---
title: Particle Decays and Two-Body Kinematics
module: Relativistic Dynamics
moduleNumber: 3
lessonNumber: 2
order: 302
summary: >
  Conservation of four-momentum fixes the kinematics of a decay from the masses
  alone. In the center-of-momentum frame a parent breaks into two daughters with
  equal and opposite momenta and energies set by the Kallen triangle function.
  Boosting to the lab opens the decay into a cone, and the invariant mass built
  from the daughters reconstructs the parent as a peak. Worked cases: the
  two-photon decay of the neutral pion and a heavy two-body hadronic decay.
topics: [Relativistic Dynamics]
draft: false
sources:
  - book: PDG
    ref: "Review of Particle Physics — Kinematics review, §47.4 Two-body decays"
  - book: Hartle
    ref: "Gravity, Ch. 5 — Special Relativistic Mechanics; §5.4 Particle decays"
---

A decay is the simplest relativistic reaction: one particle in, two or more out,
with no external forces. The [conservation of total
four-momentum](/relativity/relativistic-dynamics/four-momentum-force-and-accelerated-motion)
and the invariant $p\cdot p = -m^2c^2$ are enough to fix every energy and
momentum from the particle masses, without any dynamical detail of the
interaction. This lesson works out the two-body case completely — daughter
energies, the shared momentum magnitude, the lab-frame opening angle, and the
invariant-mass reconstruction that turns raw detector hits back into a parent
mass.

The signature is $(-,+,+,+)$, so for a system of particles the total
four-momentum $P^\mu = \sum_i p_i^\mu$ has $P\cdot P = -M^2c^2$, where $M$ is the
**invariant mass** of the system.

## The invariant mass and the center-of-momentum frame

For any collection of particles the total four-momentum is a four-vector, and its
Minkowski length defines a mass:

$$
(Mc^2)^2 = -\,(P\cdot P)\,c^2
        = \Bigl(\sum_i E_i\Bigr)^{2} - \Bigl(\sum_i \vec p_i\Bigr)^{2}c^2.
$$

Unlike the individual energies and momenta, $M$ is the same in every inertial
frame. The **center-of-momentum (CM) frame** is the frame in which the total
three-momentum vanishes, $\sum_i \vec p_i = \vec 0$; there the total energy is
purely the invariant mass energy,

$$
E_{\text{CM}} = \sum_i E_i^\ast = Mc^2 .
$$

For a single particle $M$ is just its rest mass and the CM frame is its rest
frame. For a system, $M$ exceeds the sum of the constituent masses by the
kinetic and binding energy carried in the CM frame. A decay conserves $P^\mu$, so
the invariant mass of the final state equals the parent's rest mass.

> **Definition (Invariant mass).** The invariant mass of a set of particles is
> $Mc^2 = \sqrt{(\sum_i E_i)^2 - (\sum_i \vec p_i c)^2}$, the Minkowski length of
> the total four-momentum. It is Lorentz-invariant and, for a decay, equals the
> rest energy of the parent regardless of the frame the daughters are measured in.

## Two-body decay in the CM frame

A parent of mass $M$ at rest decays into daughters of masses $m_1$ and $m_2$.
Because the parent is at rest, this is already the CM frame. Conservation of
energy and momentum reads

$$
E_1^\ast + E_2^\ast = Mc^2,
\qquad
\vec p_1^{\,\ast} + \vec p_2^{\,\ast} = \vec 0 .
$$

The momenta are equal and opposite, so the daughters emerge **back to back** with
a common momentum magnitude $p^\ast = |\vec p_1^{\,\ast}| = |\vec p_2^{\,\ast}|$.
Each daughter is on shell, $E_i^\ast = \sqrt{(p^\ast c)^2 + (m_i c^2)^2}$. Two
equations in the two unknowns $E_1^\ast, E_2^\ast$ (with $p^\ast$ shared) solve
cleanly. Subtract the squared on-shell relations to eliminate $p^\ast$,

$$
(E_1^\ast)^2 - (E_2^\ast)^2 = (m_1^2 - m_2^2)c^4,
$$

and combine with $E_1^\ast + E_2^\ast = Mc^2$ to get the daughter energies

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

The heavier daughter takes the larger share; the split is fixed entirely by the
three masses. The shared momentum follows from either on-shell relation and can
be written with the **Källén triangle function** $\lambda(a,b,c) = a^2 + b^2 +
c^2 - 2ab - 2bc - 2ca$:

$$
p^\ast = \frac{c}{2M}\,\sqrt{\lambda\!\left(M^2, m_1^2, m_2^2\right)}
       = \frac{c}{2M}\sqrt{\bigl[M^2-(m_1{+}m_2)^2\bigr]\bigl[M^2-(m_1{-}m_2)^2\bigr]} .
$$

