---
title: Relativistic Collisions and Threshold Energies
module: Relativistic Dynamics
moduleNumber: 3
lessonNumber: 3
order: 303
summary: >
  Two-body collisions run on the same conserved four-momentum as decays. The
  invariant s sets the total energy available in the center-of-momentum frame and
  therefore the threshold for producing new particles. Fixed-target energy grows
  only as the square root of beam energy while a collider grows linearly, which is
  why colliders reach high energy. Compton scattering follows as a worked
  photon-electron collision giving the wavelength shift.
topics: [Relativistic Dynamics]
draft: false
sources:
  - book: PDG
    ref: "Review of Particle Physics — Kinematics review, §47.4.3 Inclusive reactions; invariant s"
  - book: Taylor & Wheeler
    ref: "Spacetime Physics, 2nd ed., Ch. 7 — Momenergy; §7.6 collisions"
---

A collision conserves the total four-momentum of the incoming particles just as a
[decay](/relativity/relativistic-dynamics/particle-decays-and-two-body-kinematics)
does, but now two particles come in and (in general) different particles go out.
The invariant mass of the initial state — its total four-momentum length — sets a
hard ceiling on what the collision can produce, because it equals the total
energy available in the center-of-momentum frame. This single invariant, written
$s$, decides the threshold for creating a heavy particle, and its very different
scaling for fixed-target versus colliding beams is the reason particle physics is
done with colliders. The lesson closes with Compton scattering, the cleanest
worked photon-electron collision.

Signature $(-,+,+,+)$; for the two-particle initial state $P^\mu = p_1^\mu +
p_2^\mu$ and the invariant $s$ is defined below.

## The invariant s as available energy

For a reaction with incoming four-momenta $p_1$ and $p_2$, the Mandelstam
invariant $s$ is the squared invariant mass of the pair, written in energy-squared
units as

$$
s = -\,(p_1 + p_2)\cdot(p_1 + p_2)\,c^2
  = \bigl(E_1 + E_2\bigr)^2 - \bigl|\vec p_1 + \vec p_2\bigr|^2 c^2 .
$$

Its square root is the total energy in the CM frame, $\sqrt{s} = E_{\text{CM}}$,
because there $\vec p_1 + \vec p_2 = \vec 0$ and $\sqrt{s} = E_1^\ast + E_2^\ast$.
Since $s$ is built from the total four-momentum, it is conserved and
frame-invariant: whatever the collision produces, the final state has the same
$s$, so $\sqrt{s}$ is the total energy budget for making rest mass and kinetic
energy in the CM frame.

> **Definition (Total available energy).** The invariant $\sqrt{s} = E_{\text{CM}}$
> is the total energy in the center-of-momentum frame. A final state of particles
> with rest masses $M_f$ is kinematically accessible only if $\sqrt{s} \ge \sum_f
> M_f c^2$; the excess $\sqrt{s} - \sum_f M_f c^2$ is the kinetic energy shared
> among the products in the CM frame.

An **elastic** collision keeps the same particles ($m_i$ unchanged, kinetic
energy conserved in the CM frame); an **inelastic** collision converts kinetic
energy into rest mass, producing heavier or additional particles. Relativity
removes the classical distinction between "kinetic" and "rest" energy: the CM
kinetic energy above threshold becomes the raw material for new mass.

## Fixed target versus collider

Evaluate $s$ two ways. In a **collider**, two beams of equal energy $E$ meet head
on with opposite momenta, so $\vec p_1 + \vec p_2 = \vec 0$ and

$$
\sqrt{s}_{\text{collider}} = E_1 + E_2 = 2E .
$$

The available energy is the full sum of beam energies. In a **fixed-target**
setup, a beam particle of mass $m_1$ and lab energy $E_{\text{beam}}$ strikes a
target of mass $m_2$ at rest. Then $P^\mu = (E_{\text{beam}}/c + m_2 c,\ \vec
p_{\text{beam}})$ and

