University Physics IV · Special Relativity II · 3.8

Energy–Momentum Four-Vector

Energy and momentum stop being separate ledgers here. Stack them into one four-component object, boost it with the same matrix that moves (ct, x), and its Minkowski length hands you a number every frame agrees on — the mass. Most relativistic problems are easier read off that length than solved frame by frame.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Divide a four-vector by an invariant and the result is forced to be a four-vector again. That is the whole trick: P = m dX/dτ, built from X = (ct, x) and the two frame-independent quantities m and dτ. Work out its components and mechanics' own quantities fall out — γmc = E/c in the time slot, γmv = p in the space slots.

So energy and momentum are not two conservation laws that happen to travel together; they are one object, and a boost mixes them with the same matrix that mixes t and x. The payoff is the Minkowski length. Squaring P gives E² − (pc)² = (mc²)², one number every inertial observer computes and every one of them agrees on.

Read one way it converts a tracker's momentum and a calorimeter's energy into a mass, which is how a detector names a particle; read the other way it returns β = pc/E and γ = E/mc² from that same pair. The cost is that mass stops behaving like an amount of stuff. Applied to a system the same construction gives (Mc²)² = (ΣEᵢ)² − (Σpᵢc)², which counts the kinetic energies and relative directions of the parts — so two massless photons flying apart carry a mass between them, and a proton weighs far more than the quarks inside it.

Simple definition: The energy-momentum four-vector is the single object (E/c, p) whose components mix under a Lorentz boost exactly as (ct, x) do, and whose invariant length E² − (pc)² equals (mc²)² in every inertial frame.
Example: An electron with total energy E = 5.00 MeV has pc = √(5.00² − 0.511²) = 4.974 MeV, so β = pc/E = 0.9948 and γ = E/mc² = 9.78 — three numbers from one measured energy plus the invariant.

Four-momentum from the worldlineP = m dX/dτ = (E/c, p), E = γmc², p = γmv

Only invariants multiply X here, so P must transform like X — the boost law is inherited, not assumed.

m and dτ are invariants and X = (ct, x) is a four-vector; E in J or MeV, p in kg m s⁻¹ or MeV/c

How a boost mixes energy and momentumE′ = γ(E − vpₓ), p′ₓ = γ(pₓ − vE/c²), p′_y = p_y, p′_z = p_z

One matrix does both columns, so energy never needs a transformation rule of its own.

v and γ belong to the new frame; substituting ct → E/c and x → p reproduces the (ct, x) boost exactly

The invariant: mass as a Minkowski lengthE² − (pc)² = (mc²)²

Two lab numbers give the mass, so a detector identifies a particle without ever entering its rest frame.

Same value in every inertial frame; with E and pc in MeV, mc² comes out in MeV

Speed and Lorentz factor read off Pβ = pc/E, γ = E/(mc²), K = E − mc²

The ratio cancels γ and m at once, so β follows from the two lab numbers without either being known first.

β and γ dimensionless; K = (γ − 1)mc² is the kinetic energy, and only K collapses to ½mv² when β ≪ 1

The massless branchm = 0 ⇒ E = pc and β = 1E ≈ pc + (mc²)²/2pc when E ≫ mc²

A photon has no rest frame at all, while a massive particle only ever approaches this branch — how closely is set by mc²/E.

m = 0 makes P a null four-vector; the expansion is the ultrarelativistic limit, good once pc ≫ mc²

Invariant mass of a system(Mc²)² = (ΣEᵢ)² − (Σpᵢc)²two photons: (Mc²)² = 2E₁E₂(1 − cos θ)

The cross terms carry the parts' relative motion, which is why a spread-out system outweighs its pieces.

