Lorentz's Transform:

$S'$ move at the speed of $u$(direction $+x$) relative to $S$

$$ (x',y',z',t')=L(x,y,z,t):\\ x'=\frac{x-ut}{1-u^2/c^2} \\ y'=y\\ z'=z\\ t'=\frac{t-ux/c^2}{1-u^2/c^2}\\ $$

let $\beta=u/c<1,\gamma=\frac{1}{1-u^2/c^2}=\frac{1}{1-\beta^2}\geq1$:

$$ x'=\gamma(x-ut)\\ y'=y\\ z'=z\\ t'=\gamma(t-ux/c^2)\\ $$

in time $\Delta t$:

$$ \Delta x'=\gamma(\Delta x -u\Delta t)\\ \Delta y'=\Delta y\\ \Delta z'=\Delta z\\ \Delta t'=\gamma(\Delta t -u\Delta x/c^2) $$

Think:
what if $\Delta x=0$ or $\Delta t=0$?

  1. $\Delta x=0\Rightarrow\Delta t'=\gamma\Delta t\geq\Delta t$, time dilation(时间膨胀).
    When we measure the time at the same place($\Delta x=0$), we observe a time dilation(not a fact).
  2. $\Delta t=0\Rightarrow\Delta x'=\gamma\Delta x\geq\Delta x$,length contraction(长度收缩).
    When we observe(measure) the length of an object, we get the position of beginning($x_0$) and end($x_1$) in the same time($\Delta t=0$), so we observe a length contraction(not a fact).

thus,the velocity:

$$ v'_x=\frac{dx'}{dt'}=\frac{\gamma(dx-udt)}{\gamma(dt-udx/c^2)}=\frac{v_x-u}{1-uv_x/c^2}\\ v'_y=\frac{dy'}{dt'}=\frac{dy}{\gamma(dt -udx/c^2)}=\frac{v_y}{\gamma(1-uv_x/c^2)}\not=v_y\\ v'_z=\frac{dz'}{dt'}=\frac{dz}{\gamma(dt -udx/c^2)}=\frac{v_z}{\gamma(1-uv_x/c^2)}\not=v_z\\ $$

Momentum & Energy

$\bold{p}=\frac{m\bold{v}}{\sqrt{1-v^2/c^2}}=m_{rel}\bold{v}$,where $m_{rel}=\frac{m}{\sqrt{1-v^2/c^2}}$
$K = \int\bold{F}\cdot d\bold{r}=\int[\frac{d}{dt}(m_{rel}\bold{v})]\cdot\bold{v}dt=\int v^2dm_{rel}+m_{rel}d(v^2)$
$(dm_{rel}=md\frac{1}{\sqrt{1-v^2/c^2}}=m\frac{vdv}{c^2(1-v^2/c^2)^{\frac{3}{2}}}=m_{rel}\frac{vdv}{c^2-v^2}\Rightarrow m_{rel}d(v^2)=m_{rel}vdv=(c^2-v^2)dm_{rel})$
$=\int_m^{m_{rel}}c^2dm_{rel}=(m_{rel}-m)c^2=\Delta m\ c^2$
$E=K+m_0c^2=\sqrt{(pc)^2+(m_0c^2)^2}$