Kinematic Invariants; Systems of Interacting Particles

One number — the invariant mass of whatever you collected — decides what could have produced it, and the same number decides how big your machine has to be.

🎯 Why this matters

Because the invariant can be computed from any subset of the particles you caught, you can ask “did these three come from one object?” without a model of what the object was. Every resonance search is that question, asked over all the combinations.

§1.2 gave one particle an invariant. Now do it for a system — and the result is the single most-used quantity in experimental particle physics.

The mass of a system

Add up the 4-momenta of $n$ non-interacting particles and take the norm:

\htmlClass{t-s}{s} \equiv \htmlClass{t-m}{m^2} = \left(\sum_{i=1}^{n}\htmlClass{t-E}{E_i}\right)^{\!2} - \left(\sum_{i=1}^{n}\htmlClass{t-p}{\mathbf{p}_i}\right)^{\!2} \;\geq\; 0

(1.35–1.37)

The mass of the system, and the quantity every experiment in this book plots on its x-axis.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

Go to the frame where the total momentum vanishes — the CM frame — and the second bracket dies:

s = \left(\sum_i E_i^*\right)^2 .

The mass of a system of non-interacting particles is its energy in the CM frame. That is Eq. (1.38), and it is why “√s” and “centre-of-mass energy” are used interchangeably for the rest of the book.

For two particles, expanding the brackets and using $\boldsymbol{\beta} = \mathbf{p}/E$ :

s = m_1^2 + m_2^2 + 2E_1E_2 - 2\mathbf{p}_1\cdot\mathbf{p}_2 = m_1^2 + m_2^2 + 2E_1E_2\left(1 - \boldsymbol{\beta}_1\cdot\boldsymbol{\beta}_2\right).

🔢 Worked example — two photons, and the end of “mass = amount of stuff” (Example 1.2)

Two photons, each of energy

E

. Photons are massless, so

p = E

for each.

Same direction: $E_\text{tot} = 2E$ , $p_\text{tot} = 2E$ , so $m^2 = 4E^2 - 4E^2 = 0$ . The system is massless.
Opposite directions: $E_\text{tot} = 2E$ , $p_\text{tot} = 0$ , so $m = 2E$ .
General angle θ: $p_\text{tot}^2 = 2E^2(1+\cos\theta)$ , hence

m^2 = 2E^2\left(1 - \cos\theta\right).

Two 1 GeV photons: 0 GeV when they fly together, 1.414 GeV at 90°, 2 GeV back-to-back. Nothing about the photons changed — only the angle between them.

The system contains no matter at all, only energy, and it has mass. So mass is not a measure of how much matter a body contains. It is the norm of a 4-vector, and nothing more.

m₁0 GeVm₂0 GeVE₁1 GeVE₂1 GeV

angle θ between them180 °

s = 4 GeV² → m = √s = 2 GeVvs m₁ + m₂ = 0 GeV

mass of the two-particle system

Two massless particles flying the same way have zero mass; the same two flying apart have mass 2E. Nothing about the particles changed — only the angle. This is the clearest possible demonstration that mass is not a count of matter.

💡 What this really says — mass is conserved and mass is not additive, and both are true

Two statements that look contradictory and are not:

Mass is conserved. Energy and momentum are conserved, so their invariant combination is too: the mass of the initial system always equals the mass of the final system.
The sum of the masses is not conserved. Adding up the individual masses of the products generally gives a different number from adding up the reactants’.

Both are true because $m_\text{system} \neq \sum m_i$ . Chapter 4 lives on this: a resonance is precisely a bump in the distribution of $\sqrt{s}$ computed from final-state particles whose individual masses are perfectly well known.

Collisions, and the argument for colliders

A collision is $a + b \to c + d + e \ldots$ ; if the final state contains only the initial particles it is elastic. The L frame is the one where the target $b$ is at rest before the collision.

In L, with beam energy $E_a$ :

s = m_a^2 + m_b^2 + 2m_b E_a \;\;\xrightarrow{\;E_a \gg m\;}\;\; 2 m_b E_a .

