Special Relativity

Relativistic Energy and Momentum

The total energy of a moving body — E = γmc² — and the invariant that ties energy to momentum

Relativistic energy is the total energy of a moving body, E = γmc², where γ = 1/√(1 − v²/c²) is the Lorentz factor. It splits into a rest energy E₀ = mc² that a body carries even at rest and a kinetic part (γ − 1)mc². Energy and momentum p = γmv obey the invariant relation E² = (pc)² + (mc²)², which defines the invariant mass m, reduces to E = pc for massless particles, and unites into the four-momentum. Einstein published mc² in 1905; the full relation is the backbone of nuclear physics and particle accelerators.

  • Total energyE = γmc²
  • Rest energyE₀ = mc²
  • Kinetic energyK = (γ − 1)mc²
  • Energy-momentumE² = (pc)² + (mc²)²
  • Massless limitE = pc (m = 0)
  • Speed of light c2.998 × 10⁸ m/s

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Why relativistic energy matters

Newtonian mechanics says kinetic energy is ½mv² and momentum is mv, and that you can keep adding energy to keep speeding up without limit. Both statements fail badly near the speed of light. Special relativity replaces them with a single, self-consistent picture in which energy and momentum are two faces of one four-dimensional object, and in which mass itself is a form of energy.

The consequences are enormous and everyday-real. The Sun shines because fusing four protons into a helium-4 nucleus leaves the products about 0.7% lighter, and that missing mass Δm becomes radiant energy via E = Δmc². Particle accelerators like the LHC push protons to γ ≈ 7460 (7 TeV each), so almost all their energy is kinetic rather than rest. Your GPS receiver, PET scan, and every nuclear reactor are engineered around these equations. Get a factor of c wrong and the physics is off by 17 orders of magnitude.

The governing equations

Everything follows from the Lorentz factor and two definitions. The Lorentz factor is

γ = 1 / √(1 − v²/c²) = 1 / √(1 − β²),    β ≡ v/c

The total energy and the relativistic momentum are

E = γ m c²          (total energy)
p = γ m v           (relativistic momentum, a vector)

Eliminating v and γ between these two gives the frame-independent energy-momentum relation:

E² = (p c)² + (m c²)²

Symbols and SI units:

  • E — total energy, in joules (J). 1 electronvolt (eV) = 1.602 × 10⁻¹⁹ J.
  • E₀ = mc² — rest energy, the value of E when v = 0.
  • m — invariant (rest) mass, in kilograms (kg). It does not change with speed.
  • v — speed of the body, in m/s; p is its momentum in kg·m/s.
  • c — speed of light in vacuum, exactly 299,792,458 m/s ≈ 2.998 × 10⁸ m/s.
  • γ — Lorentz factor, dimensionless, ranging from 1 (rest) to ∞ (as v → c).
  • β = v/c — speed as a fraction of c, dimensionless.

The relativistic kinetic energy is the total minus the rest energy:

K = E − E₀ = (γ − 1) m c²

For small speeds (β ≪ 1) expand γ ≈ 1 + ½β², and K → ½mv² — Newton's kinetic energy is the low-speed limit, exactly as it must be.

How it works, step by step

  1. Start at rest. With v = 0, γ = 1, so E = mc² and p = 0. Even a motionless body has energy locked in its mass. For an electron, E₀ = m_e c² = 0.511 MeV; for a proton, 938.3 MeV.
  2. Give it speed. As v grows, γ grows and E = γmc² climbs above the rest energy. The surplus (γ − 1)mc² is kinetic energy — the work you did on it.
  3. Track the momentum. p = γmv also grows, but faster than Newton predicts because of the γ. Near c, tiny gains in v produce huge gains in p and E.
  4. Hit the invariant. No matter the frame, E² − (pc)² always equals (mc²)². Observers who disagree about E and p agree exactly on the rest mass.
  5. Approach c. √(1 − β²) → 0, so γ → ∞ and E → ∞. Infinite energy would be needed to reach c, so massive particles never get there.
  6. Take m → 0. The rest term vanishes and E = pc. Massless particles (photons, gluons) must travel at exactly c and carry momentum p = E/c despite having no mass.

