1. Sep 3, 2008

### duordi

If I am watching spherical mass A that is traveling away from me at close to light speeds it lorenz contracts like I am looking down at a plate.
Does the distance mass A is from me contract also or does it stay the same distance away ?

2. Sep 3, 2008

### JesseM

It depends what distance (and in whose frame) you are comparing it to. For example, suppose we have two ships which are moving at the same speed, and in their own rest frame they are 10 light years apart. If in my frame they are moving at 0.6c, then in my frame the distance between them is shrunk to just 8 light years. So, in the frame of the ships, at the moment I am passing next to the rear ship, the front ship is 10 light years away; but in my frame, at the moment I am passing next to the rear ship, the front ship is 8 light years away, so that's obviously a contracted distance when compared to the ship's frame.

On the other hand, suppose a ship is 10 light years away in my frame, and initially at rest relative to me. Then it instantaneously accelerates to 0.8c in my frame. In this case the distance between me and the ship does not contract immediately after the acceleration; if the ship is 10 light years away in my frame immediately before the acceleration, it will still be 10 light years away in my frame immediately after the acceleration.

3. Sep 3, 2008

### duordi

If I am the one who accelerated then the ship would appear to be closer just after the acceleration?

If I took a camera and accelerated it to 99% the speed of light and took a picture at stars and then had it send the information to me I would get a photo of distant stars that look very close?

Is this correct?

4. Sep 3, 2008

### JesseM

Length contraction is not about what you see visually, it's about what you measure using rigid rulers and clocks at rest relative to yourself (and only non-accelerating objects can remain rigid in relativity, so it's problematic to talk about what is 'observed' in this non-visual sense by a non-inertial observer). See this thread for more on the difference between measurements and visual appearances in SR. In fact, visually the length of a moving object appears exactly the same as the same object at rest--see this thread.

5. Sep 3, 2008

Staff Emeritus
At the risk of being pedantic, it's Lorentz with a T, not Lorenz. It was named after Hendrik Lorentz and not Ludvig Lorenz.

6. Sep 3, 2008

7. Sep 3, 2008

### duordi

So there is a lot of distortion and delay due to light travel.

Suppose a distant star was traveling 99% the speed of light away from me and it slowed down to 90% the speed of light. Eventually when the light reached me the lorentz contraction would decrease and the star would appear farther away.

As the distant star was slowing down it would appear to be accelerating away from me.
Am I right in this?

Last edited: Sep 3, 2008
8. Sep 3, 2008

### JesseM

No, as I said, Lorentz contraction is not a matter of what is seen visually. Anyway, in this case neither the apparent visual difference nor the actual distance in my frame would change before and after the acceleration (assuming the acceleration was instantaneous).
No, that's definitely wrong.

9. Sep 3, 2008

### granpa

its your velocity that determines how far away you perceive the star to be. not the stars velocity.

10. Sep 3, 2008

### duordi

Thanks for the reference Granpa
It is weird how the light focuses in the direction of motion.

JesseM you lost me.

The two ships are 1 foot and 10 light years away from me and traveling 95% of light velosity away from me.
Now I accelerate to 95% light speed toward them.
Our velocities are identical, in my frame of reference.
No LorenTz contraction, the furthest ship must be farther away then it was before the distances between the ships were contracted before.

Is this not true because I accelerated?

11. Sep 3, 2008

### granpa

in the frame of the moving star light spreads outwards evenly in all directions. so when its contracted in the frame of a moving observer its light appears to mostly go outward perpendicular to its motion.

if you accelerate to 95% of c then the whole universe will appear to contract.

12. Sep 3, 2008

### duordi

Yes the whole universe will contract but the ship will un-contract because we are moving the same velocity.

I think.

13. Sep 3, 2008

### granpa

yes it will.

but the distance between you and the ship is part of the universe not part of the ship.

14. Sep 3, 2008

### JesseM

If you accelerate, you're no longer an inertial observer. All the standard equations of SR, like the time dilation equation and length contraction equations, only apply in inertial frames. There is no standard definition of the "frame" of a non-inertial observer--distances and times in inertial frames are defined in terms of rigid rulers and clocks at rest in that frame, but rulers cannot remain rigid when they accelerate.

15. Sep 3, 2008

### JesseM

Are you claiming the universe will visually appear to contract, or are you claiming that in the coordinate system where you remain at rest throughout the acceleration, the rest of the universe will contract in that coordinate system? If the first I don't think that's correct, if the second, then see what I said above about the fact that there's no standard definition of the "frame" of a non-inertial observer, you could pick a variety of different coordinate systems where you're at rest and which disagree about what happens to the distance to other objects.

16. Sep 3, 2008

### duordi

Ok I got it.

17. Sep 4, 2008

### granpa

I think I blew it. the distance between the ships would increase. its no different than if there was a solid rod connecting them.

if on the other hand the ships stopped then the distance between them would be the same. it gets confusing.

18. Sep 4, 2008

### granpa

acceleration isnt really a factor here. when you are already moving at 95% of c then the universe will obviously seem to you to be moving past you at that same speed and therefore it will seem to be contracted. thats all I am saying