Nature of Particles and Their Interactions

Lessons

Senior 6 · Physics

Nature of Particles and Their Interactions

5 lessons

Lesson 4 of 5

Conservation Laws in Particle Reactions and Decays

25 min

Why doesn't the proton decay?

A proton could, it seems, turn into a positron and a neutral pion: $\mathrm{p} \rightarrow \mathrm{e^+} + \pi^0$ . Charge is conserved ( $+ 1 = + 1 + 0$ ), and the products are lighter than the proton, so energy is available. Yet huge underground detectors, watching tens of thousands of tonnes of water for years, have never seen it. The Particle Data Group (2026) gives this decay a partial lifetime of more than $2.4 \times 10^{34}$ years, about $10^{24}$ times the age of the universe. Something else must be conserved. This lesson sets out the conservation laws of particle physics and uses them to predict which reactions and decays can happen, and through which interaction.

Quantities that are always conserved

Energy (including rest energy $m c^{2}$ ) and momentum. A particle can decay only into products whose total rest mass is LESS than its own.
Charge $Q$ .
Baryon number $B$ : each quark has $B = +\frac{1}{3}$ and each antiquark $-\frac{1}{3}$ , so baryons have $B = + 1$ , antibaryons $- 1$ , and mesons and leptons 0.
Lepton number, separately for each generation: $L_{e}$ for $\mathrm{e^-}$ and $\nu_e$ , $L_\mu$ for $μ^{-} \mu^-$ and $\nu_\mu$ , $L_\tau$ for $\tau^-$ and $\nu_\tau$ . Each lepton has $+ 1$ , each antilepton $- 1$ , every other particle 0.

A quantity conserved only in strong and electromagnetic interactions

Strangeness $S$ : each strange quark has $S = - 1$ , each anti-strange quark $+ 1$ . The strong and electromagnetic interactions never change a quark's flavour, so they conserve $S$ . The weak interaction can change a strange quark into an up quark, so it can change $S$ by $\pm 1$ .

Particle	Quarks	Q	B	$L_{e}$	$L_\mu$	S
p	uud	+1	+1	0	0	0
n	udd	0	+1	0	0	0
$\Lambda^0$	uds	0	+1	0	0	−1
$π^{+} \pi^+$	$\mathrm{u}\bar{\mathrm{d}}$	+1	0	0	0	0
$\mathrm{K}^+$	$\mathrm{u}\bar{\mathrm{s}}$	+1	0	0	0	+1
$\mathrm{K}^0$	$\mathrm{d}\bar{\mathrm{s}}$	0	0	0	0	+1
$\mathrm{e^-}$ , $\nu_e$	none	−1, 0	0	+1	0	0
$μ^{-} \mu^-$ , $\nu_\mu$	none	−1, 0	0	0	+1	0

For each antiparticle, reverse every sign: the positron has $Q = + 1$ , $L_{e} = - 1$ ; the $\mathrm{K}^-$ ( $\mathrm{s}\bar{\mathrm{u}}$ ) has $S = - 1$ .

The proton is the lightest particle with $B = 1$ . Any decay would need lighter products with $B = 1$ , and there are none: baryon number conservation makes the proton stable. In $\mathrm{p} \rightarrow \mathrm{e^+} + \pi^0$ , $B$ would change from 1 to 0 and $L_{e}$ from 0 to −1.

Definition

Baryon number $B$ is +1 for baryons, −1 for antibaryons and 0 for all other particles ( $+\frac{1}{3}$ per quark). Lepton number is +1 for leptons and −1 for antileptons, counted separately for the electron, muon and tau families. Strangeness $S$ is −1 for each strange quark (+1 for each anti-strange quark); it is conserved in strong and electromagnetic interactions but not in weak ones. Example: in $\mathrm{n} \rightarrow \mathrm{p} + \mathrm{e^-} + \bar{\nu}_e$ , $B$ is 1 on both sides and $L_{e}$ is 0 on both sides.

Checking a reaction

Write each quantity under each particle, add up each side, and compare. If $Q$ , $B$ or any lepton number fails, the reaction is impossible. If they all balance but $S$ changes, only the weak interaction can do it, so it will be slow. For a decay, also check that the parent is heavier than the products.

