1. Oct 4, 2015

### valentin mano

A highly accelerated object near the speed of light,should be seen contracting faster,than the speed of light.Maybe the contraction could not be observed?

2. Oct 4, 2015

### bcrowell

Staff Emeritus
Lorentz contraction doesn't describe what we see. When we see things, that's an optical measurement. Relativistic optics is a whole separate subject. Lorentz contraction describes the results of the kind of elaborate surveying process that we have to undertake in order to lay out a coordinate system.

Lorentz contraction also doesn't describe the reduction in length of an accelerated object. To prove the equation for Lorentz contraction, we assume an inertial world-tube, which we then slice with a surface of simultaneity. The derivation doesn't hold if the world-tube is noninertial.

3. Oct 4, 2015

### valentin mano

O.K.It'snot about Lorentz contraction,but the contraction near the speed of light with acceleration.

4. Oct 4, 2015

### bcrowell

Staff Emeritus
What do you consider to be the distinction?

5. Oct 4, 2015

### valentin mano

Lorentz contraction also doesn't describe the reduction in length of an accelerated object.
So,describe it.

6. Oct 4, 2015

### Strilanc

What happens depends on how the object is decelerated. Usually ships accelerate with an engine and that acceleration is transferred to the rest of the ship by atomic binding forces or whatever. So if you decelerate with an actual engine at the back of the ship, it won't de-contract faster than light simply because it takes time for the force-signal to be passed from atom to atom.

But if you had a strange ship that could decelerate uniformly, like each atom could be scheduled to experience a force at a particular time, then you might be able to cause effects akin to sweeping a laser pointer across the face of the moon, where a choreographed pattern looks like it's moving faster than light but the pieces making it up never exceed light speed and there's no transfer of information.

If you scheduled all the ship's atoms to decelerate at the same time in the ship's frame, then I think the ship would be stretched (in its final frame) and torn apart due to length-contraction shrinking its individual atoms without shrinking the space between them.

If you scheduled all the ship's atoms to decelerate at the same time in the final frame (the "rest" frame the ship is initially moving very fast w.r.t to), then the opposite problem would happen. The ship would go from being length-contracted to actually literally being squished.

I have no idea what horrible things would happen if you actually significantly squished or broke-by-stretching all those atomic bonds at the same time. Probably gigantic explosions. You'll have to [ask xkcd](http://what-if.xkcd.com/). But FTL effects? Nope.

Last edited: Oct 4, 2015
7. Oct 4, 2015

### bcrowell

Staff Emeritus
You're welcome.

8. Oct 5, 2015

### valentin mano

Maybe the contraction would not be faster than light because of the time dilation?

9. Oct 5, 2015

### SlowThinker

You are saying that, with enough acceleration, an object can shrink at any speed, even that exceeding speed of light. That is wrong.
You can view the front and rear end of the ship as 2 independent entities. None of them will move faster than light.
Lorentz contraction only works for objects that are not accelerating, and thus can't be used to find the "speed of shrinking".

10. Oct 5, 2015

### valentin mano

Are You saying that,if an object is accelerated,it doesn't seem to be contracted?

11. Oct 5, 2015

### harrylin

Assuming that it didn't permanently deform due to a too fast acceleration, and after acceleration effects have died out, it is predicted to be Lorentz contracted according to measurements done with the "rest" system.

Coincidentally your question corresponds to the following thought experiment: "Now to the origin of one of the two systems (k) let a constant velocity v be imparted in the direction of the increasing x of the other stationary system (K), and let this velocity be communicated to the axes of the co-ordinates, the relevant measuring-rod, and the clocks." - intro of §3 of http://fourmilab.ch/etexts/einstein/specrel/www/

You can read the following SR prediction in §4 here, in plain English: http://fourmilab.ch/etexts/einstein/specrel/www/

12. Oct 5, 2015

### SlowThinker

Objects are shorter if they move fast, not if they just accelerate but are still moving slowly.
And, you can't use Lorentz contraction during acceleration, because it is not valid in accelerated frames. It gives incorrect answers, such as ship contracting faster than light, which does not happen if the computation is performed correctly.

13. Oct 5, 2015

### Fredrik

Staff Emeritus
If a solid object is accelerated gently enough to ensure that its shape isn't significantly distorted by shock waves traveling through the material, then the rear of the object will automatically (due to internal forces) have a higher coordinate acceleration than the front (and therefore also a different velocity), in the inertial coordinate system where the object started out at rest. So an observer using that coordinate system would say that the object is shrinking. Still, he would describe each part of the object as having a speed less than c.

14. Oct 5, 2015

### Demystifier

Here the Lorentz contraction with acceleration is described in detail:
http://lanl.arxiv.org/abs/physics/9810017 [Am.J.Phys.67:1007,1999]

15. Oct 5, 2015

### valentin mano

Let's change the velocity of 1-metre measuring-rod from 0.9c to 0.9999c for one nano-second.What kind of a contraction shall we observe from
our stationary system?

16. Oct 5, 2015

### Fredrik

Staff Emeritus
It seems that what you have in mind is to start changing the velocities of the component particles all at the same time. But if two events are simultaneous in one inertial coordinate system, they're not simultaneous in another (unless the velocity difference between them is zero). So what inertial coordinate system will you use to define what "the same time" means? I'm guessing that what you have in mind is "the one in which the object was originally at rest".

I'm also guessing that your intention is that at each time t (in that coordinate system), all component particles will have the same velocity v(t).

In that scenario, there will be no contraction or expansion. What you have done is to subject the component atoms to enormous forces that ensure that the length of the object stays the same in the coordinate system in which it was originally at rest. This is impossible to do in practice, because you would have to push each atom individually.

17. Oct 5, 2015

### valentin mano

I'm thinking about the kinematics only.

18. Oct 5, 2015

### Fredrik

Staff Emeritus
And you described a scenario that involves enormous forces.

If you prefer to consider a scenario in which all the component particles are independent (not interacting with each other), then you might as well consider only two non-interacting point particles. If they both have the same velocity v(t) at each time t, then the distance between them will never change.

19. Oct 5, 2015

### valentin mano

Last edited by a moderator: May 7, 2017
20. Oct 6, 2015

### Demystifier

Last edited by a moderator: May 7, 2017