1. This is a variant of the Einstein train experiment. There is a human body suspended horizontally by two wires in a train car moving on the tracks. The body is positioned symmetrically wrt the center of the car. Two flashes of light are emitted simultaneously from the car center activating two devices that cut the wires simultaneousy (in the frame of the car). Obviously, the body falls parallel to the car floor hitting the floor simultaneously with all its points (in the car frame).
But in the track frame the two wires are not cut simultaneously and the body does not fall parallel to the car floor : one end of the body will hit the floor before the other end. This will result into one end (say, the head) absorbing the full impact. This cannot be since the same experiment has differing results in the two frames. What is the resolution of the puzzle?

2.

3. Originally Posted by xyzt
But in the track frame the two wires are not cut simultaneously and the body does not fall parallel to the car floor
It is not clear from your post why you think this is true.

Could to expand a little to explain why you think the wires are not cut simultaneously in the track frame?

4. Originally Posted by RedPanda
Originally Posted by xyzt
But in the track frame the two wires are not cut simultaneously and the body does not fall parallel to the car floor
It is not clear from your post why you think this is true.

Could to expand a little to explain why you think the wires are not cut simultaneously in the track frame?
Sure:

In the frame of the car , the wires are cut simultaneously:

meaning that in the frame of the tracks:

where is the length of the body. This is what relativity of simultaneity is all about.

5. but that is for the length contraction and time dilation, and (v+c)/(1+v*c/c^2) = c" will show that the light will travel at c in both frames. so what both would disagree on is to when the body hit the ground due to time dilation and length contraction, and not which part of the body hit the ground first.

6. Either way, the passenger will still have a headache before the days over.

7. Originally Posted by curious mind
but that is for the length contraction and time dilation, and (v+c)/(1+v*c/c^2) = c" will show that the light will travel at c in both frames. so what both would disagree on is to when the body hit the ground due to time dilation and length contraction, and not which part of the body hit the ground first.
I don't think you understood the exercise. The exercise has nothing to do with "time dilation, length contraction". Nor does it have anything to do with c being frame invariant.

8. Originally Posted by shlunka
Either way, the passenger will still have a headache before the days over.
Yes :-)

9. Originally Posted by xyzt
This is a variant of the Einstein train experiment. There is a human body suspended horizontally by two wires in a train car moving on the tracks. The body is positioned symmetrically wrt the center of the car. Two flashes of light are emitted simultaneously from the car center activating two devices that cut the wires simultaneousy (in the frame of the car). Obviously, the body falls parallel to the car floor hitting the floor simultaneously with all its points (in the car frame).
But in the track frame the two wires are not cut simultaneously and the body does not fall parallel to the car floor : one end of the body will hit the floor before the other end. This will result into one end (say, the head) absorbing the full impact.
Here's your problem, Just because one end hits before the other does not mean that it will absorb the full impact. The body does not rotate (you'll never get a situation where it is falling straight head first, for example). You have to take into account how the information that one wire is cut travels through the body. It is not instantaneous. What the track frame observer will see is the body falling straight down but skewed, with the force of impact distributed equally along the whole length of the body just as it is in the train frame. quote]
This cannot be since the same experiment has differing results in the two frames. What is the resolution of the puzzle?[/QUOTE]

10. Originally Posted by Janus
Originally Posted by xyzt
This is a variant of the Einstein train experiment. There is a human body suspended horizontally by two wires in a train car moving on the tracks. The body is positioned symmetrically wrt the center of the car. Two flashes of light are emitted simultaneously from the car center activating two devices that cut the wires simultaneousy (in the frame of the car). Obviously, the body falls parallel to the car floor hitting the floor simultaneously with all its points (in the car frame).
But in the track frame the two wires are not cut simultaneously and the body does not fall parallel to the car floor : one end of the body will hit the floor before the other end. This will result into one end (say, the head) absorbing the full impact.
Here's your problem, Just because one end hits before the other does not mean that it will absorb the full impact. The body does not rotate (you'll never get a situation where it is falling straight head first, for example). You have to take into account how the information that one wire is cut travels through the body. It is not instantaneous. What the track frame observer will see is the body falling straight down but skewed, with the force of impact distributed equally along the whole length of the body just as it is in the train frame. quote]
This cannot be since the same experiment has differing results in the two frames. What is the resolution of the puzzle?
[/QUOTE]

Can you put all the above in mathematical form? And what is with "the information that one wire is cut travel through the body"? In the frame of the car, the wires are cut simultaneously, in the frame of the track, they aren't cut simultaneously. The removal of the wire tension from the end that is disconnected first propagates through the body at the speed of sound while the disparity in wire cutting is , as I showed . The effects don't cancel each other, in case that this is what you are intimating.

11. if the train had a length of 600km relative to the track observer and a velocity of 0.7 he would say the person at the center of the train started falling after 1 sec, where from the trains view it would be 1.4 sec.

12. Originally Posted by curious mind
if the train had a length of 600km relative to the track observer and a velocity of 0.7 he would say the person at the center of the train started falling after 1 sec, where from the trains view it would be 1.4 sec.
I very much doubt that the above is correct. Moreover, I am quite sure that it isn't relevant in solving the puzzle.

13. well what's your result then?

