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Conservation laws

Particle physics polices itself with ledgers. Charge, baryon number, the two lepton numbers and strangeness must all balance, and energy and momentum always get a vote. If an interaction fails one audit, it does not happen.

Year 12AQA 3.2.1.7

Builds on Quarks and antiquarks and Stable and unstable nuclei.

IN THIS TOPIC

  • Describe the quark character change in β− and β+ decay.
  • Audit interactions against conservation of charge, baryon number, lepton number and strangeness.
  • Recognise that energy and momentum are conserved in every interaction.

WHAT YOU PROBABLY THINK

In particle physics, anything can happen.

Beta decay as a change of flavour

The quark character change in beta decay: d to u in beta-minus, u to d in beta-plus, the W carrying the differenceβ⁻duW⁻ leavesβ⁺udW⁺ leavesthe nucleon keeps two spectator quarks; one changes flavour
FIG. 1β⁻ is d becoming u with a W⁻ leaving; β⁺ is u becoming d with a W⁺. Two spectator quarks watch in each case.

Both beta decays are single-quark events. In β−, a down quark becomes an up: d → u, the W− carrying off the charge and becoming e− plus an electron antineutrino. In β+, an up becomes a down: u → d, via a W+, giving a positron and an electron neutrino. State the quark change, name the W, and the mark scheme is satisfied.

The audit

Whether any proposed interaction is allowed comes down to a line-by-line audit. Charge must balance. Baryon number must balance. Lepton number must balance, for the electron family and the muon family separately. Strangeness must balance in strong interactions, and may shift by 0 or ±1 in weak ones.

Auditing an interaction: charge, baryon number, electron lepton number and strangeness each balance across the arrowν+ pn + e⁺charge Q+1 = +1baryon number B+1 = +1lepton number L-1 = -1strangeness S0 = 0four ledgers, four balances: the interaction is allowed
FIG. 2An interaction audited: antineutrino plus proton to neutron plus positron passes all four ledgers, a tick on every line.

Work in columns, one quantum number at a time, totalling each side of the arrow. AQA promises that for any particle outside the specified set, the necessary data will be provided in the question, so the skill is pure bookkeeping, not recall. A single failed line kills the interaction; nature has never once been caught cooking these books.

The two laws that always vote

Beyond the quantum numbers, energy and momentum are conserved in every interaction, and they can veto a process the ledgers would allow. The showpiece: a free proton decaying by p → n + e+ + νe balances charge, baryon number and electron lepton number perfectly, yet it never happens, because the products' rest energies (about 939.6 MeV plus 0.511 MeV) exceed the proton's 938.3 MeV. There is no energy to pay for the difference, so the decay is forbidden, and the proton's stability, promised two lessons ago, is finally explained. Inside a nucleus the energy accounts differ, which is why bound protons can undergo β+ decay while free ones cannot.

THE EXAM BIT

  • Audit format earns marks by itself: a row per quantum number, totals both sides, a verdict per row. Show the table even when the answer is “not allowed”.
  • Lepton numbers are audited per family. An interaction turning a muon neutrino into an electron fails, however neatly the totals seem to add.
  • Strangeness gets a conditional rule: conserved in strong interactions; ΔS of 0 or ±1 permitted in weak ones. State which interaction you are auditing before applying it.
  • Unfamiliar particle in the question? Its quantum numbers will be printed there. Read them; never guess from the symbol.
  • When every ledger balances and the process still cannot occur, look to energy: compare rest energies from the data booklet. That is the free proton's alibi.

CHECK YOURSELF

A student proposes that a free proton decays: p → n + e+ + νe. Audit the conservation laws, then explain what actually forbids the decay. (Rest energies: p 938.3 MeV, n 939.6 MeV, e 0.511 MeV.)

Show a hint

Every ledger passes. So check the one thing left.

Show the answer

Charge: +1 → 0 + 1 + 0 ✓. Baryon number: +1 → +1 ✓. Electron lepton number: 0 → 0 − 1 + 1 = 0 ✓. Strangeness: 0 → 0 ✓. Every quantum number balances.

Energy does not. The products carry at least 939.6 + 0.511 = 940.1 MeV of rest energy, more than the proton's 938.3 MeV, and a free proton has nothing extra to pay with.

Energy conservation forbids the decay, and the proton stays the only stable baryon. Inside a nucleus the binding-energy accounts can differ, which is why β+ decay happens there.

Audit charge, baryon and lepton numbers, and strangeness.

Energy and momentum always get a vote.

No animated video for this topic yet; these notes stand alone. InkPhysics on YouTube.