Worked example 1: neutron decay

Show that $\mathrm{n} \rightarrow \mathrm{p} + \mathrm{e^-} + \bar{\nu}_e$ obeys the conservation laws, and find the energy released ( $m_nc^2 = 939.565\,\mathrm{MeV}$ , $m_pc^2 = 938.272\,\mathrm{MeV}$ , $m_ec^2 = 0.511\,\mathrm{MeV}$ ; the antineutrino's mass is negligible).

Solution

	n	p	$\mathrm{e^-}$	$\bar{\nu}_e$	check
Q	0	+1	−1	0	0 = 0 ✓
B	+1	+1	0	0	1 = 1 ✓
$L_{e}$	0	0	+1	−1	0 = 0 ✓

Energy released: $939.565 - 938.272 - 0.511 = 0.782\,\mathrm{MeV}$ , shared as kinetic energy. Without the antineutrino, $L_{e}$ would be 0 before and +1 after: this is why it must be there.

Worked example 2: allowed or forbidden?

Decide whether each process can happen: (a) $\mu^- \rightarrow \mathrm{e^-} + \bar{\nu}_e + \nu_\mu$ ; (b) $\mu^- \rightarrow \mathrm{e^-} + \gamma$ ; (c) $\pi^- + \mathrm{p} \rightarrow \mathrm{n} + \pi^0$ ; (d) a free proton decaying as $\mathrm{p} \rightarrow \mathrm{n} + \mathrm{e^+} + \nu_e$ .

Solution

(a) Q: $- 1 = - 1 + 0 + 0$ ✓. $L_{e}$ : $0 = 1 - 1 + 0$ ✓. $L_\mu$ : $1 = 0 + 0 + 1$ ✓. Allowed; this is how every muon decays.

(b) Q ✓, but $L_\mu$ : $1 \rightarrow 0$ and $L_{e}$ : $0 \rightarrow 1$ . Forbidden. Searches have found no such decay: PDG 2026 gives a branching fraction below $1.5 \times 10^{-13}$ .

(d) Q, B and $L_{e}$ all balance, but $m_{n} > m_{p}$ : a free proton cannot supply the rest energy of the heavier neutron. Forbidden for a FREE proton. Inside some nuclei it happens (beta-plus decay, Unit 5), because the change in nuclear binding energy supplies the difference.

Worked example 3: strangeness and the speed of a decay

(a) Show that $\pi^- + \mathrm{p} \rightarrow \mathrm{K}^0 + \Lambda^0$ can occur through the strong interaction. (b) The $\Lambda^0$ then decays: $\Lambda^0 \rightarrow \mathrm{p} + \pi^-$ . Which interaction is responsible, and why does the $\Lambda^0$ live for $2.6 \times 10^{-10}$ s, much longer than particles that decay strongly (about $10^{-23}$ s)?

Solution

(a) Q: $- 1 + 1 = 0 + 0$ ✓. B: $0 + 1 = 0 + 1$ ✓. S: $0 + 0 = (+ 1) + (- 1)$ ✓. Strangeness is conserved: the strong interaction makes strange particles in PAIRS, one with $S = + 1$ and one with $S = - 1$ (called associated production).

(b) S: $-1 \rightarrow 0 + 0$ . Strangeness changes, so only the weak interaction can do it, and weak processes are slow. The $\Lambda^0$ is the lightest baryon containing a strange quark, so it has no strong decay route, and it must wait for the weak interaction. (Energy: $1115.68 - 938.27 - 139.57 = 37.8\,\mathrm{MeV}$ released.)

Decays and their interactions

The interaction responsible for a decay shows in its lifetime: the stronger the interaction, the faster the decay.

