Deep Inelastic Scattering (DIS)

Home , Elastic scattering, Inelastic scattering

Ingo Schienbein

Laboratoire de Physique UGA/LPSC Grenoble Campus Subatomique et de Cosmologie Universitaire

CTEQ School on QCD and Electroweak Phenomenology

Monday, October 8, 12 EAOM LPSC, Serge Kox, 20/9/2012

Mayaguez, Puerto Rico, USA 18-28 June 2018

Tuesday 19 June 18 Disclaimer

This lecture has proﬁted a lot from the following resources:

• The text book by Halzen&Martin, Quarks and Leptons

• The text book by Ellis, Stirling & Webber, QCD and Collider Physics

• The lecture on DIS at the CTEQ school in 2012 by F. Olness

• The lecture on DIS given by F. Gelis Saclay in 2006

Tuesday 19 June 18 Lecture 1

1. Kinematics of Deep Inelastic Scattering

2. Cross sections for inclusive DIS (photon exchange)

3. Longitudinal and Transverse Structure functions

4. CC and NC DIS

5. Bjorken scaling

6. The Parton Model

7. Which partons?

8. Structure functions in the parton model

Tuesday 19 June 18 1. Kinematics of Deep Inelastic Scattering

Tuesday 19 June 18 What is inside nucleons?

‣ Basic idea: smash a well known probe on a nucleon or nucleus in order to try to ﬁgure out what is inside ‣ Photons are well suited for that purpose because their interactions are well understood ‣ Deep inelastic scattering: collision between an electron and a nucleon or nucleus by exchange of a virtual photon

Note: the virtual photon is spacelike: q 2 < 0 e- • q 2 2 2 2 • Deep: Q = -q >> MN ~1 GeV X 2≣ 2 2 p , A • Inelastic: W MX > MN ‣ Variant: collision with a neutrino, by exchange of a Z0 or W±

Tuesday 19 June 18 18 Masses in deep inelastic scattering

Kinematic variables Figure 2.1: Kinematics of DIS in the single exchange boson approximation. Let’s consider inclusive DIS where a sum over all • l′ brokenhadronic up such final that states the final X state is performed: consists of the scattered lepton andl a hadronic finalθ state X. Here, l and l0 are the four-momenta of the incoming and outgoing leptons, - - e (l)+N(p) → e (l’)+X(pX) q = l l0 is the four-momentum of the exchange boson (, Z,orW )andp theq four- momentum of the hadron. The hadronic final state X carries the four-momentum pX . 2 2 2 2 2 •In DISOn-shell processes, conditions: an inclusive p =M sum, l =l’ over=m all hadronic final states is performed. Therefore,

only the initial state momenta l and p and the ﬁnal lepton momentum l0, which has to be X θ •measured,Measure are energy available and polar to describe angle of the scattered kinematics electron of DIS. (E’, ) P It is useful to introduce the following Lorentz-invariant quantities to describe the •kinematicsOther invariants of a DIS of process: the reaction:

2 2 2 Q = q = (l l0) > 0, the square of the momentum transfer, • ⌫ = p q/M lab= E E , • · l l0 0 x = Q2/(2p q)=Q2/(2M⌫) 1, the (dimensionless) Bjorken scaling variable, •  ·  0 y = p q/p l lab=(E E )/E 1, the inelasticity parameter, •  · · l l0 l  * Heres ‘lab’=( designatesp + l)2 ,the the proton square rest frame of the p=(M,0,0,0) lepton–hadron which coincides energy with the in thelab frame center-of-mass for ﬁxed target system,experiments • Tuesday 19 June 18 S =2p l = s M 2 m2, where M is the hadron mass and m the lepton mass, • · W 2 = p2 =(p + q)2, the square of the invariant mass of the hadronic ﬁnal state. • X Kinematic variables

• There are two independent variables to describe the kinematics of inclusive DIS (up to trivial φ dependence): l′ l θ (E’,,θ) or (x,Q2) or (x,y) or ... q • Relation between Q2, x, and y:

Q2 p q X Q2 = (2p l)( )( · ) S =2p l · 2p q p l · · · =(p + l)2 p2 l2 P = Sxy = 2MExy

• Invariant mass W of the hadronic ﬁnal state X: (also called missing mass since only outgoing electron measured)

W 2 M 2 =(p + q)2 = M 2 + 2p q+q2 elastic scattering: W =MN, x=1 ⌘ X N · Q2 Q2 = M 2 + Q2 = M 2 + (1 x) N x N x inelastic: W ≥ MN + m�, x<1

Tuesday 19 June 18 Electron-proton scattering

The scattering picture used so far needs to be extended for a composite object The invariant mass spectrum shows the elastic peak, excited baryons followed by an inelastic smooth distribution The ep→eX cross section as function of W

Halzen&Martin, Quarks&Leptons, Fig. 8.6 - - e ki kf e Data from SLAC; The elastic peak at W=M has been reduced by a factor 8.5 q

1 Invariant 2 Elastic a resonance Inelastic

... mass W n } + 0 peak ep→ea →epé region p Elastic peak: W=M, x=1 (proton doesn’t break up: ep→ep) The missing mass of the• hadronic system can be derived from measuring electron 2 2 2 2 2 2 2 • WResonances:=(p + W=Mq) =R, ωM=1/x=1+(M+2M⌫ R+-Mq )/Q (Note that there is also a non-resonant background in the resonance region!) B. Kilminster (Uni. Zürich) – Phenomenology of Particle Physics, FS2015 13 • ‘Continuum’ or ‘inelastic region‘: W>~1.8 GeV complicated multiparticle ﬁnal states resulting in a smooth distribution in W (Note there are also charmonium and bottonium resonances at W~3 and 9 GeV)

