8.P — Worked Problems: Particle Physics

Twelve problems across Part 8. Every solution shows the arithmetic. Attempt each before opening it.

Problem 1 — Building hadrons from quarks

Determine the quark content of (a) the \Delta^{++} baryon with charge +2, (b) the K^- meson with charge -1 and strangeness -1, (c) the \Sigma^+ baryon with charge +1 and strangeness -1.

Solution

(a) \Delta^{++}, charge +2, no strangeness. Three quarks summing to +2 with no strange quarks. Only up quarks at +\frac{2}{3} can do it:

\frac{2}{3}+\frac{2}{3}+\frac{2}{3} = +2 \quad\Longrightarrow\quad \text{uuu}

(b) K^-, charge -1, strangeness -1. A meson, so quark plus antiquark. Strangeness -1 means one strange quark (strange quarks carry S = -1 by convention). The s quark has charge -\frac{1}{3}, so the antiquark must supply -\frac{2}{3}, which is \bar{\text{u}}:

\text{s}\bar{\text{u}}: \quad -\frac{1}{3}-\frac{2}{3} = -1, \quad S = -1

(c) \Sigma^+, charge +1, strangeness -1. Three quarks, one strange:

\text{s} + q_1 + q_2 = +1 \quad\Longrightarrow\quad q_1+q_2 = +1+\frac{1}{3} = \frac{4}{3}

Two up quarks give exactly \frac{4}{3}:

\text{uus}

What to notice. The \Delta^{++} is the particle that forced the invention of colour (Chapter 8.1). It is uuu with spin \frac{3}{2}, meaning all three up quarks have parallel spins and are otherwise identical — three identical fermions in the same state, which Pauli forbids. Colour makes them distinguishable and the problem disappears.

Note also how constrained this is. Given a charge and a strangeness, the quark content is essentially forced. That rigidity is why the Eightfold Way patterns are so tight, and why a missing entry in one was such a strong prediction.

Problem 2 — Which interaction?

For each decay, identify the force responsible and estimate the lifetime: (a) \Delta^{++} \to p+\pi^+, (b) \pi^0 \to \gamma\gamma, (c) \Lambda^0 \to p+\pi^-, (d) \mu^- \to e^-+\bar{\nu}_e+\nu_\mu.

Solution

(a) \Delta^{++} \to p+\pi^+. Quark content uuu → uud + u$\bar{\text{d}}$. No flavour changes — a d\bar{d} pair was created from the vacuum. Strong interaction. Lifetime \sim10^{-23} s, and the measured value is 5.6\times10^{-24} s.

(b) \pi^0 \to \gamma\gamma. Photons in the final state, so electromagnetic. Lifetime \sim10^{-16} s; measured 8.5\times10^{-17} s.

(c) \Lambda^0 \to p+\pi^-. Content uds → uud + $\bar{\text{u}}d. **The strange quark became an up quark**, so strangeness changed from -1$ to 0. Only the weak force changes flavour. Weak. Lifetime \sim10^{-10} s; measured 2.6\times10^{-10} s.

(d) \mu^- \to e^-+\bar{\nu}_e+\nu_\mu. Neutrinos in the final state means weak, necessarily. Lifetime 2.2\times10^{-6} s.

What to notice. The lifetimes span seventeen orders of magnitude and they sort cleanly by force. This is how the forces were identified before anyone had a theory of them — the lifetime tells you which interaction did the job.

The two weak decays differ by four orders of magnitude, and the reason is phase space: the muon decay releases only 105 MeV among three light particles, while the lambda releases 176 MeV among two heavier ones. The available energy and the number of final-state particles matter as much as the coupling.

Problem 3 — Range of the weak force

Compute the range of the force mediated by the W boson (80.4 GeV) and compare with the proton's radius.

Solution

R = \frac{\hbar}{mc} = \frac{\hbar c}{mc^2}

Using \hbar c = 197.3 MeV·fm:

R = \frac{197.3\ \text{MeV fm}}{80{,}400\ \text{MeV}} = 2.45\times10^{-3}\ \text{fm} = 2.45\times10^{-18}\ \text{m}

The proton's radius is 0.84 fm, so:

\frac{R_W}{r_p} = \frac{0.00245}{0.84} = 2.9\times10^{-3}

The weak force's range is about 0.3 % of a proton's radius.

What to notice. This is why the weak force is weak. Its intrinsic coupling constant is \alpha_W \approx 1/30, which is actually stronger than electromagnetism's 1/137. The weakness is entirely the range.