In the CM, with two beams of energy $E^*$ each:

s = \left(E_a^* + E_b^*\right)^2 \;\;\xrightarrow{\;E^* \gg m\;}\;\; \left(2E^*\right)^2 .

⚙️ Engineer’s bridge — √s is your energy budget, and a fixed target wastes it

Only the CM energy is available for making new particles; the rest is locked up in the motion of the centre of mass, because momentum must be conserved. Think of it as a power budget with a fixed overhead: at a fixed target most of your beam energy goes into pushing the wreckage forwards, not into the collision.

The scaling is the punchline. Fixed target gives $\sqrt{s} \propto \sqrt{E_a}$ ; a collider gives $\sqrt{s} \propto E$ . On the log–log plot below, slope ½ against slope 1. To gain a decade in reach you buy ten times the beam energy in a collider and a hundred times at a fixed target — and beam energy is roughly what a machine costs.

Where it breaks: $\sqrt s$ is a ceiling on what you can make, not a description of what you typically get. At a hadron collider the colliding objects are partons carrying a fraction of the proton momentum, so a 14 TeV machine delivers 14 TeV to a hard scatter essentially never — the useful parton luminosity falls steeply with the mass being produced, which is why the LHC’s reach for a heavy particle is a few TeV rather than fourteen. The budget analogy is right about the constraint and silent about the spend, and the spend is governed by §6.2’s parton distributions.

beam energy (log₁₀ of GeV)3

beam = 1000 GeV
fixed target (on a proton): √s = 43.339 GeV
collider (two such beams): √s = 2000 GeV
collider is 46.15× better at the same beam energy

collider: √s = 2E
fixed target: √s ≈ √(2m_p E)

On a log–log plot the collider line has slope 1 and the fixed-target line slope ½. To gain a factor of ten in √s you buy ten times the beam energy in a collider — or a hundred times at a fixed target. That gap is the entire reason colliders exist.

Put differently: to match the LHC's 14 TeV with a fixed-target machine you would need a beam of 104.4×10⁶ GeV = 0.104 EeV — an energy only the rarest cosmic rays reach.

🔢 Worked example — how big would a fixed-target LHC be?

The LHC collides 7 TeV on 7 TeV, so

\sqrt{s} = 14

TeV. To reach the same

s

with a beam on a stationary proton you would need

E_a = \frac{s - 2m_p^2}{2m_p} = \frac{(14\,000)^2}{2 \times 0.938}\ \text{GeV} = 1.04\times10^{8}\ \text{GeV} = 0.10\ \text{EeV}.

That is not merely expensive, it is above almost the entire cosmic-ray spectrum — a flux of roughly one particle per square kilometre per century. Meanwhile the 7 TeV beam gets you a fixed-target $\sqrt{s}$ of only 115 GeV, worse than LEP. The collider wins by a factor of 122 at the same beam energy.

t and u: the other two invariants

$s$ says how much energy went in. For two-body scattering $a + b \to c + d$ there are two more invariants, both momentum transfers:

\htmlClass{t-t}{t} = \left(E_c - E_a\right)^2 - \left(\mathbf{p}_c - \mathbf{p}_a\right)^2, \qquad \htmlClass{t-u}{u} = \left(E_d - E_a\right)^2 - \left(\mathbf{p}_d - \mathbf{p}_a\right)^2

(1.48, 1.50)

Both are the squared norm of a difference of 4-momenta — how much 4-momentum the projectile handed over, and to which product.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

They are not independent. Energy–momentum conservation forces

s + t + u = m_a^2 + m_b^2 + m_c^2 + m_d^2 ,

so a 2 → 2 process has exactly two free kinematic variables, not three.

⚙️ Engineer’s bridge

(s, t, u)

are a coordinate system for a scattering process, and the identity above is a constraint plane through that space: the physical states live on a 2-D surface embedded in 3-D. Exactly like a system with three state variables and one conservation law — you may parameterise it however you like, but only two degrees of freedom are real.

That is why an experimentalist quotes $d\sigma/dt$ at fixed $s$ and considers the job done. Two numbers describe the event completely, so a plot against two numbers is a complete description.

