# Thread: trains and length contraction -do trains really get shorter?

1. Here is an interesting thought experiment: two identical trains are positioned on a rail in such a way that the front of the second touches the back of the first.
The trains are identical in every way, with identical computers with identical trajectories programmed in them. They have perfect clocks that are perfectly synchronised between the trains.
The trains start moving, and at every instant, they move at exactly the same speed.

Question: do the trains still touch?

2.

3. The clocks will not remain synchronized between trains. Each train will mantain that all its clocks remained in sync, but that the clocks on the other train go out of sync. Thus according to each train, while every part of itself started moving at the same instant, every part of the other train did not. As a result, each train determines that the rear of the other train starts moving before the front does and the train contracts along its length.

4. I think I see what you're getting at, but just to be sure: At the exact instant the trains start to move, the clocks are still synchronised.

5. The trains start moving, and at every instant, they move at exactly the same speed.

Question: do the trains still touch?
Absolutely not. The trains, if moving at the same speed, will measure the same time, but will still contract to an observer outside the trains. An observer inside will measure the same length of the train they are on, but measure a shorter length for the other train.

6. What if instead of touching they're connected to each other. It shouldn't make a difference, should it? Whether the first train is pulling the second or the second is moving on its own power. But if one train length contracts away from the other, you could apply that to mean that all matter, when accelerated to near c, disintegrates as one molecule length contracts away from the other.

For observers on the first train, the second train is essentially in its inertial frame all the time, so my guess is that the trains never stop touching to observers on the trains. For outside observers, I'm not sure what the answer is. Does the length contract about the center of mass of each train? If that's the case, it would appear that the trains separate. But if the length contraction is around the center of mass of the two train system, they'll stay together. But if one train is more massive than the other, and the center of mass is inside one train, they'd still appear to separate.

My gut says that length contraction multiplies distances, so since the distance between trains is 0 it stays 0 for all observers in all inertial frames. I just can't back that up with math or reasoning.

7. Ah, somehow I mis-read the question. I thought you were talking about two trains on parallel tracks pointing in opposite directions.

What you are proposing is akin to the Bell spaceship problem. So let's frame it a little differently:

Let's say that you have two locomotives tied together by a string. Both trains start off simultaneously at the same speed in the same direction maintaining a constant distance between them. What happens to the string? As it starts moving, the String should contract. But if it does and the trains stay the same distance apart, then the string will break. But from the frame of the locomotives, the string doesn't contract and doesn't break.

The trick is in when we say "Both trains start off simultaneously at the same speed in the same direction maintaining a constant distance between them."

According to who?, the embankment or the trains?

If it is according to the trains themselves, then in that frame, nothing unusual happens, the string does not contract as the locomotives maintain a constant distance. And from the embankment, the string contracts but so doe the distance between the locomotives. So the string does not break in either frame.

If it is according to the embankment, then from the embankment frame the string contracts as the trains maintain a constant distance and it snaps.
From the frame of the trains the string also snaps for the following reason.

In order to go from at reast to the embankmen to moving with respect to it, the trains must accelerate. In GR acceleration is equivalent to gravity. An clock higher in a gravity field will run slower than one lower. Due to the trains acceleration, a clock in the rear locomotive acts like a clock lower in a gravity field compared to a clock in the front of the train. So in the accelerated frame of the train, the rear locomotive runs slower than the front locomotive and starts to fall behind, this increases the distance between the locomotives and breaks the string. So in this case the string breaks in both frames.

8. Originally Posted by Janus
In order to go from at reast to the embankmen to moving with respect to it, the trains must accelerate. In GR acceleration is equivalent to gravity. An clock higher in a gravity field will run slower than one lower. Due to the trains acceleration, a clock in the rear locomotive acts like a clock lower in a gravity field compared to a clock in the front of the train. So in the accelerated frame of the train, the rear locomotive runs slower than the front locomotive and starts to fall behind, this increases the distance between the locomotives and breaks the string. So in this case the string breaks in both frames.
If this is true, it means accelerating near the speed of light will cause all matter to disintegrate (molecules can be thought of as tied together with little pieces of string).

9. Originally Posted by Liongold
Absolutely not. The trains, if moving at the same speed, will measure the same time, but will still contract to an observer outside the trains. An observer inside will measure the same length of the train they are on, but measure a shorter length for the other train.
I would think an observer inside one train, would also observe the other train as having the same length when they are moving at constant speeds.
Originally Posted by Numsgil
What if instead of touching they're connected to each other.
The connection would get strained/stressed/break. Janus explained this very nicely.
He gave a very nice explanation overall, I think, especially the last paragraph, something I was not entirely sure about.

10. Originally Posted by Numsgil
Originally Posted by Janus
In order to go from at reast to the embankmen to moving with respect to it, the trains must accelerate. In GR acceleration is equivalent to gravity. An clock higher in a gravity field will run slower than one lower. Due to the trains acceleration, a clock in the rear locomotive acts like a clock lower in a gravity field compared to a clock in the front of the train. So in the accelerated frame of the train, the rear locomotive runs slower than the front locomotive and starts to fall behind, this increases the distance between the locomotives and breaks the string. So in this case the string breaks in both frames.
If this is true, it means accelerating near the speed of light will cause all matter to disintegrate (molecules can be thought of as tied together with little pieces of string).
No.
This situation occurs only in the unusual circumstance of having two motive forces which we arbitrarily decided maintained a constant distant as measured from the embankment frame. In real life, this would not be the case, it would be more like the case where the ends maintain a constant distance in the frame of the train, In this case, the string doesn't break in either frame.

Also, any strain on the string would be due to the rate of acceleration not speed. The strain would be the same at any speed as long as the acceleration remained constant. It would not increase as you got closer to the speed of light.

As long as the object was able to withstand the stresses of acceleration (which would need to be true in any case) it would not be torn apart as it increased speed.

11. Originally Posted by Janus
Ah, somehow I mis-read the question. I thought you were talking about two trains on parallel tracks pointing in opposite directions.

What you are proposing is akin to the Bell spaceship problem. So let's frame it a little differently:

Let's say that you have two locomotives tied together by a string. Both trains start off simultaneously at the same speed in the same direction maintaining a constant distance between them. What happens to the string? As it starts moving, the String should contract. But if it does and the trains stay the same distance apart, then the string will break. But from the frame of the locomotives, the string doesn't contract and doesn't break.

The trick is in when we say "Both trains start off simultaneously at the same speed in the same direction maintaining a constant distance between them."

According to who?, the embankment or the trains?

If it is according to the trains themselves, then in that frame, nothing unusual happens, the string does not contract as the locomotives maintain a constant distance. And from the embankment, the string contracts but so doe the distance between the locomotives. So the string does not break in either frame.

If it is according to the embankment, then from the embankment frame the string contracts as the trains maintain a constant distance and it snaps.
From the frame of the trains the string also snaps for the following reason.

In order to go from at reast to the embankmen to moving with respect to it, the trains must accelerate. In GR acceleration is equivalent to gravity. An clock higher in a gravity field will run slower than one lower. Due to the trains acceleration, a clock in the rear locomotive acts like a clock lower in a gravity field compared to a clock in the front of the train. So in the accelerated frame of the train, the rear locomotive runs slower than the front locomotive and starts to fall behind, this increases the distance between the locomotives and breaks the string. So in this case the string breaks in both frames.
I think this is fairly easy if you simply look at it from the point of view of the trains. Let's say they are simply coupled together, whether with a string or a normal railroad coupling does not matter. They really are just one single train, a number of cars hooked together. From the perspective of the train, not much happens, the cars maintain their original coupling distance. From the perspective of an observer at rest with respect to the tracks, the train shrinks, and that means that the coupling also shrinks, so the separation distance shrinks. Points that were touching remain touching (a shrinkage of a zero length remains a zero length).

