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grav
2008-Nov-14, 01:15 AM
I've decided to start this thread again fresh. I'm still trying to get the simultaneity thing down pat. In the thread "Twin paradox (relativity)", SeanF provided a scenario very similar to the one I'm about to present.

Alice and Bob are born at the same time in the same reference frame (stationary to each other) at some distance from each other. Carl is born in a ship that is moving directly past Alice where their births coincide. Danielle is born in a ship that is moving directly past Bob where their births coincide. Carl and Danielle are in the same frame of reference and moving along the line from Alice toward Bob.

So let's say Alice and Bob observe that they, Carl, and Danielle were all born at the same time. That means that Alice and Bob measure the same distance between themselves as the distance between Carl and Danielle. Due to a simultaneity effect, Carl and Danielle cannot agree that all four were born simultaneously. That also means that Carl and Danielle cannot measure the same distance between Alice and Bob as themselves. So a length contraction must take place for the distance between observers in each frame. If Alice and Bob measure the same distance between observers in each frame, then Carl and Danielle must see their own distance greater and Alice's and Bob's as shorter. In that case, Carl would say that Danielle coincides with Bob first, so they are born at the same time and both older than himself, and then Alice and Carl coincide later.

Okay, now here's another scenario somewhat similar to the one from the end of "Lorentz contraction 2" after some discussion with KenG, which so far seems to be about the only way things can work out, but I'm still examining it. All four observers are originally stationary to each other. Alice and Carl coincide at some distance from where Bob and Danielle coincide. At T=0, Carl and Danielle then instantly accelerate to some relative speed to Alice and Bob. Or instead of instant acceleration, one can figure the acceleration is tremendous but acts over infinitesimal time, so there is virtually no distance gained and no time dilation over the time of acceleration.

According to Alice and Bob, then, Carl and Danielle will still coincide with them after the acceleration takes place and therefore the distance between them remains the same before and after the acceleration, so no contraction has taken place of the distance between them. Carl, on the other hand, has experienced a simultaneity shift so that he now views Bob and Danielle as future forward. So Carl says that Danielle and Bob have already separated, while Bob and Alice remain in the same places, so their distance stays the same, but since Danielle has already moved, Carl and Danielle's distance has elongated.

Danielle has also experienced a simultaneity shift so she views Alice and Carl as past. Yet in the past, Carl had not yet accelerated so he still coincides with Alice. Then Danielle should still see the distance between Carl and herself as the same as that between Alice and Bob.

So what's going on?

1) Why do each of the observers in the first scenario see the distance between observers in the other reference frame contracted while things occur differently in the second scenario?

2) Why is Carl the only one who will see an elongated distance of his own frame in the second scenario while the other three observe the same distances between observers in each frame as before, or do they?

3) How could we tell the true length of a ship or the distance between observers in another frame as the observers in that other frame observe it if each sees different lengths as in the second scenario where Alice and Bob, for instance, might not even see a contraction between Carl and Danielle after they have attained a relative speed, even if Carl and Danielle were within a ship that accelerated all at once?

Ken G
2008-Nov-14, 02:48 AM
So let's say Alice and Bob observe that they, Carl, and Danielle were all born at the same time. I presume when you say this, you are adopting the Einstein simultaneity convention. That is not part of the physics here, but it is a very reasonable and commonly used convention.

That means that Alice and Bob measure the same distance between themselves as the distance between Carl and Danielle. Correct.

Due to a simultaneity effect, Carl and Danielle cannot agree that all four were born simultaneously. That also means that Carl and Danielle cannot measure the same distance between Alice and Bob as themselves. No, it just means they get a smaller answer than did Alice and Bob. They still think the distance between Carl and Danielle is the same as the distance between Alice and Bob.
If Alice and Bob measure the same distance between observers in each frame, then Carl and Danielle must see their own distance greater and Alice's and Bob's as shorter. I'll stop here, because this is not true.

grav
2008-Nov-14, 02:57 AM
No, it just means they get a smaller answer than did Alice and Bob. They still think the distance between Carl and Danielle is the same as the distance between Alice and Bob. Yes, I've been thinking about that since I posted. If Alice and Carl coincide, and Alice actually physically observes the same distance, including flight of light effects and all, between herself and Bob as between Carl and Danielle, then Carl will actually physically observe the same distance between observers in each frame as well. That would be because the photons that travelled from the event of Bob and Danielle coinciding will now coincide with Alice so that she can observe and it will also coincide with Carl at the same place as Alice, so Carl physically observes the exact same thing as Alice when they coincide. So it would appear that a Lorentz contraction is not the solution to the first scenario after all, and the so the simultaneity shift is still absent. There should always be a simultaneity shift between frames, right? So what is going on?

