Elementary Particle Physics - uni-tuebingen.de · Elementary Particle Physics Valery Lyubovitskij Institute of Theoretical Physics, Tubingen University,¨ Kepler Center for Astro

Elementary Particle Physics

Valery Lyubovitskij

Institute of Theoretical Physics, Tubingen University,

Kepler Center for Astro and Particle Physics, Germany

Department of Quantum Field Theory, Tomsk State University, Russia

Mathematical Physics Department, Tomsk Polytechnic University

[email protected], [email protected]

[email protected], [email protected]

http://www.tphys.physik.uni-tuebingen.de/lyubovitskij/

Course “Elementary Particle Physics”

for students, PhDs and PostDocs at TSU and TPU

– p. 1

I . Introduction

• Particle Physics — Structure of Atomic Nucleus

(bound state of protons and neutrons)

• Proton p (Q = +1 particle) Rutherford (1919)

(experiments with Helium ions 4He (α-particles) bombarding 14N atoms)4He+14N →17O + p

• Neutron n (Q = 0 particle) Chadwick (1932)

(experiments with α-particles bombarding Beryllium Be)9Be+4He→12C + n

• Electron e (Q = −1 particle) Thomson (1897)

(experiments with negatively charged cathode rays)

– p. 2

I . Introduction

• Two main problems of particle physics in the beginning of XX century

• 1. Nature of nuclear forces

• 2. Nature of β-decay: neutrons converse to protons

• Elementary Particles — structureless particles

with quantum numbers: mass m, charge Q, spin S, isospin I, etc.

• Typical size — nucleon size 1 fm = 10−15 m

• Typical energy — nucleon mass 1 GeV = 109 eV = 1.6× 10−10 J

• Planck constant ~ = h/2π = 6.582× 10−25 GeV sec

• Conversion constant ~c = 0.197 GeV fm

• 1 fm = 5.068 GeV−1 , 1 s = 1.519× 1024 GeV−1 at ~ = c = 1

– p. 3

I . Introduction• Prediction of photon γ — Planck (1900)

electromagnetic radiation of black body is quantized

it comes in discrete amounts

• Quantum of light (the photon) — Einstein (1905),

which behaves like a particle

Electromagnetic radiation — flow of quanta (photons)

Particle-wave duality of photons

• Discovery of nucleus — Geiger, Marsden (1909)

Atoms have a small, dense, positively charged nucleus

• Theory of atomic structure — Bohr (1913)

• Experimental proof of Einstein theory — Millikan (1912-1915)

• Discovery of proton — Rutherford (1919)

• Strong forces bounding nucleus — Chadwick and Bieler (1921)

• Another experimental proof of photon — Compton (1922)

Compton effect (light-electron scattering) γ + e− → γ + e−

– p. 4

I . Introduction• Exclusion principle for electrons in an atom — Pauli (1925)

• The term “photon” — Lewis (1926)

• Dirac equation for electron — Dirac (1928)

Dirac equation combining Quantum Theory and Special Relativity

describes the behavior of an electron moving at relativistic speed

• Neutrino — Pauli conjecture (1930)

massless, does not interact with matter

• Antiparticle — Dirac (1931)

Positrons — like electrons but positively charged

• Solution of β-decay puzzle

n→ p+ e− + νe

νe has spin J = 1/2 and antiparticle νe

• Discovery of positron — Anderson (1932)

in cosmic rays

• Discovery of neutron — Chadwick (1932)

– p. 5

I . Introduction

• Meson theory of nuclear forces — Yukawa (1933)

Exchange of new particles (mesons called “pions”) between protons and neutrons

mπ ≃ mN/7 = 140 MeV

3 types of pions:

Charge transitions n+ π+ → p , n→ p+ π−

Neutral transitions p+ π0 → p , n+ π0 → n

Isospin I 1/2 for nucleons, 1 for pions

• Theory of beta decay based on weak interaction — Fermi (1933)

– p. 6

I . Introduction

• Discovery of muons — Anderson, Neddermeyer (1936)

the same charge -1, spin 1/2,

magnetic moments in Bohr magnetons µB = e~/2me

µe = 1.001159 . . .± 10−11 µB , µµ = 1.001159 . . .± 10−19 µB

µe ≃ µµ ≃ µB

muon is heavier — mµ/me ≃ 207

e− is stable with τe > 2.7× 1023 years

µ− is nonstable with τ ≃ 2.2× 10−6 s

dominant decay µ− → e− + νe + νµ

• Discovery of pions — Powell (1947)

in cosmic rays

• Discovery of νe and νe — Cowan, Rynes (1956)

Inverse β-decay

n+ νe → p+ e− , p+ νe → n+ e+

– p. 7

I . Introduction

• Interactions of elementary particles

• Gravitational Interaction

VGR(r) = −γm1m2

r, γ - gravitational constant

αGR =γm2

p

~c= 5.76× 10−36

Radius is infinitely large, effects are negligible for m ∼ 1 GeV

Comparable at Planck scale 10−33 cm

• Weak Interaction

β-decay n→ p+ e− + νe, µ-capture p+ µ− → n+ νµ

Flavor transitions in nuclei, atoms

αweak = GFm2p ≃ 10−5

where GF is the Fermi constant

– p. 8

I . Introduction

• Electromagnetic Interactions

Vem(r) = αemq1q2

r, αem =

e2

4π=

1

137.036

• Strong Interactions

Vstr(r) = −αstre−Mπr

r, αstr =

g2str4π

= 10

r ∼ 1/Mπ ∼ 1 fm = 10−13 cm

– p. 9

I . Introduction

• Hadrons — strong interacting particles: Mesons and Baryons

– p. 10

I . Introduction

– p. 11

I . Introduction

• Discovery of strange particles (end of 1940th)

K mesons, Σ+, Σ0, Σ−, Λ0 hyperons

• Violation of isospin 4 Kaons decay into 3 pions

K+ → π+π0 (∆I = 1/2 rule)

• Discovery of quarks — Gell-Mann, Zweig (1964)

• Supported by SLAC exp — (1967)

Feynman Parton model

• Discovery of color — Bogoliubov, Greenberg, Nambu et al. (1964-1965)

• Discovery of c quark hadrons, mc = 1.2− 1.7 GeV — (1974)

• Discovery of τ lepton, mτ = 1.776 GeV — (1975)

• Discovery of b quark hadrons, mb ≃ 4.5− 5.2 GeV — (1977)

• Discovery of t quark, mt ≃ 172 GeV — (1994)

– p. 12

I . Introduction• Quantum numbers (QN) of elementary particles

• Fundamental QN: Mass m

Electric Charge Q (+2/3 for u, c, t and -1/3 for d, s, b),

-1 for leptons e−, µ−, τ−, 0 for neutrinos)

Spin J (0, 1, 2, . . . for bosons and 1/2, 3/2, 5/2, . . . for fermions)

Baryon Charge: B = 1 (baryons), B = −1 (antibaryons),

B = 0 mesons, B = 1/3 (quarks), B = 0 (leptons)

• Specific QN: Isospin I — quantum number of u, d quarks (for others is 0)

Hypercharge Y = B (for light u, d quarks)

Gell-Mann-Nishijima (GMN) formula Q = I3 + Y/2

Y = B + S (for s quark), S — strangeness (-1 for s and +1 for s)

Y = B + S + C (for c quark), C — charm (+1 for c and -1 for c)

Y = B + S + C + b (for b quark), b — beauty (-1 for b and +1 for b)

Y = B + S + C + b+ T (for t quark), T — truth (+1 for t and -1 for t)

Generalized GMN formula

Q = I3 + Y/2, Y = B + S + C + b+ T

– p. 13

I . Introduction• P -parity under ~x→ −~x

Pure Scalar mesons JP = 0+, Pseudoscalar JP = 0−,

Vector mesons JP = 1−, Axial-vector JP = 1+,

Nucleons JP = 12

+, etc.

• C-parity under charge conjugation

pure Scalars JPC = 0++,

Pseudoscalar mesons JPC = 0−+ good for π0 with π0 → π0, π± → π∓

Vectors JPC = 1−− good for ρ0 with ρ0 → −ρ0, ρ± → −ρ∓

Axial-vectors JPC = 1++

Baryons (no C-parity, because always particle → antiparticle)

• G-parity results from the product of

C parity and operator eiπI2 under charge conjugation

Scalars G = +1 or IG = 0+, Pseudoscalars (pions) G = −1 or IG = 1−,

Vectors (ρ mesons) G = +1 or IG = 1+, Axials (a1(1270) meson) G = −1 or

IG = 1−

Not for baryons

– p. 14

I . Introduction

• Towards to Standard Model UY (1)× SUL(2)× SUc(3)

electroweak + strong interactions

• Electroweak theory — Weinberg-Salam (1967)

γ and Z0 — mixing of singlet hypercharge B and neutral isotriplet boson state W 3

• W± = W1∓iW2√2

—- charged weak bosons

Color charge of quarks — Fritsch, Gell-Mann, Leutwyler (1973)

Term “Quantum Chromodynamics (QCD)” — gauge theory of strong interactions

Quarks — real particles carrying a color charge

Gluons — massless quanta of the strong-interaction field

Asymptotic freedom — Politzer, Gross, Wilczek (1973)

Running of αs(Q2) = g2s(Q2)/(4π) on scale

αs(Q2) ∼ 1 at Q2 = m2N and αs(Q2) ≪ 1 at Q2 ≫ m2

N

• Discovery of Higgs H with mH ≃ 125 GeV — (2012)

Mass to charged fermions (quarks and leptons) and W±, Z0 weak boson

– p. 15

I . Introduction

• Fundamental Fermions and Bosons

– p. 16

I . Introduction

• Elementary Particles Dictionary

– p. 17

I . Introduction

• Structure of the Atom (updated view)

– p. 18

I . Introduction

• Typical Processes

– p. 19

I . Introduction

• Higgs Discovery

– p. 20

I . Introduction

• Current Experiments on Particle Physics

• From linear accelerators (Van de Graaf high-voltage) to modern colliders

• Collider — linear accelerators of two colliding beams

electron-positron (Novosibirisk, Russia)

electron-proton (DESY, Hamburg, Germany)

proton-antiproton (Tevatron, Fermilab, USA)

heavy ion (Pb) collider (BNL, USA)

proton-proton (LHC, CERN).

• beams inside the LHC are made to collide at four locations around the accelerator

ring — positions of four particle detectors — ATLAS, CMS, ALICE, LHCb.

– p. 21

I . Introduction

– p. 22

I . Introduction

• Hydrogen nuclei → protons

Linac2 — linear accelerator (due electric field) of protons up to v = c/3 and

E = 50 MeV

Proton Synchrotron Booster (PSB) — cyclic accelerator with L = 157 m;

v/c = 0.916; 4 particle beams to get maximal density; E = 1.4 GeV

Proton Synchrotron (PS) — particle beam in cyclic accelerator with L = 628 m;

speed v/c = 0.999; E = 450 GeV.

Proton Supersynchrotron (SPS) — 1 particle beam in cyclic accelerator with

L = 6.9 km; speed v/c = 0.999998; E = 450 GeV

Big loss of energy Erad ∼ E4

R.

Large Hadron Collider (LHC) — L = 27 km; 2 vacuum tubes with proton beams in

opposite directions

Beam E = 7 TeV 2 vacuum tubes with proton beams in opposite directions;

for 1 sec more 11000 times 27 km ring

4 colliding points — 4 detectors (ALICE, ATLAS, LHCb, SMS)

2 sec — typical time after colliding

– p. 23

I . Introduction

• ALICE — A Large Ion Collider Experiment

Study of quark-gluon plasma (strongly interacting matter at extreme energy

densities) — a state of matter thought to have formed just after the Big Bang

Temperature is hotter > 100 000 times than the centre of the Sun

– p. 24

I . Introduction

• ALICE detector — 10 000 tonne, 26 m long, 16 m high, 16 m wide.

• Location: Vast cavern 56 m below ground close to the village of St. Genis-Pouilly

(France), receiving beams from the LHC.