12. Originally Posted by DrRocket
I think this is fairly easy if you simply look at it from the point of view of the trains. Let's say they are simply coupled together, whether with a string or a normal railroad coupling does not matter. They really are just one single train, a number of cars hooked together. From the perspective of the train, not much happens, the cars maintain their original coupling distance. From the perspective of an observer at rest with respect to the tracks, the train shrinks, and that means that the coupling also shrinks, so the separation distance shrinks. Points that were touching remain touching (a shrinkage of a zero length remains a zero length).
Right, but where does it shrink? Does the center of mass stay in one spot and the train shrink around it? If so, then in the case of two trains just touching, one would expect the trains to separate as they each shrink around their own center of masses.

I'm sure there's a nifty trick I'm not getting. My gut says the trains stay connected in all reference frames.

Maybe the shrinkage "origin" is the point furthest aft from the direction of motion?

13. Originally Posted by Numsgil
Originally Posted by DrRocket
I think this is fairly easy if you simply look at it from the point of view of the trains. Let's say they are simply coupled together, whether with a string or a normal railroad coupling does not matter. They really are just one single train, a number of cars hooked together. From the perspective of the train, not much happens, the cars maintain their original coupling distance. From the perspective of an observer at rest with respect to the tracks, the train shrinks, and that means that the coupling also shrinks, so the separation distance shrinks. Points that were touching remain touching (a shrinkage of a zero length remains a zero length).
Right, but where does it shrink? Does the center of mass stay in one spot and the train shrink around it? If so, then in the case of two trains just touching, one would expect the trains to separate as they each shrink around their own center of masses.

I'm sure there's a nifty trick I'm not getting. My gut says the trains stay connected in all reference frames.

Maybe the shrinkage "origin" is the point furthest aft from the direction of motion?
What you need to remember is that it is not a "shrinking" it is a change in the measurement of length along the direction of motion. You can equally visualize what is going on as a change in the ruler that you use to make the length measurement. The effect is precisely like looking at the train from a long way off, or through the wrong end of a telescope. If they were touching before they are touching afterwards.

14. Originally Posted by DrRocket
What you need to remember is that it is not a "shrinking" it is a change in the measurement of length along the direction of motion. You can equally visualize what is going on as a change in the ruler that you use to make the length measurement. The effect is precisely like looking at the train from a long way off, or through the wrong end of a telescope. If they were touching before they are touching afterwards.
So you say the train does not "really" shrink, but it just appears to shrink?
What about the the rails? If the trains remain in contact, it means the front wheel of the last train has approached the last wheel of the first train (either that or the rail, which is not moving, also (appears to) shrink). What causes the first wheel of the last train to go faster? Doesn't the reverse make more sense?

15. Originally Posted by Bender
Originally Posted by DrRocket
What you need to remember is that it is not a "shrinking" it is a change in the measurement of length along the direction of motion. You can equally visualize what is going on as a change in the ruler that you use to make the length measurement. The effect is precisely like looking at the train from a long way off, or through the wrong end of a telescope. If they were touching before they are touching afterwards.
So you say the train does not "really" shrink, but it just appears to shrink?
What about the the rails? If the trains remain in contact, it means the front wheel of the last train has approached the last wheel of the first train (either that or the rail, which is not moving, also (appears to) shrink). What causes the first wheel of the last train to go faster? Doesn't the reverse make more sense?
No, I did not say that. The length depends on the reference frame of the observer. In the reference frame of the train there is no change.

In the reference frame of the tracks the length of trains is less than in the reference frame of the train.

My deacription was an attempt to describe the kinematics of the shrinkage from the perspective of an observer on the tracks. The length contraction is real, though dependent on the reference frame. But points that touch continue to touch since a contraction of zero length is still of zero length.

The shrinkage is simply a contraction of the length scale in the direction of motion, which gives you precisely the same effect that I tried to describe optically although only in one direction.

Try thinking about this (warning this may hurt your head). The shrinkage is only in one direction. So from the perspective of the an observer on the tracks the wheels are not round but rather are oval, with the major axis vertical and the minor axis horizontal -- all the time. Moreover, due to effects related to the speed of light they will appear optically to be round, but they are not.

16. Suppose you had two trains, set up so that, ignoring length contractions, the two will cross a finish line at exactly the same moment to an observer on the tracks. The first one will be traveling at relativistic speeds, but the second one is traveling at 100kips (obviously it gets a substantial head start :P)

Will length contraction cause them to still appear to cross at the same time? What if they're set up so that their geometric centers cross at the same time? What if they're set up so their cabooses cross at the same time? Is their some magic point where the length contraction doesn't matter in this respect?

17. There's one major flaw in your question. Part of relativity states that "at exactly the same moment" doesn't make sense globally. What appears to be exactly the same moment for someone in one train, will be different for someone in the other. (The slower train is moving slow enough w.r.t. the ground that any relativistic differences can be ignored.)

18. I know, that's why I said "with respect to the tracks".

19. Ah. Sorry. I think I overlooked that.

20. Originally Posted by Numsgil
Suppose you had two trains, set up so that, ignoring length contractions, the two will cross a finish line at exactly the same moment to an observer on the tracks. The first one will be traveling at relativistic speeds, but the second one is traveling at 100kips (obviously it gets a substantial head start :P)

Will length contraction cause them to still appear to cross at the same time?
If you set them up to cross at the same time, then by definition they cross at the same time. "Ignoring length contractions" is not a meaningful statement. You need to make that statement precise, using mathematics.

What if they're set up so that their geometric centers cross at the same time?
If the geometric centers cross at the same time, then without working out the mathematics in detail you don't know if the fronts cross at the same time, and the answer is likely to be different depending whether the observer is on the tracks, on the fast train or on the slow train. This has to do with the relativity of simultaneity.

What if they're set up so their cabooses cross at the same time?
See the previous answer and the question of relativity of simulaneity.

Is their some magic point where the length contraction doesn't matter in this respect?
Again you need to make your question precise.

You probably need to take the time to read a good book on special relativity and see some of the problems with relativity of simultaneity discussed in some detail, with the relevant mathematics. Introduction to Special Relativity by Wolfgang Rindler is a good book for that purpose.

21. Originally Posted by DrRocket
No, I did not say that. The length depends on the reference frame of the observer. In the reference frame of the train there is no change.

In the reference frame of the tracks the length of trains is less than in the reference frame of the train.

My deacription was an attempt to describe the kinematics of the shrinkage from the perspective of an observer on the tracks. The length contraction is real, though dependent on the reference frame. But points that touch continue to touch since a contraction of zero length is still of zero length.
Sure, if there is a connection to keep the length zero, but there is no connection.
To an external viewer, the speed of the rear of both trains is the same at all times, how can the distance between them have changed?

How do you address Janus' explanation from the point of view of one of the trains? The other train is higher or lower in the inertial field, causing the clocks to go out of sync and the first train goes faster during acceleration. If the first train goes faster, there is a gap.

