University Physics IV · Particle Physics · 14.1
Relativistic Kinematics and Invariant Mass
Detectors report energies and angles in one frame only. This is the arithmetic that turns those frame-bound numbers into a mass every observer agrees on, says what a collision can afford to make, and prices a fixed target against a collider before either is built.
Build the model
Connect the measurement to the mechanism.
Energy and momentum are worth more together than apart. Stack them into a single four-vector p = (E/c, p) and the combination E² − |p|²c² is untouched by any boost, so it belongs to the system rather than to the observer: for one particle it equals m²c⁴, and for a collection it defines the invariant mass of the whole. Conservation then pays twice.
Four-momentum is conserved because energy and momentum separately are, so its square is conserved too, and that licenses the one move this topic rests on — evaluate the invariant s = (Σp)² in whichever frame makes it trivial, then use it in the frame you actually measure. Everything follows. √s is the energy available in the centre-of-momentum frame, so it fixes what a reaction can make; a resonance shows up as a bump in the invariant mass rebuilt from its decay products, sitting at the parent's mass however fast the parent was moving; and the same algebra prices machines, because a stationary target buys √s only as the square root of the beam energy while a symmetric collider buys it linearly. Two costs come with it.
The bookkeeping recognises only free, on-shell particles long before and long after the event, so it says nothing about the interaction itself. And an unstable particle has no sharp mass: it lives a proper time τ, so its peak carries a width Γ = ħ/τ and what you fit is a Breit–Wigner, not a spike.
- Simple definition
- The invariant mass of a system is M = √(Eₜₒₜₐₗ² − |pₜₒₜₐₗ c|²)/c², assembled from the total energy and the total momentum, and every inertial observer computes the same value for it.
- Example
- Two photons of 3.00 GeV each, opening angle 2.58°, give M²c⁴ = 2E₁E₂(1 − cos θ) = 0.0182 GeV², so M = 135 MeV/c² — a π⁰, although neither photon has any mass at all.
Turns two frame-dependent measurements into one permanent label for the particle.
E in GeV, |p| in GeV/c, m in GeV/c²; the right-hand side is the same number in every inertial frame
One number per event that ignores how fast the parent was moving — the whole basis of a mass histogram.
Sum energies as scalars and momenta as vectors; M ≥ Σmᵢ, with equality only if all the parts are mutually at rest
√s is the budget: a reaction can only run if √s ≥ the sum of the final-state rest energies.
s in GeV²; the centre-of-momentum frame is the one where Σp = 0, so there √s is simply the total energy
√s rises as √Eb on a target but as 2E in a collider — the entire case for colliding beams, in one line.
Eb the beam energy onto a target of mass mₜ at rest; E the energy of each beam of a head-on symmetric collider
π⁺ → μ⁺ν fixes Eμ at 109.78 MeV, so the muon always emerges with exactly 4.12 MeV of kinetic energy.
M the parent mass and m₁, m₂ the daughters, all in GeV/c²; the shared momentum p* follows from E₁² − p*²c² = m₁²c⁴
The Z's 2.495 GeV width means τ = 2.64 × 10⁻²⁵ s — a lifetime measured with a histogram, not a clock.
Γ the full width at half maximum of the mass peak in MeV, τ the mean proper lifetime in s
Energy and momentum, welded into one object
Under a boost, energy and momentum mix into each other exactly as time and space do, so they belong in one four-component object, p = (E/c, pₓ, py, pz). Its Minkowski square carries a minus sign, and that is the combination every frame agrees on: E² − |p|²c² = m²c⁴. Test it on a proton. At rest it has E = 938.272 MeV and no momentum at all. Boost until γ = 2 and it carries E = 1876.5 MeV with pc = γβmc² = 1625.1 MeV; then E² − p²c² = 3.5214 − 2.6411 = 0.8804 GeV², whose square root is 938.3 MeV again. Neither E nor p survived the boost. Their difference of squares did, which is why mass is a useful label for a particle and speed never is.
The mass of a system is not the sum of its masses
Extend the definition to several particles by adding the four-momenta first: M²c⁴ = (ΣEᵢ)² − |Σpᵢ|²c². Energies add as scalars, momenta as vectors, and that asymmetry is where the surprise lives. Take two photons of 3.00 GeV each, both with m = 0 exactly. Sent back to back, ΣE = 6.00 GeV and Σp = 0, so M = 6.00 GeV/c². Sent in the same direction, ΣE = 6.00 GeV but |Σp|c = 6.00 GeV, and M = 0. Same particles, same energies, and the invariant mass ranges from zero to 6 GeV on the opening angle alone. What M counts is the energy that no boost can remove — the energy of relative motion — so M ≥ Σmᵢ always, with equality only when every constituent is at rest in one common frame.