Tuesday 19 June 18 Kinematic phase-space Phase Space in (ν,Q2) plane Elastic scattering Halzen&Martin, x2=1 q2 =2M⌫ W 2 = M 2 Q =(2MEx)! y ! Quarks&Leptons, Fig. 9.3

Allowed kinematic region 2ME

Line of constant invariant mass

Line of constant momentum fraction

2 2 2 2 2 2 2 2 Fixed W=MR: Q =2MExy, x=Q /(Q +MR -M ) → Q = M -MR + 2MEy Hence: ﬁxed W curves are parallelB. Kilminster to W=M(Uni. Zürich) curve! – Phenomenology of Particle Physics, FS2015 14

Tuesday 19 June 18 Kinematic phase-space Phase Space in (ν,Q2) plane Elastic scattering Qx2=1=(2MEx)q2 =2yM⌫ W 2 = M 2 ! ! • The phase space is separated into a resonance region (RES) and the inelastic region at W~1.6 ... 1.8 GeV Allowed kinematic region 2ME (red line)

• LineThe phaseof constant space is separated into a W~1.8 GeV invariantdeep and amass shallow region at Q2 ~1 GeV2 (blue horizontal line)

• In global analyses of DIS data often the ME DIS cuts Q2>4 GeV2, W>3.5 GeV are employed Resonances Line of constant momentum fraction deep inelastic • The W-cut removes the large x region: ~1 GeV2 W2= M2 + Q2/x (1-x) > 3.5 GeV shallow inelastic • The Q-cut removes the smallest x: Q2 = S x y > 4 GeV2 B. Kilminster (Uni. Zürich) – Phenomenology of Particle Physics, FS2015 14

Tuesday 19 June 18 Kinematic phase-space Phase Space in (ν,Q2) plane Elastic scattering x2=1 q2 =2M⌫ W 2 = M 2 Q =(2MEx)! y !

Allowed kinematic region 2ME W~1.8 GeV Line of constant invariant mass With increasing energy E the deep inelastic region ME dominates the phase space!

deep inelastic Line of constant momentum fraction

~1 GeV2

B. Kilminster (Uni. Zürich) – Phenomenology of Particle Physics, FS2015 14

Tuesday 19 June 18 Neutrino cross sections at atmospheric ν energies NeNeWithuu trtrincreasingiinnoo crcr energyoossss Ess theeectct deepiioonn inelasticss aatt aatmtm regionooss dominatespphheerriicc nn the ee nnphaseeerrgg space!iieess

Paschos,JYY,PRD65(2002)033002

P. Lipari, hep-ph/0207172

Tuesday 19 June 18 Homework Problems

1. Recap that the allowed kinematic region for ep→eX is 0≤x≤1 and 0≤y≤1. Construct the phase space in the (ν,Q2)-plane yourself. [Ex. 8.11 in Halzen]

2. Show that Q2 = 2 E E‘ (1 - cos(θ)) = 4 E E‘ sin2(θ/2) neglecting the lepton mass. Here, the z-axis coincides with the incoming lepton direction and θ is the polar angle of the outgoing lepton with respect to the z-axis

3. Show that in the target rest frame x= [2 E E‘ sin2(θ/2)]/[M(E-E’)] still neglecting the lepton mass and the energies E, E’ are now in the target rest frame

Tuesday 19 June 18 II. Cross section for inclusive DIS (photon exchange)

Tuesday 19 June 18 The cross section for inclusive ep→eX

• Let’s consider inclusive DIS where a sum over all l′ hadronic ﬁnal states X is performed: l θ

- - e (l)+N(p) → e (l’)+X(pX) q

• The amplitude (A) is proportional to the interaction of a leptonic current (j) with a hadronic current (J): X P 1 A jµJ ⇠ q2 µ

The leptonic current is well-known µ ˆµ µ • j = l0,sl0 j l, sl =¯u(l0,sl0 ) u(l, sl) perturbatively in QED: h | | i

The hadronic current is non-pert. • J µ = X, spins Jˆµ p, s and depends on the multi-particle ﬁnal h | | pi state over which we sum:

Tuesday 19 June 18 The cross section for inclusive ep→eX

The cross section which is proportional to the amplitude squared can be factored into a leptonic and a hadronic piece: d A 2 L W µ⌫ ⇠ | | ⇠ µ⌫

MMA A** l l γµ l' γν µν Leptonic tensor L calculable in pert. theory Q2 Q2

µ ν µν Hadronic tensor J J W not calculabe in pert. theory

pX p p

Tuesday 19 June 18 The cross section for inclusive ep→eX

3 4 3 1 2 d l0 1 e µ⌫ d l0 d = AX spindQX 3 = 2 2 Lµ⌫ W 4⇡ 3 F h| | i (2⇡) 2E0 F (q ) (2⇡) 2E0 XX  • With the Møller ﬂux:

F =4 (l p)2 l2p2 =4 (l p)2 m2M 2 2S · · ' p p

• The phase space of the hadronic ﬁnal state X with NX particles:

NX 3 4 (4) d pk 4 (4) dQX =(2⇡) (p + q pX ) 3 = (2⇡) (p + q pX )dX (2⇡) 2Ek kY=1

• The amplitude with ﬁnal state X: 2 2 e µ e ⌫ A = [¯u(l0) u(l)] X J (0) N(p) A⇤ = [¯u(l) u(l0)] N(p) J †(0) X X q2 h | µ | i X q2 h | ⌫ | i

Tuesday 19 June 18 The cross section for inclusive ep→eX

3 4 3 1 2 d l0 1 e µ⌫ d l0 d = AX spindQX 3 = 2 2 Lµ⌫ W 4⇡ 3 F h| | i (2⇡) 2E0 F (q ) (2⇡) 2E0 XX  AMMA** l l In the amplitude squared appears the leptonic tensor: γµ l' γν 1 µν L = u¯(l0) u(l)¯u(l) u(l0) L µ⌫ 2 µ ⌫ Q2 Q2 sl s X Xl0 1 µ ν J J µν = Tr[ (l/ + m) (/l0 + m)] W 2 µ ⌫ p 2 X = 2[lµl⌫0 + l⌫ lµ0 gµ⌫ (l l0 m )] p p ·

The hadronic tensor is deﬁned as:

4 (4) 4⇡Wµ⌫ = dX (2⇡) (p + q pX ) N(p) J⌫†(0) X X Jµ(0) N(p) h | | ih | | i spin statesX XZ D E

Note that the factor 4� is a convention. In this case the hadronic tensor is dimensionless (Exercise!). Halzen&Martin, for example, use a factor 4�M and the hadronic tensor has dimension mass-1.

