这是一份southampton南安普敦大学GENG0005W1-01作业代写的成功案例
$$
m=m_{0} \sqrt{1-\frac{v^{2}}{c^{2}}}
$$
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.