θ is the opening angle; M ≥ Σmᵢ, with equality only if every part shares one velocity

Differentiate the worldline by proper time

Newton's p = mv fails relativistically for a structural reason: dt is frame-dependent, so m dx/dt divides a four-vector component by somebody's coordinate time and transforms into a mess. Replace dt with the proper time dτ, which every frame agrees on, and the repair is automatic. Define P = m dX/dτ with X = (ct, x). Since m and dτ are invariants and X is a four-vector, P is a four-vector — no assumption, just a four-vector divided by a scalar. Now read the components. With dt/dτ = γ, the time part is m c (dt/dτ) = γmc and the space part is m (dx/dτ) = γmv. Identify γmc² as the total energy and γmv as the momentum and you have P = (E/c, p). Energy and momentum were never two separate conservation laws; they are the four components of one.

The boost is the (ct, x) matrix with new labels

Because P transforms like X, no new derivation is needed — substitute ct → E/c and x → pₓ into the standard boost and read off E′ = γ(E − v pₓ) and p′ₓ = γ(pₓ − vE/c²), with p_y and p_z passing through unchanged. Try it on a 1400 MeV proton, whose momentum is 1039.0 MeV/c, seen from a frame moving at v = 0.500c along its direction of travel. Here γ = 1/√0.750 = 1.1547, so E′ = 1.1547(1400 − 0.500 × 1039.0) = 1016.7 MeV and p′c = 1.1547(1039.0 − 0.500 × 1400) = 391.5 MeV. Both fell, and by different factors — that is the point. The transverse rule matters as much as the longitudinal one: a boost along x cannot touch p_y, so a particle's momentum can swing in direction under a boost even though one of its components never changes value.

Square it: the length no boost can change

Every four-vector has a Minkowski square, and for P that square is E² − (pc)². Compute it directly: (γmc²)² − (γmvc)² = γ²m²c⁴(1 − β²) = m²c⁴, because γ²(1 − β²) = 1 by definition. So E² − (pc)² = (mc²)², and mc² is the Minkowski length itself — the mass, written in energy units. Two pictures help. The first is a right triangle with legs pc and mc² and hypotenuse E. The second is what a boost does to it: in the (pc, E) plane the states available to a given particle lie on the hyperbola E = √((pc)² + (mc²)²), and a boost slides the state along that curve. The minus sign in the Minkowski square is why the curve is a hyperbola rather than a circle, and why E can be pushed up without bound while the mass sits still. Check it on the boosted proton: 1016.7² − 391.5² = 8.804 × 10⁵ MeV² = (938.3 MeV)², exactly the lab value.

Two lab numbers give speed, γ, and identity

Dividing the space part of P by the time part kills γ and m together: pc/E = (γmvc)/(γmc²) = v/c, so β = pc/E. The same pair gives γ = E/mc² and the kinetic energy K = E − mc². This is how detectors actually work: a magnetic tracker returns p from the curvature of the track, a calorimeter returns E, and the invariant turns that pair into a mass. Suppose the tracker gives pc = 629.5 MeV and the calorimeter E = 800.0 MeV. Then (mc²)² = 800.0² − 629.5² = 640 000 − 396 270 = 243 730 MeV², so mc² = 493.7 MeV and the track belonged to a charged kaon. The same two numbers finish the description: β = 629.5/800.0 = 0.7869, γ = 800.0/493.7 = 1.620, and K = 306.3 MeV. Nothing was timed and nothing was measured twice. It is also why particle physicists quote masses, momenta and energies all in electronvolts: the invariant relates the three with nothing but factors of c.

The massless branch, and how close to c counts

Set m = 0 in the invariant and it gives E = pc with β = pc/E = 1 exactly. The four-vector is then null: its Minkowski length is zero although neither E nor p is. There is no rest frame to boost into and γ = E/mc² is undefined, so any argument beginning "in the photon's frame" has already gone wrong. For a massive particle the same branch is a limit rather than a state. Expanding E = pc√(1 + (mc²/pc)²) gives E ≈ pc + (mc²)²/2pc, and equivalently 1 − β ≈ (mc²)²/2E². At E = 5.00 GeV an electron has mc²/E = 1.02 × 10⁻⁴, so 1 − β ≈ 5.2 × 10⁻⁹ and treating it as massless costs almost nothing. A proton at that same 5.00 GeV has mc²/E = 0.1877, giving β = √(1 − 0.1877²) = 0.982 — same energy, nowhere near the limit. "Ultrarelativistic" is a statement about E/mc², never about energy alone.