Tuesday 19 June 18 The hadronic tensor and structure functions

• Wμν(p,q) cannot be calculated in perturbation theory. q q It parameterizes our ignorance of the nucleon.

• Goal: write down most general covariant expression for Wμν(p,q)

• Other symmetries (current conservation, parity, time-reversal inv.) p p have to be respected as well, depending on the interaction

• All possible tensors using the independent momenta p, q and the metric g are:

gµ⌫ , pµp⌫ , qµq⌫ , pµq⌫ + p⌫ qµ, ✏ p⇢q, p q p q µ⌫⇢ µ ⌫ ⌫ µ

• For a (spin-averaged) nucleon, the most general covariant expression for Wμν(p,q) is: pµp⌫ p q W µ⌫ (p, q)= gµ⌫ W + W i✏µ⌫⇢ ⇢ W 1 M 2 2 M 2 3 qµq⌫ pµq⌫ + p⌫ qµ pµq⌫ p⌫ qµ + W + W + W M 2 4 M 2 5 M 2 6

2 2 2 • The structure functions Wi can depend only on the Lorentz-invariants p =M , q , and p.q

Tuesday 19 June 18 The hadronic tensor and structure functions

pµp⌫ p q W µ⌫ (p, q)= gµ⌫ W + W i✏µ⌫⇢ ⇢ W 1 M 2 2 M 2 3 qµq⌫ pµq⌫ + p⌫ qµ pµq⌫ p⌫ qµ + W + W + W M 2 4 M 2 5 M 2 6 2 2 d W4 ml d W5 ml d W6 =0 | ⇠ | ⇠ |

2 2 • Instead of p.q use ν or x as argument: Wi = Wi(ν,q ) or Wi=Wi(x,Q )

• W6 doesn’t contribute to the cross section! No (lμ qν - lν qμ) in the leptonic tensor

• W4 and W5 terms are proportional to the lepton masses squared in the cross section μ 2 since q Lμν ~ ml . Only place where they are relevant is charged current ντ-DIS.

• Parity and Time reversal symmetry implies Wμν=Wνμ

• W3=0 and W6=0 for parity conserving currents (like the e.m. current)

Tuesday 19 June 18 The hadronic tensor and structure functions

pµp⌫ p q W µ⌫ (p, q)= gµ⌫ W + W i✏µ⌫⇢ ⇢ W 1 M 2 2 M 2 3 qµq⌫ pµq⌫ + p⌫ qµ pµq⌫ p⌫ qµ + W + W + W M 2 4 M 2 5 M 2 6 2 2 d W4 ml d W5 ml d W6 =0 | ⇠ | ⇠ |

μ ν • Current conservation at the hadronic vertex requires q Wμν = q Wμν=0 implying

p q p q 2 M 2 W = · W , W = · W + W 5 q2 2 4 q2 2 q2 1 ✓ ◆

• With current conservation+parity symmetry we are left with 2 independent sfs:

qµq⌫ 1 p q p q W µ⌫ = gµ⌫ + W + pµ · qµ p⌫ · q⌫ W q2 1 M 2 q2 q2 2 ✓ ◆ ✓ ◆✓ ◆

Tuesday 19 June 18 The cross section for inclusive ep→eX

4 3 * 1 e d l0 AMMA* µ⌫ d = 2 2 Lµ⌫ W 4⇡ 3 l l F (q ) (2⇡) 2E0 γµ l' γν 

µν 2 L Lµ⌫ = 2[lµl⌫0 + l⌫ lµ0 gµ⌫ (l l0 m )] Q2 Q2 · µ⌫ µ⌫ 1 µ ⌫ µ ν W = g W + p p W J J µν 1 2 2 W ? M ? ?

µ ⌫ pX µ⌫ µ⌫ q q µ µ q p µ g = g 2 ,p = p ·2 q p p ? q ? q

µ⌫ 2 2 Show that (m=0): Lµ⌫ W = 4(l l0)W1 + 2(p l)(p l0) M l l0 W2 · M 2 · · · ⇥ ⇤ µ⌫ 2 2 Giving in the nucleon rest frame: Lµ⌫ W =4EE0 2 sin (✓/2)W1 + cos (✓/2)W2 ⇥ ⇤ The DIS cross section in the nucleon rest frame reads (photon exchange, neglecting m): d2 ↵2 = em [2W (x, Q2) sin2(✓/2) + W (x, Q2) cos2(✓/2)] 2 4 1 2 dE0d⌦0 M4E sin (✓/2)

Tuesday 19 June 18 The cross section for inclusive ep→eX

The DIS cross section in the nucleon rest frame reads (photon exchange, neglecting m): d2 ↵2 = em [2W (x, Q2) sin2(✓/2) + W (x, Q2) cos2(✓/2)] 2 4 1 2 dE0d⌦0 M4E sin (✓/2)

It is customary to deﬁne Q2 Q2 F , F , F = W , W , W “scaling” structure functions: 1 2 3 1 2xM 2 2 xM 2 3 ⇢ ⇢