14. But in the track frame the two wires are not cut simultaneously
I don't know if I understand your scenario correctly, but I fail to see the issue. Both the notion of simultaneity and mechanical forces are frame-dependent and thus not invariant under Lorentz transformations. The fact that the two observers disagree on the sequence of events, especially on what forces act where, is therefore not a paradox at all, but fully expected.

Can you explain once again where exactly you see the "puzzle" or the "paradox" ?

15. Originally Posted by Markus Hanke
But in the track frame the two wires are not cut simultaneously
I don't know if I understand your scenario correctly, but I fail to see the issue. Both the notion of simultaneity and mechanical forces are frame-dependent and thus not invariant under Lorentz transformations. The fact that the two observers disagree on the sequence of events, especially on what forces act where, is therefore not a paradox at all, but fully expected.

Can you explain once again where exactly you see the "puzzle" or the "paradox" ?
Sure,

In the train car frame the rod touches the ground with all its points simultaneously, such that the reaction force of the ground is distributed evenly.
In the platform frame, the farthest end touches the platform first, taking the full impact while the other points do not touch until latter. So, the platform observer will measure the rod as either:

-shattered at one end (the end that touches first)
-bent

Either of these observations contradicts the observation of the train car observer. Same experiment, two different physical outcomes in two fully equivalent inertial frames, this cannot be. So, the whole problem statement , the premise of the exercise, has a hidden flaw. Where is the flaw?

16. Where is the flaw?
To compare the outcome in any physically meaningful way, the light pulse needs to be emitted simultaneously in both frames. That is going to be kind of difficult if they are in relative motion, don't you think
Once you account for relativity of simultaneity for the point in time when the light pulse is emitted ( in the embankment frame ), the paradox vanishes.

So, the platform observer will measure the rod as either:
What rod ?

17. I think I understand xyzt's problem (and don't see the solution - I didn't really follow Janus's argument) so let me try and explain it again.

The famous train has a laboratory set up with a metal rod suspended by two wires. Exactly equidistant from the two wires (i.e. lined up with the centre of the rod) there is a flash bulb. The two wires are made of photosensitive material and will break as soon as they are hit by the light.

The flash goes off. In the train, the experimenter sees that the light reaches the two wires at the same time, the wires break and the rod falls, and remains parallel to the train floor as it falls. The whole length of the rod hits the floor at the same time.

An observer watching the train go past just as the flash happens, sees the light reach one wire before the other. One wire breaks and then the other and so the rod does not fall parallel to the floor. One end hits the floor first.

The argument is that the rod hitting the floor parallel could have different physical effects to one end of the rod hitting the floor first (e.g. it could bend the rod).

(I thought I had spotted the flaw as I wrote that, but I was wrong ...)

18. An observer watching the train go past just as the flash happens
This is exactly the issue. How do you determine the simultaneity of this event in both frames ?

19. Originally Posted by Markus Hanke
Where is the flaw?
To compare the outcome in any physically meaningful way, the light pulse needs to be emitted simultaneously in both frames.
Obviously, this is not possible. But this has nothing to do with the difference in the mechanical deformation observed by the two observers, so, I do not understand why you are introducing this subject.

Once you account for relativity of simultaneity for the point in time when the light pulse is emitted ( in the embankment frame ), the paradox vanishes.
How? Can you present a mathematical formalism? this is what I'm after.

What rod ?
Sorry, I meant the falling body, you can replace it with a rod, the puzzle becomes less violent this way :-)

20. It doesn't have to be simultaneous in both frames. There is only one flash. The simultaneity comes in because they both see the light reach the wires at different times (relative to the flash). This is where I think the problem lies but I can't quite put my finger on it.

Lets say the rod is long enough that it takes 1ms (in the train's frame) for the light to get from the flash to each wire (both wires).

And the train is travelling fast enough that the external observer sees it take 1.1ms to reach the wire towards the front of the train (1) and 0.9ms to reach the wire towards the back of the train (2). As there is only one flash, it must reach the rear wire before the front one. Mustn't it ...

(1) because that is travelling away from the source of the flash and so the light has to travel further
(2) because that is travelling towards the source of the flash and so the light has less far to travel

21. Originally Posted by Strange
I think I understand xyzt's problem (and don't see the solution - I didn't really follow Janus's argument) so let me try and explain it again.

The famous train has a laboratory set up with a metal rod suspended by two wires. Exactly equidistant from the two wires (i.e. lined up with the centre of the rod) there is a flash bulb. The two wires are made of photosensitive material and will break as soon as they are hit by the light.

The flash goes off. In the train, the experimenter sees that the light reaches the two wires at the same time, the wires break and the rod falls, and remains parallel to the train floor as it falls. The whole length of the rod hits the floor at the same time.

An observer watching the train go past just as the flash happens, sees the light reach one wire before the other. One wire breaks and then the other and so the rod does not fall parallel to the floor. One end hits the floor first.

The argument is that the rod hitting the floor parallel could have different physical effects to one end of the rod hitting the floor first (e.g. it could bend the rod).

(I thought I had spotted the flaw as I wrote that, but I was wrong ...)
Yes, you have the puzzle correct.

22. Originally Posted by Strange
It doesn't have to be simultaneous in both frames.
It isn't simultaneous in both frames, it is simultaneous only in the train car frame. This is why one end of the body hits the ground before the rest in the platform frame. Hence, the puzzle.

