Relativistic Momentum Calculator
Solve the relativistic-momentum relation p = γmv = mv / √(1 − v²/c²) for any of its three unknowns — momentum, rest mass, or velocity — with step-by-step math, a live momentum-comparison schematic, and companion energy metrics including γ, total energy, and kinetic energy.
Momentum grows without bound
In Newtonian mechanics, momentum p = mv grows linearly with velocity and there is no ceiling. In relativity, momentum is p = γmv, and γ = 1/√(1 − v²/c²) diverges as v approaches c. So the momentum of a particle grows without bound as it is pushed faster and faster, even though its speed never quite reaches c. To accelerate a particle from 0.99c to 0.999c requires far more energy than to accelerate it from rest to 0.99c — this is why particle accelerators grow to kilometres in length.
Mass itself does not increase
The idea of "relativistic mass" γm was popular in early textbooks but has fallen out of favour. Modern physics treats the rest mass m as a Lorentz-invariant — the same in every frame — and puts the entire velocity dependence into the γ factor that multiplies the momentum and energy. When people say "mass increases with speed," they are really describing the fact that it takes more and more force to produce the same acceleration, which is captured correctly by p = γmv combined with the relativistic version of Newton's second law F = dp/dt.
The energy–momentum triangle
Momentum and energy are two faces of the same relativistic four-vector. They are related by the fundamental identity E² = (pc)² + (mc²)², which is the Minkowski analogue of the Pythagorean theorem. For a massless particle like the photon, m = 0 so E = pc; for a slow-moving massive particle, the identity reduces to the familiar E ≈ mc² + p²/(2m). When physicists quote momenta in "GeV/c," they are really quoting energies that have been divided by c to make the units of momentum — a practice rooted in this identity.
Why nothing can reach the speed of light
Because the momentum diverges at v = c, an infinite amount of impulse — and therefore an infinite amount of energy — would be needed to accelerate any massive particle to the speed of light. The formula makes this concrete: at v = 0.9999c, γ ≈ 70.7, so the momentum is 70 times larger than the Newtonian prediction and the kinetic energy is 69mc². At v = 0.999999c, γ ≈ 707, and the required energy has grown by another factor of ten. This is the deepest reason why c is a speed limit, and it is baked into the formula for relativistic momentum.
How to Use This Calculator
The relativistic-momentum relation links three quantities: p, m, and v, with c as a fundamental constant. Pick which one you want the tool to solve, then fill in the other two.
1. Choose what to solve for
Pick p, m, or v from the toggle. The calculator rearranges p = γmv = mv/√(1 − v²/c²) automatically for whichever variable you choose.
2. Use realistic presets
Momentum presets span 1 eV/c to 100 GeV/c (particle-physics natural units); mass presets cover the electron, muon, proton, alpha particle, C₆₀ and 1 kg; velocity presets span 0.01c to 0.99c.
3. Match units carefully
Momentum is in kg·m/s, mass in kg, velocity in m/s. In particle physics, momentum is often quoted in eV/c, keV/c, MeV/c or GeV/c — the hint line under the momentum field shows the value in these units automatically. The conversion factor is 1 eV/c = 5.344286 × 10⁻²⁸ kg·m/s.
4. Respect the speed limit
Velocity must be strictly less than c = 299,792,458 m/s. At v = c the Lorentz factor diverges and the momentum is infinite — no massive particle can reach it. The slider uses a log-odds scale in β = v/c so the whole range from non-relativistic to ultra-relativistic is explorable.
5. Read the result and verdict
The result card shows the solved quantity plus β, γ, and the value in eV/c for momentum. A colour-coded verdict classifies the regime from non-relativistic through mildly, significantly, strongly, extremely, and ultra-relativistic.
6. Check the schematic and charts
The schematic shows the particle with its velocity arrow and two momentum arrows — one classical and one relativistic — side by side, plus the γ badge and β gauge. The basic chart plots p vs. β for three masses; the advanced chart plots p vs. m on a log–log scale for three velocities.
7. Switch to Advanced for depth
Advanced mode adds the p vs. m chart, a sensitivity check on a +1% input change, and companion metrics including γ, the classical momentum p_cl = mv, total energy E = γmc², rest energy mc², kinetic energy K = (γ−1)mc², and the momentum-to-energy ratio pc/E.
