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Relativistic Kinetic Energy
13,927,755,897,018,748
joules (J)
Lorentz factor (γ) 1.154967
Speed ratio (v/c = β) 0.500346
Classical KE (½mv²) 11,250,000,000,000,000 J

What is relativistic kinetic energy?

In special relativity, a moving object's kinetic energy grows faster than Newton's \(\frac{1}{2}mv^2\) predicts as its speed approaches the speed of light \(c \approx 299{,}792{,}458\ \text{m/s}\). The exact expression is $$KE = (\gamma - 1)\,mc^2,$$ where \(\gamma\) (the Lorentz factor) equals \(\frac{1}{\sqrt{1 - (v/c)^2}}\). As \(v \to c\), \(\gamma \to \infty\), so an infinite amount of energy would be required to reach light speed — which is why no massive object can.

Graph comparing relativistic and classical kinetic energy versus velocity fraction
Relativistic kinetic energy diverges toward infinity as v approaches c, while the classical formula underestimates it.

How to use this calculator

Enter the object's rest mass in kilograms and its velocity in metres per second. The calculator returns the relativistic kinetic energy in joules, the Lorentz factor \(\gamma\), the speed ratio \(\beta = v/c\), and the classical \(\frac{1}{2}mv^2\) value so you can see how large the relativistic correction is.

The formula explained

First compute \(\beta = v/c\). Then \(\gamma = \frac{1}{\sqrt{1 - \beta^2}}\). The relativistic kinetic energy is the total relativistic energy \(\gamma mc^2\) minus the rest energy \(mc^2\), giving \((\gamma - 1)mc^2\). At low speeds \(\gamma \approx 1 + \frac{1}{2}\beta^2\), so \((\gamma - 1)mc^2 \approx \frac{1}{2}mv^2\), recovering the familiar Newtonian formula.

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Lorentz factor gamma rising sharply as velocity approaches the speed of light
The Lorentz factor γ equals 1 at rest and grows without bound as v approaches c.

Worked example

Take \(m = 1\ \text{kg}\) moving at \(v = 150{,}000{,}000\ \text{m/s}\). Then \(\beta = 150{,}000{,}000 / 299{,}792{,}458 \approx 0.50035\), \(\beta^2 \approx 0.25035\), \(\gamma = \frac{1}{\sqrt{0.74965}} \approx 1.15490\). $$KE = (1.15490 - 1) \times 1 \times (299{,}792{,}458)^2 \approx 1.39 \times 10^{16}\ \text{J}.$$ The classical estimate \(\frac{1}{2} \cdot 1 \cdot 150{,}000{,}000^2 = 1.125 \times 10^{16}\ \text{J}\) understates it noticeably at this high speed.

FAQ

Why does it differ from \(\frac{1}{2}mv^2\)? The classical formula is only the first term of the relativistic series; the difference becomes significant above roughly 10% of light speed.

What if I enter \(v \geq c\)? No massive object can reach or exceed c, so the calculator returns zero for impossible inputs.

What units are used? SI throughout: mass in kg, velocity in m/s, energy in joules.

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