Decay	Interaction	Mean life (PDG 2026)
$\Delta^{++} \rightarrow \mathrm{p} + \pi^+$	strong	about $6 \times 10^{-24}$ s
$\pi^0 \rightarrow \gamma + \gamma$	electromagnetic	$8.4 \times 10^{-17}$ s
$\Lambda^0 \rightarrow \mathrm{p} + \pi^-$	weak	$2.6 \times 10^{-10}$ s
$\pi^+ \rightarrow \mu^+ + \nu_\mu$	weak	$2.6 \times 10^{-8}$ s
$\mu^- \rightarrow \mathrm{e^-} + \bar{\nu}_e + \nu_\mu$	weak	$2.2 \times 10^{-6}$ s
$\mathrm{n} \rightarrow \mathrm{p} + \mathrm{e^-} + \bar{\nu}_e$	weak	878 s

Any decay that produces a neutrino, or changes a quark's flavour, must be weak. Figure 4.1 shows three weak decays at the level of quarks and leptons. Check the conservation laws at each vertex, as in Lesson 3.

Figure 4.1: Feynman diagrams for three weak decays (time runs upward).

Worked example 4: reading a pion decay

Use Figure 4.1(b) to explain how a $π^{+} \pi^+$ decays, and find the energy released ( $m_{\pi^+}c^2 = 139.57\,\mathrm{MeV}$ , $m_\mu c^2 = 105.66\,\mathrm{MeV}$ ).

Solution

The up quark and the anti-down quark annihilate into a $\mathrm{W^+}$ : charge $+\frac{2}{3} + \frac{1}{3} = +1$ ✓, baryon number $\frac{1}{3} - \frac{1}{3} = 0$ ✓. The $\mathrm{W^+}$ becomes $μ^{+} \mu^+$ and $\nu_\mu$ : charge $+ 1 = + 1 + 0$ ✓, $L_\mu$ : $0 = - 1 + 1$ ✓.

Energy released: $139.57 - 105.66 = 33.91\,\mathrm{MeV}$ , shared between the antimuon and the neutrino.

Warning

Common mistakes. (1) Adding all lepton numbers together: each family ( $L_{e}$ , $L_\mu$ , $L_\tau$ ) is conserved SEPARATELY. (2) Giving mesons baryon number 1: a meson has $\frac{1}{3} - \frac{1}{3} = 0$ . (3) Forgetting the sign of strangeness: the s quark has $S = - 1$ , so the $\mathrm{K}^+$ ( $\mathrm{u}\bar{\mathrm{s}}$ ) has $S = + 1$ . (4) Calling a reaction impossible because strangeness changes: it is possible, but only through the weak interaction. (5) Forgetting energy: a decay needs a parent heavier than its products.

Tip

Use a Q–B–L–S table: one column per particle, one row per quantity, then add the rows on each side. If Q, B or L fails: impossible. If only S fails: weak. If everything balances: allowed (if the energy is there), and the fastest interaction available will take it.

Lepton Number Conservation (Pretty Much Physics). Watch how electron and muon lepton numbers are counted separately to decide whether a decay is allowed.

✦

Practice

Write down Q, B, $L_{e}$ and S for: $\mathrm{e^+}$ , $\bar{\mathrm{p}}$ , $π^{-} \pi^-$ , $\mathrm{K}^-$ , $\bar{\nu}_e$ .
Check whether each is allowed: (a) $\mathrm{p} + \bar{\mathrm{p}} \rightarrow \pi^+ + \pi^- + \pi^0$ ; (b) $\mathrm{n} \rightarrow \mathrm{p} + \mathrm{e^-}$ ; (c) $\pi^+ \rightarrow \mu^+ + \nu_\mu$ ; (d) $\mathrm{p} + \mathrm{p} \rightarrow \mathrm{p} + \mathrm{n} + \pi^+$ .
The reaction $\mathrm{p} + \mathrm{p} \rightarrow \mathrm{p} + \Lambda^0 + \mathrm{K}^+$ is observed. Show that strangeness is conserved and say which interaction is responsible.
Explain why $\Lambda^0 \rightarrow \mathrm{n} + \pi^0$ must be a weak decay.
Find the energy released in $\Lambda^0 \rightarrow \mathrm{n} + \pi^0$ ( $m_\Lambda c^2 = 1115.68$ , $m_{n} c^{2} = 939.57$ , $m_{\pi^0}c^2 = 134.98$ , all in MeV).
Explain why the $\pi^0$ decays about $10^{8}$ times faster than the $π^{+} \pi^+$ .
Describe the Feynman diagram for $\mu^- \rightarrow \nu_\mu + \mathrm{e^-} + \bar{\nu}_e$ and check lepton numbers at each vertex.
Keza says: " $\mathrm{p} \rightarrow \mathrm{e^+} + \gamma$ conserves charge and energy, so the proton must eventually decay this way." Evaluate her claim.