Ken G
2008-Nov-14, 03:15 AM
You are invoking the Einstein simultaneity convention, so simply study that convention and see how it plays out differently for the two sets of observers. That is all you need to understand why the convention gives a different answer when the observer using it is doing something different.

grav
2008-Nov-14, 03:22 AM
So it would appear that a Lorentz contraction is not the solution to the first scenario after all, and the so the simultaneity shift is still absent. There should always be a simultaneity shift between frames, right? So what is going on?Oh wait. If Carl and Alice actually physically observe the same thing while they are at the same place, and let's say they were both to observe that Bob and Danielle coincide at the same time they do, then from Carl's perspective, Bob has already travelled closer and Danielle remains the same distance away, being in the same frame and travelling at the same relative speed. In other words, if Carl sees Bob and Danielle coincide, then that was in the past since it took time for the light to reach him, and Bob is currently closer to him and Danielle remains the same distance. From Alice's perspective, however, Danielle would currently be further away but Bob would remain at the same distance from Alice as observed. So both would say that the distance between Carl and Danielle is greater than that between Alice and Bob. Carl would say the distance between Alice and Bob is presently less than Alice would say, as she would say it is the same as it is actually presently observed by both Carl and Alice, and Alice would say the distance between Carl and Danielle is greater than they both observe. I'm still not sure if that is by the Lorentz contraction, though. I'll have to rearrange how they coincide in a better way or something, probably after I get some rest from all this. Geez, Grant was right, this is confusing. Looks like it's probably going to take a while to get straight.

grav
2008-Nov-14, 03:35 AM
You are invoking the Einstein simultaneity convention, so simply study that convention and see how it plays out differently for the two sets of observers. That is all you need to understand why the convention gives a different answer when the observer using it is doing something different.Like that last post? I hope so. :) I'm probably going to have to run quite a few just to get the jest of it.

grav
2008-Nov-14, 03:41 AM
Like that last post? I hope so. :) I'm probably going to have to run quite a few just to get the jest of it.Jest is right. This stuff's like a cosmic joke. :lol:

grav
2008-Nov-15, 04:10 AM
Oh wait. If Carl and Alice actually physically observe the same thing while they are at the same place, and let's say they were both to observe that Bob and Danielle coincide at the same time they do, then from Carl's perspective, Bob has already travelled closer and Danielle remains the same distance away, being in the same frame and travelling at the same relative speed. In other words, if Carl sees Bob and Danielle coincide, then that was in the past since it took time for the light to reach him, and Bob is currently closer to him and Danielle remains the same distance. From Alice's perspective, however, Danielle would currently be further away but Bob would remain at the same distance from Alice as observed. So both would say that the distance between Carl and Danielle is greater than that between Alice and Bob. Carl would say the distance between Alice and Bob is presently less than Alice would say, as she would say it is the same as it is actually presently observed by both Carl and Alice, and Alice would say the distance between Carl and Danielle is greater than they both observe. I'm still not sure if that is by the Lorentz contraction, though.Okay, I've got a way to tell if it is Lorentz contraction. If two ships pass each other at some relative speed, observers on the first sees the length of their own ship as d1 and the other as L*d2, while passengers on the second sees their own length as d2 and the first's as L*d1, where L is the Lorentz contraction of L = sqrt[1 - (v/c)^2]. Now, to find the ratios of the lengths in respect to the Lorentz contraction and regardless of the actual lengths of the ships, we can just multiply the ratios of how observers on each ship view the other as compared to their own. That gives us [(L*d2) / d1] * [(L*d1) / d2] = L^2. So if we do this with other examples, regardless of the convention used, and get L^2 = 1 - (v/c)^2 as the result, then a Lorentz contraction is taking place between frames (or else it is an extremely unlikely coincidence it works out the same).

Now in the example, if Alice and Carl coincide at the same moment they observe Bob and Danielle coinciding, then each measures an optical distance between each set of observers as d. I was worried at first that with different relative speeds, and including time of flight effects of light over a distance, that some sort of beaming effect might take place, where the angles of observation might be different for Alice and Carl in different reference frames while observing the same event, almost like one might be looking through binoculars while the other isn't, but I do get the correct result for the answer so I guess not, and I'm not sure if that effect would apply over the same line of observation as the line of motion anyway. Such an effect couldn't take place from the point of coinciding of Bob and Danielle either, because the light from each would travel at the same speed over the same path to the observers. The speed of light would have to be ballistic to travel at different speeds from each, with the emitters having different relative speeds.

So if Alice sees Bob and Danielle coincide at a distance of d, she would conclude that Bob, remaining stationary to herself, would still be at a distance of d currectly, while Danielle, travelling away from her at v over a time of t = d / c since the light she observes was originally emitted, has travelled a distance of (d / c) v since then, so is currently at a distance of d + d v / c = d (1 + v/c) from Alice. Carl, on the other hand, while observing the same thing, would conclude that Danielle, being stationary to him, remains at a current distance of d, while Bob, travelling at a relative speed of v toward Carl, is currently at a distance of d - (d / c) v = d (1 - v/c) from Carl and Alice.

Therefore, Alice will measure a distance of d between herself and Bob, and a distance of d (1 + v/c) between Carl and Danielle, while Carl measures a distance d between himself and Danielle, and d (1 - v/c) between Alice and Bob. Multiplying the ratios of how the observers in each frame views the other's distance as compared to their own, we get [d (1 + v/c) / d] * [d (1 - v/c) / d] = (1 + v/c) (1 - v/c) = 1 - (v/c)^2 = L^2. So it appears that observers in each frame would indeed see the other Lorentz contracted, even over empty distances.

