Relativistic Momentum Calculator
Compute the relativistic momentum p = γmv of a particle from its rest mass and velocity. Reduces to classical mv at low speeds but grows without bound as velocity approaches c.
Last updated: May 2026
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About this calculator
The formula is p = m·v / √(1 − (v/c)²) = γ·m·v, where m is the rest mass in kg, v is the velocity in m/s, c = 299,792,458 m/s, and γ is the Lorentz factor. The output is momentum in kg·m/s. Classical (Newtonian) momentum p = mv is recovered in the limit v ≪ c (γ ≈ 1); at relativistic speeds γ amplifies the momentum well above the classical value. As v approaches c, γ diverges and so does p, reflecting that arbitrary momentum requires arbitrary energy. Edge cases: at v = 0, p = 0. For v ≥ c the formula gives NaN. For massless particles (photons), p = E/c = h/λ — the relativistic energy-momentum relation E² = (pc)² + (mc²)² reduces to E = pc when m = 0. Relativistic momentum (unlike classical) is conserved in all inertial frames; using classical mv in particle-physics conservation laws gives wrong answers above v ≈ 0.1c. The four-momentum (E/c, p_x, p_y, p_z) is a four-vector under Lorentz transformations, and its squared magnitude (E/c)² − p² = (mc)² is invariant — the same in all frames. This invariant is what defines the 'invariant mass' of a system, central to particle identification in collider experiments.
How to use
Example 1 — proton at 0.5c. m = 1.673 × 10⁻²⁷ kg, v = 1.499 × 10⁸ m/s (0.5c). Step 1: classical p = m·v = 1.673e−27 × 1.499e8 ≈ 2.508e−19 kg·m/s. Step 2: γ = 1/√(1 − 0.25) ≈ 1.1547. Step 3: relativistic p = γ × classical = 1.1547 × 2.508e−19 ≈ 2.897e−19 kg·m/s. Verify: at v = 0.5c, γ ≈ 1.155, and the relativistic momentum exceeds classical by 15.5% ✓. Convert to natural units: p × c ≈ 8.683e−11 J ≈ 542 MeV/c — a half-light-speed proton carries ~542 MeV/c of momentum, comparable to its 938 MeV rest-mass-energy. Example 2 — electron at 0.99c. m = 9.109 × 10⁻³¹ kg, v = 2.969 × 10⁸ m/s. Step 1: classical p = 9.109e−31 × 2.969e8 ≈ 2.704e−22 kg·m/s. Step 2: γ at 0.99c ≈ 7.089. Step 3: relativistic p = 7.089 × 2.704e−22 ≈ 1.917e−21 kg·m/s. Verify: at 0.99c, relativistic momentum is ~7× classical ✓. Convert: pc ≈ 5.744e−13 J ≈ 3.585 MeV/c — a 0.99c electron carries ~3.585 MeV/c of momentum, about 7× its 511 keV rest mass-energy. This is the kind of regime electron-beam therapy and accelerators reach.
Frequently asked questions
Why doesn't classical momentum work at high speeds?
Classical momentum p = mv is conserved only in the non-relativistic limit; in special relativity, conservation requires the relativistic form p = γmv. The mathematical derivation comes from requiring that momentum be a component of a four-vector (E/c, p) transforming covariantly under Lorentz transformations. Equivalently, requiring that the laws of conservation of momentum and energy be valid in all inertial frames forces γ into the formula. Experimentally, particle collisions at relativistic speeds have momentum and energy distributions that match the relativistic formulas precisely — using classical mv would predict wrong scattering angles, wrong threshold energies for particle production, and wrong decay product spectra. At everyday speeds (v < 10⁶ m/s, v/c < 10⁻²), γ ≈ 1 + 5 × 10⁻⁵, so the classical formula is accurate to about one part in 100,000 — perfectly fine for engineering but not for particle physics or precision measurements.
What is four-momentum and the invariant mass relation?
