# Rindler coordinates - signals which never arrive

1. Sep 21, 2014

### Cancer

Hi everybody,

I know that there are a lot of threads in this forum about Rindler coordinates but none of them have helped me
I'll explain you my problem. First of all, my coordinates $(x^0,x)$ (Cartesian coord., where $x^0=ct$) are related to the Rindler coordinates $(\omega ^0,\omega)$ (where $\omega ^0$ is the proper time) as follows:
$$x = \frac{c^2}{g}\left[ \cosh \left( \frac{g\omega^0}{c^2}\right)-1\right]+\omega \cosh \left( \frac{g\omega^0}{c^2}\right)$$
$$x^0 = \sinh \left( \frac{g\omega^0}{c^2}\right) \left( \frac{c^2}{g}+\omega \right)$$
Which are defined in an exercise I'm trying to solve, and c is the speed of light and g is the constant acceleration. It says that I have to proof that some signals sent from an inertial frame I won't never reach an observer which is at rest in the frame R i.e., which is at constant $\omega$.
So the first thing I do is isolate $\omega^0$ from the second equation, getting:
$$\frac{g\omega^0}{c^2}=\sinh^{-1} \left( \frac{x^0}{\frac{c^2}{g}+\omega}\right)$$
We put this into the first equation, using this mathematical relation:
$$\cosh (\sinh^{-1}(x))=\sqrt{1+x^2}$$
Finally we arrive to
$$x= \left(\left(\frac{c^2}{g}+\omega\right)^2 +(x^0)^2\right)^{1/2} -\frac{c^2}{g}$$

If I have properly done the math, that's the path which will follow an observer at rest in the R frame in the $(x,x^0)$ coordinates. BUT I don't see why light signals won't reach that observer! If I take x as ct (i.e. the diagonal of the light cone, a photon) I will always be able to find a solution for $x^0$, i.e., the light will eventually reach the observer.

I don't know what's happening here, I've read all I've found in Google about Rindler coordinates but nothing helps me, because nobody uses this definition for the Rindler coordinates...

I hope some of you could help me with this!

Thanks,
Victor

2. Sep 21, 2014

### ShayanJ

Your last equation is equivalent to the following:

$x^0=\sqrt{(x+\frac{c^2}{g})^2-(\omega+\frac{c^2}{g})^2}$

And if a signal is going to reach somewhere, it should reach there at a real time($t\epsilon\mathbb{R}$), right?

3. Sep 21, 2014

### sweet springs

Hi. It is a kind of Zeno's paradox of Achilles and the Tortoise. Achille reaches the tortoise in finite time in the original version. In your case the tortoise keeps accelerating and relativistic effects prohibit Achiles of light speed to reach the tortoise.
Best.

4. Sep 21, 2014

### sweet springs

PS Event horizon in Rildler coordinate or just Rindler horizon is of your interest.

5. Sep 21, 2014

### Staff: Mentor

You might want to specifically make a beam of light from some arbitrary event and explicitly calculate when it reaches the Rindler observer. Then look and see if that equation becomes infinite under any circumstances.

6. Sep 21, 2014

### Cancer

Yes. In my view, if I replace that $x$ by $ct$ (i.e. a photon sent from $x=0$) then the t I find will tell me when are they going to coincide... And I am always able to find a solution, which is:
$$t=\frac{g\omega^2}{2c^3}+\frac{\omega}{c}$$
Am I doing something stupid maybe? xD
I understand the theory more or less, but I can't see such things in my calculations... :(

7. Sep 21, 2014

### ShayanJ

You're not doing something stupid. You're just not seeing something easy to see.
I don't know how you got that formula, but in the formula $x^0=\sqrt{(x+\frac{c^2}{g})^2−(\omega+\frac{c^2}{g})^2}$, for t to be real, the difference should be positive, so:
$(x+\frac{c^2}{g})^2\geq(\omega+\frac{c^2}{g})^2\Rightarrow \left[ (x+\frac{c^2}{g}\geq\omega+\frac{c^2}{g}) or (-x-\frac{c^2}{g} \leq -\omega-\frac{c^2}{g}) \right]$
Which are conditions on your coordinates and any signal not obeying conditions above, is not going to reach the observer!

8. Sep 21, 2014

### Cancer

But, those two conditions are in fact just one:
$$x\geq \omega$$
As I understand, this $x$ will tell me when will the rest observer will reach that point.
The observer will never pass through a point $x$ which is less than $\omega$ because is accelerating to the other direction (to $x$ positive), that's why we have that condition. Then I can ask for instance to that equation when will it reach a point $x'$ (whatever it is) and I will get a certain time $x^0$.
So if I want to ask the function to tell me whether it'll see the light coming from $x=0$ then I just have to substitute $x$ for $ct$...

Oo

9. Sep 21, 2014

### Staff: Mentor

Try explicitly calculating when a signal is received.

10. Sep 21, 2014

### Staff: Mentor

This is an equation for $x$ as a function of $x_0$. Note that, at $x_0 = 0$, we have $x$ positive--i.e., $x_0$ starts out being less than $x$. Will $x_0$ ever become equal to $x$? And if not, what does that tell you about the relationship between this curve (the worldline of the observer) and a light beam launched from $x_0 = 0$, $x = 0$ (for which you will always have $x_0 = x$)?

Edit: What I wrote above actually isn't quite right, because the coordinate transformation you give is different from the usual one, for example as given in this Wikipedia page:

http://en.wikipedia.org/wiki/Rindler_coordinates

That transformation, in your notation, would be:

$$x = \omega \cosh \left( \frac{g \omega_0}{c^2} \right)$$

$$t = \omega \sinh \left( \frac{g \omega_0}{c^2} \right)$$

The questions I posed apply directly to this case as I posed them. Your transformation, however, shifts $x$ relative to $\omega$, which makes the logic somewhat more complicated; the correct comparison for the questions I posed would be with a light beam launched from $x_0 = 0$, $x = - c^2 / g$ (rather than $x_0 = 0$, $x = 0$).

(Note also that your equations evidently have a problem at $\omega = - c^2 / g$, since some of them become undefined there. That's a clue that your $\omega_0$, $\omega$ coordinates only cover a limited range of the $x_0$, $x$ coordinates.)

Last edited: Sep 21, 2014
11. Sep 21, 2014

### Cancer

I think I did it in #6, is anything wrong in that calculation?...

I think you might be right, but still I can't see it.
Why do I have to emit the photon from $x=-\frac{-c^2}{c}$?
The observer at the begining ($x^0 = 0$) is at a positive $x$, and another observer sends the beam from $x = 0$.

The problem is this one:
http://www.staff.science.uu.nl/~proko101/GR_homework2014_1.pdf
(Maybe it can be understood better reading this...)

Thanks a lot!

12. Sep 21, 2014

### Staff: Mentor

(You mean $- c^2 / g$, correct?)

Because the problem asks you to show that there are *some* spacetime locations in the $x^0$, $x$ coordinates that cannot send light signals to any spacetime location in the $\omega^0$, $\omega$ coordinates. If $x^0 = 0$, $x = - c^2 / g$ is such a location (and as you can show, it is), then that answers the problem.

13. Sep 21, 2014

### Cancer

Hahahaha yeah, I meant that.
Ok, I'll try that, then!
Thanks!! :)

14. Sep 21, 2014

### Cancer

Yes, it works!
Thank you a lot man, I was really lost with this ;)