A system has a mass of its own

Four-momenta add, and a sum of four-vectors is a four-vector, so any collection of particles has a total P whose Minkowski square is again an invariant: (Mc²)² = (ΣEᵢ)² − (Σpᵢc)². That M is the system's invariant mass; its energy equivalent Mc² is what collider work calls √s, and it is the total energy the system shows in the centre-of-momentum frame — the frame where Σp = 0, which exists for every system that is not entirely massless and collinear. M is not the sum of the parts' masses. Expanding for two particles gives (Mc²)² = m₁²c⁴ + m₂²c⁴ + 2(E₁E₂ − p₁⋅p₂c²), and those cross terms are never smaller than 2m₁m₂c⁴, so M ≥ m₁ + m₂ with equality only when the two move together. Two 500 MeV photons crossing at 90.0° have (Mc²)² = 2E₁E₂(1 − cos θ) = 2(500)(500)(1) = 500 000 MeV², so M = 707 MeV/c² — built entirely out of two massless things. Reconstructing that number from decay products is how resonances are found.

Change one variable at a time

Make the relationship visible.

Interactive model

Rest energy mc²240 MeV

Lab momentum pc320 MeV

Speed of the new frame0.40 c

Set the boost to 0 and the dashed triangle lands on the solid one. Push it to 0.60 and both the base and the hypotenuse shrink — but the apex never changes height, because mc² is what the two frames must agree on. Slide the mass to 0 and the triangle flattens onto the axis: E = pc.

LAB E400 MeV

BOOSTED E′297 MeV

BOOSTED p′c175 MeV

INVARIANT √(E′²−(p′c)²)240 MeV

Live interpretationLAB E: 400 MeV. BOOSTED E′: 297 MeV. BOOSTED p′c: 175 MeV. INVARIANT √(E′²−(p′c)²): 240 MeV

Catch the common trap

Explain before calculating.

Misconception check

The error that costs marks here is treating invariant mass as a sum. Total energy adds and total momentum adds as a vector, but mass does not: M is what survives squaring the total four-momentum, (Mc²)² = (ΣEᵢ)² − (Σpᵢc)², and the cross terms 2(E₁E₂ − p₁⋅p₂c²) are exactly what break additivity. Two 511 keV photons flying apart from positron annihilation each have m = 0, yet together they carry an invariant mass of 1.022 MeV/c²; send those same two photons the same way instead and the pair's mass is precisely zero. Neither photon was altered — only the angle between them. The inequality runs one way only: M can exceed Σmᵢ but never fall below it, and the two coincide just when the parts share a single common velocity. That is why a proton's 938 MeV/c² is mostly the energy of nearly massless quarks and gluons rather than a tally of their masses.

A muon (mc² = 105.7 MeV) is measured in the lab with total energy E = 1000 MeV. A second observer, moving along the muon's direction of travel, measures its total energy as E′ = 400 MeV. What momentum does that second observer measure?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA charged pion (mc² = 139.6 MeV) is measured with total energy E = 250.0 MeV. Find its momentum in MeV/c, its speed as a fraction of c, its Lorentz factor, and its kinetic energy.