23. Originally Posted by Markus Hanke
An observer watching the train go past just as the flash happens
This is exactly the issue. How do you determine the simultaneity of this event in both frames ?
Correct, there is no experimental confirmation of Relativity of Simultaneity (see the FAQ). Of all the relativistic effects, RoS has eluded experimental verification (so far). This puzzle would provide (a very unrealistic, experimentally unrealizable) verification of RoS. Unfortunately, it also brings about a nasty paradox.

24. Originally Posted by xyzt
Originally Posted by Strange
It doesn't have to be simultaneous in both frames.
It isn't simultaneous in both frames, it is simultaneous only in the train car frame. This is why one end of the body hits the ground before the rest in the platform frame. Hence, the puzzle.
Markus's comment was about "the flash". There is only one flash.

That one event isn't (necessarily) simultaneous in both frames. But it doesn't matter whether it is or not because the other events (the ones we care about) are both relative to the flash - as seen in the appropriate frame of reference. In one they must both be simultaneous with one another. In the other they cannot be.

25. There is an animation of an equivalent setup on this page: Discovering the Relativity of Simultaneity (see section 4)

26. How? Can you present a mathematical formalism? this is what I'm after.
I was starting to type this, but then realized that I had made an algebraic error, and the effects didn't actually cancel out. So it appears I was wrong.
Back to the drawing board...

27. Originally Posted by Strange
There is an animation of an equivalent setup on this page: Discovering the Relativity of Simultaneity (see section 4)
Yes, I know, the person who wrote the puzzle must have modeled after chapter 4 in John Norton's writeup. The effect (Thomas Precession) is well known, the outcome presented in the puzzle is not known and, to my best knowledge, there is no known solution.

28. Originally Posted by xyzt
Yes, I know, the person who wrote the puzzle must have modeled after chapter 4 in John Norton's writeup. The effect (Thomas Precession) is well known, the outcome presented in the puzzle is not known and, to my best knowledge, there is no known solution.
Are you referring to Strange's link, or to the puzzle in the OP ?

29. Originally Posted by Markus Hanke
Originally Posted by xyzt
Yes, I know, the person who wrote the puzzle must have modeled after chapter 4 in John Norton's writeup. The effect (Thomas Precession) is well known, the outcome presented in the puzzle is not known and, to my best knowledge, there is no known solution.
Are you referring to Strange's link, or to the puzzle in the OP ?
Both, I thought it was quite obvious, the puzzle in the OP is a variant of the Thomas Precession effect (with a collision added at the intersection of the y=0 axis).

30. Originally Posted by xyzt
Both, I thought it was quite obvious, the puzzle in the OP is a variant of the Thomas Precession effect (with a collision added at the intersection of the y=0 axis).
I got to be honest, I associate Thomas Precession with the spin of particles, and the dynamics of gyroscopes. I would not associate it with a suspended rod.

31. I think this is just a problem of intuition rather than physics. (Well, obviously it is )

Here is an informal argument that sort of resolves the paradox (but doesn't explain why it still seems to be one).

Let's say the rod has a micro-switch in each end to detect impact, and the signals are sent to a receiver in the exact centre of the rod, which records arrival time of the signals.

The experimenter sees the rod fall parallel to the floor and assume both micro-switches get triggered at the same time. When he checks the received data, that is confirmed.

The external observer sees one end hit before the other. However, that signal takes longer to reach the centre of the rod. And so concludes that the two signals will reach the receiver at the same time (which is confirmed).

So there is no difference in the physical result.

I think what Janus was saying was that you can apply that same argument to every segment of the rod, and it still ends up receiving the impact uniformly.

32. Originally Posted by Markus Hanke
Originally Posted by xyzt
Both, I thought it was quite obvious, the puzzle in the OP is a variant of the Thomas Precession effect (with a collision added at the intersection of the y=0 axis).
I got to be honest, I associate Thomas Precession with the spin of particles, and the dynamics of gyroscopes. I would not associate it with a suspended rod.
Sure, this is the most common association. But, it is nothing else than the perceived rotation of a parallel transport of a line segment when viewed from a frame moving at an angle. See the Norton example and the better example by Brown.

33. Originally Posted by Strange

I think what Janus was saying was that you can apply that same argument to every segment of the rod, and it still ends up receiving the impact uniformly.
Can't be: the endpoint farthest to the platform observer hits the ground first (according to him), all the other points "feel" the impact LATER ( in order to accomodate the fact that the force propagates at the finite speed of sound). If anything, Janus' argument makes things worse, far from explaining the puzzle. I already countered Janus' argument immediately after his post. This is one nasty puzzle.

34. I think the problem lies with assuming the rigidity of the rod. Isn't this effectively a variation of Bell's spaceship paradox?

35. how you come to that result if it's t=0 for both frames?

36. Originally Posted by Strange
I think the problem lies with assuming the rigidity of the rod. Isn't this effectively a variation of Bell's spaceship paradox?
No, it isn't. It is just some nasty animal of its own, quite original and unique, I must confess. The only thing it has in common with Bell's paradox is the role played by RoS.

37. Originally Posted by curious mind
how you come to that result if it's t=0 for both frames?

38. Originally Posted by curious mind
how you come to that result if it's t=0 for both frames?
Who is "you"?

Which result?

