Chapter 1.1 by Manu Verma

Chapter 1 Interactions And Field

Chapter 1: Interactions and Field 1.1 Introduction 1.2 The Electromagnetic Interaction 1.3 The Strong Interaction 1.4 The Weak and Gravitational Interaction 1.5 Vacuum Polarization 1.6 Interactive Exercise

Interactions and fields

Particle Physics

Introduction One of the fundamental goals of physics is the discovery of the basic building blocks of matter. The systematization of the elements through their electronic configuration was one step in this process. After the realization that atoms were made up of a nucleus surrounded by electron clouds, it was natural to ask about the structure of nuclei and the composition of the "elementary" particles that make up the nucleus. The question about the fundamental nature of nucleons soon became confused by the plethora of additional particles that were discovered, beginning in the 1930's.

Fig 1 Dayalbagh Educational Institute

Interactions and fields Particle Physics Rather than reciting a list of these particles, we begin with a discussion of the four fundamental interactions. This allow us to classify the particles according to the type of interaction they experience. Particle physics studies the elementary “building blocks” of matter and interactions between them. Matter consists of particles and fields. Particles interact via forces caused by fields. Forces are being carried by specific particles, called gauge bosons. Name

Acts on:

Carrier

Range

Strength

Stable systems

Induced reaction

Gravity

all particles

gravitation

long F ∝ 1/r2

-10-39

Solar system

Object falling

Weak force

fermions

boson W and Z

<10-17 m

-10-5

None

β-decay

Electro magnetism

particles with electric charge

photon

long F ∝1/r2

1/137

Atoms, stones

Chemical reactions

Strong force

quarks and gluons

gluons

10-15 m

Hadrons, nuclei

Nuclear reactions

Table 1

The Standard Model Electromagnetic and weak forces can be described by a single theory ⇒ the “Electroweak Theory” was developed in 1960s (Glashow,Weinberg, Salam). Theory of strong interactions appeared in 1970s: “Quantum Chromodynamics” (QCD). Dayalbagh Educational Institute

Interactions and fields

Particle Physics

Natural Units I kg m s

GeV

SI units: [M] [L] [T]

Natural units: [Energy] [velocity] [action]

For everyday physics SI units are a natural choice: M(SH student)~75kg. Not so good for particle physics: Mproton~10-27kg PP chooses a different basis - Natural Units, based on: quantum mechanics (ħ); relativity (c); appropriate unit of energy 1 GeV = 109 eV = 1.60 × 10-10 J Energy GeV

Time (GeV/ħ)-1

Momentum GeV/c

Length (GeV/ħc)-1

Mass GeV/c

Area (GeV/ħc)-2

Units of elementry particle physics = q relative units ħ = c = 1. energy [E] = eV, KeV, MeV, GeV, TeV. eV = 1.6 X 10

-19

mass: m = E/C2 = E = [m] = MeV Vel: v = v/c : dimensionless momentum p = mv = [p] = [m] = MeV Distance l = ħ/p = 1/p = [X] = 1/MeV time : t = ħ/E = 1/E = 1/MeV [l] = [t] = 1/ [E] = 1/[m] = 1/[p] Dayalbagh Educational Institute

Interactions and fields

Particle Physics

What is length in MeV. Relation between units: 1fm = 10-13 cm = 10-15 m. 1MeV = 106 eV = 106 x 1.6 x 10-19 J = 1.6 x 10-13 J

To write length in terms of GeV-1: [L] = [E]-1 [Ń&#x203A;]m [c]n = [E]-1 [E]m [T]m [L]n [T]-n m=1 , n=1 so length is measured in Ń&#x203A;c/GeV. Dayalbagh Educational Institute

Interactions and fields Particle Physics To write time in units of inverse energy [T] = [E]-1 [ћ]m [c]n =[E]-1 [E]m [T]m [L]n [T]-n [T] = [E]m-1 [T]m-n [L]n 0 = m-1 , l = m-n , n=0 m=1 , n =0 1sec =[E]-1 ћ = ћ/[E]