$$
s = m_1^2 c^4 + m_2^2 c^4 + 2\,m_2 c^2 E_{\text{beam}} .
$$

At high beam energy the last term dominates, so

$$
\sqrt{s}_{\text{fixed}} \approx \sqrt{2\,m_2 c^2\,E_{\text{beam}}}\, .
$$

The available energy grows only as the square root of the beam energy: doubling
$E_{\text{beam}}$ multiplies $\sqrt{s}$ by $\sqrt{2}$, not by $2$. A collider's
$\sqrt{s}$ grows linearly with beam energy. Reaching $\sqrt{s} = 14\ \text{TeV}$
head-on requires two $7\ \text{TeV}$ beams; matching it on a fixed target would
need a beam of $E_{\text{beam}} = s/(2m_p c^2) \approx 10^{17}\ \text{eV}$, far
beyond any accelerator. Most of a fixed-target beam's energy goes into moving the
center of momentum forward, energy that cannot be tapped to make new particles.

$$
% caption: Available center-of-momentum energy versus beam energy. A collider
% rises linearly (root-s equals twice the beam energy) while a fixed target rises
% only as the square root, so the two curves diverge at high energy.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (6.0,0) node[right] {beam energy};
  \draw[->, black] (0,0) -- (0,4.2) node[above] {available energy $\sqrt{s}$};
  % collider linear: y = 0.72 x
  \draw[acc, very thick, domain=0:5.4, samples=2, variable=\x] plot ({\x},{0.72*\x});
  \node[acc, above, rotate=36] at (4.4,{0.72*4.4+0.05}) {collider};
  % fixed target sqrt: y = 1.55*sqrt(x)
  \draw[black, very thick, dashed, domain=0:5.6, samples=90, variable=\x]
    plot ({\x},{1.55*sqrt(\x)});
  \node[black, above] at (5.0,{1.55*sqrt(5.0)+0.05}) {fixed target};
  \node[black, right] at (2.6,1.05) {gap grows};
\end{tikzpicture}
$$

## Threshold energy for particle production

At **threshold**, the reaction has just enough energy to make the final-state
particles, which are all produced at rest in the CM frame. The threshold
condition is therefore

$$
\sqrt{s} = \sum_f M_f c^2
\quad\Longleftrightarrow\quad
s_{\text{th}} = \Bigl(\sum_f M_f\Bigr)^2 c^4 .
$$

For a fixed target, insert $s = m_1^2c^4 + m_2^2c^4 + 2m_2c^2 E_{\text{beam}}$ and
solve for the beam energy:

$$
E_{\text{beam}}^{\text{th}}
  = \frac{\bigl(\sum_f M_f\bigr)^2 - m_1^2 - m_2^2}{2\,m_2}\,c^2 .
$$

The threshold scales with the **square** of the produced mass — a heavy final
state is expensive on a fixed target because the invariant mass must be built out
of an inefficient $\sqrt{s} \sim \sqrt{E_{\text{beam}}}$.

> **Worked example.** Antiproton production, $p + p \to p + p + p + \bar p$, needs
> a final state of four nucleons (baryon number and charge conservation forbid
> making a lone $\bar p$). The final rest mass is $\sum_f M_f = 4 m_p$, so
> $s_{\text{th}} = (4 m_p c^2)^2 = 16\,m_p^2 c^4$. With $m_1 = m_2 = m_p$,
> $$
> 2 m_p^2 c^4 + 2 m_p c^2 E_{\text{beam}} = 16\,m_p^2 c^4
> \;\Longrightarrow\;
> E_{\text{beam}} = 7\,m_p c^2 .
> $$
> The threshold kinetic energy is $T_{\text{th}} = E_{\text{beam}} - m_p c^2 =
> 6\,m_p c^2 = 6 \times 938.3\ \text{MeV} = 5.63\ \text{GeV}$. Six proton rest
> energies of kinetic energy are needed to create one proton's worth of new mass:
> the Bevatron was built above this threshold and discovered the antiproton in
> 1955.

