Consider a particle P that is described by its rest mass $m$ and momentum $p$. And please notice that these numbers have been defined by a methodical description of sensation. Definition: the **mechanical energy** of P is

$E \equiv \sqrt{ c^{2}p^{2} + m^{2}c^{4} \vphantom{\sum^{2}} \ }$

where $c$ is a constant. This statement comes from Albert Einstein's theory of special relativity. For material particles, we can divide the foregoing definition by $m c^{2}$ to get$E = m c^{2} \sqrt{ \ 1 + \left( p/mc \right)^{2} \ }$

This square root can be expanded in a binomial series as$\begin{align} 1 + \frac{p^{2}}{2m^{2}c^{2}} - \frac{p^{4}}{8m^{4}c^{4}} + \frac{3p^{6}}{48m^{6}c^{6}} + \ldots \end{align}$

And if motion is not relativistic we can ignore the smaller terms to approximate the mechanical energy by

$\begin{align} E \simeq m c^{2} \left( 1 + \frac{p^{2}}{2m^{2}c^{2}} \right) \end{align}$

But for slow motion, remember that $\gamma$ the Lorentz factor can also be approximated as

$\begin{align} \gamma \simeq 1 + \frac{p^{2} }{2m^{2} c^{2}} \end{align}$

So that overall

$E \simeq \gamma m c^{2}$