Special Relativity
Relativistic Momentum
p = γmv — diverges at light speed, conserved in every frame, equals E/c for photons
Relativistic momentum p = γmv with γ = 1/√(1−v²/c²). Conserved in all inertial frames. Reduces to Newtonian p = mv at low v. Photons carry p = E/c.
- Formulap = γmv ; γ = 1/√(1 − v²/c²)
- Low-v limitγ → 1, so p → mv (Newtonian)
- High-v limitγ → ∞ as v → c — momentum diverges
- Electron at 0.99cγ = 7.09, p ≈ 1.93 MeV/c
- Photonp = E/c = h/λ — massless, finite momentum
- InvariantE² − (pc)² = (mc²)² — same in every frame
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Why momentum needs fixing
In Newtonian mechanics, the momentum of a particle of mass m moving with velocity v is p = mv. The total momentum of an isolated system is conserved over time, and the same conservation law holds in every inertial frame because Galilean velocity addition is linear.
In special relativity, velocities do not add linearly. Two velocities each less than c, added relativistically, give a result still less than c. If you apply the Newtonian momentum formula in one inertial frame and then look at the same collision from a frame moving at v relative to the first, conservation breaks: the two sides of the equation no longer balance.
The fix is unique. Replace p = mv with p = γmv, where γ = 1/√(1−v²/c²). With that one change, momentum conservation holds in every inertial frame — and reduces to Newtonian conservation at low speeds.
The numbers
p = γmv, with γ = 1/√(1 − v²/c²)
| v / c | γ | p / (mc) | Newtonian mv / (mc) | Error |
|---|---|---|---|---|
| 0.01 | 1.00005 | 0.01000 | 0.01000 | 0.005% |
| 0.1 | 1.0050 | 0.1005 | 0.1000 | 0.5% |
| 0.3 | 1.0483 | 0.3145 | 0.3000 | 4.83% |
| 0.5 | 1.1547 | 0.5774 | 0.5000 | 15.47% |
| 0.7 | 1.4003 | 0.9802 | 0.7000 | 40.03% |
| 0.866 | 2.0000 | 1.7321 | 0.8660 | 100% |
| 0.95 | 3.2026 | 3.0425 | 0.9500 | 220% |
| 0.99 | 7.0888 | 7.0179 | 0.9900 | 609% |
| 0.999 | 22.366 | 22.344 | 0.999 | 2137% |
By the time v reaches 0.5c the Newtonian estimate is off by 15%; at 0.9c it's off by a factor of 2.3; at 0.99c it's off by a factor of 7. Above 0.999c the Newtonian formula is essentially useless. Particle physicists abandoned Newtonian momentum for relativistic kinematics in the 1930s.
Worked example — electron at 0.99c
An electron has rest mass m = 9.109 × 10⁻³¹ kg, or in natural units, m_e c² = 0.511 MeV. At v = 0.99c:
- γ = 1/√(1 − 0.99²) = 1/√0.0199 = 7.089.
- Total energy E = γmc² = 7.089 × 0.511 = 3.624 MeV.
- Relativistic momentum p = γmv = 7.089 × m_e × 0.99c. Expressed in MeV/c: p = 7.089 × 0.511 × 0.99 ≈ 3.59 MeV/c.
- In SI: p = 3.59 MeV/c × (1.602 × 10⁻¹³ J/MeV) / (3 × 10⁸ m/s) ≈ 1.92 × 10⁻²¹ kg·m/s.
- Newtonian momentum mv = m_e × 0.99c ≈ 0.506 MeV/c — wrong by a factor of γ ≈ 7.09.
- Invariant check: E² − (pc)² = 3.624² − 3.59² ≈ 0.262 MeV² ; √0.262 ≈ 0.512 MeV ≈ mc². ✓
In a particle detector with B = 4 T, this electron follows a circular arc with radius r = p/(qB) = (1.92 × 10⁻²¹) / (1.6 × 10⁻¹⁹ × 4) ≈ 3.0 mm. The radius reads off the relativistic momentum directly.
