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Derivation

The Energy-Momentum Four-Vector

Statement

The energy E and three-momentum p of a particle of rest mass m are the timelike and spacelike parts of a single four-vector pμ = (E/c, p), whose Minkowski norm is a Lorentz invariant fixed by the mass. Evaluating that invariant in two frames — the particle's rest frame and a general frame — yields the exact dispersion relation E2 = (pc)2 + (mc2)2.

Why it matters

This single relation replaces the two separate Newtonian formulas for kinetic energy and momentum with one exact, frame-independent statement valid at every speed up to c. It is the equation you actually use in particle physics: to convert between a measured momentum and a measured energy, to compute reaction thresholds, and to describe massless particles, for which it collapses to E = pc.

More deeply, it demonstrates that energy and momentum are not independent quantities that merely happen to be conserved — they are components of one geometric object. Conservation of the four-momentum in every frame is then a single covariant law, and the invariance of its norm is the physical content of the particle's mass.

Assumptions
Spacetime is flat (special relativity).In curved spacetime there is no global frame in which to add four-vectors at different events; energy–momentum becomes a local quantity and the global conservation law is lost. The particle has a definite, frame-independent rest mass m ≥ 0.If m were frame-dependent the norm would not be invariant and the relation would not hold; the whole construction rests on pμpμ being the same number in all frames. The four-momentum transforms as a genuine Lorentz four-vector.If (E/c, p) did not transform under the same Lorentz matrix as (ct, x), its "norm" would be frame-dependent and carry no invariant meaning; this property is what earns m2c2 the status of an invariant. The metric signature is fixed (here +−−−).The sign of the invariant tracks the convention; with −+++ the same physics gives pμpμ = −m2c2. The measurable relation is unchanged, but every intermediate sign must be kept consistent.
Derivation
1
uμ = dxμ/dτ = γ(c, v)
The four-velocity is the derivative of the worldline with respect to proper time; dt/dτ = γ supplies the factor (prior result: proper time and four-velocity). A
2
pμ ≡ m uμ = (γmc, γmv)
Multiply the four-velocity by the invariant scalar m. A scalar times a four-vector is a four-vector, so pμ transforms correctly. A
3
p = γmv,   E = γmc2
Identify the spatial part with the relativistic momentum (prior result: momentum from conservation) and the time part with total energy via E = γmc2 (prior result: mass–energy equivalence). A
4
pμ = (E/c, p)
Rewrite the time component γmc = γmc2/c = E/c. This is pure substitution using step 3. A
5
pμpμ = ημνpμpν = (E/c)2 − |p|2
Contract with the Minkowski metric ημν = diag(+1,−1,−1,−1). The contraction of a four-vector with itself is a Lorentz invariant. B
6
pμpμ = m2(uμuμ) = m2γ2(c2 − |v|2)
Same invariant computed from the definition pμ = m uμ of step 2; expand the four-velocity norm. Because both sides are the same scalar, the two expressions are equal in every frame. B
7
γ2(c2 − v2) = c2(c2 − v2)/(c2 − v2) = c2
Insert γ2 = c2/(c2 − v2) and cancel. Hence uμuμ = c2 identically — the four-velocity has fixed norm. B
8
pμpμ = m2c2
Substitute step 7 into step 6. This is the invariant, most cleanly read as the value in the rest frame where p = 0 and E = mc2. A
9
(E/c)2 − |p|2 = m2c2
Equate the frame-explicit form (step 5) with the invariant value (step 8): the same scalar written two ways. A
10
E2 = (pc)2 + (mc2)2
Multiply through by c2 and rearrange — algebra only, no new physics. A
Result
E2 = (pc)2 + (mc2)2

Reading. Total energy, momentum, and rest mass sit at the three sides of a right triangle: E is the hypotenuse, pc the "motion" leg, and mc2 the "rest" leg. The rest energy is the momentum-independent floor: at p = 0, E = mc2. Nothing here refers to a frame — m is the invariant length of the four-momentum, so all observers agree on it even while disagreeing on E and p separately.

Units check. [pc] = (kg·m·s−1)(m·s−1) = kg·m2·s−2 = J; [mc2] = kg·m2·s−2 = J. Both legs, and therefore E, carry units of joules; squaring gives J2 throughout, so the equation is dimensionally homogeneous.