Four-momentum: energy and momentum unified

Relativity packages E and p into a single four-vector, the four-momentum:

pᵘ = (E/c, pₓ, p_y, p_z) = m (γc, γvₓ, γv_y, γv_z)

Its invariant square (using the (+,−,−,−) metric) is

pᵘpᵤ = (E/c)² − |p|² = (mc)²

which is the energy-momentum relation again. Under a Lorentz boost, E and p transform into each other exactly the way time t and position x do — energy is to time what momentum is to space. In a collision, conservation of four-momentum encodes conservation of energy (time component) and conservation of momentum (spatial components) in one covariant statement, and it holds in every inertial frame automatically.

For a system of particles you add the four-momenta component by component first, then take the magnitude. That system magnitude is the invariant mass. Two 0.5 GeV photons flying in opposite directions each have m = 0, yet the pair has invariant mass 1 GeV/c² — this is exactly how detectors reconstruct particles like the Higgs boson from their decay products.

Key values: how γ, E, and K scale with speed

Speed v (as β = v/c)Lorentz factor γTotal E / mc²Kinetic K / mc²
0 (rest)1.0001.0000.000
0.10c1.0051.0050.005
0.50c1.1551.1550.155
0.90c2.2942.2941.294
0.99c7.0897.0896.089
0.999c22.3722.3721.37
0.99999c223.6223.6222.6

Notice the runaway: going from 0.99c to 0.99999c is a tiny change in speed but a 30-fold jump in energy. This is why the last few percent toward c cost essentially unbounded energy.

Rest energies of common particles

ParticleRest energy E₀ = mc²Rest mass m
Photon (massless)00
Electron (e⁻)0.511 MeV9.109 × 10⁻³¹ kg
Muon (μ⁻)105.7 MeV1.884 × 10⁻²⁸ kg
Proton (p)938.3 MeV1.673 × 10⁻²⁷ kg
Neutron (n)939.6 MeV1.675 × 10⁻²⁷ kg
W boson80.4 GeV1.43 × 10⁻²⁵ kg
Z boson91.19 GeV1.63 × 10⁻²⁵ kg
Higgs boson125.25 GeV2.23 × 10⁻²⁵ kg

Worked example: a proton at the LHC

The LHC accelerates protons to a total energy of E = 7 TeV = 7 × 10¹² eV. The proton rest energy is E₀ = 938.3 MeV ≈ 0.9383 × 10⁹ eV. So the Lorentz factor is

γ = E / E₀ = 7 × 10¹² / 0.9383 × 10⁹ ≈ 7460

Inverting γ = 1/√(1 − β²) gives β = √(1 − 1/γ²) ≈ 1 − 1/(2γ²) ≈ 1 − 9 × 10⁻⁹. The proton moves at 99.9999991% of c — just 2.7 m/s slower than light. Its momentum is

p = √(E² − (mc²)²) / c ≈ E/c   (since E ≫ mc²)
p ≈ 7 TeV/c = 3.7 × 10⁻¹⁵ kg·m/s

because at ultra-relativistic energies the rest term is negligible and E ≈ pc, just like a massless particle. Almost all of that 7 TeV is kinetic energy: K = (γ − 1)mc² ≈ 6999 GeV.

A short history

In September 1905 Einstein published the three-page paper "Does the Inertia of a Body Depend Upon Its Energy Content?" concluding that a body radiating energy L loses mass L/c² — the seed of E = mc². Max Planck and others refined the argument, and by 1907–1908 Hermann Minkowski had recast relativity in four-dimensional spacetime, making four-momentum natural. The full relation E² = (pc)² + (mc²)² became the springboard for Paul Dirac's 1928 relativistic electron equation, which "square-rooted" it and predicted antimatter. Cockcroft and Walton gave the first direct laboratory confirmation of mass-energy equivalence in 1932 by splitting lithium-7 with protons and balancing the energy books to within experimental error.

