University Physics IV · Special Relativity II · 3.7

Work, Kinetic Energy & Rest Energy

Push a particle harder and harder and it never reaches c — so where does the work go? Here is the integral that answers it, the K = (γ − 1)mc² it lands on, the constant of integration relativity pins at mc² rather than zero, and the exact speed at which the ½mv² you have used since first year stops paying its way.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Newtonian kinetic energy is not a definition you can carry into relativity; it is the output of an integral, and once momentum is p = γmv that integral returns something else. Start from the only survivor of the classical statement — net work is ∫F⋅dx with F = dp/dt — change the variable to get W = ∫v dp, and integrate by parts. What falls out is K = (γ − 1)mc², a quantity that agrees with ½mv² to within a fraction ¾β² at low speed and diverges as v → c.

That divergence is the entire content of the light barrier: nothing forbids applying the force, it is that the work needed to reach c is infinite, so c is a ceiling priced in energy rather than a rule posted about speed. The integration also leaves a constant, and relativity fixes it at mc² rather than zero, giving E = γmc² with a rest energy that does not vanish when the particle stops. Inside mechanics that choice costs nothing and buys nothing: add the same constant to every energy in a closed problem and every prediction is unchanged.

The bill comes due only in processes where the invariant mass of the system changes — binding, decay, annihilation — where the missing mc² reappears as kinetic energy or photons, and where writing the constant as zero would break conservation of energy outright.

Simple definition: The relativistic kinetic energy of a particle is the net work needed to take it from rest to speed v, K = (γ − 1)mc², and its total energy E = γmc² is that work plus a rest energy mc² the particle carries even at v = 0.
Example: A proton (mc² = 938.3 MeV) at β = 0.800 has γ = 1/√(1 − 0.640) = 1.6667, so K = (γ − 1)mc² = 625.5 MeV and E = γmc² = 1563.8 MeV, while the Newtonian ½mv² = ½(0.640)mc² claims only 300.3 MeV — low by a factor 2.08.

Net work, rewrittenW = ∫F⋅dx = ∫(dp/dt)dx = ∫v dp, with p = γmv

The one classical statement relativity keeps. F = ma is gone: along the motion F = γ³ma, across it γma.

F in N, p in kg m s⁻¹, v in m s⁻¹; one particle of fixed invariant mass m, taken from rest.

Relativistic kinetic energyK = (γ − 1)mc², γ = (1 − v²/c²)^(−1/2)

The value of the work integral. It vanishes at v = 0 and diverges as v → c, which is what puts c out of reach.

m in kg and invariant, c = 2.998 × 10⁸ m s⁻¹, K in J; or put mc² in MeV and K comes out in MeV.

Total energy and rest energyE = K + mc² = γmc²

Splits the energy into a part motion can change and a part it cannot. γ = E/mc² reads straight off a stated beam energy.

mc² is 0.511 MeV for the electron, 938.3 MeV for the proton and 1875.6 MeV for the deuteron; K, E and mc² must share one unit.

Speed from an energyγ = 1 + K/mc², β = √(1 − 1/γ²)

Turns an accelerator's quoted energy into a speed without ever solving for v inside a square root of itself.

K and mc² in the same unit; β = v/c is dimensionless. Exact at every speed, not an approximation.

Low-speed expansionK = ½mv²(1 + ¾β² + ⅝β⁴ + …), β = v/c

Says precisely when ½mv² is good enough, and shows the Newtonian form is the first term of this series, not a rival law.

β is dimensionless, so the bracket is a pure number and ½mv² carries the units. Every term is positive, so any truncation underestimates K.

Energy released by a change of rest massQ = (Σmᵢₙᵢₜᵢₐₗ − Σm_final)c²

The only place mc² stops being an arbitrary additive constant. Zero for a particle merely being pushed; nonzero for binding, decay, annihilation.

1 u × c² = 931.494 MeV; 1 kg × c² = 8.988 × 10¹⁶ J. These are invariant masses of the whole system.

Only ∫F⋅dx survives, and it turns into ∫v dp

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.

Doing the integral: by parts, then one substitution

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 constant of integration is where mc² enters

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.

Where ½mv² gives out: the ¾β² correction

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.

What the approach to c actually costs

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.

When the rest term stops being bookkeeping

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.

Change one variable at a time

Make the relationship visible.

Interactive model

Speed β = v/c0.60

Series terms kept1

Leave the series at one term — that is exactly ½mv² — and walk β up from 0.10: the gap opens at 0.75%, reaches 28% by β = 0.60 and 88% by 0.98. Add the ⅜β⁴ term and the low-β end is rescued while the right edge still gives way, because no polynomial can follow a curve that diverges at β = 1.

LORENTZ FACTOR γ1.250

K / mc² EXACT0.250

K / mc² SERIES0.180

SERIES SHORTFALL28.0 %

Live interpretationLORENTZ FACTOR γ: 1.250. K / mc² EXACT: 0.250. K / mc² SERIES: 0.180. SERIES SHORTFALL: 28.0 %

Catch the common trap

Explain before calculating.