Units And Subatomic Physics Physics and technology → SI units usual Relative Physics cgs/ Gaussian Natural System “4πϵ0” in EM. Length

Meter

Time

Second

Energy

Electron Volt

Mass

Electron Volt

eV/c2

Momentum

Electron Volt

eV/c

106eV = MeV 109 eV = GeV ≈ Mass energy of proton 1012eV = TeV ≳ mass energy of Higgs boson

10-15 m = Femtometer = 1 Fermi = Diameter of proton Time for light signal to cross proton

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Interactions and fields

Particle Physics

Why use Electron Volts? Use of electron volt very convenient as energy unit. Most experimental work in subatomic physics is done by accelerating beams of particles in electrostatic fields and scattering the beam off of a target. One electron volt(eV) is the kinetic energy an electron acquires by being accelerated thru a potential of one volt. -1V Fig 2

1eV = e x 1 Volt = 1.60 x 10-19 (Coulomb)x 1 Volt = 1.60 x 10-19 Joules = 1.60 x 10-12 ergs The energy is measured in electron-volts: 1 eV ≈ 1.602 ✕ 10-19 J 1 keV = 103eV; 1 MeV = 106 eV; 1 GeV = 109 eV The Planck constant (reduced) is then: ħ ≡ h / 2π = 6.582 ✕ 10-22 Mev s and the “conversion constant” is: ħc = 197.327 ✕ 10-15 MeV m For simplicity, the natural units are used: ħ=1

and

c=1

so that the unit of mass is eV/c2, and the unit of momentum is eV/c Dayalbagh Educational Institute

Interactions and fields

Particle Physics

Quantum Forces and Feynman diagrams In our simple model of Rutherford scattering, we used the classical idea of a force. What is a force? It's something which changes the momentum of a particle. Newton invented the idea Force = mass x Acceleration . What one measures physically is a momentum change. One can have different abstract models of what the force is that â&#x20AC;&#x153;causesâ&#x20AC;? the momentum change Classically we think of fields causing potentials

Fig 5

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Interactions and fields

Particle Physics

Interaction force between particles

4-momentum transfer between e1 and e2

Fig 6

Note that the 4-momentum transfer, transfers both energy and momentum The two electrons in initial and final state ensure that energy and momentum are conserved

They remain “on mass shell” E2= p2+ m2 (E2=p2c2 + m2c4) In relativistic quantum mechanics, we think of forces being caused by the exchange of virtual particles which are cause boson How can exchanging a particle be a “force”? answer this on two levels. Dayalbagh Educational Institute

Interactions and fields

Particle Physics

Simple answer is that it causes a momentum change. That is what we have defined a force to be.

Fig 7

According to Heisenberg's uncertainly principle. A freely propagating electron can exit and reabsorb a photon. Not True; because off mass shell Re-absorption has to occur within a time Heisenberg

Fig 8

The γ will carry 4-momentum away from the electron If

Ee2 = pe2 c2 + me2 c4 me2 c4 = Ee2 - pe2 c2

Before the emission of the γ, then me2 c4 ≠ E'e2 – p'e2 c2 = m*e2 c4 After the emission

massless

For a real γ E γ2 - p γ2 c 2 = 0 For virtual γ E'γ2 – p'γ2 c2 = m*γ2 c4 Dayalbagh Educational Institute

Interactions and fields Particle Physics Both γ and e will have nonphysical masses for short time ΔE → “ off mass shell” Virtual particles Transfer momentum

Transit forces

Classical ↔ Quantum Analogy Since ΔEΔt ≲ ħ Propagation time of virtual particle is: t ~ Δt ≲ ħ/mc

~ΔE

“mass” of virtual particle Assume virtual particle propagates at velocity of light Range of force R = C.Δt ≲ ħ/mc coulomb

mγ=0

For m → 0 ; R → ∞ For M → large ; R → small

The mass of the exchanged virtual Particle, determines the range of the force

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Interactions and fields Classically

