3.2.1.7 - Applications of Conservation Laws

What must be conserved

Every particle interaction must conserve energy, momentum, charge, baryon number, and the relevant lepton numbers. Strangeness is the special case: it is conserved in strong interactions, but it can change in weak interactions.

For this topic, you should check:

• energy
• momentum
• charge, $Q$
• baryon number, $B$
• electron lepton number, $L_e$
• muon lepton number, $L_\mu$
• strangeness, $S$

Energy includes rest energy as well as kinetic energy. Momentum is always conserved too, so if the total momentum before an interaction is zero, the vector sum of the momenta after the interaction must also be zero.

Strangeness behaves differently from the other particle-number rules. In a strong interaction the total strangeness must stay the same. In a weak interaction it can change by $0$, $+1$, or $-1$.

After you know which quantities matter, the rest of the method is bookkeeping: write the reaction, total each quantity on the left-hand side, total it again on the right-hand side, and compare the two sets of totals.

Assigning quantum numbers

To use the conservation laws, you need the quantum numbers of the particles involved. For familiar particles you should know them directly. For unfamiliar hadrons, you may be given the quark composition and asked to work the values out from that.

The key assignments for this lesson are:

• charges in this table are in units of $e$; $u$ has charge $+\frac{2}{3}$, $d$ and $s$ each have charge $-\frac{1}{3}$
• antiquarks have opposite charge and opposite strangeness to the matching quark
• the strange quark has $S = -1$, while the anti-strange quark has $S = +1$
• leptons have baryon number $0$, and hadrons have lepton number $0$
• baryons are three-quark states, mesons are quark-antiquark states

Here is a compact reference table for some common particles:

The electron family has $L_e=+1$ for $e^-$ and $\nu_e$, and $L_e=-1$ for $e^+$ and $\bar\nu_e$. The muon family has $L_\mu=+1$ for $\mu^-$ and $\nu_\mu$, and $L_\mu=-1$ for $\mu^+$ and $\bar\nu_\mu$. Each has zero lepton number of the other family; photons and hadrons have both zero.

The exam will supply the necessary particle data when it goes beyond the small set you are expected to know. Your job is to use that information accurately, not to memorise every hadron ever discovered.

Systematic checking

The cleanest exam method is to use a before-and-after table. This prevents missed signs and makes it obvious which quantity fails to balance.

For $\beta^-$ decay,

the totals are:

Every listed quantity is conserved; with the neutron’s larger rest energy and conserved total momentum, the decay is allowed. Because a quark changes flavour, the interaction is weak.

You should also remember that momentum must be conserved. For example, in a head-on collision of two identical particles with equal and opposite momenta, the initial total momentum is zero, so the final vector sum of momenta must also be zero. In a symmetric collision the laboratory is the centre-of-momentum frame. At threshold, the products can have no kinetic energy in that frame, so the minimum total initial kinetic energy is the increase in total rest energy. If a projectile instead hits a stationary target, initial laboratory momentum is non-zero: some final kinetic energy is necessary to carry it. Do not apply the same simple threshold subtraction to that laboratory projectile energy. For example, $p+\bar p\rightarrow K^++K^-$ at rest has zero initial momentum. With rest energies $938.3\,\mathrm{MeV}$ and $493.7\,\mathrm{MeV}$, the total product kinetic energy is $2(938.3)-2(493.7)=889.2\,\mathrm{MeV}$. The two kaons have equal and opposite momenta. Their equal masses then imply equal energies; momentum conservation alone would not give equal kinetic energies for unequal masses.

Quark changes in beta decay

Beta decay is a weak interaction because one quark changes flavour.

In $\beta^-$ decay, a neutron becomes a proton:

One down quark changes into an up quark.

Charge is still conserved because

and baryon number is conserved because the quark still has baryon number $+\frac{1}{3}$ before and after the change.

In $\beta^+$ decay inside a proton-rich nucleus, one proton changes into a neutron:

This time one up quark changes into a down quark.

Again, charge is conserved because

and the baryon number of the changing quark remains $+\frac{1}{3}$.

It is worth being precise here: this is a change inside a nucleus, not a description of a free proton spontaneously decaying in empty space. At A-level, the important idea is the quark flavour change and the conservation laws, not an isolated free-proton process.

The two-panel quark sketch below shows the key feature to notice in both decays: only one quark changes flavour, while the other two quarks stay the same and the emitted lepton pair carries away the extra charge and lepton number.

[DIAGRAM: asset_name: 2.1.7 - Applications of Conservation Laws - Diagram 1; asset_slug: 2.1.7 - Applications of Conservation Laws - Diagram 1; recommended_method: imagegen; caption: Beta-minus changes d to u, beta-plus changes u to d. The other two quarks keep their flavours.; description: Two stacked quark diagrams. Top: neutron udd becoming proton uud, with one d quark changing to u and an electron plus electron antineutrino emitted. Bottom: proton uud becoming neutron udd, with one u quark changing to d and a positron plus electron neutrino emitted.]

Allowed reactions and quark descriptions

A reaction is forbidden if any always-conserved quantity fails to balance. For example,

cannot occur because the baryon number changes from $0$ on the left to $+1$ on the right.

By contrast,

passes the charge, baryon, lepton and strangeness checks. These necessary checks alone do not prove that a process occurs: energy, momentum and the interaction mechanism must also allow it.

When you are given a strange particle, strangeness is often the deciding test. Consider:

Charge is conserved: $0 + 1 = 0 + 1$.

Baryon number is conserved: $0 + 1 = 1 + 0$.

Strangeness is not conserved: before the interaction the $K^0$ contributes $+1$, but after the interaction both products have $S = 0$. That means the reaction can only proceed via the weak interaction.

At the quark level,

At A-level, the useful quark description is that the products are a neutron ($udd$) and a pion ($u\bar{d}$), and that the change in strangeness tells you a weak flavour change must occur somewhere in the process. You do not need a detailed intermediate-boson mechanism unless the question explicitly asks for one.