---
title: Decay, Scattering, and Mandelstam Variables
module: Units and Kinematics
moduleNumber: 2
lessonNumber: 3
order: 203
summary: >
  Two-body decay in the rest frame fixes the daughter momenta from the three masses
  alone; production thresholds follow from the minimum invariant mass. This lesson
  works both, then introduces the Mandelstam invariants $s$, $t$, $u$ for $2\to2$
  scattering, proves the identity $s+t+u=\sum m_i^2$, and maps the physical regions
  and the crossing that relates channels.
topics: [Units and Kinematics]
draft: false
sources:
  - book: Griffiths
    ref: "Ch. 3 — Relativistic Kinematics, §3.3–3.5"
  - book: Thomson
    ref: "§2.4 Mandelstam variables"
  - book: Halzen & Martin
    ref: "§4 — kinematics of two-body reactions"
---

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, $\hbar = c = 1$ 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

$$
E_1 + E_2 = M, \qquad \vec p_1 + \vec p_2 = 0,
$$

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,

$$
E_1 = \frac{M^2 + m_1^2 - m_2^2}{2M}, \qquad
E_2 = \frac{M^2 + m_2^2 - m_1^2}{2M},
$$

and the momentum magnitude is

$$
p^\ast = \frac{1}{2M}\sqrt{\lambda(M^2, m_1^2, m_2^2)},
$$

where the Källén function

$$
\lambda(a,b,c) = a^2 + b^2 + c^2 - 2ab - 2bc - 2ca
$$

is symmetric in its arguments. The decay is possible only when $p^\ast$ is real
and non-negative, which requires $M \geq m_1 + m_2$: 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.

$$
% caption: Two-body decay in the parent rest frame. The parent sits at rest at the
% origin; the two daughters leave back-to-back with equal and opposite momenta of
% magnitude p-star, set entirely by the three masses through the Kallen function.
\begin{tikzpicture}[font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  \fill[black!70] (0,0) circle (2.6pt);
  \node[black, above, font=\scriptsize] at (0,0.12) {parent at rest};
  \draw[-{Stealth[length=2.6mm]}, acc, very thick] (0,0) -- (3.0,0.9);
  \draw[-{Stealth[length=2.6mm]}, acc, very thick] (0,0) -- (-3.0,-0.9);
  \node[right, font=\scriptsize] at (3.0,0.9) {daughter 1};
  \node[left, font=\scriptsize] at (-3.0,-0.9) {daughter 2};
  \node[black, right, font=\scriptsize] at (1.5,0.15) {momentum};
  \node[black, left, font=\scriptsize] at (-1.5,-0.15) {momentum};
\end{tikzpicture}
$$

> **Worked example.** The charged pion decays $\pi^+ \to \mu^+ \nu_\mu$ with $M =
> m_\pi = 139.6$ MeV, $m_1 = m_\mu = 105.7$ MeV, and $m_2 = m_\nu \approx 0$. The
> muon momentum is
>
> $$
> p^\ast = \frac{M^2 - m_\mu^2}{2M} = \frac{(139.6)^2 - (105.7)^2}{2(139.6)}\ \text{MeV}
> = 29.8\ \text{MeV}.
> $$
>
> The monoenergetic muon momentum is a signature of a two-body decay; a three-body
> decay would spread the muon over a continuous spectrum.

## 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:

$$
\sqrt{s_{\min}} = \sum_{\text{final}} m_i.
$$

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

$$
p + p \longrightarrow p + p + p + \bar p.
$$

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

$$
s = (p_{\text{beam}} + p_{\text{target}})^2 = 2 m_p^2 + 2 E_{\text{beam}} m_p,
$$

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

$$
E_{\text{beam}} = \frac{16 m_p^2 - 2 m_p^2}{2 m_p} = 7 m_p,
\qquad
T_{\text{beam}} = E_{\text{beam}} - m_p = 6 m_p \approx 5.63\ \text{GeV}.
$$

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.

$$
% caption: The antiproton production threshold. Below 5.63 GeV of beam kinetic
% energy no antiproton can be made; the available invariant energy root-s rises as
% the square root of the beam energy and first reaches the four-nucleon final-state
% mass, four proton masses, at the threshold marked.
\begin{tikzpicture}[font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, -{Stealth[length=2mm]}] (0,0) -- (6.6,0)
    node[right, black, font=\scriptsize] {beam energy};
  \draw[black, -{Stealth[length=2mm]}] (0,0) -- (0,3.6)
    node[above, black, font=\scriptsize] {available energy};
  % sqrt curve
  \draw[very thick]
    (0,0.3) .. controls (1.4,1.55) and (3.0,2.25) .. (6.3,3.1);
  % threshold line at 4 m_p
  \draw[black, dashed] (0,2.5) -- (6.3,2.5);
  \node[black, left, font=\scriptsize] at (0,2.6) {$4 m_p$};
  \draw[black, dashed] (3.55,0) -- (3.55,2.5);
  \node[black, below, font=\scriptsize] at (3.55,-0.12) {threshold};
  \fill[acc] (3.55,2.5) circle (2.0pt);
  \node[acc, right, font=\scriptsize] at (3.6,2.15) {5.63 GeV};
\end{tikzpicture}
$$