Where it breaks: “two numbers describe the event completely” is exact for a 2 → 2 collision of spinless particles and stops being true immediately afterwards. Spin adds polarization observables that $s$ and $t$ cannot carry; a 2 → 3 final state needs five independent invariants, which is why §4.3 has to introduce the Dalitz plot rather than plotting against $t$ . The counting is a statement about a specific kinematics, not a general licence to describe events with two numbers.

Elastic proton–proton scattering, so all four masses are m_p.

√s10 GeVCM scattering angle θ*60 °

s	100 GeV²	squared CM energy — how much is available
t	-24.1196 GeV²	squared 4-momentum transfer a → c — how hard the hit was
u	-72.3589 GeV²	the other transfer, a → d
s+t+u	3.52142 GeV²	must equal Σm² = 3.52142 GeV² — residual 2.58e-14

t (≤ 0 always)
u (≤ 0 always)
s + t + u = Σm², flat

t and u trade off against each other exactly so that their sum with s never moves. Forward scattering (θ* → 0) means t → 0: barely any momentum was exchanged, which is why forward cross-sections are enormous and why detectors leave a hole for the beam.

§1.5 When the particles interact, the field joins in

Everything above assumed non-interacting particles. Once there is a force between them, Eq. (1.35) no longer gives the system’s mass, because the field itself carries energy and momentum. The total energy is not the sum of the particles’ energies; the total momentum is not the sum of theirs.

The book’s honest summary: the concept of potential is non-relativistic. It works when speeds are small compared with $c$ — an electron in an atom, to first approximation — and fails when they are not. For positronium you cannot use a potential at all, because the electron and positron can annihilate into photons and photons can materialise back into pairs. There is no fixed number of particles to write a potential between.

How much mass a binding energy costs — three scales (Examples 1.3, 1.5, 1.6)

System↕	Binding energy↕	Mass of the system↕	Fractional mass defect↕	How you would notice
Two colliding wax balls at 300 m/s	½β² of the rest energy	—		You cannot weigh it. You measure the temperature rise instead.
Hydrogen atom	13.6 eV	938.8 MeV	1.4×10⁻⁸	Far below any mass measurement — which is exactly why chemistry can pretend mass is additive.
⁴He nucleus		3727.41 MeV	7.6×10⁻³	Nearly 1 % — measurable, and the energy source of the Sun.

The pattern: the fractional mass defect measures the strength of the interaction holding the system together. Chapter 6 pushes this to its limit — in a proton, the quark masses account for only about 1 % of the total.

🔢 Worked example — three processes that cannot happen (Example 1.4)

Invariants make impossibility proofs trivial: compute

s

before and after, and if they differ, the process is forbidden — no dynamics needed.

$e^+e^- \to \gamma$ . Final state: one photon, $s = 0$ . Initial state:

s = 2m_e^2 + 2\left(E_+E_- - \mathbf{p}_+\!\cdot\mathbf{p}_-\right) > 0 .

Not equal. Forbidden — a single photon can never be the whole final state. The inverse, $\gamma \to e^+e^-$ , is forbidden for the same reason.

$\gamma + e^- \to e^-$ . With the electron initially at rest, $s = 2m_eE_\gamma + m_e^2$ before and $s = m_e^2$ after. Equality demands $2m_eE_\gamma = 0$ : forbidden.

So why do pair production and bremsstrahlung happen constantly in a detector? Because they happen near a nucleus, which absorbs the leftover momentum. That is why §1.11 writes them as $\gamma + N \to e^+e^- + N$ and $e^- + N \to e^- + N + \gamma$ — the $N$ is not decoration, it is what makes the books balance.