What This Tool Offers
Built around the relativistic-momentum relation p = γmv = mv/√(1 − v²/c²), the fundamental formula that replaces Newtonian momentum at high speeds.
Three-way solving
Solve for momentum, rest mass, or velocity. The closed-form solution for velocity is v = pc/√(p²c² + m²c⁴) — no numerical iteration needed.
Classical vs relativistic schematic
A live diagram showing the particle, its velocity arrow, and two side-by-side momentum arrows — one showing the Newtonian mv, one showing the relativistic γmv — plus a γ badge and β gauge.
Relativity-class verdict
Each result is classified from non-relativistic through barely, mildly, significantly, strongly, extremely, and ultra-relativistic, up to the extreme β > 0.999999 regime.
Particle presets
A dedicated mass row with one-click presets for the electron, muon, proton, alpha particle, C₆₀ buckyball, and a macroscopic 1 kg reference.
Dual comparison charts
Momentum vs. β for three masses on a linear β axis showing the divergence as β → 1, plus an advanced log–log p vs. m chart for three velocities (a family of slope-1 lines).
Reference tables
A momentum ↔ energy converter, a lookup grid of γ and p against β, a table of famous particles and their relativistic momenta, and a full variable reference.
Reference Tables & Conversions
A momentum converter, a γ/p lookup grid, and famous particles at relativistic speeds.
Momentum ↔ energy ↔ γ converter
Lorentz Factor & Momentum Lookup
Computed from γ = 1/√(1 − β²) and p = γmv with m = mₑ (electron) unless noted.
| β = v/c | γ | pₑ (kg·m/s) | pₑ (MeV/c) | pₚ (MeV/c) | Kₑ (MeV) |
|---|
Famous Particles at Relativistic Speeds
Representative momenta and energies for real particle-physics scenarios.
| Scenario | β = v/c | γ | p (MeV/c) | K (MeV) | Notes |
|---|---|---|---|---|---|
| Electron in CRT (30 kV) | 0.34 | 1.06 | 180 | 30 | Classical electron-beam energy |
| Electron in TEM (100 kV) | 0.55 | 1.20 | 339 | 100 | High-resolution imaging |
| Electron in TEM (300 kV) | 0.78 | 1.59 | 822 | 300 | Atomic-resolution imaging |
| Electron at 0.9c (LHC injector) | 0.90 | 2.29 | 1,171 | 661 | Linear-accelerator exit |
| Electron at 0.99c | 0.99 | 7.09 | 3,622 | 3,111 | Beta spectroscopy regime |
| Electron in LEP (104 GeV) | 1 − 5e−11 | 2.0 × 10⁵ | 1.04 × 10⁵ | 1.04 × 10⁵ | Electron–positron collider |
| Proton in LHC (7 TeV) | 1 − 3e−9 | 7,463 | 7.0 × 10⁶ | 7.0 × 10⁶ | World's highest-energy collider |
| Cosmic-ray proton (10²⁰ eV) | 1 − 3e−21 | 1.07 × 10¹¹ | 1.07 × 10¹¹ | 1.07 × 10¹¹ | The Oh-My-God particle |
| Muon at 0.98c (sea level) | 0.98 | 5.03 | 1,015 | 1,058 | Cosmic-ray shower |
| Thermal neutron (300 K) | 2.2 × 10⁻⁶ | 1.000000000002 | 1.4 × 10⁻⁶ | 2.5 × 10⁻⁸ | Essentially Newtonian |
| Alpha particle from ²³⁸U decay | 2.1 × 10⁻² | 1.0002 | 8.5 × 10³ | 4,270 | Radioactive-decay energy |
Variable Reference
| Symbol | Meaning | Typical range | Units |
|---|---|---|---|
| p | Relativistic momentum | 10⁻⁴⁰ – 10⁸ | kg·m/s |
| m | Rest mass | 10⁻³¹ – 10³ | kg |
| v | Velocity | 0 – c | m/s |
| c | Speed of light in vacuum | 299,792,458 (exact) | m/s |
| β | Velocity as fraction of c | 0 – 1 | — |
| γ | Lorentz factor = 1/√(1 − β²) | 1 – 10¹¹ | — |
| E | Total energy = γmc² | — | J |
| E₀ | Rest energy = mc² | — | J |
| K | Kinetic energy = (γ − 1)mc² | — | J |
| eV/c | Momentum unit in particle physics | — | 5.344286 × 10⁻²⁸ kg·m/s |
What Does p = γmv = mv / √(1 − v²/c²) Measure?