Limiting cases
  • Massless particle (m = 0): E = pc exactly — photons and (to excellent approximation) ultra-relativistic particles. Energy and momentum magnitude are locked together.
  • Rest (p = 0): E = mc2, recovering mass–energy equivalence as the special case of a stationary particle.
  • Non-relativistic (pc « mc2): expand E = mc2√(1 + (p/mc)2) ≈ mc2 + p2/2m, restoring rest energy plus the Newtonian kinetic term.
  • Ultra-relativistic (pc » mc2): E ≈ pc + (mc2)2/2pc, so mass survives only as a vanishing correction.
Breaks when
  • Spacetime is curved. In a gravitational field there is no single inertial frame spanning the region, so distant four-momenta cannot be added and the global norm is not defined; one must use the local metric gμν and covariant derivatives. The relation holds only locally, in a freely-falling frame.
  • The "particle" is not a single free quantum with definite mass. For a virtual particle in a Feynman diagram E2 − (pc)2 ≠ (mc2)2 — it is "off shell." Likewise an unstable resonance has a mass spread Γ, so the norm is smeared rather than sharp.
  • Interaction or medium effects dress the dispersion. A photon in a dielectric, or a quasiparticle in a solid, obeys an effective E(p) set by the medium, not the vacuum relation; the invariant-mass form is a vacuum, free-particle statement.
Failure modes
  • Adding the legs linearly: writing E = pc + mc2. It is the squares that add; the linear sum overestimates E at every finite momentum.
  • Using p = mv: the momentum here is γmv. Substituting the Newtonian momentum into E2 = (pc)2 + (mc2)2 gives the wrong energy and breaks consistency with E = γmc2.
  • Treating "relativistic mass" as the m in the formula: the m is the invariant rest mass. Plugging in γm double-counts the motion already carried by p.
  • Confusing total energy with kinetic energy: E includes rest energy. The kinetic energy is K = E − mc2, not E itself.
  • Sign slips from the metric: writing (E/c)2 + |p|2 for the norm. The spatial part enters with a minus sign; getting it wrong turns the invariant into something frame-dependent.
  • Applying it per-particle to a system's invariant mass: for a collection, the invariant mass comes from the total four-momentum M2c2 = (Σpμ)2, which is not the sum of individual masses.
Discussion

The construction is a template that recurs throughout relativistic physics: take two four-vectors, contract them with the metric, and the result is a number every observer agrees on. Here the two four-vectors are the same one, pμ, and its self-contraction is m2c2. Mass is thereby demoted from a "quantity of matter" to a purely geometric attribute — the Minkowski length of the energy–momentum vector. Two observers in relative motion assign different E and different p to the same electron, yet compute the identical m, exactly as two people using rotated axes disagree on a vector's components but agree on its length.

Because pμ is a four-vector, conservation of energy and conservation of momentum are not two laws but one: in a collision Σpμ is conserved, and this single statement holds in every frame simultaneously. This is why the "invariant mass" of a set of decay products — the norm of their summed four-momentum — reconstructs the mass of the parent particle, and it is the working principle behind every peak in a collider's invariant-mass spectrum, from the J/ψ to the Higgs boson.

The massless limit deserves emphasis. Setting m = 0 gives E = pc with no rest frame at all: a photon moves at c for every observer, and γ is undefined, so pμ = m uμ fails as a definition. Yet E2 = (pc)2 + (mc2)2 remains valid because it is a statement about the four-vector's norm, not about how the four-vector was built. The relation is thus more fundamental than the γm route used to derive it.

At the level of field theory the relation is the on-shell condition. The relativistic wave equations enforce it as an operator identity: substituting E → iħ∂t and p → −iħ∇ into E2 = (pc)2 + (mc2)2 yields the Klein–Gordon equation (□ + (mc/ħ)2)φ = 0, and Dirac's linearization of the same constraint forces the introduction of spin and antimatter. In the path integral, only on-shell external legs satisfy the relation; internal propagators are deliberately off-shell, carrying four-momenta with pμpμ ≠ m2c2, which is precisely what lets a virtual particle mediate a force.

Common misconceptions. "Energy has mass, so heavier when moving" confuses total energy with invariant mass — the rest mass m in this formula does not change with speed; only E and p do. And "E = mc2 is the full story" omits the momentum leg: that famous equation is only the p = 0 corner of the triangle.

Worked examples

Example 1 — Energy of a relativistic proton from its momentum. A proton (mc2 = 938.3 MeV) is measured with momentum p = 1.00 GeV/c. Find its total energy and kinetic energy.

1
E = √[(pc)2 + (mc2)2]
Solve the result for E, taking the positive root (a physical particle has E > 0). A
2
pc = 1.00 GeV = 1000 MeV,  mc2 = 938.3 MeV
Express both legs in the same energy unit (MeV) so the sum is meaningful. Using GeV/c for p makes pc come out directly in GeV. A
3
E = √(10002 + 938.32) = √(1.000×106 + 8.804×105) MeV
Square and add the two legs. A
4
E = √(1.8804×106) MeV = 1371 MeV
Take the root. A
5
K = E − mc2 = 1371 − 938.3 = 433 MeV
Kinetic energy is total minus rest energy. A
E ≈ 1371 MeV = 1.371 GeV,   K ≈ 433 MeV

Reading. Even with pc exceeding the rest energy, the total energy is well below the naive linear sum (1938 MeV): the Pythagorean combination is always smaller than the arithmetic sum.

Units check. Every term carried MeV; the square root of MeV2 returns MeV. Consistent.

