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Notes on Elementary Particle

Physics

Instructor: MVN Murthy

Teaching Assistant: Dibyakrupa Sahoo

August–December 2012

Chapter 1

Natural Units, Scales, Notations &

Conventions

In this chapter we survey the various units, scales, notations and conventions we shall be using in our discussions ahead. More over this is done keeping in mind the current practices followed by the modern literatures in high energy physics.

1.1 Natural units

As is true for any branch of physics, particle physics is based on experiments. And these experiments look for the most elementary constituents of matter. These necessarily involve probing at extremely small length scales (typically of the order of 10−^15 m or less). And the typical masses involved are 10−^27 kg. The standard system of units in physics, in general, is the International System of Units (SI). In particle physics we use a system of units known as the ‘Natural Units’. There are two fundamental constants in relativistic quantum mechanics:

1. Reduced Planck’s constant ℏ (dimension M L^2 T −^1 ): ℏ ≡ 2 hπ = 1.054 571 628(53) × 10 −^34 J s.
2. Speed of light in vacuum c (dimension LT −^1 ): c = 299 792 458 m s−^1.

In natural units, we specify the energy in units of GeV (1 GeV = 10^9 eV)^1. This choice is motivated by the fact that the rest energy of a nucleon (proton or neutron) is approximately 1 GeV. In the natural units, we put ℏ = c = 1, (1.1)

so that we won’t have to worry about the pesky ℏ and c that appear in most of the equations in particle physics. Instead of mass, length and time, we use mass, action (ℏ) and speed (c). In these units. In (^1) One eV (electron-Volt) is the energy gained by an electron when accelerated by a potential difference of one volt.

have considered another set of units (known as Heaviside-Lorentz units) where

ǫ 0 = μ 0 = ℏ = c = 1, (1.4)

keeping the value of α unchanged (as it should be since this is a constant)

α = e

2 4 π ≃^

137.^ (1.5)

In the Heaviside-Lorentz units the Maxwell’s equations take the following form:

∇ ·^ ~ E~ =ρ, (1.6a) ∇ ·^ ~ B~ =0, (1.6b) ∇ ×^ ~ E~ = − ∂ B~ ∂t ,^ (1.6c) ∇ ×^ ~ B~ = ∂ ∂tE~ + J,~ (1.6d)

where E~ is the electric field, B~ is the magnetic field, ρ is the charge density and J~ is the current density. Electric charge is also a constant in the Heaviside-Lorentz units, the charge of electron being

e ≃ 0. 3028. (1.7)

For sake of completeness we provide the SI prefixes in the following table. Factor Prefix Symbol Factor Prefix Symbol 1024 yotta- Y 10 −^24 yocto- y 1021 zetta- Z 10 −^21 zepto- z 1018 exa- E 10 −^18 atto- a 1015 peta- P 10 −^15 femto- f 1012 tera- T 10 −^12 pico- p 109 giga- G 10 −^9 nano- n 106 mega- M 10 −^6 micro- μ 103 kilo- k 10 −^3 milli- m 102 hecto- h 10 −^2 centi- c 101 deka- da 10 −^1 deci- d

Example 1.1. de Broglie wavelength: The de Broglie wavelength associated with a 1 GeV photon is given by: λdBγ = h p =^2 π Eℏ c= 2π GeV−^1 = 2π × 0. 1975 × 10 −^15 m = 0. 395 π fm.

Example 1.2. Compton wavelength: The compton wavelength of a particle of mass m is defined as

λC^ = (^) mcℏc 2 = (^) mc ℏ.

In natural units the Compton wavelength is

λC^ = m^1.

So the Compton wavelength for a pion (whose mass is approximately 140 MeV) is

λCπ = (^) 140 MeV^1 = 10

3 140 GeV =

0. 1975 × 103

140 fm = 1.411 fm.

Similarly the Compton wavelength for an electron (whose mass is approximately 0.5 MeV) is

λCe = (^0) .5 MeV^1 = 10

3 0 .5 GeV =

0. 1975 × 103

1. 5 fm = 395 fm.

If we approximate the proton’s mass to be 1 GeV, then its Compton wavelength is

λCp = 1 GeV−^1 = 0.1975 fm.