Just for the record, Bob would agree with the distances Alice measures and Danielle would agree with Carl. For instance, Bob would agree that he and Alice are separated by d because they are stationary in the same frame of reference. The light from when Bob and Danielle coincide takes t = d / c to reach Alice, at the same moment that Alice and Carl coincide, and that light also takes t = d / c to travel back to Bob. So during that time, Bob says Danielle has travelled 2 d / c past him, and Carl has travelled d / c toward him. The distance he measures between Carl and Danielle, then, is d - d v /c for Carl plus 2 d v /c for Danielle, which gives d' = d - d v / c + 2 v /c = d + d v / c = d (1 + v/c), the same as Alice says.

grav
2008-Nov-15, 04:56 AM
I've noticed a couple of strange things about the last post. For one, it provides kind of a strange twist for Relativity that I've never seen before. I've always thought one frame would view distances in another frame Lorentz contracted from what observers in the other frame would measure for the same distance. The result of the ratios of observed distances in the scenario is still L^2 as it should be, but Alice and Bob would actually observe a larger distance between Carl and Danielle than Carl and Danielle measure themselves. However, Carl and Danielle also measure a much smaller distance between Alice and Bob, so things still work out, but I never would have imagined that occuring. It is the only way I can see things working out for SeanF's scenario, though, and it also ties in with KenG's explanation as well. In the second scenario, Alice and Bob wouldn't even notice a contraction between Carl and Danielle at all after they accelerate to some relative speed, which also seems contrary to my thinking about a Lorentz contraction between frames, but that appears to be the only way that scenario can work out as well.

The other thing that seems strange is that even if the speed of light did not travel at c to all observers in any frame, some form of Relativity must still persist. For instance, let's say light travelled at c through a medium in space, and therefore measured at some different speed to an observer, depending upon the speed of the receiver to the medium and the angle between the emitter and receiver and the receiver's line of travel. Now let's look at our example again. From what Alice and Carl observe, Alice would say that Bob remains at d since he is stationary to her. Carl, on the other hand, would still say that Bob has moved some distance during the time the light has taken to travel from Bob to Carl. So whatever speed of light Carl measures, say c', Carl says Bob is currently at a distance of d - (d / c') v from Alice while Alice says he remains at d. So unless Carl measures the light as travelling instantaneously between emitter and receiver, which he would not, then Carl and Alice cannot agree upon the distance between Alice and Bob. So even classically, without a formal representation of Relativity, it can still be shown that some form of Relativity, namely contraction of measured distances between frames of observation, still exists. I haven't performed this ballistically yet, though, and would be curious about what that would say classically as well. I think it's also strange that different distances should be measured between frames even classically, though, so I'm still going over it, but I cannot identify an error yet. Does anybody else see one?

grav
2008-Nov-20, 01:46 AM
Okay, I've got a way to tell if it is Lorentz contraction. If two ships pass each other at some relative speed, observers on the first sees the length of their own ship as d1 and the other as L*d2, while passengers on the second sees their own length as d2 and the first's as L*d1, where L is the Lorentz contraction of L = sqrt[1 - (v/c)^2]. Now, to find the ratios of the lengths in respect to the Lorentz contraction and regardless of the actual lengths of the ships, we can just multiply the ratios of how observers on each ship view the other as compared to their own. That gives us [(L*d2) / d1] * [(L*d1) / d2] = L^2. So if we do this with other examples, regardless of the convention used, and get L^2 = 1 - (v/c)^2 as the result, then a Lorentz contraction is taking place between frames (or else it is an extremely unlikely coincidence it works out the same).Welp, looks like that was indeed coincidence after all. I hate it when that happens. It seems to happen often enough that I probably can't rely too much on that type of logic as much as I would like to. Anyway, this thread (http://www.bautforum.com/questions-answers/81312-distance-sound-emitter.html) shows that an observer will see an observed distance become greater by sqrt[(1 + v/c) / (1 - v/c)] when he accelerates toward an object according to Relativity and it will become greater by 1 + v/c classically. Classically, then, Carl and Alice will both agree on the current positions of Bob and Danielle and the current distances between two observers in each frame after all, as outlined in the other thread. With Relativity, each will observe a distance between the current positions of the two observers in the other frame after acceleration takes place that is just the Lorentz contracted distance that the two observers in the other frame measure between themselves, and the multiple of the ratios of distances is still L^2. All of that seems perfectly consistent with Relativity so far.

But now what I want to do is to find how the current positions and observed positions vary before and after acceleration takes place in the same beginning and resulting frame. For instance, let's say Carl observes Danielle at some distance from him. When Carl observes Danielle to accelerate, Carl accelerates also in the same way, whereby both observers are once again in the same frame after the acceleration is over. The acceleration is instantaneous, so we don't need to figure in any time dilation for the time of acceleration. I want to pick that scenario apart to see what takes place.

