# Homework Help: General relativity: Rindler space problems

1. Oct 5, 2012

### Funzies

1. The problem statement, all variables and given/known data
I have derived the metric in for 2D Rindler space in a previous problem and it is explicitly given again here:
$$ds^2 = dx^2 - (dx^0)^2 = dw^2 - (1+gw/c^2)^2(dw^0)^2$$,
where $(x^0, x)$ are Minkowski coordinates in an intertial system I and $(w^0,w)$ the Rindler coordinates of a system of reference R with constant acceleration g relative to I.
In the question I have furthermore derived the solution for a particle starting at w=w~ and with velocity zero for w^0=0:
$$w(w^0) = \frac{c^2}{g}\left( (1+g\tilde w/c^2)\frac{1}{\cosh(gw^0/c^2)} -1 \right)$$

2. Question
Calculate the velocity $$v = c \frac{dw}{dw^0}$$. Find the maximum velocity of the particle and compare with the speed of light (calculate that too). What happens for $w^0 \rightarrow \infty$?

3. The attempt at a solution
I have solved all the questions and I am sure they are correct; only the interpretation is lacking:
$$v_{\mathrm{particle}} = -c(1+g\tilde w/c^2)\frac{\sinh(gw^0/c^2)}{\cosh^2(gw^0/c^2)}$$
$$v_{\mathrm{particle, max}} = \pm c/2(1+g\tilde w/c^2)$$
$$v_{\mathrm{light}} = \pm c(1+g\tilde w/c^2)\frac{1}{\cosh(gw^0/c^2)}$$
$$v_{\mathrm{light, max}} = \pm c(1+g\tilde w/c^2)$$

Now for the interpretation:
-First of all: I presume the minus sign in the equation for the speed of the particle just reflects the fact that for w^0<0 it travels in one direction and for $w^0>0$ in another direction?
-I find it weird that the speed of light is not a fixed c. I know that the for light $ds^2 \equiv 0$, but I still don't find this answer rather comforting. Can anyone elaborate on this?
-In the limit $w^0 \rightarrow \infty$ the speed of the particle and of the light go to zero. I completely do not understand what is happening here.
-What does it mean that the maximum speed of light is two times larger than the maximum particle velocity?

2. Oct 8, 2012

### clamtrox

Yeah that would seem to make sense; in particular it depends on whether the directions of initial velocity and acceleration are same or opposite.

But the Rindler coordinates are not inertial, so there's no reason to expect things like velocities to follow usual rules of special relativity. For example, imagine a coordinate system which is rotating, in an ordinary Minkowski space. Now, test particles which are stationary in ordinary Minkowski coordinates appear to be moving at a velocity proportional to their distance from you.

Can you show the coordinate transformations? What's w0 in terms of x and t of Minkowski coordinates?

Probably nothing. Note that the maximum is evaluated at different coordinate values; maximum for speed of light is at w0=0 while maximum for massive particles is when w0 has some finite value.

3. Oct 9, 2012

### Funzies

What do you mean by this? Before I saw Rindler space I always believed that special relativity could not cope with accelerating frames, but apparently it can, if you define the frame at every proper time tau. I am a bit confused by all this.

I have the following relations:
$$x = c^2/g(\cosh(gw^0/c^2)-1) + w\cosh(gw^0/c^2)$$
$$x^0 = c^2/g\sinh(gw^0/c^2) + w\sinh(gw^0/c^2)$$.

4. Oct 11, 2012

### clamtrox

Well, there's nothing stopping you from making a coordinate transformation to an accelerating coordinate system, but the transformation is different from the usual Lorentz transformations. Or, as you say, you need to do an infinite number of them. This means that the accelerating frames do not respect same symmetries as inertial frames in SR. For example, speed of light need not be constant. I think the rotating frame serves as a good example. There all photons which are sufficiently far away move at an arbitrarily high speed. This should be pretty obvious once you think about it for a while.

I don't have a good answer here, but you can see that the coordinate transformation is pretty tricky when w0→∞. In particular, you see that the hypersurfaces w=const. approach the null surface x=x0 in this limit. In some sense, this corresponds to an infinitely accelerated reference frame. Perhaps you can think this result so, that all particles (regardless of their initial velocity) travel at the same speed once they've been accelerated so much.

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