Reproduce it

import numpy as np
mp, me = 0.938272, 0.000511

for th in (0, 90, 180):                                   # Example 1.2
    m2 = 2 * 1.0**2 * (1 - np.cos(np.radians(th)))
    print(f"two 1 GeV photons at {th:3d} deg: m = {np.sqrt(m2):.6f} GeV")

E = 7000.0                                                # fixed target vs collider
s_f = 2*mp*E + 2*mp**2
print(f"LHC beam {E:g} GeV: fixed-target sqrt(s) = {np.sqrt(s_f):.4f} GeV, "
      f"collider = {2*E:g} GeV, ratio {2*E/np.sqrt(s_f):.1f}x")
Eeq = (14000**2 - 2*mp**2) / (2*mp)
print(f"fixed-target beam needed for sqrt(s) = 14 TeV: {Eeq:.4e} GeV = {Eeq*1e9/1e18:.3f} EeV")

rs, th = 10.0, np.radians(60)                             # Mandelstam, elastic pp
s = rs**2; ps = np.sqrt(s/4 - mp**2)
t, u = -2*ps**2*(1-np.cos(th)), -2*ps**2*(1+np.cos(th))
print(f"elastic pp, sqrt(s)=10 GeV, theta*=60 deg: s={s:.4f} t={t:.4f} u={u:.4f}")
print(f"  s+t+u = {s+t+u:.6f}  vs  4 mp^2 = {4*mp**2:.6f}   residual {s+t+u-4*mp**2:.2e}")

rng = np.random.default_rng(1); ok = 0
for _ in range(100000):
    S = rng.uniform(2*mp+0.01, 200)**2; a = rng.uniform(0, np.pi)
    P = np.sqrt(S/4 - mp**2)
    ok += abs(S - 2*P**2*(1-np.cos(a)) - 2*P**2*(1+np.cos(a)) - 4*mp**2) < 1e-9
print(f"s+t+u = sum m^2 held in {ok}/100000 random configurations")
print(f"mass defects: H atom {13.6/938.8e6:.2e}, He-4 {28.3/3727.41:.2e}")

prints

two 1 GeV photons at   0 deg: m = 0.000000 GeV
two 1 GeV photons at  90 deg: m = 1.414214 GeV
two 1 GeV photons at 180 deg: m = 2.000000 GeV
LHC beam 7000 GeV: fixed-target sqrt(s) = 114.6192 GeV, collider = 14000 GeV, ratio 122.1x
fixed-target beam needed for sqrt(s) = 14 TeV: 1.0445e+08 GeV = 0.104 EeV
elastic pp, sqrt(s)=10 GeV, theta*=60 deg: s=100.0000 t=-24.1196 u=-72.3589
s+t+u = 3.521417  vs  4 mp^2 = 3.521417   residual 2.58e-14
s+t+u = sum m^2 held in 100000/100000 random configurations
mass defects: H atom 1.45e-08, He-4 7.59e-03

🔑 If you remember only three things

A fixed target loses the race by arithmetic. The energy you must supply grows as s/2m, so doubling the reach costs four times the beam, while a collider pays twice.
A bound system’s mass counts the field. Add up the constituents and you get the wrong answer unless the interaction energy is included, which is why a nucleus weighs less than its nucleons.
Two invariants, not three. s, t and u are locked by a sum fixed by the masses, so a scattering process has exactly two free ones — that is why differential cross-sections are always plotted at fixed s.

Where this goes next

§1.6–1.7 turns $s$ into rates: cross-sections, luminosity and decay widths.
§1.8 shows why $|t|$ is the resolving power — the reason the whole field chases energy.
Chapter 4 plots invariant mass against event count and finds bumps; every resonance in the book is discovered exactly that way.

✅ Check yourself — invariants and systems

0/5 answered · 0 correct

1.Two 1 GeV photons fly off at 90° to each other. What is the mass of the system, and what does the answer establish?
2.You want to double your reach in $\sqrt{s}$ . By what factor must the beam energy rise?
3.For $a+b \to c+d$ the identity $s+t+u = \sum m^2$ holds. What does that tell you about the process?
4.Why can a photon not simply convert into an $e^+e^-$ pair in empty space, when it plainly does so all the time inside a detector?
5.The ⁴He nucleus has a fractional mass defect of 0.76 %, the hydrogen atom 1.4×10⁻⁸. What does the ratio of those two numbers measure?