Change of variables (E’,Ω’)→(x,y) and Wi →Fi:

d2 2⇡My d2 Show that = PDG’17, Eq. (19.1) dxdy 1 y dE d⌦ 0 0

The DIS cross section in terms of Lorentz-invariants x, y (photon exchange, neglecting m): d2 4⇡↵2 S = em xy2F (x, Q2)+(1 y xyM 2/S)F (x, Q2) dxdy Q4 1 2 ⇥ ⇤ PDG’17, Eq. (19.8)

Tuesday 19 June 18 Homework Problems

1. Show that the phase space for the outgoing lepton takes the following form in the variables x and y (without any approximation), where F is the ﬂux and S=2 p.l:

3 2 d l0 2S y 3 = 2 dxdy (2⇡) 2E0 (4⇡) F

2. Derive the following general expression for the doubly differential cross section:

d2 2S2y e4 4S2 2⇡↵2 = L W µ⌫ 4⇡ = y L W µ⌫ dxdy (4⇡)2F 2 Q4 µ⌫ F 2 Q4 µ⌫ 

(Note that the factor 4S2/F2 = 1+ O(m2/S * M2/S) and the mass term is negligibly small for incoming neutrinos, electrons, and muons even if the nucleon mass is taken into account.)

3. Show that the hadronic tensor in terms of the structure functions F1, F2 is given by:

µ⌫ µ⌫ 2 1 µ ⌫ 2 W = g F1(x, Q ) + p p F2(x, Q ) ? p q ? ? ·

Tuesday 19 June 18 Homework Problems

Show that the hadronic tensor can be brought in the following forms:

4 (4) 4⇡Wµ⌫ = dX (2⇡) (p + q pX ) N(p) J⌫†(0) X X Jµ(0) N(p) h | | ih | | i spin statesX XZ D E 4 iqy = dX d ye N(p) J⌫†(y) X X Jµ(0) N(p) h | | ih | | i spin statesX XZ Z D E 4 iqy = d ye N(p) J⌫†(y)Jµ(0) N(p) h | | i spin Z D E 4 iqy = d ye N(p) [J⌫†(y),Jµ(0)] N(p) h | | i spin Z D E • Use the integral representation for the delta-distribution:

4 (4) i(p+q p )y iqy i(p p )y (2⇡) (p + q p )= dy e X = dy e e X X Z Z • The space-time translation of an operator in QM is generated by the 4-momentum operator: iPˆ y iPˆ y Oˆ(y):=e · Oˆ(0)e · • Use the completeness relation: d X X = 1 X | ih | statesX X Z • The second term in the commutator leads to q+pX-p=0 violating mom. cons. q+p-pX=0!

Tuesday 19 June 18 Hadronic tensor

Optical theorem: W Im T µ⌫ / µ⌫

Im /

4 iqx T = i d xe N T [J †(x)J (0)] N µ⌫ h | µ ⌫ | i Z

Tuesday 19 June 18 III. Longitudinal and transverse structure functions

Tuesday 19 June 18 Structure functions

µ⌫ µ⌫ 2 1 µ ⌫ 2 W = g F1(x, Q ) + p p F2(x, Q ) ? p q ? ? · • The sfs are non-perturbative objects which parameterize the structure of the target as ‘seen’ by virtual photons �T* �T* • They are obtained with the help of projection operators: μν Pi Wμν = Fi ~FT

• The projectors are rank-2 tensors formed out of the independent momenta p, q and the metric g (similar to Wμν) �L* �L* • One can introduce transverse and longitudinal structure functions by contracting the hadronic tensor with the ~FL polarization vectors for transversely/longitudinally polarized virtual photons: FT, FL

• It turns out that: FT = 2xF1, F2 = FL + FT (neglecting M)

Tuesday 19 June 18 Homework Problems Halzen, Chap. 8.5

Chosing the z-axis along the three-momentum ~q, such that qµ =(q0, 0, 0, ~q ), the polarisation vectors of spacelike photons with helicity =0, 1 can| be| written as: ± 1 = 1:✏ (q)= (0, 1, i, 0) ± ± ⌥p2 ± 1 =0:✏ (q)= ( ⌫2 q2, 0, 0, ⌫) ± q2 p p

1. Verify that q ✏ =0for each , and show the following completeness relation for a space· like photon (q2 < 0): µ ⌫ +1 µ ⌫ µ⌫ q q ( 1) ✏⇤ (q)✏ (q)= g + q2 =0, 1 X± 2. Neglecting terms of order O(M 2/Q2) show that:

µ ⌫ 1 a) ✏⇤ (q)✏ (q)W = F with F = F 2xF = F F 0 0 µ⌫ 2x L L 2 1 2 T 1 µ ⌫ µ ⌫ 1 b) 2 [✏+⇤ (q)✏+(q)+✏⇤ (q)✏ (q))]Wµ⌫ = 2x FT with FT =2xF1 It is useful to do the calculation in the nucleon rest frame p =(M,0, 0, 0).

Tuesday 19 June 18 IV. CC and NC DIS

Tuesday 19 June 18 Cross section for CC and NC DIS

The differential cross section for DIS mediated by ** AMMA interfering gauge bosons B,B’ can be written as: l l γµ l' γν d2 d2BB0 = Lµν dxdy dxdy B B’ B.B0 Q2 Q2 X

µ ν B,B0 ,Z in the case of NC DIS J J Wµν • 2 { } B = B0 = W in the case of CC DIS • pX p p dBB0 LBB0 W µ⌫ ⇠ µ⌫ BB0

Each of the terms dσBB’ can be calculated from the general expression: PDG’17, Eq. (19.2)

