Relativistic Energy Calculator
Compute the total relativistic energy E = γmc² of a particle from its rest mass and velocity. Reduces to E = mc² at rest and grows without bound as velocity approaches the speed of light.
Last updated: May 2026
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About this calculator
The formula is E = m·c² / √(1 − (v/c)²) = γ·m·c², where m is the rest mass in kg, v is the velocity in m/s, and c = 299,792,458 m/s. The output is the total energy in joules, including both rest energy (mc²) and kinetic energy (γ−1)·mc². At v = 0, E = mc² — the famous mass-energy equivalence. As v approaches c, γ diverges and so does E, reflecting that infinite energy would be needed to accelerate a massive particle to light speed. Kinetic energy KE = (γ − 1)mc² reduces to ½mv² in the non-relativistic limit (v ≪ c, γ ≈ 1 + ½(v/c)²) and grows without bound at relativistic speeds. Edge cases: rest mass is invariant (frame-independent), but total energy is frame-dependent — different observers measure different E for the same particle. For v ≥ c the formula gives NaN; massive objects cannot reach c. For massless particles (photons), the formula breaks down — they have m = 0 and rest energy 0, but their total energy is E = pc = hν (photon energy from frequency), and γ is undefined because they always travel at c. The relationship E² = (pc)² + (mc²)² unifies massive and massless particles: massive particles have mc² > 0; photons have m = 0 so E = pc.
How to use
Example 1 — proton at 0.99c. m = 1.673 × 10⁻²⁷ kg (proton rest mass), v = 2.969 × 10⁸ m/s (0.99c). Step 1: mc² = 1.673e−27 × (2.998e8)² ≈ 1.673e−27 × 8.988e16 ≈ 1.504e−10 J. Step 2: (v/c)² = 0.9801. Step 3: 1 − 0.9801 = 0.0199. Step 4: √0.0199 ≈ 0.1411. Step 5: E = 1.504e−10 / 0.1411 ≈ 1.066e−9 J. Verify: γ = 7.089, and γ·mc² = 7.089 × 1.504e−10 ≈ 1.066e−9 J ✓. Convert to eV: 1.066e−9 / 1.602e−19 ≈ 6.65 GeV — a proton at 0.99c carries about 6.65 GeV total energy, with rest energy ~938 MeV and kinetic energy ~5.71 GeV. Example 2 — electron at 0.5c. m = 9.109 × 10⁻³¹ kg, v = 1.499 × 10⁸ m/s. Step 1: mc² = 9.109e−31 × 8.988e16 ≈ 8.187e−14 J = 511 keV (the electron rest mass-energy). Step 2: (v/c)² = 0.25. Step 3: 1 − 0.25 = 0.75. Step 4: √0.75 ≈ 0.8660. Step 5: E = 8.187e−14 / 0.8660 ≈ 9.453e−14 J ≈ 590 keV. Verify: γ at 0.5c is 1.1547, and γ·mc² = 1.1547 × 511 ≈ 590 keV ✓. Kinetic energy is γ·mc² − mc² ≈ 79 keV, contrasting with the non-relativistic value ½mv² ≈ 64 keV — the relativistic kinetic energy is ~24% higher even at this moderate v.
Frequently asked questions
What's the difference between rest energy, kinetic energy, and total energy?
Rest energy E₀ = mc² is the energy a particle has at rest — purely from its mass. Kinetic energy KE = (γ − 1)·mc² is the additional energy from motion; at low speeds it reduces to the classical ½mv², but at high speeds it diverges as v approaches c. Total energy E = γ·mc² is their sum, E = E₀ + KE. Mass-energy equivalence (E₀ = mc²) means that mass can be converted to energy: nuclear reactions release energy because the products have less total mass than the reactants, with the missing mass radiated as energy (γ rays, kinetic energy of products). Conversely, energy can become mass: in pair production a gamma ray of E > 2·mc² of an electron-positron pair (~1.02 MeV) can convert to a particle-antiparticle pair. Notice that mc² for an electron is 511 keV, for a proton 938 MeV, for a neutron 940 MeV — these are the 'rest energies' that appear in particle physics tables and are the threshold energies for creating these particles in collisions.