Example 1.3. Classical electron radius: The classical electron radius, also known as the Lorentz radius or the Thomson scattering length, is given by

re = (^4) πǫ^10 e 2 mc^2.^ (1.8)

In the ‘Natural-Heaviside-Lorentz units’ we have ǫ 0 = μ 0 = ℏ = c = 1 and e ≃ 0. 3028. Using these values we have

re = e 2 4 πm ≃^

4 π(0.5 MeV) =

0. 30282 × 0. 1975 × 103

4 π × 0. 5 fm = 2.88 fm^ (1.9)

One can compare it with the more accurate value is re = 2.817 940 289 4(58) fm.

Example 1.4. The Bohr radius: The Bohr radius is the radius of the lowest energy stable orbit of the atomic electron. The energy of the electron in a Hydrogen atom is given by

E ≃ p

2 2 m −^

α r ,^ (1.10)

where the first term is kinetic energy and the second term is the electrostatic energy. The momentum p

‘graviton’ (still undiscovered, so only speculated) is also massless (if it exists), and thus leads to the infinite range of the gravitational interaction that it mediates. The W ±^ and Z bosons, which mediate the weak interaction are massive (approximately of 90 GeV), so the range of weak interaction is given by: Rweak = 180 1 GeV−^1 =^0. 1801975 × 10 −^15 m = 1. 097 × 10 −^18 m. (1.14)

The strong interaction is mediated by massless gluons, so it might be thought to be of infinite range. However, in fact it is confined to nuclear dimensions only. This is because of the property of ‘confinement’. We can however find out the range of strong interaction by analysing the force between two nucleons, which is mediated by pions (having mass around 140 MeV). This is a residual strong force^3. If we consider the mass of pion then the range of strong interaction is given by:

Rstrong = (^) 280 MeV^1 =^197280.^5 × 10 −^15 m = 0.7053 fm. (1.15)

This is exactly half of the Compton wavelength for pion.

If you are absorbing or producing a force

carrier particle, you’d better be

affected by the force carried!

Violators will be prosecuted for

undermining reality.

WARNING

4

Thus far we have talked about the ranges of the fundamental interactions. However, there is another aspect to these interactions and that is their relative strengths.

1.2.2 Strengths of Fundamental Interactions

We shall consider the four fundamental interactions one by one.

Gravitational Interaction Let us consider two protons of mass Mp, separated by a distance r. The gravitational potential energy between them is given by Newton’s law:

V = GN^ M^ p^2 r ,^ (1.16) (^3) Residual means that it is a consequences of gluonic exchanges amongst the consituting quarks of the nucleons. The idea of pion exchange is due to the Japanese physicist Hideki Yukawa. (^4) The content and essential design is taken from one such picture in the beautiful website ‘The Particle Adventure’: www.particleadventure.org

where GN is Newton’s gravitational constant:

GN = 6.674 28(67) × 10 −^11 m^3 kg−^1 s−^2 = 6.708 81(67) × 10 −^39 GeV−^2. (1.17)

Suppose we consider the two protons to be inside a nucleus, i.e. say r = 10−^15 m = 5.07 GeV−^1. Then V ≃ 6. 708 × (^5).^107 × 10 −^39 GeV = 1. 323 × 10 −^39 GeV. (1.18) This energy is thus found to be extremely negligible in comparison to the mass of proton itself. However, it must be noted that √G N =0.^819 ×^10 −^19 GeV−^1 , =0. 819 × 0. 1975 × 10 −^34 m = 1. 617 × 10 −^35 m = LPlanck, =0. 819 × 6. 59 × 10 −^44 s = 5. 397 × 10 −^44 s = TPlanck, and √G^1 N = (^0). 8191 × 1019 GeV = 1. 221 × 1019 GeV = MPlanck,

where LPlanck, TPlanck, MPlanck are called as the Planck length, the Planck time and the Planck mass respectively. At these Planck scales gravitation becomes the dominant interaction of the four in the realm of elementary particles. Assuming that at the Planck mass scale the gravitational interaction has as much strength as electromagnetic interaction at the proton mass scale, we obtain the relative strength of gravitational interaction as follows:

αgr = M^ p^2 M (^) Planck^2 αem.^ (1.19) However, we know that αem = 1/ 137. Therefore

αgr = (^137) × (1.221)^12 × 1038 = 0. 4896 × 10 −^40. (1.20)

Looking at these extremely small numbers, it must be clear that gravity has no measure bearings in the realm of elementary particles at our currently available energies.

Weak Interaction Weak interaction is the driving mechanism behind most of radioactive decays, e.g.

n → p + e−^ + νe, O^14 → N^14 + e−^ + νe.

