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I have been thinking about how length contraction should work between two identical objects. Suppose object 1 is stationary relative to object 2, and as measured by object 1, object 2 is a distance of l away. Then, 2 starts moving towards 1 with relativistic velocity v. Because the length l is relative to 1, it does not contract in length in the reference frame of 1, only object 2 itself contracts. However, 2 is moving relative to the length l, so it sees the length itself contracted, as well as object 1. This makes sense to me, and after reading through this post: Special Relativity: Length Contraction Confusion it agrees with what I was thinking.

But then I wondered, why does one of them experience the contraction of the length between them and not the other? What would happen if these were the only two objects in space and had always been moving towards each other at relative velocity v, with no information about which one accelerated; how do you know which one is which? Where and how is the information itself stored which says 'Object 2 should experience the length between them to be contracted whereas object 1 shouldn't', when this seems like such a symmetrical situation? Or am I interpreting this incorrectly?

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    $\begingroup$ No object ever "experiences" length contraction (or time dilation for that matter). It is purely an effect of "observing" other objects (which in turn is not the same as "seeing" other objects). Which one is which is immaterial. You can choose either as your "reference object". $\endgroup$ – m4r35n357 Apr 15 at 12:36
  • $\begingroup$ @m4r35n357 So if for example you were told that object 1 at an instant in time (relative to itself) observes object 2 to be a distance of l away, and you wanted to know how far away object 2 observes object 1 to be, surely it is not immaterial to know whether this length is contracted or not? Because 2 will either observe 1 to be gamma*l away or l/gamma away, which seems very important. $\endgroup$ – Alex Ghorbal Apr 15 at 13:14
  • $\begingroup$ @Wolphram jonny Does this mean that without any information about who has accelerated or not, there is no way to tell what they will measure the distances between them to be? $\endgroup$ – Alex Ghorbal Apr 15 at 17:05
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No, there is no privileged reference system after the acceleration. The issue is "when" each observer measures the distance. Let us say they synchronize their clocks right after the acceleration event, and then let each observer measure the distance at t=t'=0 (where t' is the time in the ship). The line of "present", that is, t'=0, disagrees with the "stationary" observer's "present" line, t=0. So that the ship at t'=0 will measure the distance at a time when the stationary observer clock marks t>0, let us say t=1. That is because to the ship, his clocks are synchronized in a line t'=0 that cuts the stationary observer at t=1. Even is the ship thinks that she is measuring the distance right at the end of acceleration, she will measure the distance when the stationary observer is closer to him than he "initially" was.

The same will be measured by the stationary observer if you had synchronized the clocks in a symmetrical way. For instance, if both clocks are set to 0 when a bomb than moves at the same speed in between them explodes when their distance is the same.

If you do not synchronize your clocks symmetrically, then the "clock time" of each observer when he measures the distance will not be symmetric either. That is what happens when you implicitly assume that both clocks are set to zero at the end of the acceleration. You are implicitly assuming that for both observers that moment will be zero, and because of that the times you chose for each observer to measure the distance is non symmetrical, resulting in the apparent paradox that you asked.

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But then I wondered, why does one of them experience the contraction of the length between them and not the other? What would happen if these were the only two objects in space and had always been moving towards each other at relative velocity v, with no information about which one accelerated; how do you know which one is which? Where and how is the information itself stored which says 'Object 2 should experience the length between them to be contracted whereas object 1 shouldn't', when this seems like such a symmetrical situation? Or am I interpreting this incorrectly?

We know that right after the acceleration the static object will point to the accelerated object and say: "That object is now claiming to be very close to me, but I disagree, at this moment the distance is not so small"

The accelerated object must say that same thing about the static object, so that there is no preferred frame.

So that's what it says. Problem solved. :)

But let's consider that in some more detail:

When the static object utters "the accelerated object is now very close" the static object's clock reads something, let's say 9.

So we can conclude that right after the acceleration the accelerated object considers the current time at the static object to be 9 or slightly less than 9 ... let's say 8.59. (accelerated object says it takes just a small amount of time to travel to static object according to his normal clock, and static object's clock ticks more slowly - that's why the static object's clock must read 9 or just a little bit less according to accelerated observer)

But the static object says "no, right after the acceleration the time was 3. My clock proceeded 6 hours while the accelerated object traveled from the position of acceleration to the position very close to me"

That above thing may be hard to follow, but the accelerated object says that the static object's clock jumps forwards during the acceleration, and then the static object starts saying what it says at that time, like for example that the accelerated object is now very close.

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  • $\begingroup$ Thanks, but this is not everything I was asking. If you began to observe the situation after any acceleration had ever happened, could you tell any difference? $\endgroup$ – Alex Ghorbal Apr 16 at 9:42

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