Particles change into other particles all the time: neutrons decay, pions decay, collisions in accelerators make showers of new particles. Some changes never happen, though, even when charge balances. A handful of conservation laws, found by watching which reactions occur and which don't, sorts the allowed from the forbidden. They build on particle families.
Key idea
In every reaction and decay, these totals are the same before and after:
- electric charge $Q$;
- baryon number $B$ ($+1$ for baryons, $-1$ for antibaryons, $\frac{1}{3}$ per quark, 0 for mesons and leptons);
- lepton numbers $L_e$, $L_\mu$, $L_\tau$, separately ($+1$ for each lepton of that family and its neutrino, $-1$ for their antiparticles);
- energy and momentum (a particle at rest can only decay into lighter particles).
Why it works
These laws are summaries of experiments, and they are very well tested. Baryon number explains why the proton, the lightest baryon, does not decay: there is no lighter baryon for it to become. Experiments watching enormous tanks of water have found no proton decay at all, putting its lifetime beyond about $10^{34}$ years. Lepton numbers explain the neutrinos' bookkeeping: in neutron decay, $n \to p + e^- + \bar{\nu}_e$, the electron has $L_e = +1$, so an antineutrino with $L_e = -1$ must come out with it. Without that antineutrino, energy and momentum would not balance either; that is how Wolfgang Pauli predicted the neutrino in 1930, decades before it was detected.
Worked examples
Worked example: Neutron decay
Check $n \to p + e^- + \bar{\nu}_e$, and find the energy released. ($m_nc^2 = 939.565$ MeV, $m_pc^2 = 938.272$ MeV, $m_ec^2 = 0.511$ MeV.)
- Charge: $0 \to +1 - 1 + 0$. ✓
- Baryon number: $1 \to 1 + 0 + 0$. ✓ Lepton number $L_e$: $0 \to 0 + 1 - 1$. ✓
- Energy: $Q = 939.565 - 938.272 - 0.511 = 0.782$ MeV (the antineutrino's mass is negligible).
Answer: Allowed, releasing 0.782 MeV
Worked example: A forbidden decay
Could a proton decay as $p \to e^+ + \gamma$?
- Charge: $+1 \to +1 + 0$. ✓
- Baryon number: $1 \to 0 + 0$. ✗
- Lepton number $L_e$: $0 \to -1 + 0$. ✗
Answer: No: it breaks baryon number and lepton number
Worked example: Find the missing particle
In the decay $\pi^+ \to \mu^+ + \text{?}$, what is the missing particle?
- Charge: $+1 \to +1 + q$, so $q = 0$.
- $L_\mu$: $0 \to -1 + L$, so the missing particle has $L_\mu = +1$: a muon neutrino.
- Baryon number: $0 \to 0$; nothing else changes.
Answer: $\nu_\mu$, a muon neutrino
Common mistake
Allowing any reaction whose charges balance. Charge is only the first check: also count baryon number and each lepton number separately, and make sure there is enough energy. $p \to e^+ + \gamma$ balances charge but is forbidden; $\mu^- \to e^- + \gamma$ balances charge and total lepton number but breaks $L_\mu$ and $L_e$ separately, and it has never been seen.