---
title: Four-Vectors and Invariant Mass
module: Units and Kinematics
moduleNumber: 2
lessonNumber: 2
order: 202
summary: >
  The energy and momentum of a particle form a four-vector whose square is the
  frame-independent quantity $p^2 = m^2$. This lesson develops the metric and
  four-vector products, the invariant mass of a multiparticle system, the
  center-of-momentum and laboratory frames, and the description of collinear
  boosts by rapidity, whose additivity replaces the awkward velocity-addition law.
topics: [Units and Kinematics]
draft: false
sources:
  - book: Griffiths
    ref: "Ch. 3 — Relativistic Kinematics, §3.1–3.2"
  - book: Thomson
    ref: "§2.3 Four-vector notation"
  - book: Halzen & Martin
    ref: "§3 — the relativistic kinematics of scattering"
---

Relativistic kinematics is the bookkeeping of energy and momentum in reactions
where speeds approach $c$ and particles are created and destroyed. Its central
object is the four-vector, and its central quantity is an invariant that every
observer agrees on: the mass. This lesson sets up the four-vector formalism and
the invariants built from it, working in the natural units of the previous lesson,
where energy, momentum, and mass share the unit GeV.

## The energy-momentum four-vector

A particle's energy $E$ and three-momentum $\vec p = (p_x, p_y, p_z)$ combine into
a single four-component object,

$$
p^\mu = (E,\, \vec p) = (E,\, p_x,\, p_y,\, p_z), \qquad \mu = 0,1,2,3,
$$

with the energy as the time component $p^0 = E$ and the momentum as the space
components. Under a Lorentz transformation between two inertial frames, the four
components mix exactly as the time and space coordinates $x^\mu = (t, \vec x)$ do.
A boost of speed $\beta$ along the $z$-axis, with $\gamma = (1-\beta^2)^{-1/2}$,
acts as

$$
E' = \gamma(E - \beta p_z), \qquad
p_z' = \gamma(p_z - \beta E), \qquad
p_x' = p_x, \qquad p_y' = p_y.
$$

The transverse components are untouched; energy and longitudinal momentum rotate
into each other. Any four quantities that transform by this rule form a
four-vector, and the machinery below applies to all of them equally.

## The metric and invariant products

The scalar product of four-vectors is not the Euclidean sum of component products.
It is defined with the Minkowski metric

$$
g_{\mu\nu} = \operatorname{diag}(1,\,-1,\,-1,\,-1),
$$

so that for two four-vectors $a^\mu = (a^0, \vec a)$ and $b^\mu = (b^0, \vec b)$,

$$
a \cdot b \equiv g_{\mu\nu}\, a^\mu b^\nu = a^0 b^0 - \vec a \cdot \vec b.
$$

The single minus sign carrying the space components is the entire content of
special relativity for kinematics. This product is a Lorentz invariant: every
inertial observer computes the same number for $a \cdot b$, because the metric is
built to cancel the frame-dependence introduced by the boost. The square of a
single four-vector,

$$
p^2 \equiv p \cdot p = E^2 - |\vec p|^2,
$$

is therefore the same in every frame.

$$
% caption: The energy-momentum relation as a right triangle. The energy E is the
% hypotenuse, the momentum magnitude and the mass are the legs, and the Pythagorean
% relation E squared equals p squared plus m squared is the frame-independent
% content. Boosting a particle slides it along the hypotenuse without changing the
% mass leg.
\begin{tikzpicture}[font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  \coordinate (O) at (0,0);
  \coordinate (P) at (4.2,0);
  \coordinate (E) at (4.2,2.6);
  \draw[acc, very thick] (O) -- (P) -- (E) -- cycle;
  \draw[black] (P) ++(-0.32,0) -- ++(0,0.32) -- ++(0.32,0);
  \node[acc, below, font=\small] at (2.1,-0.05) {momentum $p$};
  \node[black!70, right, font=\small] at (4.25,1.3) {mass $m$};
  \node[acc, above left, font=\small] at (2.1,1.3) {energy $E$};
  \node[black, anchor=west, align=left, font=\scriptsize] at (0.15,2.3)
    {$E^2 = p^2 + m^2$};
\end{tikzpicture}
$$

## Mass as the invariant square

The energy and momentum of a free particle satisfy the relativistic relation

$$
E^2 = |\vec p|^2 + m^2,
$$

which restores in SI to $E^2 = (pc)^2 + (mc^2)^2$. Substituting into the invariant
square gives the fundamental identity

$$
p^2 = E^2 - |\vec p|^2 = m^2.
$$

The square of the energy-momentum four-vector equals the mass squared, in every
frame. This is the mass-shell condition. Its power is that mass, an intrinsic
property, is computed from the frame-dependent $E$ and $\vec p$ by a combination
that removes all frame-dependence. Two limits fix intuition:

- **massless particle** ($m = 0$): $E = |\vec p|$, so a photon's energy equals its
  momentum magnitude. Its four-vector squares to zero and is called null.
- **particle at rest** ($\vec p = 0$): $E = m$, the rest energy, which restores to
  the familiar $E = mc^2$.

