Energy–momentum four-vector

Newton's p = mv fails relativistically for a structural reason: dt is frame-dependent, so m dx/dt divides a four-vector component by somebody's coordinate time and transforms into a mess. Replace dt with the proper time dτ, which every frame agrees on, and the repair is automatic. Define P = m dX/dτ with X = (ct, x). Since m and dτ are invariants and X is a four-vector, P is a four-vector — no assumption, just a four-vector divided by a scalar. Now read the components. With dt/dτ = γ, the time part is m c (dt/dτ) = γmc and the space part is m (dx/dτ) = γmv. Identify γmc² as the total energy and γmv as the momentum and you have P = (E/c, p). Energy and momentum were never two separate conservation laws; they are the four components of one.

Because P transforms like X, no new derivation is needed — substitute ct → E/c and x → pₓ into the standard boost and read off E′ = γ(E − v pₓ) and p′ₓ = γ(pₓ − vE/c²), with p_y and p_z passing through unchanged. Try it on a 1400 MeV proton, whose momentum is 1039.0 MeV/c, seen from a frame moving at v = 0.500c along its direction of travel. Here γ = 1/√0.750 = 1.1547, so E′ = 1.1547(1400 − 0.500 × 1039.0) = 1016.7 MeV and p′c = 1.1547(1039.0 − 0.500 × 1400) = 391.5 MeV. Both fell, and by different factors — that is the point. The transverse rule matters as much as the longitudinal one: a boost along x cannot touch p_y, so a particle's momentum can swing in direction under a boost even though one of its components never changes value.

Every four-vector has a Minkowski square, and for P that square is E² − (pc)². Compute it directly: (γmc²)² − (γmvc)² = γ²m²c⁴(1 − β²) = m²c⁴, because γ²(1 − β²) = 1 by definition. So E² − (pc)² = (mc²)², and mc² is the Minkowski length itself — the mass, written in energy units. Two pictures help. The first is a right triangle with legs pc and mc² and hypotenuse E. The second is what a boost does to it: in the (pc, E) plane the states available to a given particle lie on the hyperbola E = √((pc)² + (mc²)²), and a boost slides the state along that curve. The minus sign in the Minkowski square is why the curve is a hyperbola rather than a circle, and why E can be pushed up without bound while the mass sits still. Check it on the boosted proton: 1016.7² − 391.5² = 8.804 × 10⁵ MeV² = (938.3 MeV)², exactly the lab value.

Dividing the space part of P by the time part kills γ and m together: pc/E = (γmvc)/(γmc²) = v/c, so β = pc/E. The same pair gives γ = E/mc² and the kinetic energy K = E − mc². This is how detectors actually work: a magnetic tracker returns p from the curvature of the track, a calorimeter returns E, and the invariant turns that pair into a mass. Suppose the tracker gives pc = 629.5 MeV and the calorimeter E = 800.0 MeV. Then (mc²)² = 800.0² − 629.5² = 640 000 − 396 270 = 243 730 MeV², so mc² = 493.7 MeV and the track belonged to a charged kaon. The same two numbers finish the description: β = 629.5/800.0 = 0.7869, γ = 800.0/493.7 = 1.620, and K = 306.3 MeV. Nothing was timed and nothing was measured twice. It is also why particle physicists quote masses, momenta and energies all in electronvolts: the invariant relates the three with nothing but factors of c.

Set m = 0 in the invariant and it gives E = pc with β = pc/E = 1 exactly. The four-vector is then null: its Minkowski length is zero although neither E nor p is. There is no rest frame to boost into and γ = E/mc² is undefined, so any argument beginning "in the photon's frame" has already gone wrong. For a massive particle the same branch is a limit rather than a state. Expanding E = pc√(1 + (mc²/pc)²) gives E ≈ pc + (mc²)²/2pc, and equivalently 1 − β ≈ (mc²)²/2E². At E = 5.00 GeV an electron has mc²/E = 1.02 × 10⁻⁴, so 1 − β ≈ 5.2 × 10⁻⁹ and treating it as massless costs almost nothing. A proton at that same 5.00 GeV has mc²/E = 0.1877, giving β = √(1 − 0.1877²) = 0.982 — same energy, nowhere near the limit. "Ultrarelativistic" is a statement about E/mc², never about energy alone.

Four-momenta add, and a sum of four-vectors is a four-vector, so any collection of particles has a total P whose Minkowski square is again an invariant: (Mc²)² = (ΣEᵢ)² − (Σpᵢc)². That M is the system's invariant mass; its energy equivalent Mc² is what collider work calls √s, and it is the total energy the system shows in the centre-of-momentum frame — the frame where Σp = 0, which exists for every system that is not entirely massless and collinear. M is not the sum of the parts' masses. Expanding for two particles gives (Mc²)² = m₁²c⁴ + m₂²c⁴ + 2(E₁E₂ − p₁⋅p₂c²), and those cross terms are never smaller than 2m₁m₂c⁴, so M ≥ m₁ + m₂ with equality only when the two move together. Two 500 MeV photons crossing at 90.0° have (Mc²)² = 2E₁E₂(1 − cos θ) = 2(500)(500)(1) = 500 000 MeV², so M = 707 MeV/c² — built entirely out of two massless things. Reconstructing that number from decay products is how resonances are found.