I can do that at least a couple of different ways. First, let's say Carl accelerates a moment before he sees Danielle accelerate. In that case, the new observed distance that Carl sees for Danielle jumps from d to sqrt[(1 + v/c) / (1 - v/c)] d. Danielle is also thrown future-forward due to the simultaneity shift for her current position, but what is actually observed by Carl doesn't change over zero time of acceleration, so a moment later, he sees Danielle accelerate to the same relative speed as Carl and they remain stationary thereon. So Carl will observe the new distance between Carl and Danielle might be just sqrt[(1 + v/c) / (1 - v/c)] d. A second scenario would have Carl observe Danielle accelerate a moment before he does. In that case, the time the light takes to travel over the distance of d would place Danielle further ahead since she has already been moving away from Carl over that time. Then Carl accelerates also and the observed distance becomes greater.

Now here's the thing. Since Carl and Danielle are in the same frame of reference after they both have accelerated, then whatever distance Carl observes between himself and Danielle should be the actual distance between them at that point, because they are both once again stationary to each other. This will be the key point in the solution, I think. Also, Danielle should see the same distance between herself and Carl and calculate the same thing according to the events that took place from her point of view. The events I have described in the last paragraph are not precise, and may be altered greatly, since I have not worked it all out yet, but it provides the gist of what I am attempting to do.

SeanF
2008-Nov-20, 03:52 PM
Due to a simultaneity effect, Carl and Danielle cannot agree that all four were born simultaneously. That also means that Carl and Danielle cannot measure the same distance between Alice and Bob as themselves.
No, it just means they get a smaller answer than did Alice and Bob. They still think the distance between Carl and Danielle is the same as the distance between Alice and Bob.
Uh, Ken?

If, in any given reference frame, the Alice-Bob distance is the same as the Carl-Danielle distance, then that would mean the Alice-Carl crossing is simultaneous with the Bob-Danielle crossing. That's simple geometry - unless Danielle is at Bob's location when Carl is at Alice's location, the Alice-Bob distance must be different than the Carl-Danielle distance*.

But since the two crossings are separated in space-time, they cannot be simultaneous in two different reference frames.

So Grav is right. If the Carl-Danielle distance and the Alice-Bob distance are the same in the Alice-Bob reference frame, they must be different in all others.

*The is not necessarily true in four-dimensional space-time, of course, but we're really dealing with only two dimensions in this thought experiment.

grav
2008-Nov-22, 03:08 AM
Okay. Time to run a few scenarios.

Alice and Carl are stationary in the same place and Bob and Danielle are stationary in the same place but a distance of d from Alice and Carl. Danielle instantly accelerates to v. As soon as Alice and Carl optically observe that Danielle has accelerated, Carl accelerates to v also. From the "Lorentz contraction 2" thread, we found that with Relativity, whatever distance is optically observed between observers must also be the actual distance between them when the observers are stationary to each other. We also found in the "distance of a sound emitter" thread (which became light instead), the optical distance between two observers changes for an observer that instantly accelerates toward the other, from d to d sqrt[(1 + v/c) / (1 - v/c)]. So let's look at what the observers view with this scenario.

From Carl's point of view, if Carl accelerates to v a mere moment before he sees Danielle accelerate, he would then observe Danielle at a distance of d sqrt[(1 + v/c) / (1 - v/c)]. A moment later, Danielle also accelerates to v and they are now in the same frame of reference, so that is their actual distance from each other.

Alice optically observes both Carl and Danielle accelerate at the same time, so Carl is at his actual position since it coincides with Alice, but Danielle has already been travelling forward over the time it took the light to reach Alice, which is d / c, so Alice says Danielle is currently d v / c further ahead than her observed position. Therefore, Alice says the current distance between Bob and Danielle according to her frame is d + d v / c = d (1 + v/c). Alice sees that distance Lorentz contracted from Carl and Danielle's measure of the distance in their own frame, so Alice says Carl and Danielle would measure the distance as d (1 + v/c) / L = d sqrt[(1 + v/c) / (1 - v/c)], the same as Carl did measure.

From Danielle's point of view, she accelerated first while the others remained stationary and now measures the actual distance between Alice and Bob as Lorentz contracted to L d, so measures the same distance between herself and Carl. Danielle accelerated at T=0 and Carl will accelerate when he sees the light from Danielle accelerating, which will reach him t = d / c later, so when his clock reads T = d / c. Danielle's simultaneity shift puts his clock at T = - Z L d v / (c^2 - v^2) = -d v / c^2 after she accelerates. Therefore, Danielle continues to travel away from Carl until a time of t = d / c + d v / c^2 passes on Carl's clock, where by t = (d / c + d v / c^2) / Z passes on Danielle's clock, and she has travelled a distance of (d / c + d v / c^2) v / Z away from Carl during this time, providing a total distance of L d + (d /c + d v / c^2) v / Z that she measures between herself and Carl when he accelerates, giving us

L d + (d / c + d v / c^2) v / Z
= [Z L d + (d / c + d v / c^2) v] / Z
= [(1 - (v/c)^2) d + d (v/c) + d (v/c)^2] / Z
= d [1 - (v/c)^2 + (v/c) + (v/c)^2] / Z
= d (1 + v/c) / Z
= d sqrt[(1 + v/c) / (1 - v/c)]

