University Physics IV · Special Relativity II · 3.9

Collisions, Decays & Thresholds

Conservation hands you four equations per reaction and one number every frame agrees on. Learn to pick the frame that makes the algebra vanish, read a parent's mass off its daughters, and price the beam energy a new particle costs.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

A collision or a decay is bookkeeping on a single object: the total four-momentum P = (E/c, p), whose four components all carry across the event because spacetime is homogeneous. Two readings of that ledger do almost all the work. Square the sum of the final particles' four-momenta and you get s = (ΣE)² − (Σpc)², a Lorentz invariant, so √s/c² is the mass of whatever produced them — a parent reconstructed from its debris without ever being seen directly.

Square the initial state instead and √s is the largest mass the reaction could possibly make, because it is the total energy in the frame where the momenta cancel. That second reading comes with a bill. Against a stationary target, s = m_b² + mₜ² + 2mₜ E_lab, so the useful energy climbs only as √E_lab while the remainder stays locked in the products' shared motion; a collider, where the momenta already cancel, gets √s = 2E instead.

And the framework has a hard limit built into it: four-momentum conservation is a constraint, not a mechanism. It says what a reaction may not do, and it fixes every magnitude once a reaction does happen, but it will never hand you a lifetime, a branching ratio, or a cross-section.

Simple definition: The invariant mass of a system is √s/c², where s = (ΣE)² − (Σpc)² is the square of its total four-momentum: conserved through any reaction, and the same number in every inertial frame.
Example: Two photons of 0.60 GeV and 0.40 GeV flying 15.8° apart give s = 2E₁E₂(1 − cos θ) = 0.01814 GeV², so √s = 134.7 MeV — the π⁰ (135.0 MeV) that made them, rebuilt from two particles with no mass at all.

Four-momentum conservationΣ Pᵢₙ = Σ Pₒᵤₜ, with P = (E/c, p)

Holds in every inertial frame and through any change of particle content. Mass is not on the list.

Four equations per reaction — one for energy, three for momentum; E in MeV, p in MeV/c.

The invariant ss = (ΣE)² − (Σpc)² = (Mc²)²

Evaluate s in whichever frame is easiest — the number you get then holds in all of them.

s in MeV²; √s is the total energy in the frame where Σp = 0. Masses written as rest energies, c = 1.

Two-body decay, parent at restE₁* = (M² + m₁² − m₂²)/(2M), p*c = √(E₁*² − m₁²)

Three masses fix every magnitude in the decay; only the emission direction is left free.

M, m₁, m₂ are rest energies in MeV; the daughters leave back to back with the same |p*|.

Lab decay lengthℓ = βγcτ, βγ = pc/mc², N(d) = N₀e^(−d/ℓ)

Converts a lifetime no clock can follow into a flight path a tracker can actually measure.

τ is the proper lifetime in s, p the lab momentum, ℓ and the flight path d in m.

Fixed targets = m_b² + mₜ² + 2mₜ E_lab ≈ 2mₜ E_lab

√s grows as √E_lab, so quadrupling the beam energy only doubles the mass you can make.

Rest energies m_b (beam) and mₜ (target), beam total energy E_lab, all in MeV.

Threshold, and the collider comparisonE_lab = [(Σm_f)² − m_b² − mₜ²]/(2mₜ) against √s = 2E

The fixed-target cost is quadratic in the mass produced; the collider's is only linear.

Σm_f is the sum of the final-state rest energies; the right-hand form is a head-on collider.

Four equations, and mass is not one of them

A reaction is a ledger with four entries. Total energy is conserved and so is each component of total momentum: ΣEᵢₙ = ΣEₒᵤₜ and Σpᵢₙ = Σpₒᵤₜ, which is what it means to say the total four-momentum P = (E/c, p) carries across the event. Rest energy sits inside E, so it converts freely in either direction. A π⁰ at rest has E = 134.98 MeV and no momentum; it decays to two photons, which must therefore leave back to back with 67.49 MeV each. Read as the invariant of the pair, the system's mass is 134.98 MeV/c² before and after — but the sum of the individual particles' masses fell from 134.98 MeV/c² to exactly zero. Conservation of mass is not a relativistic law. Conservation of energy and momentum is, and in this topic it is the only law you are allowed to use.

