Relativistic Total Energy

Relativistic energy: E = γmc²

Einstein's total relativistic energy is E = γmc². Here m is the rest mass (kg), c ≈ 3·10⁸ m/s, and γ = 1/√(1 − v²/c²) is the Lorentz factor. Put the object at rest (v = 0) and it falls back to the famous E₀ = mc², the rest energy. The relativistic kinetic energy works out to KE = (γ − 1)mc², which reduces to ½mv² once v ≪ c. Example: a proton has a rest energy of 938 MeV and an electron 511 keV. At the LHC, protons reach 7 TeV, so γ ≈ 7460, meaning they carry around 7460× their rest energy.

Applications: accelerators, nuclear reactors, stars

E = γmc² is what governs particle accelerators like the LHC and RHIC. It explains the energy let loose in nuclear fission (U-235 reactors and atomic bombs; Hiroshima turned about 0.7 g of mass into roughly 70 TJ) and in stellar fusion, where 4 H → He releases 26.7 MeV per cycle and keeps the Sun burning. The same equation sits behind PET scans, where positron–electron annihilation produces 511 keV photons, and it fixes the energy scale for cosmic-ray physics.

FAQ

What does γ mean? It's the Lorentz factor, a measure of how much time dilates and energy grows as v approaches c. At rest γ = 1, and it blows up as v → c.

Why can't massive objects reach c? As v → c, γ → ∞, which sends E → ∞. You'd need infinite energy to get there, so only massless particles such as photons travel at c.

Does ½mv² still work? Only when v ≪ c. Once you're past about 10% of c, the classical formula starts to undershoot the real energy by a noticeable margin, so switch to (γ − 1)mc².

Where does the energy in a nuclear bomb come from? A tiny slice of the mass, around 0.1%, gets converted into energy through E = mc². That enormous c² factor is what turns a few grams into terajoules.