The relativistic-momentum relation connects three quantities at the heart of special relativity: the momentum of a particle (p), its rest mass (m), and its velocity (v), with the speed of light c setting the scale. Together they replace Newton's classical p = mv with a formula that correctly describes motion at any speed up to — but never reaching — c. The relation was introduced by Einstein in his 1905 paper on special relativity, and it is the foundation on which all of high-energy physics is built: every accelerator experiment, every cosmic-ray measurement, and every relativistic scattering calculation uses this formula.
Relativistic Momentum Formula
The defining relationship, where γ is the Lorentz factor, is:
Rearranged to solve for the other two unknowns:
v = p · c / √( p² + m² · c² )
The velocity formula is worth emphasising: it is the closed-form solution of a nonlinear equation, and it can be written even more compactly using the energy–momentum identity. With total energy E = √((pc)² + (mc²)²), the velocity is simply:
The classical limit at low speeds is recovered by expanding γ for small β:
which shows that the relativistic correction to momentum is second-order in β — the reason Newtonian mechanics works so well for everyday motion.
Relativistic Momentum Variables
| Symbol | Meaning | Notes |
|---|---|---|
| p | Relativistic momentum | Grows without bound as v → c |
| m | Rest mass | Lorentz-invariant — the same in every frame |
| v | Velocity | Relative to the observer; must be < c for massive particles |
| c | Speed of light in vacuum | 299,792,458 m/s, exact by definition |
| β | v/c | Dimensionless speed; the natural variable of relativity |
| γ | Lorentz factor | γ = 1/√(1 − β²); approaches ∞ as β → 1 |
| E | Total energy | E = γmc² = √((pc)² + (mc²)²) |
| K | Kinetic energy | K = (γ − 1)mc²; reduces to ½mv² at low speeds |
How Each Variable Affects the Momentum
Momentum is directly proportional to rest mass: double the mass and you double the momentum at the same velocity. This part is unchanged from Newtonian mechanics. The velocity dependence, however, is radically different. In Newtonian mechanics p grows linearly with v and there is no limit; in relativity p grows as γv, and γ diverges as v approaches c. The first relativistic correction is small — at β = 0.1, γ ≈ 1.005, so the momentum is only 0.5% larger than Newtonian. At β = 0.5, γ ≈ 1.155, so the momentum is 15% larger. At β = 0.9, γ ≈ 2.294, so the momentum is 129% larger — more than double the Newtonian value. At β = 0.99, γ ≈ 7.089, and the momentum is more than seven times the Newtonian prediction. At β = 0.999, γ ≈ 22.37. And at β = 0.999999, γ ≈ 707.
This divergence is what makes c a speed limit. To accelerate a particle from 0.99c to 0.999c requires adding an impulse Δp = γ(0.999)mv − γ(0.99)mv ≈ 15.28mv, where m is the rest mass. For an electron that is about 1.4 × 10⁻²⁰ kg·m/s. But to accelerate the same electron from 0.999c to 0.9999c, the required impulse is γ(0.9999)mv − γ(0.999)mv ≈ 48.4mv — over three times larger for a change in speed that is ten times smaller. Extrapolating, the impulse required to reach c exactly is infinite. This is the deepest reason why c is a speed limit: the formula p = γmv makes it so.
In particle physics, momenta are usually quoted in eV/c (or keV/c, MeV/c, GeV/c, TeV/c), because these units connect directly to the energies that accelerators deliver. The conversion is 1 eV/c = 5.344286 × 10⁻²⁸ kg·m/s, so a proton with momentum 7 TeV/c carries 7 TeV of energy in the ultra-relativistic limit — the same as the beam energy of the LHC. This is why particle physicists use natural units in which c = 1: the momentum and energy become interchangeable, and the formula simplifies to E² = p² + m².