• 1000 scientists from over 100 institutes in 30 countries

– p. 25

I . Introduction

• ATLAS — A Toroidal LHC ApparatuS

• Search for Higgs, Physics beyond Standard Model, Supersymmetric Particles

• L = 45 m, Diameter = 25 m

• Cavern 100 metres below a small Swiss village, the 7000-tonne ATLAS detector is

probing for fundamental particles

• 3000 scientists from over 175 institutions in 38 countries

– p. 26

I . Introduction

• CMS — Compact Muon Solenoid Collider Experiment

• Study of Higgs Boson, Extra Dimensions, particles making up Dark Matter

• Built around a huge solenoid magnet

• Form of a cylindrical coil of superconducting cable

• Generates a field of 4 tesla, about 100,000 times the magnetic field of the Earth

• Field is confined by a steel “yoke” that forms the bulk of the detector (12,5 tonne

weight); H = 15 m, L = 21 m, W = 15 m

• Constructed in 15 sections at ground level before being lowered into an

underground cavern near Cessy in France and reassembled.

• 4300 scientists from 182 institutes in 42 countries

– p. 27

I . Introduction

• CMS detector

– p. 28

I . Introduction

• LHCb — LHC-beauty

• Study a differences between matter and antimatter by studying a b-quark hadrons

• Uses a series of subdetectors to detect mainly forward particles - those thrown

forwards by the collision in one direction

• The first subdetector is mounted close to the collision point, with the others

following one behind the other over a length of 20 metres

• An abundance of different types of quark are created by the LHC before they

decay quickly into other forms

• To catch the b quarks, LHCb has developed sophisticated movable tracking

detectors close to the path of the beams circling in the LHC

• 5600 tonne, L = 21 m, H = 10 m, W = 13 m, sits 100 metres below ground near

village of Ferney-Voltaire (France)

• About 700 scientists from 66 different institutes and universities from 16 countries

– p. 29

I . Introduction

• LHCb detector

– p. 30

I . Introduction

Plan of the Course

1. Introduction

2. Particle Kinematics

3. Quantum Chromodynamics (QCD)

4. Weinberg-Salam Theory

5. Effective Theories of QCD

6. Phenomenological Approaches to QCD

– p. 31

I . Introduction

– p. 32

II . Particle Kinematics

• Particles with relativistic velocities based on Special Theory of Relativity

• Basic characteristics

frame systems, phase space, decay rates, differential cross sections, angular

distributions

• Basic Blocks ~ = c = 1

Vectors: 4-momentum pµ = (E, ~p ); E – energy, ~p – three-momentum

Four-velocity uµ = pµ

m=

(Em, ~pm

)

State at the rest ~p = 0

Free-moving particle (mass-shell): p2 = m2 = E2 − ~p 2

Physical momentum is always time-like: p2 > 0

Light-like: p2 = 0 (massless particles, like neutrino)

Space-like: p2 < 0 (virtual particles)

E.g. pions in Yukawa exchange between nucleons

– p. 33


• Invariants composed of momenta p1, p2, . . . pn

Scalar product pipj = EiEj − ~p i~p j

Square of invariant mass (full energy) s12 = (p1 + p2)2 = m21 +m2

2 + 2p1p2

Invariant square of transversed moment t12 = (p1 − p2)2 = m21 +m2

2 − 2p1p2

• At fixed m1 and m2 the invariants s12, t12 and p1p2 reach their extremum at the

same conditions

• At fixed ~p 1 the condition of the extremum reads∂(p1p2)

∂~p 2= E1

~p 2

E2− ~p 1 = 0 when ~v 1 =

~p 1

E1is equal to ~v 2 =

~p 2

E2

Here∂(E2)∂~p 2

=∂(

√

m22+~p 2

2)

∂~p 2=

~p 2

E2

Equal velocities ~v 1 = ~v 2 in any system (Lorentz-invariant condition) — minimum

for p1p2 and s12 and maximum for t12

Reason: second derivative is positive

∂2(p1p2)

∂~p 2∂~p 2

=E1

E2

(

1− ~p 22

m22 + ~p 2

2

)

= m22

E1

E32

> 0

– p. 34


• To calculate the extremum we put ~v 1 = ~v 2 = 0, therefore (please check)

p1p2 = E1E2 = m1m2

s12 = (m1 +m2)2

t12 = (m1 −m2)2

• Finally, p1p2 ≥ m1m2 , s12 ≥ (m1 +m2)2 , t12 ≤ (m1 −m2)2

• Product of Levi-Cevita tensor of 4th rang with four vectors is invariant

ǫ(p1, p2, p3, p4) = ǫα1α2α3α4pα1 pα2 pα3 pα4

= ǫ(Lp1, Lp2, Lp3, Lp4) = detL︸︷︷︸

=1

ǫ(p1, p2, p3, p4)

where Lµν is the matrix of Lorentz transformation

– p. 35


• Frame Systems

Laboratory (Lab) — system, where the experiment occurs. Index L

Center-of-Mass (CM) — system, where ~p 1 + ~p 2 = 0. Index ⋆.

System of Target (T) — system, where ~p T = 0 or target at rest.

System of Beam (B) — system, where ~pB = 0 or beam at rest.

Kinematically equivalent to the (T).

System of colliding beams (CB) — system, where the particles of equal

three-momentum |~pCB1 | = |~pCB

2 |.

Normaly equal masses e+e−, pp, pp colliders

Also e−p colliders

When the angle between 3-vectors is zero — identical to CM frame.

– p. 36


• Target Frame

Momenta components of target ET = mT , ~pT = 0

Momenta components of bombarding particle EA, ~pA with |~pA| ≡ PA

s ≡ sAT = (EA + ET )2 − P 2A = E2

A − P 2A

︸︷︷︸

=m2A

+E2T + 2EAET = m2

A +m2T + 2EAmT

Therefore

EA =s−m2

A −m2T

2mT,

PA =√

E2A −m2

A =λ1/2(s,m2

A,m2T )

2mT

where λ(x, y, z) = x2 + y2 + z2 − 2xy − 2xz − 2yz is the triangle Källen function

– p. 37


• Geometric interpretation S = 1/4√

−λ(x2, y2, z2) is the square of the triangle

with sides x, y and z.

Triangle function can be rewritten in the form

λ(s,m2A,m

2T ) = [s− (mA +mT )2] [s− (mA −mT )2]

Therefore, the momentum PA is real when√s ≥ mA +mT .

The minimal value√s = mA +mT is the threshold of reaction

The value√s = mA −mT is the pseudothreshold of reaction

– p. 38


• CM Frame

Momenta components pA = (E∗A, ~p ), pB = (E∗

B ,−~p ) with |~p | = P ∗

√s = E∗

A + E∗B =

√

m2A + (P ∗)2 +

√

m2B + (P ∗)2

Making square of√s− E∗

A = E∗B we get

s+ (E∗A)2 − 2

√sE∗

A = (E∗B)2

Then using (E∗)2 = m2 + (P ∗)2 one gets

E∗A =

s+m2A −m2

B

2√s

, E∗B ↔ E∗

A(mA ↔ mB)

Finally

P ∗ =√

(E∗A)2 −m2

A =√

(E∗B)2 −m2

B =λ1/2(s,m2

A,m2B)

2√s

is symmetric under mA ↔ mB .

– p. 39


• Frame of Colliding Beams

Momenta components pA = (E, ~pA), pB = (E, ~pB) with mA = mB = m,

~pA = ~pB = P and π − θ is the angle between 3-vectors of colliding beams

s = 4E2 − (~pA + ~pB)2

= E2 − (~pA)2︸︷︷︸

=m2

+E2 − (~pB)2︸︷︷︸

=m2

+2E2 − 2P 2 cos(π − θ)

= 4m2 + 2P 2(1 + cos θ) = 4(

m2 + P 2 cos2θ

2

)

At the modern colliders m≪ P,E or E ≃ P

s ≃ 4P 2 cos2θ

2≃ 4E2 cos2

θ

2

Therefore

√s ≃ 2E cos

θ

2

– p. 40


• Normally, the θ ≃ 0 and√s = 2E.

At LHC√s = 2× 7 TeV = 14 TeV

• Problem 1

Without approximation (exact result for finite mass m)

E =

√

s− 4m2 sin2(θ/2)

2 cos θ/2, P =

√s− 4m2

2 cos θ/2,

• Problem 2

Explain why the colliders are more useful for study of elementary particles.

Hint: Compare the total energy√s in CB and System of Target.

PCB ≃√

s2

PA =λ1/2(s,m2

A,m2T )

2mT≃ s

2mT

To reach the same s we get an estimate PA =2P2

CBmT

≃ 105 TeV for mT ≃ 1 GeV

– p. 41


• Geometrical variables — absolute values of momenta and angles

in terms of invariants

θij is angle between ~pi and ~pjφ2 is angle between (~p2, ~p3) and (~p2, ~p4) planes

z~p2

y

x

~p3 ~p4

~p1 = 0

φ2

φ3 φ4

θ23 θ24

– p. 42


• Reference Frame – rest frame of time-like vector p1 = (m,~0 )

• Calculate:

1. P2 — length of ~p2

2. θ23 — angle between vectors ~p2 and ~p3

3. φ2 — angle between (~p2, ~p3) and (~p2, ~p4)

We choose ~p2 along z,

~p3 in (xz) plane with positive x component

p1 = (m1, 0, 0, 0)

p2 = (E2, 0, 0, P2)

p3 = (E3, P3 sin θ23, 0, P3 cos θ23)

p4 = (E4, P4 sin θ24 cosφ, P4 sin θ24 sinφ, P4 cos θ24)

Notation:

θij — angle between momenta (~pi and ~pj)

φk — angle between planes, composed by pairs of vectors (~pk, ~pi) and (~pk, ~pj) ,

0 ≤ θ ≤ π and 0 ≤ φ ≤ 2π

– p. 43


• Calculate P2 = |~p2|• Using ~p1 = 0 we get p1p2 = m1E2 and E2 = p1p2

m1

Next

P 22 = E2

2 −m22 =

(p1p2)2

m21

−m22

=(p1p2)2 − p21p

22

p21= − 1

p21

∣∣∣∣∣∣

p21 p1p2

p1p2 p22

∣∣∣∣∣∣

= − 1

p21∆2(p1, p2)

where ∆2(p1, p2) is symmetric Gram determinant

• Definition

Gram determinant of vectors p1, . . . , pn; q1, . . . , qn is the determinant of matrix

composed of scalar product piqj

G

p1, . . . , pn

q1, . . . , qn

= det||piqj || =

∣∣∣∣∣∣∣∣

p1q1 p1q2 . . . p1qn

· · · · · · · · · · · ·pnq1 pnq2 . . . pnqn

∣∣∣∣∣∣∣∣

– p. 44


• Symmetric Gram determinant of vectors

∆n(p1, . . . , pn) = G

p1, . . . , pn

p1, . . . , pn

= det||piqj || =

∣∣∣∣∣∣∣∣

p21 p1p2 . . . p1pn

· · · · · · · · · · · ·pnp1 pnp2 . . . pnpn

∣∣∣∣∣∣∣∣

As soon as the scalar products of the four-momentum are Lorentz-invariant, then

the Gram determinant is also Lorentz-invariant

• ∆2(p1, p2) = − 14λ(s, p21, p

22) where s = (p1 + p2)2

• Another interpretation of the Källen function — Gram determinant

• P2 = 1m1

√−∆2(p1, p2) =

12m1

λ1/2(s,m21,m

22) , s = (p1 + p2)2

– p. 45


• Polar angle θ23 between ~p2 and ~p3

p2p3 = E2E3 − P2P3 cos θ23 =p1p2

m1

p1p3

m1−

√∆2(p1, p2)∆2(p1, p3)

p21cos θ23

=1

p21

[

(p1p2)(p1p3)−√

∆2(p1, p2)∆2(p1, p3) cos θ23

]

Therefore,

cos θ23 =(p1p2)(p1p3)− p21(p2p3)√

∆2(p1, p2)∆2(p1, p3)= −

G

p1, p2

p1, p3

√∆2(p1, p2)∆2(p1, p3)

Or

sin2 θ23 =∆1(p1)∆3(p1, p2, p3)

∆2(p1, p2)∆2(p1, p3)

– p. 46


• Azimutal angle φ between (~p2, ~p3) and (~p2, ~p4) planes

cosφ =(~p2 × ~p3) · (~p2 × ~p4)

|~p2 × ~p3| |~p2 × ~p4|

Because of 0 ≤ φ ≤ 2π we need also sin of this angle

sinφ =P2 ~p2 · (~p3 × ~p4)

|~p2 × ~p3| |~p2 × ~p4|

• Lets obtain invariant form for cosφ in terms of Gram determinants

• Using p21 = m21, p1p2 = m1E2, p1p3 = m1E3, p1p4 = m1E4 we get

G

p1, p2, p3

p1, p2, p4

=

∣∣∣∣∣∣∣∣

m21 m1E2 m1E4

m1E2 p22 p2p4

m1E3 p2p3 p3p4

∣∣∣∣∣∣∣∣

• Now we multiply 1st line with E2/m1 and reduce the result from the 2nd line. Then

multiply the 1st line with E3/m1 and reduce it from the 3rd line.