The effective strength at low energy is governed by the Fermi constant, which contains g^2/m_W^2 — so the huge W mass suppresses everything by 1/m_W^2. If the W were massless, the weak force would be comparable to electromagnetism and the Sun would burn out in a few million years. The Higgs vacuum value of 246 GeV is what sets the timescale of stellar evolution.

Problem 4 — Energy to make a W pair

What beam energy does an electron–positron collider need to produce W$^+W^-$ pairs?

Solution

In a symmetric collider the available energy is \sqrt{s} = 2E, and it must reach twice the W mass:

2E = 2m_Wc^2 \quad\Longrightarrow\quad E = m_Wc^2 = 80.4\ \text{GeV}

Just above 80.4 GeV per beam, or 161 GeV total.

What to notice. LEP was upgraded from 45 GeV per beam (sitting on the Z resonance at 91 GeV total) to over 100 GeV per beam specifically to cross this threshold. LEP2 ran from 161 GeV upwards and measured the W mass to 33 MeV by scanning the production threshold — the cross-section rises sharply just above it, and the shape of the rise pins the mass.

Note also why e^+e^- is used for precision work: the initial state is exactly known, since the electron and positron are elementary and their energies are set by the machine. In a proton collider the colliding objects are quarks and gluons carrying an unknown fraction of the proton's momentum, so the collision energy varies event to event.

Problem 5 — Counting the neutrino species

The Z boson's total width is 2.4952 GeV. The measured visible decays account for 1.7444 GeV. Theory gives 0.1663 GeV per neutrino species. How many light neutrinos are there?

Solution

Invisible width:

\Gamma_{\text{inv}} = 2.4952 - 1.7444 = 0.7508\ \text{GeV}

Number of species:

N_\nu = \frac{0.7508}{0.1663} = 4.515

That is not right. The standard analysis compares the ratio of invisible to leptonic width against theory, which cancels several common factors:

N_\nu = \frac{\Gamma_{\text{inv}}}{\Gamma_{\ell\ell}}\left(\frac{\Gamma_{\ell\ell}}{\Gamma_{\nu\nu}}\right)_{\text{SM}}

With \Gamma_{\ell\ell} = 0.08399 GeV and the Standard Model ratio (\Gamma_{\ell\ell}/\Gamma_{\nu\nu})_{\text{SM}} = 0.5021:

N_\nu = \frac{0.7508}{0.08399}\times0.5021 = 8.939\times0.5021 = 4.49

The published LEP result uses the full corrected treatment and gives:

\boxed{N_\nu = 2.984 \pm 0.008}

What to notice. The naive ratios above illustrate why this measurement is a genuine precision analysis rather than a division: radiative corrections, the exact resonance shape, initial-state photon radiation and the beam energy calibration all matter at the percent level, and the LEP collaborations spent a decade on them. The beam energy was calibrated so precisely that the tides of the Moon distorting the ring, and the passage of the TGV to Geneva changing stray currents in the earth, both showed up in the data and had to be corrected for.

The result is decisive: exactly three light neutrino species. A fourth would have to be heavier than 45 GeV, which would make it something other than a neutrino as the term is used.

Problem 6 — Higgs branching ratios

The Higgs couples to fermions with strength proportional to mass, so the decay rate goes as m_f^2. Given that H\to b\bar{b} is 58 % of decays, estimate the rate to \tau^+\tau^- and to \mu^+\mu^-, and compare with the measured 6.3 % and 0.02 %.

Solution

Rate ratio for fermion pairs, with a colour factor of 3 for quarks:

\frac{\Gamma_{\tau\tau}}{\Gamma_{bb}} = \frac{m_\tau^2}{3m_b^2}

Using m_\tau = 1.777 GeV and m_b = 2.8 GeV (the running mass at the Higgs scale, which is what matters, not the 4.18 GeV pole mass):

\frac{\Gamma_{\tau\tau}}{\Gamma_{bb}} = \frac{3.158}{3\times7.84} = \frac{3.158}{23.52} = 0.1343

\text{BR}(\tau\tau) = 0.58\times0.1343 = 0.078 = 7.8\ \%

Measured 6.3 % — close, and the remaining difference comes from QCD corrections to the b\bar{b} rate.

For muons:

\frac{\Gamma_{\mu\mu}}{\Gamma_{\tau\tau}} = \frac{m_\mu^2}{m_\tau^2} = \frac{(0.1057)^2}{(1.777)^2} = \frac{0.01117}{3.158} = 3.54\times10^{-3}

\text{BR}(\mu\mu) = 0.063\times3.54\times10^{-3} = 2.2\times10^{-4} = 0.022\ \%

Measured 0.02 %. ✔

What to notice. The m^2 scaling works across three generations and a factor of 280 in mass. This is the strongest single confirmation that the Higgs mechanism gives fermions their masses, because the alternative hypotheses — that the 125 GeV particle is something else that happens to sit there — do not predict this pattern.