Why does kinetic energy not reduce to ½mv² at high speeds?
Classical kinetic energy KE = ½mv² is derived from Newtonian mechanics, which is the low-velocity limit of relativity. The full relativistic kinetic energy is KE = (γ − 1)·mc². Expanding γ in a Taylor series for small v/c: γ ≈ 1 + ½(v/c)² + ⅜(v/c)⁴ + ..., so KE ≈ ½mc²·(v/c)² + ⅜mc²·(v/c)⁴ + ... ≈ ½mv² + corrections of order v²/c² and higher. At v = 0.1c, the classical formula under-estimates KE by ~0.4%; at v = 0.5c, by ~15%; at v = 0.9c, by ~60%. So the classical formula is excellent for everyday speeds (everything we encounter is v/c < 10⁻⁶) but fails dramatically for particle physics. The same expansion shows why relativistic momentum p = γmv differs from classical p = mv only at high speed. Energy-momentum relations are governed by the relativistic four-vector E² = (pc)² + (mc²)², which contains all the above as special cases.
How does E = mc² relate to nuclear reactions and energy production?
Nuclear reactions release energy when the products have less total rest mass than the reactants — the missing mass is converted to energy via E = Δm·c². In a typical fission reaction (e.g., uranium-235 splitting), the products have about 0.1% less mass than U-235; this 0.001 of the original mass-energy is released as the energy that powers nuclear reactors and weapons. For fusion, hydrogen → helium loses about 0.7% of its mass, which is why fusion is so much more efficient per kg of fuel than fission. Conversely, chemical reactions (combustion, batteries) release at most about 10⁻⁹ of their reactants' mass as energy — about 10⁷ times less than nuclear. This is why nuclear reactions, despite involving the same chemical elements rearranged at the nuclear level, release millions of times more energy per kg than chemical reactions. Mass-energy equivalence also explains why the sun continuously loses ~4 million tonnes per second to power its luminosity — that's about 4 × 10²⁶ joules per second.
What are the common mistakes when computing relativistic energy?
The biggest mistake is using non-SI units without conversion — m must be in kg, v in m/s, and c in m/s. Mixing in eV or g without rescaling gives wildly wrong numbers. The second is confusing rest energy (mc²) with total energy (γmc²) or kinetic energy ((γ−1)mc²); the formula here returns total energy. The third is applying the formula to massless particles like photons, which have m = 0 and need E = pc = hν instead. People also use classical kinetic energy ½mv² at relativistic speeds, under-estimating energy by factors of 2–10× at v > 0.5c. Floating-point precision matters for ultra-relativistic particles: at γ > 10⁸, double precision loses digits in 1 − (v/c)². Finally, mixing 'rest mass' (the invariant scalar) with 'relativistic mass' (γm, an older convention) leads to expressions like E = m_rel·c² where m_rel = γm — true but inconsistent with the modern convention where m always means rest mass.
When should I not use this calculator?
Do not use it for massless particles (photons, gluons, gravitons) — they have m = 0 and the formula gives 0 regardless of velocity. Use E = hν (Planck-Einstein) or E = pc instead. It is not appropriate for accelerating particles or in gravitational fields — these need general relativity or instantaneous-rest-frame analysis. Do not use it for systems of particles (atoms, molecules) without considering binding energies; the total mass of a bound system is less than the sum of constituent rest masses by the binding energy divided by c². It is unsuitable for v ≥ c (unphysical for massive objects) — the formula gives NaN. For very high γ (above ~10⁸), use extended precision to avoid losing digits in 1 − (v/c)²; alternatively express E in terms of momentum via E² = (pc)² + (mc²)². Avoid using it as a substitute for full quantum-relativistic treatment in particle physics (Dirac equation, quantum field theory) — the simple γmc² is the classical relativistic limit and misses spin, antiparticle creation, and quantum effects.