The decay is kinematically allowed only when $p^\ast$ is real, which requires $M
\ge m_1 + m_2$: a particle cannot decay into daughters heavier than itself. When
$M = m_1 + m_2$ exactly, $p^\ast = 0$ and the daughters are produced at rest —
the threshold of the decay.

$$
% caption: In the parent rest frame the two daughters leave back to back with
% equal momentum magnitude p-star and energies fixed by the masses; the heavier
% daughter carries the larger energy.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  % parent
  \draw[very thick] (0,0) circle (0.42);
  \node at (0,0) {parent};
  \node[black!70, above] at (0,0.5) {at rest, mass $M$};
  % daughter 1 (heavier) to right
  \draw[->, acc, very thick] (0.5,0) -- (3.1,0);
  \draw[black!75, fill=black!8] (3.4,0) circle (0.34);
  \node[black!75] at (3.4,0) {$m_1$};
  \node[above] at (2.0,0.05) {momentum $p^\ast$};
  % daughter 2 (lighter) to left
  \draw[->, acc, very thick] (-0.5,0) -- (-2.4,0);
  \draw[black!75, fill=black!8] (-2.7,0) circle (0.30);
  \node[black!75] at (-2.7,0) {$m_2$};
  \node[above] at (-1.5,0.05) {momentum $p^\ast$};
  \node[black!70, below] at (3.4,-0.5) {heavier: more energy};
  \node[black!70, below] at (-2.7,-0.5) {lighter: less energy};
\end{tikzpicture}
$$

## The neutral pion to two photons

The neutral pion decays to two photons, $\pi^0 \to \gamma\gamma$, more than
$98\%$ of the time.[^pdg] Both daughters are massless, $m_1 = m_2 = 0$, so the
symmetric formulas give each photon half the parent rest energy in the CM frame:

$$
E_\gamma^\ast = \frac{Mc^2}{2} = \frac{m_{\pi^0}c^2}{2}
             = \frac{134.98\ \text{MeV}}{2} = 67.49\ \text{MeV},
$$

with the two photons back to back. Their invariant mass is

$$
(Mc^2)^2 = (E_1 + E_2)^2 - (\vec p_1 + \vec p_2)^2 c^2
        = 2 E_1 E_2\,(1 - \cos\theta),
$$

using $E = |\vec p|c$ for each photon and $\theta$ for the opening angle between
them. In the pion rest frame $\theta = 180^\circ$ and $E_1 = E_2 = m_{\pi^0}c^2/2$
reproduces $Mc^2 = m_{\pi^0}c^2$. In the lab the pion moves with Lorentz factor
$\gamma = E_\pi/(m_{\pi^0}c^2)$, and the two photons are beamed forward into a
cone. The opening angle has a minimum, reached when the decay is symmetric
(each photon perpendicular to the boost in the CM frame):

$$
\sin\!\frac{\theta_{\min}}{2} = \frac{1}{\gamma} = \frac{m_{\pi^0}c^2}{E_\pi}.
$$

A fast pion produces two nearly collinear photons; measuring their opening angle
and energies recovers the pion energy and confirms the parent mass. This is the
basic signature by which the $\pi^0$ is detected in a calorimeter.

$$
% caption: The neutral-pion decay in the rest frame (photons back to back, each
% carrying half the rest energy) and boosted to the lab, where the same two
% photons are beamed forward into a cone of opening angle at least 2 arcsin of
% one over gamma.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  % rest frame (left)
  \fill (0,0) circle (1.6pt);
  \node[black!70, above] at (0,0.15) {rest frame};
  \draw[->, very thick] (0,0) -- (1.3,0) node[right] {photon};
  \draw[->, very thick] (0,0) -- (-1.3,0) node[left] {photon};
  \node[black!70, below] at (0,-0.3) {back to back};
  % divider
  \draw[black] (2.3,-1.6) -- (2.3,1.6);
  % lab frame (right)
  \fill (3.4,0) circle (1.6pt);
  \node[black!70, above] at (3.4,0.9) {lab frame};
  \draw[->, black, thick] (2.9,0) -- (3.4,0);
  \node[black, below] at (3.15,-0.05) {boost};
  \draw[->, acc, very thick] (3.4,0) -- (5.4,0.55) node[right] {photon};
  \draw[->, acc, very thick] (3.4,0) -- (5.4,-0.55) node[right] {photon};
  \draw[black] (4.6,0.30) arc (16:-16:1.25);
  \node[black!70, right] at (4.75,0) {opening angle};
\end{tikzpicture}
$$