Square the ledger and the frame drops out

Add the four-momenta of any set of particles and square the sum: s = (ΣE)² − (Σpc)². Because four-momentum boosts exactly like (ct, x), that combination is a Lorentz invariant — every inertial observer computes the same number — and √s is precisely the total energy in the frame where the momenta cancel. Two consequences make this the workhorse of the subject. First, √s/c² is the invariant mass of the system, and it is emphatically not the sum of the parts. Take a μ⁺μ⁻ pair, each muon of total energy 5.00 GeV, so pc = √(5.00² − 0.1057²) = 4.9989 GeV, with an opening angle of 36.0° between them: s = 2m_μ² + 2E₁E₂ − 2p₁p₂c²cos θ = 0.0223 + 50.0000 − 40.4328 = 9.5895 GeV², so √s = 3.097 GeV and the pair has rebuilt a J/ψ from two tracks whose masses total only 0.211 GeV/c². Second, you may compute s wherever the arithmetic is easiest — usually the lab for an initial state, the parent's rest frame for a decay — and then use that number in the other frame without transforming anything. It is how a resonance is found: histogram √s for every candidate pair, and a real parent stands up as a peak at its own mass while unrelated pairs spread into a smooth background.

The centre-of-momentum frame solves a two-body decay outright

Choose the frame where Σp = 0. For a decay that is the parent's rest frame, and the problem collapses: the daughters leave back to back with a common momentum magnitude p*, and energy conservation together with E² = (pc)² + (mc²)² fixes it from the three masses alone. Rearranging gives E₁* = (M² + m₁² − m₂²)/(2M), everything written as rest energies. Take Λ → p + π⁻, with M = 1115.683 MeV, mₚ = 938.272 MeV and m_π = 139.570 MeV: Eₚ* = (1244748.6 + 880354.3 − 19479.8)/2231.366 = 943.648 MeV, so p*c = √(943.648² − 938.272²) = 100.58 MeV and the pion takes E_π* = 1115.683 − 943.648 = 172.04 MeV. Nothing about the strong interaction entered. Kinematics fixed every magnitude and left only the emission direction free. Boosting each daughter into the lab with E = γ(E* + βp*c cos θ*) then spreads that single rest-frame energy into a band, which is why lab decay products are never monoenergetic unless the parent was at rest.

A lifetime you cannot watch becomes a length you can measure

A decay rate belongs to the parent in its own rest frame and is quoted as a proper lifetime τ. In the lab the parent's clock runs slow by γ and the parent is also moving, so the mean distance it covers before decaying is ℓ = βγcτ, with the surviving fraction after a flight path d equal to e^(−d/ℓ). The convenient handle is βγ = pc/mc², straight off the lab momentum with no separate γ needed. A charged pion has cτ = 7.80 m: at 2.00 GeV/c, βγ = 2000/139.57 = 14.33 and ℓ = 112 m, so a 20 m beamline still delivers e^(−20/112) = 84% of them. The Λ of the previous section has cτ = 7.89 cm, so even at 5.00 GeV/c, where βγ = 5000/1115.683 = 4.48, it covers only 35 cm, and a tracker registers it not as a track of its own but as a V of charged daughters opening from a vertex displaced from the interaction point. Notice what conservation did not supply: τ itself. Kinematics tells you where a particle decays once you know how long it lives, never how long it lives.