$$
% caption: Threshold configuration in the center-of-momentum frame: the incoming
% momenta are equal and opposite, and at exactly threshold every product is
% created at rest, so all the available energy has gone into rest mass.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  % before: two incoming
  \node[black!70, above] at (1.6,1.2) {before (CM frame)};
  \draw[->, very thick] (0.1,0.5) -- (1.4,0.5);
  \draw[->, very thick] (3.1,0.5) -- (1.8,0.5);
  \draw[black!75, fill=black!8] (1.6,0.5) circle (0.05);
  \node[black!75, below] at (0.6,0.42) {beam};
  \node[black!75, below] at (2.6,0.42) {target};
  % arrow
  \draw[->, black, thick] (3.6,0.5) -- (4.4,0.5);
  \node[black, above] at (4.0,0.55) {at threshold};
  % after: cluster at rest
  \node[black!70, above] at (6.0,1.2) {after};
  \foreach \dx/\dy in {-0.25/0.18, 0.22/0.2, -0.2/-0.2, 0.24/-0.18}
    \draw[black!75, fill=acc!14] (6.0+\dx,0.5+\dy) circle (0.16);
  \node[black!70, below] at (6.0,0.1) {all products at rest};
\end{tikzpicture}
$$

## Compton scattering

A photon of energy $E_\gamma = hc/\lambda$ strikes an electron at rest and
scatters at angle $\theta$, the electron recoiling. Write the four-momenta before
as $p_\gamma = (E_\gamma/c)(1,\hat n)$ and $p_e = (m_e c, \vec 0)$, and after as
$p_\gamma'$ and $p_e'$. Conservation gives $p_e' = p_\gamma + p_e - p_\gamma'$.
Square both sides using the invariant $p_e'\cdot p_e' = -m_e^2 c^2$ and the null
condition $p_\gamma\cdot p_\gamma = p_\gamma'\cdot p_\gamma' = 0$:

$$
-m_e^2c^2 = -m_e^2c^2 + 2\,p_e\cdot(p_\gamma - p_\gamma') - 2\,p_\gamma\cdot p_\gamma' .
$$

The electron terms give $p_e\cdot(p_\gamma - p_\gamma') = -m_e(E_\gamma -
E_\gamma')$, and the photon cross term is $p_\gamma\cdot p_\gamma' = -(E_\gamma
E_\gamma'/c^2)(1 - \cos\theta)$. Substituting and cancelling,

$$
m_e c^2\,(E_\gamma - E_\gamma')
  = E_\gamma E_\gamma'\,(1 - \cos\theta) .
$$

Dividing by $E_\gamma E_\gamma' m_e c^2$ and using $E_\gamma = hc/\lambda$ turns
this into the wavelength shift

$$
\lambda' - \lambda = \frac{h}{m_e c}\,(1 - \cos\theta),
\qquad
\frac{h}{m_e c} = 2.426\ \text{pm},
$$

the **Compton wavelength** of the electron. The shift depends only on the
scattering angle, not on the incident wavelength — a purely kinematic result of
four-momentum conservation. Back-scattering ($\theta = 180^\circ$) gives the
maximum shift $2h/m_e c = 4.85\ \text{pm}$; forward scattering ($\theta = 0$)
gives none. That the shift is measurable for X-rays but negligible for visible
light is a matter of scale: $2.4\ \text{pm}$ against an X-ray's $\sim100\
\text{pm}$ is a percent-level effect, against visible light's $\sim500\
\text{nm}$ it is one part in $10^8$.

$$
% caption: Compton scattering geometry: an incident photon scatters at angle
% theta off an electron initially at rest, emerging with longer wavelength while
% the electron recoils to conserve momentum.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  % incident photon (wavy-ish straight)
  \draw[->, acc, very thick] (-2.6,0) -- (-0.25,0);
  \node[above] at (-1.5,0.05) {incident photon};
  \node[black!70, below] at (-1.5,-0.1) {wavelength short};
  % electron at rest
  \draw[black!75, fill=black!10] (0,0) circle (0.14);
  \node[black!75, below] at (0,-0.3) {electron at rest};
  % scattered photon
  \draw[->, acc, very thick] (0.2,0.1) -- (2.3,1.4);
  \node[above] at (1.5,1.1) {scattered photon};
  \node[black!70, right] at (1.9,0.75) {wavelength longer};
  % recoil electron
  \draw[->, black!70, very thick] (0.15,-0.05) -- (2.1,-1.1);
  \node[black!75, below] at (1.5,-0.9) {recoil electron};
  % angle arc
  \draw[black] (0.9,0) arc (0:33:0.9);
  \node[black!70, right] at (0.95,0.28) {angle};
\end{tikzpicture}
$$

