Particle and Nuclear Physics
Handout #1 Introduction Problem Sheets Appendices
Lent/Easter Terms 2021 Dr. Tina Potter 1. Introduction Particle and Nuclear Physics
Dr. Tina Potter
Dr. Tina Potter 1. Introduction 1 In this section...
Course content Practical information Matter Forces
Dr. Tina Potter 1. Introduction 2 Course content These lectures will cover the core topics of Particle and Nuclear physics. Particle Physics is the study of Nuclear Physics is the study of Matter: Elementary particles Matter: Complex nuclei (protons & neutrons) Forces: Basic forces in nature Electroweak (EM & weak) Forces: Strong “nuclear” force Strong (underlying strong force) + weak & EM decays Current understanding is embodied in the Complex many-body problem, Standard Model requires semi-empirical approach. which successfully describes all current data∗. Many models of Nuclear Physics. Historically, Nuclear Physics preceded and led to Particle Physics. Our course will discuss Particle Physics first, and then Nuclear Physics.
∗ with some interesting exceptions! Dr. Tina Potter 1. Introduction 3 Practical information
Website holds course information, notes, appendices and problem sheets www.hep.phy.cam.ac.uk/~chpotter/particleandnuclearphysics/mainpage.html
Books Introduction to High Energy Physics, Perkins Introductory Nuclear Physics, Krane
Lecturing material provided as three handouts. Lectures will cover additional examples – please attend!! Problem sets in 4 parts Part 1: Chapter 1-2 Part 2: Chapter 3-8 Part 3: Chapter 9-12 Part 4: Chapter 13-16
My availability: before/after lectures, via email ([email protected]), in-person chats are always welcome
Dr. Tina Potter 1. Introduction 4 Zooming into matter
Atom Binding energy Rydberg 10 eV Electrons bound to atoms by EM∼ force ∼ 10 19 Size: Atom 10− m, e− < 10− m ∼ Charge: Atom is neutral, electron e − Mass: Atom mass nucleus, m = 0.511MeV /c2 ∼ e Chemical properties depend of Atomic Number, Z
Nucleus Binding energy 10 MeV/nucleon ∼ Nuclei held together by strong “nuclear” force 15 Size: Nucleus (medium Z) 5 fm (1 fm= 10− m) ∼ Nucleon Binding energy 1 GeV Protons & neutrons held together∼ by the strong force Size: p, n 1 fm ∼ Charge: proton +e, neutron is neutral Mass: p, n = 939.57 MeV/c2 1836 m ∼ e
Dr. Tina Potter 1. Introduction 5 Matter
1 In the Standard Model, all matter is made of spin 2 fundamental particles. There are two types, each with 3 generations:
Consequence of relativity and quantum mechanics (Dirac equation) Antiparticle for every existing particle: identical mass, spin, energy, momentum, but has the opposite sign of interaction (e.g. electric charge). Particles and antiparticles + electron e− & positron e up quark u (Q = +2) & antiupu ¯ (Q = 2) 3 −3 proton udu & antiprotonu ¯d¯u¯
Dr. Tina Potter 1. Introduction 6 Matter The first generation Almost all the matter in the universe is made up from just four of the fermions.
Particle Symbol Type Charge [e]
Electron e− lepton 1 − Neutrino νe lepton 0 2 Up quark u quark +3 Down quark d quark 1 −3 The proton and neutron are simply the lowest energy bound states of a system of three quarks: essentially all an atomic or nuclear physicist needs.
Proton Neutron (p) (n)
Dr. Tina Potter 1. Introduction 7
Matter Three generations Nature is not so simple. There are 3 generations/families of fundamental fermions (and only 3).
1st generation 2nd generation 3rd generation
Electron e− Muon µ− Tau τ −
Electron Neutrino νe Muon Neutrino νµ Tau Neutrino ντ Up quark u Charm quark c Top quark t Down quark d Strange quark s Bottom quark b
Each generation is a replica of (e−, νe, u, d). The mass of the particles increases with each generation: the first generation is lightest and the third generation is the heaviest. The generations are distinct i.e. µ is not an excited e, or µ− e−γ would be allowed – this is not seen. → There is a symmetry between the generations, but the origin of 3 generations is not understood! Dr. Tina Potter 1. Introduction 8 Matter Leptons
Leptons are fermions which do not interact via the strong interaction. Flavour Charge [e] Mass Strong Weak EM
1st generation 1 2 Spin 2 fermions e− 1 0.511 MeV/c − 2 νe 0 < 2 eV/c 6 distinct flavours 2nd generation 3 charged leptons: e−, µ−, τ −. 2 µ− 1 105.7 MeV/c 3 neutral leptons: νe, νµ, ντ . − ν 0 < 0.19 MeV/c2 + µ Antimatter particles e , ν¯e etc 3rd generation e is stable, 2 τ − 1 1777.0 MeV/c − 2 µ and τ are unstable. ντ 0 < 18.2 MeV/c
Neutrinos are stable and almost massless. Only know limits on ν masses, but have measured mass differences to be < 1 eV/c2. Not completely true, see later... Charged leptons experience only the electromagnetic & weak forces. Neutrinos experience only the weak force.
Dr. Tina Potter 1. Introduction 9
Matter Quarks
Quarks experience all the forces (strong, electromagnetic, weak).
Flavour Charge [e] Mass Strong Weak EM 1 st 1 generation Spin 2 fermions 2 2 u +3 2.3 MeV/c 6 distinct flavours d 1 4.8 MeV/c2 Fractional charge leptons: −3 2nd generation 2 u c t +3 c +2 1.3 GeV/c2 3 d s b 1 s 1 95 MeV/c2 −3 −3 Antiquarksu ¯, d¯ etc 3rd generation t +2 173 GeV/c2 Quarks are confined within hadrons, 3 + ¯ b 1 4.7 GeV/c2 e.g. p=(uud), π =(ud) −3 Quarks come in three colours (colour charge) Red, Green, Blue. Colour is a label for the charge of the strong interaction. Unlike the electric charge (+ ), the strong charge has three orthogonal colours (RGB). −
Dr. Tina Potter 1. Introduction 10 Matter Hadrons
Single, free quarks have never been observed. They are always confined in bound states called hadrons. Macroscopically, hadrons behave as almost point-like composite particles. Hadrons have two types: Mesons (qq¯): Bound states of a quark and an antiquark. Mesons have integer spin 0, 1, 2... bosons. + ¯ 2 1 e.g. π (ud), charge = (+3 + +3)e = +1e ≡ 2 1 + π− (¯ud), charge = ( + )e = 1e; antiparticle of π ≡ −3 −3 − π0 (uu¯ dd¯)/√2, charge = 0; is its own antiparticle. ≡ − Baryons (qqq): Bound states of three quarks. 1 3 Baryons have half-integer spin 2, 2... fermions. e.g. p (udu), charge = (+2 + 1 + +2)e = +1e ≡ 3 −3 3 n (dud), charge = ( 1 + +2 + 1)e = 0 ≡ −3 3 −3 Antibaryons e.g.p ¯ (¯ud¯u¯),n ¯ (d¯u¯d¯) ≡ ≡
Dr. Tina Potter 1. Introduction 11
Matter Nuclei
A nucleus is a bound state of Z protons and N neutrons. Protons and neutrons are generically referred to as nucleons. A (mass number) = Z (atomic number) + N (neutron number). A nuclide is a specific nucleus, characterised by Z, N.
A Notation: Nuclide Z X. 1 e.g. 1H or p: Z=1, N=0, A=1 2 1H or d: Z=1, N=1, A=2 4 2He or α: Z=2, N=2, A=4 208 82 Pb: Z=82, N=126, A=208
In principle, antinuclei and antiatoms can be made from antiprotons, antineutrons and positrons – experimentally challenging!
Dr. Tina Potter 1. Introduction 12 Matter The Periodic Table Periodic table classifies elements according to their chemical properties.
Only hydrogen, helium and lithium were formed in the Big Bang. All other elements are formed in stars. Natural elements, H(Z=1) to U(Z=92). Dr. Tina Potter 1. Introduction 13
Matter Chart of the nuclides
Many more nuclides than elements.
Colour coded according to decay mode.
Dr. Tina Potter 1. Introduction 14 Forces Classical Picture
A force is ’something’ which pushes matter around and causes objects to change their motion. In classical physics, the electromagnetic forces arise via action at a distance through the electric and magnetic fields, E~ and B~. q q ~r F~ = 1 2 r 2 Newton: “...that one body should act upon another at a distance, through a vacuum, without the mediation of anything else, by and through which their force may be conveyed from one to another, is to me so great an absurdity that I believe no man who has, in philosophical matters, a competent faculty of thinking, can ever fall into it. Gravity must be caused by an agent, acting constantly according to certain laws, but whether this agent be material or immaterial, I leave to the consideration of my reader.”
Dr. Tina Potter 1. Introduction 15
Forces Quantum Mechanics Matter particles are quantised in QM, and the electromagnetic field should also be quantised (as photons). Forces arise through the exchange of virtual field quanta called Gauge Bosons. This process is called “second quantisation”.
This process violates energy/momentum conservation (more later). However, this is permissible for sufficiently short times owing to the Uncertainty Principle The exchanged particle is “virtual” – meaning it doesn’t satisfy E 2 = p2c2 + m2c4. Uncertainty principle: ∆E∆t ~ range R c∆t ~c/∆E i.e. larger energy transfer∼ (larger⇒ force) ∼smaller∼ range. ↔ Prob(emission of a quantum) q1, Prob(absorption of a quanta) q2 Coulomb’s law can be regarded∝ as the resultant effect of all virtual∝ exchanges. Dr. Tina Potter 1. Introduction 16 Forces The four forces
All known particle interactions can be explained by four fundamental forces.
Carried by the gluon. Carried by the W and Holds atomic nuclei Z bosons. Responsible together. for radioactive decay.
Carried by the photon. Carried by the graviton. Acts between charged Acts between massive particles. particles.
Dr. Tina Potter 1. Introduction 17
Forces Gauge bosons
Gauge bosons mediate the fundamental forces Spin 1 particles i.e. Vector Bosons Interact in a similar way with all fermion generations The exact way in which the Gauge Bosons interact with each type of lepton or quark determines the nature of the fundamental forces. This defines the Standard Model.
Force Boson Spin Strength Mass Strong 8 gluons g 1 1 massless 2 Electromagnetic photon γ 1 10− massless + 7 Weak W and ZW , W −, Z 1 10− 80, 91 GeV 39 Gravity graviton ? 2 10− massless
Gravity is not included in the Standard Model. The others are.
Dr. Tina Potter 1. Introduction 18 Forces Range of forces The maximum range of a force is inversely related to the mass of the exchanged bosons.
2 ∆E∆t ~, E = mc ∼ ~ ~c ~ mc2 r ⇒ ∼ ∆t ∼ r ⇒ ∼ mc
Force Range [m] Strong inf 15 Strong (nuclear) 10− Electromagnetic inf 18 Weak 10− Gravity inf
Due to quark confinement, nucleons start to experience the strong interaction at 2 fm. ∼ Dr. Tina Potter 1. Introduction 19 Summary
Particle vs nuclear physics Matter: generations, quarks, leptons, hadrons, nuclei Forces: classical vs QM, fundamental forces, gauge bosons, range
Up next... Section 2: Kinematics, Decays and Reactions.
Dr. Tina Potter 1. Introduction 20 Glossary
Strong force - force which binds quarks into hadrons; mediated by gluons. Electromagnetic Force - force between charged particles, mediated by photons. Weak force - force responsible for β-decay. Mediated by W and Z bosons. Gauge boson - particle which mediates a force. Lepton - fermion which does not feel the strong interaction. Neutrino - uncharged lepton which experiences only weak interactions. Quark - fundamental fermion which experiences all forces. Hadron - bound state of quarks and/or antiquarks. Baryon - hadron formed from three quarks. Meson - hadron formed from quark+antiquark. Generations/Families - three replicas of the fundamental fermions. Nucleus - massive bound state of neutrons and protons at centre of an atom. Strong nuclear force - strong force between nucleons which binds atomic nucleus. Mediated by mesons, such as the pion. Nucleon - proton or neutron. Nuclide - specific nuclear species with N neutrons and Z protons. Mass number - total number of nucleons in nucleus, A. Atomic Number - number of protons in nucleus, Z. Neutron Number - number of neutrons in nucleus, N. Isobars - nuclides with the same Mass Number A. Isotopes - nuclides with the same Atomic Number Z. Isotones - nuclides with the same Neutron Number N. Dr. Tina Potter 1. Introduction 21 2. Kinematics, Decays and Reactions Particle and Nuclear Physics
Dr. Tina Potter
Dr. Tina Potter 2. Kinematics, Decays and Reactions 1 In this section...
