Kinetic energy for a moving object is the integral of force with respect to distance, often given as:


This would imply that for mass moving at the speed of light, the kinetic energy would be:


This puts it off from the Einstein result by a factor of two. Why the discrepancy?


3 Answers 3


As the other answers point out, the full relativistic energy expression is $$E^2=m^2c^4+p^2c^2$$ where $E$ is the energy, $m$ is the rest mass of the particle, $c$ is the speed of light and $p$ is the momentum of the particle.

If the particle isn't moving (i.e. has $p=0$) then this expression reduces to the famous $$E=mc^2$$ which describes the rest energy of the particle (note this has nothing to do with the kinetic energy, which is $0$ for a particle at rest).

We can obtain the classical formula $E_\text{kinetic}\approx\frac{1}{2}mv^2$ in the following way...

Write the above general energy expression as $$E^2=m^2c^4\left(1+\frac{p^2}{m^2c^2}\right)$$ and then take the square root: $$E=mc^2\left(1+\frac{p^2}{m^2c^2}\right)^{1/2}$$ For a slow moving particle (i.e. one that isn't relativistic, $v\ll c$), one finds that $p$ is quite a lot smaller than the quantity $mc$. This is because the relativistic momentum is given by $p=\gamma mv\approx mv$, where $\gamma=\left(1-\frac{v^2}{c^2}\right)^{-1/2}\approx1$ for a non-relativistic particle, and so the ratio $\frac{p^2}{m^2c^2}\approx\frac{v^2}{c^2}\ll1$.

This means we can binomially expand the above energy expression (valid if $\frac{p^2}{m^2c^2}\approx\frac{v^2}{c^2}\ll1$), giving \begin{align} E&\approx mc^2\left(1+\frac{p^2}{2m^2c^2}+...\right)\\&\approx mc^2\left(1+\frac{m^2v^2}{2m^2c^2}\right)\\&\approx mc^2\left(1+\frac{v^2}{2c^2}\right)\\&\approx mc^2 + \frac{1}{2}mv^2 \end{align}

We see we've obtained the total energy of the particle as the sum of the rest energy (which it always has), and the non-relativistic kinetic energy (valid if $v\ll c$).

  • $\begingroup$ Interestingly, the term, $$\left(1+\frac{v^{2}}{2c^{2}}\right)$$ Describes the Relativistic Doppler effect. In which, to calculate the actual energy of an object, from a moving perspective, you would multiply the observed energy, by this term. $\endgroup$
    – Matrix001
    Commented Jan 10, 2020 at 7:25

$E=mc^2$ isn't supposed to be the body's kinetic energy. In that equation, $c$ is the speed of light, but in the formula for kinetic energy, $c$ (or preferably $v$ is the velocity of the body. Furthermore, there's no body which can be accurately described as having kinetic energy $E=\frac{1}{2}mc^2$, because that implies a massive body travelling at the speed of light.


You cannot make this 1:1 correspondence between the classic kinetic energy of a particle and the rest energy of a relativistic particle. $E = mc^2$ does not apply only to objects moving at the speed of light, but instead to all objects, moving or not. As such, it isn't a relativistic correction of the kinetic energy, but describes the rest energy of a particle.

The relativistic energy of a moving particle is given (from special relativity) by $$E^2 = m^2 c^4 + p^2 c^2$$

This can further be simplified to $$E = \gamma m c^2,$$ where $\gamma = \frac{1}{\sqrt{1-\beta^2}}$ (the Lorentz factor) and $\beta = \frac{v}{c}$.

$\gamma$ is $1$ if the object is at rest, which reduces this to the famous $$E = mc^2$$

  • 2
    $\begingroup$ Given the relative nature of velocity, what does at rest mean? $\endgroup$ Commented Nov 24, 2018 at 10:38
  • 2
    $\begingroup$ It means at rest WRT you, who measures the KE. $\endgroup$
    – m4r35n357
    Commented Nov 24, 2018 at 12:12

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