2 2 BB0 2 4 (Q )=1 d 2S y e BB0 µ⌫ = L W 4⇡ 2 2 2 2 2 2 4 B B0 µ⌫ BB0 2 g Q GF MZ Q dxdy (4⇡) F Q Z (Q )= 2 2 2 2 = 2 2  (2 cos ✓w) e Q + M p2 2⇡↵ Q + M 2 2 Z Z 4S 2⇡↵ 2 2 2 2 BB0 µ⌫ 2 g Q GF MW Q = yBB L W W (Q )= = 2 4 0 µ⌫ BB0 2 2 2 2 2 2 F Q (2p2) e Q + MW p2 4⇡↵ Q + MW

Tuesday 19 June 18 CC ντ-DIS Albright, Jarlskog’75 Paschos, Yu’98 Kretzer, Reno’02 d2⌫(¯⌫) G2 M E m2 y = F N ⌫ (y2x + ⌧ )F W ± dx dy ⇡(1 + Q2/M 2 )2 2E M 1 W ⇢ ⌫ N m2 M x y m2 y + (1 ⌧ ) (1 + N )y F W ± xy(1 ) ⌧ ) F W ± 4E2 2E 2 ± 2 4E M 3  ⌫ ⌫  ⌫ N m2(m2 + Q2) m2 + ⌧ ⌧ FW ± ⌧ FW ± 4E2M2 x 4 E M 5 ⌫ N ⌫ N Albright-Jarlskog relations: (derived at LO, extended by Kretzer, Reno)

valid at LO [ (↵0)], M =0 O s N F =0 (even for mc = 0) 4 6

valid at all orders in ↵s, F2 =2xF5 for MN = 0, mq =0

Full NLO expressions (M =0,m = 0): Kretzer, Reno’02 N 6 c 6 Tuesday 19 June 18 SENSITIVITY TO F4 AND F5

The SHiP experiment has the unique capability of being sensitive to F4 and F5! ! Sensitivity to F4 and F5 F4 = F5 = 0 hypothesis ➙ increase of the ντ and ντ CC DIS cross sections! ➙ increase of the number of expected ντ and anti-ντ ! interactions!

F4 = F5 = 0 F4 = F5 = 0

SM prediction SM prediction

ντ CC DIS cross-section ντ CC DIS cross-section

A. Di Crescenzo - DIS 2015 17

Tuesday 19 June 18 Homework Problems

* Little research project:

Work out the cross sections for NC and CC DIS

(Find typos in the following expressions, Compare with expressions in PDF review)

Tuesday 19 June 18 Cross section for CC and NC DIS

Show that for an incoming electron with general (V A ) current the leptonic µ 5 tensor is given by (neglecting the lepton masses m1 and m2):

BB0 1 B0 B L = u¯(l, ) u(l0, 0)¯u(l0, 0) u(l, ) µ⌫ 2 ⌫ µ X X0 1 B0 B = Tr[(l/ + m ) (/l0 + m ) ] 2 1 ⌫ 2 µ µ ⌫ ⌫ µ µ⌫ ⇢ =2L [l l0 + l l0 (l l0)g ]+4iR ✏ l l0 + · l,+ µ⌫⇢

B B B B B0 B B0 B B0 B0 B Here µ = µ(Ve = Ae 5), L = Ve Ve Ae Ae , Re, = Ve Ae Ve Ae . ± ± ± ±

see Halzen&Martin

Tuesday 19 June 18 Cross section for CC and NC DIS

The weak currents are not conserved (*) and parity is violated. Therefore, one has to assume the most general structure for the hadronic tensor. In particular ⇢ one has to include a parity violating piece i✏µ⌫⇢p q : ε0123 ⇠ convention: =+1 p p p⇢q W BB0 = g F BB0 (x, Q2)+ µ ⌫ F BB0 (x, Q2) i✏ F BB0 (x, Q2) µ⌫ µ⌫ 1 p q 2 µ⌫⇢ 2p q 3 q q p· q + p q p· q p q + µ ⌫ F BB0 (x, Q2)+ µ ⌫ ⌫ µ F BB0 (x, Q2)+ µ ⌫ ⌫ µ F BB0 (x, Q2) p q 4 2p q 5 2p q 6 · · ·

The terms proportional to F4, F5 will be proportional to the lepton masses squared and are usually neglected (F6 will not contribute to the cross section at all). Of course, these terms have to be kept in the hadronic tensor when projecting out structure functions.

µ (*) With Jw =¯u(p0)µ(v a5)u(p) and using the Dirac equation one ﬁnds µ 2 2 2 2 µ⌫ q J a(m + m0) with p = m , p0 = m0 . Therefore q L lepton mass. µ w ⇠ µ ⇠

Tuesday 19 June 18 Cross section for CC and NC DIS

We are now in a position to calculate the cross section:

2 BB0 2 d 4⇡↵ S 2 BB0 2 BB0 BB0 = 0 xy L F +(1 y xyM /S)L F y(1 y/2)2R xF dxdy Q4 B B + 1 + 2 l,+ 3 h i

Introducing generalized structure functions we can form the Neutral Current (NC) cross section:

d2NC 4⇡↵2S = xy2 NC +(1 y xyM 2/S) NC y(1 y/2)x NC dxdy Q4 F1 F2 F3 ⇥ ⇤ with

Tuesday 19 June 18 V. Bjorken scaling

Tuesday 19 June 18 Expectations from elastic ep scattering

Pointlike proton without spin, neglecting recoil:

dMott ↵2 = cos2(✓/2) d⌦ 4E2 sin4(✓/2)

electron spin Rutherford scattering

Tuesday 19 June 18 Expectations from elastic ep scattering Electron-proton elastic scattering: Pointlike proton without spin, neglecting recoil:

dMott ↵2 = cos2(✓/2) d⌦ 4E2 sin4(✓/2)

Pointlike proton with spin:

d dMott E = 0 [1 + 2⌧ tan2(✓/2)] d⌦ d⌦ E

HEP Lecture 8 7

2 Q 2 2 ⌧ = ,Q =4EE0 sin (✓/2) 4M 2

Tuesday 19 June 18 Expectations from elastic ep scattering Electron-proton elastic scattering: Pointlike proton without spin, neglecting recoil:

dMott ↵2 = cos2(✓/2) d⌦ 4E2 sin4(✓/2)