Natural units Symmetries and conservation laws Relativistic kinematics Particle properties Decays Cross-sections Scattering Resonances
Dr. Tina Potter 2. Kinematics, Decays and Reactions 2 Units
The usual practice in particle and nuclear physics is to use Natural Units. Energies are measured in units of eV: Nuclear keV(103 eV), MeV(106 eV) Particle GeV(109 eV), TeV(1012 eV) Masses are quoted in units of MeV/c2 or GeV/c2 (using E = mc2) 31 31 8 2 2 e.g. electron mass me = 9.11 10− kg = (9.11 10− )(3 10 ) J/c 14 19× 2 ×5 2 × 2 = 8.20 10− /1.602 19− eV/c = 5.11 10 eV/c = 0.511 MeV/c × × × Atomic/nuclear masses are often quoted in unified (or atomic) mass units 12 1 unified mass unit (u) = (mass of a 6 C atom) / 12 27 2 1 u = 1 g/N = 1.66 10− kg = 931.5 MeV/c A × 28 2 Cross-sections are usually quoted in barns: 1b = 10− m .
Dr. Tina Potter 2. Kinematics, Decays and Reactions 3
Units Natural Units
Choose energy as the basic ...and simplify by unit of measurement... choosing ~ = c = 1
Energy GeV GeV Momentum GeV/c GeV Mass GeV/c2 GeV 1 1 Time( GeV/~)− GeV− 1 1 Length( GeV/~c)− GeV− 2 2 Cross-section( GeV/~c)− GeV−
Reintroduce “missing” factors of ~ and c to convert back to SI units. ~c = 0.197 GeV fm = 1 Energy Length 25 ←→ ~ = 6.6 10− GeV s = 1 Energy Time × 8 1 ←→ c = 3.0 10 ms− = 1 Length Time × ←→ Dr. Tina Potter 2. Kinematics, Decays and Reactions 4 Units Examples
6 2 1 cross-section σ = 2 10 GeV− change into standard units × − Need to change units of energy to length. Use ~c = 0.197 GeVfm = 1. 1 GeV− = 0.197 fm 6 32 2 1 15 σ = 2 10− (3.89 10− m ) GeV− = 0.197 10− m 2 × 32 2 × ×38 2 × GeV− = 3.89 10− m = 7.76 10− m × × 28 2 And using 1 b = 10− m , σ = 0.776 nb
1 2 lifetime τ = 1/Γ = 0.5 GeV− change into standard units 1 25 Need to change units of energy− to time. Use ~ = 6.6 10− GeV s = 1. 1 × 25 GeV− = 6.6 10− s 25 25 × τ = 0.5 (6.6 10− s) = 3.3 10− s × × ×
Also, can have Natural Units involving electric charge: 0 = µ0 = ~ = c = 1 3 Fine structure constant (dimensionless) 2 2 α = e 1 becomes α = e 1 i.e. e 0.30(n.u.) 4π0~c ∼ 137 4π ∼ 137 ∼ Dr. Tina Potter 2. Kinematics, Decays and Reactions 5 Symmetries and conservation laws
The most elegant and powerful idea in physics Noether’s theorem: every differentiable symmetry of the action of a physical system has a corresponding conservation law.
Symmetry Conserved current Time, t Energy, E Translational, x Linear momentum, p Rotational, θ Angular momentum, L Probability Total probability always 1 Lorentz invariance Charge Parity Time (CPT) Gauge charge (e.g. electric, colour, weak)
Lorentz invariance: laws of physics stay the same for all frames moving with a uniform velocity. Gauge invariance: observable quantities unchanged (charge, E, v) when a field is transformed. Dr. Tina Potter 2. Kinematics, Decays and Reactions 6 Relativistic Kinematics Special Relativity
Nuclear reactions Low energy, typically K.E. (10 MeV) nucleon rest energies. non-relativistic formulaeO ok Exception:⇒ always treat β-decay relativistically (m 0.5 MeV < 1.3 MeV m m ) e ∼ ∼ n − p Particle physics High energy, typically K.E. (100 GeV) rest mass energies. relativistic formulae usuallyO essential. ⇒
Dr. Tina Potter 2. Kinematics, Decays and Reactions 7
Relativistic Kinematics Special Relativity
Recall the energy E and momentum p of a particle with mass m 1 v E = γm, p~ = γβm γ = , β = = v | | 1 β2 c − E p~ p or γ = , β = | | 2 2 2 m E and these are related by E = p~ + m
Interesting cases when a particle is at rest, p~ = 0, E = m, when a particle is massless, m = 0, E = p~ , | | when a particle is ultra-relativistic E m, E p~ . ∼ | | Kinetic energy (K.E., or T ) is the extra energy due to motion T = E m = (γ 1)m in the non-relativistic limit−β 1, −T = 1mv 2 2
Dr. Tina Potter 2. Kinematics, Decays and Reactions 8 Relativistic Kinematics Four-Vectors The kinematics of a particle can be expressed as a four-vector, e.g. p = (E, p~), pµ = (E,p ~) and x = (t, ~x), xµ = (t, ~x) µ − µ − µ : 0 3 multiply by a metric tensor to raise/lower indices 1 0 0 0 → v µ µv µv 0 1 0 0 gµv = g = − pµ = gµv p , p = g pv 0 0 1 0 − 0 0 0 1 − Scalar product of two four-vectors Aµ = (A0, A~), Bµ = (B0, B~) is invariant: AµB = A.B = A0B0 A~.B~ µ − or µ µ v µ v 2 2 2 2 p pµ = p gµv p = p gµv p = g00p0 + g11p1 + g22p2 + g33p3 µ=0,3 v=0,3 X X = E 2 p~ 2 = m2 invariant mass − | | (t, ~x) and (E,p ~) transform between frames of reference, but d 2 = t2 ~x2 Invariant interval is constant m2 = E 2− p~2 Invariant mass is constant Dr. Tina Potter− 2. Kinematics, Decays and Reactions 9
Relativistic Kinematics Invariant Mass A common technique to identify particles is to form the invariant mass from their decay products. Remember, for a single particle m2 = E 2 p~2. − For a system of particles, where X 1 + 2 + 3...n: → n 2 n 2 M2 = ((E ,p ~ ) + (E ,p ~ ) + ...)2 = E p~ X 1 1 2 2 i − i i=1 ! i=1 ! X X In the specific (and common) case of a two-body decay, X 1 + 2, this reduces to → M2 = m2 + m2 + 2 (E E p~ p~ cos θ) X 1 2 1 2 − | 1|| 2| n.b. sometimes invariant mass M is called “centre-of-mass energy” ECM, or √s
Dr. Tina Potter 2. Kinematics, Decays and Reactions 10 Relativistic Kinematics Decay Example
Consider a charged pion decaying at rest in the lab frame π− µ−ν¯ → µ Find the momenta of the decay products
Dr. Tina Potter 2. Kinematics, Decays and Reactions 11 How do we study particles and forces?
Static Properties What particles/states exist? Mass, spin and parity (JP), magnetic moments, bound states Particle Decays Most particles and nuclei are unstable. Allowed/forbidden decays Conservation Laws. → Particle Scattering Direct production of new massive particles in matter-antimatter annihilation. Study of particle interaction cross-sections. Use high-energies to study forces at short distances. Force Typical Lifetime [s] Typical cross-section [mb] 23 Strong 10− 10 20 2 Electromagnetic 10− 10− 8 13 Weak 10− 10−
Dr. Tina Potter 2. Kinematics, Decays and Reactions 12 Particle Decays Reminder
Most particles are transient states – only a few live forever (e−, p, ν, γ...). Number of particles remaining at time t λt N(t) = N(0)p(t) = N(0)e− where N(0) is the number at time t = 0.
dN λt Rate of decays = λN(0)e− = λN(t) dt − − Assuming the nuclei only decay. More complicated if they are also being created. dN Activity A(t) = = λN(t) dt
Its rather common in nuclear physics to use the half-life (i.e. the time over which 50% of the particles decay). In particle physics, we usually quote the mean life. They are simply related:
N(0) λτ ln 2 N(τ ) = = N(0)e− 1/2 τ = = 0.693τ 1/2 2 ⇒ 1/2 λ Dr. Tina Potter 2. Kinematics, Decays and Reactions 13
Particle Decays Multiple Particle Decay Decay Chains frequently occur in nuclear physics
N1 λ1 N2 λ2 N3 ... −→ Parent−→ Daughter−→ Granddaughter e.g. 235U 231Th 231Pa 235→ → 8 τ1/2( U) = 7.1 10 years 231 × τ1/2( Th) = 26 hours
Activity (i.e. rate of decay) of the daughter is λ2N2(t). Rate of change of population of the daughter dN (t) 2 = λ N (t) λ N (t) dt 1 1 − 2 2 Units of Radioactivity are defined as the number of decays per unit time. Becquerel (Bq) = 1 decay per second Curie (Ci) = 3.7 1010 decays per second. Dr. Tina Potter 2.× Kinematics, Decays and Reactions 14 Particle Decays
A decay is the transition from one quantum state (initial state) to another (final or daughter). The transition rate is given by Fermi’s Golden Rule:
2 Γ(i f ) = λ = 2π Mfi ρ(Ef ) ~ = 1 → | | where λ is the number of transitions per unit time Mfi is the transition matrix element ρ(Ef ) is the density of final states. λ dt is the (constant) probability a particle will decay in time dt. ⇒
Dr. Tina Potter 2. Kinematics, Decays and Reactions 15
Particle Decays Single Particle Decay Let p(t) be the probability that a particle still exists at time t, given that it was known to exist at t = 0. Probability for particle decay in the next time interval dt is = p(t)λ dt Probability that particle survives the next is = p(t + dt) = p(t)(1 λ dt) − dp p(t)(1 λ dt) = p(t + dt) = p(t) + dt − dt dp = p(t)λ dt − p dp t = λ dt p − Z1 Z0 λt p(t) = e− Exponential Decay Law ⇒ Probability that a particle lives until time t and then decays in time dt is λt p(t)λ dt = λe− dt Dr. Tina Potter 2. Kinematics, Decays and Reactions 16 Particle Decays Single Particle Decay
The average lifetime of the particle
∞ λt λt ∞ λt 1 λt ∞ 1 τ = t = tλe− dt = te− ∞ + e− dt = e− = h i − 0 −λ λ Z0 Z0 0 1 t/τ τ = p(t) = e− λ Finite lifetime uncertain energy ∆E, (c.f. Resonances, Breit-Wigner) ⇒ Decaying states do not correspond to a single energy – they have a width ∆E ~ ∆E.τ ~ ∆E = ~λ = 1 (n.u.) ∼ ⇒ ∼ τ ~ The width, ∆E, of a particle state is therefore Inversely proportional to the lifetime τ Proportional to the decay rate λ (or equal in natural units)
Dr. Tina Potter 2. Kinematics, Decays and Reactions 17 Decay of Resonances
QM description of decaying states Consider a state formed at t = 0 with energy E0 and mean lifetime τ iE t t/2τ 2 2 t/τ ψ(t) = ψ(0)e− 0 e− ψ(t) = ψ(0) e− | | | | i.e. the probability density decays exponentially (as required). The frequencies (i.e. energies, since E = ω if ~ = 1) present in the wavefunction are given by the Fourier transform of ψ(t), i.e.