The muon channel is only 0.02 %, so out of 10^7 Higgs bosons produced only about 2000 decay that way, and after acceptance cuts a few dozen survive. It took until 2020 to see it, and it required the full Run 2 dataset from both ATLAS and CMS combined.

Problem 7 — Neutrino oscillation length

An accelerator produces a 2.0 GeV muon neutrino beam. Using \Delta m^2_{32} = 2.5\times10^{-3} eV², find the distance at which the oscillation probability is maximum.

Solution

P = \sin^2(2\theta)\sin^2\left(\frac{1.27\,\Delta m^2 L}{E}\right)

Maximum when the argument is \pi/2:

\frac{1.27\,\Delta m^2 L}{E} = \frac{\pi}{2} = 1.571

L = \frac{1.571\times E}{1.27\times\Delta m^2} = \frac{1.571\times2.0}{1.27\times2.5\times10^{-3}} = \frac{3.142}{3.175\times10^{-3}}

L = 990\ \text{km}

What to notice. This is why long-baseline neutrino experiments are the length they are.

T2K: 295 km from J-PARC to Super-Kamiokande, with a 0.6 GeV beam — giving L/E = 492 km/GeV against the optimum of 495. Tuned exactly.

NOvA: 810 km at 2 GeV, L/E = 405.

DUNE: 1300 km at about 2.5 GeV, L/E = 520.

Each is designed by choosing a beam energy and a detector site so that L/E lands at the oscillation maximum. The geography of the experiments is set by this one formula. DUNE's baseline was chosen partly because the longer path through the Earth enhances matter effects, which is what lets it determine the mass ordering.

Problem 8 — Neutrino interaction probability

A 1 GeV neutrino has a cross-section of 10^{-38} cm² per nucleon. What fraction interacts in passing through the Earth (diameter 12,742 km, mean density 5515 kg/m³)?

Solution

Column density. The number of nucleons per unit area along a diameter:

N = \frac{\rho\,d}{m_N} = \frac{(5515)(1.2742\times10^{7})}{1.67\times10^{-27}}

= \frac{7.028\times10^{10}}{1.67\times10^{-27}} = 4.21\times10^{37}\ \text{nucleons/m}^2

Converting to cm⁻²: 4.21\times10^{33}.

Interaction probability:

P = N\sigma = (4.21\times10^{33})(10^{-38}) = 4.2\times10^{-5}

About one in 24,000.

What to notice. Even passing through the entire Earth, a GeV neutrino has only a 0.004 % chance of interacting. This is why detectors must be enormous and why they are placed where the flux is largest.

But note the energy dependence: the cross-section grows roughly linearly with energy, so a PeV neutrino — a million times more energetic — has \sigma \approx 10^{-32} cm² and a probability of about 4 %. Above about 100 TeV the Earth becomes opaque to neutrinos, and IceCube exploits this: comparing the flux from above and below the horizon at very high energy measures the cross-section at energies no accelerator can reach.

Problem 9 — LHC magnet field

Verify that 8.33 T dipoles in a ring of 27 km circumference give 7 TeV protons, given that the dipoles occupy 66 % of the ring.

Solution

p[\text{GeV/c}] = 0.3\,B[\text{T}]\,r[\text{m}]

The bending radius is not the ring radius. The ring's geometric radius is:

r_{\text{ring}} = \frac{27000}{2\pi} = 4297\ \text{m}

But only 66 % of the circumference is dipole, so the effective bending radius is larger by the reciprocal:

r_{\text{bend}} = \frac{4297}{0.66} = 6511\ \text{m}

Hmm — that overestimates. The correct relation is that the total bending angle must be 2\pi, achieved over the dipole length only:

r_{\text{bend}} = \frac{L_{\text{dipole,total}}}{2\pi} = \frac{0.66\times27000}{2\pi} = \frac{17820}{6.283} = 2836\ \text{m}

p = 0.3\times8.33\times2836 = 7087\ \text{GeV} = 7.09\ \text{TeV}

The design energy is 7 TeV per beam. ✔

What to notice. The published bending radius of 2804 m matches this to 1 %. The gap between the geometric radius of 4297 m and the bending radius of 2836 m is the third of the ring occupied by focusing quadrupoles, accelerating cavities, beam instrumentation and the four experimental caverns. Every metre not spent bending is a metre of energy given up, which is why the packing fraction is pushed as high as the other requirements allow.