Well, one thing's for sure. Relativity is nothing if not mathematically consistent. So now I want to explore KenG's example. In that scenario, Carl and Danielle accelerate simultaneously to v at T=0. Say Danielle accelerates a mere moment before Carl. After acceleration, Danielle measures herself a distance of L d from Carl. Carl will accelerate when his clock reads T=0 also. Danielle's simultaneity shift puts Carl's clock at T = - Z L d v / (c^2 - v^2) = - d v / c^2. Therefore, Danielle continues to travel away from Carl until a time of t = d v / c^2 passes on Carl's clock, wehreby t = d v / (c^2 Z) passes on Danielle's clock. During that time, she travels an extra distance of d v^2 / (c^2 Z) from Carl, making the total distance between them L d + d v^2 / (c^2 Z) when Carl then also accelerates to the same frame. That gives us

L d + d v^2 / (c^2 Z)
= [Z L d + d (v/c)^2] / Z
= d [(1 - (v/c)^2) + (v/c)^2] / Z
= d / Z

So if Danielle measures thier distance as d / Z, then Alice and Bob will measure it as L d / Z = d in the original frame. So KenG's explanation for that in the "Lorentz contraction 2" thread is correct. If Carl and Danielle accelerate simultaneously (or all parts of a rocket), then the stationary frame will continue to measure the same distance between them (or the same length of a rocket).

Sam5
2008-Nov-22, 03:59 AM
I've decided to start this thread again fresh.

First, do you understand why Lorentz said there would be a contraction?

grav
2008-Nov-22, 04:21 AM
First, do you understand why Lorentz said there would be a contraction?Of course. Due to the result of the M-M experiment. I always keep that foremost in my mind when working through these types of things, to see how it might ever be applied differently or what might be implied and to make sure nothing contradicts the result of that sole experimental result. It is the ground level that all of this is based upon, so without it everything crumbles.

Sam5
2008-Nov-22, 04:26 AM
Of course. Due to the result of the M-M experiment. I always keep that foremost in my mind when working through these types of things, to see how it might ever be applied differently or what might be implied and to make sure nothing contradicts the result of that sole experimental result. It is the ground level that all of this is based upon, so without it everything crumbles.

What did Lorentz say caused the contraction?

grav
2008-Nov-22, 05:07 AM
What did Lorentz say caused the contraction?I've never read Lorentz's paper or anything, but as far as I've ever read otherwise, he didn't. He just noted that if the M-M apparatus was contracted by a certain amount in the direction of travel, then the mathematics would work out. But of course, those mathematics were already based upon motion through a medium which propagates light, if that's what you mean.

grav
2008-Nov-22, 05:15 AM
Now let's invoke my sorta variation on the Equivalence Principle, that what is true of all frames is also true between frames. That is, if stationary observers see a rocket ship that has accelerated along all parts remain the same length according to what they measure, then it will also remain the same length while constantly accelerating away, really just a constant in the integration for the acceleration. This means the passengers of the ship will measure their own length as d / L, depending upon the instantaneous relative speed. This relates back to what Richard discussed about Rindler observers. I had almost forgotten that that was part of the reason I started the "Lorentz contraction 2" thread in the first place. So things have finally come full circle. This type of elongation would be quite a bit different, though, I think, because the Rindler horizon and such describes optical effects, like how light is trying to catch up to the accelerating observer, which includes time of flight of light, while this is purely physical.

grav
2008-Nov-22, 04:54 PM
Going back to the original scenario, if the ship accelerates along all parts to v at T = 0, then observers in its new frame would also measure the length of the ship as d / L, the same as the passengers do. Now let's say the ship again accelerates to v from the new frame. Then the observers in the new frame should say it has again elongated by a factor of 1 / L. It shouldn't matter which way it accelerates, even back into the original frame, so Alice and Bob in the original frame should say it has now elongated to d / L^2. Let's find out.

After Carl and Danielle accelerate to v, Carl's clock will lag behind Danielle's by tl = d v / (c^2 Z), although ticking at the same rate, as we saw a few posts back with that scenario. Alice and Bob will see no time lag between Carl and Danielle, so would say both Carl's and Danielle's clocks read the same since they accelerated simultaneously according to the stationary frame, although ticking at a different due to the time dilation between the new and original frames. We could continue using these new times on Carl and Danielle's clocks, but let's make things easier by allowing Carl and Danielle to resynchronize. It will not affect anything physically. Alice and Bob will then see the usual simultaneity shift between them.