∞ iEt ∞ t(iE + 1 ) iEt f (ω) = f (E) = ψ(t)e dt = ψ(0)e− 0 2τ e dt Z0 Z0 ∞ t(i(E E)+ 1 ) iψ(0) = ψ(0)e− 0− 2τ dt = (E E) i Z0 0 − − 2τ Probability of finding state with ψ(0) 2 P(E) = f (E) 2 = | | energy E = f (E) f (E) is | | (E E)2 + 1 ∗ 0 − 4τ 2 Dr. Tina Potter 2. Kinematics, Decays and Reactions 18 Decay of Resonances Breit-Wigner
Probability for producing the decaying state has this energy dependence, i.e. resonant when E = E0 1 P(E) ∝ (E E)2 + 1/4τ 2 0 − Consider full-width at half-maximum Γ P(E = E ) 4τ 2 0 ∝ 1 1 1 P(E = E0 Γ) = ± 2 ∝ (E E 1Γ)2 + 1/4τ 2 Γ2 + 1 0 − 0 ∓ 2 4 4τ 2 1 1 1 P(E = E Γ) = P(E = E ), = 2τ 2 0 ± 2 2 0 ⇒ Γ2 1 4 + 4τ 2 1 Total width (using natural units) Γ = = λ τ
Dr. Tina Potter 2. Kinematics, Decays and Reactions 19 Partial Decay Widths
Particles can often decay with more than one decay mode + + e.g. Z e e−, or µ µ−, or qq¯ etc, each with its own transition rate, → i.e. from initial state i to final state f : λ = 2π M 2 ρ(E ) f | fi | f
The total decay rate is given by λ = f λf P1 This determines the average lifetime τ = λ
The total width of a particle state Γ = λ = f λf
is defined by the partial widths Γf = λf P The proportion of decays to a particular Γf decay mode is called the branching fraction Bf = Γ , f Bf = 1 or branching ratio P
Dr. Tina Potter 2. Kinematics, Decays and Reactions 20 Reactions and Cross-sections
The strength of a particular reaction between two particles is specified by the interaction cross-section. Cross-section σ – the effective target area presented to the incoming particle for it to cause the reaction. 28 2 Units: σ 1 barn (b) = 10− m Area
σ is defined as the reaction rate per target particle Γ, per unit incident flux Φ Γ = Φσ where the flux Φ is the number of beam particles passing through unit area per second. Γ is given by Fermi’s Golden Rule (previously used λ).
Dr. Tina Potter 2. Kinematics, Decays and Reactions 21 Scattering with a beam Consider a beam of particles incident upon a target:
Target of n nuclei per Beam of N particles per unit volume unit time in an area A Target thickness dx is small
Number of target particles in area A, NT = nA dx Effective area for absorption = σNT = σnA dx Incident flux Φ = N/A Number of particles scattered per unit time = dN = ΦσN = N σnA dx − T A dN σ = − nN dx Dr. Tina Potter 2. Kinematics, Decays and Reactions 22 Attenuation of a beam Beam attenuation in a target of thickness L:
Thick target σnL 1: N dN L = σn dx − N ZN0 Z0 σnL N = N0e− This is exact. i.e. the beam attenuates exponentially.
σnL Thin target σnL << 1, e− 1 σnL ∼ − N = N (1 σnL) 0 − Useful approximation for thin targets. Or, the number scattered = N N = N σnL 0 − 0 Mean free path between interactions = 1/nσ often referred to as “interaction length”.
Dr. Tina Potter 2. Kinematics, Decays and Reactions 23 Differential Cross-section The angular distribution of the scattered particles is not necessarily uniform ** n.b. dΩ can be considered in position space, or momentum space **
Number of particles scattered per unit time into dΩ is dN dΩ = dσ ΦNT
Differential cross-section dσ dN dΩ = units: area/steradian dΩ (Φ N dΩ) × T × The differential cross-section is the number of particles scattered per unit time and solid angle, divided by the incident flux and by the number of target nuclei, NT , defined by the beam area. Most experiments do not cover 4π solid angle, and in general we measure dσ/ dΩ. Angular distributions provide more information than the total cross-section about the mechanism of the interaction, e.g. angular momentum.
Dr. Tina Potter 2. Kinematics, Decays and Reactions 24 Partial Cross-section
Different types of interaction can occur between particles + + e.g. e e− γ, or e e− Z... → → σtot = σi i X where the σi are called partial cross-sections for different final states.
Types of interaction Elastic scattering: a + b a + b only the momenta→ of a and b change Inelastic scattering: a + b c + d final state is→ not the same as initial state
Dr. Tina Potter 2. Kinematics, Decays and Reactions 25 Scattering in QM
Consider a beam of particles scattering from a fixed potential V (r):
q~ = p~f p~i “momentum− transfer”
NOTE: using natural units p~ = ~~k p~ = ~k etc → The scattering rate is characterised by the interaction cross-section Γ Number of particles scattered per unit time σ = = Φ Incident flux How can we calculate the cross-section? Use Fermi’s Golden Rule to get the transition rate Γ = 2π M 2ρ(E ) | fi | f where Mfi is the matrix element and ρ(Ef ) is the density of final states.
Dr. Tina Potter 2. Kinematics, Decays and Reactions 26 Scattering in QM 1st order Perturbation Theory using plane wave solutions of form i(Et p~.~r) ψ = Ne− − Require: 1 Wave-function normalisation 2 Matrix element in perturbation theory Mfi 3 Expression for incident flux Φ 4 Expression for density of states ρ(Ef )
1 Normalisation Normalise wave-functions to one particle in a box of side L: ψ 2 = N2 = 1/L3 | | N = (1/L)3/2
Dr. Tina Potter 2. Kinematics, Decays and Reactions 27 Scattering in QM
2 Matrix Element This contains the interesting physics of the interaction:
3 ip~ .~r ip~ .~r 3 M = ψ Hˆ ψ = ψ∗Hˆψ d ~r = Ne− f V (~r)Ne i d ~r fi h f | | ii f i Z Z 1 iq~.~r 3 M = e− V (~r) d ~r where q~ = p~ p~ fi L3 f − i Z 3 Incident Flux Consider a “target” of area A and a beam of particles travelling at velocity vi towards the target. Any incident particle within a volume viA will cross the target area every second. v A Φ = i n = v n A i where n is the number density of incident particles = 1 per L3 Flux = number of incident particles crossing unit area per second 3 Φ = vi/L Dr. Tina Potter 2. Kinematics, Decays and Reactions 28 Scattering in QM
4 Density of States also known as “phase space” For a box of side L, states are given by the periodic boundary conditions: 2π p~ = (p , p , p ) = (n , n , n ) x y z L x y z Each state occupies a volume (2π/L)3 in p space (neglecting spin). Number of states between p and p + dp in solid angle dΩ L 3 L 3 dN = d3p~ = p2 dp dΩ( d3p~ = p2 dp dΩ) 2π 2π dN L 3 ρ(p) = = p2 dΩ ∴ dp 2π Density of states in energy E 2 = p2 + m2 2E dE = 2p dp dE = p ⇒ ⇒ dp E dN dN dp L 3 E ρ(E) = = = p2 dΩ dE dp dE 2π p L 3 For relativistic scattering (E p) ρ(E) = E 2 dΩ 2π ∼ Dr. Tina Potter 2. Kinematics, Decays and Reactions 29 Scattering in QM
Putting all the parts together:
3 2 3 1 2 L 1 iq~.~r 3 L dσ = 2π M ρ(E ) = 2π e− V (~r) d ~r p E dΩ Φ | fi | f v L3 2π f f i Z
2 dσ 1 iq~.~r 3 = e− V (~r) d ~r p E dΩ (2π)2v f f i Z
For relativistic scattering, vi = c = 1 and p E Born approximation for the differential∼ cross-section 2 2 dσ E iq~.~r 3 = e− V (~r) d ~r dΩ (2π)2 Z 2 2 n.b. may have seen the non-relativistic version, using m instead of E
Dr. Tina Potter 2. Kinematics, Decays and Reactions 30 Rutherford Scattering Consider relativistic elastic scattering in a Coulomb potential Ze2 Zα V (~r) = = −4π0r − r Special case of Yukawa potential V = ge mr /r 16π2Z 2α2 − M 2 = with g = Zα and m = 0 (see Appendix C) | if | q4
q~ = p~ p~ f − i 2 2 2 q~ = p~i + p~f 2p~i.~pf | | | | elastic| | scattering,− p~ = p~ = p~ | i | | f | | | θ = 2 p~ 2(1 cos θ) = 4E 2 sin2 | | − 2 dσ 4E 2Z 2α2 4E 2Z 2α2 Z 2α2 = = = dΩ q4 4 4 θ 2 4 θ 16E sin 2 4E sin 2 Dr. Tina Potter 2. Kinematics, Decays and Reactions 31 Cross-section for Resonant Scattering
Some particle interactions take place via an intermediate resonant state which then decays a + b Z ∗ c + d → →
Two-stage picture: (Bohr Model)
Formation a + b Z ∗ Decay Z ∗ c + d → → Occurs when the collision energy The decay of the resonance Z ∗ is ECM the natural frequency (i.e. independent of the mode of mass)∼ of a resonant state. formation and depends only on the properties of the Z ∗. May be multiple decay modes.
Dr. Tina Potter 2. Kinematics, Decays and Reactions 32 Resonance Cross-Section The resonance cross-section is given by Γ 2 σ = with Γ = 2π Mfi ρ(Ef ) ** same as Born Approx. Φ | | v incident flux Φ = i L3 1 2 dN L 3 dσ = 2π Mfi ρ(Ef ) ∗∗ density of states ρ(p) = = p2 dΩ dp 2π Φ | | L3 p2L3 2 f dN dp L 3 E = 2π Mfi dΩ ρ(E) = = p2 dΩ 3 → dp dE 2π p vi | | vf (2π) L 3 1 2 = p2 dΩ dσ pf 2 factors of L cancel 2π v = M 2 fi as before, M L3 dΩ (2π) vivf | | ∝ nd The matrix element Mfi is given by 2 order Perturbation Theory
MiZ MZf nd Mfi = n.b. 2 order effects are large since E EZ E E is small large perturbation Z − − Z → where the sum runsX over all intermediate states. Near resonance, effectively only one state Z contributes. Dr. Tina Potter 2. Kinematics, Decays and Reactions 33 Resonance Cross-Section Consider one intermediate state described by
Γ iE0t t/2τ i(E0 i )t ψ(t) = ψ(0)e e− = ψ(0)e− − 2 this describes a states with energy = E iΓ/2 0 − 2 2 2 MiZ MZf Mfi = | | | | | | (E E )2 + Γ2 − 0 4 Rate of decay of Z: 2 2 2 2 4πpf 2 pf ΓZ f = 2π MZf ρ(Ef ) = 2π MZf 3 = MZf → | | | | (2π) vf | | πvf Rate of formation of Z: 2 2 2 2 4πpi 2 pi Γi Z = 2π MiZ ρ(Ei) = 2π MiZ 3 = MiZ → | | | | (2π) vi | | πvi nb. M 2 = M 2. | Zi | | iZ | Hence MiZ and MZf can be expressed in terms of partial widths. Dr. Tina Potter 2. Kinematics, Decays and Reactions 34 Resonance Cross-Section
2 dσ pf 2 Putting everything together: = 2 Mfi dΩ (2π) vivf | |
2 4πpf πvf πvi ΓZ iΓZ f π ΓZ iΓZ f σ = → → = → → ⇒ (2π)2 v v p2 p2 (E E )2 + Γ2 p2 (E E )2 + Γ2 i f f i − 0 4 i − 0 4 We need to include one more piece of information to account for spin...
Dr. Tina Potter 2. Kinematics, Decays and Reactions 35 Resonance Cross-Section
πg ΓZ iΓZ f Breit-Wigner Cross-Section σ = . → → p2 (E E )2 + Γ2 i − 0 4 The g factor takes into account the spin 2JZ + 1 a + b Z∗ c + d, g = → → (2Ja + 1)(2Jb + 1) and is the ratio of the number of spin states for the resonant state to the total number of spin states for the a+b system, i.e. the probability that a+b collide in the correct spin state to form Z∗.
Useful points to remember:
pi is calculated in the centre-of-mass frame (σ is independent of frame of reference!) p lab momentum if the target is heavy (often true in nuclear physics, but not in i ∼ particle physics).