Pointlike proton with spin:

d dMott E = 0 [1 + 2⌧ tan2(✓/2)] d⌦ d⌦ E

HEP Lecture 8 7 Extended proton with spin (Rosenbluth formula):

Mott 2 2 Q2 d d E0 GE + ⌧GM 2 2 ⌧ = ,Q2 =4EE sin2(✓/2) = +2⌧G tan (✓/2) 2 0 d⌦ d⌦ E 1+⌧ M 4M 

2 • Elastic form factor GE(Q ), GE(0)=1 Steeply falling form factors: 2 2 GM (Q ) 2 2 2 GE(Q )= =(1+Q /a ) 2 µ Magnetic form factor GM(Q ), GM(0)=μp=2.79 p • 2 2 μp=2.79: proton anomalous magnetic moment a =0.71 GeV

Tuesday 19 June 18 Expectations from elastic ep scattering

Pointlike proton without spin, neglecting recoil: Note that the idea of a point-like strongly interacting particle is rather academic! dMott ↵2 = cos2(✓/2) d⌦ 4E2 sin4(✓/2) Due to quantum corrections we have to generalize the ‘point-like current’ by the most general current respecting all symmetries of the Pointlike proton with spin: interaction and introduce form factors.

Mott d d E0 This is even the case in QED. However, here the = [1 + 2⌧ tan2(✓/2)] d⌦ d⌦ E Dirac and Pauli form factors are calculable in perturbation theory.

Extended proton with spin (Rosenbluth formula):

Mott 2 2 Q2 d d E0 GE + ⌧GM 2 2 ⌧ = ,Q2 =4EE sin2(✓/2) = +2⌧G tan (✓/2) 2 0 d⌦ d⌦ E 1+⌧ M 4M 

Tuesday 19 June 18 What do we expect for a point-like particle?

Point-like proton without spin, neglecting recoil: Point-like proton/muon with spin:

Mott dMott ↵2 d d E0 2 = cos2(✓/2) = [1 + 2⌧ tan (✓/2)] d⌦ 4E2Whatsin4( ✓do/2) we expect for a point like particled⌦ d⌦ E

Fred Olness, CTEQ school 2012

Dimensional considerations Structure Function

Tuesday 19 June 18 Expectations from elastic ep scattering

HEP Lecture 8 17 Fig. 23. Summary of results on nuclear form factors presented by the Stanford group at the 1965 “International Symposium on Electron and Photon Interactions at High Energies”. (A momentum transfer of 1 GeV2 is equivalent to 26 Fermis-2.)

increasing data was a powerful way to check on the progress of the experiments (Fig.The 22). results formed the prejudice that the proton was a soft “mushy” extended object, In the summerpossibly of 1966 with there a was hard a call core for proposals surrounded to use the beam by aat cloud of mesons, mainly pions. SLAC. The accelerator was nearing completion, and some early tests of the accelerator with beam were being done with considerable success. Although the initial programsThe SLAC-MIT in End Station team A were saw built intoits objectivethe design of thein searching for the hard core of the proton. facility, it wasFirst now DIS necessary experiments to parcel out beam(>=1967). time and arrange the sequence of experiments for the first year of operation. The Cal Tech-MIT- SLAC collaboration prepared a proposal that consisted of three parts: a. Elastic electron-proton scattering measurements (8 GeV spectrometer) Tuesdayb. Inelastic 19 June 18electron-protron scattering measurements (20 GeV spectrometer) c. Comparison of positron and electron scattering cross sections (8 GeV spectrometer) It is clear from the proposal that the elastic experiment was the focus of interest at this juncture. “We expect that most members of the groups in the collaboration will be involved in the e-p elastic scattering experiment, and that the other experiments will be done by subgroups.” During the construction of SLAC and the experimental facilities a lot of progress had been made on the measurements of nucleon form factors at other laboratories. The program at HEPL had continued to produce a great Bjorken scaling

Early data on DIS from the SLAC-MIT experiment Elastic scattering (Rosenbluth formula): [PRL23(1969)935]

d dMott E G2 + ⌧G2 = 0 E M +2⌧G2 tan2(✓/2) d⌦ d⌦ E 1+⌧ M 

The DIS cross section resembles the elastic one:

d dMott[W +2W tan2(✓/2)] DIS ⇠ 2 1 σDIS depends only mildly on Q2

The form factors had been know to fall rapidly as a function of Q2.

Therefore, the general expectation for σDIS before its measurement was that it also would be a fast falling function of Q2.

Tuesday 19 June 18 Bjorken scaling

Scaling hypothesis (Bjorken 1968):

In the limit Q2→∞, ν→∞, such that x=Q2/(2Mν) is ﬁxed (‘Bjorken limit’) 2 2 the structure functions Fi(x,Q ) are insensitive to Q : Fi=Fi(x)

This behaviour is called scaling and x is called the scaling variable

Tuesday 19 June 18 Bjorken scaling

Scaling hypothesis (Bjorken 1968):

In the limit Q2→∞, ν→∞, such that x=Q2/(2Mν) is ﬁxed (‘Bjorken limit’) 2 2 the structure functions Fi(x,Q ) are insensitive to Q : Fi=Fi(x)

This behaviour is called scaling and x is called the scaling variable

Scaling implies that the nucleon appears as a collection of point-like constituents when probed at very high energies (Q2 large).

The possible existence of such point-like constituents was also proposed by Feynman from a different theoretical perspective and he gave them the name ‘partons’.