22. Originally Posted by DrRocket
Originally Posted by Numsgil
Suppose you had two trains, set up so that, ignoring length contractions, the two will cross a finish line at exactly the same moment to an observer on the tracks. The first one will be traveling at relativistic speeds, but the second one is traveling at 100kips (obviously it gets a substantial head start :P)

Will length contraction cause them to still appear to cross at the same time?
If you set them up to cross at the same time, then by definition they cross at the same time. "Ignoring length contractions" is not a meaningful statement. You need to make that statement precise, using mathematics.

What if they're set up so that their geometric centers cross at the same time?
If the geometric centers cross at the same time, then without working out the mathematics in detail you don't know if the fronts cross at the same time, and the answer is likely to be different depending whether the observer is on the tracks, on the fast train or on the slow train. This has to do with the relativity of simultaneity.

What if they're set up so their cabooses cross at the same time?
See the previous answer and the question of relativity of simulaneity.

Is their some magic point where the length contraction doesn't matter in this respect?
Again you need to make your question precise.

You probably need to take the time to read a good book on special relativity and see some of the problems with relativity of simultaneity discussed in some detail, with the relevant mathematics. Introduction to Special Relativity by Wolfgang Rindler is a good book for that purpose.
Again, I said "at the exact same moment for an observer on the tracks". Where is the ambiguity there? You're the second person to stop reading as soon as you saw the words "at the exact same moment" and balk. I know enough to put that qualifier in there.

Look, length contraction means that the recorded distance of the trains snow plow to caboose will shrink as the train approaches relativistic speeds for an observer on the tracks. Suppose I set up an experiment where, according to Newtonian mechanics, two trains will cross the finish line at the exact same moment for an observer on the tracks. That is, I just used the SUVAT equations (google if you aren't familiar with the term). I didn't take in to account length contraction. Will length contraction change the results? Why?

Or here's another experiment: suppose I have a perfectly spherical cloud of particles in space above the Moon, and they uniformly accelerate through some means to near the speed of light. Ignoring obvious issues with gravity fields and the like, will length contraction make the cloud an ellipsoid to an observer on the Moon? Or does it just contract the length of the individual particles and the cloud itself remains spherical?

23. The equivalent gravity field:
Both frames experience equal acceleration, thus equal fields, thus equal time dilation. Both frames agree that the forward frame clock is ahead of the rearward clock. This is only because of reassigning events (by calculation) in each frame. You can't have equal dilation and still gain time on one clock. Postulate 2: the physics in one frame is the same as in the other.
In an equivalent gravity field, both objects are at rest on the surface. They don't have to touch, but could have an arbitrary separation. (refer: S.P.Boughn-accelerated twins-1989)

Length contraction:
Given two observers moving in the x direction at different speeds. An object of length d passes both in the x direction.
The slower observer will measure a length d' and the faster d''. If the contraction is real, say d', then this will effect d''. The observations are independent of each other, therefore there is no contraction.

24. Originally Posted by Numsgil
Originally Posted by DrRocket
Originally Posted by Numsgil
Suppose you had two trains, set up so that, ignoring length contractions, the two will cross a finish line at exactly the same moment to an observer on the tracks. The first one will be traveling at relativistic speeds, but the second one is traveling at 100kips (obviously it gets a substantial head start :P)

Will length contraction cause them to still appear to cross at the same time?
If you set them up to cross at the same time, then by definition they cross at the same time. "Ignoring length contractions" is not a meaningful statement. You need to make that statement precise, using mathematics.

What if they're set up so that their geometric centers cross at the same time?
If the geometric centers cross at the same time, then without working out the mathematics in detail you don't know if the fronts cross at the same time, and the answer is likely to be different depending whether the observer is on the tracks, on the fast train or on the slow train. This has to do with the relativity of simultaneity.

What if they're set up so their cabooses cross at the same time?
See the previous answer and the question of relativity of simulaneity.

Is their some magic point where the length contraction doesn't matter in this respect?
Again you need to make your question precise.

You probably need to take the time to read a good book on special relativity and see some of the problems with relativity of simultaneity discussed in some detail, with the relevant mathematics. Introduction to Special Relativity by Wolfgang Rindler is a good book for that purpose.

Again, I said "at the exact same moment for an observer on the tracks". Where is the ambiguity there? You're the second person to stop reading as soon as you saw the words "at the exact same moment" and balk. I know enough to put that qualifier in there.

Look, length contraction means that the recorded distance of the trains snow plow to caboose will shrink as the train approaches relativistic speeds for an observer on the tracks. Suppose I set up an experiment where, according to Newtonian mechanics, two trains will cross the finish line at the exact same moment for an observer on the tracks. That is, I just used the SUVAT equations (google if you aren't familiar with the term). I didn't take in to account length contraction. Will length contraction change the results? Why?

Or here's another experiment: suppose I have a perfectly spherical cloud of particles in space above the Moon, and they uniformly accelerate through some means to near the speed of light. Ignoring obvious issues with gravity fields and the like, will length contraction make the cloud an ellipsoid to an observer on the Moon? Or does it just contract the length of the individual particles and the cloud itself remains spherical?
If you are trying to sound like a snotty little kid you are succeeding admirably. Not only did I read your entire post, I apparently understood what you actually said, and failed to say, better than did you.

As I stated, you statement ot setting up the problem by ignoring length contraction and having the trains cross at the same time from the perspective of an observer on the tracks is not well defined. When you invoke simplified Newtonian mechanics in the form of SUVAT equation, you have then defined the problem. But then too the answer is quite clear. The trains will cross at the same time from the perspective of an observer in reference frame of the rails -- because you set up the problem to have just that precise outcome.

If the trains are traveling such that the fronts cross the finish line simultaneously in the frame of the observer on the tracks then the geometric center of the faster train will cross first, since it is shorter and traveling faster. Ditto for the caboose.

If you wan to "google" something try googling special relativity and go read about the Lorentz transformation. Or better yet, go read Rindler's book on special relativity. You might just learn something.

25. Originally Posted by DrRocket
If you are trying to sound like a snotty little kid you are succeeding admirably. Not only did I read your entire post, I apparently understood what you actually said, and failed to say, better than did you.
I am not trying to be snotty, but I get intemperate when I feel as if someone hasn't read a post of mine carefully. Doubly so if I feel it happens twice. If you did indeed see that I had put "for an observer on the tracks" in my post, then I apologize. But I also fail to see how there's any ambiguity as to which inertial frame I'm measuring simultaneity from (which is presumably why your post contained an explanation on how simultaneity is dependent on inertial frame).

As I stated, you statement ot setting up the problem by ignoring length contraction and having the trains cross at the same time from the perspective of an observer on the tracks is not well defined. When you invoke simplified Newtonian mechanics in the form of SUVAT equation, you have then defined the problem. But then too the answer is quite clear. The trains will cross at the same time from the perspective of an observer in reference frame of the rails -- because you set up the problem to have just that precise outcome.
So you're saying that even with relativistic speeds, SUVAT can accurately describe which train wins for an observer on the tracks? That the phenomena of length contraction doesn't modify any of the SUVAT results? If so, I don't see how that's possible.

If the trains are traveling such that the fronts cross the finish line simultaneously in the frame of the observer on the tracks then the geometric center of the faster train will cross first, since it is shorter and traveling faster. Ditto for the caboose.
That would imply that length contraction doesn't change the expected position of a point forward most on the fast moving train, right? Which would make that forward most point the "origin" for the contraction, right? (Meaning that before and after the Lorentz transformation, it's position will remain unchanged).

