Structure of Hadrons

Electron-Muon / Proton Scattering

QED is a well understood theory as we have seen and can be used as useful probes of the hadron
structures. As it turns out, the best probe to explore the structure of proton is the electron-proton
scattering. Whether proton is considered as point-like or not, the starting point is the QED process
e− e− → µ− µ− , which is related to the process e+ e− → µ+ µ− through crossing symmetry.
Crossing symmetry implies making the e+ e− → µ+ µ− Feynman diagram to stand on its feet,

The idea here is to study first the scattering of electrons from massive point-like particle (here muon),
M ≡ mµ  me . From the Feynman diagram above, the Feynman amplitude is

−igµν ie2
iMf i = i [ūs2 (k2 )(−ieγ µ )us1 (k1 )]e [ūr 2
(p 2 )(−ieγ ν
)u r 1
(p1 )]µ ≡ je · jµ (1)
q2 q2
The unpolarized scattering cross-section can be obtained by taking modulus square of Mf i , as is
done before in other QED processes but this time writing explicitly in terms of tensors associated
with the electron and muon vertex,

e4
|Mf i |2 = Le µν Lµν
µ (2)
4q 4
where, Lµν
e = (ū2 γ u1 ) (ū1 γ ν u2 )
µ

= Tr [γ µ (k/1 + me )γ ν (k/2 + me )]
= 4 k1µ k2ν + k1ν k2µ + (m2e − k1 · k2 ) g µν
 
(3)
Lµν = 4 pµ1 pν2 + pν1 pµ2 + (m2µ − p1 · p2 ) g µν .
 
similarly, µ (4)

Hence the invariant amplitude for the e− µ− → e− µ− is,

8e4 
|Mf i |2 = (k1 · p1 )(k2 · p2 ) + (k1 · p2 )(k2 · p1 ) − m2e (p1 · p2 ) − m2µ (k1 · k2 ) + 2m2e m2µ

(5)
q4
Relevant for electron-proton scattering, assuming proton acting like a point particle at the scattering
energy, mµ is replaced with M and in the limit M ≡ mµ  me , in which case we get,

8e4 
|Mep |2 = (k1 · p1 )(k2 · p2 ) + (k1 · p2 )(k2 · p1 ) − m2µ (k1 · k2 )

4
(6)
q
The 4-momentum conservation and momentum transfer q gives,

k1 + p1 = k2 + p2 ⇒ p2 = k1 − k2 + p1 = q + p1
q = k1 − k2 ⇒ q 2 = −2k1 · k2 , where k12 = k12 = 0 (7)

since me = 0 is considered and mµ = M . Using this (7) in (6) yields,

8e4
|Mep |2 (k1 · p1 )(k2 · {k1 − k2 + p1 }) + (k1 · {k1 − k2 + p1 })(k2 · p1 ) − M 2 (k1 · k2 )
 
=
q4
8e4
 2 
q 1
= − {(k1 · p1 ) − (k2 · p1 )} + 2(k1 · p1 )(k2 · p1 ) + q 2 M 2 (8)
q4 2 2

1
For e p scattering consider Lab frame kinematics, where the initial proton is at rest.

p1 = (M, ~0), k1 = (E, ~k), k2 = (E 0 , ~k 0 ) where E = |~k|, E 0 = |~k 0 |

q2 θ
− = k1 · k2 = EE 0 − ~k · ~k 0 = EE 0 (1 − cos θ) ⇒ q 2 = −4EE 0 sin2 (9)
2 2
also p2 = q + p1 ⇒ q 2 = −2q · p1 = −2(k1 − k2 ) · p1 = −2M (E − E 0 ).

Evaluating |Mep |2 in (8) in the Lab frame, we find,

8e4
 2
q2 M 2

2 q 0 2 0
|Mep | = − M (E − E ) + 2M EE +
q4 2 2
4 2
q M (E − E 0 )
2
 
8e 2 0 q
= 2M EE 1 + −
q4 4EE 0 2M 2 2EE 0
4 2
 
8e 2 0 2 θ q 2 θ
= 2M EE 1 − sin − sin
q4 2 2M 2 2
4 2
 
8e θ q θ
= 4
2M 2 EE 0 cos2 − 2
sin2 (10)
q 2 2M 2

Differential scattering cross-section, evaluated in relativistic kinematics handout eqn. (115), is

2
E0

dσ 1
= |M |2 (11)
dΩ 64π 2 ME
 0 2 4
q2
 
1 E 8e 2 0 2 θ 2 θ
= 2M EE cos − sin
64π 2 M E q4 2 2M 2 2
4 03 2
 
e E 1 θ q θ
= 2 0 2 θ
cos2 − 2
sin2
4π E (−4EE sin 2 ) 2 2 2M 2
2 0 2
 
α E θ q θ
= cos2 − sin2 , (12)
4E 2 sin4 θ2 E 2 2M 2 2

where α = e2 /4π. The factor E 0 /E is due to recoil of the target proton, which is at rest in the Lab
frame we have considered. In the case where there is no recoil i.e. no momentum transfer q = 0
implying E 0 = E and M  E meaning the heavier proton target is not moving at all, we obtained
the so-called Mott scattering formula,

α2
 
dσ 2 θ
= 4 θ cos 2 . (13)
dΩ Mott 2
4E sin 2

In the non-relativistic limit, q = 0 → E = E 0 still but additionally we have me  E, E 0 . In the

Lab frame,
p1 = p2 = (M, ~0), k1 = (E, ~k), k2 = (E, ~k 0 ) where |~k| = |~k 0 | = k (14)
Hence the expression of |M |2 of e − p scattering in non-relativistic limit, using (5) in Lab frame as
defined above (14), is

2
 8e4 
|Mep |2 2M 2 E 2 − m2e M 2 − M 2 (E 2 − k 2 cos θ) + 2m2e M 2

= 4
q
8e4  2 2
M E + m2e M 2 + M 2 k 2 cos θ

= 2 2
[(k1 − k2 ) ]
8e4  2 2
M (me + k 2 ) + m2e M 2 + M 2 k 2 cos θ

= 2 2 2
[2k − 2k cos θ]
8M 2 e4  2
2me + k 2 (1 + cos θ)

= 4 2
4k (1 − cos θ)
M 2 e4
 
2 2 2 θ
= m e + k cos
k 4 sin4 θ2 2
m2e M 2 e4
= in the non − relativistic limit me  k. (15)
k 4 sin4 θ2

Thus in the non-relativistic limit of e − p scattering we obtained the famous Rutherford formula,
 0 2
dσ 1 E
= |M |2
dΩ 64π 2 M E
α2
= , where Ek = k 2 /2me . (16)
16Ek2 sin4 θ2

Looking back, the differential cross-section in (12) depends only on a single variable which is the
scattering angle θ for a given energy of the incident electron. When proton recoils E 0 /E 6= 1, instead
we have
E0 1
= . (17)
E 1 + M sin2 θ2
2E

Just to mention, both Mott and Rutherford scattering formula can also be obtained from scattering
of electron in electrostatic Coulomb potential.
Furthermore, consider the generic current that goes in determination of the fundamental QED
processes,
e e−ipx
J µ = e ψγ µ ψ = ūγ µ u where, ψ = p u (18)
2EV 2Ep V
The Dirac equation can be manipulated to obtain the following identity (sum over spin is assumed),

ūγ µ × (γ ν pν − m)u = 0 and ū(γ ν pν − m) × γ µ u = 0 (19)

Adding the above two equations ⇒
ū(γ µ γ ν + γ ν γ µ ) pν u − 2mūγ µ u = 0
or, ū 2g µν pν u = 2mūγ µ u ⇒ pµ ūu = mūγ µ u. (20)

