Work, kinetic energy and rest energy

In Newtonian mechanics you may define K as ½mv² and prove W = ΔK, or start from work and derive K. Relativity closes the first route: F = ma is gone, because along the motion F = γ³ma and across it F = γma, so there is no single effective mass to hide the difference in. What survives is the definition of work itself, W = ∫F⋅dx, with F = dp/dt and p = γmv. Swap the order of the differentials: F dx = (dp/dt) dx = (dx/dt) dp = v dp. So the net work needed to take a particle from rest to speed v is W = ∫₀^p v dp′, where v is itself a function of p through p = γmv. That is the whole model for this topic. Everything below is that one integral — evaluated, read, expanded, and then asked what its constant means.

Integrate by parts: ∫₀^p v dp′ = [v p′]₀^p − ∫₀^v p dv′ = γmv² − ∫₀^v γmv′ dv′. Kill the remaining integral with w = 1 − v′²/c², so dw = −2v′ dv′/c² and ∫₀^v mv′(1 − v′²/c²)^(−1/2) dv′ = −mc²[√w]₀^v = mc²(1 − 1/γ). That leaves W = γmv² − mc² + mc²/γ. Now use 1/γ = γ(1 − β²) to recombine the first and last terms: γmc²β² + γmc²(1 − β²) = γmc². Hence W = γmc² − mc², or K = (γ − 1)mc². Two checks are free. As β → 0, γ − 1 → 0 and K → 0, which any kinetic energy must do. As β → 1, γ → ∞ and K diverges. Nothing at all was assumed about the force that did the work — only that m is fixed and the particle started at rest.

The integral produced K, and K alone. Writing E = K + mc² = γmc² is a choice about where to put the zero of energy, and inside mechanics that choice is free: shift every energy in a closed problem by the same constant and every prediction survives, because only differences are measured. Two later demands force the constant to be mc² rather than 0. First, (E/c, p) transforms as a four-vector only with E = γmc², and the invariant E² − (pc)² = (mc²)² then holds the same value in every frame. Second, energy conservation across a process that changes the system's invariant mass fails outright unless the rest term is carried through. Numerically mc² is enormous — 0.511 MeV for an electron, 938.3 MeV for a proton, 8.99 × 10¹⁶ J for a kilogram — and essentially none of it is available to mechanics, which is exactly why mechanics never noticed it was there.

Expand γ = (1 − β²)^(−1/2) = 1 + ½β² + ⅜β⁴ + (5/16)β⁶ + …, so K = mc²(½β² + ⅜β⁴ + …) = ½mv²(1 + ¾β² + ⅝β⁴ + …). The Newtonian form is not a separate law that relativity corrects; it is the first term of this series, which is why it worked. The leading fractional error is ¾β²: 0.75% at β = 0.10, 1.0% at β = 0.115, about 7% at β = 0.30, and hopeless by β = 0.60, where ½mv² returns 0.180 mc² against a true 0.250 mc² — 28% low. Translate the 1% mark into an energy and it is K = 0.00673 mc², a fixed fraction of the rest energy, so the same speed threshold lands wherever mc² puts it. That is why a 100 keV electron microscope runs at β = 0.548 and is thoroughly relativistic, while a proton accelerated through the same 100 kV reaches only β = 0.0146 and is not.

Because K = (γ − 1)mc² diverges as β → 1, no finite amount of work reaches c. Read it as a price list, in units of mc²: β = 0.5 costs 0.155, β = 0.9 costs 1.294, β = 0.99 costs 6.09, β = 0.999 costs 21.4. Each factor-of-ten cut in the gap (1 − β) costs roughly √10 ≈ 3.16 times more energy, because 1 − β ≈ 1/(2γ²) at high speed. Nothing here forbids applying a force. Under a constant F the momentum p = Ft grows without bound and the speed v = Ft/√(m² + (Ft/c)²) climbs forever toward c without arriving; it is the work, not the push, that runs out. A 50 GeV electron has γ = 9.78 × 10⁴ and 1 − β = 5.2 × 10⁻¹¹, so after all that energy it is still travelling about 1.6 cm s⁻¹ slower than light.

mc² becomes physical the moment the system's invariant mass changes. Bind two protons and two neutrons into a helium-4 nucleus — 3727.379 MeV against 3755.674 MeV for the free four — and 28.30 MeV must leave; weigh the α particle afterwards and it is short by exactly that energy divided by c², 0.75% of what went in. Let an electron and a positron annihilate at rest and 2 × 0.511 MeV departs as two photons with no matter left behind to carry kinetic energy. The traffic runs both ways: heat a 1 kg block by 1 kJ and its invariant mass rises by 1 kJ/c² = 1.11 × 10⁻¹⁴ kg, and a compressed spring is heavier than a relaxed one. Mechanics never sees any of this because it holds m fixed by assumption — which is precisely the assumption K = (γ − 1)mc² was derived under.