这是一份ucl伦敦大学学院 math0055作业代写的成功案
问题 1.
\begin{aligned}
\boldsymbol{p} &=m \boldsymbol{v}\left(1-\beta^{2}\right)^{-1 / 2} \
& \approx m v+\frac{1}{2} m \frac{v^{2}}{c^{2}} \boldsymbol{v} \
& \approx m \boldsymbol{v}
\end{aligned}
证明 .
\begin{aligned}
E &=m c^{2}\left(1-\beta^{2}\right)^{-1 / 2} \
& \approx m c^{2}+\frac{1}{2} m v^{2}
\end{aligned}
MATH0055 COURSE NOTES ：
A very useful equation suggested by the new, correct expressions for $E$ and $\boldsymbol{p}$ is
$$
\boldsymbol{v}=\frac{\boldsymbol{p} c^{2}}{E}
$$
By taking the magnitude-squared of $\vec{p}$ we get a relation between $m, E$ and $p \equiv|\boldsymbol{p}|$,
$$
|\vec{p}|^{2}=m^{2} c^{2}=\left(\frac{E}{c}\right)^{2}-p^{2}
$$
which, after multiplication by $c^{2}$ and rearrangement becomes
$$
E^{2}=m^{2} c^{4}+p^{2} c^{2}
$$