So after resynchronization, Alice and Bob will see Carl's clock read t = - Z x v / (c^2 - v^2) less than their own, where x is Carl's current distance according to Alice and Bob, and Danielle's read t = - Z (x + d) v / (c^2 - v^2), for a simultaneity difference between them of tl = -Z d v / (c^2 - v^2) observed by Alice and Bob, Danielle's time lagging behind Carl's, since they are moving away. Carl and Danielle then instantly accelerate back into the original frame. Alice and Bob see Carl come back to rest first and Danielle t = d v / (c^2 - v^2) later according to Alice and Bob's clocks. After they see that Carl has stopped, they will continue to see Danielle travel for the additional time, travelling an extra distance of d v^2 / (c^2 - v^2). So according to Alice and Bob, that will make a total distance between Carl and Danielle when they come to a halt of

d + d v^2 / (c^2 - v^2)
= d [(c^2 - v^2) + v^2] / (c^2 - v^2)
= d c^2 / (c^2 - v^2)
= d / [1 - (v/c)^2]
= d / L^2

v used for L here would still be the relative speed they originally accelerated to, not the new relative speed when they come back to rest again, of course. If Carl and Danielle continue to accelerate to v and stop, accelerate to v and stop, accelerating and decelerating simultaneously in their own frame each time, then the distance between them (or the length of their ship) will become d / L^(2n), where n is the number of times they accelerate to v and stop.

grav
2008-Nov-22, 05:58 PM
Now I want to know what the distance between Carl and Danielle will be if Carl and Danielle accelerate simultaneously to v, then simultaneously in their own frame to -v, then come to a stop where Danielle once again coincides with Bob. For simplicity, since it doesn't make a difference with the distance, let's say Carl and Danielle always resynchronize their clocks when they change frames and become inertial and stationary to each other.

So Carl and Danielle instantly accelerate to v at T=0. Alice and Bob sees the distance between them remain d. Carl and Danielle travel some distance togather inertially, x for Carl and x + d for Danielle according to Alice and Bob. Then they instantly accelerate to v in the other direction simultaneously in their own frame, so Carl will be seen to accelerate first to Alice and Bob and Danielle t = d v / (c^2 - v^2) later. During this time, Alice and Bob see Carl travel travel a distance of -d v^2 / (c^2 - v^2) in the opposite direction and Danielle d v^2 / (c^2 - v^2) in the same direction, making a total distance between them of d + 2 d v^2 / (c^2 - v^2) according to Alice and Bob when Danielle accelerates to -v also.

Now the simultaneity shift that Alcie and Bob sees occurs oppositely since Carl and Danielle are now travelling toward them. So when Carl and Danielle come to a halt in the original frame, Alice and Bob will see Danielle stop first and Carl stop (d + 2 d v^2 / (c^2 - v^2)) v / (c^2 - v^2) later. Danielle instantly stops at Bob's position and Alice and Bob will see Carl continue to travel an additional distance of (d + 2 d v^2 / (c^2 - v^2)) v^2 / (c^2 - v^2), making a total distance between Carl and Danielle of

d + 2 d v^2 / (c^2 - v^2) + (d + 2 d v^2 / (c^2 - v^2)) v^2 / (c^2 - v^2)

= [d (c^2 - v^2) + 2 d v^2 + (d + 2 d v^2 / (c^2 - v^2)) v^2 ] / (c^2 - v^2)

= [ d (c^2 - v^2)^2 + 2 d v^2 (c^2 - v^2) + d v^2 (c^2 - v^2) + 2 d v^4] / (c^2 - v^2)^2

= d [ c^4 - 2 v^2 c^2 + v^4 + 2 v^2 c^2 - 2 v^4 + v^2 c^2 - v^4 + 2 v^4] / (c^2 - v^2)^2

= d [c^4 + v^2 c^2] / (c^2 - v^2)^2

= d [1 + (v/c)^2] / [1 - (v/c)^2]^2

= d [1 + (v/c)^2] / L^4

It actually wouldn't matter if Danielle came to a stop at Bob's position again, though. This is really just the elongation that would occur if something were to instantly accelerate simultaneously along all parts to v from rest, then -v, then stop.

grav
2008-Nov-22, 07:29 PM
Looking back at the last couple of posts, it almost seems like they should lead to the same results, but I have rechecked the math and don't see any errors. In the first scenario, Carl and Danielle accelerate to v and stop, making their length d / L^2. If they were to then accelerate again in the opposite direction and stop, where Danielle once again coincides with Bob, with a simultaneity shift between them after acceleration of (d / L^2) v / (c^2 - v^2) after accelerating the second time according to Alice and Bob, we would have a total distance between them when they stop again of

d / L^2 + (d / L^2) v^2 / (c^2 - v^2)
= [(d / L^2 (c^2 - v^2) + (d / L^2) v^2] / (c^2 - v^2)
= (d / L^2) c^2 / (c^2 - v^2)
= d / L^4

With the second scenario, after running back through the scenario to recheck it, and performing the math for the final distance in a slightly different way this time, we get

d + 2 d v^2 / (c^2 - v^2) + (d + 2 d v^2 / (c^2 -v^2)) v^2 / (c^2 - v^2)

= d + 2 d v^2 / (c^2 - v^2) + d v^2 / (c^2 - v^2) + 2 d v^4 / (c^2 - v^2)^2

= [d (c^2 - v^2)^2 + 3 d v^2 (c^2 - v^2) + 2 d v^4] / (c^2 - v^2)

