Partice Physics: Problem Sheet 4 QCD, Gluons, Partons and the Weak Force

2010 — Subatomic: Particle Physics 1

Partice Physics: Problem Sheet 4 QCD, gluons, partons and the weak force

1. At very high beam momenta, p ∼ 100GeV/c, the total cross sections for π+p and pp scattering is dominated by the exchange of a gluon between quarks inside the pions and protons.

(a) Draw a Feynman diagram for pp and π+p scattering. (b) Use the number of possible diagrams to calculate the ratio of cross sections σ(π+p)/σ(pp). Hint: every (diﬀerent) diagram you can draw is equally likely.

For π+p scattering there are two (anti-)quarks in π+ that can exchange a gluons with any of the three quarks in the proton. Therefore the number of allowed interactions is proportional to 2 × 3. For pp scattering there are three quarks in the ﬁrst proton that can exchange a gluons with any of the three quarks in the second proton. Therefore the number of allowed interactions is proportional to 3 × 3.

Prediction Experiment σ(π+p) 2 × 3 2 σ(π+p) 24 mb = = ≈ = 0.63 σ(pp) 3 × 3 3 σ(pp) 38 mb 2010 — Subatomic: Particle Physics 2

2. Draw a Feynman diagram for the process pπ+ → ∆++ → pπ+.

You have to think of the constituents quarks. We have uud+du¯ → uuu → uud+du.¯ The down and anti-down quarks annihilate by producing a gluon. The gluon can be absorbed by any of the quarks. Remember as long as the ﬂavour of the quark doesn’t change, a gluon can be emitted or absorbed by any quark. After about 10−24 s (any) one of the quarks emits a gluon which turns into a dd¯ pair. It decays so quickly as the strong coupling constant is large, so the quarks are very likely to emit an energetic enough gluon to make a dd,¯ and because the minimum energy state is favoured. The mass of the pion and the proton is smaller than the mass of the ∆++. This looks like a pretty silly Feynman diagram, but the uuu state actually lives for slightly longer than just three quarks would if they didn’t form a bound state.

3. The φ meson decays via the strong force as follows:

φ → K+K− 49% → K0K¯ 0 34% → π+π−π0 17%

What is the amount of kinetic energy produced in the decay, for these decays (also know as the Q-value of the decay). Draw Feynman diagrams of the above decays and explain why the decays to kaons is favoured despite the low Q-value? Hint: You have to consider the colour charge of the gluons.

4. What is meant by colour conﬁnement? Colour conﬁnement describes the phenomena where quarks are only found in hadrons. The gluons exchanged between the quarks self interact. The potential between two quarks, due to the gluon interactions is V (r) = kr. The energy stored between the quarks increases as they are pulled apart! 2010 — Subatomic: Particle Physics 3

MSTW 2008 NLO PDFs (68% C.L.)

) 1.2 ) 1.2 2 2

Q2 = 10 GeV2 Q2 = 104 GeV2

xf(x,Q 1 xf(x,Q 1 g/10 g/10 0.8 0.8

0.6 u 0.6 b,b u

d 0.4 d 0.4 c,c

c,c s,s 0.2 s,s d 0.2 d u u

0 0 -3 -3 10-4 10 10-2 10-1 1 10-4 10 10-2 10-1 1 x x

5. The figureFigure 1:shows MSTW the 2008 (predicted) NLO PDFs parton at Q2 =10GeV distribution2 and Q function2 =104 GeV for2 the. proton for a momentum transfer (boson momentum) of Q2 = 104 GeV, withwhich broader is grid roughly coverage appropriate in x and Q2 forthan proton-proton in previous sets. collisions at the LHC. In this paper we present the new MSTW 2008 PDFs at LO, NLO and NNLO. These sets are (a) At this value of Q2, what is the most commonly found parton in the amajorupdatetothecurrentlyavailableMRST2001LO[15],MRST 2004 NLO [18] and MRST proton? 2006 NNLO [21] PDFs. The “end products” of the present paper are grids and interpolation code for theThe PDFs, gluon - whichxf for can the be gluon found is larger at Ref. than [27]. forAn any example other parton. isgiveninFig.1,which shows(b) theThe NLO ”hard PDFs scatter” at scales of isQ the2 =10GeV term used2 and forQ the2 =10 collision4 GeV2,includingtheassociated of two partons. one-sigmaIn (68%) one confidence particular level collision (C.L.) eventuncertainty at the bands. LHC, a gluon with x = 0.1 The contentscollides of with this paper an up are quark as√ follows. with Thex = new 0.2. experiment What isal the information effective iscollsion summarised in Section 2.energy An overview of this of theevent, theoreticalsˆ. framework is presented in Section 3 and the treatment