## 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

$$
s = (p_a + p_b)^2, \qquad
t = (p_a - p_c)^2, \qquad
u = (p_a - p_d)^2.
$$

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:

> **Theorem (Mandelstam identity).** For any $2 \to 2$ reaction,
>
> $$
> s + t + u = m_a^2 + m_b^2 + m_c^2 + m_d^2.
> $$

> **Proof.** Expand each invariant using $p_i^2 = m_i^2$ and conservation. Summing,
>
> $$
> s + t + u = (p_a+p_b)^2 + (p_a-p_c)^2 + (p_a-p_d)^2.
> $$
>
> The three squares give $3 p_a^2 + p_b^2 + p_c^2 + p_d^2 + 2 p_a\cdot(p_b - p_c -
> p_d)$. Momentum conservation makes $p_b - p_c - p_d = -p_a$, so the cross term is
> $-2 p_a^2$, leaving $p_a^2 + p_b^2 + p_c^2 + p_d^2 = \sum_i m_i^2$.

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.

$$
% caption: The three Mandelstam invariants on a two-to-two reaction. The two
% incoming lines a and b meet the two outgoing lines c and d; s is the squared
% energy of the a-plus-b combination, t the squared momentum transfer from a to c,
% and u the transfer from a to d.
\begin{tikzpicture}[font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  \coordinate (L) at (0,0);
  \coordinate (R) at (3.2,0);
  \draw[-{Stealth[length=2.4mm]}, very thick] (-1.8,1.2) -- (L);
  \draw[-{Stealth[length=2.4mm]}, very thick] (-1.8,-1.2) -- (L);
  \draw[acc, very thick] (L) -- (R);
  \draw[-{Stealth[length=2.4mm]}, very thick] (R) -- (5.0,1.2);
  \draw[-{Stealth[length=2.4mm]}, very thick] (R) -- (5.0,-1.2);
  \node[left, font=\scriptsize] at (-1.8,1.2) {$a$};
  \node[left, font=\scriptsize] at (-1.8,-1.2) {$b$};
  \node[right, font=\scriptsize] at (5.0,1.2) {$c$};
  \node[right, font=\scriptsize] at (5.0,-1.2) {$d$};
  \node[black, above, font=\scriptsize] at (1.6,0.1) {$s$ channel};
  \node[black, font=\scriptsize] at (0.4,1.35) {$t$: $a$ to $c$};
  \node[black, font=\scriptsize] at (0.4,-1.4) {$u$: $a$ to $d$};
\end{tikzpicture}
$$

## 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 $s \geq (m_a + m_b)^2$ and $s \geq (m_c + m_d)^2$, 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.

$$
% caption: The Mandelstam plane. With s plus t plus u fixed, the invariants are the
% distances to the three sides of a triangle; the three physical regions, one per
% channel, occupy the outer sectors where the energies and angles are real, while
% the central region between them is unphysical for every channel.
\begin{tikzpicture}[font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  \coordinate (A) at (0,0);
  \coordinate (B) at (5.4,0);
  \coordinate (C) at (2.7,2.6);
  \draw[black, thick] (A) -- (B) -- (C) -- cycle;
  \node[black, below, font=\scriptsize] at (2.7,-0.15) {$u = 0$};
  \node[black, left, font=\scriptsize] at (1.0,1.5) {$s = 0$};
  \node[black, right, font=\scriptsize] at (4.4,1.5) {$t = 0$};
  % s-channel region shaded near top
  \draw[thick] (2.7,2.6) -- (1.95,1.35) -- (3.45,1.35) -- cycle;
  \node[font=\scriptsize] at (2.7,1.68) {$s$};
  % t-channel region bottom right
  \draw[thick] (5.4,0) -- (4.2,0) -- (4.65,0.9) -- cycle;
  \node[font=\scriptsize] at (4.7,0.35) {$t$};
  % u-channel region bottom left
  \draw[thick] (0,0) -- (1.2,0) -- (0.75,0.9) -- cycle;
  \node[font=\scriptsize] at (0.7,0.35) {$u$};
\end{tikzpicture}
$$

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.[^mand]

[^mand]: Griffiths, _Introduction to Elementary Particles_, Ch. 3, §3.3–3.5, works two-body decay, thresholds, and the Mandelstam variables; Thomson, _Modern Particle Physics_, §2.4, and Halzen & Martin, _Quarks and Leptons_, §4, give the invariant formulation and crossing. The kinematic identities and physical-region boundaries follow the Particle Data Group kinematics review, [pdg.lbl.gov](https://pdg.lbl.gov).