All these decays are characterised by their half-life. The average life-time of the neutron is about

Note. It must be kept in mind that the values given here and those that will be given in the next section are not always true. In some particular particle event deviations from these numbers can be found. Here we are trying to make only a few gross quantitative comparisons amongst the four fundamental interactions. Our estimates are not valid for all processes, but these are typical values (that is, most particle events will testify these numbers to be approximately correct). Since we have found that gravity plays absolutely no role in particle events at the currently achieved (or near future) energy regimes, we shall henceforth drop gravity from our discussion.

1.3 Typical Cross-sections and Mean Life-times

The cross-section is directly proportional to the square of the coupling constant of the relevant underlying interaction. So we expect that

σst : σem : σwk :: 1-10^2 : 10−^4 : 10−^14 (1.25)

The mean life-time of a particle decaying via a particular channel (involving mediation by a particular interaction) is inversely proportional to the coupling constant of the mediating interaction. Thus we expect that τst : τem : τwk :: 1-10−^2 : 10^4 : 10^14 (1.26)

These are, however, only approximate ratios. We have to look at some specific examples to find out the typical values of cross-sections and mean life-times.

1.3.1 Typical Cross-sections

Let us consider the electromagnetic scattering of electron and positron to muon and antimuon:

e−(p 1 ) + e+(p 2 ) → μ+(k 1 ) + μ−(k 2 ).

The cross-section for this process is given by

σe− (^) e+ = α^2 f (s, me, mμ), (1.27)

where f is a function of the center-of-momentum energy (√s where s = (p 1 + p 2 )^2 = (k 1 + k 2 )^2 ), and the masses: me and mμ. Note that the total cross-section is in general a function of Lorentz invariant variables, which in this case are the square of the sum of the two four momenta in the initial state (or final state) and the masses. At very high energies we may neglect the masses of the particles, and purely by dimensional consid-

erations the cross section must be given by

σe− (^) e+ ≃ α

2 s.^ (1.28)

The exact expression is in fact σe− (^) e+ =^4 πα 2 3 s.^ (1.29)

At a center-of-momentum energy of 1 GeV, we have

σe−^ e+^ =^4 πα

2 3 s =^

4 π 3 × 1372 GeV

− (^2) = 4 π 3 × 1372 ×^0.^1975

(^2) × 10 − (^30) m (^2) = 87.05 nb. (1.30)

If we consider that the energy of each colliding beam is Eb, then √s = 2Eb. Therefore

σe− (^) e+ = (^) E^212 .7625 nb b (in GeV^2 )^

Another example of electromagnetic scattering is the scattering of low-energy photon on proton. At low-energy (which corresponds to long-wavelength photons) we can use the Thomson formula for the scattering cross-section. In our case

σγp =^83 π

( (^) α mp

≃ 83 π

137 GeV

− 1

= 1. 741 × 10 −^35 m^2 = 174.1 nb. (1.32)

Let us now consider the strong scattering of proton and proton. The charge radius of the proton as measured by experiments (electron- proton scattering) is about 0.81 fm. This is infact larger than the compton wavelength of the proton. Because the strong interaction strength is close to unity, the cross-section for proton-proton scattering is given by

σpp = πr^2 p ≃ 3. 141 × 0. 812 × 10 −^30 m^2 = 2. 061 × 10 −^30 m^2 = 20.61 mb

using the classical analogy for the cross-section. Indeed the experimental value is close to this, about 45 mb at close to 1 GeV energy. Therefore, the ratio of electron-positron and proton-proton scattering cross-sections is given by σe−^ e+ σpp^ =

87 .05 nb 45 mb = 1.^934 ×^10

Similarly σγp σpp^ =

174 .1 nb 45 mb = 3.^869 ×^10

These are consistent with our observation

=Eν (in GeV) × 0. 44256 × 0. 19752 × 10 −^43 m^2 = Eν (in GeV) × 1. 726 × 10 −^45 m^2 =Eν (in GeV) × 1. 726 × 10 −^17 b = Eν (in GeV) × 17 .26 atto-barn (ab).

So if Eν = 1 GeV, then σνe e = 17.26 ab. Therefore, for this case

σνe e σpp^ =

17 .26 ab 45 mb = 3.^83 ×^10

This result is consistent with our observation

σst : σwk :: 1-10^2 : 10−^14 which implies σst : σwk :: 1 : 10−^14 -10−^16. So we conclude that our observations for the relative cross-sections are approximately correct.

