这是一份uwa西澳大学PHYS1200的成功案例

A particle is moving with a speed of modulus $v$ and components $\left(v_{x}, v_{y}, v_{z}\right)$. What is the modulus $v^{\prime}$ of the velocity for an observer moving with a speed $w$ along the $x$ axis? Comment the result as $v$ and/or $w$ approach $c$.
Answer: Applying the relativistic laws for the addition of velocities we find
$$
v^{\prime 2}=v_{x}^{\prime 2}+v_{y}^{\prime 2}+v_{z}^{\prime 2}=\frac{\left(v_{x}-w\right)^{2}+\left(1-w^{2} / c^{2}\right)\left(v_{x}^{2}+v_{y}^{2}\right)}{\left(1-v_{x} w / c^{2}\right)^{2}}
$$
and after some simple algebra
$$
v^{\prime 2}=c^{2}\left(1-\frac{\left(1-v^{2} / c^{2}\right)\left(1-w^{2} / c^{2}\right)}{\left(1-v_{x} w / c^{2}\right)^{2}}\right)
$$
It is interesting to notice that as $v$ and/or $w$ approach $c$, also $v^{\prime}$ approaches $c$ (from below): that verifies that a body moving with $v=e$ moves with the same velocity in every reference frame (invariance of the speed of light).

PHYS1200 COURSE NOTES ：

To derive the equation of motion, let us notice that, in the frame instantaneously at rest with the particle, the velocity goes from 0 to adt in the interval $d \tau$, so that $v$ changes into $(v+a d \tau) /\left(1+\operatorname{vad} \tau / c^{2}\right)$ in the same interval, meaning that $d v / d \tau=a\left(1-v^{2} / c^{2}\right)$. Using previous equations, it is easy to derive
$$
\frac{d u}{d t}=\frac{d u}{d v} \frac{d v}{d \tau} \frac{d \tau}{d t}=a
$$
which can immediately be integrated, with the initial condition $u(0)=v(0)=0$, as $u(t)=a t$. Using the relation between $u$ and $v$ we have
$$
v(t)=\frac{a t}{\sqrt{1+a^{2} t^{2} / c^{2}}} .
$$
That gives the variation of velocity, as observed in the laboratory, for a uniformly accelerated motion: for $t \ll c / a$ one recovers the non-relativistic result, while for $t \gg c / a$ the velocity reaches asymptotically that of light. The dependence of $v$ on $t$ can be finally integrated, using the initial condition $x(0)=0$, giving
$$
x(t)=\frac{c^{2}}{a}\left(\sqrt{1+\frac{a^{2} t^{2}}{c^{2}}}-1\right) .
$$