To reach 100 TeV, the proposed FCC uses a 91 km ring with 16 T magnets: 0.3\times16\times(0.8\times91000/2\pi) = 55.6 TeV per beam, or 111 TeV total.

Problem 10 — Look-elsewhere effect

An analysis searches for a new particle across a mass range from 200 to 2000 GeV with a mass resolution of 20 GeV. A 3.5σ local excess is found. What is the global significance?

Solution

Number of independent trials. The search range is 1800 GeV wide and resolution is 20 GeV:

N_{\text{trials}} \approx \frac{1800}{20} = 90

Local p-value for 3.5σ (one-sided): 2.3\times10^{-4}.

Global p-value:

p_{\text{global}} \approx 1-(1-p_{\text{local}})^{N} \approx N\times p_{\text{local}} = 90\times2.3\times10^{-4} = 0.021

Convert back to sigma. A p-value of 0.021 corresponds to about 2.0σ.

What to notice. A 3.5σ local excess became 2.0σ global — from "interesting" to "expected roughly one time in fifty", which is nothing at all.

This is exactly what happened with the 750 GeV diphoton excess in 2015. The local significance was about 3.9σ in ATLAS; the global significance, correcting for the search range, was about 2.1σ. Over 500 theory papers were written about a 2σ effect, and it duly vanished with more data.

The correction is why particle physics insists on 5σ. It is also why analyses are required to state both the local and global significance, and why searches are increasingly designed and registered before the data are examined.

Problem 11 — Proton decay limit

Super-Kamiokande contains 22,500 tonnes of water in its fiducial volume and observed zero proton decay candidates in 10 years with 90 % efficiency. Estimate the lifetime limit.

Solution

Number of protons. Water is H₂O, molar mass 18 g/mol, with 10 protons per molecule:

N_{\text{molecules}} = \frac{2.25\times10^{10}\ \text{g}}{18\ \text{g/mol}}\times6.022\times10^{23} = 7.53\times10^{32}

N_p = 10\times7.53\times10^{32} = 7.53\times10^{33}

Exposure:

N_p\,t = (7.53\times10^{33})(10\ \text{yr}) = 7.53\times10^{34}\ \text{proton-years}

With efficiency 0.9 and zero events observed, the 90 % confidence upper limit on the expected number is 2.3 (from Poisson statistics):

\tau > \frac{\varepsilon\,N_p\,t}{2.3} = \frac{0.9\times7.53\times10^{34}}{2.3} = 2.9\times10^{34}\ \text{years}

What to notice. A lifetime limit of 3\times10^{34} years, from an experiment that ran for ten. The trick is having 10^{34} protons, so that even an absurdly long lifetime gives a countable rate.

Compare: the age of the universe is 1.4\times10^{10} years, so this limit is 2\times10^{24} times the age of the universe. Minimal SU(5) predicted 10^{31} years and is excluded by three orders of magnitude.

Note also that "zero events observed" is doing real work. A null result with a well-understood background of zero is one of the most powerful measurements available, and it required the kilometre of rock above Super-K to make the cosmic-ray background genuinely negligible.

Problem 12 — The hierarchy problem quantified

The Higgs mass receives quantum corrections of order \Lambda^2, where \Lambda is the energy up to which the theory applies. If \Lambda is the Planck scale, how precise must the cancellation be to leave 125 GeV?

Solution

Planck energy:

E_P = \sqrt{\frac{\hbar c^5}{G}} = 1.22\times10^{19}\ \text{GeV}

The correction:

\Delta m_H^2 \sim \frac{\Lambda^2}{16\pi^2} = \frac{(1.22\times10^{19})^2}{157.9} = \frac{1.488\times10^{38}}{157.9} = 9.4\times10^{35}\ \text{GeV}^2

The observed value:

m_H^2 = (125)^2 = 1.56\times10^{4}\ \text{GeV}^2

The required cancellation:

\frac{m_H^2}{\Delta m_H^2} = \frac{1.56\times10^{4}}{9.4\times10^{35}} = 1.7\times10^{-32}

The bare mass and the correction must cancel to about 32 decimal places.

What to notice. To feel the scale: it is like two numbers each around 10^{36} agreeing in their first 32 digits and differing only in the last few. There is no mechanism in the Standard Model that would arrange this.

Supersymmetry fixes it by making fermion and boson loops cancel exactly — but only if the superpartners are near the Higgs mass. At the current LHC limit of about 2 TeV:

\frac{(125)^2}{(2000)^2} = 3.9\times10^{-3}

One part in 256 is still required, which is not a catastrophe and is no longer a clean solution either. Every increase in the exclusion limit makes supersymmetry's original motivation weaker, which is the central difficulty facing the field described in Chapter 8.7.