Evaluate s where it is easy, use it where you measure
The working method is a single line: s = (Σp)² is both conserved and frame-independent, so evaluate it wherever it is trivial and equate the two evaluations. Threshold problems are the standard case. For p + p → p + p + p + p̄ the cheapest final state has all four particles at rest in the centre-of-momentum frame, where Σp = 0 and √s = 4mpc² = 3.753 GeV — one line there, and a miserable simultaneous solve in the lab frame. Now write the same s in the lab, where the target proton sits still: s = (Eb + mpc²)² − pb²c² = 2mₚ²c⁴ + 2Eb mpc². Setting the two equal gives Eb = 7mpc² = 6.568 GeV immediately. Nothing was Lorentz-transformed; the invariant did the transforming for you.
Why a stationary target stops paying
That same lab expression is the economics of accelerators. For Eb ≫ mtc² the mass terms drop and s ≈ 2mtc²Eb, so √s ≈ √(2mtc²Eb): the useful energy grows only as the square root of what you pay for. A symmetric collider has Σp = 0 by construction, so s = (2E)² exactly and √s = 2E, linear in the beam energy. The numbers are brutal. Each LHC beam carries 6500 GeV; fired at a stationary proton it would deliver √s = √(2 × 0.938 × 6500) = 110 GeV, against 13 000 GeV when the two beams are brought together. Turned around, reaching 13 TeV on a fixed target needs a beam of 9 × 10⁷ GeV, some 14 000 times what the machine actually stores. Fixed targets survive only where high rate, a well-understood stationary target, or beam intensity matters more than √s.
A decay reconstructs its parent, whatever the parent was doing
Run the invariant backwards. A parent of mass M decaying to masses m₁ and m₂ has, in its own rest frame, completely fixed daughter energies: E₁ = (M² + m₁² − m₂²)c²/(2M). For π⁺ → μ⁺ν that is Eμ = (139.570² + 105.658²)/(2 × 139.570) = 109.78 MeV, so the muon always leaves with 4.12 MeV of kinetic energy and the neutrino with 29.79 MeV — a monoenergetic line, and the reason a two-body decay is such a clean signature. In the lab the parent is moving and both daughter energies are boosted all over the place, but rebuild M from the pair and the same number comes back. That is exactly what an invariant-mass histogram plots: one boost-proof number per event.
A width is a lifetime, read straight off the peak
Nothing unstable has a sharp mass. A state that survives a proper time τ is not an energy eigenstate, and the time–energy relation turns that into a full width at half maximum Γ = ħ/τ, with ħ = 6.582 × 10⁻²² MeV s. So a lifetime far too short to time can be read off a histogram: the Z's Γ = 2.495 GeV gives τ = 2.64 × 10⁻²⁵ s, a flight of 0.08 fm even at the speed of light, and the ρ⁰'s 149 MeV gives 4.4 × 10⁻²⁴ s. The catch is that the detector adds a width of its own. The J/ψ's true Γ is 92.6 keV, meaning τ = 7.1 × 10⁻²¹ s, yet a dimuon peak is typically tens of MeV wide — what is measured there is resolution, not physics. Only when the fitted width clears the resolution does Γτ = ħ return a lifetime, and even then it assumes one isolated, non-overlapping resonance.
Change one variable at a time
Make the relationship visible.
Leave the beam at 46 GeV: the collider reaches 92 GeV, enough for a Z, while the target curve is stuck at 9.4 GeV and would need a 4501 GeV beam to catch up. Now drag the target mass down to 0.1 GeV/c² and watch the curve sag further — a light target is a poor anvil.
CM ENERGY, COLLIDER92.0 GeV
CM ENERGY, FIXED TARGET9.39 GeV
TARGET BEAM TO MATCH4501 GeV
BEAM ENERGY PENALTY98 times
Live interpretationCM ENERGY, COLLIDER: 92.0 GeV. CM ENERGY, FIXED TARGET: 9.39 GeV. TARGET BEAM TO MATCH: 4501 GeV. BEAM ENERGY PENALTY: 98 times
Catch the common trap
Explain before calculating.
An e⁺e⁻ collider runs each beam at 45.6 GeV head-on, so √s = 91.2 GeV, enough to make a Z. A colleague proposes reaching the same √s by firing electrons at a stationary electron target instead. What beam energy would that need? (mec² = 0.511 MeV)
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA tracker measures a charged particle's momentum as p = 1.500 GeV/c and the calorimeter records its total energy as E = 1.579 GeV. Identify the particle from its rest mass, then find its speed and Lorentz factor.