Example 2 — Invariant mass of two photons (a π0 decay). A neutral pion decays to two photons. In the lab each photon has energy E1 = E2 = 100 MeV, emerging with an opening angle θ = 90° between them. Reconstruct the parent's invariant mass.

1
M2c2 = (p1μ + p2μ)2 = p12 + p22 + 2p1·p2
The invariant mass of a system is the norm of the total four-momentum; expand the square with the metric. B
2
p12 = p22 = 0
Each photon is massless, so each four-momentum has zero norm (on-shell, m = 0). A
3
p1·p2 = E1E2/c2 − p1·p2
Minkowski dot product of the two four-momenta, using piμ = (Ei/c, pi) and metric +−−−. B
4
|pi| = Ei/c,  p1·p2 = (E1E2/c2)cosθ
For massless quanta momentum magnitude equals E/c; the spatial dot product carries the opening angle. A
5
M2c4 = 2E1E2(1 − cosθ)
Combine steps 1–4 and multiply by c2 to express as an energy squared. A
6
M2c4 = 2(100)(100)(1 − cos90°) = 2×104×(1 − 0) MeV2
Insert numbers; cos90° = 0. A
7
Mc2 = √(2×104) MeV = 141 MeV
Take the root. A
Mc2 ≈ 141 MeV

Reading. The two massless photons combine into a system with nonzero invariant mass — close to the true π0 mass of 135 MeV, the residual gap reflecting the idealized equal-energy, right-angle geometry. Mass emerges from the relative motion of massless constituents, the same mechanism that gives most of a proton's mass.

Units check. E1E2 has units MeV2; the root returns MeV, correctly an energy for Mc2.

Problems
  1. (A) An electron (mc2 = 0.511 MeV) has total energy E = 5.00 MeV. Find its momentum in MeV/c.
    Solution

    From E2 = (pc)2 + (mc2)2, pc = √(E2 − (mc2)2) = √(5.002 − 0.5112) = √(25.0 − 0.261) = √24.74 = 4.974 MeV. So p = 4.97 MeV/c. The electron is highly relativistic, so pc ≈ E as expected.

  2. (A) Show that for a massless particle the relation forces v = c, using p = Ev/c2.
    Solution

    The general relations give p = γmv and E = γmc2, so p/E = v/c2, i.e. p = Ev/c2 (this holds even in the massless limit as a statement about the four-vector). Setting m = 0 in the dispersion relation gives E = pc. Substitute p = Ev/c2: E = (Ev/c2)c = Ev/c. Cancelling E (≠ 0) gives 1 = v/c, so v = c. A massless particle must travel at the speed of light in every frame.

  3. (B) A proton (mc2 = 938.3 MeV) has kinetic energy K = 200 MeV. Find p (in MeV/c) and the Lorentz factor γ.
    Solution

    Total energy E = K + mc2 = 200 + 938.3 = 1138.3 MeV. Then pc = √(E2 − (mc2)2) = √(1138.32 − 938.32) = √(1.2957×106 − 8.804×105) = √(4.153×105) = 644.5 MeV, so p = 644 MeV/c. The Lorentz factor follows from E = γmc2: γ = 1138.3/938.3 = 1.213.

  4. (B) Two protons each of total energy E = 6.5 TeV collide head-on (equal and opposite momenta). Find the invariant mass Mc2 of the pair, and compare with a fixed-target collision of a 6.5 TeV proton on a stationary proton (mc2 = 0.938 GeV).
    Solution

    Collider: total four-momentum is (2E/c, 0) since the spatial momenta cancel. So Mc2 = 2E = 13.0 TeV — all the energy is available. Fixed target: the invariant is M2c4 = (Ebeam + mc2)2 − (pbeamc)2 = m2c4 + m2c4 + 2Ebeammc2 ≈ 2Ebeammc2 for Ebeam » mc2. Numerically Mc2 ≈ √(2×6500×0.938) GeV = √(1.219×104) = 110 GeV. The collider yields ~13000 GeV versus ~110 GeV fixed-target — roughly a factor 100 more reach, which is why colliders are built.

  5. (C) A particle of mass M at rest decays into two fragments of masses m1 and m2. Show that the energy of fragment 1 is E1 = (M2 + m12 − m22)c2/(2M).
    Solution

    Work in units where c = 1 and restore at the end. In the rest frame of M, conservation gives E1 + E2 = M and the momenta are equal and opposite, |p1| = |p2| ≡ p. Apply the dispersion relation to each fragment: E12 − p2 = m12 and E22 − p2 = m22. Subtract: E12 − E22 = m12 − m22. Factor the left side: (E1 − E2)(E1 + E2) = (E1 − E2)M = m12 − m22, so E1 − E2 = (m12 − m22)/M. Add this to E1 + E2 = M: 2E1 = M + (m12 − m22)/M = (M2 + m12 − m22)/M. Hence E1 = (M2 + m12 − m22)/(2M). Restoring c, each mass carries c2, giving E1 = (M2 + m12 − m22)c2/(2M). The fragment energies are fixed entirely by the three masses — the signature of a two-body decay.