1.3.2 Typical Mean life-times

Let us consider a particle decay that proceeds via strong interaction. One such decay is: ∆++^ → π+^ + p. Experimentally, the peak corresponding to ∆++^ has a full width at half maximum of Γ∆++ ≃ 100 MeV. Now the life-time of ∆++^ is given by

τ∆++ = (^) Γ∆^1 ++ = 100 1 MeV−^1 = 10 GeV−^1 = 6. 59 × 10 −^24 s. (1.43)

Let us now consider a decay that proceeds via electromagnetic interaction, e.g. π^0 → γ + γ. The life-time of π^0 is found to be about 10−^16 s. So

τπ^0 τ∆++^ ≃^

10 −^16

6. 59 × 10 −^24 = 1.^52 ×^10

Now let us look at one decay that proceeds via weak interaction: Σ → n + π. The life-time of Σ is about 10 −^10 s. So τΣ τ∆++^ ≃^

10 −^10

10 −^24 = 10

As we can see these ratios confirm that our observation for mean life-times is approximately corrct:

τst : τem : τwk :: 1-10−^2 : 10^4 : 10^14 which implies τst : τem : τwk :: 1 : 10^4 -10^6 : 10^14 -10^16. We would like to stress that all these ratios about cross-sections and life-times are given here only to make out a gross quantitative comparisons amongst the various fundamental interactions.

1.4 Notation and Conventions

We shall follow the notation and conventions as given below. If there is any deviation from these, then they would be specified at their place of usage. Also not all the conventions and notations are listed

below. The most general ones have been listed, to make the presentation of the latter concepts clear and smoother.

Lorentz indices μ, ν, ρ, σ, etc. Three-vector indices i, j, k, l, etc. Three-vector ~x Unit vector xˆ = (^) |~x~x|

Kronecker delta δij =

1 , i = j 0 , i 6 = j

Levi-Civita tensor ǫijk : totally antisymmetric. ǫ 123 = 1. Contractions ǫijk ǫijk = 6, ǫijk ǫijm = 2δkm, ǫijk ǫimn = δjmδkn − δjnδkm. Dot product A~ · B~ = ∑ i

AiBi Cross product ( A~ × B~)i = ǫijk Aj Bk metric gμν = gμν^ = diag(1, − 1 , − 1 , −1)

gνμ = gνσgμσ = δνμ =

1 , μ = ν 0 , μ 6 = ν

Contravariant four-vector Aμ^ = (A^0 , A~), xμ^ = (t, ~x) Covariant four-vector Aμ = gμν Aν^ = (A^0 , − A~), xμ = (t, −~x) Lorentz invariant A · B ≡ AμBμ^ = gμν AμBν^ = A^0 B^0 − A~ · B~ antisymmetric tensor ǫμνρσ. ǫ 0123 = +1 and ǫ^0123 = − 1 contractions ǫμνρσ^ ǫμνρσ = −24, ǫμνρσ^ ǫμνρτ = − 6 gστ , ǫμνρσ^ ǫμντ ω = −2(gρτ gωσ − gωρgστ ) derivatives ∂μ ≡ (^) ∂x∂μ =

∂t , ∇~

, ∂μ^ ≡ (^) ∂x∂μ =

∂t ,^ −∇~

∂μ∂μ^ = (^) ∂t∂^22 − ∇~^2 ∂ · A = ∂μAμ^ = ∂A 0 ∂t +^ ∇ ·~^ A~ A← ∂→μ^ B = A(∂μB) − (∂μA)B

1.5 Particle Nomenclature

With discovery of large number of particles, it has been really troublesome to name and denote them. During the early days of particle physics, both Greek and Latin characters were used. They are stil used, though there has been some modification of names for hadrons. The only particle which uses both Greek and Latin characters is the particle J/Ψ. There is one convention we shall follow in these notes. An anti-particle can be denoted by putting a horizontal line above the particle name, e.g. anti-proton ≡ proton, positron ≡ electron, anti-neutrino ≡ neutrino etc. The practice of denoting the anti-particle by putting a horizontal line over the particle

for these mesons by specifying their quark-antiquark content and by the value of 2 S+1LJ , where S^6 , L, J stand for spin, orbital and total angular momenta of the qq system.

qq content

2 S+1LJ

(^1) (L even)J 1 (L odd)J 3 (L even)J 3 (L odd)J ud, uu − dd, du (I = 1) π ba^ ρ a dd + uu and/or ss

(I = 0) η, η′^ h, h′^ ω, φ f, f ′

cc ηc hc ψb^ χc bb ηb hb Υ χb tt ηt ht θ χt aDo not confuse this with bottom quark. bThis is the same J/ψ particle. Note. Although there are names for the tt mesons, such bound states are unlikely to be formed and found. Top quark is evidently so heavy that even before it can pair with top, it decays in laboratory. We have not shown any electric charges of the mesons in the above table, but they are usually placed on the top-right corner of the symbol.