of heavy flavoursThe energy is explained of a proton in Section in the 4. LHC In Section is Ep = 5 7 we TeV. present the results of the global fits and in SectionWe 6 we can explain write the the improvements four-momentum made ofthe in the gluon error as: propagation of the experimental data to the PDF uncertainties, and their consequences. Then we present a more detailed discussion of the description of different datap sets= (E includedg/c, ~pg) in= (thexgE globalp/c, 0,fit:0, x inclusivegpp) DIS structure functions g (Section 7), dimuon cross sections from neutrino–nucleon scattering (Section 8), heavy flavour DIS structurewhere functionsxg = 0.1 (Section and pp ≈ 9),Ep low-energy/c = 7 TeV/c Drell–Yanand we haveproduction chosen (Section the direction 10), W of and Z productionthe at proton the Tevatron beam to (Section be along 11), the positive and inclusivez-axis. jet production at the Tevatron and at HERASimilarly (Section we 12). can In write Section the 13energy we discuss of the up the quark low-x as:gluon and the description of the longitudinal structure function, in Section 14 we compare our PDFs with other recent sets, p = (xuEp/c, 0, 0, −xupp) and in Section 15 we present predictionsu for W and Z total cross sections at the Tevatron and LHC. Finally, we conclude in Section 16. Throughout the text we will highlight the numerous where x = 0.2. Note the minus sign in the third entry as the up quark and refinements and improvementsu made to the previous MRST analyses. gluon are travelling in different directions. √ The effective collision energy, sˆ, is defined as: 5 2 sˆ = p + p g u = p2 + p2 + 2p p g u g u 2 2 2 2 2 = muc + mgc + 2(Eg Eu/c − ~pg · ~pu) 2 2 2 2 ≈ 2(xg xuEp /c + xg xuEp /c ) 6 41. Plots of cross sections and related quantities

+ σ and R in e e− Collisions

-2 10 ω φ J/ψ -3 10 ψ(2S) ρ Υ -4 ρ ! 201010 — Subatomic: Particle Physics Z 4

-5 10 Where we have set the masses of the up quark and gluon to be zero (as they [mb]

σ -6 are much smaller than the energy of the proton and substituted pp = Ep/c (i.e. 10 ignoring the proton mass). We can write this in terms of the centre of mass energy of the proton-proton collision, s = 14 TeV: -7 10 2 2 sˆ = 4 xg xu Ep /c -8 = xg xu s 10 2 1 10 10

Υ 3 10 J/ψ ψ(2S) Z

10 2 φ R ω 10 ρ!

1 ρ -1 10 2 1 10 10 √s [GeV]

+ + + + Figure 41.6: World data on the total cross section of e e− hadrons and the ratio R(s)=σ(e e− hadrons, s)/σ(e e− µ µ−,s). + → → → + σ(e e− hadrons, s)istheexperimentalcrosssectioncorrectedforinitialstateradiationandelectron-positronvertexloops,σ(e e− + → 2 → µ 6.µ−,sThe)=4πα above(s)/3s.Dataerrorsaretotalbelow2GeVandstatisticalabove2GeV.Thecurvesareaneducativeguide:thebrokenone plot shows the measured value of cross section ratio, R, defined (green)as: is a naive quark-parton model prediction, and the solid one (red) is 3-loop pQCD prediction (see “Quantum Chromodynamics” section of this Review,Eq.(9.7) or, for more details,σ(e K.+e G.− Chetyrkin→ hadrons)et al.,Nucl.Phys.B586σ(e,56(2000)(Erratum+e− → q¯q) ibid. B634,413(2002)).Breit-Wigner parameterizations of J/ψ, ψ(2RS),= and Υ(nS),n =1, 2, 3, 4arealsoshown.Thefulllistofreferen≈ ces to the original data and the details of the R ratio extraction from them canσ be(e found+e− in →[arXiv:hep-ph/0312114]µ+µ−) σ(.Correspondingcomputer-readabledatafilesareavailableate+e− → µ+µ−) http://pdg.lbl.gov/current/xsect/.(CourtesyoftheCOMPAS(Protvino)andHEPDATA(Durham)Groups,May2010.)Seefull-color√ version on color pages at end of book. √as a function of the centre of momentum energy, s. The jumps at s ∼ 3 GeV and ∼ 10 GeV are due to the production of extra quarks. In the range: √ • 2 < s/GeV < 4, when u, d and s quarks can be produced, R ∼ 2. √ • 4 < s/GeV < 10, u, d, s and c quarks can be produced , R ∼ 3.3 √ • 10 < s/GeV < 30, u, d, s, c and b quarks can be produced, R ∼ 3.7 √ • (For s > 30 GeV weak force processes also take place, we won’t therefore consider this here.) Can you explain the observed values of R? Remember that e+e− → q¯q √should include all possible q¯qfinal states that can be produced for a given s. This questions follows up from Q5 on problem sheet 5. The answer to that question is that the ratio of the cross section: 2 4 4 σ(e+e− → qq) Qqe q ∝ = Q2 σ(e+e− → µ+µ−) e4q4 q 2010 — Subatomic: Particle Physics 5

where Qq = +2/3 or −1/3, depending on the charge of the quark. This is for one ﬂavour of quark–anti-quark pair. A key point here is that it is possible to make more than one ﬂavour of quark–anti-quark pair. If there is enough energy to make u, d and s quarks (but not any heavier quarks), then u, d and s quarks will all be produced. Therefore you have to sum up the probability to make u quarks and d quarks and s quarks. Moreover, the fundamental objects are not quarks, but red quarks, blue quarks and green quarks. Therefore you actually have to sum up the probability to make red up quarks, blue up quarks, green up quarks, red down quarks, blue down quarks, green down quarks, red strange quarks, blue strange quarks and green strange quarks, or Nc = 3 times the number you originally thought of. Therefore the answers are:

• For u, d and s quarks: " # 12 22 12 R = N + + = 2 c 3 3 3

• For u, d, s and c quarks: " # 12 22 12 22 10 R = N + + + = c 3 3 3 3 3

• For u, d, s, c and b quarks: " # 12 22 12 22 12 11 R = N + + + + = c 3 3 3 3 3 3

Where Nc = 3 is the number of colours. These agree with very well with the plot. This was one observation that was made that supported the idea of the colour quantum number, and that the number of colours is three.