Decay, Scattering, and Mandelstam Variables

The invariant-mass machinery of the previous lesson determines the kinematics of the two reactions that dominate particle physics: a single particle decaying into two, and two particles scattering into two. This lesson works both cases in full, then packages the invariants of a $2\to 2$ reaction into the three Mandelstam variables, which are the natural arguments of every scattering amplitude. Throughout, $\hslash =c=1$ and masses, energies, and momenta share the unit GeV.

Two-body decay in the rest frame

A particle of mass $M$ decaying at rest into two daughters of masses $m_1$ and $m_2$ has completely determined kinematics. Work in the parent rest frame, where the total four-vector is $P=(M,\vec{0})$. Conservation of energy and momentum gives

so the daughters emerge back-to-back with equal and opposite momenta of common magnitude $p^{\ast }\equiv |{\vec{p}}_1|=|{\vec{p}}_2|$. With $E_i=\sqrt{p^{\ast 2}+m_i^2}$, the energy equation fixes everything. Solving for the daughter energies,

and the momentum magnitude is

where the Källén function

is symmetric in its arguments. The decay is possible only when $p^{\ast }$ is real and non-negative, which requires $M\ge m_1+m_2$: the parent must be at least as heavy as its products. The daughter momenta depend on the three masses alone; no other feature of the decay changes them. The angular direction of the back-to-back axis is unconstrained by kinematics and is fixed by the decay dynamics, uniform over solid angle for a spinless parent.

Production thresholds

A reaction that creates new particles proceeds only above a minimum beam energy, the threshold, set by requiring the final-state invariant mass to reach the sum of the product masses. At threshold the products are all at rest in the CM frame, so $\sqrt{s}$ equals the total product mass:

The classic case is antiproton production. Baryon-number conservation forbids making an antiproton alone; it must be produced with a proton, and the lowest-mass final state reachable from a proton beam on a hydrogen target is

The threshold has four nucleons in the final state, so $\sqrt{s_{\min }}=4m_p$. In the fixed-target frame with the target proton at rest,

and setting $s=(4m_p{)}^2=16m_p^2$ gives the threshold beam energy

The kinetic threshold is six proton masses, not the naive two ($p+\bar{p}$) that one might expect from the created pair, because momentum conservation forces the final state to retain kinetic energy in the lab. This inefficiency of fixed-target production restates the CM-versus-lab argument of the previous lesson: only the invariant $\sqrt{s}$ is available for mass, and it grows slowly with beam energy.

Elastic and inelastic scattering

A $2\to 2$ reaction $a+b\to c+d$ is elastic when the outgoing particles are the incoming ones, $c=a$ and $d=b$, so kinetic energy is conserved and only directions change. It is inelastic when the final particles differ, converting kinetic energy into new masses or internal excitation. Both are governed by the same invariants. The most useful description does not track the six final momentum components directly but the Lorentz-invariant products of the four external four-vectors, of which only three are independent combinations.

The Mandelstam variables

For $a+b\to c+d$, define the three Mandelstam invariants

Each is a Lorentz scalar, computable in any frame. Their physical meanings are distinct:

• $s$ is the squared total CM energy, $s=(\text{CM energy}{)}^2$, the same invariant as in the previous lesson; it labels the total energy of the collision.
• $t$ is the squared four-momentum transfer between $a$ and $c$; small $|t|$ means a glancing collision with little deflection, and $t$ controls the exchange of a particle in the $t$-channel.
• $u$ is the squared momentum transfer between $a$ and $d$, the alternative exchange assignment relevant when $c$ and $d$ are identical or interchanged.

By four-momentum conservation $p_a+p_b=p_c+p_d$, the three are related. Their sum telescopes to the total mass:

Only two of the three invariants are independent; the third is fixed by the identity. A scattering amplitude for a $2\to 2$ process is therefore a function of two variables, conventionally $s$ and $t$, and the differential cross section is expressed through them in the next lesson.

Physical regions and crossing

The identity $s+t+u=\sum m_i^2$ means the three invariants live on a plane. A convenient picture uses symmetric coordinates so that $s$, $t$, and $u$ are the perpendicular distances to the three sides of an equilateral triangle; every point of the plane has $s+t+u$ constant automatically. Not every point corresponds to a real reaction. The physical region for a given channel is the part of the plane where the energies and scattering angles are real, bounded by curves where the CM scattering angle reaches $0$ or $\pi$. For the direct reaction $a+b\to c+d$ this region has $s\ge (m_a+m_b{)}^2$ and $s\ge (m_c+m_d{)}^2$, with $t$ and $u$ confined to a band set by the angle.

The same algebraic amplitude describes three related reactions, obtained by moving particles between initial and final states and using an antiparticle in place of a particle:

• the $s$-channel reaction $a+b\to c+d$, physical for large $s$;
• the $t$-channel reaction $a+\bar{c}\to \bar{b}+d$, physical for large $t$;
• the $u$-channel reaction $a+\bar{d}\to c+\bar{b}$, physical for large $u$.

This is crossing symmetry: one amplitude $\mathcal{M}(s,t,u)$, analytically continued, serves all three physical regions, which occupy disjoint parts of the Mandelstam plane. Crossing links, for example, electron-muon scattering to electron-positron annihilation into a muon pair, a relation used repeatedly in the QED module.

The Mandelstam variables and the invariant kinematics of decay and scattering are the inputs to the dynamical quantities of the next lesson: the cross section that measures how often a scattering happens, and the decay rate that measures how fast an unstable particle disintegrates.¹