In my GR lectures we've derived Rindler coordinates by first showing that the four velocity, which we defined as $$u^{\mu} = (\gamma c, 0, 0, \gamma u),$$ as a function of proper time can be written as $$ u^{\mu}(\tau) = B^{\mu}\sinh{(\frac{g}{c}\tau)} + C^{\mu}\cosh{(\frac{g}{c}\tau)}.$$ Here, $B^{\mu}$ and $C^{\mu}$ are integration constants. By integrating again and using the initial conditions we get $$x^{\mu}(\tau) = \frac{c}{g}B^{\mu}\cosh{(\frac{g}{c}\tau)} + \frac{c}{g}C^{\mu}\sinh{(\frac{g}{c}\tau)} + A^{\mu}$$ where $$C^{\mu} = (c,0,0,0) \text{; } B^{\mu} = (0,0,0,c) \text{; } A^{\mu} = -\frac{c}{g}B^{\mu}.$$ Eventually showing that the $(c\tau,0,0,0)$ coordinate transforms as $$\begin{align} x^{\mu}(\tau) &= \begin{pmatrix} \frac{c^2}{g}\sinh(\frac{g}{c}\tau) \\ 0 \\ 0 \\ \frac{c^2}{g}(\cosh(\frac{g}{c}\tau) - 1) \end{pmatrix} \end{align}$$ Then the lecturer made an argument that don't fully understand. He said that this represents a Lorentz transformation with parameter $\frac{g}{c}\tau$ and, therefore, the coordinates $\xi^{\mu}$ of any other point in the inertial frame transform as $$\begin{align} x^{\mu}(\tau) &= \begin{pmatrix} (\frac{c^2}{g} + \xi^3)\sinh(\frac{g}{c}\tau) \\ \xi^1 \\ \xi^2 \\ (\frac{c^2}{g} + \xi^3)(\cosh(\frac{g}{c}\tau) - 1) \end{pmatrix} \end{align}$$
There are a couple things I don't understand:
1) In the derivation the only thing that applies only to the $(c\tau,0,0,0)$ coordinate we've used is the initial condition $x^{\mu}(0) = 0$. As far as I understand $u^{\mu}(0) = (c,0,0,0)$ and $\frac{d}{d\tau}u^{\mu}(0) = (0,0,0,g)$ apply to all points. So how come we can't just use the initial condition $x^{\mu}(0) = (0,\xi^1,\xi^2,\xi^3)$ to get the transformation for other points? I know this gives a wrong result but I don't see what I am missing.
2) Is there any good justification for the step to get from the transformation for $(c\tau,0,0,0)$ to a transformation for any coordinate in the inertial frame? I know a Lorentz boost can be written as a sort of hyperbolic rotation but how does this justifies changing $\frac{c^2}{g}$ to $\xi^3+\frac{c^2}{g}$?