What is "t"?

How is "t=0" relevant?

39. Originally Posted by xyzt
Originally Posted by Markus Hanke
But in the track frame the two wires are not cut simultaneously
I don't know if I understand your scenario correctly, but I fail to see the issue. Both the notion of simultaneity and mechanical forces are frame-dependent and thus not invariant under Lorentz transformations. The fact that the two observers disagree on the sequence of events, especially on what forces act where, is therefore not a paradox at all, but fully expected.

Can you explain once again where exactly you see the "puzzle" or the "paradox" ?
Sure,

In the train car frame the rod touches the ground with all its points simultaneously, such that the reaction force of the ground is distributed evenly.
In the platform frame, the farthest end touches the platform first, taking the full impact while the other points do not touch until latter. So, the platform observer will measure the rod as either:

-shattered at one end (the end that touches first)
-bent

Either of these observations contradicts the observation of the train car observer. Same experiment, two different physical outcomes in two fully equivalent inertial frames, this cannot be. So, the whole problem statement , the premise of the exercise, has a hidden flaw. Where is the flaw?
This is where you are going wrong; in assuming that somehow all or even most of the force of the impact will be taken by the point that hits first.

You can't treat the rod as if it is a perfectly rigid object. While, in the embankment frame one end of the rod does begin to fall before the other, it takes time for this information to travel through the rod. So what happens is that from the embankment frame the rod doesn't fall "tilted" but "skewed".

I'd include an image, but I just had to get a new computer and I am having problems getting my drawing software to play nice with Windows 8. So imagine a slinky where you have pushed one end to the side. the ends are no longer even with each other, but are still parallel to each other.

Now consider what happens when the rod hits the train surface. One end does hit first, but it doesn't absorb the whole force of the rod's impact, To do that, the whole rod would have to stop instantly at the moment the end of the rod hit. However, just as above, it takes time for the information that the end hit to propagate through the rod and more time for the reaction force to propagate back through the rod to the initial impact point.

But by that time, the other parts of the rod have already impacted. IOW, each individual point along the length of the rod hits and absorbs its force of impact before it ever even gets the information that the initial impact happened. Each point of the rod will experience exactly the same force of impact. There will be no difference in the distribution of impact force for the embankment observer from what the train observer measures.

40. your experiment starts with both observers seeing the light from the bulb simultaneously, this sets t=0 for both.

41. I still think that the answer lies in rigidity. For example when the first string breaks (from the stationary FoR) the rod will not immediately start rotating about the attachment of the other wire. It will take time for the information about the break to reach that end at the speed of sound. So the rod will deform as the free end starts to move and ... and then something similar happens when that end hits the carriage floor, the rod will deform allowing it to flatten against the floor.

...

Or something.

42. Originally Posted by curious mind
your experiment starts with both observers seeing the light from the bulb simultaneously, this sets t=0 for both.
No, they don't (and probably won't) see that flash simultaneously. But it doesn't matter either way. They both time all subsequent events relative to the time of the flash (in their frame of reference). And it is the subsequent events where the relativity of simultaneity becomes important. Maybe that is what you mean by setting t=0; but that is a red herring.

43. Originally Posted by Janus
Originally Posted by xyzt
Originally Posted by Markus Hanke
But in the track frame the two wires are not cut simultaneously
I don't know if I understand your scenario correctly, but I fail to see the issue. Both the notion of simultaneity and mechanical forces are frame-dependent and thus not invariant under Lorentz transformations. The fact that the two observers disagree on the sequence of events, especially on what forces act where, is therefore not a paradox at all, but fully expected.

Can you explain once again where exactly you see the "puzzle" or the "paradox" ?
Sure,

In the train car frame the rod touches the ground with all its points simultaneously, such that the reaction force of the ground is distributed evenly.
In the platform frame, the farthest end touches the platform first, taking the full impact while the other points do not touch until latter. So, the platform observer will measure the rod as either:

-shattered at one end (the end that touches first)
-bent

Either of these observations contradicts the observation of the train car observer. Same experiment, two different physical outcomes in two fully equivalent inertial frames, this cannot be. So, the whole problem statement , the premise of the exercise, has a hidden flaw. Where is the flaw?
Each point of the rod will experience exactly the same force of impact.
Yes, so what? In the platform frame the farthest point gets the first impact, bending the rod a little, the neighboring point gets the impact next, bending it a little MORE, so the whole rod ends up BENT, not straight.

There will be no difference in the distribution of impact force for the embankment observer from what the train observer measures.
Sure it is, in the car train frame all points of the rod get impacted SIMULTANEOUSLY. The rod remains STRAIGHT.
In the platform frame the rod points get impacted SEQUENTIALLY, first the farthest from the observer, next the neighboring one and so forth. The rod gets progressively BENT.

Sorry for the caps, I am not shouting, I am just pointing out the areas of difference.

44. Originally Posted by Strange
I still think that the answer lies in rigidity. For example when the first string breaks (from the stationary FoR) the rod will not immediately start rotating about the attachment of the other wire. It will take time for the information about the break to reach that end at the speed of sound.
I think you are on to something, I tried this avenue (prior to posting here) but I quickly realized that this approach is easily nullified by assuming the suspension wires to be very short, close to zero length.