Why a fixed target wastes most of the beam

Fire a beam of total energy E_lab at a stationary target and square the initial four-momentum: s = m_b² + mₜ² + 2mₜ E_lab, so once the beam energy dominates the masses, √s ≈ √(2mₜ c²E_lab). The available energy grows only as the square root of what you paid for. The reason is momentum, not energy: the products must carry the beam's momentum away, and that shared forward motion can never become mass. A 100 GeV proton on a hydrogen target carries 100.94 GeV in the lab yet offers √s = 13.76 GeV, with the approximation √(2mₜ E_lab) = 13.70 GeV already good to half a percent. Collide two beams of energy E head-on instead and the momenta cancel from the start, giving √s = 2E, linear and with nothing thrown away. Inverting the fixed-target relation gives the threshold beam energy E_lab = [(Σm_f)² − m_b² − mₜ²]/(2mₜ), quadratic in the mass wanted. Doubling your mass reach costs four times the beam energy at a fixed target and only twice at a collider, which is why every energy-frontier machine since the 1970s has collided beams rather than aimed them at a block of metal.

Conservation says permitted, never how often

Every result above is a constraint. It rules reactions out, and it fixes magnitudes once a reaction happens, but it contains no rate, no cross-section and no branching fraction. Charged pions and charged kaons are both allowed to decay to μν; the pion lives 2.60 × 10⁻⁸ s and the kaon 1.24 × 10⁻⁸ s, and no kinematic argument separates them, because those numbers come from the coupling of the interaction and from how much phase space the final state offers. Kinematics is equally blind to the other conservation laws: p → π⁺ + π⁰ is energetically wide open, releasing 664 MeV, and it simply does not occur, because baryon number forbids it. So use four-momentum for what it is good at — thresholds, allowed energies and angles, masses reconstructed from debris — and reach for the interaction itself the moment the question becomes how likely, how fast, or how often.

Change one variable at a time

Make the relationship visible.

Interactive model

Beam energy per particle E6.6 GeV

Beam and target rest energy m0.94 GeV

Mass to be produced M3.75 GeV

Put the dashed line at 3.75 GeV, four proton masses, and slide the beam: the collider crosses it at 1.9 GeV, the fixed-target curve not until 6.5 GeV. Raise the line to 30 GeV and the readout asks 477.8 GeV of a fixed target against 15 GeV per collider beam.

√s FIXED TARGET3.76 GeV

√s COLLIDER13.20 GeV

FIXED-TARGET E FOR M6.5 GeV

COLLIDER SAVING3.5 times

Live interpretation√s FIXED TARGET: 3.76 GeV. √s COLLIDER: 13.20 GeV. FIXED-TARGET E FOR M: 6.5 GeV. COLLIDER SAVING: 3.5 times

Catch the common trap

Explain before calculating.

Misconception check

The energy available for making new particles is not the beam energy. A 100 GeV proton striking a proton at rest brings 100.94 GeV of total lab energy, yet only √s = √(2mₚ c²E_lab + 2mₚ²c⁴) = 13.8 GeV of it is convertible; the balance is held in the forward drift of the whole final state, which momentum conservation will not let it give up, and the convertible part climbs only as the square root of the beam energy. The same error in its other form is adding rest energies together: the invariant mass of a system is not the sum of its parts' masses. Two photons of 0.50 GeV each have exactly zero mass apiece, yet the pair carries √s = 1.00 GeV when they fly back to back and exactly zero when they fly parallel — same particles, same energies, different invariant mass, because it depends on the relative directions. That is why squaring the total four-momentum, rather than adding masses, is the move that works.

A proton beam of total energy 30.0 GeV strikes a hydrogen target at rest (mₚ c² = 0.938 GeV). What is the largest invariant mass the collision can create?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA charged pion at rest decays by π⁺ → μ⁺ + ν_μ. Using m_π c² = 139.570 MeV, m_μ c² = 105.658 MeV and treating the neutrino as massless, find the momentum shared by the daughters, the muon's total energy, and its kinetic energy.

The parent is at rest, so ΣE = 139.570 MeV and Σp = 0. The daughters must leave back to back with equal momentum magnitudes: p_μ c = p_ν c = p*c, and E_μ + E_ν = 139.570 MeV.
Use the rest-frame result E₁* = (M² + m₁² − m₂²)/(2M) with m_ν = 0: E_μ = (139.570² + 105.658²)/(2 × 139.570) = (19479.8 + 11163.6)/279.140 = 30643.4/279.140 = 109.778 MeV.
The neutrino takes the remainder, E_ν = 139.570 − 109.778 = 29.792 MeV, and being massless its momentum is p*c = E_ν = 29.792 MeV.
Check against the invariant: √(p*²c² + m_μ²c⁴) = √(29.792² + 105.658²) = √(887.6 + 11163.6) = √12051.2 = 109.778 MeV ✓. Kinetic energy is K_μ = 109.778 − 105.658 = 4.120 MeV.