Using the identity (20) in (18) and supposing p ≈ p0 in the non-relativistic limit, we get
e pµ e pµ e µ
J µ = eψγ µ ψ = ū u = 2m = p . (21)
2EV m 2EV m EV
Let this current J µ to couple to an external electromagnetic field Aµ . Take the case of pure electric
field when A~ = 0 and A0 = ϕ, then
e 0 e
J µ Aµ = p ϕ = ϕ ≡ ρϕ (22)
EV V
which is the pure electro-static interaction with the potential ϕ i.e. the contribution to the scattering
cross-section due to the electric charges of the fermions. If the current J µ is made to couple with a

3
more general photon field Aµ , then we make use of the Gordon identity,
1
ū(p0 )γ µ u(p) = ū(p0 ) [(p + p0 )µ − iσ µν qν ] u(p), q = p − p0 (23)
2m
to write the interaction term as,
0 0
e e−i(p−p )x e−i(p−p )x e 0
µ
J Aµ = p p ū0 γ µ u Aµ = p p ū [(p + p0 )µ − iσ µν qν ] u(p) Aµ (24)
2Ep V 2Ep0 V 2Ep V 2Ep0 V 2m

In the non-relativistic limit, p ≈ p0 → 0, the dominant contribution from the (p + p0 )µ is due to

electric charge as we have seen above. The σ µν in the second term is anti-symmetric, which can be
used to write
e−iqx e 0 µν e−iqx e 0 µν
J µ Aµ = ū σ (qν Aµ − qµ Aν )u = ū σ Fµν u (25)
2EV 4m 2EV 4m
In the non-relativistic limit, the Dirac spinor u is

0
      
1 1 0
√  √ 1 √  √
 
 0 
 , E + m  px −ipy  →
  0
 , 2m  1  ≡
  χs
u = E + m pz 2m 
 0  0  .
 E+m  
E+m
  0
px +ipy −pz
E+m E+m
0 0
(26)
From the form of u in (26) it is evident that χ† σ 0i χ = 0 and only χ† σ ij χ 6= 0 will contribute

σi
   
µν i µ ν 0 1 0 i 0
σ = [γ , γ ] where, γ = , γ =
2 0 −1 −σ i 0
σi σj −σ i σ j
       k 
i j 0 0 0 σ 0
γ γ = = = ijk (27)
−σ i 0 −σ j 0 0 −σ i σ j 0 σk

Hence, the expression in (25) becomes,

e−iqx 2me † k
J µ Aµ = χ 0 σ χs ijk Fij
2EV 4m s
e−iqx e † k e−iqx
= χs0 σ χs × 2Bk = e χ†s0 ~σ · Bχ
~ s
2EV 2 2EV
e ~ = 1 e ~σ · B ~ ∝ µ ~
= ~σ · B ~D · B (28)
2mV V 2m

where µ ~ D = (e/m)S ~ is the Dirac magnetic momentum. If, however, e = 0 then µD = 0. From the
expression of differential scattering cross-section of point like particles (12), it is now evident that
the (q 2 /2M 2 ) sin2 θ/2 is the magnetic interaction or intrinsic magnetic moment term. The cos2 θ/2
is, however, a mixed contribution term.

Elastic scattering from particle of finite size

In all the discussion above, the target proton is considered as point like object. But the protons
have finite size, which has to be considered in the scattering process. It is important to do so since,
electrons scattering off from target provides a powerful technique for exploring the internal structure
of the target. This is mostly done with scattering high energy electron and observing the angular
distribution and energy of the scattered electrons. This can lead to disintegration of the proton
which is a study in inelastic scattering. Presently, we assume that the momentum transfer q 2 is not
high enough to break the proton and the scattering is elastic. The electrons and protons currents
are (without the kinematical factors), respectively,

jeµ = −ieū(k 0 ) γ µ u(k) and Jpµ = −ieū(p0 ) Γµ u(p) (29)

4
where Γµ depends explicitly on the proton state but otherwise must be a 4-vector. The most general
4-vector can be formed from combination of p, p0 , q = p − p0 and γ matrices. One is certainly γ µ
and the other σ µν qν , whereas the combination (p + p0 )µ is dependent on the first two by virtue
of Gordon identity (23). For parity reasons there cannot be any γ 5 term involved. The q 2 is the
only independent scalar at the proton vertex, thus the coefficients can be functions of q 2 . Because,
p · q = −q 2 /2 we may not need to consider p · q as independent scalar. Therefore,
κ
Γµ = F1 (q 2 ) γ µ + F2 (q 2 ) iσ µν qν (30)
2M
where F1 (q 2 ) and F2 (q 2 ) are called form factors. If we run through the whole of previous argument,
then it follows trivially that eF1 (0) is now the charge of the proton implying F1 (0) = 1. The
same argument implies that F2 term contributes an additional magnetic moment , called anomalous
magnetic moment,
e κ F2 (0)
µ
~A = ~σ , (31)
2M
which thus arises entirely due to finite size of the target particles, with F2 (0) = 1. If we are dealing
with point particle F1 (q 2 ) = 1 and F2 (q 2 ) = 0 for all q 2 . Now using the currents jeµ and Jpµ in (29)
along with (30), the differential scattering cross section for electron-proton elastic scattering turns
out to be,
α2 E0 κ2 q 2 2 q2
  
dσ 2 2 θ 2 2 θ
= F1 − F cos − (F1 + κF2 ) sin . (32)
dΩ lab 4E 2 sin4 θ2 E 4M 2 2 2 2M 2 2

In order to avoid interference terms like F1 F2 , it is customary to define,

κq 2
GE (q 2 ) ≡ F1 (q 2 ) + F2 (q 2 ) and GM (q 2 ) ≡ F1 (q 2 ) + κF2 (q 2 ) (33)
4M 2
where GE is known as electric form factor and GM as magnetic form factor. This leads to Rosenbluth
formula,
α2 E 0 G2E + τ G2M q2
 
dσ 2 θ 2 2 θ
= cos + 2τ GM sin where τ = − > 0. (34)
dΩ lab 4E 2 sin4 θ2 E 1+τ 2 2 4M 2

In the limit q 2 /4M 2  1, the GE and GM have got simpler interpretation in terms of Fourier
transform of charge and magnetic moment distribution respectively,

q2 (E − E 0 )2 − ~q2 = −2M (E − E 0 ) ⇒
=
  q 2 
~q2 = q 2 1 − → q 2 ≈ ~q2 when q 2 /4M 2  1 (35)
2M
Z
GE (~q2 ) = GE (q 2 ) ≈ ei~q·~r ρ(~r) d3~r (36)
Z
2 2
GM (~q ) = GM (q ) ≈ ei~q·~r µ(~r) d3~r (37)

In order to measure GE (q 2 ) and GM (q 2 ), we rewrite the above scattering cross section in Lab frame,

α2 E0
 2
GE + τ G2M
  
dσ 2 θ 2 2 θ
= cos + 2τ G M tan
dΩ 4E 2 sin4 θ E 2 1+τ 2
  2  2 2

dσ GE + τ G M θ
= + 2τ G2M tan2 (38)
dΩ Mott 1+τ 2

Therefore, at very low q 2 ⇒ τ ≈ 0,

 
dσ dσ
/ ≈ G2E (q 2 ) (39)
dΩ dΩ Mott

5
At high q 2 where τ  1, we have
   
dσ dσ 2 θ
/ ≈ 1 + 2τ tan G2M (q 2 ). (40)
dΩ dΩ Mott 2

For instance, the electron beam energies are chosen to give certain values of q 2 ,

E0 M θ 2M E 2 (1 − cos θ)
= , q 2 = −2M (E − E 0 ) = −4EE 0 sin2 = (41)
E M + E(1 − cos θ) 2 M + E(1 − cos θ)

Measure the cross-section to 2-3% versus electron beam energy at various detector angles, which is
dσ dσ


shown schematically in the figure below. Once the q 2 are extracted, the ratio dΩ / dΩ Mott
is plotted
against tan2 θ/2 whose slope and intercept gives respectively the 2τ G2M and (G2E + τ G2M )/(1 + τ ).
The GE and GM thus obtained is plotted against q 2 to determine their dependence on q 2 . The
entire strategy is shown in the series of plot below.