Relativistic Momentum Calculation Examples
β = 0.8, β² = 0.64, 1 − β² = 0.36
γ = 1/√0.36 = 1.66667
p = γmv = 1.66667 × 9.109 × 10⁻³¹ × 2.398 × 10⁸
p = 3.641 × 10⁻²² kg·m/s = 681.4 keV/c
γ = 1.66667
m = p / (γv) = 3.641 × 10⁻²² / (1.66667 × 2.398 × 10⁸)
m = 9.109 × 10⁻³¹ kg
pc = 3.641 × 10⁻²² × 2.998 × 10⁸ = 1.0918 × 10⁻¹³
mc² = 9.109 × 10⁻³¹ × (2.998 × 10⁸)² = 8.187 × 10⁻¹⁴
E = √((pc)² + (mc²)²) = 1.3646 × 10⁻¹³
v = pc² / E = 2.398 × 10⁸ m/s = 0.800c
pc = 7 × 10¹² eV = 1.1216 × 10⁻⁶ J
mc² = 1.673 × 10⁻²⁷ × (2.998 × 10⁸)² = 1.5036 × 10⁻¹⁰ J
γ = E/(mc²) = √((pc)² + (mc²)²)/(mc²) ≈ pc/(mc²) = 7,463
β = √(1 − 1/γ²) ≈ 1 − 9 × 10⁻⁹ — that is, v = 0.999999991c
Frequently Asked Questions
The classical formula p = mv cannot be correct at high speeds, because it would allow a particle to be accelerated past the speed of light by any finite impulse. The relativistic formula p = γmv was constructed to satisfy two requirements: (1) momentum must be conserved in every inertial frame, and (2) the speed of light must be the same in every frame. The Lorentz factor γ = 1/√(1 − v²/c²) is exactly the factor needed to make momentum conservation compatible with the Lorentz transformation between frames. In the limit v ≪ c, γ ≈ 1 and the classical formula is recovered — so Newtonian mechanics is not wrong, it is simply the low-speed approximation of a deeper theory.
The idea of "relativistic mass" γm was popularised by Einstein's early followers and appears in many older textbooks, but it has been almost entirely abandoned in modern physics. The reason is that if you define mass as γm, you get a quantity that depends on the observer's frame — and mass, by its very nature, should be a property of the particle alone. Modern treatments keep m as the rest mass (a Lorentz-invariant) and put all the velocity dependence into γ. When you read "mass increases with speed," you should mentally substitute "momentum and energy increase with speed." The invariant mass of a system is still a meaningful quantity — it is what an observer at rest with respect to the system would measure.
The Lorentz factor γ = 1/√(1 − v²/c²) is the central quantity of special relativity. It appears in time dilation (Δt = γΔt₀), length contraction (L = L₀/γ), relativistic momentum (p = γmv), total energy (E = γmc²), and the Lorentz transformation itself. It equals 1 when v = 0, and approaches infinity as v → c. For everyday speeds it is indistinguishable from 1 — a jet at 900 km/h has γ = 1.0000000000003 — but for relativistic particles it can be enormous: the LHC proton beam has γ ≈ 7,463, and the highest-energy cosmic ray ever recorded had γ ≈ 10¹¹. To calculate it: γ = 1/√(1 − β²), where β = v/c.
Because the relativistic momentum formula diverges at v = c. If you try to accelerate a massive particle closer and closer to c, the impulse required grows without bound: pushing the particle from 0.99c to 0.999c requires a certain impulse; going from 0.999c to 0.9999c requires a larger impulse, and so on. The total impulse needed to reach c is infinite. This is not a mechanical or engineering limitation — it is a structural feature of spacetime described by the Lorentz transformation. Only massless particles like the photon can travel at c, because their momentum p = E/c is finite and does not require a Lorentz factor. Massive particles are forever confined to v < c.
Particle physicists work in natural units where c = 1, so momentum and energy have the same dimensions and are quoted in the same units — eV, keV, MeV, GeV, TeV. To convert: 1 eV/c = e/c kg·m/s = 1.602176634 × 10⁻¹⁹ / 299792458 = 5.344286 × 10⁻²⁸ kg·m/s. Going the other way, multiply kg·m/s by 1.8685 × 10²⁷ to get eV/c. The relation is exact and depends only on the elementary charge and the speed of light. For quick reference: a proton at rest has mass 938 MeV/c², an electron has 0.511 MeV/c², and a 1 kg object at 1 m/s has momentum 5.34 × 10²¹ keV/c.