Or something.

45. i wait for a detailed answer or link then. because i can't see/undrstand why the embarkment observer would see one end hitting the ground first.

46. Originally Posted by curious mind
i wait for a detailed answer or link then. because i can't see/undrstand why the embarkment observer would see one end hitting the ground first.
It is simply the reverse of the classic Einstein thought experiment to demonstrate relativity of simultaneity.

In that, there are two lightning flashes, one at each end of the train as it passes through a station. The observer on the platform, midway between the points of impact, sees the lightning reach him at the same time and concludes the lightning strikes were simultaneous.

The observer on the train, also midway between the points where the lightning strikes, is moving towards one flash and way from the other. As a result, he sees one flash before the other. However, he knows they light travelled the same distance from each end of the train and therefore one flash must have happened before the other.

Relativity of simultaneity - Wikipedia, the free encyclopedia

47. Originally Posted by xyzt
I think you are on to something, I tried this avenue (prior to posting here) but I quickly realized that this approach is easily nullified by assuming the suspension wires to be very short, close to zero length.
Does the length of the wires matter?

48. Originally Posted by xyzt
Originally Posted by Janus
Originally Posted by xyzt
Originally Posted by Markus Hanke
But in the track frame the two wires are not cut simultaneously
I don't know if I understand your scenario correctly, but I fail to see the issue. Both the notion of simultaneity and mechanical forces are frame-dependent and thus not invariant under Lorentz transformations. The fact that the two observers disagree on the sequence of events, especially on what forces act where, is therefore not a paradox at all, but fully expected.

Can you explain once again where exactly you see the "puzzle" or the "paradox" ?
Sure,

In the train car frame the rod touches the ground with all its points simultaneously, such that the reaction force of the ground is distributed evenly.
In the platform frame, the farthest end touches the platform first, taking the full impact while the other points do not touch until latter. So, the platform observer will measure the rod as either:

-shattered at one end (the end that touches first)
-bent

Either of these observations contradicts the observation of the train car observer. Same experiment, two different physical outcomes in two fully equivalent inertial frames, this cannot be. So, the whole problem statement , the premise of the exercise, has a hidden flaw. Where is the flaw?
Each point of the rod will experience exactly the same force of impact.
Yes, so what? In the platform frame the farthest point gets the first impact, bending the rod a little, the neighboring point gets the impact next, bending it a little MORE, so the whole rod ends up BENT, not straight.

There will be no difference in the distribution of impact force for the embankment observer from what the train observer measures.
Sure it is, in the car train frame all points of the rod get impacted SIMULTANEOUSLY. The rod remains STRAIGHT.
In the platform frame the rod points get impacted SEQUENTIALLY, first the farthest from the observer, next the neighboring one and so forth. The rod gets progressively BENT.

Sorry for the caps, I am not shouting, I am just pointing out the areas of difference.

No, it does not get bent. All that happens is that the "deformation" seen by the Embankment observer caused by the ends starting their fall at different times is cancelled out after impact. There will be no end result difference between the frames.

49. Originally Posted by Janus
Originally Posted by xyzt
Originally Posted by Janus
Originally Posted by xyzt
Originally Posted by Markus Hanke
But in the track frame the two wires are not cut simultaneously
I don't know if I understand your scenario correctly, but I fail to see the issue. Both the notion of simultaneity and mechanical forces are frame-dependent and thus not invariant under Lorentz transformations. The fact that the two observers disagree on the sequence of events, especially on what forces act where, is therefore not a paradox at all, but fully expected.

Can you explain once again where exactly you see the "puzzle" or the "paradox" ?
Sure,

In the train car frame the rod touches the ground with all its points simultaneously, such that the reaction force of the ground is distributed evenly.
In the platform frame, the farthest end touches the platform first, taking the full impact while the other points do not touch until latter. So, the platform observer will measure the rod as either:

-shattered at one end (the end that touches first)
-bent

Either of these observations contradicts the observation of the train car observer. Same experiment, two different physical outcomes in two fully equivalent inertial frames, this cannot be. So, the whole problem statement , the premise of the exercise, has a hidden flaw. Where is the flaw?
Each point of the rod will experience exactly the same force of impact.
Yes, so what? In the platform frame the farthest point gets the first impact, bending the rod a little, the neighboring point gets the impact next, bending it a little MORE, so the whole rod ends up BENT, not straight.

There will be no difference in the distribution of impact force for the embankment observer from what the train observer measures.
Sure it is, in the car train frame all points of the rod get impacted SIMULTANEOUSLY. The rod remains STRAIGHT.
In the platform frame the rod points get impacted SEQUENTIALLY, first the farthest from the observer, next the neighboring one and so forth. The rod gets progressively BENT.

Sorry for the caps, I am not shouting, I am just pointing out the areas of difference.

No, it does not get bent. All that happens is that the "deformation" seen by the Embankment observer caused by the ends starting their fall at different times is cancelled out after impact. There will be no end result difference between the frames.
How? can you express this in a mathematical form? I don't disbelieve you but I am a physicist and I believe only in mathematical formalisms. Can you write it down in a mathematical form that we could all examine?