Take an example e− p → e− p at Ebeam = 529 MeV and look at scattered electron at θ = 75o ,
ME 938 × 529
E0 = = = 373 MeV
M + E((1 − cos θ) 938 + 529(1 − cos 75o )
2 × 938 × 5292 (1 − cos 75o )
|q 2 | = = 294 MeV2
938 + 529(1 − cos 75o )

The form factors fall rapidly with q 2 which is a clear indication that proton is not point-like. It is
found that electric and magnetic form factors GE and GM have the same distribution and a fit to
the data gives,
2
GM (q 2 ) q2

2
GE (q ) ≈ ≈ 1/ 1 + (42)
2.79 0.71
Taking Fourier transformation of GE (36), the spatial charge distribution is

ρ(r) ≈ ρ0 e−r/a where a = 0.24 fm (43)

If the q 2 is not too large, we can expand the exponential in GE (q 2 ) (36) and the corresponding r.m.s.
charge radius is,
dGE
hr2 i = −6 ⇒ rrms ≈ 0.8 fm (44)
dq 2 q2 =0
The experimental results of electron-proton elastic scattering is thus suggestive of finite size of
proton. It is due to this finite size, the elastic scattering at very high q 2 is unlikely and the proton

6
tends to break up. From the empirical formula (42), that the data suggests, it follows from (40) that
at high q 2 ,
q2
   
dσ dσ θ
GM ∝ q −4 ⇒ = G2 tan2 ∝ q −6 (45)
dΩ elastic dΩ Mott 2M 2 M 2
Therefore, the scattering cross section is supposed to fall as q −6 with increasing q 2 but in reality it
is almost independent of the q 2 , as is shown in the experimental result below.

This is a strong hint of composite nature i.e. presence of sub-structure of proton.

Inelastic e-p scattering

At high q 2 , i.e. at high momentum transfer, the electron scatters inelastically from proton, the
later ended up breaking into number of particle species with at least one baryon (because of baryon
number conservation). The final state hadronic system does not have proton mass M but rather the
invariant mass W where W > M . The e p → e X inelastic scattering is shown in the figure below.

Going over from point-point scattering to point-finite size scattering, we simply replaced γ µ with Γµ
and constructing the most general form of Γµ (30). But such approach is inadequate for inelastic
scattering since it is not obvious what do ū and u stand for, since the final state is not a single
fermion. The canonical approach is to parameterize the invariant amplitude |M |2 as (without an
extra 1/2),

e4 µν X
|M |2 = L Wµν (46)
2q 4 e
= 4 k k + k ν k 0µ + (m2 − k · k 0 ) g µν
 µ 0ν
where Lµν

e (47)
X W2 W4 W5
and Wµν = −W1 gµν + 2 pµ pν + 2 qµ qν + 2 (pµ qν + pν qµ ) (48)
M M M
where the Wµν is the most general 2nd rank tensor built out of p and q. The antisymmetric
term pµ qν − pν qµ does not contribute since, Lµν being symmetric, its contribution vanishes when
contracted with Lµν . The factor of 2 in the denominator is due to spin averaging only over initial
electron. The current conservation at electron vertex requires,

qµ Lµν = (kµ − kµ0 ) k µ k 0ν + k ν k 0µ + (m2 − k · k 0 ) g µν

 

= k 2 k 0ν − (k · k 0 )k 0ν + (k · k 0 )k ν − k 02 k ν + (m2 − k · k 0 )(k ν − k 0ν )
or, qµ Lµν = 0 (49)

7
Similarly, current conservation at the hadronic vertex demands q µ Wµν = 0 leading to,
W2 W4 W5
q µ Wµν = −W1 qν + 2 (p · q)pν + 2 q 2 qν + 2 (p · qqν + q 2 pν )
 M M  M 
W4 2 W5 W2 W5 2
= qν −W1 + 2 q + 2 p · q + pν p · q + q =0
M M M2 M2
p·q
⇒ W5 = − 2 W2 (50)
q
2
M p·q M2 (p · q)2
W4 = 2 W1 − 2 W5 = 2 W1 + W2 (51)
q q q q4
Substituting the results of (50) and (51) back in (48), we get
    
qµ qν W2 p·q p·q
Wµν = W1 −gµν + 2 + 2 pµ − 2 qµ pν − 2 qν . (52)
q M q q
The W1 and W2 are known as structure functions. This expression of Wµν thus obtained can be
used in (46). The condition qµ Lµν = 0 at the electron vertex renders
    
µν qµ qν µν W2 p·q
L × W1 2 = 0 or, L × − 2 qµ =0 (53)
q M2 q
and similary with the qν terms in W2 . In the limit m → 0, we obtain
 
W2
Lµν Wµν = Lµν −W1 gµν + 2 pµ pν
M
 
W2
= 4 [k µ k 0ν + k ν k 0µ − (k · k 0 ) g µν ] −W1 gµν + 2 pµ pν
M
 
W2
= 4 2W1 (k · k 0 ) + 2 2(p · k)(p · k 0 ) − M 2 (k · k 0 ) .

(54)
M

In the Lab frame, p = (M, ~0), k = (E, E k̂), k 0 = (E 0 , E 0 k̂ 0 ) implying,

θ
p · k = EM, p · k 0 = E 0 M, k · k 0 = EE 0 − EE 0 k̂ · k̂ 0 = EE 0 (1 − cos θ) = 2EE 0 sin2 (55)
2
Using the above Lab frame kinematics, we obtain
   
θ 4W2 2 θ
Lµν Wµν = 8W1 2EE 0 sin2 + 2EE 0
M 2
− 2EE 0
M 2
sin
2 M2 2
 
θ θ
= 8EE 0 W2 cos2 + 2W1 sin2 (56)
2 2
e4
 
0 2 θ 2 θ
|M |2 = 4EE W 2 cos + 2W 1 sin . (57)
q4 2 2
Before we get into the complication of writting the differential cross section for inelastic scattering,
let us determine the kinematic variables on which the structure functions W1,2 might depend. We
define the following four variables but not all independent,
q2 p·q p·q
x≡− , y≡ , ν≡ , Q2 ≡ −q 2 (58)
2p · q p·k M
x is a Lorentz invariant quantity and is known as Bjorken x. To determine its value or range,
consider the invariant mass of the final state X as MX ,
2
MX = (p + q)2 = p2 + q 2 + 2p · q = M 2 − Q2 + 2p · q
or, Q2 = 2p · q + M 2 − MX
2
⇒ Q2 ≤ 2p · q since M < MX (59)
Hence 0 ≤ x ≤ 1 and x = 1 when MX = M (60)

