Relativistic Energy Calculator
Enter mass m and velocity v to compute total energy E=γmc², rest energy E0=mc² and kinetic energy K=(γ−1)mc². v=0.9c, m=1 kg → E≈2.29×10¹⁷ J, K≈1.94×10¹⁷ J.
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At a glance:Relativistic energy (Einstein, 1905, special relativity): the total energy of a particle is E=γmc², where m is rest mass, c the speed of light, and γ=1/√(1−v²/c²) the Lorentz factor. Rest energy E0=mc² (mass-energy equivalence); kinetic energy K=E−E0=(γ−1)mc². As v→c, γ→∞ so energy diverges — no massive particle can reach light speed (would need infinite energy). At low speed (v≪c) K≈½mv² (classical limit). Momentum is p=γmv, and the energy-momentum relation E²=(pc)²+(mc²)². History: Einstein's 1905 paper 'Does the inertia of a body depend on its energy content?' established E=mc²; Minkowski (1908) and Planck (1907) developed the 4-vector formalism. Classic example: m=1 kg at v=0.9c → γ=2.294, E=2.294×9×10¹⁶=2.06×10¹⁷ J, E0=9×10¹⁶ J, K=1.16×10¹⁷ J (huge — 1 kg mass-energy = 21.5 megatons TNT). Applications: (1) nuclear fission/fusion (mass defect → energy); (2) particle accelerators (LHC protons γ~7000, E~7 TeV); (3) PET scans (positron annihilation E=mc²); (4) GPS (relativistic time correction); (5) mass of binding energy.
Formula
Lorentz factor: γ = 1 / √(1 − v²/c²)
Total energy: E = γ·m·c²
Rest energy: E0 = m·c²
Kinetic energy: K = (γ − 1)·m·c²
Energy-momentum: E² = (p·c)² + (m·c²)²
$$E = \gamma m c^2, \quad \gamma = \frac{1}{\sqrt{1-v^2/c^2}}, \quad K = (\gamma-1)mc^2, \quad E^2 = (pc)^2 + (mc^2)^2$$How to Use
- Enter mass m (kg) and velocity v (m/s) or v/c fraction.
- The tool computes γ, E0=mc², E=γmc², K=(γ−1)mc².
- Example: v=0.9c, m=1 kg → E≈2.06e17 J, K≈1.16e17 J.
Relativistic Energy vs Speed
| v/c | γ | K/(mc²) |
|---|---|---|
| 0.1 | 1.005 | 0.005 |
| 0.5 | 1.155 | 0.155 |
| 0.9 | 2.294 | 1.294 |
| 0.99 | 7.089 | 6.089 |
| 0.999 | 22.37 | 21.37 |
γ=1/√(1−v²/c²). As v→c, γ and K diverge. At v=0.9c kinetic energy already exceeds rest energy (K/mc²=1.29). The classical ½mv² underestimates K badly at high v.
Case Studies
LHC Proton Acceleration
LHC accelerates protons to 6.5 TeV. Rest energy mc²=938 MeV, so γ=6500/0.938≈6930.
v = c·√(1−1/γ²) ≈ c·(1−1×10⁻⁸), i.e. 3 m/s slower than c — essentially light speed but never reaching it.
The kinetic energy 6.5 TeV >> rest 0.938 GeV (7000×) shows how relativistic the beam is; collisions reach 13 TeV center-of-mass.
Nuclear Fission Energy Release
A uranium-235 fission releases ~200 MeV. The mass defect Δm=200 MeV/c²≈3.6×10⁻²⁸ kg per fission.
In 1 kg of U-235, ~8×10²⁴ fissions release E=Δm·c²≈2×10¹⁴ J — equivalent to 48 kilotons TNT.
This is pure E=mc²: a tiny mass loss becomes enormous energy because c²=9×10¹⁶. Nuclear power exploits this mass defect.
FAQ
Why can nothing with mass reach light speed?
Total energy E=γmc² with γ=1/√(1−v²/c²). As v→c, γ→∞ so E→∞. To accelerate a massive particle to c would require infinite energy, impossible. Only massless particles (photons) travel at c, and they always do. This is a direct consequence of special relativity's postulates. Particle accelerators approach c but never reach it (LHC protons at 0.999999991c need 7 TeV for γ≈7000).
How does relativistic kinetic energy differ from classical?
Classical K=½mv² holds only for v≪c. Relativistically K=(γ−1)mc². At low speed γ≈1+½(v/c)² so K≈½mv² (matches). At high speed K grows faster and diverges as v→c. At v=0.9c, K/mc²=1.29 (classical would give 0.405 — a 3× error). The relativistic formula is exact; classical is the low-speed approximation.
What is rest energy E0=mc²?
Even at rest, a mass m contains energy E0=mc² (e.g. 1 kg = 9×10¹⁶ J ≈ 21.5 megatons TNT equivalent). This is 'locked' energy released in nuclear reactions via mass defect (ΔE=Δmc²). It is not 'stored fuel' but the equivalence of mass and energy — mass is a form of energy. In particle physics, rest energy is the threshold to create particles (e.g. pair production needs 2mc²=1.022 MeV for e⁺e⁻).
What is the energy-momentum relation?
E²=(pc)²+(mc²)² unifies total, momentum and rest energy. For massless particles (m=0) it gives E=pc (photons). For a particle at rest (p=0) it gives E=mc². It is the invariant magnitude of the energy-momentum 4-vector: E²−(pc)²=(mc²)² is the same in all frames. This is why photons have momentum p=E/c despite zero rest mass.
Why does GPS need relativity?
GPS satellites move at ~3.9 km/s and sit ~20,200 km up. Special relativity slows their clocks by ~7 μs/day (γ>1); general relativity speeds them by ~45 μs/day (weaker gravity). Net +38 μs/day. At light speed this is a 11 km position error per day. The GPS system applies relativistic corrections to clock rates — without E=mc²-type relativistic timing, navigation would fail within minutes. Relativity is practical engineering here.
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References
Content reviewed by the Calculatorism editorial team. Results are for reference only; please refer to the relevant authorities for the official figures.