# Why is $E=mc^2$ and not $E=m\frac{c^2}{2}$?

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

$$E=m\frac{v^2}{2}.$$

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

$$E=m\frac{c^2}{2}.$$

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

## 3 Answers

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 $$v\ll c$$).

$$E=mc^2$$ 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 $$E = mc^2$$

• Given the relative nature of velocity, what does at rest mean? – TheEnvironmentalist Nov 24 '18 at 10:38
• It means at rest WRT you, who measures the KE. – m4r35n357 Nov 24 '18 at 12:12