Common misconceptions

  • "Mass increases with speed." The outdated "relativistic mass" γm has been retired. Modern physics keeps mass invariant; it is the energy and momentum that grow via γ. Say "the energy increases," not "the mass increases."
  • "E = mc² applies to moving objects." E = mc² is only the rest energy. A moving body has E = γmc². Using mc² alone for a fast particle omits its kinetic energy.
  • "Photons have no momentum because p = mv and m = 0." p = mv is Newtonian. The relativistic momentum is the spatial part of the four-momentum; for m = 0 it gives p = E/c ≠ 0. Photons carry momentum, as radiation pressure and solar sails demonstrate.
  • "Nothing can move faster than light because of energy." Partly right: massive particles can't reach c because E → ∞. But the deeper reason is causal — the invariant structure of spacetime forbids it regardless of energy budget.
  • "Kinetic energy is ½mv² in relativity." Only as a low-speed approximation. The exact form is (γ − 1)mc², which diverges as v → c whereas ½mv² stays finite and wrong.
  • "Invariant mass is just the sum of the parts' masses." No — you sum four-momenta and then take the magnitude. Binding energy and kinetic energy contribute, so a bound system can weigh less (nuclei) or a massless pair can weigh more (two photons) than a naive sum suggests.

Frequently asked questions

What is the difference between E = mc² and E = γmc²?

E = mc² is the rest energy E₀ of a body at rest (v = 0, so γ = 1). E = γmc² is the total energy of the same body moving at speed v, where γ = 1/√(1 − v²/c²) is the Lorentz factor. The two agree only when the object is at rest. The extra energy γmc² − mc² = (γ − 1)mc² is the relativistic kinetic energy. Here m is the invariant (rest) mass — modern usage does not treat mass as increasing with speed.

Why does energy go to infinity as an object approaches the speed of light?

The total energy is E = γmc² with γ = 1/√(1 − v²/c²). As v → c, the denominator √(1 − v²/c²) → 0, so γ → ∞ and E → ∞. Reaching c would require infinite energy, which is why no massive particle can ever be accelerated to c. At v = 0.99c, γ ≈ 7.09; at v = 0.999c, γ ≈ 22.4; at v = 0.99999c, γ ≈ 223.6. Each extra digit of nines buys far less speed for far more energy.

What is the energy-momentum relation E² = (pc)² + (mc²)²?

It is the relativistic dispersion relation linking total energy E, momentum magnitude p, and rest mass m: E² = (pc)² + (mc²)². The right-hand quantity mc² is a Lorentz invariant — every observer measures the same rest mass even though they disagree on E and p. It reduces to E = mc² when p = 0 (rest) and to E = pc when m = 0 (massless). Squaring the four-momentum p^μp_μ = (mc)² is the covariant statement of exactly this relation.

How can a photon have momentum if it has no mass?

For a massless particle, m = 0, so E² = (pc)² + 0 gives E = pc, and momentum p = E/c. A photon of energy E = hf therefore carries momentum p = hf/c = h/λ. Newtonian p = mv fails here because it was never the fundamental definition — momentum is the spatial part of the four-momentum, p = γmv, which stays finite as m → 0 while v = c. Radiation pressure and Compton scattering are direct experimental confirmations.

What is invariant mass and why is it 'invariant'?

Invariant (rest) mass m is defined by m²c⁴ = E² − (pc)². Because E² − (pc)² is the squared magnitude of the energy-momentum four-vector, it has the same value in every inertial frame — it is Lorentz-invariant. For a system of particles you sum the four-momenta first, then take the magnitude: the invariant mass of two photons flying apart is nonzero even though each photon is massless. This is the quantity particle physicists reconstruct to find resonances like the Z boson at 91.19 GeV.

What is four-momentum?

Four-momentum is the four-vector pᵘ = (E/c, p_x, p_y, p_z) that unites energy and momentum. Its time component is energy (over c) and its spatial components are the relativistic momentum p = γmv. Under a Lorentz transformation, E and p mix exactly as time and space do. Conservation of four-momentum in a collision packs energy conservation and momentum conservation into a single covariant law, and its invariant square gives the rest mass: p^μp_μ = (mc)².

How much energy is stored in one kilogram of rest mass?

E₀ = mc² = 1 kg × (2.998 × 10⁸ m/s)² ≈ 8.99 × 10¹⁶ J — about 90 petajoules, or roughly 21.5 megatons of TNT. Everyday reactions release only a tiny fraction of this: burning gasoline converts about 10⁻¹⁰ of the mass, fission about 0.1%, and hydrogen fusion in the Sun about 0.7%. Only matter-antimatter annihilation converts 100% of rest mass to energy.