# Does Length Contraction Imply Superluminal Speeds?

1. Nov 19, 2014

### Fantasist

Assume a rocket sets off from earth accelerating to 0.8c within 1 second. The distance to a star originally 1000 light years away contracts thus to 600 lightyears within 1 second. This means the star moves 400 lightyears in one second in the travelers frame. I make this an average speed of more than 10 billion times the speed of light during the acceleration phase.

Last edited: Nov 19, 2014
2. Nov 19, 2014

### Staff: Mentor

While the rocket is accelerating, its reference frame is non-inertial. In a non-inertial frame, the speed of light is not necessarily c, and the coordinate speed of other objects can exceed c.

3. Nov 19, 2014

### mathman

The star didn't move. The distance is dependent on the reference frame. That is the basis of the twin paradox.

It has nothing to do with non-inertial.

4. Nov 19, 2014

### Staff: Mentor

This "traveler's frame" is not an inertial frame, so "speeds" in it are not limited to the speed of light. Light itself travels much "faster than light" in this frame. A light beam emitted by the star toward earth an instant before the rocket sets off from earth would move *more* than 400 light-years in 1 second in the traveler's frame.

5. Nov 19, 2014

### Staff: Mentor

Yes, it does. The "reference frame" in which the star has v>c is non inertial, and v>c is permitted in non inertial coordinates.

6. Nov 20, 2014

### Fantasist

Come to think of it, I am actually having problems to understand how the distance to the star should contract by 400 lightyears at all:

assume you have a number of spaceships distributed as distance markers between the earth and the star, with the last spaceship just 1 ly from the star. Now with all their clocks synchronized, the ships all begin to accelerate for 1 second to 0.8c at an agreed time. This means the distance of the last spaceship to the star will merely be reduced from 1 to 0.6 ly, so the star won't even pass the last spaceship in the line. So it seems paradoxical that the distance of the star to the first ship should be reduced from 1000 to 600 ly, considering that all spaceships are accelerated the same and thus should maintain their mutual distance.

7. Nov 20, 2014

### Khashishi

The spaceship maintain their mutual distance in the "star" frame, which is taken to be rest. They don't maintain their mutual distance in any of the spaceships' noninertial reference frames. In an accelerating frame, there are all sorts of fictitious forces which are allowed to bend spacetime like gravity.

8. Nov 20, 2014

### Fantasist

They are supposed to accelerate synchronously in the spaceships' frame, not the star's frame (the acceleration is started and stopped simultaneously according to the spaceships' clocks)

9. Nov 20, 2014

### Khashishi

Before they start accelerating, the spaceship frame is the same as the star's frame. If they all start their boosters (each providing the same accelerating) at the same time in the star's frame, they will maintain their distance in the star's frame. To maintain their distance in one of the accelerating spaceship's frames is impossible.

I'm not sure if it is possible to even define what synchronous means on each of the spaceships during acceleration.

10. Nov 20, 2014

### Fantasist

They only start their boosters simultaneously in the star's frame when they still have zero velocity. If you consider the acceleration consisting of discrete boosts, the following boosts will not be simultaneous in the star's frame anymore. They are by assumption simultaneous in the spaceships' frame. Your scenario is not the scenario I suggested.

I defined it above: the acceleration is kept the same on each ship as a function of the ships' time (assumed to be synchronized)

11. Nov 20, 2014

### Staff: Mentor

This is a bad assumption. Each ship has a different non inertial frame and there is no a priori reason to expect that the mutual distances will be constant in any of the non inertial frames. After all, they are non inertial!

12. Nov 20, 2014

### Staff: Mentor

I agree with Khashishi, I don't think this is possible. Certainly I don't know of any easy non inertial coordinates where this is true.

Could you please post the explicit coordinate transform for the reference frame you are envisioning?

Last edited: Nov 20, 2014
13. Nov 20, 2014

### jartsa

Are the spaceships perhaps identical in every way, including the programmed acceleration?

Let's see ... We know that according to the first spaceship the star accelerates from zero speed to a superluminal speed, this happens when the first ship accelerates, when the motors are turned off, the superluminal speed changes to a normal speed.

What does the last spaceship do according to the first spaceship, during the acceleration? It accelerates a lot, for a short time, reaches a large speed, may fly past the star.

During acceleration all speeds at distance x are multiplied by factor k. Accelerations at that distance are multiplied by factor k*k.