Four-momentum P = (E/c, p_x, p_y, p_z) is the relativistic generalisation of momentum: a four-component object that transforms as a Lorentz four-vector. Its temporal component E/c (the energy divided by c) and spatial component p (the ordinary three-momentum) mix between frames according to the Lorentz transformation. The squared magnitude P·P = (E/c)² − p² = (mc)² is invariant — the same in all inertial frames — and equals (mc)², the squared rest mass times c². This identity is the relativistic energy-momentum relation E² = (pc)² + (mc²)². For massless particles (m = 0), it reduces to E = pc; for massive particles at rest (v = 0, p = 0), E = mc². The invariant mass is what particle physicists compute to identify decay products: if you sum the four-momenta of all decay products and compute the magnitude, you get the invariant mass of the original particle — a frame-independent quantity that doesn't depend on the lab frame's velocity.
How does relativistic momentum apply to photons?
Photons have zero rest mass (m = 0) but nonzero momentum: p = E/c = h/λ = ℏk, where λ is wavelength, k is wavenumber, and h, ℏ are Planck constants. They always travel at c so γ is undefined, but the relativistic energy-momentum relation E² = (pc)² + (mc²)² with m = 0 gives E = pc, which is consistent with E = hν = hc/λ and p = h/λ. Photon momentum is real and observable: radiation pressure pushes solar sails and dust grains away from the sun (radiation pressure on a fully absorbing surface is P_rad = I/c where I is the intensity); Compton scattering shows X-ray photons exchanging momentum with electrons; laser cooling uses photon momentum to slow atoms. The formula p = h/λ shows that short-wavelength (high-frequency) photons carry more momentum than long-wavelength ones: an X-ray photon carries roughly 10⁵× the momentum of a visible-light photon at the same intensity, which is why X-rays can knock electrons out of atoms while visible light typically cannot. In quantum field theory, the photon four-momentum k_μ = (ω/c, k_x, k_y, k_z) is a null four-vector (k·k = 0) because m = 0, while massive particle four-momenta have k·k = (mc)² > 0 — this is the formal mathematical statement of the rest-mass distinction.
What are the common mistakes when computing relativistic momentum?
The biggest mistake is using classical p = mv instead of γmv at relativistic speeds — the classical formula under-estimates momentum by factor γ. The second is mixing units — m must be in kg, v in m/s. The third is forgetting that momentum is a vector — p_x, p_y, p_z each get a factor γ from the magnitude of v, not from the components individually (in 3D γ depends on the speed |v|, not on each component). People also use the relativistic mass concept m_rel = γm to write p = m_rel·v, which is technically equivalent but obscures that mass is invariant while γ is frame-dependent. Floating-point precision matters at ultra-relativistic speeds (γ > 10⁸); use 1 − v/c subtraction carefully. Finally, in collisions and decays, conservation of momentum applies to the four-momentum vector, not just its spatial part — both E and p must be conserved across the interaction, and both transform under Lorentz boosts.
When should I not use this calculator?
Do not use it for massless particles (photons, gluons) — they have m = 0 and the formula gives 0 regardless of velocity. Use p = h/λ for photons. It is not appropriate for accelerating particles or in curved spacetime — these need instantaneous-rest-frame analysis or general-relativistic four-momentum. Do not use it for v ≥ c (unphysical for massive objects); the formula returns NaN. It is unsuitable for compound systems (atoms, nuclei) where binding energies and internal degrees of freedom matter; for those, use the system's invariant mass via four-momentum. For ultra-relativistic γ (> 10⁸), use extended-precision arithmetic to avoid catastrophic cancellation. Avoid using classical conservation laws (∑mv = const) at relativistic speeds — use ∑P_4 = const (four-momentum conservation), which automatically incorporates energy. Finally, for quantum-mechanical treatment (de Broglie waves, scattering amplitudes), use the relativistic Dirac equation or quantum field theory; the simple γmv is the classical limit.