– p. 47


G

p1, p2, p3

p1, p2, p4

= m21

∣∣∣∣∣∣∣∣

1 E2 E4

0 −~p 22 − ~p2~p4

0 − ~p2~p3 − ~p3~p4

∣∣∣∣∣∣∣∣

= m21(~p2 × ~p3) · (~p2 × ~p4) = m2

1P22 P3P4 sin θ23 sin θ24 cosφ

Therefore

cosφ =

G

p1, p2, p3

p1, p2, p4

m1P 22 P3P4 sin θ23 sin θ24

Taking into account the expression sin2 θ23 =∆1(p1)∆3(p1,p2,p3)∆2(p1,p2)∆2(p1,p3)

we get

cosφ =

G

p1, p2, p3

p1, p2, p4

(∆3(p1, p2, p3)∆3(p1, p2, p4)1/2

– p. 48


• Consider reaction pa + pb → p1 + . . .+ pn with 4-momentum conservation

Ea + Eb =

n∑

i=1

Ei , ~pa + ~pb =

n∑

i=1

~pi ,

E2j = m2

j + ~p 2j , j = a, b, 1, . . . , n .

• Momentum Space — space of the dimension D = 3n not constrained by the

4-momentum conservation 3n

• Phase Space — space of the dimension D = 3n− 4 deduced from momentum

space with taking into account the 4-momentum conservation.

.........

a

b

1

2

n

Exclusive Reaction

...

a

b

1

2

m

X

Inclusive Reaction

– p. 49


• Exclusive processes (all particles and their momenta are known)

1. Decay p0 → p1 + p2 + . . .+ pn (process 1 → n)

2. Colliding (or scattering) of two particles pa + pb → p1 + p2 + . . .+ pn

(process 2 → n)

• Inclusive processes (only some of the particles and their momenta are known)

• Probability of the transition from initial state |i〉 = |pa,pb〉 to final state

|f〉 = |p1, . . . ,pn〉 is given by

Pfi =|〈f |T |i〉|2〈i|i〉〈f |f〉

where T = S − 1; S and T are the S- and T -operators, respectively

〈f |T |i〉 = iMfi (2π)4 δ4(Pf − Pi)

2∏

j=1(2π)3/2

n∏

l=1

(2π)3/2, Pi = pa + pb , Pf =

n∑

i=1

pi

where Mfi =Mfi(pa,pb,p1, . . . ,pn) is the matrix element.

– p. 50


• To measure/determine Mfi is one the main purposes of the experiment.

• Calculation of observables — integration of |Mfi|2 over all possible values of

momenta

• Lifetime τ or Γtot = 1/τ (total width) for nonstable particles

• Breit-Wigner propagator

D(p2) =1

m2 − q2 − imΓ=

m2 − q2 + imΓ

(m2 − q2)2 +m2Γ2

• The pole position is complex (matrix element is real). Important to keep imaginary

part for q2 > m2. When m≫ Γ we get D(p2) = 1m2−q2

1

τ=

1

2m

1

(2π)3n−4In(m

2)

where

In(m2) =

∫ n∏

i=1

d3pi

2Eiδ4

(

p−∑

i

pi

)

|Mfi|2

is integral over phase space or LIPS (Lorentz Invariant Phase Space)

– p. 51


• Derivation of Breit-Wigner propagator

• Consider some unstable system, like radioactive isotopes with life time τ ,

proportional to the conventional half-life. The number of isotopes at some time t is

given by

N(t) = N(0) e−t/τ = N(0)e−Γt

with “width” or “decay rate” Γ = 1/τ .

• In Quantum Mechanics, the number of “surviving” isotopes is given by absolute

value squared of their wave function

N(t) =

∫

d3x |ψ(~x, t)|2

• To generate decreasing amplitudes we should construct the following w.f.

ψ(~x, t) ∼ e−Γ/2te∓iMt

We do not specify a position dependence.

A simple check proves that this w.f. give the correct e−Γt behavior.

– p. 52


• Consider now the energy dependence of a process, where such a wave function

enters. The Fourier-transform over time to energy reads

ψ(~x,E) =

∫

dtψ(~x, t) eiEt ∼∫

dt e[i(E−M)−Γ/2]t ∼ 1

E −M + iΓ/2

and therefore

|ψ(~x,E)|2 ∼ 1

(E −M)2 + (Γ/2)2

• This looks as the resonance stucture of dampened linear oscillator with an

eigenfrequency of Ω and a driving frequency ω yielding an intensity of oscillations

given by

I(ω) ∼ 1

(ω − Ω)2 + (Γ/2)2

The above structure is known as Breit-Wigner resonance

– p. 53


• Relativistic generalization can be obtained by replacing

ψ(~p,E) =1

E2 − ~p 2

︸︷︷︸

= p2

−M2 + iMΓ

leading to a factor

1

(p2 −M2)2 +M2Γ2

in the squared amplitude

• Clearly, when p2 →M2 of such an internal line, the amplitude square increses

drastically, it “resonanates”.

– p. 54


• Total cross section

σn =1

FIn(s) , F = 2λ1/2(s,m2

a,m2b)(2π)

3n−4 ,

In(s) =

∫ n∏

i=1

d3pi

2Eiδ4

(

pa + pb −∑

i

pi

)

|Mfi|2

• Differential cross section over variable x = x(pi)

dσn

dx=

1

F

∫ n∏

i=1

d3pi

2Eiδ4

(

pa + pb −∑

i

pi

)

δ(

x− x(pi))

|Mfi|2

• One can see that the condition

σn =

∫

dxdσn

dx

is fulfilled automatically. Higher-order differential cross sections are obtained by

inserting corresponding number of delta-functions.

– p. 55


• Distributions

ωn(x) =1

σn

dσn

dx

It is normalized to 1

∫

dxωn(x) =1

σn

∫

dxdσn

dx= 1

Higher-order distributions ω(x, y), ω(x, y, z), etc. are derived via corresponding

higher-order differential cross sections.

• Integration measure d3pE

is Lorentz-invariant.

– p. 56


• Using the property of the Dirac delta-function

δ(f(x)) =1

|f ′(x0)|δ(x− x0) , f(x0) = 0

• One gets

d3p

2E=

∫

d4p δ(p2 −m2) θ(p0)

where p0 = +√

m2 + ~p 2

• Finally

In(s) =

∫ n∏

i=1

d4pi δ(p2i −m2

i ) θ(p0i ) δ

4(

pa + pb −∑

i

pi

)

|Mfi|2

• Phase space In(s) → Rn(s) when |Mfi|2 ≡ 1

Rn(s) =

∫ n∏

i=1

d3pi

2Eiδ4

(

p−∑

i

pi

)

– p. 57


• Problem 3

(a) Show, that the decay e− → e− + γ kinematically is not allowed

(b) In Grand Unification Theory (GUT) the proton is not stable and can decay via

p→ e+ + π0, where e+ is electron and π0 is neutral pion. Show that this

process is kinematically allowed.

Hint: Calculate momenta of the positron and electron in rest frame of proton.

Solution

(a) With pe = p′e + pγ follows

p2e︸︷︷︸

=m2e

= p′2e

︸︷︷︸

=m2e

+ p2γ︸︷︷︸

=0

+ 2p′epγ︸︷︷︸

=2Eγ(Ee−|~pe| cos θ)

.

Therefore, Ee = |~pe| cos θ ≤ |~pe|.On the other hand, it is in contradiction with Ee =

√~p 2e +m2

e > |~pe|.

– p. 58


(b) We start with pp = pe + pπ .

In rest frame of the proton:

~pe = −~pπ = ~P , M = Ee + Eπ

Therefore,

M2 = E2e

︸︷︷︸

=P2+m2e

+ E2π

︸︷︷︸

=P2+m2π

+ 2EeEπ︸︷︷︸

=2√

P2+m2e

√P2+m2

π

where P ≡ |~P |. It comes:

1

2(M2 −m2

e −m2π)− P 2 =

√

P 2 +m2e

√

P 2 +m2π

Squaring:

1

4(M2 −m2

e −m2π)

2 + P 4 − P 2(M2 −m2e −m2

π) = P 4 + P 2(m2e +m2

π) +m2em

2π

– p. 59


We have

P =λ1/2(M2,m2

e,m2π)

2M,

where λ(x, y, z) = x2 + y2 + z2 − 2xy − 2xz − 2yz is the Källen function.

For m2e ≃ 0:

P =M

2

√

1−m2π/M

2

Because mπ < M , the decay is allowed.

– p. 60


• Problem 4

Consider the process pp→ pppp: two protons collide producing in addition

proton-antiproton pair.

(a) Calculate the energy of colliding protons in their center-of-mass frame.

(b) Consider the laboratory system, where one of the protons is in rest. What is

the value of the energy of the other colliding proton ?

Solution

(a) CM frame is defined as

p1 = (E∗, ~p ) ,

p2 = (E∗,−~p ) ,p′1 + p′2 + p′3 + p′4 = (4M,~0 )

where p1 + p2 = p′1 + p′2 + p′3 + p′4 (momentum conservation).

Therefore, E∗ = 2M due to energy conservation.

– p. 61


(b) In Lab-system

p1 = (E, ~p) ,

p2 = (M,~0) ,

p′1 + p′2 + p′3 + p′4 = (E +M, ~p)

we have

E =p1p2

M=

(p1 + p2)2 − p21 − p222M

=s− 2M2

2M

With√s ≥ 4M comes E ≥ 7M .