A photon has zero rest mass, so the formula p = γmv does not directly apply — γ would be infinite and m zero, an indeterminate form. Instead, the photon's momentum is determined by its energy: p = E/c = h/λ, where h is Planck's constant and λ is the wavelength. Equivalently, p = ħk where k = 2π/λ is the wave number. This is the de Broglie relation, and it applies to all particles — massive or massless — when the momentum is expressed in terms of wavelength. A visible photon at 550 nm has momentum about 1.2 × 10⁻²⁷ kg·m/s, roughly 1.2 eV/c.
By bending the particle in a magnetic field. A charged particle moving through a magnetic field B experiences a Lorentz force qv × B that is perpendicular to its velocity, so its trajectory becomes circular with radius R satisfying p = qBR (in SI units with p in kg·m/s). By measuring the curvature of the track and knowing the magnetic field, physicists can compute the momentum directly. This is how the LHC measures proton momenta: superconducting dipoles generate fields up to 8.3 T, and the radius of curvature of the ring is 2,800 m, giving a proton momentum of 7 TeV/c. The same technique works for any charged particle, from electrons in a cathode-ray tube to muons in a cosmic-ray shower.
They are the two components of the relativistic four-momentum (E/c, p), a four-vector whose invariant length squared is (mc)². This gives the fundamental identity E² = (pc)² + (mc²)², sometimes called the energy–momentum triangle. For a particle at rest, p = 0 so E = mc². For a massless particle, m = 0 so E = pc. For a slow massive particle, expanding the square root gives E ≈ mc² + p²/(2m), which is rest energy plus classical kinetic energy. Every relativistic collision problem can be solved using this identity plus conservation of energy and momentum — the two quantities are locked together by the geometry of Minkowski space.
Variables
| Symbol | Quantity | Unit |
|---|---|---|
| KE_rot | Rotational kinetic energy | J |
| I | Moment of inertia about the rotation axis | kg.m2 |
| omega | Angular velocity | rad/s |
What it means
Rotational kinetic energy is the energy possessed by an object due to its rotation. It is analogous to translational kinetic energy (½mv²), replacing mass with moment of inertia and linear velocity with angular velocity. This energy is stored in rotating systems like flywheels, turbines, and spinning wheels. The conservation of energy applies: rotational kinetic energy can be converted to other forms. In engineering, flywheels store energy for smoothing power output, and wheels store energy in vehicles. In physics, it is used to analyse rotating bodies and to solve problems involving both translation and rotation (rolling motion). The formula is derived from summing the kinetic energies of all mass elements. Understanding this energy is essential for designing efficient rotational systems.
Worked example
Rotational Kinetic Energy – Two Examples
Real‑World| Parameter | Value |
|---|---|
| I | 5 kg·m² |
| ω | 10 rad/s |
| Parameter | Value |
|---|---|
| I | 2 kg·m² |
| ω | 20 rad/s |
Common mistakes
- Missing ½ factor: KE_rot = ½ I ω² – do not omit the ½.
- Moment of inertia I: About the axis of rotation – must match the axis for the motion.
- Angular velocity ω: In rad/s – not rpm.
- Units: I in kg·m², ω in rad/s → Joules.
- Translational vs. rotational: Total KE may include both translational and rotational parts.
Applications
Rotational kinetic energy, KE_rot = ½Iω², is the energy of a rotating body. It is used to analyse energy storage in flywheels, the motion of wheels and gears, and the dynamics of spinning objects. Engineers use this formula to design energy‑efficient rotating machinery, to calculate the energy required to spin up turbines, and to evaluate the performance of gyroscopes. In automotive engineering, it is applied to the design of rotating parts like crankshafts and wheels. In renewable energy, it helps assess the energy content of wind turbine rotors. By understanding rotational kinetic energy, professionals can optimise the design of rotating systems, improve energy efficiency, and predict the behaviour of spinning objects in various applications.
- Flywheel energy storage system design
- Performance analysis of turbines, motors, and generators
- Design of gyroscopes and inertial navigation systems
- Automotive drivetrain and wheel dynamics
- Wind turbine rotor energy assessment
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