50. Originally Posted by Strange
Originally Posted by xyzt
I think you are on to something, I tried this avenue (prior to posting here) but I quickly realized that this approach is easily nullified by assuming the suspension wires to be very short, close to zero length.
Does the length of the wires matter?
If you make them to be zero length then the transmission of the "cut" signal becomes virtually instantaneous. I think Janus is on to something, if he can put it in math, we may have a very interesting solution.

51. I have been thinking about it a little more over night, and unless I am missing something crucial, the animation which Strange has linked to earlier provides the answer, doesn't it ? The embankment observer sees the rod to be slightly rotated in his frame during its fall ( here's your Thomas precession ); at the same time we know that the wire at the end that appears closer to him was cut earlier. These two effects should cancel each other out exactly, so that net total rotation of the rod at the time of impact is actually zero in both frames.
Leaving aside all effects of bending and rigidity, I think the rod must contact the cabin floor in the exact same way in both frames, or else one could conceive of an experimental setup on the floor which could in principle distinguish between the two outcomes, even if the rod doesn't bend. The puzzle would thus remain unresolved, even if the bending can be explained away.

I am trying to put this into mathematical form, but I am very much struggling with the magnitude of the Thomas precession in this case. I have never done any calculations with this, so it is unfamiliar territory.

52. This is an interesting puzzle.

53. xyzt, with your permission I would like to open a thread on this over on The Physics Forum, and cop-and-paste your OP. I'd be interested to hear what some participants over there have to say to this puzzle.

54. Originally Posted by Markus Hanke
xyzt, with your permission I would like to open a thread on this over on The Physics Forum, and cop-and-paste your OP. I'd be interested to hear what some participants over there have to say to this puzzle.
I am not a member over there and I do not have time to participate, so there will be nobody to express my objections. Do you understand my objections to the simplistic attempts at solving this puzzle?

55. Originally Posted by Markus Hanke
I have been thinking about it a little more over night, and unless I am missing something crucial, the animation which Strange has linked to earlier provides the answer, doesn't it ? The embankment observer sees the rod to be slightly rotated in his frame during its fall ( here's your Thomas precession ); at the same time we know that the wire at the end that appears closer to him was cut earlier. These two effects should cancel each other out exactly, so that net total rotation of the rod at the time of impact is actually zero in both frames.
Leaving aside all effects of bending and rigidity, I think the rod must contact the cabin floor in the exact same way in both frames, or else one could conceive of an experimental setup on the floor which could in principle distinguish between the two outcomes, even if the rod doesn't bend. The puzzle would thus remain unresolved, even if the bending can be explained away.

I am trying to put this into mathematical form, but I am very much struggling with the magnitude of the Thomas precession in this case. I have never done any calculations with this, so it is unfamiliar territory.
I tried that exact calculation, long ago. The problem I ran into is the fact that I couldn't prove that the floor of the train car, as it is moving parallel with itself in the frame of the platform rotates exactly the same amount as the falling body, such that they are parallel in the frame of the platform. If you manage this last step, you have a valid solution to the problem.

56. Originally Posted by xyzt
I tried that exact calculation, long ago. The problem I ran into is the fact that I couldn't prove that the floor of the train car, as it is moving parallel with itself in the frame of the platform rotates exactly the same amount as the falling body, such that they are parallel in the frame of the platform. If you manage this last step, you have a valid solution to the problem.
Actually this is not the direction I was going into; the rod rotates anyway due to the fact that one of the wires is cut before the other, producing a torque around the endpoint that is still attached. The Thomas precession should then induce a second rotation of the same magnitude, but opposite sign, which acts once the rod is released. The effect is that both rotations cancel, and, even though one of the wires is cut before the other, the rod actually falls straight down in both frames, thereby avoiding any paradoxes altogether.

This is highly counterintuitive, so maths are definitely needed. How did you calculate the Thomas precession ?

57. Originally Posted by Markus Hanke
Originally Posted by xyzt
I tried that exact calculation, long ago. The problem I ran into is the fact that I couldn't prove that the floor of the train car, as it is moving parallel with itself in the frame of the platform rotates exactly the same amount as the falling body, such that they are parallel in the frame of the platform. If you manage this last step, you have a valid solution to the problem.
Actually this is not the direction I was going into; the rod rotates anyway due to the fact that one of the wires is cut before the other, which induces a torque into the system. The Thomas precession should then induce a second rotation of the same magnitude, but opposite sign. The effect is that both rotations cancel, and, even though one of the wires is cut before the other, the rod actually falls straight down in both frames.
I tried this avenue as well: the wire farthest from the platform observer is detected as being cut first, so the farthest end would drop first, yet the Thomas precession (see the pictures in chapter 4 of the Norton link and the link to the Mathpages) show the body rotating in the opposite direction, so, the two effects may cancel each other. I remember that I got the calculations to agree with the above guess but I was still unhappy with them (I am very critical of my own work) , the difficulty was caused that this puzzle mixes in the presence of a gravitational field, so, it isn't a true SR problem. Your approach is surely tempting, maybe you could use the work I've done already.

This is highly counterintuitive, so maths are definitely needed. How did you calculate the Thomas precession ?
I will need to do serious digging in order to find my notes. I remember I was quite disappointed when I found that my guess (identical with the one expressed by you above) proved to be wrong.

