Rindler space

We proved the hyperbolic motion viewed in the lab frame of a relativistic particle with constant proper acceleration as in this post: \begin{equation}t(\tau) = \frac{c}{a}\sinh\left(\frac{a\tau}{c}\right)\,,\quad x(\tau) = \frac{c^2}{a}\cosh\left(\frac{a\tau}{c}\right)\,,\tag{1} \end{equation} where $\tau$ is the proper time of the particle. The next question is: What is the time and space coordinates transformation between the lab frame $(t, x)$ and the particle frame?

The time in the particle frame is just the proper time $\tau$ and we denote the space coordinate in the particle frame by $\xi$. The transformation between $(t, x)$ and $(\tau, \xi)$ is obviously not Lorentz transformation since the particle frame is not an inertial frame because of the acceleration. Instead, Eq. (1) is a special case of the transformation at $\xi=0$.

The key insight is that although the particle frame is not an inertial frame, there exists a series of instantaneous co-moving inertial frames at each time $\tau$ from which the particle looks rest, and the coordinates transformation between the lab frame and a co-moving inertial frame is Lorentz transformation. Furthermore, in an instantaneous co-moving frame that is only valid at $\tau_0$, we can only describe events $(\tau_0, \xi)$ with arbitrary $\xi$ but fixed $\tau_0$. As a result, we can establish a Lorentz transformation between $(t-t(\tau_0), x-x(\tau_0))$ and $(0, \xi)$ as \begin{equation} x - x(\tau_0) = \frac{\xi}{\sqrt{1-\frac{v(\tau_0)^2}{c^2}}}\,,\quad t-t(\tau_0) = \frac{v(\tau_0)\xi/c^2}{\sqrt{1-\frac{v(\tau_0)^2}{c^2}}}\,.\tag{2}\end{equation} Using Eq. (1) as well as $v(\tau)=c\tanh(a\tau/c)$, we have \begin{equation}t =\left( \frac{c}{a}+\frac{\xi}{c}\right)\sinh\left(\frac{a\tau_0}{c}\right)\,,\quad x = \left(\frac{c^2}{a}+\xi\right)\cosh\left(\frac{a\tau_0}{c}\right)\,. \end{equation} Finally, applying to any $\tau=\tau_0$, we obtain our transformation \begin{equation}\boxed{t(\tau) =\left( \frac{c}{a}+\frac{\xi}{c}\right)\sinh\left(\frac{a\tau}{c}\right)\,,\quad x(\tau) = \left(\frac{c^2}{a}+\xi\right)\cosh\left(\frac{a\tau}{c}\right)\,.} \tag{3}\end{equation} As a result, the metric in Rindler coordinates becomes \begin{equation}\boxed{ds^2 = -c^2dt^2 + dx^2 = -\left(1+\frac{a\xi}{c^2}\right)^2c^2d\tau^2 + d\xi^2}\,. \tag{4}\end{equation}

Rindler horizon

The Rindler horizon is at $x=ct$. For lights in the region at $x < ct$, they can never catch up with the particle, i.e., the region of $x<ct$ is not accessible to the accelerating particle.

Fig. 1 light at x<ct can never catch up 

Using Eq. (3), $x(\tau)=ct(\tau)$ gives the solution $\xi=-\frac{c^2}{a}$. As a result, the Rindler horizon in the particle frame is located at $\xi = -\frac{c^2}{a}$. Indeed, in the metric (4), $g_{\tau\tau}=0$ at $\xi = -\frac{c^2}{a}$.

Rindler coordinates

Since the region of $x<ct$ is not accessible by the accelerating particle, by Eq. (3), the constraint of  $x>ct$ requires $\xi>-c^2/a$ and thus $x > 0, x>c|t| $. That is, the Rindler coordinates $(\tau, \xi)$ only covers the patch of Minkowski space with $x>c|t|$ as shown in Fig. 2:

• $-\infty < \tau <+ \infty$: constant $\tau$ corresponds to a series of straight lines with slopes from -1 to 1.  
• $-c^2/a < \xi <+ \infty$: constant $\xi$ corresponds to a series of hyperbolic lines.

Fig. 2 Rindler coordinates $(\tau, \xi)$

Speed of light in non-inertial frame

The speed of light is the constant $c$ in all inertial frames, which is one of the hypotheses in special relativity. However, the speed of light may vary from $c$ in a non-inertial frame.

Consider the metric (4): since the light ray follows the null path $ds=0$, the speed of light in accelerating the particle frame is\begin{equation}v\equiv\left|\frac{d\xi}{d\tau}\right|= c \left(1+\frac{a\xi}{c^2}\right)\,,\end{equation}

which depends on the Rindler space coordinate $\xi$. $v>c$ when $\xi>0$ and $v<c$ when $\xi<0$. Especially, at the Rindler horizon $\xi = -\frac{c^2}{a}$, light even stops with $v=0$.

Note:

• In the above, we measure the speed of light in the accelerating particle frame using at least two different spacetime points: the propagation of light starts at $(\xi, \tau)$ and end $(\xi+d\xi, \tau+d\tau)$ and the speed of light is $d\xi/d\tau$ by the definition of speed. The speed of light measured in such non-local way can vary from $c$.
• In contrast, the speed of light measured locally should always be $c$. For example, $v=c$ at the point where the accelerating particle locates ($\xi=0$). This is because at a single point, there always exists an equivalent co-moving inertial frame and the speed of light remains to be $c$ in all inertial frames.

Surface gravity of black hole 

For the Schwarzschild black hole \begin{equation}ds^2=-\left(1-\frac{2GM}{c^2r}\right)c^2dt^2+\frac{dr^2}{1-\frac{2GM}{c^2r}}+r^2(d\theta^2+\sin^2\theta d\phi^2)\,,\end{equation} there is an event horizon at $r_H=\frac{2GM}{c^2}$. Near the horizon, $r=r_H + \epsilon$ with $\epsilon \ll r_H$, the Schwarzschild metric along $(t, r)$ can be approximated as \begin{equation}ds^2 \approx -\frac{\epsilon}{r_H}c^2dt^2 + \frac{r_H}{\epsilon}d\epsilon^2 =  -\frac{\rho^2}{4 r_H^2}c^2dt^2 + d\rho^2\,,\tag{5}\end{equation} where we make the variable transformation $\epsilon \equiv \rho^2/4r_H$.

The near-horizon metric (5) is of the same form as the Rindler metric (4) with the relation \begin{equation}a = \frac{c^2}{2r_H}=\frac{c^4}{4GM}\,.\end{equation} That is, the proper acceleration at the event horizon of the Schwarzschild black hole is $\frac{c^4}{4GM}$. For this reason, the quantity $\frac{c^4}{4GM}$ is also defined as the surface gravity of the Schwarzschild black hole.