– p. 62


• Problem 5

Prove the formulae for decay width and integral cross section

Solution

Lets consider a general exclusive reaction k1 + . . .+ km → p1 + . . .+ pn with

Pi =m∑

j=1kj – total initial momentum and Pf =

m∑

i=1pi – total final momentum

Put system in finite volume V and consider the finite time T of reaction T

∫

d3p→ (2π)3

V

∑

p

With the use of

|i〉 = |k1, . . .km〉 , |f〉 = |p1, . . .pn〉 , 〈p′|p〉 = 2EpV

(2π)3δpp′

〈i|i〉 =m∏

j=1

2EjV

(2π)3, 〈f |f〉 =

n∏

i=1

2EiV

(2π)3

– p. 63


Pfi =|〈f |T |i〉|2〈i|i〉〈f |f〉 =

(2π)4 δ4(Pf − Pi)V T |Mfi|2n∏

i=j(2EjV )

n∏

i=1(2EiV )

Here we use

[(2π)4 δ4(Pf − Pi)]2 = (2π)4 δ4(Pf − Pi) (2π)3δ(~0)

︸︷︷︸

→V

(2π)δ(0)︸︷︷︸

→T

→ (2π)4 δ4(Pf − Pi)V T

Transition rate is defined as

Γ ≡ Γfi =∑

p1

. . .∑

pn

dPfi

dT

=∑

p1

. . .∑

pn

︸︷︷︸

→ (V/(2π)3)n∫

n∏

i=1

d3pi

(2π)4 δ4(Pf − Pi)V |Mfi|2m∏

j=1(2EjV )

n∏

i=1(2EiV )

→ 1

(2π)3n−4V 1−m

m∏

j=1

1

2EjV

∫ n∏

i=1

d3pi

2Eiδ4(Pf − Pi) |Mfi|2

– p. 64


Γfi =V 1−m

(2π)3n−4

m∏

j=1

1

2Ej

∫ n∏

i=1

d3pi


Decay 1 → n or lifetime

Γ =1

τ=

1

2m

1

(2π)3n−4

∫ n∏

i=1

d3pi


where Ek = m

Cross section

σn =Γfi

ρ|~v |

where ρ|~v | is the current density with ρ = 1/V and |~v | = |~va − ~vb | is the relative

velocity of scattering particles. Møller flow-factor f = 4EaEb|~va − ~vb| in covariant form

is given by

f = 4√

(ka · kb)2 −m2am

2b = 2λ(s,m2

a,m2b)

– p. 65


Proof: f = 4√

(ka · kb)2 −m2am

2b = 2λ(s,m2

a,m2b)

In CM frame

ka = (E∗1 ,~k∗) ,

kb = (E∗2 ,−~k∗)

we get

ka + kb = (√s = E∗

1 + E∗2 ,~0) ,

E∗1 =

s+m21 −m2

2

2√s

,

E∗2 =

s+m22 −m2

1

2√s

,

|~k∗ | =√

(E∗1 )

2 −m21 =

√

(ka · kb)2 −m21m

22√

s.

– p. 66


Therefore

f = 4EaEb|~va − ~vb|

= 4E∗1E

∗2

∣∣∣∣∣

~k∗

E∗1

+~k∗

E∗2

∣∣∣∣∣

= 4|~k∗| (E∗1 + E∗

2 )

= 4√

(k1 · k2)2 −m21m

22

Finally

σn =1

FIn(s) , F = 2λ1/2(s,m2

a,m2b)(2π)

3n−4 ,

In(s) =

∫ n∏

i=1

d3pi

2Eiδ4

(

pa + pb −∑

i

pi

)

|Mfi|2

– p. 67


• Two-body final states: Decay 1 → 2

• The decay properties depend on the time-like 4-momentum p = (E, ~p).

• Two-particle phase integral

R2(s) =

∫d3p1

2E1

d3p2

2E2δ4

(

p− p1 − p2)

Use

d3p2

2E2=

∫

d4p2 δ(p22 −m2

2) θ(E2)

where E2 = +√

m22 + ~p 2

2

R2(s) =

∫d3p1

2E1d4p2 δ

4(

p− p1 − p2)

δ(p22 −m22) θ(E2)

Integrate over d4p2 with the use δ4(p− p1 − p2) and

use p22 = (p− p1)2 = s− 2pp1 +m21

– p. 68


R2(s) =

∫d3p1

2E1θ(E2) δ(s− 2pp1 +m2

1 −m22) θ(E2)

=

∫dP1P 2

1

2E1

∫

dθ1 sin θ1

∫

dφ

︸︷︷︸

=∫

dΩ1

δ(

s− 2√sE1 +m2

1 −m22

)

θ(E2)

where P1 = |~p 1 |.

Using P 21 = E2

1 −m21 and dP1P1 = dE1E1 we get

R2(s) =1

2

∫

dE1P1dΩ1δ(

s− 2√sE1 +m2

1 −m22

)

Here E1 and P1 are the variables of integration, while the on-shell energy and

magnitude of the 3-momentum are

P ∗ = P ∗1 = P ∗

2 =λ1/2(s,m2

1,m22)

2√s

,

E∗1 =

s+m21 −m2

2

2√s

, E∗2 =

s+m22 −m2

1

2√s

, E∗1 + E∗

2 =√s

– p. 69


From θ function we get s ≥ (m1 +m2)(m1 −m2) or√s ≥ m1 +m2.

Simplify to new theta-function θ(√s−m1 −m2), which is symmetric on m1 ↔ m2

Finally

R2(s) =π P ∗√sθ(√s−m1 −m2) =

π λ1/2(s,m21,m

22)

2sθ(√s−m1 −m2)

Here we use the property of the delta function

δ(

s− 2√sE1 +m2

1 −m22

)

=1

2√sδ

(

E1 − s+m21 −m2

2

2√s

︸︷︷︸

=E∗

1

)

– p. 70


The formula for R2(s) can be derived in arbitrary frame

p = (E,p), p1 = (E1,p1), p2 = (E2,p2) ,

where P = |p| , P1 = |p1| , P2 = |p2|.

R2(s) =

∫d3p1

2E1

d3p2

2E2δ4

(

p− p1 − p2)

=

∫d3p1

4E1E2δ(

E − E1 − E2

)

=

∫

dΩ1dP1P 2

1

4E1E2δ(

E −√

P 21 +m2

1 −√

P 22 +m2

2

)

=

∫

dΩ1dP1P 2

1

4E1E2δ(

E −√

P 21 +m2

1 −√

P 2 + PP1 − 2PP1 cos θ1 +m22

)

where θ1 is the angle between p and p1.

– p. 71


Using the property of the Dirac delta-function

δ(f(x)) =1

|f ′(x0)|δ(x− x0) , f(x0) = 0

where f(P1) = E −√

P 21 +m2

1 −√

P 2 + P 21 − 2PP1 cos θ +m2

2;

P±1 – roots of f(P1) = 0

In particular, the equation f(P1) = 0 can be written in the form

P 21 − 2P1α+ β = 0 ,

α =P∆cos θ1

E2 − P 2 cos2 θ1,

β =E2m2

1 −∆2

E2 − P 2 cos2 θ1

∆ =m2 +m2

1 −m22

2

– p. 72


• Therefore, the solutions are

P±1 = α±

√

α2 − β

• One can see, that both solutions are positive

• The√α2 − β gives constraint α2 ≥ β

• From the constraint we get ∆ ≥ m1

√

E2 − P 2 cos2 θ1

• Substituting ∆ = (m2 +m21 −m2

2)/2 and using E2 = m2 + P 2

• We get

sin θ1 ≤ λ1/2(m2,m21,m

22)

2Pm1=

m

m1

P ∗

P

where P ∗ = λ1/2(m2,m21,m

22)/(2m) is the magnitude of the 3-momenta of the

decay products in the rest frame of decaying particle

– p. 73


• Rest Frame: θ1 → 0, E → m, P → 0

• Therefore

α =P∆cos θ1

E2 − P 2 cos2 θ1→ 0 ,

β =E2m2

1 −∆2

E2 − P 2 cos2 θ1→ m2

1 − (m2 +m21 −m2

2)2

4m2

= −λ(m2,m2

1,m22)

4m2= −(P ∗)2

• The equation for the magnitude of the 3-momenta (which is positive) is trivial:

P 21 + β = 0

• Or P 21 = (P ∗)2

• Finally P1 ≡ P ∗ (one solution)

– p. 74


R2(s) =

∫

dΩ1dP1P 2

1

4E1E2

δ(P1 − P+1 ) + δ(P1 − P−

1 )

P1/E1 + (P1 − P cos θ1)/E2︸︷︷︸

= (P1E−PE1 cos θ)/(E1E2)

=1

4

∫

dΩ1

∑

i=+,−

(P i1)

2

EP i1 − Ei

1P cos θ1

where P 22 = p2

2 = (p− p1)2 = P 2 + P 21 − 2PP1 cos θ1 , E = E1 + E2

In the rest frame p = (√s,0) we get

R2(s) =π P ∗√s

=π λ1/2(s,m2

1,m22)

2s

– p. 75


Lifetime (1 → 2), integral and differential cross section 2 → 2 in arbitrary frame

1

τ2=

1

32mπ2

∫

dΩ1

∑

k=+,−

(Pk1 )2

EPk1 − PEk

1 cos θ1|Mfi|2 ,

σ2 =1

32π2 λ1/2(s,m21,m

22)

√s

∫

dΩ1

∑

k=+,−

(Pk1 )2

EPk1 − PEk

1 cos θ1|Mfi|2 ,

dσ2

dΩ1=

1

32π2 λ1/2(s,m21,m

22)

√s

∑

k=+,−

(Pk1 )2

EPk1 − PEk

1 cos θ1|Mfi|2 .

– p. 76


• Two-particle final states

pa

pb

p1

p2

Mandelstam variables:

s = (pa + pb)2 = (p1 + p2)2 — square of full energy

t = (pa − p1)2 = (pb − p2)2 — square of transversed momentum

u = (pa − p2)2 = (pb − p1)2 — crossing variable

s+ t+ u = (pa + pb)2 + (pa − p1)

2 + (pb − p1)2

= p2a + p2b + p21 + (pa + pb − p1)2

︸︷︷︸

= p22

= p2a + p2b + p21 + p22 = m2a +m2

b +m21 +m2

2

– p. 77


• Channels

a

b

1

2

s-channel

pa + pb → p1 + p2

a

b

1

2

t-channel

pa + p1 → p2 + pb

a

b

1

2

u-channel

pa + p2 → p1 + pb

a

b

1

2

decay channel

pb → pa + p1 + p2

Here pi = −pi (i = a, 1, 2) are the momenta of antiparticles

– p. 78


• Frames

~p1∗

~p2∗

~pa∗

~pb∗

θ ∗

a1= θ ∗

1

θ ∗

a12= θ ∗

2

Center-of-mass frame:

~pa∗+ ~pb

∗= 0 and ~p1

∗+ ~p2

∗= 0

~p1

~p2

~pa~pb = 0

θT

a1= θ1

θT

a2= θ2

Rest frame of particle b:

~pb = 0

• θ∗1 = 0 — forward scattering

• θ∗1 = π — backward scattering

• s = sab = (pa + pb)2 — full energy

• t = ta1 = (pa − p1)2 — invariant square of transverse momenta

– p. 79


• Invariants in center-of-mass (CM) and rest or target (T ) frames

s = (pa + pb)2 = (p1 + p2)

2 = (E∗a + E∗

b )2 = (E∗

1 + E∗2 )

2

= m2a +m2

b + 2mbETa ,

t = (pa − p1)2 = (pb − p2)

2 = m2a +m2

1 − 2E∗aE

∗1 + 2P ∗

aP∗1 cos θ∗a1

= m2b +m2

2 − 2mbET2 ,

u = (pa − p2)2 = (pb − p1)

2 = m2a +m2

2 − 2E∗aE

∗2 + 2P ∗

aP∗2 cos θ∗a2

= m2b +m2

1 − 2mbET1 ,

where we use p2i = m2i (mass shell) and papb = mbE

Ta (b is at rest pb = 0)

• Center-of-mass (CM) frame

E∗a =

s+m2a −m2

b

2√s

, E∗b =

s+m2b −m2

a

2√s

,

E∗1 =

s+m21 −m2

2

2√s

, E∗2 =

s+m22 −m2

1

2√s

,

P ∗a = P ∗

b =λ1/2(s,m2

a,m2b)

2√s

, P ∗1 = P ∗

2 =λ1/2(s,m2

1,m22)

2√s

– p. 80


• For polar angle

cos θ∗a1 =t−m2

a −m21

2P ∗aP

∗1

=t−m2

a −m21 + (s+m2

a −m2b)(s+m2

b −m2a)/(2s)

λ1/2(s,m2a,m

2b)λ

1/2(s,m21,m

22)/(2s)

=

s (s+ 2t−m2a −m2

b −m21 −m2

2)︸︷︷︸

=t−u

+(m2a −m2

b)(m21 −m2

2)

λ1/2(s,m2a,m

2b)λ

1/2(s,m21,m

22)

=s(t− u) + (m2

a −m2b)(m

21 −m2

2)

λ1/2(s,m2a,m

2b)λ

1/2(s,m21,m

22)

– p. 81


• Four-particle kinematical function G(x, y, z, u, v, w)

G(x, y, z, u, v, w) = x2y + y2x+ z2u+ zu2 + v2w + vw2

+ xzw + xuv + yzv + yuw − xy(z + u+ v + w)

− zu(x+ y + v + w)− vw(x+ y + z + u)

Related with Gram determinant

G(s, t,m22,m

2a,m

2b ,m

21) = −4∆3(pa, pb, p1)

Shows up in polar angle

sin2 θ∗a1 = −4sG(s, t,m2

2,m2a,m

2b ,m

21)

λ(s,m2a,m

2b)λ(s,m

21,m

22)

Kinematical constraint

G(s, t,m22,m

2a,m

2b ,m

21) ≤ 0

– p. 82


• At ma = m1 and mb = m2

G(x, y, z, u, v, w) = y(

xy + λ(x, y, z))

and

sin2 θ∗a1 =2

λ(s,m2a,m

2b)

(

−st(

st+ λ(s,m2a,m

2b)))1/2

• Geometric meaning

G(x, y, z, u, v, w) = (−144)(

Volume of tetrahedron)2

with edges (√x,

√y), (

√z,

√u), (

√v,

√w). Tetrahedron function.

