Rest energy, kinetic energy & the energy-momentum invariant

Kinetic energy is the work a net force does from rest, K = ∫F dx = ∫(dp/dt) v dt = ∫v dp, and nothing in that definition is Newtonian except the momentum you put in. With p = γmv along the line of motion, dp = γ³m dv, because d(γv)/dv = γ³. So K = ∫₀^v γ³mv dv, and the substitution c² dγ = γ³v dv turns it into mc² ∫₁^γ dγ = (γ − 1)mc². Try it on an electron pushed through 1.00 MV: K = 1.00 MeV and mc² = 0.511 MeV, so γ = 1 + 1.00/0.511 = 2.96 and β = √(1 − 1/γ²) = 0.941. Newton's ½mv² would have given β = √(2 × 1.00/0.511) = 1.98, nearly twice c. Read the integral backwards and dK/dp = v: plotted against momentum, the energy's slope is the velocity. Newton's parabola p²/2m has slope p/m and steepens for ever; the relativistic curve bends over towards slope c, so past about 0.9c more work buys more energy and momentum but almost no more speed.

Every relativistic result must return the classical one when v ≪ c, and this one does through the binomial series. With β = v/c, γ = 1/√(1 − β²) = 1 + ½β² + ⅜β⁴ + …, so K = (γ − 1)mc² = ½mv² + ⅜mv⁴/c² + … = ½mv²(1 + ¾β² + …). At β = 0.10 the correction is ¾ × 0.010 = 0.75%, invisible in most laboratory work; at β = 0.50 the exact factor γ − 1 = 0.1547 against ½β² = 0.125 makes K 24% larger than Newton's; at β = 0.90, γ − 1 = 1.294 against 0.405, a factor of 3.2. Dimensions check as well: γ is a pure number, so (γ − 1)mc² carries the units of mc², kg m² s⁻² = J. The series is the correspondence check in its sharpest form: not only does the relativistic formula tend to Newton's, it tells you the size of the first correction, ¾β², before you have measured anything.

Write E = γmc² and split it as E = K + mc². The second term is there at v = 0, so a body at rest is not at zero energy: it holds mc², and 1.00 kg of anything holds (1.00)(2.998 × 10⁸)² = 8.99 × 10¹⁶ J, roughly the output of a 1 GW power station over three years. Chemistry never noticed because its energies are minute on that scale: burning 1.0 kg of petrol releases 4.6 × 10⁷ J, so the products are lighter than the reactants by 4.6 × 10⁷/(2.998 × 10⁸)² = 5.1 × 10⁻¹⁰ kg, and conservation of mass looked exact to every balance Lavoisier owned. Nuclear binding is where the term shows: a deuteron's mass is 2.224 MeV/c² below that of a free proton plus a free neutron, 0.12% of the total, and that deficit is the energy released when they bind. Particle physics works in the units this invites, quoting masses in MeV/c² so that mc² is read off directly: 0.511 MeV for the electron, 938.3 MeV for the proton, 931.5 MeV for one atomic mass unit.

Square E = γmc² and pc = γmvc and subtract: E² − (pc)² = γ²m²c⁴(1 − v²/c²) = (mc²)², because γ²(1 − β²) = 1. The speed has vanished, and with it every frame-dependent quantity. An observer moving along the particle's line at speed u measures E′ = γᵤ(E − up) and p′c = γᵤ(pc − uE/c), the same transformation that (ct, x) obeys, so E and p both change while E² − (pc)² does not, just as Δt and Δx change under a boost while c²Δt² − Δx² is fixed. Plot E against pc and every observer's (pc, E) for this particle lands on one hyperbola, vertex at mc² and asymptotes E = ±pc; a boost slides the point along it. Read as a right triangle, the same equation has legs pc and mc² and hypotenuse E, so β = pc/E is the cosine of the angle between E and the momentum leg and 1/γ = mc²/E is its sine. At p = 0 the triangle collapses to E = mc²; as m → 0 it flattens to E = pc with β = 1, and a photon needs nothing more.

Accelerator units make the invariant an arithmetic identity: quote E and pc in MeV and m in MeV/c², and E² = (pc)² + (mc²)² uses the same numbers with no powers of ten. A proton spectrometer returns pc = 800 MeV. Then E = √(800² + 938.3²) = √(1 520 407) = 1233 MeV, K = E − mc² = 295 MeV, β = 800/1233 = 0.649 and γ = 1233/938.3 = 1.314: four results and not one square root of 1 − β². Now let a second observer ride along the beam at u = 0.500c, so γᵤ = 1.155: E′ = 1.155 × (1233 − 0.500 × 800) = 962 MeV and p′c = 1.155 × (800 − 0.500 × 1233) = 212 MeV. Different energy, different momentum, and √(962² − 212²) = 938 MeV, the proton mass again. Newton would have read K = p²/2m = 800²/(2 × 938.3) = 341 MeV from the same spectrometer, 16% too high; at pc = 2500 MeV his estimate would be nearly double the true value.

Add the energies and momenta of several particles in one frame and form M²c⁴ = (ΣE)² − |Σp|²c². ΣE and Σp transform together exactly as E and p do, so M is the same in every frame, and since both sums are conserved in a reaction, M is conserved too. It is not the sum of the rest masses. Two 1.0 MeV photons flying apart back to back have m = 0 each, Σp = 0 and ΣE = 2.0 MeV, so the pair has M = 2.0 MeV/c²; the same two photons flying side by side have ΣE = |Σp|c and M = 0. A Σ⁺ hyperon (1189.4 MeV) at rest decays to a proton (938.3 MeV) and a π⁰ (135.0 MeV): the system's M stays 1189.4 MeV/c², while the products' rest masses sum to 1073.3 MeV/c². The missing Q = 116.1 MeV is their kinetic energy. The two-body formula gives Eₚ = (1189.4² + 938.3² − 135.0²)/(2 × 1189.4) = 957.1 MeV, so Kₚ = 18.8 MeV and the pion carries 97.3 MeV: at equal momentum the lighter particle always has the larger K. Nothing was destroyed: the system's M is unchanged, and 116.1 MeV of energy moved from the rest-energy column to the kinetic one.