8
The W = M i.e. x = 1 corresponds to elastic scattering where proton remains intact. The variable
y is also Lorentz invariant and in the Lab frame it becomes,

p = (M, ~0), k = (E, E k̂), k 0 = (E 0 , E 0 k̂ 0 ) ⇒ q = k − k 0 = (E − E 0 , E k̂ − E 0 k̂ 0 )

p·q M (E − E 0 ) E0
thus y= = =1− ⇒ 0 ≤ y ≤ 1. (61)
p·k ME E
that is fractional energy loss of the incoming electron. Finally, the ν = p · q/M in the lab frame
is simply ν = E − E 0 , that is the energy lost by the incoming electron. From the definition it is
obvious that there are only two independent variables out of the four and usually they are chosen
to be either (x, Q2 ) or (ν, Q2 ). For eleastic scattering x = 1 and so the only independent variable is
Q2 . Therefore, the structure functions depends on this two variables: W1,2 (x, Q2 ). For future use,
let us express the elastic scattering in terms of y and Q2 . First we note that,
θ
−Q2 = q 2 = (k − k 0 )2 = −2EE 0 (1 − cos θ) = −4EE 0 sin2 ⇒ dQ2 = −2EE 0 d(cos θ)
2
dσ dσ dQ2 dσ π dσ
dΩ = 2πd(cos θ) and = ⇒ = . (62)
dQ2 dQ2 dΩ dQ 2 EE 0 dΩ
Next we find that the various terms in (34) can be rewritten as,
E0 θ E0 Q2 Q2 1 Q4 y2
2τ sin2 = ×2× × = = (63)
E 2 E 4M 2 4EE 0 2 4M 2 E 2 2
2 2 2
2 (p · q) (Q /2) Q4
since x = 1 and y = = = .
(p · k)2 (M E)2 4M 2 E 2
E0 E0 E0 Q2 y2 M 2
   
2 θ 2 θ
cos = 1 − sin = 1− 0
=1−y− . (64)
E 2 E 2 E 4EE Q2
The differential cross section for elastic scattering is thus,
α2 E 0 G2E + τ G2M
 
dσ 2 θ 2 2 θ
= cos + 2τ G M sin
dΩ lab 4E 2 sin4 θ2 E 1+τ 2 2
2 2 2 2 2 2
   
dσ 4πα GE + τ GM y M y
equivalently, = 1−y− + G2
2
dQ lab Q 4 1+τ Q 2 2 M
4πα2 y2 M 2 y2
  
2 2
≡ 1 − y − f2 (Q ) + f 1 (Q ) (65)
Q4 Q2 2
G2 (Q2 ) + τ G2M (Q2 )
where, f1 (Q2 ) = G2M (Q2 ) and f2 (Q2 ) = E .
1+τ
For elastic e− µ− → e− µ− , f1,2 (Q2 ) = 1 for all Q2 . But for inelastic scattering the structure
functions depend on both x and Q2 and the corresponding expression for e p → e X is
d2 σ 4πα2 x2 y 2 M 2 F2 (x, Q2 )
  
2 2
= 1−y− + y F1 (x, Q ) (66)
dx dQ2 Q4 Q2 x
where F1 (x, Q2 ) = M W1 (x, Q2 ) and F2 (x, Q2 ) = ν W2 (x, Q2 ).

The structure functions W1,2 or F1,2 cannot any longer be interpreted as the Fourier transforms of the
charge and magnetic moment distributions, rather they are interpreted as momentum distribution
of the quarks within the proton.
In order to derive the expression for inelastic scattering cross section (66), we need to re-visit and
re-work the derivation of dσ/dΩ. The starting point is the equation (93) in Relativistic Kinematics
handout, written without the integral sign,

(2π)4 d3~k 0 d3 p~0

d2 σ = |M |2 δ(E1 + E2 − E3 − E4 ) δ 3 (~k + p~ − ~k 0 − p~0 ) (67)
4E1 E2 vr (2π) 2E3 (2π)3 2E4
3

9
The cross section σ or dσ/dΩ is obtained first by integrating over the final state momentum p~0
followed by integration over ~k 0 or just |~k 0 | i.e. E 0 . This supposes that |M |2 does not depend either
on ~k 0 or p~0 . If it, however, does then we cannot perform this integration, for instance when the
details of the final state X in inelastic scattering is not known. In Lab frame,

k = (E, ~k), k 2 = 0 ⇒ E1 = E = |~k| and k 0 = (E 0 , ~k 0 ), k 02 = 0 ⇒ E3 = E 0 = |~k 0 |

q
p = (M, ~0) ⇒ E2 = M and p0 = (E4 , p~0 ), p02 = MX 2
⇒ E 4 = MX 2 +p
~02 .
δ(E1 + E2 − E3 − E4 ) = δ(E + M − E 0 − E4 ) and E 0 dE 0 = |~k 0 | d|~k 0 |
q 2 = (k − k 0 )2 = −2EE 0 (1 − cos θ) and p0 = q + p ⇒ q 2 = −2q · p = −2M (E − E 0 ). (68)

Generally the energy E 0 of the outgoing electron is measured, hence we can take k 0 to be independent
variable and consider variation of cross section against E 0 . The integration over p~0 can be performed
thus taking away the momentum conserving δ-finction if the final state is known, as in elastic
scattering. Otherwise, it can be included in the |M |2 through Wµν along with the momentum δ-
funtion, as in inelastic scattering. Therefore, the scattering cross section is written without the p~0
integral and δ 3 (~k + p~ − ~k 0 − p~0 ),

(2π)4 E 02 dE 0 dΩ
d2 σ = |M |2 δ(E + M − E 0 − E4 )
4EM vr (2π)3 2E 0 (2π)3 2E4
0 0
1 E dE dΩ
= |M |2 δ(f (E 0 )) (69)
(2π)2 4EM vr 4E4
0 ~ ~0 0 0
√ can take vr = c = 1 and p~ = k − k , therefore, f (E 0 ) = E + E4 − E − M =
Since m = 0, we
0 2 02 0 2
E − E − M + E + E − 2EE cos θ + M . The delta function δ(f (E )) can be evaluated as,

δ(E 0 − E00 )
δ(f (E 0 )) =
|df (E 0 )/dE 0 |E00
df (E 0 ) d h 0 p i
= E − E − M + E 2 + E 02 − 2EE 0 cos θ + M 2
dE 0 dE 0

E 0 − E cos θ E4 + E 0 − E cos θ
= 1+ =
E4 E4
df (E 0 ) E4 + E + M − E4 − E cos θ M + E(1 − cos θ)
= =
dE 0 E00 E4 E4
M + M (E − E 0 )/E 0 ME
= = 0
E4 E E4
E 0 E4
⇒ δ(f (E 0 )) = 0 0
δ(E − E0 ) (70)
ME
Using this result in (69) we get,
0 0 0
1 2 E E4 0 0 E dE dΩ
d2 σ = |M | δ(E − E 0 )
(2π)2 4EM ME 4E4
02
d2 σ 1 E
or, = |M |2 δ(E 0 − E00 ) (71)
dE 0 dΩ (2π)2 (4M )2 E 2

First consider e− µ− → e− µ− elastic scattering whose |M |2 is given in (10) with M replaced by

10
mµ (same would be true if we consider proton to be point-like),
d2 σ E 02 8e4 q2
 
1 0 2 θ 2 θ
= 2m 2
µ EE cos − sin δ(E 0 − E00 ) (72)
dE 0 dΩ (2π)2 (4mµ )2 E 2 q 4 2 2m2µ 2
E E
where, E0 = 2 θ ≡ A
1 + 2E
M sin 2
d2 σ 4α2 E 02 E 0 q2
 