√w

√v

√z √

y

√x

√u

– p. 83


• Expressions for G(x, y, z, u, v, w)

• Determinant of 3× 3 matrix

G(x, y, z, u, v, w) = −1

2

∣∣∣∣∣∣∣∣

2u x+ u− v u+ w − y

x+ u− v 2x x− z + w

u+ w − y x− z + w 2w

∣∣∣∣∣∣∣∣

• Kelly determinant (determinant of the traceless 5× 5 matrix)

G(x, y, z, u, v, w) = −1

2

∣∣∣∣∣∣∣∣∣∣∣∣∣

0 1 1 1 1

1 0 v x z

1 v 0 u y

1 x u 0 y

1 z y w 0

∣∣∣∣∣∣∣∣∣∣∣∣∣

– p. 84


• Rest or Target (T) frame

pa = (ETa , ~p

Ta ) , pb = (mb,~0 )

p1 = (ET1 , ~p

T1 ) , p2 = (ET

2 , ~pT2 )

Therefore

ETa =

papb

mb=

(pa + pb)2 − p2a − p2b2mb

=s−m2

a −m2b

2mb,

ET1 =

p1pb

mb=p2b + p21 − (pb − p1)2

2mb=m2

b +m21 − u

2mb,

ET2 =

p2pb

mb=p2b + p22 − (pb − p2)2

2mb=m2

b +m22 − t

2mb,

PT1 = |~pM

1 | =√

(ET1 )2 −m2

1 =λ1/2(u,m2

b ,m21)

2mb,

PT2 = |~pM

2 | =√

(ET2 )2 −m2

2 =λ1/2(t,m2

b ,m22)

2mb.

– p. 85


• Polar angles

cos θTa1 =(s−m2

a −m2b)(m

2b +m2

1 − u) + 2m2b(t−m2

a −m21)

λ1/2(s,m2a,m

2b)λ

1/2(u,m2b ,m

21)

,

cos θTa2 =(s−m2

a −m2b)(m

2b +m2

1 − t) + 2m2b(u−m2

a −m21)

λ1/2(s,m2a,m

2b)λ

1/2(t,m2b ,m

21)

– p. 86


• Differential cross sections

Using

dσ

dΩ1=

1

32π2√s λ1/2(s,m2

1,m22)

∑

k=+,−

(Pk1 )2

EPk1 − PEk

1 cos θ1|Mfi|2 .

we get

dσ

dΩ∗1

=1

64π2 s

P ∗1

P ∗a

|Mfi|2

dσ2

dΩT1

=1

64π2mb PTa

(PT1 )2

(ETa +mb)P

T1 − PT

a ET1 cos θTa1

|Mfi|2

– p. 87


• Now we get covariant cross sections

Using dt = 2P ∗aP

∗1 d cos

∗a1 =

1

πP ∗a P

∗1 dΩ

∗1

We getdσ

dt=

dσ

dΩ∗1

dΩ∗1

dt=

|Mfi|264πs(P ∗

a )2=

|Mfi|216πλ(s,m2

a,m2b)

Integral cross section σ(s) =1

16πλ(s,m2a,m

2b)

t+∫

t−

dt |Mfi|2

where t± are the limiting values of t at fixed s found from

t = m2a +m2

1 − 2E∗aE

∗1 + 2P ∗

aP∗1 cos θ∗a1

Choosing cos θ∗a1 = ±1 we get

t± = m2a +m2

1 − 2E∗aE

∗1 + 2P ∗

aP∗1

= m2a +m2

1 − 1

2s((s+m2

a −m2b)(s+m2

1 −m22)∓ λ1/2(s,m2

a,m2b)λ

1/2(s,m21,m

22))

– p. 88


• Three-body final states: decay 1 → 3

p

p1

p2

p3

• Invariant variables

s12 = s1 = (p1 + p2)2 = (p− p3)

2 ,

s23 = s2 = (p2 + p3)2 = (p− p1)

2 ,

s13 = s3 = (p1 + p3)2 = (p− p2)

2 ,

s1 + s2 + s3 = s+m21 +m2

2 +m23 , s = m2

– p. 89


• Rest frame of decaying particle p = p1 + p2 + p3 = 0

E1 =s+m2

1 − s2

2√s

, P1 =λ1/2(s,m2

1, s2)

2√s

,

E2 =s+m2

2 − s3

2√s

, P2 =λ1/2(s,m2

2, s3)

2√s

,

E3 =s+m2

3 − s1

2√s

, P3 =λ1/2(s,m2

3, s1)

2√s

From

cos θ12 =p1 · p2

|p1||p2|=

(s+m21 − s2)(s+m2

2 − s3) + 2s(m21 +m2

2 − s1)

λ1/2(s,m21, s2)λ

1/2(s,m22, s3)

– p. 90


• Center-of-mass frames — Gottfried-Jackson frames

p1 + p2 = p− p3 = 0

p2 + p3 = p− p1 = 0

p1 + p3 = p− p2 = 0

For example, p2 + p3 = p− p1 = 0

E23 =s+ s2 −m2

1

2√s2

, E231 =

s− s2 −m21

2√s2

,

E232 =

s2 +m22 −m2

3

2√s2

, E233 =

s2 +m23 −m2

2

2√s2

,

P 23 = P 231 =

λ1/2(s, s2,m21)

2√s2

, P 232 = P 23

3 =λ1/2(s,m2

2,m23)

2√s2

– p. 91


• Polar angle

For θ2312 we find

s1 = (p1 + p2)2 = m2

1 +m22 + 2E23

1 E232 − 2P 23

1 P 232 cos θ2312

and

cos θ2312 =(s− s2 −m2

1)(s2 +m22 +m2

3) + 2s2(m21 +m2

2 − s1)

λ1/2(s, s2,m21)λ

1/2(s,m22,m

23)

– p. 92


• Phase space

R3(s) =

∫ 3∏

i=1

d3pi

2Eiδ4(p− p1 − p2 − p3)

• First integrate over d3p2 using δ3(p− p1 − p2 − p3)

R3(s) =

∫d3p1d3p3

8E1E2E3δ(√s− E1 − E2 − E3)

Then

d3p1d3p3 = P 2

1 dP1P23 dP3dΩ1dΩ3 θ(1− cos2 θ13)

where θ function restricts to physical values of θ

• Change variables (E1, E3) to (s1, s2). Jacobian is 1/(4s)

• Ω3 = (cos θ13, φ3) orientation of p3 with respect to p1. Integral over φ3 gives 2π.

• Ω1 orientation of p1 with respect to a general axe. Integral over Ω1 gives 4π.

• Change of variables d cos θ13 = dE2E2/(P1P3) and integrate over E2

– p. 93


• We get

R3(s) =π2

4s

∫

ds1

∫

ds2 θ(1− cos θ213)

• Theta-function constrains the variation of s1 and s2

s1 = m21 +m2

2 +1

2s2(s− s2 −m2

1)(s2 +m22 −m2

3)

− 1

2s2cos θ13 λ

1/2(s, s2,m21)λ

1/2(s2,m22,m

23)

• Therefore (corresponding to cos θ13 = ∓1)

s±1 = m21 +m2

2 +1

2s2(s− s2 −m2

1)(s2 +m22 −m2

3)

± 1

2s2λ1/2(s, s2,m

21)λ

1/2(s2,m22,m

23)

– p. 94


• Variation of variable s2 is fixed from limits of s±1 (must be real)

• λ1/2(s, s2,m21) and λ1/2(s2,m2

2,m23) must be real

• Therefore (m2 +m3)2 ≤ s2 ≤ (√s−m1)2

• Finally

R3(s) =π2

4s

(√s−m1)

2∫

(m2+m3)2

ds2

s+1∫

s−1

ds1

=π2

4s

(√

s−m1)2

∫

(m2+m3)2

ds2

s2λ1/2(s, s2,m

21)λ

1/2(s2,m22,m

23)

– p. 95


• Limiting cases

• Ultrarelativistic limit (UR) Ei → Pi or mi → 0

RUR3 (s) =

π2

4s

s∫

0

ds2

s2λ1/2(s, s2, 0)︸︷︷︸

=s−s2

λ1/2(s2, 0, 0)︸︷︷︸

=s2

=π2

8s

• Nonrelativistic limit (NR)√s→ m1 +m2 +m3

Expand in powers of√s− (m1 +m2 +m3)

RNR3 (s) =

π3

2

(m1m2m3)1/2

(m1 +m2 +m3)3/2(√s−m1 −m2 −m3)

2

• Lifetime of the particle decaying into 3 particles

1

τ=

1

256π3m3

(√s−m1)

2∫

(m2+m3)2

ds2

s+1∫

s−1

ds1 |Mfi|2

– p. 96


• Problem 6

Consider the Møller scattering e−e− → e−e−, which at the leading order O(α) is

described by the t- and u-channel diagrams shown in Figure.

t-channel

γ

e−(p1)

e−(p2)

e−(p3)

e−(p4)

u-channel

γ

e−(p1)

e−(p2)

e−(p3)

e−(p4)

Figure: Leading-order diagrams contributing to the Møller scattering

– p. 97


(a) Check that the corresponding matrix element is given by

Mfi = e2 u(p3, s3)γµu(p1, s1)

(

−gµν

t

)

u(p4, s4)γνu(p2, s2)

− e2 u(p4, s4)γµu(p1, s1)

(

−gµν

u

)

u(p3, s3)γνu(p2, s2) ,

where s, t and u are the Mandelstam variables with s+ t+ u = 4m2; m is the

electron mass; where s1 and s2 are the spins of the initial electrons; s3 and s4 are

the spins of the final electrons

(b) Calculate square of the matrix element |Mfi|2 using the formula

|Mfi|2 =1

2s1 + 1

1

2s2 + 1

∑

s1,s2,s3,s4

|Mfi|2

(c) Derive the formula for the differential cross section dσ/dΩ as function of scattering

(polar angle) θ in the center-of-mass frame of initial (or final electrons) in two

specific limits: i) Nonrelativistic limit; ii) Ultrarelativistic limit

– p. 98


Solution

(a) Using Feynman rules

Mfi = e2 u(p3)γµu(p1)

(

−gµν

t

)

u(p4)γνu(p2)

− e2 u(p4)γµu(p1)

(

−gµν

u

)

u(p3)γνu(p2) ,

where

s = (p1 + p2)2 = (p3 + p4)

2 ,

t = (p1 − p3)2 = (p2 − p4)

2 ,

u = (p1 − p4)2 = (p2 − p3)

2

are the Mandelstam variables satisfy to s+ t+ u = 4m2, and m is the electron

mass. Note, the u-amolitude has relative sign minus due to the Pauli principle.