But now take the case of a point aft most on the train, near the caboose. It will appear, to an observer on the tracks, to have traveled a greater distance than the nose of the train when the nose of the train crosses the finish line, right? (Otherwise there wouldn't be any length contraction at all). Which means the SUVAT equations can't hold for any points aft of the very nose of the train. You need to factor in the length contraction to get an accurate result (which clearly the regular SUVAT equations do not do).

So now put a second train with its nose right on the tail of the first's caboose. And have it, unconnected, accelerate at the same speed as the train ahead of it such that at any infinitely small instant the two are in the same inertial frame. Its nose will have appeared to have contracted towards the nose of the first train (if not, then the two trains will disconnect, which I believe you said wasn't the case). But you should be able to evaluate that second train in isolation to the first, in which case it'd be an identical case to the first train just with a slightly larger distance. Which would mean its nose wouldn't contract towards the nose of the first train, but be perfectly predictable using the SUVAT equations.

Which means the path of a point connected to the first train can't be calculated using SUVAT, but the path of that exact same point, from a second train, can be calculated using SUVAT, which leads to a contradiction, which should not be possible. This isn't a complicated paradox, so someone has already posed it and someone else has already solved it, or length contraction wouldn't be an accepted scientific theory. Which means somewhere I made an invalid assumption in my train (pun intended) of logic, and I don't understand what that is. Please tell me where, I honestly don't see it. (Note, in any cases of ambiguity for frame of reference, assume the tracks).

I can draw some pictures if this last little bit was confusing.

If you wan to "google" something try googling special relativity and go read about the Lorentz transformation. Or better yet, go read Rindler's book on special relativity. You might just learn something.
I understand the math involved, that's not the issue. And I understand the time dilation and added mass effects. It's just the exact nature of the length contraction which has me curious. A few days ago I would tell you I understood it, but the more I think about it, the more I realize I couldn't program it in to a simulation because its exact mapping is still a bit of a mystery.

Or here's an even simpler example: take a singularity and accelerate it to .9c relative to yourself. Are there any differences in its position at some given time from what SUVAT would tell us? What if it's a long rod traveling along its long axis? What if its two singularities the same distance from each other as the long rod is long, but entirely unconnected?

While reading Rindler's book might be a fine way of my learning the answer, so to is asking pointed questions to learned people. This method also (hopefully) takes less time.

26. Originally Posted by phyti
The equivalent gravity field:
Both frames experience equal acceleration, thus equal fields, thus equal time dilation. Both frames agree that the forward frame clock is ahead of the rearward clock. This is only because of reassigning events (by calculation) in each frame. You can't have equal dilation and still gain time on one clock. Postulate 2: the physics in one frame is the same as in the other.
As I understood, the time dilation is equal for both frames with respect to an observer in rest. To an observer on one of the trains, the time dilation would be different, as long as the trains would be accelerating. The physics in both frames is the same, as one perceives the other to be slower, and the other will perceive the first to go faster.
Originally Posted by phyti
Length contraction:
Given two observers moving in the x direction at different speeds. An object of length d passes both in the x direction.
The slower observer will measure a length d' and the faster d''. If the contraction is real, say d', then this will effect d''. The observations are independent of each other, therefore there is no contraction.
Interesting. But isn't this a bit like the twin paradox: suppose they all started at rest, the observers and the object all have a different history of accelerations. I thought if you do the calculations, it should all fit.

27. Originally Posted by Bender
Originally Posted by DrRocket
No, I did not say that. The length depends on the reference frame of the observer. In the reference frame of the train there is no change.

In the reference frame of the tracks the length of trains is less than in the reference frame of the train.

My deacription was an attempt to describe the kinematics of the shrinkage from the perspective of an observer on the tracks. The length contraction is real, though dependent on the reference frame. But points that touch continue to touch since a contraction of zero length is still of zero length.
Sure, if there is a connection to keep the length zero, but there is no connection.
To an external viewer, the speed of the rear of both trains is the same at all times, how can the distance between them have changed?

How do you address Janus' explanation from the point of view of one of the trains? The other train is higher or lower in the inertial field, causing the clocks to go out of sync and the first train goes faster during acceleration. If the first train goes faster, there is a gap.
The physical effect of the coupling is irrelevant. I was simply using that to help make the kinematics easier to see.

When you start talking about effects of the gravitational field you have left the realm of special relativity.

28. Originally Posted by Numsgil
Originally Posted by DrRocket
If you are trying to sound like a snotty little kid you are succeeding admirably. Not only did I read your entire post, I apparently understood what you actually said, and failed to say, better than did you.
I am not trying to be snotty, but I get intemperate when I feel as if someone hasn't read a post of mine carefully. Doubly so if I feel it happens twice. If you did indeed see that I had put "for an observer on the tracks" in my post, then I apologize. But I also fail to see how there's any ambiguity as to which inertial frame I'm measuring simultaneity from (which is presumably why your post contained an explanation on how simultaneity is dependent on inertial frame).

As I stated, you statement ot setting up the problem by ignoring length contraction and having the trains cross at the same time from the perspective of an observer on the tracks is not well defined. When you invoke simplified Newtonian mechanics in the form of SUVAT equation, you have then defined the problem. But then too the answer is quite clear. The trains will cross at the same time from the perspective of an observer in reference frame of the rails -- because you set up the problem to have just that precise outcome.
So you're saying that even with relativistic speeds, SUVAT can accurately describe which train wins for an observer on the tracks? That the phenomena of length contraction doesn't modify any of the SUVAT results? If so, I don't see how that's possible.

If the trains are traveling such that the fronts cross the finish line simultaneously in the frame of the observer on the tracks then the geometric center of the faster train will cross first, since it is shorter and traveling faster. Ditto for the caboose.
That would imply that length contraction doesn't change the expected position of a point forward most on the fast moving train, right? Which would make that forward most point the "origin" for the contraction, right? (Meaning that before and after the Lorentz transformation, it's position will remain unchanged).

But now take the case of a point aft most on the train, near the caboose. It will appear, to an observer on the tracks, to have traveled a greater distance than the nose of the train when the nose of the train crosses the finish line, right? (Otherwise there wouldn't be any length contraction at all). Which means the SUVAT equations can't hold for any points aft of the very nose of the train. You need to factor in the length contraction to get an accurate result (which clearly the regular SUVAT equations do not do).

So now put a second train with its nose right on the tail of the first's caboose. And have it, unconnected, accelerate at the same speed as the train ahead of it such that at any infinitely small instant the two are in the same inertial frame. Its nose will have appeared to have contracted towards the nose of the first train (if not, then the two trains will disconnect, which I believe you said wasn't the case). But you should be able to evaluate that second train in isolation to the first, in which case it'd be an identical case to the first train just with a slightly larger distance. Which would mean its nose wouldn't contract towards the nose of the first train, but be perfectly predictable using the SUVAT equations.

Which means the path of a point connected to the first train can't be calculated using SUVAT, but the path of that exact same point, from a second train, can be calculated using SUVAT, which leads to a contradiction, which should not be possible. This isn't a complicated paradox, so someone has already posed it and someone else has already solved it, or length contraction wouldn't be an accepted scientific theory. Which means somewhere I made an invalid assumption in my train (pun intended) of logic, and I don't understand what that is. Please tell me where, I honestly don't see it. (Note, in any cases of ambiguity for frame of reference, assume the tracks).

I can draw some pictures if this last little bit was confusing.

