Do the distance between two objects contract in length contraction

[Trojan666ru]

Imagine two asteroids which are separed by 1Ly distance. They are in uniform velocity (.9c) with respect to an observe in space. So from the observers point of view (rest frame), does the distance between the asteroids (ie. 1Ly) appear contracted?

 

[dEdt]

If the distance between two asteroids is measured to be $d_0$ in the asteroids' rest frame, then in a reference frame where the asteroids are moving at $0.9c$ the distance will be about $0.44d_0$.

 

[Trojan666ru]

So i can see the asteroid just behind the first one within a less amount of time after seeing the first asteroid. But what if there was no asteroid in front? Will that take more time to see the last asteroid?

 

[Dale]

The presence or absence of one object does not alter the coordinates of another object unless they are exerting forces on each other or something.

 

[ghwellsjr]

Imagine two asteroids which are separed by 1Ly distance. They are in uniform velocity (.9c) with respect to an observe in space. So from the observers point of view (rest frame), does the distance between the asteroids (ie. 1Ly) appear contracted?

You've left a few unspecified parameters here to give a definitive answer. When you say they are separated by 1Ly distance, we assume you mean in their mutual rest frame. When you say they are in uniform velocity, we assume you mean in a direction in line with the two asteroids. And when you ask about what the observer observes, we assume you mean what is defined according to his rest frame and not according to what he actually sees. Then dEdt's answer is correct. Notice that he also made two of those assumptions:

If the distance between two asteroids is measured to be $d_0$ in the asteroids' rest frame, then in a reference frame where the asteroids are moving at $0.9c$ the distance will be about $0.44d_0$.

This is really a simple question that is best answered by applying the Lorentz Transform at 0.9c to the mutual rest frame of the two asteroids. You don't need any observer or any implication of his "point of view".

Here's the rest frame for the two asteroids where they are separated by 1Ly along the x-axis:

And here's a frame moving at 0.9c along the x-axis with respect to the first one:

Can you see that at the time of zero years, the red asteroid is displaced by 0.44Ly with respect to the blue asteroid?

Do you understand that nothing has changed in space simply by applying a different set of coordinates to the scenario?

 

[ghwellsjr]

So i can see the asteroid just behind the first one within a less amount of time after seeing the first asteroid. But what if there was no asteroid in front? Will that take more time to see the last asteroid?

If you insist on sticking an observer in the scenario, you have to say where he is with respect to the asteroids.

 

[Trojan666ru]

I need the equations to calculate length contraction time dilation and equation for two objects closing together and receding

 

[salvador_dali]

[STRIKE]The observed separations in position will change for different observers related by Lorentz boosts, but not by the usual length contraction relation (the simple gamma factor). The easiest way to see this is by brute force: write down expressions for x(t) = ... in some frame for each of the two objects, define x1(0) - x2(0) (the position separation) there, and then apply a Lorentz boost to both equations and subtract them to get the position separation in the new frame (for the coordinate observer).[/STRIKE]

 

[Dale]

I need the equations to calculate length contraction time dilation and equation for two objects closing together and receding

Everything comes from the Lorentz transform:
http://en.wikipedia.org/wiki/Lorentz_transformation#Boost_in_the_x-direction

 

[Mentz114]

The observed separations in position will change for different observers related by Lorentz boosts, but not by the usual length contraction relation (the simple gamma factor). The easiest way to see this is by brute force: write down expressions for x(t) = ... in some frame for each of the two objects, define x1(0) - x2(0) (the position separation) there, and then apply a Lorentz boost to both equations and subtract them to get the position separation in the new frame (for the coordinate observer).

I tried this ( back of envelope) and got γd₀. Is this correct ?

 

[ghwellsjr]

The observed separations in position will change for different observers related by Lorentz boosts, but not by the usual length contraction relation (the simple gamma factor). The easiest way to see this is by brute force: write down expressions for x(t) = ... in some frame for each of the two objects, define x1(0) - x2(0) (the position separation) there, and then apply a Lorentz boost to both equations and subtract them to get the position separation in the new frame (for the coordinate observer).

This isn't the correct way to determine "separations in position" which is why you get a different answer than the "usual length contraction". The bottom two dots in both of my diagrams above follow your scheme and end up with a separation of 2.3Ly
(the reciprocal of the correct answer, 0.44Ly) because the new time coordinates are different. Separations, distances, and lengths between two events have to be with the same time coordinate (even though the Proper Times may be different).

 

[ghwellsjr]

I tried this ( back of envelope) and got γd₀. Is this correct ?

It's the correct answer for his incorrect scheme and gives the reciprocal of the correct answer. This is one reason why drawing diagrams is so useful.

 

[Mentz114]

It's the correct answer for his incorrect scheme and gives the reciprocal of the correct answer. This is one reason why drawing diagrams is so useful.

Thanks ! I was interested because that is also the result of a certain experiment. Not on topic to talk about here.

 

[salvador_dali]

Whoops, I misread the question. ghwellsjr is absolutely right. The spatial separation in one coordinate system as measured in another is important when considering phase space distributions (otherwise p^0 d^3 x wouldn't be a relativistic invariant), but not here.