is called Bjorken variable. It is crucial in understanding of DIS, because QCD predicts that the structure functions are functions of xand independent of in the leading order2Q

DEFINITION: DIS is the study of lepton-hadron scattering in the Region of kinematics that 2 , , But is Fixed and limitedQ xν→∞ →∞

The INVARIANT mass of final state hadronic system X is

( )22 2 2 (1)2 .XM p q M p q q= + = + +

The invariant mass of X system must be at least equal to the mass of target nucleon. WHY?

The lepton energy loss E Eν ′= − is between zero and E, so, the physicallyallowed kinematic region is

0 1, 0 1x y≤ ≤ ≤ ≤(1) can be written as

2 2 2

12 2

xQ M Mxp q p q

−= = −

⋅ ⋅

The value x=1 implies that 2 2 andso,xM M=

corresponds to1 ElasticS t.catx =

2 2 2 2 2 12 .X xM M M p q Q M ⇒ ≤≥ ⇒ + − ≥

Since 2 andQ ν are both positive, x must also be positive.2

2M

Qxp q

ν

=⋅

XM

Any fixed hadron state X with invariant mass 2XM Contributes to the cross section

at the value of x

( )2 2 21

1 X

xM M Q

=+ −

gets driven to In DIS limit ( )2Q →∞ any state X with fixed mass 1x =In particular, all nucleon resonances such as N* gets pushed to 1x =

The Jacobian for converting between thses cases is

( )( )( )

( )

2 2

22

1

, 1 02

2 2,

1,0,cos

EEM

E

x Q QMExMEy MExx y y

x y EE M

ν

θ ν

′−

∂= = =

∂

∂ ′=

′ −∂

The experimental measurements give the cross section as a function of final lepton

energy and scattering angle the results often are presented instead by giving the differential cross section as a function of

2d dE dσ ′ Ω

( ) ( )2, , or , ,x Q x yφ φ

The basic Feynman graph for DIS shown. The scattering Amplitudeis given by

( ) ( ) ( )22 0 , 0 ,h

igie k j k s X j pq

μν μ ν λ⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

−= − ′

:s

:λ

is the polarization of lepton and

is the polarization of the initial hadron1 2 1 2λ = ±For spin target,

:λ Can be chosen to be the value of spin in arbitrary direction, usually beam direction

The differential cross section is obtained from by squaring it and multiplying by the phase space factors

,k E ,k E′ ′

q k k ′= −

p X

( ) ( ) ( ) ( )( )( )

( )

( ) ( )( )( )

( ) ( ) ( ) ( )

3

3

3

3

24 4

2 2

4 4 4

42 2

22 2 1

22 2

, 0 0 , , 0 0 ,

d kxE

X rel

xd kE

x

h h

d k p k pE M v

k p k p eE M Q

p j X X j p k s j k k j k s

π

π

μ νμ μ

σ π δ

π δ

λ λ

′′

′′

′= + − −=

′+ − −=

′ ′×

∑∫

∑∫