“Flavored” Mesons: By ‘flavored’ mesons we mean those mesons which have nonzero heavy flavor quantum numbers, i.e. either S 6 = 0, or C 6 = 0, or B 6 = 0, or T 6 = 0. The main symbol for such a meson is an upper-case italic letter that indicates the heavier quark (or quark) as follows:

s → K, s → K, c → D, c → D, b → B, b → B, t → T, t → T.

If the lighter quark is not an up (u) or down (d) quark, its identity is specified by keepting its symbol as a subscript. In the literature, it is also found that physicists use the subscripts u (d) also to specify that the lighter quark is u (d). Example 1.5. Following are some examples illustrating the above scheme of nomenclature. K mesons D mesons B mesons qq content Symbol qq content Symbol qq content Symbol us K+^ uc D^0 ub B+ ds K^0 dc D−^ db B^0 us K−^ uc D^0 ub B− ds K^0 dc D+^ db B^0 sc D s− sb B s^0 sc D s+ sb B^0 s (^6) Do not confuse this S with strangeness.

Baryons: For baryons the old symbols N (nucleons), ∆, Λ, Σ, Ξ, and Ω are extensively and exclusively used. The following diagram succinctly explains the current nomenclature for baryons.

Baryon

three u and/or d quarks (^) N I = (^12)

I = 32

two u and/or d quarks (^) Λ I = 0 Λ

The 3rd quark is denoted by a subscript.

I = 1

one u or d quark (^) Ξ

I = 12

one (two) subscript(s) to denote the remaining heavy quark(s).

no u or d quark

I = 0

Subscripts denote any heavy quark content.

Example 1.6. The four light quarks u, d, c and s can be combined in 20 different ways to create baryons and in 16 different ways to create mesons^7. The sixteen mesons are grouped into a 15-plet and a singlet. Figures 1.1 and 1.2 show the ground-state pseudoscalar and vector mesons respectively. Figures 1.3 and 1.4 show the ground state baryon multiplets with spins 12 and 32 respectively. The scheme of nomenclature as described above is used in all these diagrams and the quark contents are also explicitly shown.

EXERCISE

Question 1. 1. What is the energy of an electron that has a de Broglie wavelength of 10−^16 m? Question 1. 2. In units of the electron Bohr radius, what would be the Bohr radius for a muonic atom and pionic atom. Question 1. 3. The size of the proton (charge radius) is approximately 1 fm. Typically one needs a probe whose wave length is much less than this size to probe the structure of the proton. Suppose we assume that a photon probe has a wavelength less than 1/10 fm, calculate the energy of the photon required to probe the internal structure of the photon. Question 1. 4. The pions are unstable particles. Investigate the decay modes of charged and neutral pions. Assuming an equal number π±^ are enter the earths atmosphere (approximately correct), what particles are left in what ratios after all the pions and even their decay products have decayed. Question 1. 5. Suppose the proton could decay with a life time of 10^30 years, how many cubic meters of water would have to be observed if one wanted to have about 100 events in a year. Question 1. 6. Low energy neutrinos pass through a piece of solid iron- if the neutrino-nucleon cross section is about σ ≈ 10 −^47 m^2 , estimate the mean free path of the neutrinos in iron (density of iron is 8 (^7) How do we arrive at these numbers? The answer will be given in Chapter 2.

cs

us ds

du

cu

sd

cd

ud

su

sc

dc uc

K∗^0 K∗+

K∗− K∗^0

ρ− ρ+

φ ω J/ψ

ρ^0

D∗^0 D∗+

D∗−^ D∗^0

D∗ s+

D∗− s

C = +

C = 0

C = − 1

Figure 1.2: Vector meson super-multiplet with charm quark included.

dds

udd uud

uus

dss uss

ddc uuc

ssc

dcc ucc scc

Λ+ c Σ+ c

Σ−

n p

Σ+

Ξ− Ξ^0

Σ^0 c Σ++ c

Ω^0 c

Ξ+ cc Ξ++ cc Ω+ cc

Ξ+ c′

Ξ+ c

Ξ^0 c′

Ξ^0 c

Λ^0 Σ^0 uds

dsc usc

udc

C = +

C = +

C = 0

Figure 1.3: Spin- 12 baryon super-multiplet with charm quark included.