Tuesday 19 June 18 Structure of the proton Structure of the Proton 11 Fred Olness, CTEQ school 2012

L of order of the proton mass scale

Tuesday 19 June 18 Bjorken scaling for F2 The Scaling of the Proton Structure Function 12 Fred Olness, CTEQ school 2012 Varies with energy

Data is (relatively) small x independent of energy

x~0.2

Scaling large x Violations observed at extreme x Varies with energy values

Tuesday 19 June 18 Bjorken scaling for FL

ZEUS collab,ZEUS arXiv:0904.1092 The HERA combined measurement of FL 2 Q2 = 24 GeV2 Q2 = 32 GeV2

& F 1.5 1.5 is compatible with scaling 2 L F

1.0 1.0 I. INTRODUCTION H1 and ZEUS 0.5 0.5 L F x 0.000059 0.000087 0.000128 0.000168 0.000231 0.000324 0.000404 0.000534 0.000641 0.000855 0.001215 0.001599 0.002113 0.002927 0.003795 0.005420 0.006843 0.009109 0.012250 0.020170 0.026220 0.032270 0.0 A. Motivation0.0 0.4 2 -3 -3 Q2 = 45 GeV2 Q2 = 60 GeV2

& F 1.5 1.5 L The productionF of heavy quarks in high energy pro- 0.2 cesses has become1.0 an increasingly important1.0 subject of study both theoretically0.5 and experimentally.0.5 The the- 0 ory of heavy quark production in perturbative Quan- HERA preliminary ACOT full b-Sat dipole 0.0 0.0 HERAPDF1.0 ACOT-χ b-CGC dipole Q2≥ 3.5 GeV2 FFNS tum Chromodynamics (pQCD) is more challenging than 2 2 2 -3 -3 Q ≥ 5 GeV NNLO α =0.1176 2 2 2 2 -0.2 S that of light parton (jet) productionQ = 80 GeV because ofQ the = 110 new GeV RT optimized NNLO α =0.1146 & F 1.5 1.5 S HERA Inclusive Working Group April 2010 L

F F physics issues brought about by the additional2 heavy 2 F 10 10 2 2 1.0 1.0 L Q / GeV quark mass scale. The correct theory mustZEUS-JETS properly take into account the0.5 changing role of the0.5 heavy quark over the full kinematic0.0 range of the relevant0.0 process from the Figure 1: FL vs. Q from combined HERA-I inclusive threshold region (where the quark behaves like a typical 10-3 10-2 10-3 10-2 deep inelastic cross sections measured by the H1 and “heavy particle”) to the asymptoticx region (where thex ZEUS collaborations. Figure taken from Ref. [2]. same quark behaves eﬀectively like a parton, similar to

2 the well knownFigure light 3: FL quarksand F2 at 6 valuesu, d, of Q s as). a function of x.Thepointsrepresentthe ZEUS data for F ( )andF{ (!), respectively.} The error bars on the data represent With the ever-increasingWe noteL • that2 precision FL is quite of smaller experimental than F2. the combined statistical and systematic uncertainties. TheerrorbarsonF2 are Since the first non-vanishing contribution to FL data and thesmaller progression than the symbols. of theoreticalA further 2.5%correlatednormalisationuncertaintyis calculations and • not included. The DGLAP-predictions± for F and F using the ZEUS-JETS PDFs is next-to-leading order (up to mass effects), the L 2 3 parton distributionare also shown. function The bands (PDF) indicate the evolution uncertainty in tothe pre next-to-dictions. NNLO and N LO corrections are more important next-to-leadingTuesday 19 order June 18 (NNLO) of QCD there is a clear than for F2. need to formulate and also implement the heavy quark schemes at this order and beyond. The24 most important In Fig. 1 we show a comparison of different theoretical case is arguably the heavy quark treatment in inclu- calculations of FL with preliminary HERA data [2]. As 2 sive deep-inelastic scattering (DIS) since the very pre- can be seen, in particular at small Q (i.e. small x), there 2 cise HERA data for DIS structure functions and cross are considerable differences between the predictions. sections form the backbone of any modern global anal- The purpose of this paper is to calculate the leading ysis of PDFs. Here, the heavy quarks contribute up to twist neutral current DIS structure functions F2 and FL 3 30% or 40% to the structure functions at small momen- in the ACOT factorization scheme up to order (αS ) (N3LO) and to estimate the error due to approximatingO tum fractions x.Extendingtheheavyquarkschemesto 2 2 2 m 3 m higher orders is therefore necessary for extracting precise the heavy quark mass terms (α 2 ) and (α 2 ) O S × Q O S × Q PDFs and hence for precise predictions of observables at in the higher order corrections. The results of this study the LHC. However, we would like to also stress the theo- form the basis for using the ACOT scheme in NNLO retical importance of having a general pQCD framework global analyses and for future comparisons with precision including heavy quarks which is valid to all orders in per- data for DIS structure functions. turbation theory over a wide range of hard energy scales and which is also applicable to other observables than inclusive DIS in a straightforward manner. B. Outline of Paper An example, where higher order corrections are par- ticularly important is the structure function FL in DIS. The rest of this paper is organized as follows. In Sec. The leading order ( (α0 ))contributiontothisstructure O S II we review theoretical approaches to include heavy fla- function vanishes for massless quarks due to helicity con- vors in QCD calculations. Particular emphasis is put servation (Callan-Gross relation). This has several con- on the ACOT scheme which is the minimal extension sequences: of the MS scheme in the sense that the observables in the ACOT scheme reduce to the ones in the MS scheme FL is useful for constraining the gluon PDF via the • dominant subprocess γ∗g qq¯. in the limit m 0 without any finite renormalizations. → In this discussion→ we explicitly distinguish between the m2 The heavy quark mass effects of order ( 2 ) are heavy quark/heavy meson mass entering the final state • O Q relatively more pronounced.1 phase space which we will call “phase space mass” and the heavy quark mass entering the dynamics of the short

1 Similar considerations also hold for target mass corrections (TMC) and higher twist terms. We focus here mainly on the 2 An updated analysis of the H1 measurements extending down kinematic region x<0.1 where TMC are small [1]. An inclu- to even lower Q2 values has been published in Ref. [3], and a sion of higher twist terms is beyond the scope of this study. combined analysis with ZEUS is in progress. VI. The Parton Model

Tuesday 19 June 18 Naive parton model

• The historical (‘naive’) parton model describes the nucleon as a collection made of point-like constituents called ‘partons’.