E is the total energy (if two particles colliding, E = E1 + E2) Γ is the total decay rate
ΓZ i and ΓZ f are the partial decay rates → → Dr. Tina Potter 2. Kinematics, Decays and Reactions 36 Resonance Cross-Section Notes
Total cross-section πg ΓZ i ΓZ f σtot = σ(i f ) σ = → → → p2 2 Γ2 i (E E0) + 4 Xf − Replace Γf by Γ in the Breit-Wigner formula
Elastic cross-section σ = σ(i i) el → so, Γf = Γi
4πgΓiΓf On peak of resonance (E = E0) σpeak = 2 2 pi Γ 2 4πgBi 4πgBi Γi σel Thus σel = 2 , σtot = 2 , Bi = = pi pi Γ σtot
By measuring σtot and σel, can cancel Bi and infer g and hence the spin of the resonant state.
Dr. Tina Potter 2. Kinematics, Decays and Reactions 37
Resonances Nuclear Physics Example
Can produce the same resonance from different initial states, decaying into various final states, e.g.
σ[60Ni(α, n)63Zn] σ[63Cu(p, n)63Zn] ∼ n.b. common notation for nuclear reactions: a+A b+B A(a,b)B → ≡ Energy of p selected to give same c.m. energy as for α interaction. Dr. Tina Potter 2. Kinematics, Decays and Reactions 38 Resonances Particle Physics Example
The Z boson
Γ 2.5 GeV Z ∼
1 1 τ = = 0.4 GeV− ΓZ = 0.4 ~ × 25 = 2.5 10− s ×
25 (~ = 6.6 10− GeV s) ×
Dr. Tina Potter 2. Kinematics, Decays and Reactions 39
Resonances π−p scattering example Resonance observed at p 0.3 GeV, E 1.25 GeV π ∼ CM ∼
σ = σ(π−p R anything) 72 mb total → → ∼ σ = σ(π−p R π−p) 28 mb elastic → → ∼ Dr. Tina Potter 2. Kinematics, Decays and Reactions 40 Resonances π−p scattering example
Dr. Tina Potter 2. Kinematics, Decays and Reactions 41 Summary
Units: MeV, GeV, barns Natural units: ~ = c = 1 Relativistic kinematics: most particle physics calculations require this! Revision of scattering theory: cross-section, Born Approximation. Resonant scattering Breit-Wigner formula (important in both nuclear and particle physics):
πg ΓZ iΓZ f σ = → → p2 (E E )2 + Γ2 i − 0 4 Measure total and elastic σ to measure spin of resonance.
Up next... Section 3: Colliders and Detectors
Dr. Tina Potter 2. Kinematics, Decays and Reactions 42 T. Potter Lent/Easter Terms 2021
Part II Particle and Nuclear Physics Examples Sheet 1
Matter and Forces
1. Particles, Tripos A-style question. Explain the meaning of the terms quark, lepton, hadron, nucleus and boson as used in the classification of particles.
Relativistic kinematics 2. Natural Units, Tripos A-style question. Explain what is meant by natural units and the Heaviside-Lorentz system. (a) The reduced Compton wavelength of a particle can be written in natural units as 1 λ¯ = m 2 where m is the mass of the particle. Estimateλ ¯ for a pion (mπ = 139.6 MeV/c ). Quote your answer in natural units and then convert to SI units. + (b) The total cross-section for e e− annihilation can be written in natural units as 4 πα2 σ = 3 s 1 where α = 137 is the fine structure constant and √s is the centre-of-mass energy. Estimate 2 σ at a centre-of-mass energy equal to the Z mass (mZ = 91.2 GeV/c ). Calculate your answer in natural units and then convert to barns.
(c) Use dimensional analysis to add the appropriate factors of ~, 0 and c in the formulae for λ in (a) and for σ in (b), and then do the calculations directly in SI units.
28 2 [Note that ~c = 197 MeV fm; 1 barn = 10− m ] 3. Relativistic Kinematics, Tripos B-style question. Consider the decay of a particle X into two particles a and b. (a) Show that, in the rest frame of X, the energy of particle a can be written in natural units as 2 2 2 mX + ma mb Ea = − 2mX
where mi is the mass of particle i. What is the equivalent expression for the energy of particle b ? What is the energy if the final state particles are the same (or antiparticles of each other)?
1 (b) Show that the magnitude of the momentum of particle a can be written in natural units as 4 4 4 2 2 2 2 2 2 mX + ma + mb 2mX ma 2mX mb 2mamb pa = − − − . p 2mX What is the equivalent expression for the momentum of particle b ? What is the value of the momentum if the final state particles are the same (or antiparticles of each other) ? Show that if one of the final state particles is massless, e.g. mb = 0, then the expression for the momentum simplifies to 2 2 mX ma pa = − 2mX
[Note: replacing mX by the centre-of-mass energy, √s, in the above gives the equivalent expressions for collision processes.] (c) The HERA collider at DESY provided head-on collisions between an electron beam of 27.5 GeV and a proton beam of 920 GeV. What energy of electron beam colliding with a fixed target would be required to obtain the same centre-of-mass energy? Would HERA have had sufficient energy to have produced a Higgs boson of mass 125 GeV?
4. Ω Decay, Tripos A-style question. The figure shows a photograph and line diagram of the event corresponding to the first obser- 0 0 0 0 vation of the Ω− baryon (Ω− Ξ π−,Ξ Λ π ) in a K−p interaction in a liquid hydrogen bubble chamber (from Barnes→et al., Phys.→ Rev. Lett. 12 (1964) 204):
2 0 + (a) The two photons from the π γγ decay are both seen to convert to e e− pairs. Show + → that the process γ e e− is kinematically forbidden in vacuo. Explain why the con- version process can→ take place in the presence of matter, and draw a Feynman diagram representing photon conversion in material, as seen in the figure. 0 (b) The π− and Ξ from the Ω− decay have momenta of 281 MeV/c and 1906 MeV/c respec- tively. Their spatial opening angle is 71◦. Calculate the mass of the Ω− and compute its momentum.
(c) The length of the Ω− flight path is 2.5 cm. Calculate the proper lifetime of the Ω−.
2 0 2 [m(π−)=139.6 MeV/c , m(Ξ )=1315 MeV/c ]
Decays and Reactions
5. Radioactive Decay, Tripos B-style question. A sample of gold is exposed to a neutron beam of constant intensity such that 1010 neutrons per second are absorbed in the reaction
n + 197Au 198Au + γ → The nuclide 198Au undergoes β decay to 198Hg with a mean lifetime of 4 days.
(a) How many atoms of 198Au will be present after 6 days of irradiation? (b) How many atoms of 198Hg will be present after 6 days assuming that the neutron beam has no effect on the Hg? (c) What is the equilibrium number of 198Au nuclei?
6. Caesium Decay, Tripos B-style question. The decay chain 139Cs 139Ba 139La is observed for an initially pure sample of 1mCi of 139Cs. The half life of 139Cs is→ 9.5 minutes→ and that of 139Ba is 82.9 minutes; 139La is stable. Write down the rate equations for this system, and show that the number of Ba atoms present at time t is given by λCsNCs(0) λ t λ t N (t) = e− Cs e− Ba Ba λ λ − Ba − Cs in an obvious notation, where the λ values represent the corresponding decay rates. What is the maximium 139Ba activity (i.e. rate of 139Ba decay), and at what time does it occur? [1 Ci= 3.7x1010 disintegrations per second.]
7. Kaon Decay, Tripos A-style question.
(a) Calculate the branching fraction for the decay K+ π+π0, given that the partial width 8 → + 8 for this decay is 1.2 10− eV and the mean lifetime of the K meson is 1.2 10− s. × × (b) A beam of K+ mesons of momentum 10 GeV is produced. What fraction of them will remain undecayed 100 m downstream?
3 (c) When the K+ mesons decay to π+π0, what are the minimum and maximum laboratory energies of the produced π+ mesons?
8. Cross-sections, Tripos B-style question. Define the terms total cross-section and differential cross-section for scattering processes. A beam of neutrons with an intensity 105 particles per second traverses a thin foil of 235U with 1 2 1 a ”thickness” of 10− kg m− . There are three possible outcomes when a neutron interacts with a 235U nucleus: 2 (i) elastic scattering of the neutron, with a cross-section 10− b ; (ii) the capture of the neutron followed by the emission of a γ-ray., with cross-section 70 b; (iii) the capture of the neutron, followed by the resulting nucleus undergoing fission with cross- section 200 b. Using this information, determine:
(a) the intensity of the neutron beam transmitted by the foil; (b) the rate of fission reactions occurring in the foil induced by the incident beam; (c) the flux of neutrons elastically scattered out of the beam at a point 10 m from the foil, assuming that the neutrons are scattered isotropically.
9. Breit-Wigner Formula, Tripos A-style question. The Breit-Wigner formula for a reaction cross-section is given by πg Γ Γ σ(E) = i f . p2 (E E )2 + Γ2/4 i − 0 Explain the meaning of the symbols in this equation, and outline its derivation. The maximum value of the cross-section for radiative capture of neutrons in 123Te (i.e. the process n +123 Te 124 Te + γ) is 75 kb and is reached at a neutron energy of 2.2 eV, where −→ 123 the elastic width Γn is 0.0104 eV and the radiative width Γγ is 0.105 eV. The spin of Te in 1 its ground state is J = 2 . What is the elastic cross-section at resonance and what is the spin of the compound nucleus formed?
1To determine the number of particles that interact one must know the density of target particles and the thickness of the target. Instead of giving two numbers which simply have to be multiplied, it is common practice to quote the target thickness multiplied by the target density. This gives a target “thickness” in units of mass/area.
4 Numerical answers
3 1 8 2 2. (a) 7.16 10− MeV− , 1.41 fm; (b) 2.68 10− GeV− , 0.0104 nb × × 3. (c) √s = 318 GeV; 54 TeV. 4. (b) 1689 MeV/c2, 2015 MeV/c ; (c) 70 ps 5. (a) 2.7x1015; (b) 2.5x1015; (c) 3.46x1015 6. (a) 0.087 mCi; (b) 33.5 minutes 7. (a) 21.8% (b) 0.25, [0.88, 9.18] GeV. 1 1 5 2 1 8. (a) 99,311 particles s− ; (b) 510 s− ; (c) 2.03x10− particles m− s− 9. 7.4 kb; J = 1
Suggested Tripos Questions
Relativistic Kinematics: 2017 1(a), 2014 3, 2004 (3) C12(b)
Breit-Wigner resonances, production and decay rates: 2018 A1(a), 2015 3, 2005 (3) A3
5 T. Potter Lent/Easter Terms 2021
Part II Particle and Nuclear Physics Examples Sheet 2
Colliders and Detectors
10. Detector Signatures, Tripos A-style question. + For each e e− process below, sketch the signature in a typical cylindrical detector e.g.
Muon Spectrometer magnet
HCAL
ECAL
Tracker
+ + + (a) e e− µ µ−e e− → + + (b) e e− µ µ−γ → + (c) e e− ννγ¯ → + + + + 0 (d) e e− τ τ −, where the taus decay as τ − e−ν¯ ν and τ π π ν¯ → → e τ → τ + + 0 + (e) e e− π npπ¯ K K− → Draw the lowest order Feynman diagram for (a)-(d).
11. Detector Resolution, Tripos A-style question.
(a) In an experiment, the momentum measurement accuracy of the tracking detector is 1% for 1 GeV muons. What is the momentum accuracy for 20 GeV muons in the same apparatus?
1 (b) The energy resolution for 1 GeV electrons in the electromagetic calorimeter is 0.5%. What is the energy resolution for 10 GeV electrons?
Feynman Diagrams and QED
12. QED Feynman Diagrams, Tripos A-style question. Draw the lowest order Feynman diagram(s) for each of the following processes:
+ (a) γ e e− (in matter) → (b) e− + e− e− + e− → + + (c) e + e− e + e− → + + (d) e + e− µ + µ− → + (e) e + e− γ + γ → (f) γ + γ γ + γ (“Delbruck scattering” – forbidden classically) → 13. π0 Decay, Tripos A-style question.
0 P + (a) The π (J = 0−) decays predominantly to γγ but is also seen to decay to e e−γ (“Dalitz + + + 5 7 decay”), to e e−e e− and to e e− with branching fractions of 1.2%, 3.2 10− and 2 10− respectively. Draw the leading order Feynman diagrams for each of these× decays.× Based on the coupling constants involved (ignoring propagator effects etc.), give rough estimates of the branching fractions for each decay. 0 P + 5 (b) The ρ (J = 1−) decays to e e− with a branching fraction of 4 10− . Draw the Feynman × 0 + 0 diagram for this decay and comment on the difference between the π e e− and ρ + → → e e− partial widths.