If you wan to "google" something try googling special relativity and go read about the Lorentz transformation. Or better yet, go read Rindler's book on special relativity. You might just learn something.
I understand the math involved, that's not the issue. And I understand the time dilation and added mass effects. It's just the exact nature of the length contraction which has me curious. A few days ago I would tell you I understood it, but the more I think about it, the more I realize I couldn't program it in to a simulation because its exact mapping is still a bit of a mystery.

Or here's an even simpler example: take a singularity and accelerate it to .9c relative to yourself. Are there any differences in its position at some given time from what SUVAT would tell us? What if it's a long rod traveling along its long axis? What if its two singularities the same distance from each other as the long rod is long, but entirely unconnected?

While reading Rindler's book might be a fine way of my learning the answer, so to is asking pointed questions to learned people. This method also (hopefully) takes less time.
The problem is that there are misconceptions inherent in your questions. That makes for very difficult communications.

You are right. This is not all that complicated. It is not a paradox. If you formulate the problem correctly there is no problem at all. There are, in fact, no paradoxes in special relativity. Many are the result of failure to recognize that two different observers, in reference frames in relative motion, do not see simulaneity in the same way. Others are due to failure to restrict attention to inertial reference frames, as is the case with the "twin paradox". Here you set up the problem in a particular reference frame, that of the rails, and the problem is quite simple in that reference frame.

You started by stating that you were setting up the problem so that using Newtonian mechanics the two trains (at the front) would reach the finish line simultaneously. We are also looking at this in terms of special relativity. So to start, one ignores acceleration and basically uses d = rt to set the speed of the two trains. That relation continues to hold, whether you are looking at special relativity or Newtonian mechanics. In the reference frame of the track-bound observer it holds, the front of each train is simply a point, the reference point used to set up the problem, and the front of each train crosses the finish line using special relativity just as it would using Newtonian mechanics. You could set up the problem exactly the same way using any choice of points, one each train that you wish. Only those points are in question and the effects of length contraction and time dilation are irrelevant, so long as you stay in the reference frame of the tracks.

What is a singularity (physically) and how would you accelerate it ? ANS: There is no such thing, physically, as a singularity. It represents a breakdown in the theory and is a mathematical construct. You can't touch it, you can't weigh it, and you can't accelerate it.

29. Originally Posted by DrRocket
The problem is that there are misconceptions inherent in your questions. That makes for very difficult communications.
Please point out any misconceptions I have. Everything should be pretty clean and easy to follow, so any misconceptions should be glaringly obvious.

What is a singularity (physically) and how would you accelerate it ? ANS: There is no such thing, physically, as a singularity. It represents a breakdown in the theory and is a mathematical construct. You can't touch it, you can't weigh it, and you can't accelerate it.
A singularity is an object which has position, velocity, acceleration, orientation, angular velocity, angular acceleration, moment of inertia, and mass, but no volume. It is effectively a 0 dimensional point. Mathematically it's fairly easy to understand, since whenever you use the SUVAT equations you are essentially simplifying a real world example down to a case using just points anyway.

Or, if you just can't imagine such a thing, make each point a sphere with diameter of some few multiples of the plank distance.

As to how I'm accelerating it, it doesn't matter except that it's done such that special relativity can be used (ie: no warping spacetime).

Now, please read the following carefully, since it's basically a word problem and the choice of words is very deliberate.

Now that we have our concept of what a singularity is exactly, let's say we accelerate it "from rest" to .9c at a rate of 3000 km/s^2 (or roughly .001 c/s^2). Let where it started from be the origin for a Cartesian coordinate system, and let the units of distance be constructed of Lightseconds in the same "rest" inertial frame (that is, roughly 300,000 km). Using SUVAT:

t * .001 = .9 => t = 900 seconds
s = .5 * .001 * 900^2 => s = 405 Light seconds

So it takes 900 seconds to accelerate to .9c and travels a distance of 405 Lightseconds during that time.

Now apply the Lorentz length transformation (length of a singularity is 0). Does the result change at all? (the answer should be a pretty trivial: no).

Now let's replace that singularity with a long rod. Maybe 3000 km long from a "rest" frame, or .01 Lightseconds. Let the forward most point of the rod be the origin. Now it undergoes the same acceleration up to .9c. Using SUVAT, the point most forward on the rod takes 900 seconds to reach this speed, and covers a distance of 405 Lightseconds. And using SUVAT, the point most aft on the rod takes 900 seconds to reach this speed, and covers a distance of 405 Lightseconds. At the end of the journey, just as at the start, the rod is .01 Lightseconds long.

But wait, we forgot to take into account the Lorentz transformation. Using the factor, we get a new length, from a "rest" frame of reference, of .00435 Light seconds, or a little under half as long. Since length is defined as the distance between our forward and aft points, one of those points (or both) is not going to be where SUVAT predicted. And since distance traveled is defined as the difference of where the point ended (from a "rest" frame of reference) and where it started (from its "rest" frame of reference), that means that at least one point on the rod, somewhere, appeared to travel a distance greater than or less than 405 light seconds.

Now, my assumption here is that you said that the forward most point is 100% predictable using SUVAT. In which case the aft most point will appear to have traveled a distance of 405.00565 Light seconds (405 + .01 - .00435). Is this correct? Double check my math if you like.

Now replace that rod with two singularities, positioned .01 Lightseconds apart from each other. That is, the exact same setup, except our forward and aft points are singularities and there is no rod. We should get the same result, right? The forward point will travel 405 Lightseconds, and the aft point will travel 405.00565 Lightseconds.

But if we go back to our experiment with a single singularity, we found that the Lorentz length contraction does not affect (effect?) how far a singularity appears to travel. Which means if we take our aft most point in isolation, it will travel 405.00000 Lightseconds.

405.00000 Lightseconds does not equal 405.00565 Lightseconds. There's a contradiction, or paradox. Somewhere there was a failure of reasoning. Please tell me where. Or if you can't figure out where, that's fine, too. You alone are not the sole defender of special relativity, and it's okay to say "I don't know".

30. Originally Posted by Numsgil

A singularity is an object which has position, velocity, acceleration, orientation, angular velocity, angular acceleration, moment of inertia, and mass, but no volume. It is effectively a 0 dimensional point. Mathematically it's fairly easy to understand, since whenever you use the SUVAT equations you are essentially simplifying a real world example down to a case using just points anyway.

Or, if you just can't imagine such a thing, make each point a sphere with diameter of some few multiples of the plank distance.
What you have described is not what mathematicians or physicists call a singularity. It is basically a point mass. You have endowed with the characteristics of a spinning mass as well, but those have no role in the problem that you subsequently describe so you might as well just consider it a point mass. The notions of momento of inertia, angular velocity, angular acceleration usually require stipulation of a point (no necessarily within the body) to which they are referenced. The notion of orientation of a point has no meaning.

Now, please read the following carefully, since it's basically a word problem and the choice of words is very deliberate.
Fine. I will expect the precision of language that I expect of mathematicians.

Now that we have our concept of what a singularity is exactly, let's say we accelerate it "from rest" to .9c at a rate of 3000 km/s^2 (or roughly .001 c/s^2). Let where it started from be the origin for a Cartesian coordinate system, and let the units of distance be constructed of Lightseconds in the same "rest" inertial frame (that is, roughly 300,000 km). Using SUVAT:

t * .001 = .9 => t = 900 seconds
s = .5 * .001 * 900^2 => s = 405 Light seconds

So it takes 900 seconds to accelerate to .9c and travels a distance of 405 Lightseconds during that time.
You are reasonabley OK here, except as noted that you have endowed your "singularity" with unnecessary characteristics, and some of those unnecessary characteristics cannot be sensibly assigned to a point.