– p. 99


Therefore,

|Mfi|2 =1

2s1 + 1

1

2s2 + 1

∑

s1,s2,s3,s4

|Mfi|2

=e4

4t2tr[γµ( 6p1 +m)γν( 6p3 +m)] tr[γµ( 6p2 +m)γν( 6p4 +m)]

+e4

4u2tr[γµ( 6p1 +m)γν( 6p4 +m)] tr[γµ( 6p2 +m)γν( 6p3 +m)]

− e4

4tutr[γµ( 6p1 +m)γν( 6p3 +m)γµ( 6p2 +m)γν( 6p4 +m)]

− e4

4tutr[γµ( 6p1 +m)γν( 6p4 +m)γµ( 6p2 +m)γν( 6p3 +m)]

– p. 100


Next with

2p1p2 = (p1 + p2)2 − p21 − p22 = s− 2m2 ,

2p3p4 = (p3 + p4)2 − p23 − p24 = s− 2m2 ,

2p1p3 = p21 + p23 − (p1 − p3)2 = 2m2 − t ,

2p1p4 = p21 + p24 − (p1 − p4)2 = 2m2 − u ,

2p2p4 = p22 + p24 − (p2 − p4)2 = 2m2 − t ,

2p2p3 = p22 + p23 − (p2 − p3)2 = 2m2 − u ,

follows

|Mfi|2 = 128π2α2

[(s− 2m2)2 + (u− 2m2)2 + 4m2t

4t2

+(s− 2m2)2 + (t− 2m2)2 + 4m2u

4u2

+(s− 2m2) (s− 6m2)

2tu

]

with t↔ u symmetry.

– p. 101


(b) In CM frame ~p1 + ~p2 = ~p3 + ~p4 = 0 ,

p1 = (E, ~pi) ,

p2 = (E,−~pi) ,p3 = (E, ~pf ) ,

p4 = (E,−~pf ) ,

where |~pi| = |~pf | = P and θ is scattering angle

s = 4E2 = 4(m2 + P 2) ,

t = p21 + p23 − 2p1p3 = 2m2 − 2E2

︸︷︷︸

=−2P2

+2P 2 cos θ

= −2P 2 (1− cos θ) = −4P 2 sin2θ

2,

u = p21 + p24 − 2p1p4 = 2m2 − 2E2

︸︷︷︸

=−2P2

−2P 2 cos θ

= −2P 2 (1 + cos θ) = −4P 2 cos2θ

2,

s+ t+ u = 4m2

– p. 102


As we found before

dσ

dΩ

∣∣∣∣CM

=1

64π2s|Mfi|2

(i) In nonrelativistic (NR) limit P 2 ≪ m2:

s = 4m2

(

1 +O(P 2/m2)

)

,

t = −P 2 sin2θ

2∼ O(P 2/m2) ,

u = −P 2 cos2θ

2∼ O(P 2/m2) ,

(s− 2m2)2 + (u− 2m2)2 + 4m2t = 8m4

(

1 +O(P 2/m2)

)

,

(s− 2m2)2 + (t− 2m2)2 + 4m2u = 8m4

(

1 +O(P 2/m2)

)

,

(s− 2m2) (s− 6m2) = −4m4

(

1 +O(P 2/m2)

)

.

– p. 103


Therefore,

dσ

dΩ

∣∣∣∣

NR

CM

=α2m2

2

[(1

t− 1

u

)2

+1

t2+

1

u2

]

+O(1/P 2)

=α2m2

16P 4

[1

sin4 θ2

+1

cos4 θ2

− 1

sin2 θ2cos2 θ

2

]

+ O(1/P 2)

=α2m2

P 4 sin4 θ

(

1− 3

4sin2 θ

)

+ O(1/P 2)

– p. 104


(ii) In ultrarelativistic (UR) limit P 2 ≫ m2:

s = 4P 2

(

1 +O(m2/P 2)

)

,

(s− 2m2)2 + (u− 2m2)2 + 4m2t

4t2=

s2 + u2

4t2︸︷︷︸

=O(1)

+O(m2/P 2) ,

(s− 2m2)2 + (t− 2m2)2 + 4m2u

4u2=

s2 + t2

4u2︸︷︷︸

=O(1)

+O(m2/P 2) ,

(s− 2m2) (s− 6m2)

2tu=

s2

2tu︸︷︷︸

=O(1)

+O(m2/P 2) .

– p. 105


Therefore,

dσ

dΩ

∣∣∣∣

UR

CM

=α2

2s

(

s2(1

t+

1

u

)2

+

(t

u

)2

+

(u

t

)2)

+O(1/P 4)

=4α2

P 2 sin4 θ

(

1− 1

2sin2 θ +

1

16sin4 θ

)

+ O(1/P 4)

=4α2

P 2 sin4 θ

(

1− 1

4sin2 θ

)2

+ O(1/P 4)

– p. 106


• Problem 7

Consider the process of the Compton scattering γe− → γe−, which is in the

leading order O(α) is given by the diagrams

e−

γ(p2)

e−(p1)

γ(p4)

e−(p3)

s-channel

e−

γ(p2)

e−(p1)

γ(p4)

e−(p3)

u-channel

Figure: Leading-order diagrams contributing to the Compton scattering

– p. 107


(a) Check that the matrix element is given by

Mfi = e2 ǫλ2∗µ (p4) ǫ

λ1ν (p2) u(p3, s2) γ

µ m+ 6p1+ 6p2m2 − s

γνu(p1, s1)

+ e2 ǫλ2∗µ (p4) ǫ

λ1ν (p2) u(p3, s2) γ

ν m+ 6p1− 6p4m2 − u

γµu(p1, s1) ,

where ǫλ2∗µ (p4) and ǫλ1

ν (p2) are the polarization vectors of photons; where s, t

and u are the Mandelstam variables with s+ t+ u = 2m2; m is the electron mass;

where s1 (s2) are the spins of initial (final) electron and λ1 (λ2) are the helicities of

initial (final) photon

(b) Show that the matrix element squared |Mfi|2 in terms of the Mandelstam variables

is given by

|Mfi|2 = 2e4[m2 − u

s−m2+m2 − s

u−m2+ 4R(1 +R)

]

,

where R = m2/(s−m2) +m2/(u−m2) and m is the electron mass

– p. 108


Use the formula

|Mfi|2 =1

2s1 + 1

1

2

∑

s1,s2,λ1,λ2

|Mfi|2

Here factor 12

is the averaging factor over two possible helicities of the initial

photons.

(c) Prove the Klein-Nishina Formula

dσ

dΩ=

α2

2m2

(ω′

ω

)2 [ω′

ω+ω

ω′ − sin2 θ

]

,

in rest frame of initial electron, where ω and ω′ are the energies of initial and final

photons; θ is the scattering angle of the final photon (angle between

three-momenta of the final and initial photons).

– p. 109


Solution

(a)

Mfi = e2 ǫ∗µ(p4) ǫν(p2) u(p3) γµ m+ 6p1+ 6p2

m2 − sγνu(p1)

+ e2 ǫ∗µ(p4) ǫν(p2) u(p3) γν m+ 6p1− 6p4

m2 − uγµu(p1) ,

where ǫ∗µ(p4) and ǫν(p2) are photon polarisations;

s = (p1 + p2)2 = (p3 + p4)

2 ,

t = (p1 − p3)2 = (p2 − p4)

2 ,

u = (p1 − p4)2 = (p2 − p3)

2

are Mandelstam variables with s+ t+ u = 2m2.

– p. 110


Therefore

|Mfi|2 =1

2s1 + 1

1

2

∑

s1,s2,s3,s4

|Mfi|2

=e4

4(s−m2)2tr[γµ( 6p1+ 6p2 +m)γν( 6p1 +m)γν( 6p1+ 6p2 +m)γµ( 6p3 +m)]

+e4

4(u−m2)2tr[γν( 6p1− 6p4 +m)γµ( 6p1 +m)γµ( 6p1− 6p4 +m)γν( 6p3 +m)]

+e4

4(s−m2)(u−m2)

× tr[γµ( 6p1+ 6p2 +m)γν( 6p1 +m)γµ( 6p1− 6p4 +m)γν( 6p3 +m)]

– p. 111


Next with

2p1p2 = s−m2 ,

2p3p4 = s−m2 ,

2p1p3 = 2m2 − t ,

2p1p4 = m2 − u ,

2p2p4 = −t ,2p2p3 = m2 − u ,

follows

|Mfi|2 = 2e4[m2 − u

s−m2+m2 − s

u−m2+ 4R(1 +R)

]

– p. 112


(b) As we found before

dσ

dΩ=

1

64π2m2

(ω′

ω

)2

|Mfi|2

In the rest frame of initial electron we have

s = (p1 + p2)2 = m(m+ 2ω) ,

u = (p1 − p4)2 = m(m− 2ω′) ,

t = (p2 − p4)2 = −2ωω′(1− cos θ)

= (p1 − p3)2 = 2m

(

m − p03︸︷︷︸

=m+ω−ω′

)

=

= 2m(ω′ − ω) .

– p. 113


Therefore

1− cos θ = m

(1

ω′ − 1

ω

)

,

m2 − u

s−m2=

ω′

ω,

m2 − s

u−m2=

ω

ω′ ,

R = m2/(s−m2) +m2/(u−m2) =m

2

(1

ω− 1

ω′

)

=cos θ − 1

2,

1 +R =cos θ + 1

2,

4R(1 +R) = cos2 θ − 1 = − sin2 θ .

Finally,

dσ

dΩ=

α2

2m2

(ω′

ω

)2 [ω′

ω+ω

ω′ − sin2 θ

]

.

– p. 114


• Dalitz Diagram: in case of the 1 → 3 process physical region in the (s1, s2) plane

or in the plane of two other variables related linearly (Jacobian of transformation is

constant)

• Given by the four-particle kinematical function G(x, y, z, u, v, w)

G(x, y, z, u, v, w) = x2y + y2x+ z2u+ zu2 + v2w + vw2

+ xzw + xuv + yzv + yuw − xy(z + u+ v + w)

− zu(x+ y + v + w)− vw(x+ y + z + u) ≤ 0

• Symmetries G(s2, s1, s,m23,m

21,m

22) = G(s1, s2, s,m2

2,m21,m

23)

• Problem 8:

Consider decay 1 → 3 with mass parameters

m = 4 GeV, m1 = 3 GeV, m2 = 0.5 GeV, m3 = 0.2 GeV.

Calculate decay width with trivial matrix element Mif ≡ 1 using the formula

Γ(1 → 3) =1

256m3

∫

ds1

∫

ds2 θ(

−G(s1, s2,m2,m2

2,m21,m

23))

– p. 115


(Debug) In[269]:=

k@x_, y_, z_D := x^2+y^2+z^2-2 x y -2 x z-2 y z H*** Kaellen Function ***L

(Debug) In[270]:=

G@x_, y_, z_, u_, v_, w_D :=-12 Det@880, 1, 1, 1, 1<, 81, 0, v, x, z<, 81, v, 0, u, y<, 81, x, u, 0, w<,81, z, y, w, 0<<D H*** G-Funktion ***L

(Debug) In[271]:=

H*** Check of Symmetry of the G-Function ***L

(Debug) In[272]:=

G@x, y, u, z, w, vD-G@x, y, z, u, v, wD FullSimplify

(Debug) Out[272]=

0

(Debug) In[273]:=

m = 4.; m1 = 3.; m2 = 0.5; m3 = 0.2; H*** Masses Hin GeVL ***L

(Debug) In[274]:=

H*** 3-particle phase space ***L

(Debug) In[275]:=

Plot21 = ContourPlotAG@s2, s1, m^2, m2^2, m3^2, m1^2D 0,

8s2, Hm2+m3L^2, Hm-m1L^2<, 8s1, Hm1+m2L^2, Hm-m3L^2<,FrameLabel ® 9"s2 HGeV2L", "s1 HGeV2L", "", ""=E H*** Hs2, s1L plane ***L

(Debug) Out[275]=

0.5 0.6 0.7 0.8 0.9 1.0

12.5

13.0

13.5

14.0

s 2 HGeV2L

s 1HG

eV2L

– p. 116


(Debug) In[276]:=

Plot23 = ContourPlotAG@s2, s3, m^2, m3^2, m2^2, m1^2D 0, 8s2, Hm2+m3L^2, Hm-m1L^2<,

8s3, Hm1+m3L^2, Hm-m2L^2<, FrameLabel ® 9"s2 HGeV2L", "s3 HGeV2L", "", ""=E

(Debug) Out[276]=

0.5 0.6 0.7 0.8 0.9 1.0

10.5

11.0

11.5

12.0

s 2 HGeV2L

s 3HG

eV2L

(Debug) In[277]:=

H*** 1®3 Decay Width:two-dim. integral with Heaviside Θ-function ***L

(Debug) In[278]:=

s = NIntegrate@HeavisideTheta@-G@s1, s2, m^2, m2^2, m1^2, m3^2DD,8s1, Hm1+m2L^2, Hm-m3L^2<, 8s2, Hm2+m3L^2, Hm-m1L^2<D Quiet

(Debug) Out[278]=

0.776998

(Debug) In[279]:=

width = sH256 Pi^3 m^3L

(Debug) Out[279]=

1.5295´10-6

(Debug) In[280]:=

TableForm@88width<<, TableHeadings ® 88"GH1->3L"<, 8"Result Hin GeVL"<<D

(Debug) Out[280]//TableForm=

Result Hin GeVL

GH1->3L 1.5295´10-6

2 3 p a r t i c l e P h a s e . n b

– p. 117


• Problem 9

Plot Dalitz diagrams in (s1, s2) planes for specific cases:

(a) m1 6= 0, m2 = m3 = 0. Show that

G(s1, s2, s, 0,m21, 0) = s2

(

s1(s1 + s2 − s)−m21(s1 − s)

)

.