58. Originally Posted by xyzt
Originally Posted by Markus Hanke
xyzt, with your permission I would like to open a thread on this over on The Physics Forum, and cop-and-paste your OP. I'd be interested to hear what some participants over there have to say to this puzzle.
I am not a member over there and I do not have time to participate, so there will be nobody to express my objections. Do you understand my objections to the simplistic attempts at solving this puzzle?
Ok, no problem

59. Originally Posted by Markus Hanke
Originally Posted by xyzt
Originally Posted by Markus Hanke
xyzt, with your permission I would like to open a thread on this over on The Physics Forum, and cop-and-paste your OP. I'd be interested to hear what some participants over there have to say to this puzzle.
I am not a member over there and I do not have time to participate, so there will be nobody to express my objections. Do you understand my objections to the simplistic attempts at solving this puzzle?
Ok, no problem
I managed to find the calculations, it has been a long time but I remember deciding that they are not correct. Here is the link.

60. Originally Posted by xyzt
Originally Posted by Markus Hanke
Originally Posted by xyzt
Originally Posted by Markus Hanke
xyzt, with your permission I would like to open a thread on this over on The Physics Forum, and cop-and-paste your OP. I'd be interested to hear what some participants over there have to say to this puzzle.
I am not a member over there and I do not have time to participate, so there will be nobody to express my objections. Do you understand my objections to the simplistic attempts at solving this puzzle?
Ok, no problem
I managed to find the calculations, it has been a long time but I remember deciding that they are not correct. Here is the link.
Cool, I'll look at this in more detail. Where do you think is the mistake ?

61. Originally Posted by Markus Hanke
Originally Posted by xyzt
Originally Posted by Markus Hanke
Originally Posted by xyzt
Originally Posted by Markus Hanke
xyzt, with your permission I would like to open a thread on this over on The Physics Forum, and cop-and-paste your OP. I'd be interested to hear what some participants over there have to say to this puzzle.
I am not a member over there and I do not have time to participate, so there will be nobody to express my objections. Do you understand my objections to the simplistic attempts at solving this puzzle?
Ok, no problem
I managed to find the calculations, it has been a long time but I remember deciding that they are not correct. Here is the link.
Cool, I'll look at this in more detail. Where do you think is the mistake ?
I don't remember, it is so long ago, I tend to be very critical and suspicious of my own work. :-(
I made a different attempt, you may want to look at it as well.

62. Did you read through the link Strange provided in post #24? Read from section 4 all the way to the end of the page...

Originally Posted by Strange
There is an animation of an equivalent setup on this page: Discovering the Relativity of Simultaneity (see section 4)
1. The rod does not fall straight down in both frames - in the embankment frame it falls at an angle. This is the same situation as a ball being dropped perpendicular to the floor. When the train is moving, the embankment observer calculates the trajectory of the ball not to be perpendicular to the ground.

2. The rod is not rotated, it is skewed. The vertical surfaces at the ends of the rod remain perpendicular to the floor at all times. All forces that propagate vertically through the rod in the frame of the train still propagate vertically through the rod from the frame of the embankment. The force that moved from the bottom of the end of the rod to the top of the same end does not move up the rod differently, due to the rod being skewed rather than rotated.

Here is a simple diagram I made, which is only initially looking at the vertical impact forces on the rod. These forces do not propagate up the rod any differently in either frame:

63. Originally Posted by SpeedFreek
Did you read through the link Strange provided in post #24? Read from section 4 to the end of the page...

Originally Posted by Strange
There is an animation of an equivalent setup on this page: Discovering the Relativity of Simultaneity (see section 4)
1. The rod does not fall straight down in both frames - in the embankment frame it falls at an angle. This is the same situation as a ball being dropped perpendicular to the floor. When the train is moving, the embankment observer calculates the trajectory of the ball not to be perpendicular to the ground.

2. The rod is not rotated, it is skewed. The vertical surfaces at the ends of the rod remain perpendicular to the floor at all times. All forces that propagate vertically through the rod in the frame of the train still propagate vertically through the rod from the frame of the embankment. The force that moved from the bottom of the end of the rod to the top of the same end does not move up the rod differently, due to the rod being skewed rather than rotated.

Here is a simple diagram I made, which is only initially looking at the vertical impact forces on the rod. These forces do not propagate up the rod any differently in either frame:
Yes, I am fully familiar with the fact that , due to the way forces transform in SR, the reaction of the platform to the rod is oriented perpendicular to the rod in all frames. This is not the issue, the issue is that one end of the rod, by virtue of being released earlier in free fall, hits the platform before all other points (this can be seen in your drawing above). By virtue of this, the rod starts getting bent in the frame of the platform while it isn't bent in the frame of the train car. Hence, the "paradox".

64. So, did you read all the way to the end of the article posted by Strange in post #24?

65. Originally Posted by SpeedFreek
So, did you read all the way to the end of the article posted by Strange in post #24?
I am very familiar with the article, what is your point? And please, don't talk down to me, ok?

66. Has anyone taken into account the fact that the rod is also curved in the frame of the train? In that frame it is true that the ends get released at the same time, but the forces that keep the rest of the rod parallel to the floor propagate at a finite speed toward the center from each end. So each end starts falling before the center does, and the rod has a slight U-shape. In the embankment frame the same thing happens, but the bending starts at the lead end first. So what we are trying to reconcile is two different U shapes rather than one U shape and one flat one. I haven't tried any calculations, anticipating the difficulty, but could someone who has done a calculation involving the shape of the rod comment on my guess that it might be easier to reconcile differently-bent rods more easily than one bent rod compared to one straight one?