- Mass comes from the invariant, not from p/v: m²c⁴ = E² − p²c² = 1.579² − 1.500² = 2.493241 − 2.250000 = 0.243241 GeV².
- Take the root: mc² = √0.243241 = 0.4932 GeV = 493.2 MeV, which matches the charged kaon at 493.677 MeV.
- Speed follows from β = pc/E = 1.500/1.579 = 0.9500.
- Lorentz factor follows from γ = E/mc² = 1.579/0.4932 = 3.202. Check: γβ = 3.202 × 0.950 = 3.042, and pc/mc² = 1.500/0.4932 = 3.042 ✓.
Answermc² = 493 MeV, so the track is a K±; it is moving at β = 0.950 with γ = 3.20.
MediumA detector records two muon tracks from one vertex, each of momentum 2.000 GeV/c, with an opening angle of 101.0° between them. Take mμc² = 105.66 MeV. Reconstruct the invariant mass of the pair, identify the parent, and say how fast that parent was moving. Then convert the parent's known width Γ = 92.6 keV into a lifetime.
- Each muon energy: E = √(p²c² + m²c⁴) = √(4.000000 + 0.011164) = 2.002789 GeV, so ΣE = 4.005578 GeV.
- Momenta add as vectors, by the cosine rule: |Σp|²c² = p₁²c² + p₂²c² + 2p₁p₂c²cos θ = 4 + 4 + 8cos101.0° = 8 − 1.526 = 6.4735 GeV², so |Σp|c = 2.5444 GeV.
- Invariant mass: M²c⁴ = (ΣE)² − |Σp|²c² = 16.0447 − 6.4735 = 9.5711 GeV², so Mc² = 3.094 GeV — the J/ψ at 3.0969 GeV, low by 0.1%.
- Note what M is not: the muon masses sum to 0.211 GeV and the energies sum to 4.006 GeV, and neither is the mass.
- The parent was in flight: β = |Σp|c/ΣE = 2.5444/4.0056 = 0.635 and γ = ΣE/Mc² = 4.0056/3.094 = 1.295. The boost moved E and p but left M alone.
- Lifetime from the width: τ = ħ/Γ = (6.582 × 10⁻²² MeV s)/(0.0926 MeV) = 7.11 × 10⁻²¹ s. A dimuon peak tens of MeV wide is resolution; this 92.6 keV is the physics.
AnswerMc² = 3.094 GeV, a J/ψ moving at β = 0.635 (γ = 1.30); its 92.6 keV width corresponds to τ = 7.11 × 10⁻²¹ s.
HardAntiprotons were first made in 1955 by firing protons at stationary hydrogen: p + p → p + p + p + p̄, baryon number forcing the antiproton to appear alongside an extra proton. With mpc² = 0.938272 GeV, find the minimum beam kinetic energy on a fixed target, the beam energy a symmetric proton collider would need for the same reaction, and the ratio of the two energy bills.
- At threshold all four final particles are at rest in the centre-of-momentum frame, so √sₘᵢₙ = 4mpc² = 3.753088 GeV and sₘᵢₙ = 14.0857 GeV².
- Evaluate the same s in the lab: s = (Eb + mpc²)² − pb²c² = 2mₚ²c⁴ + 2Eb mpc² = 1.760708 + 1.876544 Eb (in GeV²).
- Equate: 1.876544 Eb = 14.085669 − 1.760708 = 12.324961, so Eb = 6.5679 GeV = 7mpc², and the kinetic energy is T = Eb − mpc² = 6mpc² = 5.630 GeV.
- Collider: Σp = 0, so √s = 2E and each beam needs E = 1.876544 GeV = 2mpc², i.e. T = mpc² = 0.938 GeV per beam, 1.877 GeV of kinetic energy in total.
- Ratio of kinetic energy bills: 6mpc²/2mpc² = 3.00 exactly. The Bevatron was built to 6.2 GeV of kinetic energy with this 5.63 GeV threshold in hand.
- The factor 3 is small only because √s is small. Repeat at √s = 91.2 GeV and the fixed-target beam energy is s/(2mpc²) = 8317/1.8765 = 4432 GeV against 45.6 GeV per beam — a penalty of 97×, and it keeps growing linearly with √s.
AnswerTₘᵢₙ = 6mpc² = 5.63 GeV on a fixed target; 2mpc² = 1.88 GeV of kinetic energy in a collider, a factor of exactly 3 at this threshold and far worse at high √s.