Particle Physics

Quantum mechanically, force is due to absorption of γ with momentum q Δp.Δx ~ ħ q.Γ ~ ħ The transfer of this quantum of momentum takes a time Γ /c

This “gives” the “correct” 1/Γ 2 behaviour No of γ emitted/absorbed ~Q1Q2 F ~ Q1Q2/Γ 2 Virtual particle Quantum field

Classical field Spooky action at a distance

Equally abstract (concrete?) Concepts

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Interactions and fields

Particle Physics

Feynman Diagrams Feynman diagrams are graphical representations of: Quantum mechanical amplitudes Probability α (AMP)2 of process You know that if a process can occur in two different ways Probability α (A1 + A2 ) (A1 + A2 ) A1 A2 → complex numbers ∴ interference Cf → two slit experiment with electrons A Feynman diagram is a pictorial representation of a particular process (decay or scattering) at a particular order in perturbation theory. Feynman diagrams can be used to represent and calculate the matrix elements, M, for scattering and decays. Feynman diagrams are very useful and powerful tools We will use them a Lot in this course.

Fig 10 Dayalbagh Educational Institute

Interactions and fields Particle Physics Conventions Time flows from Left to right (occasionally upwards) Fermions are solid Lines with arrows Anti-fermion are solid backward pointing arrows.

Fig 11

Lines

with

Bosons are wavy (or dashed) Lines Use Feynman Rules to calculate M at different orders in perturbation theory.

Feynman diagrams for relativistic quantum scattering Diagram represents quantum mechanical amplitude

4-momentum transfer

Fig 12

Probability of scattering thru Q P(θ) P(θ) = p(q2) = |A (q2)| 2 Quantum mechanical Simple Feynman rule scattering amplitude Aee → ee ~ √α • √α / q2 ~ e2/q2

α = e2/ ħc

P α |A|2 ~ e4/q4 Dayalbagh Educational Institute

Interactions and fields

Particle Physics

Simple Feynman rule P α |A|2 ~ e4/q4 Scattering depends on strength of interaction √α ~ e Hard to scatter through large angles Have to transfer a lot of 4-momentum Need a very virtual photon Harder In our simple model of Rutherford scattering, we had Δp2 ~ q2 ~ p2 θ2 So, we would predict that Prob of scattering through Q; P(θ) P(θ) ~ e4/p4 θ 4

We will come back to this

Apply Feynman rule to other simple processes In free space one cannot have γ → e + e-

Fig 13

This does not conserve momentum. However, a recoiling nucleus can provide momentum recoil

Fig 14 Dayalbagh Educational Institute

Interactions and fields Probability ~ |√α • √α Z √α|2

Particle Physics

~ α3 Z2 ~ e6 Z 2

Perturbation expansion Why have I only drawn one exchanged photon? If you have read Feynman’s book on “QED”, you know that in quantum mechanics you have to take account of all possible ways a process can occur. Also interference terms

Fig 15

You have to join incoming and outgoing “lines” by all possible internal “virtual” lines In electromagnetism a good approximation is just first term : born approximation only works because α « 1. If a force has “strong coupling” αs » 1 and “corrections” dominate Σ

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Interactions and fields

Particle Physics

Exercise 1 List all the fundamental fermions in the Standard Model.

Solution 1 There are six leptons and six quarks. Lepton e-, υe , μ-, υμ , τ-, υτ Quarks u, d, c, s, t, b. Plus their anti-particle counterparts i.e. 24 fermions total.

Exercise 2 What is 1fm in GeV-1? How many seconds is 1 GeV-1 ?

Solution 2 To write length in terms of inverse energy and some factors of c and ħ. We do this by balancing dimensions: [L] = [E]-1 [ħ]m [c]n = [E]-1 [E]m [T]m [L]n [T]-n Where we have used that ħ is measured in [E][T] and c is measured in [L] [T]-1. To balance the dimensions in the equation, we have to set m = 1. Thus the length is measured in GeV -1ħc.