= d [c^4 - 2 v^2 c^2 + v^4 + 3 v^2 c^2 - 3 v^4 + 2 v^4] / (c^2 - v^2)^2

= d [c^4 + v^2 c^2] / (c^2 - v^2)^2

= d [1 + (v/c)^2] / [1 - (v/c)^2]^2

= d [1 + (v/c)^2] / L^4

Now the reason it would seem these two scenarios should be the same is because in the first scenario, Carl and Danielle instantly accelerate to v and stop, then accelerate to -v and stop. In the second scenario, Carl and Danielle instantly accelerate to v, then instantly to -v, and stop. the thing is, in order to get from v to -v, they must still pass through the stationary frame speed of v=0, so it seem it should be the same as the first scenario, except that instead of stopping there momentarily and then accelerating again in the opposite direction, the acceleration from v to -v is continuous. Yet they both give different results. The only reason I can see for that would be that Carl didn't stop long enough to see Danielle decelerate to the same frame, since after decelerating back to the original frame, Carl would see a simultaneity shift for Danielle until she also accelerates according to him, since although the decelerations are simultaneous in the frame they decelerated from, they experience a momentary simultaneity shift in the new one. In order for Carl to see Danielle travelling at the same relative speed after Carl decelerates, a time of t = d' v / (c^2 - v^2) must pass for Carl after he decelerates since he now sees Danielle's clock lag behind his own by a time of tl = - Z d' v / (c^2 - v^2), for whatever distance d' that Carl now measures between them in the new frame. If that amount of time or greater is not allowed to pass upon Carl's clock before he accelerates again in the opposite direction, then he will not see Danielle decelerate to the same frame as himself and become stationary with him before Carl accelerates again, whereby they will not accelerate in the opposite direction simultaneously, so the result will be different. That's very interesting.

grav
2008-Nov-22, 09:19 PM
Alrighty then. Now I want to know how something would continue to elongate according to the stationary frame if it initially travels at v, then instantly accelerates to -v, then back to v again, over and over, oscillating back and forth. We found before that upon accelerating from v to -v, the length measured by the stationary frame increases to d + 2 d v^2 / (c^2 - v^2) = d [ (c^2 - v^2) + 2 v^2] / (c^2 - v^2) = d (c^2 + v^2) / (c^2 - v^2) = d [1 + (v/c)^2] / [1 - (v/c)^2] = d [1 + (v/c)^2] / L^2. As a quick side note, if it were then to come to a halt in the stationary frame, it would measure d [1 + (v/c)^2] / L^4, so the halting elongates it by 1 / L^2 as found in the first scenario. Anyway, once it travels in the opposite at -v, that would be the same as the original v but in the opposite direction, so an instant acceleration back to v brings its length to d ([1 + (v/c)] / [1 - (v/c)^2])^2. So that is the increase in length with each full oscillation, and it will enlongate by ([1 + (v/c)^2] / [1 - (v/c)^2])^(2n) after n full oscillations.

For small v and n, then, the enlongation approximates d [1 + 2 (v/c)^2]^(2n), and that also approximates d [1 + 4 n (v/c)^2]. So say if a ball were to bounce repeatedly between two walls, this would approximately be the measured enlongation after n bounces back and forth. I thought this might be testable with something that is made to vibrate back and forth really fast. I tried looking up "fastest oscillator" to see what might be used for this but only got some results about lasers. I also tried "fastest vibrater" but didn't quite get the results I was looking for there either. Anyway, let's say a centimeter long object travels a centimeter and back again, over and over. It does this 100 times a second. So in a second, it travels 200 meters, so v = 2 meters per second. Plugging that into the equation for the approximate enlongation, we get d' = d [1 + 4 n (v/c)^2] = d [1 + 4 (t / p) (v/c)^2], where p is the period of oscillation and t is the time the object has oscillated.

Let's say we wanted to approximately double the object's length. Then 4 (t / p) (v/c)^2 = 1 and since (v/c)^2 comes to 4.444 * 10^(-17) and p = .01 seconds, the object would have to oscillate for a time of t = (2.25 * 10^16) *.01 / 4 seconds = 5.625 * 10^13 seconds, or close to 1.8 million years. Actually, the exact result is about log(2) times smaller than this, since we used a large enough n to double the result, throwing the approximation off quite a bit, but elongating to a much smaller ratio of length would be approximated very well. The exact result for d ([1 + (v/c)^2] / [1 - (v/c)^2])^(2n) = 2d becomes 2 n log([1 + (v/c)^2] / [1 - (v/c)^2]) = log(2), so n = log(2) / (2 log([1 + (v/c)^2] / [1 - (v/c)^2])) = 3.898953*10^15, and since n = t / p, then t = n p = 3.898953*10^13 seconds, or about 1.24 million years.