Also your units of acceleration are incorrect, but you have made compensating mistakes. Since c is itself a unit of speed the correct description for the uniform acceleration that you have imposed is .001 c/s.

Now apply the Lorentz length transformation (length of a singularity is 0). Does the result change at all? (the answer should be a pretty trivial: no).
Your comment shows a misunderstanding of the Lorentz transformation. The Lorentz transformation provides a means to describe the measurement of position and time in one inertial reference frame to those of another inertial reference frame. It does not apply in a single reference frame.

But what it does tell you is how to relate position and time in the reference frame of the "stationary observer" to position and time in the reference frame of the moving point.

So after 900 seconds in the frame of the stationary observer the pont is 405 light seconds from the initial point. But in the time frame of the point something less than 900 seconds have elapsed and the distance traveled is less than that.

Uniformly accelerated motion in special relativity transforms to what is called hyperbolic motion. You can read about this in section 14 of Rindler's book where it is developed in detail. In a manner in which you ask if I can imagine something like your notion of a singularity, I add here "if you can read and comprehend the mathematics of special relativity." Enjoy.

31. Originally Posted by DrRocket
What you have described is not what mathematicians or physicists call a singularity. It is basically a point mass. You have endowed with the characteristics of a spinning mass as well, but those have no role in the problem that you subsequently describe so you might as well just consider it a point mass. The notions of momento of inertia, angular velocity, angular acceleration usually require stipulation of a point (no necessarily within the body) to which they are referenced. The notion of orientation of a point has no meaning.
Yeah, the angular components aren't really important here. What I was attempting to describe was what I use when I program physics to describe a body. It has a shape (circle, square, whatever), and a point mass (good term). The shape component gets rotated by the angular components of the point mass, so from a programming perspective the two are distinct ideas.

So yeah, ignore the angular components since it doesn't really matter.

Also your units of acceleration are incorrect, but you have made compensating mistakes. Since c is itself a unit of speed the correct description for the uniform acceleration that you have imposed is .001 c/s.
Ah, right. .001 c/s. Good catch

Uniformly accelerated motion in special relativity transforms to what is called hyperbolic motion. You can read about this in section 14 of Rindler's book where it is developed in detail. In a manner in which you ask if I can imagine something like your notion of a singularity, I add here "if you can read and comprehend the mathematics of special relativity." Enjoy.
Ah, I see. Some googling tells me you can't use SUVAT with relativistic speeds, you need to use the hyperbolic equations of motion. Specifically acceleration is constant only from the frame of reference of the accelerating object. For it to be constant to a stationary observer, it would in fact be non constant acceleration. Which I kind of guessed at but still don't know if it matters or not.

This is the only source I can find online for what the hyperbolic equivalents of SUVAT are. But I suspect there might be some typos involved, but maybe not. Any other online references anyone can give?

Something tells me the trains formulated at the top of this thread do in fact separate. And it's only the rigidity of a material which allows it to remain whole.

32. Okay, I think I understand now.

The two trains just touching in the original post for this thread will separate. The second train has to accelerate harder to keep the distance 0. This was pointed out earlier, I just wasn't understanding the math involved, which I do now (more or less). This means I was wrong earlier. This weird phenomena is known as Bell's spaceship paradox. Janus pointed this all out earlier, of course.

Likewise in the example I set up, the single point mass can be predicted using SUVAT (so long as you understand that it represents a non constant acceleration, even though it looks constant from a stationary perspective).

For the rod, it will contract about its center of mass according to the Lorentz transformation. So the aft point will appear to have traveled further than 405 Ls, and the forward point a little less. This is because the rod has Born rigidity, so its structural integrity is pulling the points together.

The two point masses, on the other hand, will maintain their .01 Ls distance from each other from the point of view of a stationary observer because they do not have Born rigidity. To the points themselves, it will seem as if they drift apart. And by quite a bit.

Which means that if you accelerate hard enough you can rip molecules apart, even if each molecule is being "accelerated uniformly" from your "at rest" frame of reference.

Or at least that is my present understanding. It's a little counter intuitive, but at least internally consistent.

33. Originally Posted by Numsgil
Okay, I think I understand now.

The two trains just touching in the original post for this thread will separate. The second train has to accelerate harder to keep the distance 0. This was pointed out earlier, I just wasn't understanding the math involved, which I do now (more or less). This means I was wrong earlier. This weird phenomena is known as Bell's spaceship paradox. Janus pointed this all out earlier, of course.

Likewise in the example I set up, the single point mass can be predicted using SUVAT (so long as you understand that it represents a non constant acceleration, even though it looks constant from a stationary perspective).

For the rod, it will contract about its center of mass according to the Lorentz transformation. So the aft point will appear to have traveled further than 405 Ls, and the forward point a little less. This is because the rod has Born rigidity, so its structural integrity is pulling the points together.

The two point masses, on the other hand, will maintain their .01 Ls distance from each other from the point of view of a stationary observer because they do not have Born rigidity. To the points themselves, it will seem as if they drift apart. And by quite a bit.

Which means that if you accelerate hard enough you can rip molecules apart, even if each molecule is being "accelerated uniformly" from your "at rest" frame of reference.

Or at least that is my present understanding. It's a little counter intuitive, but at least internally consistent.
This problem has little to do with Born ridigity, but a a lot to do with Bell's spaceship paradox.
http://math.ucr.edu/home/baez/physic...ip_puzzle.html

34. Right, that's why I linked Bell's spaceship paradox in my post.

35. i don't get it. what do you mean by contract?

I understand that if the physical object, train or not, moved in a certain direction its length would stay the same. I suppose that if you were parallel to the train, facing it, and it started to move, the back of the train would appear larger than the front, but they would both decrease in size simultaneously as they moved off into the distance. But, let's say you were stationed above the train flying at the same speed as the train, looking down at it, then the train would appear of greater size int the middle, but the perspective would remain same. OK, so what the point...I'm missing something

Hmm...so the train moves off into the distance, but the light at the front of the train takes longer to get to you then the light at the back, so the train is actually contracting slower, in regards to the vantage point, than it appears to be.

All objects are moving through time at same pace ( forget spacial contractions do to gravity in this hypothetical ) - this is a given. Ok, but if we were to look at the clock from outside the train, then the clock on the back would move faster than the one in the front because it takes light takes time to get to our eyes, and such and such....

seems simple. this isn't the cause of gravity right?

36. Originally Posted by Bender
Sure, if there is a connection to keep the length zero, but there is no connection.
To an external viewer, the speed of the rear of both trains is the same at all times, how can the distance between them have changed?

How do you address Janus' explanation from the point of view of one of the trains? The other train is higher or lower in the inertial field, causing the clocks to go out of sync and the first train goes faster during acceleration. If the first train goes faster, there is a gap.
If both were stacked in a gravity field, the upper would experience less acceleration, but then its not an equivalent field!
Thus an arbitrary separation to show the trains/ships are accelerated
independently of each other, with their own local equivalent field.