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Interactions and fields

Particle Physics

To find out how many GeV -1ħc corresponds to 1fm:

Now we use measurements of ħ and c in appropriate units: GeV sec and fm /sec. We also know ħc = 197 MeV fm = 0.197 GeV fm.

To write time in units of inverse energy: [T] = [E]-1 [ħ]m [c]n =[E]-1 [E]m [T]m [L]n [T]-n =[E]m-1 [T]m-n [L]n. to balance the dimensions, m = 1, n = 0. So Now: 1 sec = ħ = 6.58 × 10-22 MeV s = 6.58 × 10-25 GeV s. Therefore Hence, 1 GeV-1 = 6.58 × 10-25 sec.

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Interactions and fields

Particle Physics

Exercise 3

Determine the mass of the virtual photon in each of the lowest order diagrams for Bhabha scattering. What is its velocity?

Solution 3

Energy and momentum are conserved at each vertex. Thus, for the virtual photon in the “horizontal diagram”, E = 2mec2 and p = 0 ⇒ m = 2me and v = 0 and in the vertical diagram, E = 0 and p = 0 ⇒ m = 0 and v = 0.

Exercise 4 A student of chemistry lab knows all about protons, neutrons and electrons, as she sees them in action every day in the laboratory. But she is skeptical about positrons, muons, neutrinos, pions, quarks and intermediate vector bosons. Explain to her why none of these plays any direct role in chemistry. Dayalbagh Educational Institute

Interactions and fields

Particle Physics

Solution 4 This has various reasons : Anti-particles (such as positron) annihilate with the corresponding particle (the electron here) and since there are lots of electrons in the lab, positrons don’t stay long enough to have any role in ordinary chemical processes. But if we can work in total vacuum, we could make atoms and molecules of anti-matter. Most elementary particles (muons, pions and IV B) are intrinsically unstable, they disintegrate spontaneously in a tiny fraction of a second – which is small to do any chemistry. We can make short-lived “exotic atoms” (with muons (say) in orbit around the nucleus instead of electrons. Some of these systems last long enough to do spectroscopy. Neutrinos interact so feebly with matter that they have no impact on the chemistry, even though we have both of them all the time. Quarks are the basic constituents of protons and neutrons, so in an indirect way, they do play a fundamental role in chemistry. Also gluons play a fundamental role in holding the nucleus together. But due to confinement, they don’t occur as free particles, only in composite structures, so they don't act as individuals.

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Interactions and fields

Particle Physics

Exercise 5 Earlier, before discovery of neutron, the model of nucleus was thought to be made up of protons and electrons, with the atomic number equal to the excess number of protons. Beta decay supported this idea. Now use the position-momentum uncertainty relation Δx Δp ≥ ћ/2 to estimate the minimum momentum of an electron confined to a nucleus (radius 10-13 cm). From the relation E2-p2c2 = m2c4, determine the corresponding energy and compare it with that of an electron emitted in the beta decay of tritium.

Solution 5 We have ro = 10-15 m, ћ = 6.58 x 10-22 MeV sec , c = 3 x 108 m/s, me = 0.511 MeV/c2. Using Δx Δp ≥ ћ/2, we have pmin = ћ/2ro = ( ћc/2ro ) 1/c = 98.7 MeV/c Thus Emin = √(p2minc2 + me2c4) = 98.7 MeV Whereas the energy of an electron emitted in beta decay of tritium is less than 17 keV.

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Interactions and fields

Particle Physics

Exercise 6 The â&#x201E;Ś- particle has a track length of 0.5 cm, then find the lifetime.

Solution 6

= 1.67 x 10-11 sec = 2 x 10-11 sec Actually t = 0.8 x 10

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-10

sec.

This is the end of sub section of the chapter 1

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