2 θ 2 θ
thus, = cos − sin δ(E 0 − E/A)
dE 0 dΩ q4 E 2 2m2µ 2
(2αE 0 )2 q2
 
2 θ 2 θ
= cos − sin δ(AE 0 − E)
q4 2 2m2µ 2
(2αE 0 )2 q2
    
2 θ 2 θ 2E 2 θ 0
= cos − sin δ 1+ sin E −E
q4 2 2m2µ 2 M 2
(2αE 0 )2 q2 2EE 0
   
2 θ 2 θ 0 2 θ
= cos − sin δ E − E + sin
q4 2 2m2µ 2 M 2
0 2 2 2
   
(2αE ) θ q θ q
= cos2 − sin2 δ ν+
q4 2 2m2µ 2 2mµ
(2αE 0 )2 Q2 Q2
   
2 θ 2 θ
≡ cos + sin δ ν− . (73)
Q4 2 2m2µ 2 2mµ

where Q2 = −q 2 and ν is defined in (58) which is ν = p · q/mµ = E − E 0 in Lab frame for elastic
scattering. In exactly the same way, we would get the expression for e p → e p elastic scattering,
d2 σ (2αE 0 )2 G2E (Q2 ) + τ G2M (Q2 ) Q2
   
2 θ 2 2 2 θ
= cos + 2τ G M (Q ) sin δ ν − (74)
dE 0 dΩ Q4 1+τ 2 2 2M
Needless to mention that integration over E 0 will give us back the forms (12) and (34). Finally for
inelastic e p → e X scattering, we have
d2 σ (2αE 0 )2
 
2 2 θ 2 2 θ
= W2 (x, Q ) cos + 2W1 (x, Q ) sin (75)
dE 0 dΩ Q4 2 2
in which case the E 0 integration could not be carried out unless the explicit x-dependence of
W1,2 (x, Q2 ) are known. See Halzen & Martin sec. 8.4 for discussion on the missing δ-function
in the (75). For inelastic scattering, it is a standard approach to use dx instead of dE 0 in (75). It
follows from (62) that,
dσ EE 0 dσ d2 σ EE 0 d2 σ EE 0 d2 σ
= ⇒ = = − , (76)
dΩ π dQ2 dE 0 dΩ π dE 0 dQ2 π dν dQ2
where ν = p · q/M = E − E 0 in Lab frame. From the definition of Bjorken x,
Q2 Q2 Q2 4M 2 x2 2M x2
x= ⇒ dx = − dν = − × dν = − dν. (77)
2M ν 2M ν 2 2M Q4 Q2
This can be used to convert dν to dx in (76),
d2 σ 2M x2 d2 σ
= −
dν dQ2 Q2 dx dQ2
d2 σ Q2 π d2 σ
⇒ =
dx dQ2 2M x EE dE 0 dΩ
2 0

Q2 π (2αE 0 )2
 
2 θ 2 θ
= W 2 cos + 2W 1 sin
2M x2 EE 0 Q4 2 2
2 2 0
 
4πα Q E θ θ
= W2 cos2 + 2W1 sin2 . (78)
Q4 2M x2 E 2 2

11
We can elliminate the sine and cosines using the kinematic variables x, y etc. in Lab frame,
2
E0 Q4

p·q p·q θ
y= =1− , y2 = = 2 2 2
and Q2 = 4EE 0 sin2
p·k E p·k 4M E x 2
2 2 0 2 2 2 0
My Q E θ M x y E θ
⇒ = sin2 and 1 − y − = cos2 . (79)
2 2M x2 E 2 Q2 E 2

Substituting the above expressions for sine and cosine in (78) we get back (66),

d2 σ 4πα2 Q2 M 2 x2 y 2
  
2 2 2
= M y W 1 (x, Q ) + W 2 (x, Q ) 1 − y −
dx dQ2 Q4 2M x2 Q2
4πα2 νW2 (x, Q2 ) M 2 x2 y 2
  
2 2
= y {M W 1 (x, Q )} + 1 − y −
Q4 x Q2
4πα2 F2 (x, Q2 ) M 2 x2 y 2
  
= 4
y 2 F1 (x, Q2 ) + 1−y− . (80)
Q x Q2

Let us now summerize all the Lab frame scattering cross section results we have obtained so far,
including the result for ultra-relativistic limit when me = mµ = 0,

e− µ− → e− µ− :
(2αE 0 )2 Q2 Q2
   
dσ 2 θ 2 θ
= cos + sin δ ν−
dE 0 dΩ Q4 2 2m2µ 2 2mµ
α2 E0 Q2
 
dσ θ θ
= 4 θ E cos2 + 2
sin2
dΩ 2
4E sin 2 2 2mµ 2
" !#
2 2
dσ 4πα2 y 2 y mµ
= + 1−y− . (81)
dQ2 Q4 2 Q2
4πα2 y 2 2πα2 
 
1 + (1 − y)2

= 4
+ (1 − y) = 4
Q 2 Q
"  #
2 2 2

2πα Q
= 4
1+ 1− when me , mµ → 0. (82)
Q s

ep→ep:
(2αE 0 )2 G2E (Q2 ) + τ G2M (q 2 ) Q2
   
dσ 2 θ 2 2 2 θ
= cos + 2τ G M (Q ) sin δ ν −
dE 0 dΩ Q4 1+τ 2 2 2M
2 0 2 2 2 2
 
dσ α E GE (Q ) + τ GM (q ) θ θ
= cos2 + 2τ G2M (Q2 ) sin2
dΩ 4E 2 sin4 θ2 E 1+τ 2 2
4πα2 y 2 y2 M 2
   
dσ 2 2
= f1 (Q ) + 1 − y − f2 (Q ) . (83)
dQ2 Q4 2 Q2

ep→eX:
4πα2 F2 (x, Q2 ) M 2 x2 y 2
  
dσ 2 2
= y F1 (x, Q ) + 1−y−
dx dQ2 Q4 x Q2
0 2 2 2
 
dσ (2αE ) F2 (ν, Q ) 2 θ 2F1 (ν, Q ) 2 θ
= cos + sin . (84)
dE 0 dΩ Q4 ν 2 M 2
where F1 (x, Q2 ) = M W1 (x, Q2 ) and F2 (x, Q2 ) = ν W2 (x, Q2 ).

Parton Model

12
Experimentally, F1,2 (x, Q2 ) are determined from the measurements of the differential cross section of
deep inelastic scattering e p → e X at several scattering angle and incoming electron beam energies.
One of the earliest result is shown below (at x = 0.25).

It is observed that both F1 and F2 are (almost) independent of Q2 . The near independence of the
stucture functions on Q2 is known as Bjorken scaling, implying
F1 (x, Q2 ) → F1 (x) and F2 (x, Q2 ) → F2 (x). (85)
This has got a huge significance since the result strongly suggests scattering from point-like con-
stituents within the proton (recall for e µ → e µ the f1,2 (Q2 ) = 1 for all Q2 ). Feynman proposed that
the proton is made up of point-like spin-1/2 constituents called partons – this was before the quarks
and gluons became widely accepted. In parton model, the interaction is elastic scattering between
a single virtual photon and the free spin-1/2 constituents. For convenience, we will henceforth call
the partons as quarks.

It is convenient to formulate parton model in a frame where the proton has very high energy or
moving with infinite momentum where we can neglect the proton mass and p = (Ep , 0, 0, Ep ). This
frame is called infinite momentum frame. In this frame we can also neglect the masses of the partons
or quarks and any momentum transverse to the direction of the proton pT = 0. Let the quark struck
by the photon carry a fraction ξ of the proton’s 4-momentum i.e. ξp and after colliding with photon
let it carry ξp + q.