## Invariant-mass reconstruction

A detector measures the daughters, not the parent. The parent mass is recovered
by building the invariant mass of the observed daughters, which is frame- and
boost-independent. For two daughters of measured energies $E_1, E_2$, momenta
$\vec p_1, \vec p_2$, and opening angle $\theta$,

$$
(Mc^2)^2 = m_1^2c^4 + m_2^2c^4
        + 2\bigl(E_1 E_2 - |\vec p_1||\vec p_2|\,c^2\cos\theta\bigr).
$$

Every genuine decay of a given parent lands at the same $M$ regardless of how the
parent was moving, so plotting the reconstructed $M$ for many events produces a
sharp peak at the parent mass sitting on a smooth combinatorial background from
uncorrelated pairs. The peak's width is set by detector resolution and, for
short-lived states, by the natural decay width; its position is the mass. This is
how resonances are discovered: the $\rho$, the $J/\psi$, and the Higgs boson all
first appeared as invariant-mass peaks in exactly this construction.

$$
% caption: Reconstructed invariant mass of many daughter pairs: genuine decays
% pile up in a peak at the parent mass M, while random uncorrelated pairs form a
% smooth background beneath it.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (6.0,0) node[right] {reconstructed mass};
  \draw[->, black] (0,0) -- (0,3.7) node[above] {events};
  % background: slowly falling line
  \draw[black, dashed, domain=0.3:5.6, samples=40, variable=\x]
    plot ({\x},{1.5 - 0.13*\x});
  \node[black, right] at (4.2,0.75) {background};
  % peak: background + gaussian at x=3.2
  \draw[acc, very thick, domain=0.3:5.6, samples=120, variable=\x]
    plot ({\x},{1.5 - 0.13*\x + 2.0*exp(-((\x-3.2)^2)/0.10)});
  \draw[black, densely dotted] (3.2,0) -- (3.2,3.15);
  \node[black!70, below] at (3.2,0) {$M$};
  \node[acc, above] at (3.2,3.2) {parent peak};
\end{tikzpicture}
$$

## A heavy two-body hadronic decay

The lambda baryon decays to a proton and a negative pion, $\Lambda^0 \to p +
\pi^-$, about $64\%$ of the time.[^pdg] With $m_\Lambda c^2 = 1115.68\ \text{MeV}$,
$m_p c^2 = 938.27\ \text{MeV}$, $m_\pi c^2 = 139.57\ \text{MeV}$,[^pdg-mass] the
shared momentum is

$$
p^\ast c = \frac{1}{2 m_\Lambda}
  \sqrt{\bigl[m_\Lambda^2 - (m_p{+}m_\pi)^2\bigr]\bigl[m_\Lambda^2 - (m_p{-}m_\pi)^2\bigr]}\ c^4
  = 101\ \text{MeV},
$$

and the daughter energies are

$$
E_p = \frac{m_\Lambda^2 + m_p^2 - m_\pi^2}{2 m_\Lambda}\,c^2 = 943.7\ \text{MeV},
\qquad
E_\pi = \frac{m_\Lambda^2 + m_\pi^2 - m_p^2}{2 m_\Lambda}\,c^2 = 172.0\ \text{MeV}.
$$

They sum to $E_p + E_\pi = 1115.7\ \text{MeV} = m_\Lambda c^2$, as required. The
proton, being much heavier, carries $84\%$ of the total energy but shares the
same $101\ \text{MeV}/c$ momentum as the pion, which is highly relativistic
($\gamma_\pi = E_\pi/m_\pi c^2 = 1.23$) while the proton is barely moving
($\gamma_p = 1.006$). The unequal energy split with equal momentum magnitude is
the generic signature of a two-body decay, and inverting the relations from
measured daughters is the standard test of the parent hypothesis.