0 0 17 24 [The π and ρ lifetimes are 8.4 10− s and 4.4 10− s respectively.] × × 14. Drell-Yan, Tripos A-style question. The Drell-Yan process, which is the production of charged lepton pairs in hadron-hadron in- + teractions (πN µ µ−+ anything, for example), proceeds via quark-antiquark annihilation into a single virtual→ photon. Draw a typical Feynman diagram for this process. Show that the + + Drell-Yan cross-sections in π p, π n, π−p and π−n interactions would be expected to be in the ratio 1 : 2 : 8 : 4. What would you expect to be the Drell-Yan cross-sections for pp andp ¯p collisions compared with the Drell-Yan cross-section for π+p interactions ?
QCD and the Quark Model
15. Spin and Parity, Tripos A-style question. When π− mesons are stopped in deuterium they form “pionic atoms” (π−d) which usually undergo transitions to an atomic s-state (` = 0), whereupon the capture reaction π−d nn → 2 occurs and destroys them. (The fact that capture normally occurs in an s-state is established from studies of the X-rays emitted in the transitions before capture). Given that the deuteron has spin-parity J P = 1+ and the pion has spin 0, show that these observations imply that the pion has negative intrinsic parity.
16. Hadron Masses, Tripos A-style question.
(a) Verify the quark model predictions given in the lectures for the following meson masses: Meson Calculated Observed (MeV) (MeV) π 140 138 K 484 496 η 559 549 ρ 780 776 ω 780 783 K∗ 896 892 φ 1032 1020
[Assume mu=md=310 MeV, ms=483 MeV and that the spin-spin interaction coefficient A= 0.0615 GeV3 in this case.]
What would the model predict for the mass of the η0 meson (measured mass 958 MeV)? P 3 + (b) What must be the total spin of any pair of quarks in the baryons in the J = 2 decuplet? Hence predict the masses of the decuplet baryons and compare your predictions with the measured values.
[Assume in this case mu=md=360 MeV, ms=540 MeV and the spin-spin interaction coefficient A= 0.026 GeV3.]
17. p and n Moments, Tripos B-style question. Derive the magnetic moments of the proton and neutron in the quark model as follows:
(a) Assuming all the quarks are in `=0 states what must be the total spin of the two u quarks in the proton? (Give reasons for your answer). 1 (b) Hence show that the wave function for a proton in the sz=+ 2 state can be written as 1 (2u u d u u d u u d ) √6 ↑ ↑ ↓ − ↑ ↓ ↑ − ↓ ↑ ↑
and derive a similar expression for the neutron. (c) Assuming u and d quarks have equal mass, write their magnetic moments in terms of this quark mass. (d) Hence predict the ratio of the proton and neutron magnetic moments. Compare with the observed values: µp = 2.79µN , µn = 1.91µN . What value for the quark mass is needed to give these values? Is this sensible? −
3 + (e) Consider now the magnetic moments of the Σ (uus) and Σ− (dds) baryons, which are 1 also members of the spin- 2 octet. Show that 4 µ + µ = [µ µ ] . Σ − Σ− 5 p − n
Test using measured values: µ + = 2.458 0.010µ , µ = 1.160 0.025µ . Σ ± N Σ− − ± N 18. ρ0 Decay, Tripos B-style question. 0 P Consider the decay of the ρ meson (J = 1−) in the following decay modes:
(a) ρ0 π0γ → 0 + (b) ρ π π− → (c) ρ0 π0π0 → 0 + (d) ρ e e− → In each case, draw an appropriate Feynman diagram and determine whether the process is allowed or forbidden. By considering the strength of the forces involved, list the decay modes in order of expected rate. The ratio of partial widths Γ(ρ0 π0γ)/Γ(ω0 π0γ) is approximately 0.1 while the ratio 0 + 0 + → → Γ(ρ e e−)/Γ(ω e e−) is approximately 10. Suggest an explanation for these observa- tions.→ → [The quark wavefunctions for the mesons are 1 [uu¯ dd¯] for both π0 and ρ0, and 1 [uu¯ + dd¯] √2 − √2 for ω0.]
19. J/ψ Meson, Tripos B-style question. The figure shown below (from Boyarski et al., Phys. Rev. Lett. 34 (1975) 1357) shows the orig- + inal e e− annihilation cross-section measurements from the Mark II Collaboration which con- tributed to the discovery of the J/ψ meson. The measurements were made during a fine scan of the e± beam energies at the SPEAR storage ring at SLAC which consisted of oppositely + circulating e and e− beams of equal energy. Figure (a) shows the cross-section for the process + + + e e− hadrons, (b) shows the cross-section for e e− µ µ− and (c) shows the cross-section +→ + → for e e− e e−. The latter two were measured in a limited acceptance cos θ < 0.6, where θ is the polar→ angle of the produced leptons. | |
4 The observed width (a few MeV) of the J/ψ resonance peak is predominantly caused by the + energy spread inherent in the e , e− beams at each measured point. The relative centre-of- mass energy between measurement points is however known very precisely (to about 1 part in 104). The actual J/ψ width is much smaller than the observed width, but can be extracted from the data as follows:
(a) The Breit-Wigner formula for the scattering of two particles of spin s1 and s2 in the region of a resonance of spin J is: λ2 (2J + 1) Γ Γ σ(E) = i f 4π (2s + 1)(2s + 1) [(E E )2 + Γ2/4] 1 2 − 0 where λ is the de Broglie wavelength of the incoming particles in the centre of mass frame, E is the centre of mass energy, E0 is the resonance energy, Γ is the total width of the
5 resonance and Γi (Γf ) is the partial width for decay into the initial (final) state. Show + that, for the production of the J/ψ resonance in e e− collisions, the integrated elastic cross-section under the resonance peak is given by
3 2 2 σ0 σ (E)dE λ B Γ ≡ el ≈ 8 Z + where B is the branching fraction for the decay J/ψ e e−. → (b) Assume that, at each scan point, the beam energy spread produces a spread of centre of mass energies E0 distributed about the average centre of mass energy E according to a probability distribution f(E0 E). Show that the measured area under the resonance − peak, σmeas(E)dE, is equal to the true area under the peak, σ(E)dE. + + (c) GivenR that the differential cross-section dσ/dΩ for the processR e e− J/ψ e e− is 2 → → proportional to 1+cos θ where θ is the angle between the final e− and the beam direction, calculate the fraction of J/ψ decays contained within the acceptance region cos θ < 0.6 + + | | imposed for the e e− and µ−µ channels. 1 (d) Use the data in the figure to estimate the quantities σ0 and B defined above . You should obtain σ0 500 nb MeV and B 0.065 for the leptonic decays, after correcting for the ∼ ∼ limited acceptance. Hence estimate Γ and Γee for the J/ψ.
(e) The corresponding widths for the φ meson are Γ=4.4 MeV and Γee=1.37 keV. Discuss why the J/ψ and φ mesons have similar leptonic widths Γee but very different total widths Γ.
1Note that the measured cross-sections contain a significant non-resonant contribution which must be subtracted. Note also that the scales of the graphs are logarithmic.
6 Numerical answers
11. (a)20%; (b)0.16%. 5 5 13. (a) 1, 0.028, 5 10− , 5 10− (just counting powers of α); (b) 1.56 µeV, 6 keV × × 14. σ(p¯p) = 17σ(π+p) 2 2 2 16. (a) m(η0) = 349 MeV/c ; (b) S=1, m(∆) = 1.230 GeV/c , m(Σ∗) = 1.383 GeV/c , m(Ξ) = 2 2 1.535 GeV/c , m(Ω−) = 1.687 GeV/c 17. (a) S=1; (d) µ /µ = 1.5, Mass of quark = 330 MeV/c2 p n − 19. (c) 50.4%; (d) using σ0 500 nb MeV, B 0.065 should obtain Γ 70 keV, Γ 4.6 keV ≈ ≈ ≈ ee ≈ Suggested Tripos Questions
Feynman Diagrams: 2016 3(a), 2010 3, 2008 A4 2008 (3) A4, 2009 (3) A1(b), 2010 (3) A1(c)
QCD: 2018 B3 last part, 2014 3, 2013 1(b)
Hadron physics and quark model: 2018 B3, 2017 4, 2016 1(b)
7 T. Potter Lent/Easter Terms 2021
Part II Particle and Nuclear Physics Examples Sheet 3
Weak Interactions
20. Tau Decay, Tripos A-style question. Draw a Feynman diagram for each of the principal decay modes of the τ −. Show that the τ lepton decay branching fractions should be approximately in the ratios
(τ − e−ν ν¯ ):(τ − µ−ν ν¯ ):(τ − hadrons) = 1 : 1 : 3 → τ e → τ µ → The actual ratios are approximately 1.02 : 1 : 3.5. Suggest a possible explanation.
Estimate the mean lifetime of the τ lepton, given that the branching fraction for τ − e−ν¯eντ is 18%, and assume the total decay rate is given by Sargent’s Rule →
G2 E5 Γ = F 0 60π3 . 2 2 [You may use mτ = 1.777 GeV/c , mµ = 0.106 GeV/c and take the mean µ lifetime to be 6 2.197 10− s. ] × 21. Threshold Energy, Tripos A-style question. In a beam of antineutrinos, it is proposed to search forν ¯τ via their interactions on nucleons in a stationary target to produce τ-leptons.
(a) Draw the Feynman diagram for the simplest such production process.
(b) Calculate the minimum energy of theν ¯τ which would permit τ-lepton production.
(c) What is the energy of the produced τ-lepton when theν ¯τ has this threshold energy ? (d) How far will the τ-lepton travel on average before decaying, given that its mean lifetime is 290 fs ?
[The masses of τ +, proton and neutron are 1.777 GeV/c2, 0.938 GeV/c2 and 0.940 GeV/c2 respectively.]
22. Omega Decay, Tripos A-style question. The Ω− baryon (sss), produced in the event shown in Q.4 is seen to decay weakly through the 0 0 0 0 0 decay chain Ω− Ξ π−,Ξ Λ π and Λ pπ−. Draw the Feynman diagrams for the → 0 → 0 → decays of the Ω−, the Ξ and the Λ . 0 Draw a Feynman diagram for the strong decay Ω− Ξ−K and explain why the decay is not → 0 observed. With the aid of a Feynman diagram explain why the weak decay Ω− Λ π− is strongly suppressed. →
1 0 [ The strange hadrons have quark compositions and masses Ω−(sss) 1.67 GeV, Ξ (uss) 1.31 GeV, 0 0 Ξ−(dss) 1.32 GeV, K (sd¯) 498 MeV, Λ (uds) 1.12 GeV]
Electroweak Unification
23. Z Mass Peak, Tripos B-style question.
(a) In lectures we deduced that the couplings of the Z boson to fermions should be of the form: 2 e gZ [I3 Q sin θW] where gZ = . − sin θW cos θW
In this expression, Q is the electric charge (in units of e), I3 is the weak isospin of the fermion species and helicity state being considered and the weak mixing angle is given 2 by sin θW 0.23. The decay rate into ff¯ (assumed massless compared with the Z) is ≈ 2 2 proportional to (gL + gR) where gL and gR are the couplings to left-handed and right- handed fermions respectively. Compare the Z decay rates to pairs of charged lepton pairs of each species, neutrino-antineutrino pairs, and to u-like and d-like quark-antiquark pairs. + Hence predict the branching fractions for Z decay to τ τ −, to neutrinos and to hadrons. + + (b) In the OPAL experiment at LEP the cross-section for e e− τ τ − was measured at various centre-of-mass energies. Some of the results are shown→ below. Plot these data
+ + E /GeV σ(e e− τ τ −)/nb cm → 88.481 0.2769 0.0235 ± 89.442 0.4892 0.0091 ± 90.223 0.8331 0.0368 ± 91.283 1.4988 0.0213 ± 91.969 1.1892 0.0235 ± 92.971 0.7089 0.0105 ± 93.717 0.4989 0.0276 ±
and make estimates of the Z boson mass, mZ , the total width of the Z boson, ΓZ, and + the partial decay width to τ τ −,Γτ , (assuming lepton universality of the Neutral Cur- + rent). Compare the branching fraction for Z τ τ − with your predictions from (a), and comment. → Why is the measured resonance curve asymmetric? Indicate what other effects need to be taken into account when accurately determining mZ ,ΓZ and Γτ
(c) Estimate the total decay width, ΓZ, and the lifetime of the Z boson using the resonant cross-section ratio, + σ(e e− Z hadrons) → → = 20.7, σ(e+e Z µ+µ ) − → → − + and the measured values of the Z partial decay widths, Γ(Z µ µ−) = 83.3 MeV and Γ(Z ν ν¯ ) = 166.5 MeV. Make clear any assumptions you make.→ → µ µ
2 24. W Width, Tripos A-style question. The number of neutrino species can be estimated from the total width of the W boson. Using the Standard Model prediction of the partial width for W− e−ν¯ decays, → e 3 GF MW Γ(W− e−ν¯e) = , → √2 6π
2 the mass of the W boson, MW = 80.385 0.015 GeV/c and the total width, ΓW = 2.085 0.042 GeV, estimate the number of light neutrino± species. Make clear your assumptions. ± 5 2 [G = 1.2 10− GeV− .] F × 25. ν Scattering, Tripos A-style question. Draw all possible lowest order Feynman diagrams for the following neutrino scattering or an- nihilation processes:
(a) ν e− ν e− e → e (b)ν ¯ e− ν¯ e− e → e (c) ν e− ν e− µ → µ (d)ν ¯ e− ν¯ e− µ → µ (e) ν n e− p e → 26. Weak Force & Conservation, Tripos A-style question. Consider each of the groups of processes given below. In each group, with the aid of Feynman diagrams using the Standard Model vertices, determine which processes are allowed and which are forbidden. By considering the strength of the forces involved, rank the processes in each group in order of expected rate.