JavaScript — relativistic momentum calculator
const c = 299792458;
const e = 1.602176634e-19; // C
const m_e = 9.1093837015e-31; // kg
const MeV_per_c2_kg = 1.7826619e-30;
function gamma(v) {
const beta = v / c;
if (Math.abs(beta) >= 1) return Infinity;
return 1 / Math.sqrt(1 - beta * beta);
}
function relMomentum(mass_kg, v) {
return gamma(v) * mass_kg * v;
}
function totalEnergy(mass_kg, v) {
return gamma(v) * mass_kg * c * c;
}
function newtonianMomentum(mass_kg, v) {
return mass_kg * v;
}
// Electron at 0.99c
const v = 0.99 * c;
const p_rel = relMomentum(m_e, v);
const E_tot = totalEnergy(m_e, v);
const p_newt = newtonianMomentum(m_e, v);
const p_MeV = p_rel * c / (1e6 * e);
const E_MeV = E_tot / (1e6 * e);
console.log(`γ=${gamma(v).toFixed(3)}, E=${E_MeV.toFixed(3)} MeV, p=${p_MeV.toFixed(3)} MeV/c`);
// γ=7.089, E=3.624 MeV, p=3.591 MeV/c
console.log(`Newtonian mv would give: ${(p_newt * c / (1e6 * e)).toFixed(3)} MeV/c (wrong)`);
// 0.506 MeV/c
// Invariant: E² − (pc)² = (mc²)²
const invariant_MeV = Math.sqrt(E_MeV * E_MeV - p_MeV * p_MeV);
console.log(`E² − (pc)² → mc² = ${invariant_MeV.toFixed(3)} MeV (expect 0.511)`);
// Photon momentum
function photonMomentumFromEnergyEV(E_eV) {
// p = E/c, returns kg m/s
return (E_eV * e) / c;
}
const p_photon_green = photonMomentumFromEnergyEV(2.5); // visible green ~500 nm
console.log(`Visible photon p ≈ ${p_photon_green.toExponential(2)} kg m/s`);
// ~1.3e-27 kg m/s — tiny but real (radiation pressure)
// Bend radius in a magnetic field — particle physics workhorse
function bendRadius_m(p_kgms, B_tesla, q_coul = e) {
return p_kgms / (q_coul * B_tesla);
}
console.log(`Electron at 0.99c in 4 T: r = ${(bendRadius_m(p_rel, 4) * 1000).toFixed(1)} mm`);
// 3.0 mm
// LHC proton — 6.8 TeV per beam
function lhcProton() {
const proton_mass_kg = 1.6726219e-27;
const proton_E_J = 6.8e12 * e; // 6.8 TeV in joules
const proton_g = proton_E_J / (proton_mass_kg * c * c);
const proton_v = c * Math.sqrt(1 - 1 / (proton_g * proton_g));
return { gamma: proton_g, v_over_c: proton_v / c };
}
console.log(lhcProton()); // γ ≈ 7250, v ≈ (1 − 9.5e−9) c
Four-momentum and the energy-momentum invariant
The relativistic energy and momentum combine into a four-vector pμ = (E/c, p_x, p_y, p_z). Under a Lorentz boost along x with velocity u, the components transform as:
E'/c = γ_u (E/c − (u/c) p_x)
p'_x = γ_u (p_x − (u/c) E/c)
p'_y = p_y, p'_z = p_z
The invariant "length squared" of this four-vector is identical in every frame:
(E/c)² − p · p = (mc)²
E² − (pc)² = (mc²)²
This single equation contains a great deal of physics. For a massive particle at rest, p = 0 gives E = mc². For a massive particle in motion, E and p are related by E² = (pc)² + (mc²)². For a massless photon, m = 0 gives E = pc and so E = pc = hf. The relation is the workhorse of all collider kinematics.
History — Planck, Lewis, and the slow displacement of m
Einstein's 1905 paper introduced relativistic dynamics but emphasized force and energy. Max Planck wrote the explicit momentum formula p = γmv in 1906 and showed it preserves Newton's second law in covariant form. G. N. Lewis (1908) reconstructed the same formula by demanding momentum conservation across frames in a glancing collision — a tradition still used in introductory textbooks.
Through the 1910s and 1920s, many textbooks introduced the "relativistic mass" m_rel = γm so that the Newtonian-style p = m_rel v could be retained. Einstein later disliked this convention; by the 1980s most particle physicists had dropped it. Modern usage: m always denotes the invariant rest mass; the γ factor is kept explicit. Sources such as the Particle Data Group and CODATA report only rest masses.
Common misconceptions
- "Mass increases with speed." Old-school phrasing. Modern physics keeps mass invariant; momentum, energy, and inertia-along-motion increase via γ. You can't tell a "more massive" rocket from a faster one without a frame.
- "Photons are massless so they have no momentum." Wrong. Massless particles carry momentum p = E/c. Solar sails, radiation pressure, and laser cooling all exploit photon momentum.
- "Newton's law is wrong." Newton's F = dp/dt remains exactly correct in special relativity — provided you use p = γmv. The old F = ma form does not, because γ depends on v.
- "Relativistic momentum is a small correction." Below 0.1c it is. Above 0.5c the correction exceeds 15%. Above 0.9c it's a factor of 2+. Particle physics lives in this regime exclusively.
- "You can use p = mv if you replace m by relativistic mass." True in 1+1 dimensions but breaks in 3D — the longitudinal and transverse inertias scale differently (γ³ and γ respectively). Better to keep m invariant and use γmv explicitly.
- "Momentum isn't conserved relativistically." It is — exactly, in every inertial frame. The conservation just uses γmv rather than mv.