A laser could measure a much smaller elongation over a much smaller time, though, I think. There would be at least a couple of other things to consider with a test like this, though. One is that the acceleration would have to occur simultaneously along all parts of the object with each oscillation. Another is that while the object might be elongating in a true sense, the bonds that hold the atoms together might be pulling them back into place at the same time, like springs, if the oscillations even have enough strength to pull apart from the bonds of the atoms in order to elongate to begin with, and even if they did, we probably wouldn't even be able to disassociate that from the gained momentum that tries to break the bonds with each oscillation anyway. You know what? I think these last couple of statements pretty much does that particular experiment in, but I'll keep this post because I still think it's interesting. After you read it, though, just do yourself a favor and go ahead and forget about it. Done and done. :)

grav
2008-Nov-26, 06:10 PM
Okay, now for some more questions. As I said, Relativity seems to be completely mathematically consistent, but is it logically consistent as well? Let's take KenG's explanation for what stationary observer's would see of a ship that instantly accelerates along all parts to v. To the stationary observers, it would remain the same length l. To the passengers on the ship, it would enlongate to l / L. This would also be visually observed according to the "distance of a sound emitter" thread. I have verified these. It seems the only way things can work out using Relativity. Also, if the ship were to constantly accelerate, it would continue to enlongate according to the instantaneous relative speed. Now here are my questions.

1) Would rulers on the ship also enlongate in the direction of travel so that the passengers still measure the same length for the ship?

2) If so, what happens when we turn the rulers perpendicular to the line of motion? Do they shorten to the length in that direction, or do they stay enlongated so that they measure a shorter length across the ship in that direction using the rulers?

3) If the rulers and everything else enlongagated in the direction of travel, the passengers would still physically observe a longer ship, so would it look the same but just like it was "wide-lensed" or something?

4) Relativity says the elongation would be real, so would the ship start to break up at some point if it continued to accelerate along all parts?

5) If the ship were to break up according to the passengers, what would that look like to the stationary observers? The stationary observers still see the same length for the ship, so what would the breaking points look like to them?

grav
2008-Dec-05, 02:41 PM
Now let's invoke my sorta variation on the Equivalence Principle, that what is true of all frames is also true between frames. That is, if stationary observers see a rocket ship that has accelerated along all parts remain the same length according to what they measure, then it will also remain the same length while constantly accelerating away, really just a constant in the integration for the acceleration. This means the passengers of the ship will measure their own length as d / L, depending upon the instantaneous relative speed. This relates back to what Richard discussed about Rindler observers. I had almost forgotten that that was part of the reason I started the "Lorentz contraction 2" thread in the first place. So things have finally come full circle. This type of elongation would be quite a bit different, though, I think, because the Rindler horizon and such describes optical effects, like how light is trying to catch up to the accelerating observer, which includes time of flight of light, while this is purely physical.Oops. Looks like I may have misapplied my own principle there. For the same reason I got two different answers in post #20, there would be a difference between two observers or a rocketship that is constantly accelerating and one that eventually cuts off acceleration and becomes inertial, because the rear accelerating observer will actually accelerate for a longer time in the new frame while the other has already become inertial, according to what they measure, so there would be an even greater elongation while accelerating but the resulting distance is shortened while the rear observer catches up a bit before also switching off their engines at the same predesignated time according to their clocks, due to the change in simultaneity.

The principle can still be applied for what the stationary observers measure, though, both during acceleration and after, as just reading the same times on the accelerators' clocks, although time dilated, and the same distance between them at all times. I think I will need to review what is taking place according to the stationary observers and see if I can't come up with the elongation in distance between the accelerating observers while still accelerating from that point of view.

grav
2008-Dec-09, 01:10 AM
I'm afraid I'll have to give myself another -1 there. I was finding for an ever increasing length for observers that accelerate, then stop, then accelerate, then stop, and for an oscillating electromagnetic bar and such. I haven't reviewed it in full, but I think that is only if the accelerators resynchronize their clocks to the new frame every time they become inertial again, as I did in post #19 (um, and continued, I believe, in #20 and #21). The resynchronization wouldn't normally affect anything physical, unless of course they are continued to be used to determine when the accelerations will occur again in each accelerating observer's individual frame. :doh:

Anyway, one can see what occurs more clearly from the stationary frame without resynchronization. After the accelerating observers accelerate simultaneously from the stationary frame, the stationary observers will still see the same distance between them and identical times on their clocks, although time dilated to their frame. If the accelerating observers don't resynchronize in the new frame, but just decelerate again at some predesignated time according to each individual accelerating observer's clock, then the stationary frame would see them both decelerate simultaneously back into the stationary frame from what they observe, still remaining the same distance from each other as before.

mugaliens
2008-Dec-09, 06:59 PM
I've decided to start this thread again fresh. I'm still trying to get the simultaneity thing down pat.

I can well imagine that when my son is my age, he'll be saying, "Look, Dad! I'm a member on this cool message forum, and look what I found: Lorentz contraction 72, by some guy named Grav! Isn't that cool?"

:lol:

Carry on!

grav
2008-Dec-10, 12:33 AM
I can well imagine that when my son is my age, he'll be saying, "Look, Dad! I'm a member on this cool message forum, and look what I found: Lorentz contraction 72, by some guy named Grav! Isn't that cool?"

:lol:

Carry on!72, huh? No doubt. Which will probably also be how old I'll be when I finally figure all of this out. :)