Consider both frames rotated 90 deg and accelerating perpendicular to
the original direction.
The clocks run at the same rate and stay in synch.
This would position each at the same level in a gravity field.
Now remove the 2nd ship. The physics for the ship 1 should not change
due to direction. Replacing ship 2 in the tandem positions should not
alter the physics for ship 1, i.e. its clock runs the same. Repeat this
procedure for ship 2. Any difference in clocks is observer dependent,
i.e. light transit time, calculations, altered perception.

The acceleration of the objects considered does not affect distant objects.

Bell problem:
A string connecting them does not qualify all three as a rigid object.
Even if the string (or wire) was only attached to the front ship, and
both ships at the same speed, and length contraction takes place, the
ship crews cannot detect this, and will see no separation. If they
don't see it neither does anyone else. In reality the acceleration
would not be 'ideal' and the string would break due to tension.
The argument in this problem uses the (calculated) skewed axis of
simultaneity, which reassigns the local time of events for each
observer. This in turn increases distances in the direction of motion,
and erroneously concludes the ships are separating.

37. Originally Posted by phyti
Originally Posted by Bender
Sure, if there is a connection to keep the length zero, but there is no connection.
To an external viewer, the speed of the rear of both trains is the same at all times, how can the distance between them have changed?

How do you address Janus' explanation from the point of view of one of the trains? The other train is higher or lower in the inertial field, causing the clocks to go out of sync and the first train goes faster during acceleration. If the first train goes faster, there is a gap.
If both were stacked in a gravity field, the upper would experience less acceleration, but then its not an equivalent field!
Thus an arbitrary separation to show the trains/ships are accelerated
independently of each other, with their own local equivalent field.

Consider both frames rotated 90 deg and accelerating perpendicular to
the original direction.
The clocks run at the same rate and stay in synch.
This would position each at the same level in a gravity field.

Now remove the 2nd ship. The physics for the ship 1 should not change
due to direction. Replacing ship 2 in the tandem positions should not
alter the physics for ship 1, i.e. its clock runs the same. Repeat this
procedure for ship 2. Any difference in clocks is observer dependent,
i.e. light transit time, calculations, altered perception.

The acceleration of the objects considered does not affect distant objects.

Gravitational time dilation and the equivalent due to acceleration do not depend on the difference in acceleration experienced but on the difference in potential. Put two clocks in a uniform gravity field (one which does not change in strength with height) and the higher clock will run faster.
Put clocks in the nose and tail of an accelerating rocket, and the clock in the nose will run faster. Stop the acceleration and bring the clocks together at the middle of the ship and the clocks will show different times.

Put a clock in a space ship following another ship and accelerate both, and the clock in trailing ship will run slower than the one in the leading ship.

38. Gravitational time dilation and the equivalent due to acceleration do not depend on the difference in acceleration experienced but on the difference in potential. Put two clocks in a uniform gravity field (one which does not change in strength with height) and the higher clock will run faster.
Isn't it the difference in the g-field that causes clocks to run slower at the surface than at a higher altitude?

Put clocks in the nose and tail of an accelerating rocket, and the clock in the nose will run faster. Stop the acceleration and bring the clocks together at the middle of the ship and the clocks will show different times.
Why if the front and the back experience the same acceleration?

Put a clock in a space ship following another ship and accelerate both, and the clock in trailing ship will run slower than the one in the leading ship.
What if the ships are 1 lyr apart?

39. Originally Posted by phyti
Gravitational time dilation and the equivalent due to acceleration do not depend on the difference in acceleration experienced but on the difference in potential. Put two clocks in a uniform gravity field (one which does not change in strength with height) and the higher clock will run faster.
Isn't it the difference in the g-field that causes clocks to run slower at the surface than at a higher altitude?
No, it is the difference in potential. The potential is related to the amount of work it would take to lift the lower clock to the height of the upper clock. Even in a uniform field, it take work to move the lower clock to a higher point.

Imagine you have two clocks in a uniform gravity field and with a height difference of 1 micro-light second. The two clocks consist of an atom that vibrates at a fixed rate and as a result emits a light wave, one wave per vibration. We can keep track of time by counting the number of waves emitted (x number of waves per second)

Now assume that we are watching the lower clock from the upper clock. As the light waves climb against the gravity field, they lose energy and red-shift to a lower frequency. As a result, the upper clock receives few light waves per second from the lower clock than it does from itself. Since one light wave = one vibration, the lower atom is also seen as vibrating at as slower rate. IOW, as far as the upper clock is concerned, the lower clock is running slow.

Looking from the lower clock to the upper clock, the light is blue-shifted to a higher frequency, so as far as the lower clock is concerned, the upper clock is running fast.

This difference in clock rates is accumulative, the longer we keep the clocks in at their present height, the greater the time difference in their readings and can far exceed the 1 microsecond difference due to their distance apart. (If we bring the clocks back together after some time, the change in the clock's readings due to the reduction in distance can only be 1 microsecond)

Thus if we keep the clocks at different heights for some period, and then bring them together, there will be a difference in the clocks, even though there was no difference in the force of gravity felt by the clocks.

Put clocks in the nose and tail of an accelerating rocket, and the clock in the nose will run faster. Stop the acceleration and bring the clocks together at the middle of the ship and the clocks will show different times.
Why if the front and the back experience the same acceleration?
Due to the Equivalence Principle, separating two clocks in this way is the same as having them at different heights in a uniform gravity field as per the last example.

Put a clock in a space ship following another ship and accelerate both, and the clock in trailing ship will run slower than the one in the leading ship.
What if the ships are 1 lyr apart?
Then the clock in the trailing ship will run that much slower. The difference in time rate is related to the distance of separation along the line of acceleration.

40. Originally Posted by Bender
Here is an interesting thought experiment: two identical trains are positioned on a rail in such a way that the front of the second touches the back of the first.
The trains are identical in every way, with identical computers with identical trajectories programmed in them. They have perfect clocks that are perfectly synchronised between the trains.
The trains start moving, and at every instant, they move at exactly the same speed.

Question: do the trains still touch?
Yes. Of course they still touch. It is not just the length of things that contracts but the distance between them also contracts. Dr. Rocket has this one correct.

"At the same instant" is well defined for everyone on the two trains because they are all in the same intertial frame.

Originally Posted by Janus
The clocks will not remain synchronized between trains. Each train will mantain that all its clocks remained in sync, but that the clocks on the other train go out of sync. Thus according to each train, while every part of itself started moving at the same instant, every part of the other train did not. As a result, each train determines that the rear of the other train starts moving before the front does and the train contracts along its length.
This doesn't make sense Janus. There is no relative velocity between the two trains so they are both in the same inertial frame and thus those on the trains measure time and space in the same way. So all the clocks on the two trains remain synchronized for everyone on the two trains. An observer watching the two trains pass at a relativistic velocity will see an asynchronization in the clocks proportional to their distance from the front of the front train (measuring from the time of a clock on the front of the front train). Further the only distortions of space and time that observers on the trains will observe will be in things not on the train (having a velocity relative to the train).

41. Originally Posted by mitchellmckain
Originally Posted by Bender
Here is an interesting thought experiment: two identical trains are positioned on a rail in such a way that the front of the second touches the back of the first.
The trains are identical in every way, with identical computers with identical trajectories programmed in them. They have perfect clocks that are perfectly synchronised between the trains.
The trains start moving, and at every instant, they move at exactly the same speed.

