Relativistic kinematics and invariant mass

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, p_y, p_z). 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 = 3.52142 − 2.64106 = 0.88036 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.

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 6.00 GeV/c² 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.

The working method is a single line: s = (Σp)²c² 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 = 4m_pc² = 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 = (E_b + m_pc²)² − p_b²c² = 2mₚ²c⁴ + 2E_b m_pc². Setting the two equal gives E_b = 7m_pc² = 6.568 GeV immediately. Nothing was Lorentz-transformed; the invariant did the transforming for you.

That same lab expression is the economics of accelerators. For E_b ≫ m_tc² the mass terms drop and s ≈ 2m_tc²E_b, so √s ≈ √(2m_tc²E_b): 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. In Run 2 each LHC beam carried 6500 GeV; fired at a stationary proton it would have delivered √s = √(2 × 0.938 × 6500) = 110 GeV, against 13 000 GeV when the two beams were 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 stored. Fixed targets survive only where high rate, a well-understood stationary target, or beam intensity matters more than √s.

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 E_μ = (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.

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.