0 0 + 0 (a) π γγ π π−e νe and π νν¯; + → + → → + (b) e e− τ τ − ν¯µ + τ − τ − +ν ¯µ and ντ + p τ + n; 0 → + 0 +→ 0 → 0 (c) B (bd) D−(cd)π B π π− and B J/ψK 0 → + 0 → + 0 → + (d) D (cu) K−π D π π− and D K π−. → → →
Neutrino Oscillations
27. ν Oscillations, Tripos B-style question.
(a) Show that if there are two neutrino mass eigenstates ν2 and ν3 with masses m2 and m3 and energies E2 and E3, mixed so that
νµ = ν2 cos θ + ν3 sin θ ν = ν sin θ + ν cos θ τ − 2 3 then the number of muon neutrinos observed at a distance L from the muon source is E E L ν (L) 2 = ν (L = 0) 2 1 sin2 (2θ) sin2 3 − 2 . | µ | | µ | × − 2 c ~ 3 (b) If m2 and m3 are very much less than the neutrino momentum, p, show that
(m2 m2)L ν (L) 2 ν (L = 0) 2 1 sin2 (2θ) sin2 A 2 − 3 | µ | ≈ | µ | × − p where A is a constant. (c) In 2005 the MINOS experiment started to study neutrino oscillations by pointing a beam of 1-5 GeV/c muon neutrinos from Fermilab, Illinois, at the 5400 ton MINOS far dectector in the SOUDAN mine in Minnesota, 730 km away. The experiment aimed to make a precise measurement of m2 m2. 3 − 2 Sketch the expected energy spectrum of muon neutrinos at the MINOS far detector if 2 2 2 3 2 2 sin (2θ) = 0.90 and m3 m2 = 2.5 10− (eV/c ) . Assume that the energy spectum of neutrinos produced by the− beam at× Fermilab is of uniform intensity in the range 1-5 GeV and zero elsewhere (i.e. a top-hat function). (d) If muon neutrinos oscillate into tau neutrinos, will any τ leptons (produced by charged current interactions) be observed in the MINOS far detector ? Hint: You may find the result of qu.21 useful. 1 2 [A = 1.27 s− if m2 and m3 are measured in eV/c , p in GeV/c and L in km. The mass 2 of the τ − is 1.777 GeV/c .]
4 Numerical answers
20. 0.3 ps 21. (b) 3.47 GeV; (c) 2.88 GeV; (d) 110 µm; (e) 20% ∼ 23. (a) 0.034 (e±, µ±, τ ±); 0.068 (for each neutrino flavour); 0.118 (each up-type quark); 0.152 (each 25 down-type quark); 0.692 (all hadrons) ; (c) 2.47 GeV, 2.66 10− s × 24. 3
Suggested Tripos Questions
Weak interaction: 2018 B4, 2017 1(c), 2011 3
Electroweak unification: 2018 A1(b), 2015 3, 2013 1(a)
Neutrino Oscillation: 2009 (3) A4
5 T. Potter Lent/Easter Terms 2021
Part II Particle and Nuclear Physics Examples Sheet 4
Basic Nuclear Properties
28. SEMF, Tripos B-style question. The Semi-Empirical mass formula (SEMF) for nuclear masses may be written in the form
2 2 2 Z (A 2Z) 3 M(A, Z) = Zmp + (A Z)mn aVA + aSA + aC 1 + aA − + δ(A, Z) − − A 3 A
where mp and mn are the masses of the proton and neutron respectively. Fitted values for the coefficients are given at the end of the question.
(a) Explain the physical significance and functional form of the various terms. (b) Treating the nucleus as a sphere of uniform charge density, show that the constant
3e2 aC = = 0.72 MeV . 20π0R0
(c) By reference to the SEMF explain in physical terms why nuclear fission and fusion are possible. (d) Use the Semi-Empirical mass formula to predict that the value of Z of the most stable isobar of mass number A is mn mp + 4aA Z = − 1 . 2aCA− 3 + 8aA/A Predict the Z for the most stable nuclei with A = 101 and A = 191. Compare with nuclear data, which you can find on the web (e.g. http://www.nndc.bnl.gov/chart/). Predict the most stable super-heavy nucleus with mass number 300. (e) Show that the effect of gravitational binding in the nucleus may be accounted for by adding 5 a term a A 3 to the SEMF (neglecting the proton-neutron mass difference). Show that − G aG is given by 2 3Gmn 37 aG = 5.8 10− MeV. 5R0 ≈ × Use the SEMF, so modified, to estimate that the lightest “nucleus” consisting entirely of neutrons (i.e. a neutron star) has a mass approximately 10% that of the sun. Amazingly enough, this reckless extrapolation over more than 50 orders of magnitude turns out to give more or less the right answer! [Take the mass of the sun to be 2 1030 kg] ×
1 [Use the following data: mp = 938.3 MeV, mn = 939.6 MeV, me= 0.511 MeV; aV = 15.8 MeV, aS = 18.0 MeV, aA= 1/3 23.5 MeV. Nuclear radius R = R0A with R0 = 1.2 fm] 29. The Fermi Gas model, Tripos B-style question. We can predict the form of the asymmetry term, and estimate its magnitude, using the Fermi Gas model. This involves treating the N neutrons and Z protons as free fermions of mass m 4 3 moving in a box of volume V = 3 πR0A. The model therefore only accounts for the kinetic energy of the nucleons, and not their potential energy. A standard calculation, which you have done before (at least for a cubic box), gives the density of states for each species (including spin degeneracy) as
3 3 1 4√2m 2 R0 g() = BA 2 where B = . 3π~3 You need not prove this unless you want to practice. Show that the Fermi energy for the neutrons is 2 3N 3 = . F 2BA Calculate the Fermi energy F and the corresponding nucleon momentum for the symmetric 1 case N = Z = 2 A. Show that the total kinetic energy of the nucleons is given by
2 3 3 3 5 5 N 3 + Z 3 . 5 2BA 1 1 1 Expand about the symmetric point N = Z = 2 A by writing N = 2 A(1+α) and Z = 2 A(1 α) to show that the asymmetry energy has the form: −
(N Z)2 1 a − where a = . A A A 3 F Note that this underestimates the empirical value, because this model does not take account of the potential energy, which also depends on (N Z). − One contribution to the pairing energy can also be estimated from this model, reflecting the stepwise increase of the kinetic energy resulting from the exclusion principle. This would be expected to be approximately equal to the energy spacing of levels at the Fermi level, i.e. 1 1/g(F). Show that this is, for the N = Z = 2 A case: 4¯ F . 3A Evaluate and compare with the fitted value in the SEMF for a typical value of A. Note that the Fermi Gas model again gives an underestimate because it does not take account of the additional potential energy arising from the spatial overlap of two nucleons in the same energy level.
2 30. Nuclear size, Tripos B-style question. A spherically symmetric nucleus has a radial charge density ρ(r) which is normalised such that ρ(r)d3r = 1. Show that in this case the form factor is given by:
R 4π ∞ F (q2) = r sin qrρ(r)dr q Z0 which can then be approximated by 1 F (q2) 1 q2R2 + ... ' − 6 in natural units, where R2 is the mean square radius of the charge distribution. When elastic scattering of 200 MeV electrons from a gold foil is observed at 11◦, it is found that the scattered intensity is 70% of that expected for a point nucleus. Calculate the r.m.s. radius of the gold nucleus.
For larger scattering angles (> 50◦) it is found that the scattered intensity, instead of falling off monotonically with angle, exhibits definite (oscillatory) structure. What does this suggest about the form of ρ(r)?
31. Mirror Nuclei, Tripos B-style question. One method of estimating nuclear radii is from the mass difference between a pair of odd-A mirror nuclei (A, Z) and (A, Z + 1) for which Z = (A 1)/2. Use the Semi-Empirical mass ± formula to show that the difference in Coulomb energies between these nuclides, ∆EC, can be written as 3 Aα ∆E = C 5 R 2 where the fine structure constant α = e /4π0 in natural units and R is the nuclear radius. The atomic mass difference between two mirror nuclei can be determined from the β+ spectra of the (A, Z + 1) member of the pair,
M(A, Z + 1) M(A, Z) = 2m + E − e max
where me is the mass of an electron and Emax is the maximum kinetic energy of the positron. Calculate the radius of the (A, Z + 1) member of each of the following pairs of mirror nuclei
11 11 (a) (5 B, 6 C), Emax = 0.98 MeV; 23 23 (b) (11Na, 12Mg), Emax = 2.95 MeV; 39 39 (c) (19K, 20Ca); Emax = 5.49 MeV; and comment on the results. Be careful when using atomic mass vs nuclear mass.
32. Deuteron, Tripos A-style question The deuteron has spin-parity J P = 1+. Assuming that the deuteron is dominated by the orbital angular momentum state ` = 0, and noting that no excited states exist, what can be concluded about the nature of the np force and about the existence of nn and pp bound states?
3 The Nuclear Shell Model
33. Spin & Parity, Tripos A-style question. What are magic numbers ? Outline the basis of the Nuclear Shell Model and show how it accounts for magic numbers. How can the shell model be used to predict the spins and parities of nuclear ground states ? Using the shell model determine the spins and parities of the ground states of the nuclides listed below and compare them with the experimental values given. Comment on any discrepancies you find.
3 9 7 12 13 15 17 23 131 207 2He 4Be 3Li 6C 6C 7N 8O 11Na 54 Xe 82Pb + + + + 1 3 − 3 − + 1 − 1 − 5 3 3 1 − 2 2 2 0 2 2 2 2 2 2
Assume the following ordering of levels:
1s 1 1p 3 1p 1 1d 5 1d 3 2s 1 1f 7 1f 5 2p 3 2p 1 1g 9 1g 7 2d 5 2d 3 1h 11 3s 1 1h 9 2f 7 3p 3 1i 13 3p 1 2f 5 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ···
34. Energy Levels, Tripos A-style question. The diagram below shows the low-lying energy levels for the nuclides:
18 166 18 208 18 10Ne 68Er 9F 82Pb 8O
The schemes are drawn to the same scale, with energies (in MeV) with respect to the ground state and the spin and parity (J P ) values given for each level. Identify which scheme corresponds to each nuclide and explain as fully as you can which features of the levels support your choices.
4 Nuclear Decay
35. Alpha Decay, Tripos B-style question. Discuss the factors which affect the lifetimes of nuclei decaying by α-decay. You should include a description of the Geiger-Nuttall law.
