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That would be the assumption, yes. Of course, with SR, what is directly observed to take place between two observers is considered the actual physics that occurs between them since there is nothing else to relate it to such as an ether, so objects are considered to "really contract". The thing is, my assumption was that whatever contraction is observed to takes place with relative motion, being a coordinate effect, will reverse itself upon coming to rest again in the same way.So even in a Lorentz ether theory, which imagines that things "really contract" due to their velocity relative to the ether, you'd say it's just a coordinate effect since if the objects are brought to rest relative to each other (and thus have identical velocities relative to the ether) they will be the same length again?

Only if it is a coordinate effect, but I'm starting to see what you're saying in regard to the real physics.I wasn't exactly talking about SR, I was talking about applying the Lorentz transformation in a Newtonian universe where the laws of physics are not Lorentz-invariant (physicists would usually take the Lorentz-invariance of the laws of physics to be the definition of SR). You're saying that even in this case, you don't see why rigid bodies could be measured to have equal lengths in a frame where they have equal and opposite velocities, but not have equal lengths when brought to rest relative to one another?

Right, looks good.Imagine that in a Newtonian universe, we define a single unprimed frame so that it's a standard Newtonian inertial frame where Newton's laws apply. Then we define a family of other frames using the Lorentz transformation on the unprimed coordinates:

x' = gamma*(x - vt)

t' = gamma*(t - vx/c^2)

Now suppose that in the unprimed frame, we have a rigid measuring-rod that's at rest and 10 light-seconds long, and another rigid measuring rod that's moving in the +x direction at 0.8c and is 6 light-seconds long. Now consider a coordinate system, given by the transformation above, that is moving at 0.5c in the +x direction. In this coordinate system, the first rod is moving in the -x' direction at 0.5c, while the second rod is moving in the +x' direction at 0.5c (I can prove this if you like, but consider the relativistic velocity addition formula, which says that if the unprimed frame observes the primed frame to be moving at 0.5c and the primed frame observes the second measuring-rod to be moving at 0.5c, then the unprimed frame will observe it to be moving at (0.5c + 0.5c)/(1 + 0.5*0.5) = 0.8c).

Let's say that in the unprimed frame, both measuring-rods start with their left end at x=0 at t=0. Since rod #1 is at rest in the unprimed frame, rod #1's left end will have position as a function of time given by:

x(t) = 0 light-seconds

And rod #1's right end will have position as a function of time given by

x(t) = 10 l.s.

Meanwhile since rod #2 is moving at 0.8c, rod #2's left end will have:

x(t) = 0.8c*t

And since rod #2 is 6 light-seconds long in the unprimed frame, rod #2's right end will have:

x(t) = 0.8c*t + 6

Now consider two events in the unprimed frame: (x=0, t=0) and (x=10, t=5). Obviously the first event lies on the worldline of both the left end of rod #1 and the left end of rod #2 (i.e. it's the event of the left ends of both rods lining up), since we established that both their left ends started at x=0 at t=0. But the second event happens to lie on the worldline of both the right end of rod #1 and the right end of rod #2 (so it's the event of the right ends of both rods lining up), since the right end of rod #1 remains fixed at x=10, and since plugging in t=5 into the function x(t) = 0.8c*t + 6 gives x = 0.8*5 + 6 = 4 + 6 = 10.

Finally, consider what happens when you use the coordinate transformation to find the coordinates of these two events in the primed frame. The first event will become (x'=0, t'=0) while the second event will become (x'=8.66, t'=0). So the key here is that these two events aresimultaneousin the primed frame--the left ends of both rods line up at x'=0 at t'=0, while the right ends of both rods line up at x'=8.66 at t'=0. Since "length" in a given frame is just the distance between two ends of an object at a single moment in that frame, both rods must have equal lengths of 8.66 light-seconds in the primed frame. And as I said before, they also have equal and opposite velocities of 0.5c in the primed frame.

But even in Lorentz ether theory, Newton's laws do not strictly apply. Objects will still contract in the line of motion, so will "uncontract" when coming to rest again in the same way. If no contraction took place in a Newtonian universe, then all rods would remain the same lengths to all observers in the first place because no contraction took place to begin with. But once again, you do have me thinking, though, about what the physical processes are that produce contraction. If B quickly accelerates to v relative to A, then it is usually said that B will contract to A, but that isn't necessarily true, but depends upon how the acceleration took place. If B and C have some distance between them and they quickly accelerate to v simultaneously, then A will measure the same distance between them as before. Likewise, if all parts of a B's ship quickly accelerate to v simultaneously, then A will measure the ship to have the same length as before, whereas B will now measure his ship as elongated. But if all parts of B's ship quickly decelerate back to A's frame simultaneously, then since the clocks on B's ship from front to back are still synchronized to A but unsynchronized to B, then the ship will still have the same length to A, but B will say the front of his ship decelerated first and so contracted in the process. Also, if a train enters a tunnel of the same proper length, then if the tunnel observers threw spikes up all at once along the length of the tunnel, the train will quickly stop and be contracted to the tunnel, whereas the train observers say the spikes were thrown up at the front of the tunnel first and the train crunched up as it was stopped. If the train observers threw down spike simultaneously in their frame, then the train would stop all at once to them and remain longer than the tunnel, while the tunnel observers would say that the train threw down spikes at the back of the train first and stretched out as it stopped.But remember that in the unprimed frame, rod #1 is 10 light-seconds long while rod #2 is 6 light-seconds long. And the unprimed frame is just an ordinary Newtonian inertial frame, where according to Newtonian laws objects will not change length when they change velocities. So if rod #2 is brought to rest next to rod #1, they will still be different lengths.

Okay now, clocks would surely dilate and "undilate" in the same way when changing frames regardless of the process involved. That is, according to SR they would, but in regard to my presentation, there would be no reason to just assume that unless I can also assume it for the lengths regardless of the process also, so it looks like I'm losing ground again.Still doesn't make any sense to me. In each frame you can compare the tick rates of the two clocks--for example, if the two clocks are moving at 0.8c relative to one another, then in the rest frame of clock #1 it'll be true that clock #2 is ticking at 0.6 the rate of clock #1, while in the rest frame of clock #2 it'll be true that clock #1 is ticking at 0.6 the rate of clock #2. So here each frame is comparing the tick rates of the two clocks in terms of their own coordinates. I don't know what you mean when you say talk about twoclockscomparing tick rates, as opposed to using a single frame to compare the tick rates of two clocks.

The third frame would just give a third answer about the relative tick rates of the two clocks in terms of its own coordinates, no better or no worse than the answers found in either of the two clock rest frames. So like I said, I really don't understand what point you are trying to make here.