Show answersHide answers

$\mathrm{e^+}$ : $+ 1, 0, - 1, 0$ . $\bar{\mathrm{p}}$ : $- 1, - 1, 0, 0$ . $π^{-} \pi^-$ : $- 1, 0, 0, 0$ . $\mathrm{K}^-$ ( $\mathrm{s}\bar{\mathrm{u}}$ ): $- 1, 0, 0, - 1$ . $\bar{\nu}_e$ : $0, 0, - 1, 0$ .
(a) Q: $0 = 0$ ; B: $0 = 0$ : allowed. (b) $L_{e}$ : $0 \rightarrow 1$ : forbidden (the antineutrino is missing). (c) Q ✓, $L_\mu$ : $0 = - 1 + 1$ ✓: allowed. (d) Q: $2 = 1 + 0 + 1$ ✓; B: $2 = 1 + 1 + 0$ ✓: allowed if the protons have enough kinetic energy.
S: $0 + 0 = 0 + (- 1) + (+ 1)$ ✓; Q: $2 = 1 + 0 + 1$ ✓; B: $2 = 1 + 1 + 0$ ✓. Strong (associated production).
S changes from −1 to 0, and only the weak interaction can change strangeness.
$1115.68 - 939.57 - 134.98 = 41.13\,\mathrm{MeV}$ .
The $\pi^0$ decays electromagnetically ( $\pi^0 \rightarrow 2\gamma$ ); the $π^{+} \pi^+$ can only decay weakly, because its products must include a neutrino. The electromagnetic interaction is much stronger, so the decay is much faster.
The $μ^{-} \mu^-$ line becomes a $\nu_\mu$ line, emitting a $\mathrm{W^-}$ : $L_\mu$ : $1 = 1$ . The $\mathrm{W^-}$ becomes $\mathrm{e^-}$ and $\bar{\nu}_e$ : $L_{e}$ : $0 = 1 - 1$ .
Wrong. It conserves charge and energy, but not baryon number ( $1 \rightarrow 0$ ) or lepton number ( $0 \rightarrow -1$ ). These laws forbid it, and no proton decay of any kind has ever been seen: partial lifetimes are above $10^{33}$ to $10^{34}$ years.

Key points

Always conserved: energy–momentum, charge Q, baryon number B, and each lepton number $L_{e}$ , $L_\mu$ , $L_\tau$ .
Strangeness S is conserved in strong and electromagnetic interactions; the weak interaction can change it by 1.
Lifetimes: strong about $10^{-23}$ s, electromagnetic about $10^{-16}$ s, weak $10^{-10}$ s and longer.
Decays producing neutrinos or changing quark flavour are weak, via W bosons.

Sources and credits

11.3: Particle Conservation Laws, from University Physics Volume 3 by Samuel J. Ling, Jeff Sanny and William Moebs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Adapted by Ganzaa.
Particle data, lifetimes and decay limits: Review of Particle Physics, Summary Tables, F. Takahashi et al. (Particle Data Group), Int. J. Mod. Phys. A 41, 2630011 (2026).
Video: Lepton Number Conservation, Pretty Much Physics, embedded from YouTube.

The Feynman diagrams, the worked examples and the practice exercises are written by Ganzaa.

1Why doesn't the proton decay?

2Quantities that are always conserved

3A quantity conserved only in strong and electromagnetic interactions

4Checking a reaction

5Decays and their interactions

Why doesn't the proton decay?

Quantities that are always conserved

A quantity conserved only in strong and electromagnetic interactions

Checking a reaction

Decays and their interactions