Rindler coordinates

Remarks

Check of the metric: $c \, dt = \sinh\eta \, d\rho + \rho \cosh\eta \, d\eta$ and $dx = \cosh\eta \, d\rho + \rho \sinh\eta \, d\eta$. Then $dx^2 - c^2 dt^2 = (\cosh^2\eta - \sinh^2\eta)(d\rho^2 - \rho^2 d\eta^2) = d\rho^2 - \rho^2 d\eta^2$, since the cross terms cancel. The map $(\eta, \rho) \mapsto (\rho \sinh\eta, \rho \cosh\eta)$ is a bijection from $\R \times (0, \infty)$ onto the wedge $x > \abs{ct}$, with inverse $\rho = \sqrt{x^2 - c^2 t^2}$ and $\eta = \operatorname{artanh}(ct / x)$. Non-example: the event $(t, x) = (1/c, 0)$ lies outside $W$ and has no Rindler coordinates, since $x^2 - c^2 t^2 = -1 < 0$. The curve of fixed $\rho, y, z$ is the world line of an observer with constant proper acceleration $c^2 / \rho$, and its proper time is $\rho \eta / c$. A common variant uses Rindler time $T = \rho_0 \eta / c$ adapted to the observer at $\rho = \rho_0$.