Between these, a massive moving particle has $E = \gamma m$ and $|\vec p| = \gamma
m \beta$, from which $E^2 - |\vec p|^2 = \gamma^2 m^2 (1 - \beta^2) = m^2$
directly.

## The invariant mass of a system

The four-vector of a system of $n$ particles is the sum of the individual
four-vectors, because energy and momentum are additive:

$$
P^\mu = \sum_{i=1}^{n} p_i^\mu = \Big(\textstyle\sum_i E_i,\ \sum_i \vec p_i\Big).
$$

The square of this total four-vector defines the **invariant mass** $W$ of the
system,

$$
W^2 = P^2 = \Big(\sum_i E_i\Big)^2 - \Big|\sum_i \vec p_i\Big|^2,
$$

a single number that every frame agrees on. For a single particle it reduces to
the mass. For several it is generally larger than the sum of the individual
masses, the excess being the kinetic energy available in the center-of-momentum
frame.

The invariant mass is the primary reconstruction quantity in experiment. When an
unstable particle of mass $M$ decays into daughters that a detector measures, the
daughters' four-vectors reconstruct the parent through $W = M$. Combining every
candidate pair and histogramming $W$ produces a peak at the parent mass over a
smooth combinatorial background. The two-photon decay of the neutral pion is the
prototype: for $\pi^0 \to \gamma\gamma$,

$$
W^2 = (p_{\gamma 1} + p_{\gamma 2})^2 = 2\, p_{\gamma 1}\cdot p_{\gamma 2}
     = 2 E_1 E_2 (1 - \cos\theta_{12}),
$$

since each photon is massless, and the peak in $W$ sits at $m_{\pi^0} = 135$ MeV.

$$
% caption: An invariant-mass spectrum reconstructed from photon pairs. Each entry
% is the invariant mass of one photon pair; true pions from the same decay pile up
% at 135 MeV as a peak, while random pairings form the smooth combinatorial
% background beneath it.
\begin{tikzpicture}[font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, -{Stealth[length=2mm]}] (0,0) -- (7.0,0)
    node[right, black, font=\scriptsize] {pair mass};
  \draw[black, -{Stealth[length=2mm]}] (0,0) -- (0,3.6)
    node[above, black, font=\scriptsize] {counts};
  % smooth background curve
  \draw[black, dashed, thick]
    (0.2,2.7) .. controls (1.6,1.9) and (3.0,1.35) .. (4.4,1.0)
    .. controls (5.6,0.75) .. (6.7,0.55);
  % peak on top of background near x=3
  \draw[acc, very thick]
    (0.2,2.72) .. controls (1.6,1.92) and (2.3,1.55) .. (2.6,1.5)
    .. controls (2.85,2.9) and (3.15,2.9) .. (3.4,1.42)
    .. controls (3.7,1.28) and (5.4,0.75) .. (6.7,0.56);
  \draw[black] (3.0,0.08) -- (3.0,-0.08);
  \node[black, below, font=\scriptsize] at (3.0,-0.1) {135 MeV};
  \node[acc, font=\scriptsize] at (4.3,2.5) {pion peak};
  \node[black, font=\scriptsize] at (5.4,1.1) {background};
\end{tikzpicture}
$$

## Laboratory and center-of-momentum frames

Two frames organize every reaction. The **laboratory frame** is where the
experiment sits: in a fixed-target setup one particle is at rest and the other
carries the beam momentum; in a collider the two beams meet. The
**center-of-momentum frame**, written CM, is the frame in which the total
three-momentum vanishes,

$$
\sum_i \vec p_i^{\,\ast} = 0,
$$

with starred quantities denoting CM-frame values. In the CM frame the total
four-vector is purely temporal, $P^{\mu\ast} = (W, \vec 0)$, so the invariant mass
equals the total energy there:

$$
W = \sqrt{s} = \sum_i E_i^{\ast}.
$$

The symbol $s = W^2 = P^2$ is the standard name for this invariant, and $\sqrt{s}$
is the total energy available in the CM frame — the energy that can be converted
into new particle masses. For a two-body collision of particles $a$ and $b$,

$$
s = (p_a + p_b)^2 = m_a^2 + m_b^2 + 2\, p_a \cdot p_b.
$$

In a fixed-target experiment with $b$ at rest, $p_a \cdot p_b = E_a\, m_b$, so
$s = m_a^2 + m_b^2 + 2 E_a m_b$ and $\sqrt{s}$ grows only as $\sqrt{E_a}$. In a
symmetric collider the two beams have equal and opposite momenta, $p_a \cdot p_b
= 2 E_a E_b + \ldots$, giving $\sqrt{s} = 2E$ for equal beams of energy $E$. The
collider spends beam energy far more efficiently, a point developed in the lesson
on thresholds and again in the module on accelerators.