The invariant fixes the momentum from the energy alone: (pc)² = E² − (mc²)² = 250.0² − 139.6² = 62 500 − 19 488 = 43 012 MeV².
So pc = √43 012 = 207.4 MeV, i.e. p = 207.4 MeV/c. No frame change was needed — the invariant did all the work.
Speed comes straight off the four-vector: β = pc/E = 207.4/250.0 = 0.8296.
And γ = E/mc² = 250.0/139.6 = 1.791. Cross-check against the definition: 1/√(1 − 0.8296²) = 1/√0.3118 = 1.791 ✓.
Kinetic energy is what is left after the rest energy: K = E − mc² = 250.0 − 139.6 = 110.4 MeV. The Newtonian ½mv² = ½(139.6)(0.8296²) = 48.0 MeV, low by 56% — at β = 0.83 the expansion has already failed.

Answerp = 207.4 MeV/c, β = 0.8296, γ = 1.791, K = 110.4 MeV.

MediumAn electron (mc² = 0.511 MeV) travels along +x with pc = 4.00 MeV. Find its total energy, transform E and pₓ into a frame moving at v = 0.600c along +x, and verify that the invariant is unchanged.

E = √((pc)² + (mc²)²) = √(16.000 + 0.261) = √16.261 = 4.0325 MeV.
The new frame's boost factor: γ = 1/√(1 − 0.600²) = 1/√0.640 = 1.250.
Energy transforms the way ct does: E′ = γ(E − v pₓ) = 1.250(4.0325 − 0.600 × 4.00) = 1.250 × 1.6325 = 2.041 MeV.
Momentum transforms the way x does: p′ₓ c = γ(pₓ c − βE) = 1.250(4.00 − 0.600 × 4.0325) = 1.250 × 1.5805 = 1.976 MeV, so p′ₓ = 1.976 MeV/c.
Invariant check: E′² − (p′c)² = 4.1642 − 3.9031 = 0.2611 MeV² = (0.511 MeV)² ✓ — the same mass, as it must be.
The speed agrees too: β′ = p′c/E′ = 1.976/2.041 = 0.9681, which is what velocity addition returns for β = 4.00/4.0325 = 0.9919 composed with −0.600.

AnswerE = 4.033 MeV; in the moving frame E′ = 2.041 MeV and p′ₓ = 1.976 MeV/c. Both changed; E² − (pc)² = (0.511 MeV)² did not.

HardA detector records two photons: one of 600 MeV along +x, one of 400 MeV at 60.0° to +x in the xy-plane. Each photon has zero mass. Find the invariant mass of the pair, the speed of the frame in which their total momentum vanishes, and what the pair's mass becomes if the second photon is instead emitted along +x.

Add the four-momenta component by component. Total energy E = 600 + 400 = 1000 MeV. Momentum: pₓ c = 600 + 400 cos 60.0° = 600 + 200 = 800 MeV, and p_y c = 400 sin 60.0° = 346.4 MeV.
Magnitude of the total momentum: |p|c = √(800² + 346.4²) = √(640 000 + 120 000) = √760 000 = 871.8 MeV.
Square the total four-vector: (Mc²)² = E² − (|p|c)² = 1 000 000 − 760 000 = 240 000 MeV², so Mc² = 489.9 MeV and M = 490 MeV/c². Two massless particles, and the system has a mass.
The two-photon shortcut confirms it: (Mc²)² = 2E₁E₂(1 − cos θ) = 2(600)(400)(1 − 0.500) = 240 000 MeV² ✓.
The zero-momentum frame moves with the system: β = |p|c/E = 871.8/1000 = 0.8718, and γ = E/Mc² = 1000/489.9 = 2.041. In that frame the photons are back to back with Mc²/2 = 245.0 MeV each.
Set θ = 0 and the bracket (1 − cos θ) vanishes, so (Mc²)² = 0 and the pair is massless: collinear massless particles admit no zero-momentum frame to be weighed in at all.

AnswerM = 490 MeV/c² (Mc² = 489.9 MeV); the zero-momentum frame moves at β = 0.872 with γ = 2.04, each photon carrying 245 MeV there. Collinear photons (θ = 0) give M = 0.