Misconception check

The commonest error is to keep the Newtonian shape and staple γ onto it: K = ½γmv². It has the right low-speed limit, so it survives a spot check, and it is wrong everywhere else. The reason is visible in the derivation. W = ∫v dp integrated by parts is vp − ∫p dv, and ½γmv² is the vp term alone with an unexplained ½ in front; the −∫p dv piece, which supplies mc²(1 − 1/γ), has simply been dropped. At β = 0.600 it returns 0.225 mc² against the true 0.250 mc², and as β → 1 it tends to half the total energy instead of all of it minus mc². The second error travels with the first: E = γmc² is not a mass that grows with speed. m is the invariant mass and is fixed; γ belongs to the motion, not to the particle, and no frame ever measures a different m.

A particle of invariant mass m moves at β = 0.600, where γ = 1.250. What is its kinetic energy?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyAn electron (mc² = 0.511 MeV) is accelerated from rest through a potential difference of 1.50 MV. Find its kinetic energy, γ, total energy and speed, then show what ½mv² would have predicted for that speed.

All the work goes into kinetic energy, and no rest mass changes, so K = eV = 1.50 MeV with K = (γ − 1)mc².
γ = 1 + K/mc² = 1 + 1.50/0.511 = 1 + 2.935 = 3.935.
E = γmc² = K + mc² = 1.50 + 0.511 = 2.011 MeV.
β = √(1 − 1/γ²) = √(1 − 1/15.49) = √(1 − 0.06457) = √0.93543 = 0.9672, so v = 0.9672 × 2.998 × 10⁸ = 2.90 × 10⁸ m s⁻¹.
Newtonian check: ½mv² = K gives β = √(2K/mc²) = √(2 × 2.935) = √5.871 = 2.42. The classical formula does not merely lose accuracy here — it returns an impossible speed, 2.42c.

AnswerK = 1.50 MeV, γ = 3.94, E = 2.01 MeV, β = 0.967 so v = 2.90 × 10⁸ m s⁻¹. Newtonian ½mv² would have given v = 2.42c.

MediumFind the speed at which the true kinetic energy exceeds ½mv² by 1.0% of ½mv². Verify it against the exact formula, then state the kinetic energy that speed corresponds to for an electron (mc² = 0.511 MeV) and for a proton (mc² = 938.3 MeV).

Expand: K = (γ − 1)mc² = mc²(½β² + ⅜β⁴ + …) = ½mv²(1 + ¾β² + …), so (K − ½mv²)/(½mv²) = ¾β² to leading order.
Set ¾β² = 0.010: β² = 0.013333, β = 0.1155, v = 0.1155 × 2.998 × 10⁸ = 3.46 × 10⁷ m s⁻¹.
Exact check: γ = 1/√(1 − 0.013333) = 1/√0.986667 = 1.0067341, so γ − 1 = 0.0067341 while ½β² = 0.0066667. Their ratio is 1.01011, i.e. 1.011% — the ¾β² estimate was good to two figures, and better still at smaller β.
Convert to energies: electron K = 0.0067341 × 0.511 MeV = 3.44 keV; proton K = 0.0067341 × 938.3 MeV = 6.32 MeV.
The threshold is a speed, so mc² alone decides where it falls in energy: the two answers differ by 6.32 MeV / 3.44 keV = 1836, which is exactly the mass ratio 938.3/0.511.

Answerβ = 0.1155, v = 3.46 × 10⁷ m s⁻¹ (exact excess 1.011%). That is K = 3.44 keV for an electron and K = 6.32 MeV for a proton.

HardA free proton (mc² = 938.272 MeV) and a free neutron (939.565 MeV), both essentially at rest, bind into a deuteron (1875.613 MeV). (a) Find the energy released and its fraction of the initial rest energy. (b) Find the energy per mole of deuterons formed. (c) Give one deuteron that same energy as kinetic energy instead, and find its speed.

(a) Initial rest energy is 938.272 + 939.565 = 1877.837 MeV; the final state has 1875.613 MeV. The invariant mass has dropped, so Q = 1877.837 − 1875.613 = 2.224 MeV, carried off by a photon.
As a fraction, 2.224/1877.837 = 1.184 × 10⁻³: the deuteron is permanently 0.118% lighter than its parts, and that deficit is what a mass spectrometer weighs.
(b) Per mole: 2.224 MeV × 6.022 × 10²³ = 1.339 × 10²⁴ MeV, and 1 MeV = 1.602 × 10⁻¹³ J, so Q = 2.15 × 10¹¹ J mol⁻¹ — about 2.4 × 10⁵ times the roughly 9 × 10⁵ J released when a mole of methane burns.
(c) Now hand a single deuteron 2.224 MeV of kinetic energy: γ − 1 = 2.224/1875.613 = 1.1857 × 10⁻³, so γ = 1.0011857.
β = √(1 − 1/γ²) = √(1 − 0.997633) = √(2.3673 × 10⁻³) = 0.04865, giving v = 1.46 × 10⁷ m s⁻¹. Newtonian ½mv² would have given β = √(2 × 1.1857 × 10⁻³) = 0.04870, high by 0.089% — half of the ¾β² = 0.18% energy correction, as it should be.
The same 2.224 MeV therefore appears either as a permanent 0.118% deficit in invariant mass or as motion at 4.9% of c. Mechanics holds m fixed and can never touch the first; that is exactly why mc² may be dropped there and never here.

AnswerQ = 2.224 MeV, 0.118% of the initial rest energy; 2.15 × 10¹¹ J per mole. As kinetic energy the same 2.224 MeV gives only β = 0.0487, v = 1.46 × 10⁷ m s⁻¹.