$$
% caption: The same two-body reaction in the laboratory and center-of-momentum
% frames. In the lab (top) the target is at rest and the beam carries all the
% momentum; in the CM frame (bottom) the two momenta are equal and opposite and
% the total three-momentum is zero.
\begin{tikzpicture}[font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  % lab frame
  \node[black, anchor=east, font=\scriptsize] at (-0.2,1.4) {lab};
  \draw[-{Stealth[length=2.6mm]}, acc, very thick] (0,1.4) -- (2.4,1.4);
  \node[acc, above, font=\scriptsize] at (1.2,1.45) {beam};
  \fill[black!70] (3.6,1.4) circle (2.4pt);
  \node[black, below, font=\scriptsize] at (3.6,1.3) {target at rest};
  % CM frame
  \node[black, anchor=east, font=\scriptsize] at (-0.2,0) {CM};
  \draw[-{Stealth[length=2.6mm]}, acc, very thick] (1.0,0) -- (2.6,0);
  \draw[-{Stealth[length=2.6mm]}, acc, very thick] (4.6,0) -- (3.0,0);
  \fill[black] (2.8,0) circle (1.6pt);
  \node[acc, above, font=\scriptsize] at (1.8,0.05) {equal};
  \node[acc, above, font=\scriptsize] at (3.8,0.05) {opposite};
  \node[black, below, font=\scriptsize] at (2.8,-0.25) {total momentum zero};
\end{tikzpicture}
$$

## Rapidity

Collinear boosts along a fixed axis have a variable in which they add. Define the
**rapidity** $y$ of a particle by

$$
y = \frac{1}{2}\ln\frac{E + p_z}{E - p_z} = \tanh^{-1}\!\big(\beta_z\big),
$$

the second form holding when $\beta_z = p_z / E$ is the longitudinal velocity. A
boost of rapidity $\eta$ along $z$ maps

$$
y \longrightarrow y + \eta,
$$

so rapidities of collinear motions add by ordinary addition, where velocities
combine by the nonlinear relativistic law

$$
\beta = \frac{\beta_1 + \beta_2}{1 + \beta_1 \beta_2}.
$$

The two statements are the same physics: writing $\beta = \tanh y$ turns the
velocity-addition formula into the addition theorem for $\tanh$, which is exactly
$y = y_1 + y_2$. Rapidity is preferred in collider physics because differences of
rapidity are boost-invariant, so the shape of a rapidity distribution does not
depend on the longitudinal frame. For small velocities $y \approx \beta$, and the
two coincide; near the speed of light $y$ diverges while $\beta$ saturates at one.

$$
% caption: Velocity versus rapidity. The longitudinal velocity beta equals the
% hyperbolic tangent of the rapidity y, rising linearly at small y and saturating
% at one as y grows. Rapidities add under collinear boosts, so equal steps along
% the horizontal axis are equal boosts, while the velocity increments they produce
% shrink toward the light speed.
\begin{tikzpicture}[font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, -{Stealth[length=2mm]}] (0,0) -- (6.4,0)
    node[right, black, font=\scriptsize] {rapidity $y$};
  \draw[black, -{Stealth[length=2mm]}] (0,0) -- (0,3.3)
    node[above, black, font=\scriptsize] {velocity};
  \draw[black, dashed] (0,2.8) -- (6.2,2.8);
  \node[black, right, font=\scriptsize] at (5.2,3.05) {light speed};
  % tanh-like curve scaled to y in [0,3], beta*2.8
  \draw[acc, very thick]
    (0,0) .. controls (0.9,0.95) and (1.6,1.75) .. (2.4,2.25)
    .. controls (3.4,2.62) and (4.6,2.76) .. (6.2,2.8);
  \node[acc, font=\scriptsize] at (3.6,1.85) {$\tanh y$};
\end{tikzpicture}
$$

The four-vector formalism and its invariants — the mass shell $p^2 = m^2$, the
system invariant mass $\sqrt{s}$, and the additive rapidity — are the arithmetic
of every reaction in the course. The next lesson applies them to specific
processes: two-body decay, production thresholds, and the Mandelstam variables of
$2 \to 2$ scattering.[^kin]

[^kin]: Griffiths, _Introduction to Elementary Particles_, Ch. 3, §3.1–3.2, develops the four-vector formalism and the invariant mass; Thomson, _Modern Particle Physics_, §2.3, gives the same in the metric convention used here, and Halzen & Martin, _Quarks and Leptons_, §3, treats the invariants of scattering. The $\pi^0$ mass and other data are from the Particle Data Group, _Review of Particle Physics_, [pdg.lbl.gov](https://pdg.lbl.gov).