## The recoil electron and the Compton edge

Solving the shift relation for the scattered photon energy gives

$$
E_\gamma' = \frac{E_\gamma}{1 + \dfrac{E_\gamma}{m_e c^2}\,(1 - \cos\theta)},
$$

and energy conservation hands the balance to the electron as kinetic energy,

$$
T_e = E_\gamma - E_\gamma'
    = E_\gamma\,\frac{\dfrac{E_\gamma}{m_e c^2}(1 - \cos\theta)}
                      {1 + \dfrac{E_\gamma}{m_e c^2}(1 - \cos\theta)} .
$$

The electron takes the most energy when the photon back-scatters, $\theta =
180^\circ$, giving the **Compton edge**

$$
T_e^{\max} = E_\gamma\,\frac{2E_\gamma/m_e c^2}{1 + 2E_\gamma/m_e c^2} .
$$

A monoenergetic photon beam scattering in a detector therefore deposits a
continuous electron spectrum from zero up to $T_e^{\max}$, cut off sharply at the
edge — never the full photon energy, because a real photon always survives with
some energy. The gap between the edge and the full-energy photopeak is
$E_\gamma - T_e^{\max} = E_\gamma/(1 + 2E_\gamma/m_e c^2)$, a signature used to
calibrate gamma-ray spectrometers.

$$
% caption: A monoenergetic photon beam produces a continuous Compton electron
% spectrum rising to a sharp cutoff at the Compton edge, the maximum recoil
% energy at back-scatter, below the full photon energy.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (6.0,0) node[right] {electron energy};
  \draw[->, black] (0,0) -- (0,3.2) node[above] {counts};
  % continuum: gentle rise to edge at x=4.2 then cliff
  \draw[acc, very thick]
    (0.3,0.9) .. controls (2.0,1.0) and (3.2,1.35) .. (4.2,2.0)
    -- (4.2,0);
  \draw[black, densely dotted] (4.2,0) -- (4.2,2.6);
  \node[acc, above] at (4.2,2.6) {Compton edge};
  \draw[black, densely dotted] (5.3,0) -- (5.3,2.0);
  \node[black!70, above] at (5.3,1.6) {photon energy};
  \draw[<->, black] (4.3,0.5) -- (5.2,0.5);
  \node[black!70, below] at (4.75,0.5) {gap};
\end{tikzpicture}
$$

If instead a low-energy photon meets a highly relativistic electron, the same
kinematics runs in reverse: the photon leaves the collision with far more energy
than it entered, boosted by roughly $\gamma^2$. This **inverse Compton
scattering** is how relativistic electrons in astrophysical sources upscatter
starlight and microwave-background photons into X-rays and gamma rays, and it is
the same four-momentum bookkeeping viewed from the electron's rest frame, where
the process is ordinary Compton scattering.

Every result here reduced the kinematics to invariants — $s$, the squared
four-momenta, the dot products $p_i\cdot p_j$ — that any frame computes to the
same number. Systematizing those invariants for a general two-to-two process, and
the single constraint that relates them, is the [Mandelstam
variables](/relativity/relativistic-dynamics/mandelstam-variables-and-invariants).

$$
% caption: Four-momentum bookkeeping for a two-to-two collision: the sum of the
% two incoming four-momenta equals the sum of the two outgoing ones, one vector
% equation carrying both energy and momentum conservation.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[very thick] (-3.0,-1.0) rectangle (-0.4,1.0);
  \node[above] at (-1.7,1.05) {incoming};
  \node[black!75] at (-1.7,0.45) {four-momentum $p_1$};
  \node[black!75] at (-1.7,-0.45) {four-momentum $p_2$};
  \node[black!80] at (0.35,0) {$=$};
  \draw[very thick] (1.1,-1.0) rectangle (3.7,1.0);
  \node[above] at (2.4,1.05) {outgoing};
  \node[black!75] at (2.4,0.45) {four-momentum $p_3$};
  \node[black!75] at (2.4,-0.45) {four-momentum $p_4$};
  \node[black, below] at (0.35,-1.15) {one equation, four components conserved};
\end{tikzpicture}
$$

[^pdg]: Particle Data Group, _Review of Particle Physics_, Kinematics review
(invariant $s$, thresholds, and two-body scattering).
<https://pdg.lbl.gov/2023/reviews/rpp2023-rev-kinematics.pdf>