Today the Feynman partons are understood to be identical with the quarks postulated by Gell-Mann.

• At high momentum (‘inﬁnite momentum frame’) the partons are free. Therefore, the interaction of one parton with the electron does not affect the other partons. This leads to scaling in x=Q2/(2 M ν) as we will see below.

• In an IMF, the proton is moving very fast, P~(Ep,0,0,Ep) with Ep>>M.

The quark-parton is moving parallel with the proton, carrying a fraction ξ of its momentum: pˆ = ⇠P

Tuesday 19 June 18 Naive parton model

l’ l’ l l q q

⇠P p’ pˆ =

P P

1 d(e(l)+N(P ) e(l0)+X(p )) = d⇠f (⇠)dˆ(e(l)+i(⇠P ) e(l0)+i(p0)) ! X i ! i 0 X Z

Tuesday 19 June 18 Naive parton model

l’ l’ l l q q

⇠P p’ pˆ =

P P

1 d(e(l)+N(P ) e(l0)+X(p )) = d⇠f (⇠)dˆ(e(l)+i(⇠P ) e(l0)+i(p0)) ! X i ! i 0 X Z

Elastic scattering off point-like parton

We know how to calculate this!

Tuesday 19 June 18 Naive parton model

l’ l’ l l q q

⇠P p’ pˆ =

P P

1 d(e(l)+N(P ) e(l0)+X(p )) = d⇠f (⇠)dˆ(e(l)+i(⇠P ) e(l0)+i(p0)) ! X i ! i 0 X Z

Parton density: Elastic scattering off ﬁ(ξ)dξ = number of partons of point-like parton type ‘i’,carrying a momentum fraction of the proton We know how to momentum in [ξ, ξ+dξ] calculate this!

Tuesday 19 June 18 Naive parton model

l’ l’ l l q q

⇠P p’ pˆ =

P P

1 d(e(l)+N(P ) e(l0)+X(p )) = d⇠f (⇠)dˆ(e(l)+i(⇠P ) e(l0)+i(p0)) ! X i ! i 0 X Z

Incoherent sum Parton density: Elastic scattering off over all possible ﬁ(ξ)dξ = number of partons of point-like parton partonic processes type ‘i’,carrying a momentum fraction of the proton We know how to momentum in [ξ, ξ+dξ] calculate this!

Tuesday 19 June 18 Naive parton model

μ ν μ ν q q q q Partonic tensor: wˆµ⌫ calculable, not IR safe

pˆ = ⇠P pˆ = ⇠P ⌦ Parton distribution: P P fi(⇠) not calculable, but universal P P

1 d⇠ W µ⌫ (P, q) = f (⇠) wˆµ⌫ (⇠P, q) ⇠ i i i x X Z

Tuesday 19 June 18 VII. Which partons?

Tuesday 19 June 18 Which Partons?

• It is tempting to identify the partons with the quarks of Gell-Mann’s quark model 24 ProtonProton as as a a bag bag of of free free Quarks: Quarks: Part Part 2 2 24 • The proton consists of three valence quarks (uud) which carry the quantum numbers of the proton (electric charge, baryon number): u=uv, d=dv

PerfectPerfect Scaling Scaling PDFs PDFs QQ independent independent

QuarkQuark Number Number Sum Sum Rule Rule

QuarkQuark Momentum Momentum Sum Sum Rule Rule

Tuesday 19 June 18 Which Partons?

• The valence quarks are bound together by gluons which leads to a smearing of the delta distributions and the gluon can ﬂuctuate into a ‘sea’ of quark- anti-quark pairs: S(x)=ubar(x)=dbar(x)=s(x)=sbar(s)

• u(x)=uv(x)+ubar(x),Gluons d(x)=dv(x)+dbar(x), smear out PDF momentum ubar(x), dbar(x), s(x), ...26

Gluons allow partons to exchange momentum fraction

n n

o o

i i

t t

u u

b b

i i

r r

t t

s s

i i

D D

n n

o o

t t

r r

a a

P P

Momentum Fraction x Momentum Fraction x

Tuesday 19 June 18

aS is large at low Q, so it is easy to emit soft gluons Which Partons?

• Problem, experimentally it is found that the momentum fractions of the quark partons do not add up to 1!

1 dx x[q (x)+¯q (x)] 0.5 i i ⇠ i 0 X Z

• Gluons carry half of the momentum but don’t couple to photons

• We have to include a gluon distribution G(x)

Tuesday 19 June 18 Which Partons? In summary, the following picture emerges: • The proton consists of three valence quarks (uud) which carry the quantum numbers of the proton (electric charge, baryon number): uv, dv

• There is a sea of light quark-antiquark pairs: u ubar, d dbar, s sbar, ...

When probed at a scale Q, the sea contains all quark ﬂavours with mass mq<

• u(x) = uv(x) + ubar(x), d(x) = dv(x) + dbar(x)

• There is gluon distribution G(x) which carries about 50% of the proton momentum

1 1 1 Quark number sum rule: dxu (x)=2, dxd (x)=1, dx(s s¯)(x)=0 v v Z0 Z0 Z0

1 Momentum sum rule: dx x [qi(x)+¯qi(x)] + G(x) =1 0 ( i ) Z X

Tuesday 19 June 18 VIII. Structure functions in the parton model