Therefore, we get the following,

(ξp + q)2 = m2q = 0 ⇒ (ξp)2 + q 2 + 2ξp · q = 0
q2
since (ξp)2 = m2q = 0 ⇒ ξ=− =x (86)
2p · q
implying Bjorken x is essentially the fraction of the proton momentum carried by the struck quark
in infinite momentum frame. In terms of proton momentum,
p·q Q2
s = (k + p)2 = 2k · p, y= and x = (87)
p·k 2p · q

13
while for the struck quark,

xp · q Q2
sq = (k + xp)2 = 2x k · p = x s, yq = = y and xq = = 1. (88)
xp · k 2xp · q
Recall that x = 1 corresponds to elastic scattering i.e. when the target does not break up. We
encounter the same situation for parton or quark, hence the inelastic scattering with proton at high
energy shows elastic behavior. For elastic scattering of point-like particles in the ultra-relativistic
limit, we can borrow the result of e µ → e µ in (82) to apply in e q → e q,
" 2 #
2πα2 e2q Q2

dσ
= 1+ 1− , (89)
dQ2 Q4 sq

where eq is the parton / quark charge in the unit of electron charge. The expression (89) can be put
in more illuminating form by using the general result,

Q2 p · q Q2 Q2
xy = = = since s = (k + p)2 , (90)
2p · q p · k 2p · k s − M2
Q2
For quarks → Q2 = (sq − m2 ) xq yq ⇒ = yq = y (91)
sq

since m2 = 0 and xq = 1. This leads to,

2πα2 e2q   4πα2 e2q y2

 
dσ 2
= 1 + (1 − y) = (1 − y) + . (92)
dQ2 Q4 Q4 2

which is just the (82) except the quark charge factor. So the differential cross section for e q,
where the quark carries a fraction x of the proton momentum, is identical to elastic, point like
e µ scattering. This expression automatically explains Bjorken scaling – the observed constancy of
structure functions, the coefficients of (1 − y) and y 2 /2, against variation of Q2 .
Next we introduce parton distribution function qi (x) which is basically the momentum distribution
function of the partons within the proton. It reperesents the probability density to find a parton
(or equivalently the number density of partons of i-th type) carrying a momentum fraction x at a
particular Q2 . All the fractions x must add up to 1,
XZ
dx x qi (x) = 1. (93)
i

Therefore, the cross section for scattering from i-type quark having momentum fraction in the range
x and x + dx inside proton is,

4πα2 y2 2
 
dσ
= (1 − y) + eq qi (x) dx (94)
dQ2 Q4 2

which when summed up for all types of quarks within the proton gives the expression for the e − p
scattering cross section,

4πα2 y2 X 2
 
dσ
= (1 − y) + eq qi (x). (95)
dx dQ2 Q4 2 i

Compare this with the e − p scattering cross section in terms of structure functions as given in (84),
in the high Q2 limit or when Q2  M 2 y 2 ,

4πα2 F2 (x, Q2 )
 
dσ 2 2
= (1 − y) + y F1 (x, Q ) . (96)
dx dQ2 Q4 x

14
Comparison gives us parton model prediction for the structure functions, the Callen-Gross relation,
X
F2 (x, Q2 ) = 2x F1 (x, Q2 ) = e2q x qi (x). (97)

The Callen-Gross relation relate the structure functions with the parton distribution functions and
measurement of F1,2 (x, Q2 ) enable us to determine qi (x). The parton distribution functions depend
very much on strongly interacting quark-gluon system and as a result one cannot use perturbation
theory to determine it. But we can guess the expected form of qi (x), which is shown below.

To gain insight into parton distribution

P functions, we can use the Callen-Gross relation but we must
know which quarks to sum over in i e2q xqi (x). Certainly up (u) and down (d) quarks, that is what
the proton is made up of and gets its quantum numbers from. They are called valence quarks. But
vacuum polarization of the gluons can give rise to virtual q q̄ pairs in higher order perturbation
theory. They are called sea quarks and the pairs that are most likely to contribute to the above
sum are u − ū, d − d¯ and s − s̄. The heavier quark-antiquark pairs are, however, not kinematically
favourable and, therefore, less likely to pop up and negligible will be their contributions.
One point to note is that although we mentioned gluons for q q̄ production g → q q̄, we never actually
introduced it neither do we motivated its existance. But to obtain bound states out of quarks (such
as proton) some QED-like but extremely strong interaction can be envisioned and the quanta of
that interaction is what we are calling as gluons.
The Bjorken scaling is only approximate, the F1,2 is fairly independent of Q2 as long as x is large
enough, at least x & 0.1. But scaling violation i.e. F2 (x, Q2 ) 6= F2 (x) has been observed at low
x . 0.05. The plot below shows weak dependence of F2 on Q2 for x & 0.05, which is consistent with
the parton model, while at low x the dependence is significant revealing that partons are sharing
momentum with the gluons. At high Q2 it is expected to see more low x partons.

15
Hence, for electron-proton and electron-neutron scattering,
 2  2  2
1 ep 2 1  p 1
[up (x) + ūp (x)] + d (x) + d¯p (x) + [sp (x) + s̄p (x)]

F (x) = (98)
x 2 3 3 3
 2  2  2
1 ep 2 n n 1  n ¯n 1
[sn (x) + s̄n (x)]

F (x) = [u (x) + ū (x)] + d (x) + d (x) + (99)
x 2 3 3 3

where q p (x) and q̄ p (x) are distribution functions for q type quark and antiquark within proton. Now
as proton and neutron are members of isospin doublet, there are as many u quarks in proton as d
quarks in a neutron and so onr, implying

u(x) ≡ up (x) = dn (x), d(x) ≡ dp (x) = un (x) and s(x) = sp (x) = sn (x) (100)

and similarly for the antiquarks. This gives,

 2  2  2
1 ep 2 1  ¯
 1
F (x) = [u(x) + ū(x)] + d(x) + d(x) + [s(x) + s̄(x)] (101)
x 2 3 3 3
 2  2  2
1 ep 2  ¯
 1 1
F (x) = d(x) + d(x) + [u(x) + ū(x)] + [s(x) + s̄(x)] (102)
x 2 3 3 3

F2ep (x) and F2en (x) can be measured experimentally as a function of x which can be exploited
to determine the parton distribution functions. To do so, perform the following integration, now
neglecting the strangeness content since s-quarks are about 10 times heavier than u and d,
Z 1 Z 1  
4 1 4 1
F2ep (x) dx = ¯

dx x [u(x) + ū(x)] + d(x) + d(x) = fu + fd (103)
0 0 9 9 9 9
Z 1 Z 1  
4 ¯
 1 4 1
F2en (x) dx = dx x d(x) + d(x) + [u(x) + ū(x)] = fd + fu (104)
0 0 9 9 9 9
Z 1
where fu ≡ dx [x u(x) + x ū(x)] (105)
0

and similarly for fd . The fu,d are the fraction of the proton momentum carried by the u − ū and
d − d¯ quarks. The integrations on the l.h.s are basically area under the experimental curves for F2 (x)
measured against x and the corresponding experimental values are,
Z Z
ep
F2 (x) dx ≈ 0.18 and F2en (x) dx ≈ 0.12 ⇒ fu ≈ 0.36, fd ≈ 0.18. (106)

Firstly, in the proton (neutron) the u (d) quarks carry twice the momentum of the d (u) quarks as
expected from their numbers. Secondly, the momentum carried by quarks only account for a little
over 50% of the total proton momentum! This implies a substantial fraction of proton’s momentum
is carried by neutral partons and not by the quarks. These neutral partons are the gluons of QCD.

