# Reaching the Rindler horizon in a finite proper time

1. Nov 13, 2011

### teeeeee

Hi,

I am trying to show that timelike geodesics reach the Rindler horizon (X=0) in a finite proper time.

The spacetime line element is

$ds^{2} = -\frac{g^{2}}{c^{2}}X^{2}dT^{2}+dX^{2}+dY^{2}+dZ^{2}$

https://www.physicsforums.com/showpost.php?p=3316839&postcount=375

But don't understand why you have to take T=∞ in order to find X=0?

I also don't understand how to work out that the worldine is given by r=(t,sech(t),0,0).

Any help will be greatly appreciated.

2. Nov 13, 2011

### Staff: Mentor

OK, to be consistent with your notation above I will use (T,X,Y,Z) for Rindler coordinates and I will use (t,x,y,z) for Minkowski coordinates. (This is the opposite of the convention used in the thread you linked to and also the opposite of the convention in the Wikipedia page http://en.wikipedia.org/wiki/Rindler_coordinates).

In Minkowski coordinates we can easily see that the worldline $r=(t,1,0,0)$ is a geodesic. Transforming this to Rindler coordinates gives $R=(atanh(t),\sqrt{1-t^2},0,0)$. Now, this parameterizes the worldline by the Minkowski time coordinate which, although it is not wrong, it is not very aesthetically pleasing. So we solve for t in terms of T:
$T=atanh(t)$
$t=tanh(T)$

Substituting that in we get
$R=(atanh(tanh(T)), \sqrt{1-(tanh(T))^2},0,0)$
$R=(T, \sqrt{(sech(T))^2},0,0)$
$R=(T, sech(T),0,0)$

We want to find when it crosses the Rindler horizon, which is located at X=0. So, we solve $0=sech(T)$ and get $T=\infty$

3. Nov 13, 2011

### teeeeee

Still a couple of things I don't follow...

How exactly do you transform from
$\textbf{r}=(t,1,0,0)$ to $\textbf{R}=(atanh(t), \sqrt{1-t^{2}},0,0)$ ?

And how do you get from the square root expression to the sech-squared term?

4. Nov 13, 2011

### Staff: Mentor

This follows directly from the coordinate transformation equations (remember their convention is different from yours):
http://en.wikipedia.org/wiki/Rindler_coordinates#Relation_to_Cartesian_chart

Just substitute in x=1.

We have
$$\frac{d\mathbf x}{dt} = \frac{d\mathbf r}{dt} = \frac{d}{dt}(t,sech(t),0,0) = (1,-sech(t) \, tanh(t),0,0)$$
$$\mathbf g = \left( \begin{array}{cccc} -x^2 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right)= \left( \begin{array}{cccc} -sech(t)^2 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right)$$
Substituting that in we obtain
$$\tau = \int \sqrt{-g_{\mu\nu} \frac{dx^{\mu}}{dt} \frac{dx^{\nu}}{dt}} \, dt$$
$$\tau = \int \sqrt{sech(t)^2-sech(t)^2 tanh(t)^2} \, dt$$
$$\tau = \int \sqrt{sech(t)^4} \, dt$$
$$\tau = \int sech(t)^2 \, dt$$