Switching between Lorentz frames I know that for a Lorentz trasformation:
$$x' = \gamma (x-ut)$$
and
$$t'=\gamma (t-\frac{u}{c^2}x)$$
and 
$$v'=\frac{v-u}{1-\frac{uv}{c^2}}$$
If I want to switch Lorentz frames, and find x, t, and v in terms of x',t', and v', respectively, what would be the proper transitions? I am a bit confused on how this works and the intuition behind it. Thanks!
 A: So with the equations as you've written them, you're assuming that you know $x$, $t$, and $v$ in some reference frame $K$, and that you want to transform them into the corresponding values $x'$, $t'$, and $v'$ in a reference frame $K'$ that is moving with speed $u$ along the $x$-axis. You are also assuming that the primed reference frame is oriented so that its $x'$, $y'$, and $z'$ axes are all parallel to the corresponding axes in frame $K$.
There are two ways to get the inverse equations that express $x$, $t$, and $v$ in terms of $x'$, $t'$, and $v'$. The first is to algebraically rearrange the equations, which is not terribly difficult. But the second way is easier. In the $K'$ frame, an observer regards the $K$ frame as moving with speed $u$ in the $-x'$ direction. So to get the inverse equations, you just interchange the primed and unprimed symbols, and replace $u$ with $-u$:
$$x = \gamma\left(x' + ut'\right),$$
$$t = \gamma\left(t' + \frac{u}{c^2}x'\right),$$
$$v = \frac{v' + u}{1 + \frac{uv'}{c^2}}$$
