# 工程原理 Engineering Principles GENG0005W1-01

$$m=m_{0} \sqrt{1-\frac{v^{2}}{c^{2}}}$$
where
$m$ – mass of a moving body (also called the variable mass)
$m_{0}$ – rest mass of a body (velocity is zero)
$v-$ velocity of a moving body
$c$ – speed of light
The ratio, $v^{2} / c^{2}$ is usually denoted as $\beta^{2}$. Thus
$$m=m_{0} / \sqrt{1-\beta^{2}}$$
Similarly, the relativistic energy of a body moving with velocity $v$ is no more expressed as $m v^{2} / 2$, but
$$E=m_{0} c^{2} / \sqrt{1-\beta^{2}}$$

## GENG0005W1-01 COURSE NOTES ：

$$E^{2}=(p c)^{2}+\left(m_{0} c^{2}\right)^{2}$$
By definition, momentum can be described as a function of the mass and velocity of a moving body:
$$p=m v=m_{0} v / \sqrt{1-\beta^{2}}$$
Squaring Eq. (3-6)
\begin{aligned} &E^{2}=\left(m c^{2}\right)^{2}=\left(m_{0} c^{2}\right)^{2} / 1-\frac{v^{2}}{c^{2}} \rightarrow m^{2} c^{4}\left(1-\frac{v^{2}}{c^{2}}\right)=m_{0}^{2} c^{4} \ &m^{2} c^{4}=E^{2}=m_{0}^{2} c^{4}+m^{2} c^{2} v^{2} \end{aligned}
where $p=m v$, thus giving
$$E^{2}=(p c)^{2}+\left(m_{0} c^{2}\right)^{2}$$
For a massless particle (like a photon) it follows that the total energy depends on its momentum and the speed of light: $E=p c$. This aspect will be discussed in greater detail in later sections.