14. Nov 20, 2014

### ghwellsjr

I think some spacetime diagrams could help you understand what's going on. The thick lines show the earth in blue, the star in red and the rocket in green. The dots mark off 100 year increments of time. In the earth/star rest frame, the star is 1000 light-years away from the earth at all times and the rocket takes off toward the star at the Coordinate Time of 0 and takes 750 years according to its own clock to pass the star:

In the rest frame of the rocket after it accelerates, the star is 600 light-years away from the earth but it didn't just get that way when the rocket accelerated, the distance between the earth and the star was always 600 light-years in this Inertial Reference Frame:

Now if you want to consider a non-inertial rest frame for the rocket, you can't just arbitrarily say anything about it until you decide what kind of a frame you have in mind, There is no standard way to do this. There are lots of ways to do it. In some of them, the speed of light can be other than c but that isn't a requirement. I'm going to show you how the rocket can create a non-inertial rest frame from measurements and observations that it actually makes and in which it assumes that the speed of light is c in all directions, just like it is in Inertial Reference Frames. This method relies on the rocket sending out radar pulses and correlating their return echoes with the images that it sees of the star. By assuming that each radar signal, which travels at the speed of light, takes the same amount of time to get to the star as it takes for the return echo to get back, the rocket can create a non-inertial rest frame.

Here is a spacetime diagram for the earth/star rest frame showing the emitted radar signals in green and their return echoes in red:

From this information, the rocket constructs this spacetime diagram:

Notice how both diagrams show exactly the same timings (based on the dots) for the radar signals and their echoes and that they travel at c along the 45-degree diagonals. You will note that when the rocket and earth separate at time 0, the star is 500 light-years away from them but it didn't get there all of a sudden, it started "moving" 1000 years earlier at a speed of 0.5c. It isn't until much later that the rocket measures the speed of the star to be 0.8c.

So the answer to your question is no, length contraction does not imply superluminal speeds.

Last edited: Nov 20, 2014
15. Nov 23, 2014

### Fantasist

Initially, the ships are all in the same inertial frame. If you apply the same velocity boost at the same time to all the ships, how could they possibly not end up in the same inertial frame again?

For this I first have to understand the concept of a reference frame (be it inertial or not) where its parts have a non-zero velocity relatively to each other. Please explain.

16. Nov 23, 2014

### Fantasist

I don't know which transformation your diagrams are based on, but your conclusion appears to contradict the Rindler transformation formula for a uniformly accelerated observer (http://en.wikiversity.org/wiki/Rindler_coordinates ).

$$x=\left(\frac{c^2}{\alpha }+x'\right)cosh\left(\frac{\alpha t'}{c}\right)-\frac{c^2}{\alpha }$$

Here, x would be the distance of the space ship to the star when initially at rest. x' is the distance in the ship's frame as a function of the ship time t' assuming a constant proper acceleration $\alpha$
Solving this for x' and differentiating with regard to t', I obtain the velocity of the star in the accelerated frame as a function of t', which (obviously) starts at zero and reaches a maximum speed at (ignoring a factor of the order of unity (which I was too lazy to evaluate))

$$v' = -c\left(1+\frac{\alpha x}{c^2}\right)$$

This clearly can get arbitrarily large (depending on $\alpha$ and $x$) (although later it decreases again approaching zero if $t'\rightarrow\infty$)

17. Nov 23, 2014

### Staff: Mentor

Looks like you need to read up on the Bell Spaceship Paradox. Fortunately, we have a FAQ on it:

Last edited by a moderator: May 7, 2017
18. Nov 23, 2014

### Fantasist

Bell's spaceship paradox assumes the velocity boost applied simultaneously in the inertial rest frame ( http://en.wikipedia.org/wiki/Bell's_spaceship_paradox ). This is not the scenario we are considering here. Here the boosts are applied simultaneously in the accelerated frame

Last edited by a moderator: May 7, 2017
19. Nov 23, 2014

### Staff: Mentor

Here is the best overall reference I have for understanding the mathematical machinery of non-inertial reference frames:
http://preposterousuniverse.com/grnotes/ (chapters 1 and 2)

There is no one standard convention for making a reference frame for a non-inertial observer, but here is one that is commonly used due to its similar construction to inertial reference frames for inertial observers:
http://arxiv.org/abs/gr-qc/0104077

20. Nov 23, 2014

### Staff: Mentor

From your description in the OP it sounds like the observers in question are not uniformly accelerated, so any transformation that matches the OP will necessarily contradict the Rindler transformation. However, it is rather unclear which is why YOU need to be the one to specify the transformation you have in mind.

Please read the material provided and then come back with specific questions about the material or a discussion about the specific transform you are interested in learning about. When you get to that point, please send me a message and I can reopen the thread.