# Gravitational field strength and Horizon in Rindler coordinates

I came across the following statements in 't Hooft's black holes notes, but not being able to justify them.

The metric in the Rindler coordinates $x=\tilde{x}, y=\tilde{y}, z= \rho \cosh{\tau}, t= \rho \sinh{\tau}$ is $$ds^2 = -\rho^2 dt^2 + d\rho^2 + d\tilde{x}^2+d\tilde{y}^2$$

1. Gravitational Field Strength

The actual gravitational Field strength felt by the (Rindler)observer is inversely proportional to the distance from the origin.

How can I see this from the metric? How do I mathematically and physically justify this claim?

2. Horizon

Why does the surface $x=t$, act like a Horizon i.e. anything from outside the shaded area shown in the picture can't enter into the shaded region? I think, this can be seen by showing that all time-like curves starting from a point outside the shaded area don't enter the shaded region. But what are all the time-like curves in the Minkowski space-time? and would this assumption still hold for any arbitrary geometry which is locally like the Rindler metric(e.g. the schwarzchild metric near the horizon?)

-
Unless I'm missing something deep, these are quite separate questions, despite being inspired by the same source and cast into the same coordinate system. In that case I suggest breaking off one of these into a separate post. –  Chris White Jun 26 '13 at 4:40
Perhaps it's too late at night for me to be looking at this, but I use all of this (Rindler Coordinates, t'Hooft black holes, schwarzchild metric) in my daily research and this question still makes no sense to me. –  Jim the Enchanter Jun 26 '13 at 4:46
@ChrisWhite: Yeah, I agree they are two different questions. Still I don't see why one should encourage posting them as separate questions. –  user7757 Jun 26 '13 at 5:40
@Jim: Please tell me if the question is not clear. I will try to explain it more to make it clearer. But the fact that the gravitational field strength is inversely proportional to the distance, has been used to derive the Unruh effect temperature $\frac{g}{2 \pi k}$. The argument is that the proper time of the observer is related to the Rindler time by a scale factor by the distance from the centre $\rho_0$, $\tilde{t}=\frac{\tau}{\rho_0}$, which introduces the factor of $g$ in the temperature, from the statement in my question 1. –  user7757 Jun 26 '13 at 5:48
Sorry for the grammatical errors, above statement should read "by a scale factor of". –  user7757 Jun 26 '13 at 6:02

1)Gravitational Field Strength

You have to consider the analogy with a uniform accelerated move, in special relativity, that is :

$z' = \frac{1}{a} ch (a \tau'), t' = \frac{1}{a} sh (a \tau')$

Here $a$ is the acceleration and $\tau'$ is the proper time.

You get : $dz'^2 - dt'^2 = -d\tau'^2$,as wished.

Now, make the coordinate change $\tau = a \tau'$, we get :

$z' = \frac{1}{a} ch (\tau), t' = \frac{1}{a} sh (\tau)$

$dz'^2 - dt'^2 = - \frac{1}{a^2} d\tau^2$.

But this is the same result that the Rindler metric with $d\rho=d \tilde x =d \tilde y = 0$, so we make the identification between $\rho$: and $\frac{1}{a}$, so the gravity strengh (acceleration) is inversely proprotionnal to the space-like quantity $\rho$

2) Horizon

I am not sure to understand all your questions, but you have to be careful about several points :

The Minkowski metric of the form : $-dt^2 + dz^2 + d \tilde x^2 + d \tilde y^2$, linked to the Rindler metric, by the transformations $z= \rho \cosh{\tau}, t= \rho \sinh{\tau}$, is only correct :

• near the horizon
• for a small angular region, that could be considered centered at $\theta = 0$

The horizon itself is not a 3 dimensional-surface in space-time, because the horizon corresponds to $\rho=0$, that is $t=z=0$.

This is because the metrics $g_{00} = 0$ at the horizon, so the horizon has no extension in the time coordinates.

So, considering that the horizon is $z^2 - t^2 = 0$ is not correct, the horizon is $t=z=0$.

A correct global representation is then using Kruskal–Szekeres coordinates, which gives a correct global point of view.

-