(a) A nucleus with A=200 can decay by the emission of α particles. The α particles are observed with two different energies, 4.687 MeV and 4.650 MeV. Neither of these decays populates the ground state of the daughter nucleus, but each is followed by the emission of a γ ray, whose energies are 266 and 305 keV respectively. No other γ rays are seen. From this information construct an energy level diagram for the states involved, and • indicate on it the decay scheme. P P The decaying parent state has J = 1− and the daughter has ground state J = 0−. • Explain why there is no direct α decay to the ground state. (b) The values of the energy release, Q, and the measured α-decay half-lives of some isotopes of Thorium are as follows: Isotope Q / MeV Half-life / s 220 5 90 Th 8.95 1.0 10− 222 × 3 90 Th 8.13 2.8 10− 226 × 3 90 Th 6.45 1.9 10 228 × 7 90 Th 5.52 6.0 10 230 × 12 90 Th 4.77 2.5 10 232Th 4.08 4.4 × 1017 90 × 224 Estimate the α-decay half-life of 90 Th, given that the Q-value for this decay is 7.31 MeV. What is the approximate uncertainty in your estimate? (Part II 1995)
36. Beta Decay, Tripos A-style question. Outline the Fermi theory of β decay and explain the principal assumptions made. Explain the difference between Fermi and Gamow-Teller transitions and between super-allowed, allowed and forbidden decays. Explain the significance of ft values. Classify each of the following examples of β decay according to whether the decay is super- allowed, allowed, 1st forbidden etc., and whether Fermi or Gamow-Teller matrix elements are involved.
5 (i) n p → (ii) 6He(0+) 6Li(1+)(ft=830s) 2 → 3 14 + 14 + (iii) C(0 ) N∗(0 )(ft=3300s) 6 → 7 (iv) 35S( 3 +) 35Cl( 3 +)(ft=1x105s) 16 2 → 17 2 36 36 + (v) Cl(2−) Ar(0 ) 17 → 18 76 76 + (vi) Br(1−) Se(0 ) 35 → 34 (vii) 137Cs( 7 +) 137Ba( 3 +) 55 2 → 56 2
37. Fermi Theory, Tripos A-style question. Show that the electron momentum spectrum in β-decay using Fermi theory can be written as
2 dΓ GF 2 2 = 3 (E0 Ee) pe dpe 2π −
where GF is the Fermi constant, Ee and pe are the energy and momentum of the electron and E0 is the total energy released. You may treat the electron and neutrino as massless.
Show that the average kinetic energy carried off by the electron in β decay is E0/2 when the electron is highly relativistic, and E0/3 when the electron is non-relativistic. When the electron is highly relativistic, show that the total decay rate is given approximately by G2 E5 Γ = F 0 60π3 5 The E0 dependence is sometimes known as Sargent’s Rule. 38. Gamma Decay, Tripos B-style question. A nucleus in an excited state at 2.51 MeV can decay by emission of three γ rays as shown below. The transitions labelled γ1, γ2 and γ3 are known to proceed via magnetic dipole, electric dipole and electric quadrupole transitions respectively. No other transitions to the ground state, of 3 + comparable intensities, are observed. Given that the ground state has a spin of 2 what are the most probable spins and parities of these three excited states?
6 Fission and Fusion
39. Fission, Tripos B-style question.
(a) Graphite (i.e. 12C) is sometimes used as a moderator in nuclear reactors. Compute the maximum fractional energy loss which a non-relativistic neutron can undergo in a single elastic collision with a 12C nucleus. Hence calculate the minimum number of collisions which would be required in order to bring a 2.5 MeV fission neutron down to a thermal energy of 0.025 eV. (b) Experiments are being performed using a mixture of 235U and graphite. The graphite 6 10 contains a fraction of 10− by weight of B. The neutron absorption cross-sections for 12C, 10B, and 235U at thermal energies are 0.04 b, 3800 b and 700 b respectively, where fission accounts for 580 b of the cross-section in 235U. What is the maximum fraction by weight of 235U which could be allowed in the mixture if the multiplication factor at infinite volume is not to exceed unity? Assume that 2.5 neutrons are produced per fission, and that all reactions take place at thermal energies.
40. Fusion, Tripos A-style question. 16 Estimate the size of the Coulomb barrier between two 8 O nuclei which needs to be overcome before they can undergo fusion, and thus estimate the temperature needed to bring about fusion in this case.
7 Numerical answers
101 191 28. (b) 0.7 MeV; (d) 44 (44 Ru is stable); 77 (77 Ir is stable); 114 ; (e) 150 MeV. 30. 5 fm. 31. (a) 3.4 fm; (b)4.2 fm; (c)4.6 fm. 35. 10 s ∼ 36. (i) superallowed, F/GT; (ii) superallowed GT; (iii) superallowed F; (iv) allowed F/GT; (v) 1st forbidden GT; (vi) 1st forbidden F/GT; (vii) 2nd forbidden F/GT. + + 7 9 9 − 7 − 38. From the highest level downwards: 2 , 2 , 2 or 2 . Suggested Tripos Questions
Semi-Empirical Mass Formula: 2018 B2, 2010 A1(a)
Nuclear Forces & Scattering: 2016 4, 2012 1(a)
Shell Model: 2017 1(b), 2016 1(a)
Nuclear excitations: 2017 3 (last part), 2015 1(a)
Nuclear decay: 2017 3, 2013 1(c)
α-decay: 2010 A1(b), 2007 A3
β-decay: 2016 3(b)(c)(d), 2015 4
γ-decay: 2015 A1(b), 2014 4
Fission and Fusion: 2018 B2 (last part), 2011 4
8 APPENDIX A: PHYSICAL CONSTANTS
Summary of the physical constants and conversion factors used in this course:
19 Electron charge, e = 1.602 10− C h¯c = 0.197 GeV fm × 25 h¯ = 6.58 10− GeV s Fine structure× constant, α = 1/137.04 24 1 Bohr magneton, µB = 9.3 10− JT− × 27 1 Nuclear magneton, µ = 5.1 10− JT− N × 19 6 9 1 eV = 1.602 10− J, 1 MeV = 10 eV, 1 GeV = 10 eV × 15 1 fermi(fm) = 10− m 28 2 1 barn(b) = 10− m 10 1 1 Curie(Ci) = 3.7 10 decays s− × Atomic masses are often given in unified (or atomic) mass units: 12 1 unified mass unit(u) = Mass of an atom of 6 C/12 27 2 1u = 1g/N = 1.66 10− kg = 931.5 MeV/c A × APPENDIX B: PARTICLE PROPERTIES
From the Review of Particle Physics, C. Amsler et al., Phys. Lett. B667 1 (2008) http://pdg.lbl.gov/
Quarks (spin 1/2)
Name Flavour Mass Charge (e) (GeV/c2) up u 0.35 +2/3 ≈ down d md mu 1/3 charm c 1≈.5 −+2/3 strange s 0.5 1/3 top t 171.2(2.1) −+2/3 bottom b 4.5 1/3 −
Leptons (spin 1/2)
Lepton Charge Mass Mean life (s) Lepton Branching (MeV/c2) Decay Mode Fraction (%) 2 νe 0 < 2 eV/c stable νµ 0 < 0.19 stable ντ 0 < 18.2 stable e 1 0.511a stable ± b 6c µ 1 105.658 2.197 10− e−ν¯eνµ 100 ± × 15 ≈ τ 1 1776.8(2) 291(1) 10− µ−ν¯ ν 17.36(5) ± × µ τ e−ν¯eντ 17.85(5) hadrons +ν 65 τ ≈ a 8 2 The error on the e mass is 1.3 10− MeV/c . b × 6 2 The error on the µ mass is 4 10− MeV/c . c × 11 The error on the µ lifetime is 2 10− s. ×
N.B. Numbers given in brackets correspond to the error in the last digit(s). For example, m = 1776.8(2)MeV/c2 (1776.8 0.2)MeV/c2. τ ≡ ± P Gauge Bosons (J = 1−)
Force Gauge Charge (e) Mass Full Width Decay Mode Branching Boson (GeV/c2) (GeV) Fraction (%)
30 18 2 E-M γ < 5 10− < 10− eV/c stable ×
Weak W± 1 80.40(3) 2.14(4) eν 10.7(2) ± e (Charged) µνµ 10.6(2) τντ 11.2(2) hadrons 67.6(3)
Weak Z0 0 91.188(2) 2.495(2) ee 3.363(4) (Neutral) µµ 3.366(7) ττ 3.370(8) νν 20.00(6) hadrons 69.91(6)
Strong g 0 0 stable P Pseudoscalar Mesons (J = 0−)
Particle Quark Mass Mean Life (s) or Decay Mode Branching Content (MeV/c2) Width (keV) Fraction (%)
8 π± ud¯, d¯u 139.5702(4) 2.6033(5) 10− s µ−ν¯µ 100 0 × 17 ≈ π (u¯u dd)¯ /√2 134.9766(6) 8.4(6) 10− s γγ 98.80(3) η see− note a 547.85(2) 1.30(7)× keV γγ 39.3(2) π0π0π0 32.6(2) + 0 π π−π 22.7(3) + π π−γ 4.6(2) + η0 see note a 957.7(2) 0.20(2) MeV π π−η 45(2) ρ0γ 29(1) π0π0η 21(1) 8 K± u¯s, s¯u 493.677(16) 1.238(2) 10− s µ−ν¯µ 63.5(1) × 0 π−π 20.7(1) + π π−π− 5.59(4) 0 π µ−ν¯µ 3.35(4) 0 π e−ν¯e 5.08(5) 0 0 ¯ 0 10 + K , K d¯s, sd 497.61(2) KS: 0.8953(5) 10− s π π− 69.2(1) × π0π0 30.7(1) 0 8 0 0 0 KL: 5.12(2) 10− s π π π 19.5(1) × + 0 π π−π 12.5(1) π±µ∓νµ 27.0(1) π±e∓νe 40.5(1) 12 b D± cd¯, d¯c 1869.3(4) 1.040(7) 10− s e− + any 16.0(4) × K− + any 26(1) K+ + any 5.9(8) K0 + any plus K0 + any 61(5) 0 0 12 c D , D u¯c, c¯u 1864.8(2) 0.410(2) 10− s K− + any 55(3) × K+ + any 3.4(4) e+ + any 6.5(2) µ+ + any 6.7(6) K0 + any plus K0 + any 47(4) 12 Ds± c¯s, s¯c 1968.5(3) 0.500(7) 10− s seen × 12 B± ub¯, b¯u 5279.1(3) 1.64(1) 10− s many 0 0 × 12 B , B db¯, bd¯ 5279.5(3) 1.53(1) 10− s many 0 0 ¯ × 12 Bs , Bs sb, b¯s 5366.3(6) 1.47(3) 10− s many × 12 B± cb¯, b¯c 6276(4) 0.46(18) 10− s many c × ηc c¯c 2980(1) 27(3) MeV hadrons a η and η0 are linear combinations of the quark state (u¯u+ dd)¯ /√2 and s¯s(see lectures). b c 0 D− decay modes; D decay modes. P Vector Mesons (J = 1−)
Particle Quark Mass Full Width (MeV) Decay Mode Branching Content (MeV/c2) Fraction (%)
ρ± ud¯, d¯u 775.5(4) 149(1) ππ 100 ρ0 (u¯u dd)¯ /√2 − + 0 ω (u¯u+ dd)¯ /√2 782.6(1) 8.49(8) π π−π 89.2(7) π0γ 8.9(2) + π π− 1.5(1) + φ s¯s 1019.46(2) 4.26(4) K K− 49.2(6) 0 0 KLKS 34.0(5) K∗± u¯s, s¯u 891.7(3) 50.8(9) Kπ 100 0 0 ¯ ≈ K∗ , K∗ d¯s, sd 896.0(3) 50.3(6) Kπ 100 0 a ≈ D∗± cd¯, d¯c 2010.3(2) 0.096(22) D π− 67.7(5) 0 D−π 30.7(5) 0 0 0 0b D∗ , D∗ u¯c, c¯u 2007.0(2) < 2.1 D π 62(3) D0γ 38(3)
