elementary particle physics - uni-tuebingen.de · elementary particle physics valery lyubovitskij...
TRANSCRIPT
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
• 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
• 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
• 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
• 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
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
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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
(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
II . Particle Kinematics
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
II . Particle Kinematics
• 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
II . Particle Kinematics
(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
II . Particle Kinematics
• 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
II . Particle Kinematics
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
II . Particle Kinematics
Γfi =V 1−m
(2π)3n−4
m∏
j=1
1
2Ej
∫ n∏
i=1
d3pi
2Eiδ4(Pf − Pi) |Mfi|2
Decay 1 → n or lifetime
Γ =1
τ=
1
2m
1
(2π)3n−4
∫ n∏
i=1
d3pi
2Eiδ4(Pf − Pi) |Mfi|2
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
II . Particle Kinematics
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
II . Particle Kinematics
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
II . Particle Kinematics
• 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
II . Particle Kinematics
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
II . Particle Kinematics
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
II . Particle Kinematics
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
II . Particle Kinematics
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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
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
II . Particle Kinematics
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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
(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
II . Particle Kinematics
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
II . Particle Kinematics
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
II . Particle Kinematics
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
II . Particle Kinematics
(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
II . Particle Kinematics
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
II . Particle Kinematics
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 . Particle Kinematics
(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
II . Particle Kinematics
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
II . Particle Kinematics
• 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
II . Particle Kinematics
(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
II . Particle Kinematics
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
II . Particle Kinematics
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
II . Particle Kinematics
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
II . Particle Kinematics
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
II . Particle Kinematics
(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
II . Particle Kinematics
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
II . Particle Kinematics
• 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
II . Particle Kinematics
(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
II . Particle Kinematics
(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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• Using MATHEMATICA for s = 9 GeV2 and nonvanishing m2 = 1 GeV.
(Debug) In[384]:=
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[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
II . Particle Kinematics
• Using MATHEMATICA for s = 9 GeV2 and m1 = m2 = m3 = 0.
(Debug) In[390]:=
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[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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
(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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
(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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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
II . Particle Kinematics
• 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