Check that s1 varies from m21 to s;

s2 varies from 0 to (√s−m1)2.

(b) m2 6= 0, m1 = m3 = 0. Show that

G(s1, s2, s,m22, 0, 0) = (s1s2 − sm2

2) (s1 + s2 − s−m22) .

Check that s1 varies from m22 to s;

s2 varies from m22 to s.

(c) m1 = m2 = m3 = 0. Show that

G(s1, s2, s, 0, 0, 0) = −s1s2s3 = s1s2(s1 + s2 − s) .

Check that both s1 and s2 vary from 0 to s.

– p. 118


• Using MATHEMATICA for s = 9 GeV2 and nonvanishing m1 = 1 GeV.

(Debug) In[387]:=

G@x_, y_, z_, u_, v_, w_D :=-12 Det@880, 1, 1, 1, 1<, 81, 0, v, x, z<, 81, v, 0, u, y<, 81, x, u, 0, w<,81, z, y, w, 0<<D

(Debug) In[388]:=

G1@s1_, s2_D = Simplify@ G@s1, s2, 9, 0, 1, 0DD

(Debug) Out[388]=

I9+s12 +s1 H-10+s2LM s2

(Debug) In[389]:=

Plotm100 = ContourPlotAG1@s1, s2D 0, 8s1, 1, 9<,

8s2, 0, 4<, FrameLabel ® 9"s1 HGeV2L", "s2 HGeV2L", "", ""=E

(Debug) Out[389]=

2 4 6 8

0

1

2

3

4

s1 HGeV2L

s 2HG

eV2L

– p. 119


• Using MATHEMATICA for s = 9 GeV2 and nonvanishing m2 = 1 GeV.

(Debug) In[384]:=


(Debug) In[385]:=

G2@s1_, s2_D = Simplify@G@s1, s2, 9, 1, 0, 0DD

(Debug) Out[385]=

90-9 s2+s12 s2+s1 I-9-10 s2+s22M

(Debug) In[386]:=

Plotm010 = ContourPlotAG2@s1, s2D 0, 8s1, 1, 9<,

8s2, 1, 9<, FrameLabel ® 9"s1 HGeV2L", "s2 HGeV2L", "", ""=E

(Debug) Out[386]=

2 4 6 8

2

4

6

8

s1 HGeV2L

s 2HG

eV2L

– p. 120


• Using MATHEMATICA for s = 9 GeV2 and m1 = m2 = m3 = 0.

(Debug) In[390]:=


(Debug) In[391]:=

G3@s1_, s2_D = Simplify@G@s1, s2, 9, 0, 0, 0DD

(Debug) Out[391]=

s1 s2 H-9+s1+s2L

(Debug) In[392]:=

Plotm000 = ContourPlotAG3@s1, s2D 0, 8s1, 0, 9<, 8s2, 0, 9<,

PlotRange ® Full, FrameLabel ® 9"s1 HGeV2L", "s2 HGeV2L", "", ""=E

(Debug) Out[392]=

0 2 4 6 8

0

2

4

6

8

s1 HGeV2L

s 2HG

eV2L

– p. 121


• Transition 2 → 3

pa

pb

p1

p2 =

p3

pa

pb

p1

t1

s2p2

p3

• Invariants

s = sab = (pa + pb)2 = (p1 + p2 + p3)

2

s1 = s12 = (p1 + p2)2 = (pa + pb − p3)

2

s2 = s23 = (p2 + p3)2 = (pa + pb − p1)

2

t1 = ta1 = (pa − p1)2 = (p2 + p3 − pb)

2

t2 = tb3 = (pb − p3)2 = (p1 + p2 − pa)

2

– p. 122


• Phase space

R3(s) =

∫d3p1

2E1

d3p2

2E2

d3p3

2E3δ4(pa + pb − p1 − p2 − p3)

• Introduce intermediate system

1 =

∫

ds2

∫d3p23

2E23δ4(p23 − p2 − p3) , E23 =

√

~p 223 + s2

• Therefore

R3 =

∫

ds2

[∫d3p1

2E1

d3p23

2E23δ4(pa + pb − p1 − p23)

]

︸︷︷︸

=R2(s,m21,s2)

[∫d3p2

2E2

d3p3

2E3δ4(p23 − p2 − p3)

]

︸︷︷︸

=R2(s2,m22,m2

3)

• Factorization of 2 → 3 into two subprocesses 2 → 2 and 1 → 2

R3(s) =

∫

ds2 R2(s,m21, s2)R2(s2,m

22,m

23)

– p. 123


• Chew-Low Diagram: in case of the 2 → 3 process physical region in the (t1, s2)

plane or in the plane of two other variables related linearly (Jacobian of

transformation is constant)

Chew − Low condition G(s, t1, s2,m2a,m

2b ,m

21) ≤ 0

• Solutions

• Convenient to express t1 through s2

t±1 = m2a +m2

1 − 1

2s

(

(s+m2a −m2

b)(s− s2 +m21)

∓ λ1/2(s,m2a,m

2b)λ

1/2(s, s2,m21)

)

and (m2 +m3)2 ≤ s2 ≤ (√s−m1)2

• low limit for s2 — threshold for s2 ≥ m2 +m3

• upper limit for s2 — Källen function must be real

– p. 124


• Multiparticle Production

p7

p8

pa

pb

p9

p1

p2

p5

p6

p3

p4

M19

M12

M36M34

M78

R4

R2

R3

R2

R2

Cascade process

a+b→M19 →M12+M36+M78+9 → 1+2+M34+5+6+7+8+9 → 1+2+ · · ·+9

– p. 125


• Recurrence representation — chain of two-body decays

pa

pb

pn pn−1 p3 p2

p = kn kn−1 k2p1

R2 R2 R2 R2

Cascade decay

a+ b→ 1 + 2 + · · ·+ n

Rn(p) =

∫d3pn

2En

∫ n−1∏

i=1

d3pi

2Eiδ4

(

p− pn −n−1∑

i=1

)

︸︷︷︸

=Rn−1(p−pn)

=

∫d3pn

2EnRn−1(p− pn)

– p. 126


• Notations: p2i = m2i for i = 1, . . . , n

• Invariant mass of the system of n− 1 particles Mn−1

with M2n−1 = (p1 + · · ·+ pn−1)2 = (p− pn)2 = k2n−1 and

M2n = p2 = (pa + pb)

2 = k2n

• µi =i∑

k=1

mk = m1 +m2 + . . .+mi

• Insert two integrals over δ-functions

1 =

∫

dM2n−1 δ

(

M2n−1 − k2n−1

)

1 =

∫

d4kn−1 δ4(p− pn − kn−1)

and use

d3pn

2En= d4pn δ(p

2n −m2

n) θ(En)

– p. 127


• Then

Rn(M2n) =

∫

dM2n−1Rn−1(M

2n−1)

×∫

d4kn−1

∫

d4pn δ(k2n−1 −M2

n−1) δ(p2n −m2

n) δ4(p− pn − kn−1) θ(En)

︸︷︷︸

=R2(k2n,k2

n−1,p2n)

=

(Mn−mn)2∫

µ2n−1

dM2n−1R2(k

2n, k

2n−1, p

2n)Rn−1(M

2n−1)

Here we use the results for the 2-body phase space integral in general frame

R2(k2n, k

2n−1, p

2n) =

∫

dΩn−1

λ1/2(M2n,M

2n−1,m

2n)

8M2n

=1

2Mn

∫

dΩn−1Pn

2

– p. 128


• Using dM2 = 2MdM and

making iteration for Rn−1(M2n−1), Rn−2(M2

n−2), · · · , R2(M22 ) we get

Rn(M2n) =

1

2Mn

Mn−mn∫

µn−1

dMn−1 dΩn−1Pn

2· · ·

M3−m3∫

µ3

dM2 dΩ2P3

2

∫

dΩ1P2

2

where Pi = λ1/2(M2i ,M

2i−1,m

2i )/(2Mi).

– p. 129


• Problem 10

Use mathematical induction to prove that

(a) In ultrarelativistic limit (UR) mi = 0

RURn (M2

n) = Rn(M2n,m

2i = 0) =

(π/2)n−1

(n− 1)! (n− 2)!M2n−4

n

Hint: Prove and use the recurrence identity

RURn (M2

n) =π

2M2n

M2n∫

0

dM2n−1 (M

2n −M2

n−1)RURn−1(M

2n−1)

For n = 2

RUR2 (M2

2 ) =π

2

– p. 130


(b) In nonrelativistic limit (NR) Mn → µn = m1 + · · ·+mn

RNRn (M2

n) =(2π3)(n−1)/2

2Γ(

32(n− 1)

)

( n∏

i=1mi

)1/2

( n∑

i=1mi

)3/2

(

Mn −n∑

i=1

mi

)(3n−5)/2

Hint: Prove and use the recurrence identity

RNRn (Tn) = π

√mn

(2µn−1

µn

)3/2Tn∫

0

dTn−1 (Tn − Tn−1)1/2RNR

n−1(Tn−1)

where Tn =Mn − µn.

For n = 2

RNR2 (T2) = π

(2m1m2)1/2

µ3/22

T1/22

– p. 131


• Solution

(a)

RURn (M2

n) =

M2n∫

0

dM2n−1R2(M

2n,M

2n−1, 0)R

URn−1(M

2n−1)

=

M2n∫

0

dM2n−1

π(M2n −M2

n−1)

2M2n

RURn−1(M

2n−1)

=π

2M2n

M2n∫

0

dM2n−1 (M

2n −M2

n−1)RURn−1(M

2n−1)

For n = 2

RUR2 (M2

2 ) =πM2

2

2M22

=π

2

– p. 132


• For n = k + 1 using result for n = k

RURk+1(M

2k+1)

=π

2M2k+1

M2k+1∫

0

dM2k (M2

k+1 −M2k )R

URk (M2

k )

=π

2M2k+1

M2k+1∫

0

dM2k (M2

k+1 −M2k )

(π/2)k−1

(k − 1)! (k − 2)!M2k−4

k

=(π/2)k

(k − 1)! (k − 2)!