Question: do the trains still touch?
Yes. Of course they still touch. It is not just the length of things that contracts but the distance between them also contracts. Dr. Rocket has this one correct.
Actually they don't. It's one of those weird counter intuitive results of relativity. It requires a non zero force to keep two objects (trains, molecules, whatever) from moving apart from each other if they accelerate, even if the acceleration is identically programmed. If you go back and read the posts I wrote in this thread, you'll see why this has to be the case (you end up with a paradox).

42. Originally Posted by Numsgil
Actually they don't. It's one of those weird counter intuitive results of relativity. It requires a non zero force to keep two objects (trains, molecules, whatever) from moving apart from each other if they accelerate, even if the acceleration is identically programmed. If you go back and read the posts I wrote in this thread, you'll see why this has to be the case (you end up with a paradox).
Yeah? r i g h t !

Originally Posted by Numsgil
The two point masses, on the other hand, will maintain their .01 Ls distance from each other from the point of view of a stationary observer because they do not have Born rigidity. To the points themselves, it will seem as if they drift apart. And by quite a bit.
This is at least is complete nonsense. Lorentz contraction does NOT just affect objects but most certainly DOES affect the distances between them.

HOWEVER Acceleration problems like this don't really fit into the framework of SR and you probably need to use GR. There does not appear to be a consensus on Bell's spaceship paradox. But in any case, Bell's analysis as presented in Wikipedia definitely certainly looks wrong to me.(*see below) These lines of simultaneity look improperly applied. I certainly would not trust any analysis making use of them as this does, and I speculate that I would all too likely trust the results of those at CERN who solved the problem in a different manner and apparently came to a different conclusion.

But lets just SUPPOSE for a moment that this analysis of Bell is correct and look at the implications. The difficulty would seem to be imagining what can be the difference between two spaceships and a single spaceship. This answer that Bell comes up would imply that different parts of a single space ship do not accelerate in the same way. That in turn makes it difficult to believe that all parts of the ship are in the same (instantaneous) inertial reference frame - i.e. they are not actually always traveling the same velocity. So the back end of the ship always has to be traveling a little bit faster than the front of the ship in order to keep up with it??? That would be very peculiar indeed. And this is what is implied by Michael Weiss. One question that this raises in my mind is what sort of acceleration makes this alleged extra effect significant?

*Imagine space filled with synchronized clocks. Move at a relativistic speed with respect to these clocks and they will no longer be synchronized but can be said to have lines of simultaneity orthogonal to the direction of motion. But now imagine that you have an truly enormous space ship filled with synchronized clocks and the ship accelerates to a relativistic velocity. At that new velocity the clocks are certainly still synchronized because they are all in the same reference frame, thus there are no "lines of simultaneity" in regards to things which are in the same inertial frame and that is what troubles me about Wikipedia's analysis. I cannot see how these lines of simultaneity apply to the two different ships. It is possble I admit that an accelertating reference frame is a whole different ball of wax but I don't see Wikipedia deriving these "lines of simultaneity" in that way, they look misapplied to me.

Originally Posted by mitchellmckain
Originally Posted by Janus
The clocks will not remain synchronized between trains. Each train will mantain that all its clocks remained in sync, but that the clocks on the other train go out of sync. Thus according to each train, while every part of itself started moving at the same instant, every part of the other train did not. As a result, each train determines that the rear of the other train starts moving before the front does and the train contracts along its length.
This doesn't make sense Janus. There is no relative velocity between the two trains so they are both in the same inertial frame and thus those on the trains measure time and space in the same way. So all the clocks on the two trains remain synchronized for everyone on the two trains. An observer watching the two trains pass at a relativistic velocity will see an asynchronization in the clocks proportional to their distance from the front of the front train (measuring from the time of a clock on the front of the front train). Further the only distortions of space and time that observers on the trains will observe will be in things not on the train (having a velocity relative to the train).
Now I was doing this as an SR problem ignoring periods of acceleration between inertial frames and not as a constant acceleration problem. But if we accept Bell's result in regards to the constant acceleration problem then it would appear that it is no longer true that everyone on the same ship is in the same instantaeous inertial reference frame.

43. You're free to take the example I did earlier and demonstrate how there isn't a paradox involved if you assume the two trains of the initial post don't separate. Feel free to ignore any reasoning of mine you don't like and substitute your own, the important bit is demonstrating that the third case (of the two separate points) is consistent whether you approach it like the first case (that is, as two separate cases of a single point) or the second case (two points on a single large rod).

44. Originally Posted by mitchellmckain

Originally Posted by Janus
The clocks will not remain synchronized between trains. Each train will mantain that all its clocks remained in sync, but that the clocks on the other train go out of sync. Thus according to each train, while every part of itself started moving at the same instant, every part of the other train did not. As a result, each train determines that the rear of the other train starts moving before the front does and the train contracts along its length.
This doesn't make sense Janus. There is no relative velocity between the two trains so they are both in the same inertial frame and thus those on the trains measure time and space in the same way. So all the clocks on the two trains remain synchronized for everyone on the two trains. An observer watching the two trains pass at a relativistic velocity will see an asynchronization in the clocks proportional to their distance from the front of the front train (measuring from the time of a clock on the front of the front train). Further the only distortions of space and time that observers on the trains will observe will be in things not on the train (having a velocity relative to the train).
As I explained later in the thread, this response was based on a mis-read on my part of the scenerio.

45. Originally Posted by mitchellmckain
Now I was doing this as an SR problem ignoring periods of acceleration between inertial frames and not as a constant acceleration problem. But if we accept Bell's result in regards to the constant acceleration problem then it would appear that it is no longer true that everyone on the same ship is in the same instantaeous inertial reference frame.
Idd. SR does not suffice to solve the problem properly, because it can lead to different results.
After all, SR makes no difference between an observer accelerating and passing two touching trains, and two touching trains accelerating and passing an observer. GR does.

46. Ok the treatment of this subject that does make sense to me can be found at http://www.ph.utexas.edu/~gleeson/NotesChapter13.pdf and the conclusion is that in an object undergoing constant accelertation the forward and aft parts of that object are not undergoing the same acceleration, are not in the same instantaneous inertial reference frames, and their clocks are not synchronized. AND this means that two objects undergoing the same acceleration will seperate.

47. mitchellmckain;
There does not appear to be a consensus on Bell's spaceship paradox. But in any case, Bell's analysis as presented in Wikipedia definitely certainly looks wrong to me.(*see below) These lines of simultaneity look improperly applied
You're onto something here mitch.

'Boughn's accelerated twins' is similar but ends with a constant target speed.

The argument in both problems use the (calculated) skewed axis of simultaneity, which reassigns the local time of events for each observer. This in turn increases distances in the direction of motion, and erroneously concludes the ships are separating, and the lead pilot is older than the trailing pilot.
This axis is the result of the moving observer denying his motion and constructing a pseudo rest frame. SR requires a transformation of coordinates to preserve constant light speed c. It's all been calculated and can be demonstrated.

It should also be noted, the 'equivalence principle' only applies to limited spaces and short time intervals, otherwise you will detect anisotropic g-field, (paths converge). The equivalence only applies to the frame experiencing the acceleration, not any large part of the universe. The 2nd ship is just a coincidence.
As a previous post mentioned, if you observe ship 1 moving, its behavior is independent of direction. You don't change this by introducing another object with insignificant mass. If ship 2 was 1 lyr behind ship 1, when ship 1 accelerated, ship 1 could not effect ship 2 for at least 1 year.
If ship 1 is assumed on the surface of a mass, then ship 2 would be inside the mass and experience less g! (see shell theorem)

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