Relativistic momentum vs Newtonian
| Property | Newtonian | Relativistic | Identical when | Diverges when |
|---|---|---|---|---|
| Formula | p = mv | p = γmv | v ≪ c (γ ≈ 1) | v → c (γ → ∞) |
| Conservation | In every Galilean frame | In every inertial frame | Low-v limit | — |
| Force law | F = ma | F = dp/dt | v constant or v ≪ c | Near c, F ≠ ma |
| Energy | KE = ½mv² | KE = (γ−1)mc² | v ≪ c | — |
| Photon momentum | 0 (m = 0) | p = E/c | — | Newtonian fails |
| Frame transformation | p' = p − mu | 4-vector boost | v ≪ c | Near c |
Where relativistic momentum matters
- Particle accelerators. LHC protons at 6.8 TeV have γ ≈ 7250 and p ≈ 6.8 TeV/c — bending magnets, RF cavities, and beam dumps are all designed for γmv kinematics.
- Particle detectors. Track curvature in magnetic fields reads off γmv directly; combined with calorimetry, this yields the rest mass m and identifies the particle.
- Cosmic-ray physics. Ultra-high-energy primaries (10²⁰ eV protons) have γ ≈ 10¹¹; their interaction kinematics is purely relativistic.
- Photon momentum. Radiation pressure shapes comet tails, drives solar sails (LightSail-2), enables laser cooling and atom trapping, and underlies the Compton effect.
- Astrophysical jets. Relativistic plasma in AGN and microquasar jets — momentum and energy transport handled with γmv.
- Nuclear medicine. Proton therapy beams in the 70–250 MeV range have γ between 1.07 and 1.27; treatment planning uses relativistic kinematics.
- Gravitational waves. Inspiraling compact binaries reach orbital v/c near 0.5 in the last cycles; relativistic momentum is built into the post-Newtonian expansion.
Frequently asked questions
What is relativistic momentum?
Relativistic momentum is p = γmv, where m is the invariant (rest) mass, v is the velocity, and γ = 1/√(1 − v²/c²) is the Lorentz factor. It replaces Newton's p = mv. At low speeds γ ≈ 1 and the two formulas agree to a few parts in 10¹⁷. As v approaches c, γ diverges to infinity and so does p — which is why no finite force or energy can accelerate a massive object to the speed of light.
Why doesn't Newtonian p = mv work near light speed?
Imagine a glancing collision between two identical particles, analyzed in two different inertial frames. In Newtonian mechanics, momentum is conserved in both frames if you use p = mv. But under a Lorentz transformation, the velocity components do not add the Newtonian way, and Newtonian momentum is no longer conserved in the boosted frame. The fix is unique: replace p = mv by p = γmv. With that replacement, momentum conservation holds in every inertial frame.
How does relativistic momentum relate to energy?
Together they form the four-momentum pμ = (E/c, p_x, p_y, p_z), a four-vector under Lorentz transformations. Its invariant 'length squared' is E² − (pc)² = (mc²)² — the same number in every frame. For a particle at rest, p = 0 and E = mc². For a moving particle, E = γmc² and p = γmv, so the ratio p/E = v/c². For photons, m = 0, so E = pc.
Can photons have momentum if they're massless?
Yes — photons carry momentum p = E/c = h/λ = ℏk, where h is Planck's constant and λ is wavelength. Their rest mass is zero, but γm in the limit is the indeterminate form 0 × ∞ that resolves to a finite quantity. Photon momentum is responsible for radiation pressure, the design of solar sails, the recoil of atoms emitting light, and the Compton scattering effect.
Is 'relativistic mass' the same as γm?
In older textbooks 'relativistic mass' m_rel = γm was sometimes introduced so that one could write p = m_rel v in Newtonian form. Modern physics does not use that convention. 'Mass' means the invariant rest mass m (Lorentz scalar), and the γ factor is kept explicit in p = γmv and E = γmc². Particle physics papers and standards report rest masses only.
What's a worked example of relativistic momentum?
Take an electron at v = 0.99c. The rest energy is m_e c² = 0.511 MeV. γ = 1/√(1 − 0.9801) = 7.089. Total energy E = γmc² = 3.62 MeV. Relativistic momentum p = γmv ≈ 7.089 × 0.511 × 0.99 / c MeV/c ≈ 3.59 MeV/c — about 1.93 × 10⁻²¹ kg·m/s. Newtonian momentum (mv) would give only 0.506 MeV/c — wrong by a factor of γ ≈ 7.
How is relativistic momentum measured?
By magnetic deflection. The Lorentz force F = qv × B causes a charged particle to follow a circular arc in a uniform field B. The radius is r = p/(qB) where p is the relativistic momentum γmv. Measuring r, B, and q yields p directly. Modern particle detectors (CMS, ATLAS) use this method routinely: in a 4-tesla magnetic field, a 1-GeV charged particle has a bend radius of ~80 cm.