1

M2k+1

M2k+1∫

0

dM2k M

2k−4k (M2

k+1 −M2k )

=(π/2)k

(k − 1)! (k − 2)!M2k−2

k+1

(1

k − 1− 1

k

)

=(π/2)k

k! (k − 1)!M2k−2

k+1

Here we rescale the integration variable Tk → tTk+1

– p. 133


(b) After change Mn = Tn + µn we have

RNRn ((Tn + µn)

2)=2

Tn∫

0

dTn−1(Tn−1 + µn−1)R2((Tn + µn)2, (Tn−1 + µn−1)

2,m2n)

× RNRn−1((Tn−1 + µn−1)

2)

and then we expand treating Ti as small parameter and keep leading order result:

RNRn (Tn) = 2

Tn∫

0

dTn−1 µn−1 RNRn−1(Tn−1)

√

2mn(Tn − Tn−1)(µn−1)1/2

µ3/2n

= π√mn

(2µn−1

µn

)3/2Tn∫

0

dTn−1 (Tn − Tn−1)1/2RNR

n−1(Tn−1)

Here

RNR2 ((Tn + µn)

2, (Tn−1 + µn−1)2,m2

n) ≃ π√

2mn(Tn − Tn−1)(µn−1)1/2

µ3/2n

– p. 134

II . Particle Kinematics• For n = 2

RNR2 (T2) = π

√

2m2(T2 − T1)µ1

µ3/22

= π

√2m1m2T2

µ3/22

, where T1 = 0, µ1 = m1

• For n = k + 1 using result for n = k

RNRk (Tk) =

(2π3)(k−1)/2

2Γ(

32(k − 1)

)

( k∏

i=1mi

)1/2

µ3/2k

T3k−5

2

k

we get

RNRk+1(Tk+1) = π

√mk+1

(2µk

µk+1

)3/2Tk+1∫

0

dTk√Tk+1 − Tk R

NRk (Tk)

= 2π√2 T

3k−3

2

k+1

(2π3)k−1

2

2Γ(

32(k − 1)

)

( k+1∏

i=1mi

)1/2

µ3/2k

1∫

0

dt t3k−5

2 (1− t)12

︸︷︷︸

=B(3(k−1)/2,3/2)

– p. 135


• Finally using the expression for the beta-function

B(3(k − 1)/2, 3/2) =Γ(

32(k − 1)

)

Γ(

32

)

Γ(

32k)

we get

RNRk+1(Tk+1) =

(2π3)k/2

2Γ(

32k)

( k+1∏

i=1mi

)1/2

( k+1∑

i=1mi

)3/2T

(3k−3)/2k+1

– p. 136


• Decoupling formula — split group of particles (1, · · · ,m) particles into two groups

(1, · · · , l) and (l + 1, · · · , n)

.

.

.

.

.

.

1

2

lkl

l + 1

l + 2

n

Using the substitution

1 =

∫

dM2l δ(M

2l − k2l ) , 1 =

∫

d4kl δ4(

kl −l∑

i=1

pi

)

– p. 137


• and the definition for Rn(M2n)

Rn(M2n) =

∫

d4p1 · · · d4pnδ(p21 −m21) · · · δ(p2l −m2

l )

• We get

Rn(M2n) =

∫

dM2l

∫

d4kld4pl+1 · · · d4pn δ(k2l −M2

l )

× δ(p2l+1 −m2l+1) · · · δ(p2n −m2

n) δ4(

p− kl −n∑

i=l+1

pi

)

×∫

d4p1 · · · d4pl δ(p21 −m21) · · · δ(p2l −m2

l ) δ4(

kl −l∑

i=1

pi

)

=

(Mn−µn+µl)2

∫

µ2l

dM2l Rn−l+1(M

2n,M

2l ,m

2l+1, . . . ,m

2n)Rl(M

2l ,m

21, . . . ,m

2l )

where l = 2, 3, . . . , n− 1.

– p. 138


• Limits of integration over M2l are obtained from thresholds of reactions

p→ kl + pl+1 + · · ·+ pn (upper limit) and kl → p1 + · · ·+ pl (lower limit)

• Low limit

M2l ≥ (p1 + · · ·+ pl)

2 ≥ (m1 + · · ·+ml)2 = µ2l

• Upper limit

M2l ≤

(p− (pl+1 + · · ·+ pn)

)2≤ (Mn − (µn − µl))

2

Here we use p2 =M2n and take minimum for (pl+1 + · · ·+ pn) equal to µn − µl

• Case l = n− 1 reduces to the recurrence formula (chain of two-body decays)

• Comment on limits

– p. 139


• Multiparticle Production (master example)

p7

p8

pa

pb

p9

p1

p2

p5

p6

p3

p4

M19

M12

M36M34

M78

R4

R2

R3

R2

R2

R9(M219) =

∫

dM212dM

236dM

278R4(M

219,M

212,M

236,M

278,m

29)

× R2(M212,m

21,m

22)R2(M

278,m

27,m

28)

×∫

dM234R3(M

236,M

234,m

25,m

26)R2(M

234,m

23,m

24)

– p. 140


• Inclusive Reactions

...

a

b

1

2

m

X

Inclusive Reaction

Here X is unknown system of particles

• Detectors search for specific particles 1, · · · ,m

• When m = 1 we deal with one-particle distribution or one-particle spectrum

in Lab system

d3σc

d3p=

1

P 2

d3σc

dP dΩ

• d3σc/d3p is not Lorentz-invariant (LI); c is the type of particle.

– p. 141

II . Particle Kinematics• Lets define the LI distribution

Ed3σc

d3p= f(p, s)

where f(p, s) – distribution function; c is the type of particle.

• It is possible that we produce n particles of type c then

σincc (s) =

∑

n

nσnc (s) = 〈nc〉

∑

n

σnc (s) = 〈nc〉σc(s)

where σnc – inclusive cross section of n particles of type c,

σc – exclusive cross section of c particle production,

〈nc〉 – averaged number of particles of type c defined as

〈nc〉 =

∑

nnσn

c (s)

∑

nσnc (s)

,

Equivalence of exc. and inc. |〈a, b|a, b〉exc|2 =∑

n

|〈a, b;n|a, b;n〉inc|2

– p. 142


• Every event of n-particle production of type c gives the n-multiple contribution to

the cross section.

• It is clear that σincc (s) is given by the following integral

σincc (s) = 〈nc〉σc(s) =

∫d3p

Ef(p, s)

In the center-of-mass (CM) frame

pa + pb = (√s, 0) , p = (ηc

√s, p)

where ηc is the fraction of total energy of particles of type c

•

∫

d3p∗ f(p∗, s)︸︷︷︸

=E∗ d3σc/d3p∗

=

∫

d3p∗ E∗︸︷︷︸

= ηc√s

d3σc

d3p∗= ηc

√s

∫

d3p∗d3σc

d3p∗︸︷︷︸

=σc

= ηc√s σc

– p. 143


p⊥

p = pc

papL

θ

t

pa

pb

pc

sX

• Unpolarized particles (in case of spin particles: no definite orientation of the spin)

• Experiments with unpolarized particles are characterized by 3 variables:

full energy√s plus 2 variables

• Five sets of 2 variables

• 1. P (magnitude) and θ (polar angle) of the 3-momenta of the produced particle c

• 2. p⊥ (transversed) and pL (longitudinal) components of the 3-momentum p

• 3. t = (pc − pa)2 (square of transversed momentum) and sX = (pa + pb − pc)2

(square of invariant mass of X system or loosing mass)

– p. 144


• 4. t and ν = ETa − ET

c

(ν is energy of the exchange particle in the rest frame of target pb = 0)

It is easy to show that

ν = ETa − ET

c =pb(pa − pc)

mb

=(pa + pb − pc)2 − (pa − pc)2 − p2b

2mb

=sX − t−m2

b

2mb

Here we use papb = ETa mb and pcpb = ET

c mb

– p. 145

II . Particle Kinematics• 5. p⊥ and y (longitudinal rapidity)

Rapidity y is defined as

y =1

2lnEc + pL

Ec − pL= ln

Ec + pL√

m2c + p2⊥

= ln

√

m2c + p2⊥

Ec − pL

where Ec - energy and pL - longitudinal projection of the 3-momentum pc

Ec =√

m2c + p2 =

√

m2c + p2⊥ + p2L =

√

m2c + p2⊥

√

1 +p2L

m2c + p2⊥

︸︷︷︸

=cosh(y)

=√

m2c + p2⊥ cosh(y) ,

pL =√

m2c + p2⊥

pL√

m2c + p2⊥

︸︷︷︸

= sinh(y)

=√

m2c + p2⊥ sinh(y)

Here we use hyperbolic cosine and sine:

cosh(x) = (ex + e−x)/2 , sinh(x) = (ex − e−x)/2 , cosh2(x)− sinh2(x) = 1

– p. 146


• Pairs of variables (P, θ) and (p⊥, pL) are used together (Polar and Dekart

coordinates)

d3p

2E= π

P 2dPd cos θ

E= π

p⊥dp⊥dpLE

= πdp2⊥dpL

2E

Therefore invariant distribution function f(p) is given by

f(p) =E

P 2

d2σ

dPdΩ=

E

2πP 2

d2σ

dPd cos θ

=E

2πp⊥

d2σ

dp⊥dpL=E

π

d2σ

dp2⊥dpL

• Lets now consider possible values of the (P, cos θ) and (p⊥, pL) variables

– p. 147


• Smallest value of sX is sminX = min(

X∑

imi)

2

When a = c, then sminX = m2

b .

When a 6= c,

then sminX = m2

Λ in inclusive reaction π−p→ K0X, because X starts from X = Λ

then sminX = (mp +mn)2 in inclusive reaction pp→ π+X, because X starts from

X = p+ n

• In center-of-mass (CM) frame from the energy conservation√s =

√

m2c + P ∗2 +

√

s2X + P ∗2 follows

P ∗ =λ1/2(s,m2

c , sX)

2√s

Therefore, the maximal value of P ∗ is

P ∗max =

λ1/2(s,m2c , s

minX )

2√s

– p. 148

II . Particle Kinematics• Physical region is defined by inequalities

0 ≤ P ∗ ≤ P ∗max , 0 ≤ θ∗ ≤ π ,

where θ∗ is the scattering angle – angle between vectors pc and pa

• In the (p∗⊥, p∗L) plane the physical region is the circle (Peyre diagram)

(p∗⊥)2 + (p∗L)2 ≤ (P ∗

max)2

• Maximal value of the energy of particle c is

E∗max =

s+m2c − smin

X

2√s

• At large s we can expand (neglecting by sminX )

E∗max ≃

√s

2

(

1 +m2

c

s

)

, P ∗max ≃

√s

2

(

1− m2c

s

)

,

P ∗max

E∗max

≃ 1− m2c

2s

– p. 149


• Variables (t, sX) and (t, ν)

• For t and sX we have

t = m2a +m2

c − 2ETa E

Tc + 2PT

a PTc cos θT

= m2a +m2

c − 2E∗aE

∗c + 2P ∗

aP∗ cos θ∗ ,

sX = s+m2c − 2(ET

a +mb)ETc + 2PT

a PTc cos θT

= s+m2c − 2

√sE∗

c

• Invariant measure

d3p

2E=

π(P ∗c )

2dP ∗c d cos θ

∗

E∗c

= πP ∗c dE

∗c d cos θ

∗

=π

4√sP ∗

a

dtdsX =π

2λ−1/2(s,m2

a,m2b) dt dsX

where

dtds = 4√sP ∗

aP∗c dE

∗c = 4

√sλ1/2(s,m2

a,m2b)

2√s

– p. 150


• Therefore

f(t, sX) =1

πλ1/2(s,m2

a,m2b)

d2σ

dtdsX

• In (t, sX) plane the physical region is given by the Chew-Low diagram defined by

the inequalities

G(s, t, sX ,m2a,m

2b ,m

2c) ≤ 0

sX ≥ sminX

– p. 151


• Variables (p⊥, y)

• Invariant measure

d3p

E=d2p⊥dpL

E= d2p⊥dy = 2πp⊥dp⊥dy

Here we use

y =1

2lnEc + pL

Ec − pL, Ec =

√

m2c + p2⊥ + p2L

and therefore

∂y

∂pL=

1

E

• Invariant distribution

f(p⊥, y) =E

2πp⊥

d2σ

dp⊥dpL=

1

2πp⊥

d2σ

dp⊥dy=

1

π

d2σ

d2p⊥dy

– p. 152

Elementary Particle Physics - uni-tuebingen.de · Elementary Particle Physics Valery Lyubovitskij Institute of Theoretical Physics, Tubingen University,¨ Kepler Center for Astro

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