$$
% caption: Energy sharing in a two-body decay: both daughters carry the same
% momentum magnitude, but the heavier proton takes far more of the released
% energy than the light pion, with the split fixed by the masses.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (0,3.6) node[above] {energy (MeV)};
  \draw[black] (0,0) -- (5.2,0);
  % scale: 1 MeV -> 0.0031 units, so 943.7 -> 2.93, 172 -> 0.53, rest masses below
  % proton bar
  \fill[acc!14] (0.7,0) rectangle (1.9,2.93);
  \draw[acc, very thick] (0.7,0) rectangle (1.9,2.93);
  \draw[black, dashed] (0.7,2.91) -- (1.9,2.91);
  \fill[black] (0.7,0) rectangle (1.9,2.91);
  \node[black!75, below] at (1.3,-0.05) {proton};
  \node[acc, above] at (1.3,2.95) {944};
  \node[black] at (1.3,1.4) {mostly};
  \node[black] at (1.3,1.05) {rest mass};
  % pion bar
  \fill[black!8] (3.0,0) rectangle (4.2,0.53);
  \draw[very thick] (3.0,0) rectangle (4.2,0.53);
  \node[black!75, below] at (3.6,-0.05) {pion};
  \node[above] at (3.6,0.55) {172};
  \node[black!70, right] at (4.4,1.6) {equal $p^\ast$};
  \node[black!70, right] at (4.4,1.2) {unequal $E$};
\end{tikzpicture}
$$

## Decay in flight and the lab energy spectrum

A parent moving in the lab with Lorentz factor $\gamma$ and speed $\beta c$
decays isotropically in its own rest frame. A daughter emitted at CM angle
$\theta^\ast$ has rest-frame energy $E^\ast$ and momentum $p^\ast$; boosting along
the flight direction gives its lab energy

$$
E_{\text{lab}} = \gamma\bigl(E^\ast + \beta\,p^\ast c\,\cos\theta^\ast\bigr).
$$

Isotropy in the rest frame means $\cos\theta^\ast$ is distributed uniformly on
$[-1, 1]$, so $E_{\text{lab}}$ is uniform between the endpoints

$$
E_\pm = \gamma\bigl(E^\ast \pm \beta\,p^\ast c\bigr).
$$

The lab energy spectrum of the daughter is a **flat box**: constant probability
$\d N/\d E_{\text{lab}} = 1/(E_+ - E_-) = 1/(2\gamma\beta p^\ast c)$ across the
allowed band and zero outside it. The box is centered on $\gamma E^\ast$ and
widens with the parent's speed. A massless daughter ($E^\ast = p^\ast c$) has
$E_- = \gamma E^\ast(1-\beta)$ and $E_+ = \gamma E^\ast(1+\beta)$, so the
two-photon energies of a fast $\pi^0$ spread over a wide flat band whose edges fix
the pion energy without any angular measurement.

$$
% caption: The lab energy of a daughter from an isotropic decay in flight is
% uniformly distributed in a flat box between the minimum and maximum boosted
% energies, centered on gamma times the rest-frame energy.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (6.0,0) node[right] {lab energy};
  \draw[->, black] (0,0) -- (0,3.1) node[above] {events per unit energy};
  % box between E- = 1.3 and E+ = 4.7, height 1.9
  \fill[acc!12] (1.3,0) rectangle (4.7,1.9);
  \draw[acc, very thick] (1.3,0) -- (1.3,1.9) -- (4.7,1.9) -- (4.7,0);
  \draw[black, densely dotted] (3.0,0) -- (3.0,2.3);
  \foreach \x/\l in {1.3/{low edge}, 4.7/{high edge}} \draw[black] (\x,0.05)--(\x,-0.05);
  \node[black!70, below] at (1.3,-0.05) {low edge};
  \node[black!70, below] at (4.7,-0.05) {high edge};
  \node[black!70, above] at (3.0,2.3) {midpoint};
  \node[acc] at (3.0,1.0) {uniform};
\end{tikzpicture}
$$

The parent's finite lifetime turns into a measurable flight distance. If the
rest-frame mean lifetime is $\tau_0$, the parent travels a mean **decay length**

$$
L = \gamma\beta c\,\tau_0
$$

before decaying, since the lab time is dilated to $\gamma\tau_0$ and the parent
covers $\beta c$ per unit lab time. A relativistic beam therefore leaves a
decay vertex displaced from the production point, and measuring the displacement
distribution recovers $\tau_0$; this is how short-lived hadrons with
$c\tau_0$ of tens of microns are timed in a silicon vertex detector.

The same conservation law, applied to two particles colliding rather than one
decaying, gives the threshold energies of particle production and the reason
colliders beat fixed targets. That is the subject of the [next
lesson](/relativity/relativistic-dynamics/collisions-thresholds-and-the-cm-frame),
where the invariant mass of the initial state becomes the available energy $s$.

[^pdg]: Particle Data Group, _Review of Particle Physics_, Kinematics review,
two-body decay formulas; branching fractions from the particle listings.
<https://pdg.lbl.gov/2023/reviews/rpp2023-rev-kinematics.pdf>
[^pdg-mass]: Particle masses from the Particle Data Group summary tables.
<https://pdg.lbl.gov/>