Ds∗± c¯s, s¯c 2112.3(5) < 1.9 Ds±γ 94(1) 0 Ds±π 6(1) B∗ ub¯, b¯u, db¯, bd,¯ 5325.1(5) Bγ dominant sb¯, b¯s J/ψ c¯c 3096.92(1) 93(2) keV hadrons 87.7(5) + e e− 5.9(1) + µ µ− 5.9(1) + Υ(1s) bb¯ 9460.3(3) 54(1) keV τ τ − 2.6(1) + e e− 2.4(1) + µ µ− 2.48(5) a b 0 D∗− decay modes; D∗ decay modes. Baryons (J P = 1/2+)
Particle Quark Mass Mean Life (s) or Decay Mode Branching Content (MeV/c2) Full Width (MeV) Fraction (%) p uud 938.27203(8) > 2.1 1029 years × n udd 939.56536(8) 885.7(8) s pe−ν¯e 100 0 10 Λ uds 1115.683(6) 2.63(2) 10− s pπ− 63.9(5) × nπ0 35.8(5) + 10 0 Σ uus 1189.37(7) 0.802(3) 10− s pπ 51.6(3) × nπ+ 48.3(3) 0 20 0 Σ uds 1192.64(2) 7.4(7) 10− s Λ γ 100 × 10 Σ− dds 1197.45(3) 1.48(1) 10− s nπ− 99.848(5) 0 × 10 0 0 Ξ uss 1314.8(2) 2.90(9) 10− s Λ π 99.52(1) × 10 0 Ξ− dss 1321.7(1) 1.64(2) 10− s Λ π− 99.89(4) + × 13 Λc udc 2286.5(1) 2.00(6) 10− s many × 12 Λ udb 5620(2) 1.38(5) 10− s many b × Baryons (J P = 3/2+)
∆ uuu, uud 1232 118 MeV Nπ > 99 udd, ddd ≈ ≈ 0 Σ∗ uus, uds, dds 1385 36 MeV Λ π 87(2) ≈ ≈ Σπ 12(2) Ξ∗ uss, dss 1533 9 MeV Ξπ 100 ≈ ≈ 10 0 Ω− sss 1672.5(3) 0.82(1) 10− s Λ K− 67.8(7) × 0 Ξ π− 23.6(7) 0 Ξ−π 8.6(4) APPENDIX C: Scattering from a Yukawa potential
Consider relativistc elastic scattering from a Yukawa potential
g e mr V (~r) = − r The matrix element is given by
2 2 i~q.~r 3 M = e− V (~r) d ~r | if | Z
In order to perform the integral, choose the z axis to lie along ~r. Then ~q.~r = qr cos θ and −
2π π i~q.~r 3 ∞ iqr cos θ 2 e− V (~r) d ~r = V (r) e r sin θ dθ dφ dr 0 0 0 Z Z Z +1 Z ∞ = 2πV (r) eiqr cos θ r2 d(cos θ) dr 0 1 − Z Z iqr iqr ∞ e e− = 2πV (r) − r2 dr iqr 0 Z mr iqr iqr ∞ e− e e− = 2πg − r2 dr r iqr 0 Z iqr iqr ∞ mr e e− = 2πg e− − dr iq Z0 ∞ 2πg r(m iq) r(m+iq) = e− − e− dr iq − Z0 2πg 1 1 2πg 2iq = = iq m iq − m + iq iq m2 + q2 − 4πg = m2 + q2
The matrix element is then 16π2g2 M 2 = | if | (m2 + q2)2
The Yukawa potential is a general potential, and can be extended to other potentials, e.g. for the Coulomb potential Zα V (~r) = − r using g = Zα and m = 0, the matrix element for Rutherford Scattering is
16π2Z2α2 M 2 = | if | q4 Appendix D: Interaction via Particle Exchange We need to evaluate the following integral in order to determine the energy shift when in state i when a particle of mass m is exchanged between particle 1 and particle 2,
2 2 ipr ipr 1 2 g ∞ p e e− ∆E → = − dp i −2(2π)2 p2 + m2 ipr Z0 Start by rewriting
2 ipr ipr 1 2 1 g ∞ p e e− ∆E → = − dp i −2 2(2π)2 p2 + m2 ir Z−∞ using the fact that the integrand is even in p. The integrand has poles at p = im (see ipr ipr ± the figure). The integrals with the e and e− terms are performed separately. This is because one chooses an infinite semi-circular contour to do the integration over, in such a way that on the circular piece the contribution from infinity vanishes. This happens if the integrand contains a decaying exponential in p . For eipr, this happens for p = +i p | | ipr | | and so one closes the contour in the upper half plane (C1 in the figure). For e− , we want p = i p , and so close the contour in the lower half plane (C in the figure). − | | 2
The whole integral is thus: g2 p eipr p e ipr dp − dp . −2(2π)2 p2 + m2 ir − p2 + m2 ir IC1 IC2 The residue of the pole at p = im in the first integrand is:
(p im) p ipr 1 mr lim − e = e− p im (p im)(p + im) ir 2ir → − and the residue of the pole at p = im in the second integrand is: − ipr (p + im) p e− 1 mr lim − = e− . p im (p im)(p + im) ir −2ir →− − Cauchy’s residue theorem tells us that the contour integral over an anti-clockwise contour is 2πi multiplied by the sum of the residues of the poles enclosed by the contour. For a clockwise contour, there is an additional minus sign. Noting that C1 is anti-clockwise, and C2 is clockwise, one has:
2 mr mr 1 2 g e− e− ∆E → = 2πi + i −2(2π)2 2ir 2ir g2 e mr = − −8π r as given in the notes. APPENDIX E: LOCAL GAUGE INVARIANCE IN QED
Consider a non-relativistic charged particle in an electromagnetic field:
F = q (E + v B) × where E and B can be written in terms of the vector and scalar potentials, A and φ: ∂A B = A and E = φ . ∇ × −∇ − ∂t The classical Hamiltonian, 1 2 H = p qA + qφ, 2m − can be used along with Schr¨odinger’sequation to obtain
1 ∂ψ Hψ = ( i qA)2 + qφ ψ(x, t) = i (x, t). (1) 2m − ∇ − ∂t where we have substituted p i . We now need to show that Schr¨odinger’sequation is invariant under the local guage→ − transformation∇
iqα(x,t) ψ ψ0 = e ψ → A A0 = A + α → ∇ ∂α φ φ0 = φ → − ∂t
Substituting for ψ0,A0 and φ0 in equation (1):
1 ∂α ∂ ( i q(A + α))2 + q(φ ) eiqαψ = i (eiqαψ) 2m − ∇ − ∇ − ∂t ∂t 1 ∂α ∂ψ ∂α ( i qA q α)2 + qφ q eiqαψ = i eiqα + iqψ eiqα . 2m − ∇ − − ∇ − ∂t ∂t ∂t The last terms on either side of the above equation cancel. Now consider the ( i qA q α)2 term. In order to show local gauge invariance, we need to show that − ∇ − − ∇
( i qA q α)2 eiqαψ = ( i qA)2 eiqαψ − ∇ − − ∇ − ∇ − or, equivalently, 2 2 iqα ( i qA0) ψ0 = ( i qA) e ψ. − ∇ − − ∇ − Now,
( i qA q α)2 eiqαψ = ( i qA q α) ( i qA q α) eiqαψ − ∇ − − ∇ − ∇ − − ∇ · − ∇ − − ∇ and eiqαψ = eiqα ( + iq α) ψ. ∇ ∇ ∇ Therefore,
( i qA q α) eiqαψ = eiqα ( i + q α qA q α) ψ − ∇ − − ∇ − ∇ ∇ − − ∇ = eiqα ( i qA) ψ − ∇ − and
2 iqα ( i qA q α) ψ0 = ( i qA q α) e ( i qA) ψ − ∇ − − ∇ − ∇ − − ∇ − ∇ − = eiqα ( i qA)2 ψ. − ∇ − Hence, 2 iqα 2 ( i qA0) ψ0 = e ( i qA) ψ − ∇ − − ∇ − and Schr¨odinger’sequation is invariant under a local gauge transformation. APPENDIX F: NEUTRINO SCATTERING IN FERMI THEORY
Calculation of the cross-section for νe + n p + e− using Fermi theory. The cross-section is given by Fermi’s Golden Rule →
Γ = 2π M 2 ρ (E ) | fi| f where the matrix element, Mfi, is given by the 4-point interaction with a strength equal to the Fermi constant, GF ; M 2 G2 . | fi| ≈ F 1 There are a total of 4 possible spin states for the spin- 2 e and ν. These correspond to a singlet state S = 0 (Fermi transition) and three triplet states S = 1 (Gamow-Teller transition). Therefore, the matrix element becomes
M 2 4G2 . | fi| ≈ F The differential cross-section is then given by,
E2 dσ = 2π4G2 e dΩ F (2π)3 where Ee is the energy of the electron in the zero-momentum frame. It follows that
dσ G2 E2 = F e . dΩ π2
The total energy in the zero-momentum frame, √s = 2Ee. Hence, the total cross-section can be written as dσ 4G2 E2 G2 s σ = dΩ = F e = F . dΩ π π Z APPENDIX G: NEUTRINO SCATTERING WITH A MASSIVE W BOSON
From Appendix F, the differential cross-section in Fermi theory is
dσ G2 E2 = F e . dΩ π2
The correct theory involves exchange of a massive vector boson of mass MW , which leads to a propagator in the matrix element 1 . q2 M 2 − W Fermi theory is equivalent to neglecting the q2 term in the denominator. Hence, treating W-boson exchange correctly, we have
dσ G2 E2 M 2 2 = F e W . dΩ π2 M 2 q2 W − Now, for elastic scattering, q2 = 0 q 2, where − | | θ 2 1 q 2 = 2E sin = s(1 cos θ) u | | e 2 2 − ≡ and so 1 s du = s sin θ dθ = dΩ . 2 4π We can thus integrate the differential cross-section in terms of u:
G2 s M 2 2 σ = dΩ F W 4π2 M 2 + u Z W G2 M 4 s 1 = F W du π (M 2 + u)2 Z0 W G2 M 4 1 s = F W − π M 2 + u W 0 G2 M 4 1 1 = F W π M 2 − M 2 + s W W 2 2 GF MW s = 2 π(MW + s)
2 At small values of s this reduces to the Fermi theory result, while for s MW the cross-section tends towards the constant value G2 M 2 σ = F W π and is no longer divergent. APPENDIX H: GAMOW FACTOR IN ALPHA DECAY The probability for an α particle to tunnel through the Coulomb barrier can be written as P = exp 2G {− } i Y where G is the Gamow Factor,
R0 [2m (V (r) E )]1/2 G = − 0 dr. h¯ ZR R is the radius of the nucleus of mass number Z, R0 is the radius at which the α particle escapes, m is the mass of the α particle, V (r) = 2 (Z 2) e2/4π r B/r is the Coulomb − 0 ≡ potential, and E0 is the energy release in the decay. The α particle escapes the nucleus when r = R0. Hence, the potential V (R0) = E0 and R0 = B/E0. Therefore,
2m 1/2 R0 B 1/2 G = E0 dr h¯2 r − ZR 2mB 1/2 R0 1 1 1/2 = dr h¯2 r − R ZR 0 2 In order to perform the integration, let r = R0 cos θ and dr = 2R0 cos θ sin θ dθ. Then − 1 1 1/2 1 1 1/2 dr = ( 2R0 cos θ sin θ) dθ r − R R cos2 θ − R − Z 0 Z 0 0 1/2 2 = 2R0 sin θ dθ − Z 1/2 = R0 [sin 2θ θ] − 1/2 1/2 Now, using cos θ = (r/R0) , sin θ = (1 r/R0) and sin 2θ = 2 sin θ cos θ, then − R0 1/2 R0 1/2 1/2 1/2 1 1/2 R0 [sin 2θ θ]R = R0 2 (1 r/R0) (r/R0) cos− (r/R0) − − − R 1/2 h 1 1/2 1/i2 = R0 cos− (R/R0) 2 (1 R/R0)(R/R0) − { − } h i Hence, the Gamow factor
1/2 2mB 1/2 1 1/2 1/2 G = R0 cos− (R/R0) 2 (1 R/R0)(R/R0) h¯2 − { − } h i or 1/2 2m B 1 1/2 1/2 G = cos− (R/R0) 2 (1 R/R0)(R/R0) E0 h¯ − { − } h i For most practical cases, R R0, so the term in square brackets is π/2 and G becomes ≈ 2m 1/2 B π G ≈ E h¯ 2 0