Quantum Chromodynamics

16
Quantum Chromodynamics or QCD is the theory of strong interaction, which is expected to describe
all of the above phenomenon among many others. The basic degrees of freedom are quarks and
gluons, but because of confinement the final states are the hadrons – the bound states of quarks
and gluons. For most part QCD is nonperturbative i.e. αs ∼ 1 and the highly developed machinary
of perturbation theory cannot be used. QCD spectroscopy, form factors and structure functions,
fragnemtation function, decay constants etc. are all result of nonperturbative regime of QCD.
However, because of asymptotic freedom, perturbative techniques can still be used to understand
QCD phenomena in the high energy / momentum or short distance scale. Thus QCD is a complicated
stuff but its Lagrangian looks rather simple, very much like QED. In QCD, fermion fields have an
additional flavor (f ) index besides Dirac index (which is usually kept suppresed) and the gauge fields
are SU(3) non-Abelian that carry color charges (a). Besides, a quark of each flavor has an additional
quantum number which is the color i = r, g, b.
X 1 a µνa
Lqcd = / − mf ) ψf − Fµν
ψ f (x) (iD F (107)
4
f
8
X
where, Dµ = ∂µ − ig τ a Aaµ (x) and Fµν
a
= ∂µ Aaν − ∂ν Aaµ + g f abc Abµ Acν (108)
a=1
ψ(x) → U (x) ψ(x), Fµν (x) → U (x) Fµν (x) U † (x) and
i
Aµ (x) → U (x)Aµ (x)U † (x) − [∂µ U (x)]U † (x) (109)
g
8
!
X
U (x) = exp −i τ a αa (x) , Aµ (x) = Aaµ (x) τ a , Fµν (x) = Fµν
a
(x) τ a . (110)
a=1

To derive the Feynman rules from the QCD Lagrangian above, particularly to see gluons coupling
to itself, we write down the above Lagrangian density in terms of gauge fields Aaµ (x),
X X
Lqcd = ψ f (x) (i∂/ − mf ) ψf (x) + g ψ f γ µ Aaµ τ a ψf (111)
f f
1
− (∂µ Aaν − ∂ν Aaµ ) (∂ µ Aνa − ∂ ν Aµa ) (112)
4
g g 2 abc ade b c µd νe
− f abc (∂µ Aaν − ∂ν Aaµ ) Aµb Aνc − f f Aµ Aν A A . (113)
2 4
Here we ignore complications due to gauge fixing, which is needed to define gluon propagator, and
ghosts, which are needed to cancel unphyiscal degrees of freedom preserving unitarity. The Feynman
rules for QCD using covariant gauge fixing ∂µ Aaµ = 0 and Feynman gauge are,

17
From the Lagrangian above (107) it is evident that both QCD and QED are vector theory where
fermion- gauge field interaction is of the form ψγ µ ψ Aµ . But the major difference is, in QED the
photon does not carry the charge of electromagnetic interaction but in QCD gluons carry the color
charge and consequently have self interaction, (113). It is this gluon self-interaction that is believed
to give rise to color confinement. The gluonic self-interaction is also responsible for anti-screeing of
color charges which is believed to be the reason for asymptotic freedom. The value of strong coupling
αs depends on distance r i.e. energy Q ∼ r−1 as,
12π
αs (Q2 ) = . (114)
(33 − 2nf ) log(Q2 /Λ2 )
The hadronic scale Λ ∼ O(300 MeV) and for Q2 much larger than Λ2 , say around (Q2 ∼ 30 GeV)2 ,
the QCD coupling is small αs ∼ 0.1 when a perturbative description of QCD is possible. At this high
energy limit the quarks and gluons are interacting weakly. Otherwise the coupling is rather strong
and QCD is nonperturbative. An important experimental evidence of this inner working of QCD is

18
the jet events. In the e+ e− collisions, the QED events are Bhaba scattering e+ + e− → e+ + e− ,
pair annihilation e+ + e− → γ + γ and, if CM energy is high enough e+ + e− → µ+ µ− . But at
sufficient high energy they can also produced quark production e+ + e− → q + q̄ leading to hadron
jets. For instance the 2-jet event,

After e+ e− collision, the quarks separate initially at high velocity and color flux tube forms between
quarks. But once the separation reached is ∼ 1 fm or typical size of hadrons, the strong interaction
becomes so great i.e. the potential energy of the flux tubes becomes so large that new q q̄ pairs
are produced. These quarks and antiquarks get bind in multitude of combinations to form colorless
mesons and baryons – a process knwon as hadronization. These hadrons are actually recorded at the
detectors as two back-to-back jets of hadrons, with the same angular distribution (1 + cos2 θ) as in
e+ + e− → µ+ + µ− (see equation 84-85 in QED handout). One can straight away take this result to
interprete that quarks are spin-1/2 particles. Sometimes the 3-jet events e+ e− → q q̄ g can be seen,
the rate of which gives a measurement of αs and the angular distribution of the jets implies gluons
are spin-1. Even the 4-jet events e+ e− → q q̄ g g are also observed, the rate and angular distribution
of which reveals the underlying SU(3) symmetry of QCD.

Here we address the question how does QCD effect the deep inealstic scattering results, for instance
Bjorken scaling. We know that the parton model ignores the role of gluon in e p → e X and the
process is essentially γ ? q → q an O(α) process, where γ ? is the virtual gluon. But the quarks can
radiate gluons leading to the O(ααs ) process γ ? q → q g contribution to e p → e X. Moreover, gluon
can contribute to deep inelastic scattering via O(ααs ) process γ ? g → q q̄ and this is gluon initiated
hard scattering contribution to e p → e X.

19
Including these O(ααs ) diagrams has two experimentally observable consequences: (1) violation of
Bjorken scaling and (2) outgoing quark, and hence the hadron jet, will not be collinear with the
virtual photon i.e. the transverse momentum pT 6= 0.

The calculation of γ ? q → q g follows closely the Compton scattering e γ → e γ but with the initial
real photon γ replaced by virtual photon γ ? .

The invariant amplitudes for Compton scattering in the ultra-relativistic limit is (putting m = 0 in
equations 63 and 64 of QED handout),
0
2e4 0 4 k·p
|M (s) |2 = (p · k)(k · p ) = 2e , (115)
(p · k)2 p·k
0 0
2e4 0 0 0 4 k ·p
|M (u) |2 = (p · k )(k · p ) = 2e . (116)
(p · k 0 )2 p · k0

For form relevant in present discussion, we note that

s = (p + k)2 = p2 + k 2 + 2p · k = 2p · k (117)
0 2 2 02 0 0
u = (p − k ) = p + k − 2p · k = −2p · k (118)

which can be used above to write the invariant amplitudes as,

 u  s
|M (s) |2 = 2e4 − and |M (u) |2 = 2e4 − . (119)
s u

20
The crossterm (equation 65 in QED handout), however, is zero

2e4
M (s) M (u)† + M (u) M (s)† = [2(p · p0 ){(k · p) − (k · k 0 ) − (p · k 0 )}]
(k · p)(k 0 · p)
2e4
= [2(p · p0 ){s/2 − (−u/2) − (−t/2)}]
(−s/2)(−u/2)
8e4
= − [2(p · p0 )(s + u + t)] = 0 (120)
us
m2 = 0. Therefore, the total invariant amplitude is,
P
since s + u + t =
 u s
|M |2 = |M (s) |2 + |M (u) |2 = 2e4 − − . (121)
s u
For virtual photon γ ? e → γ e, however, k 2 = −Q2 6= 0 which is the mass of the incident virtual
photon and hence

s = (p + k)2 = p2 + k 2 + 2p · k = 2p · k − Q2 = 2p0 · k 0 . (122)

Refering to equations 62 in QED handout, the only term without m is (I) and it becomes for
s-channel (with p2 = m2 = 0),

Is = 32(k · p)(p·0 +k · p0 ) − 16(k 2 + 2p · k)(p · p0 )

= 32(k · p)(p·0 +k · p0 ) − 16(−Q2 + 2p · k)(p · p0 ) (123)
= 32(k · p)(k · p0 ) + 16Q2 (p · p0 ). (124)