Answerp*c = 29.79 MeV for both daughters; E_μ = 109.78 MeV, so K_μ = 4.12 MeV. Every muon from a pion at rest carries the same energy — a two-body decay is monoenergetic.

MediumA K⁰_S of momentum 3.00 GeV/c is produced in a target. Its rest energy is 497.611 MeV and its proper lifetime is 0.8954 × 10⁻¹⁰ s. Find its mean decay length in the laboratory, the fraction surviving beyond 1.00 m, and the fraction that decays within the first 5.0 cm.

Get βγ straight from the lab momentum: βγ = pc/mc² = 3000/497.611 = 6.029. (Separately E = √(3000² + 497.61²) = 3041.0 MeV gives γ = 6.111 and β = 0.9865, but only the product is needed.)
Convert the proper lifetime into the distance light covers in it: cτ = (2.998 × 10⁸ m s⁻¹)(0.8954 × 10⁻¹⁰ s) = 2.684 cm.
Mean lab decay length: ℓ = βγcτ = 6.029 × 2.684 cm = 16.2 cm. Both effects sit inside βγ — the dilated clock, and the speed at which the kaon travels while that clock runs.
Survival is exponential in flight path, N/N₀ = e^(−d/ℓ). At d = 1.00 m, e^{(−100/16.18)} = e^(−6.18) = 2.1 × 10⁻³.
Within the first 5.0 cm the decayed fraction is 1 − e^{(−5.0/16.18)} = 1 − 0.734 = 0.266, so about 27% of the kaons are gone before the beam has crossed a vertex detector.

Answerℓ = 16.2 cm; only 0.21% of the kaons survive 1.00 m, while 27% decay inside the first 5.0 cm. This is a displaced vertex, not a track that reaches the outer detector.

HardAntiprotons are made by firing protons at a hydrogen target. Find the threshold beam energy for p + p → p + p + p + p̄, using mₚ c² = 0.938272 GeV, then compare it with a proton–proton collider reaching the same √s, and extend the comparison to producing a Z boson of 91.19 GeV/c².

Baryon number and charge force the lightest allowed final state to be p + p + p + p̄, so Σm_f c² = 4mₚ c² = 3.7531 GeV. At threshold all four products move together, which means √s is exactly that: √s = 4mₚ c².
Evaluate s in the lab: s = mₚ² + mₚ² + 2mₚ E_lab. Setting s = (4mₚ)² = 16mₚ² gives 2mₚ E_lab = 14mₚ², so E_lab = 7mₚ c² = 6.568 GeV.
In kinetic terms K = E_lab − mₚ c² = 6mₚ c² = 5.630 GeV — six proton rest energies spent to make one antiproton. The Bevatron was built at 6.2 GeV kinetic for exactly this reaction.
A collider needs only √s = 2E = 3.7531 GeV, i.e. E = 1.877 GeV per beam. The ratio of beam energies is 6.568/1.877 = 3.50, and the collider wastes nothing on the products' common motion.
The gap widens as the square of the mass wanted. An e⁺e⁻ collider reaches the Z with 45.6 GeV per beam; a fixed electron target would need E_lab ≈ m_Z²/(2mₑ) = 8315.6/0.001022 = 8.1 × 10⁶ GeV, about 1.8 × 10⁵ times more. That quadratic cost is the whole case for colliders.

AnswerE_lab = 7mₚ c² = 6.57 GeV total, i.e. K = 5.63 GeV of kinetic energy; a collider reaches the same √s with 1.88 GeV per beam. The advantage grows quadratically — at the Z it is 1.8 × 10⁵.