Similar expression for u-channel is obtained by substituting k with −k 0 and taking mass of real
photon k 02 = 0,

Iu = 32(−k 0 · p)(p · p0 − k 0 · p0 ) − 16(k 02 − 2p · k 0 )(p · p0 )

= 32(p · k 0 )(p0 · k 0 ) = 32(p · k 0 )(p · k − Q2 /2]
= 32(p · k)p · k 0 ) − 16Q2 (p · k 0 ). (125)

In the ultra-relativistic limit,

Iu = 32(p · k 0 )(p0 · k 0 ) = 32(−u/2)(s/2) = −8us (126)

0 2 0
Is = 32(k · p)(k · p ) + 16Q (p · p )
= 32(p0 · k 0 + Q2 /2)(p · k 0 − Q2 /2) + 16Q2 (p · p0 )
= 32(p0 · k 0 )(p · k 0 ) + 16Q2 [(p · k 0 ) − (p0 · k 0 ) + (p · p0 )]
= 32(p · k 0 )(p0 · k 0 ) + 16Q2 (s + u + t) = −8us. (127)

Therefore, the invariant amplitudes remain the same as before (119),

 u  u  s  s
|M (s) |2 = 2e4 − = 32π 2 α2 − and |M (u) |2 = 2e4 − = 32π 2 α2 − . (128)
s s u u
But the crossterm in (120) is now nonzero,

2e4
M (s) M (u)† + M (u) M (s)† = [2(p · p0 ){(k · p) − (k · k 0 ) − (p · k 0 )}]
(k · p)(k 0 · p)
2e4 s + Q2
  
0 t u
= 2(p · p ) + +
(k 0 · p0 )(k 0 · p) 2 2 2
8e4
   2 
t Q s+t+u
= − 2 − +
us 2 2 2
 2   2 
2Q t 2Q t
= 2e4 = 32π 2 α2 (129)
us us

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The total invariant amplitude for Compton scattering with virtual photon γ ? in the initial state
thus is,
2Q2 t
 
2 (s) 2 (u) 2 (s) (u)† (u) (s)† 2 2 u s
|M | = |M | + |M | + M M +M M = 32π α − − + . (130)
s u su
We can use this result straight away to write the |M |2 for the γ ? q → q g only replacing the final
γ − e vertex involving α or with g − q vertex involving CF e2q αs , where CF is the color factor which
is 4/3 in the present case, and u ↔ t on account of different ordering of the outgoing particles,
s 2Q2 u
 
? 2 2 2 t
γ q → g q : |M | = 32CF π eq ααs − − + . (131)
s t st
One quantity of of interest is the transverse momentum pT of the outgoing quark and measuring
dσ/dp2T instead of dσ/dΩ. In the CM frame of γ ? − q, the relativistic kinematics can be used to
derive pT .

Various momentum 4-vector and the Mandelstam variables are,

q = (q0 , 0, 0, k), q1 = (k, 0, 0, −k) ⇒ q02 − k 2 = q 2 = −Q2
q2 = (k , k sin θ, 0, k 0 cos θ),
0 0
g = (k 0 , −k 0 sin θ, 0, −k 0 cos θ), q0 + k = 2k 0 . (132)
2
s = (q + q1 ) = q02 2 2
+ k + 2kq0 = 2k − Q + 2kq0 2

= (q2 + g) = (2k 0 )2 = 4k 02
2
(133)
2 2 2 0 0
s+Q = 2k + 2k(q0 ) = 2k + 2k(2k − k) = 4kk (134)
t = (q − q2 )2 = q 2 + q22 − 2qq2 = −Q2 + 0 − 2(q0 k 0 − kk 0 cos θ) = −Q2 − 2q0 k 0 + 2kk 0 cos θ
= (g − q1 )2 = g 2 + q12 2gq1 = 0 + 0 − 2kk 0 (1 − cos θ) (135)
2 0
u = (q1 − q2 ) = q12 + q22 − 2q1 q2 = −2kk (1 + cos θ) (136)
2 0 0 2 0 0 0 2 02 2 0
−t − u = Q + 2q0 k + 2kk = Q + 2k (2k − k) + 2kk = Q + 4k = s + Q = 4kk (137)
stu = 4k 02 4k 2 k 02 (1 − cos2 θ) = 16k 2 k 04 sin2 θ (138)
The transverse momentum pT of the outgoing quark and its relation with dΩ is,
stu
pT = k 0 sin θ → p2T = k 02 sin2 θ = (139)
(s + Q2 )2
= k 02 2 sin θ cos θ dθ ≈ 2k 02 d(cos θ) for θ ≈ small
dp2T
π dp2T 4π 2
dΩ = 2π d(cos θ) = = dpT , (140)
k 02 s
where we have made use of (133) s = 4k 02 . At high energy s → large and the cross section of
γ ? q → q g peaks as −t → 0 which is basically the forward scattering. The differential scattering
cross-section of γ ? q → g q is given by,
dσ 1 pf
= |M |2
dΩ 64π 2 s pi
s 2Q2 u
 
dσ 4π 1 1 4 t
or, = |M |2 = 32π 2 (e2q ααs ) − − +
dp2T s 64π 2 s 16πs2 3 s t st
8π e2q ααs 2 2
   
1 2uQ + t
= − s−
3s2 t s
8π e2q ααs 2(s + Q2 )Q2
   
1
= − s+ (141)
3s2 t s

22
where we have used −t  s and −u = s + Q2 from (137). Here we define a quantity z,

Q2 Q2 Q2
z≡ = = (142)
2pi · q (pi + q)2 − q 2 s + Q2

which can be interpreted as fractional decrease of quark momentum after it has released a gluon.
For forward scattering i.e. s large and −t → 0, using −u = s + Q2 + t, we have

stu −st(s + Q2 + t) s(−t)(s + Q2 ) s(−t)

= ≈ = . (143)
(s + Q2 )2 (s + Q2 )2 (s + Q2 )2 s + Q2

In terms of z, the forward scattering (141) can be rewritten as,

8π 2 e2q ααs 2(s + Q2 )Q2

  
dσ 1
= −s+
dp2T 3s2 t s
8π 2 e2q ααs 2(s + Q2 )Q2
 
1
= s +
3s p2T (s + Q2 ) s
α 2
 
4π αs 1 4 s 2Q
= e2q +
s 2π p2T 3 s + Q2 s
1 α s
= e2q σ0 2 Pqq (z) (144)
pT 2π
4 1 + z2
 
αs
where, Pqq (z) = and σ0 = . (145)
3 1−z 2π

The Pqq represents the probability of a quark emitting a gluon and so becoming a quark with
momentum reduced by fraction z and is also known as splitting function. There are signularities
in the forward scattering cross section at z → 1 and p2T = 0. The z → 1 singularity is associated
with emission of soft massless gluon (i.e. momentum of quark emitting gluon hardly changing). The
p2T → 0 singularity suggests that γ ? q → q does not contribute or negligible compared to γ ? q → q g.
So the experimental signature is that q-jet and g-jet neither moving in the direction of γ ? i.e. hadrons
emerge with pT 6= 0. In parton model without gluon pT = 0 always. This O(ααs ) process also leads
to scaling violation. The structure function F2 (x, Q2 ) gets contribution from γ ? q → q g,
s/4 s/4
dp2T αs
Z Z
dσ
σ(γ ? q → q g) = dp2T = e2q σ 0 Pqq (z)
µ2 dp2T µ2 p2T 2π
2
 
αs Q
= e2q σ0 Pqq (z) log 2 (146)
2π µ

where µ is the cut-off to regulate the divergence when p2T → 0. The log dependence on Q2 leads to
the Bjorken scaling violation.

23