In particular, in both frames each end may well hit the floor before before the next segment towards the center, which seems to me to be a bit of progress toward a resolution.

67. Originally Posted by mvb
Has anyone taken into account the fact that the rod is also curved in the frame of the train? In that frame it is true that the ends get released at the same time, but the forces that keep the rest of the rod parallel to the floor propagate at a finite speed toward the center from each end. So each end starts falling before the center does, and the rod has a slight U-shape. In the embankment frame the same thing happens, but the bending starts at the lead end first. So what we are trying to reconcile is two different U shapes rather than one U shape and one flat one.
Yes, I thought about it. Long ago , I wrote a paper on relativistic formalism for the theory of elasticity, believe me, you want to stay away as far as possible from this field, the math is horrendous. Besides, if we start taking into consideration the deformation of the rod (body) under its own weight, this doesn't help the fact that the deformation due to collision with the ground is in addition to it, due to the fact that one end hits ground before the other.

I haven't tried any calculations, anticipating the difficulty, but could someone who has done a calculation involving the shape of the rod comment on my guess that it might be easier to reconcile differently-bent rods more easily than one bent rod compared to one straight one?
Having done quite a few calculations on the subject (see links) , I would not touch this subject with a ten foot pole.

In particular, in both frames each end may well hit the floor before before the next segment towards the center, which seems to me to be a bit of progress toward a resolution.
Yes, it may be progress, how do we quantify it mathematically?

68. It was not my intention to "talk down to you". I may indeed be wrong, but here is how I understand the situation:

From the frame of the embankment, the concept of "now" applies differently across the length of the carriage, when compared to the frame of the train, due to the motion of the train. The further away from the embankment observer a point on the train is, the more the difference in simultaneity between them.

So, let's say the time "now" is 12:00:00 on the embankment when the rear end of the rod hits the floor, and that the rear end of the rod is directly in line with the observer on the embankment at 12:00:00. At this time, due to the relativity of simultaneity, it is only 11:59:59 at the other end of the rod - it is not "now" yet! We could prove this by having clocks that are synchronised in the train frame, spaced evenly down the carriage - from the frame of the embankment they would all show different times (after calculating out light-travel time, due purely to the relativity of simultaneity!).

So, from the embankment, the different parts of the rod hit the floor at different times, but these times all equate to the same time in the frame of the train.

But what of the forces involved? Well, if you want to use the "now" of the embankment, and say the force starts travelling up the rod at a certain time (12:00:00), you need to take the relativity of simultaneity into account in your calculations (as it is only 11:59:59 at the other end of the rod), and thus understand that because the rod is in the trains frame of reference, you need to calculate those certain times by calculating for the frame of the rod, rather than naively using the embankment frame.

From the embankment frame, the rod does not bend or break, because time across distance on the embankment does not map directly to time across distance on the train. The far end of the rod was not released later than the rear end - it is simply that from the embankment frame it is a different time of day at one end of the carriage than it is at the other end.

As far as I can tell there is no paradox in these results, in the same way that there is no paradox in two identical twins ending up having different ages. But then again, perhaps I am misunderstanding the situation.

69. Just wondering, who are the world's foremost experts on relativity? ( if there are)

The biggest authority.

70. Originally Posted by SpeedFreek
It was not my intention to "talk down to you". I may indeed be wrong, but here is how I understand the situation:

From the frame of the embankment, the concept of "now" applies differently across the length of the carriage, when compared to the frame of the train, due to the motion of the train. The further away from the embankment observer a point on the train is, the more the difference in simultaneity between them.

So, let's say the time "now" is 12:00:00 on the embankment when the rear end of the rod hits the floor, and that the rear end of the rod is directly in line with the observer on the embankment at 12:00:00. At this time, due to the relativity of simultaneity, it is only 11:59:59 at the other end of the rod - it is not "now" yet! We could prove this by having clocks that are synchronised in the train frame, spaced evenly down the carriage - from the frame of the embankment they would all show different times (after calculating out light-travel time, due purely to the relativity of simultaneity!).

So, from the embankment, the different parts of the rod hit the floor at different times, but these times all equate to the same time in the frame of the train.
Nothing new here, all standard relativity.

But what of the forces involved? Well, if you want to use the "now" of the embankment, and say the force starts travelling up the rod at a certain time (12:00:00),
Janus and I have been over this subject already. The puzzle has nothing to do with "force traveling up the rod at a certain time" . It has everything to do with different parts of the rod hitting the ground sequentially (in the embankment frame), rather than simultaneously (in the train car frame). So, the rod would bend in one frame (embankment) and not bend in the other frame (train car). Therein lies the puzzle.

The far end of the rod was not released later than the rear end - it is simply that from the embankment frame it is a different time of day at one end of the carriage than it is at the other end.
Nope, it is a different time of day of the embankment, not of the carriage.

As far as I can tell there is no paradox in these results, in the same way that there is no paradox in two identical twins ending up having different ages.
I agree that there is no "paradox", I disagree with the validity of your explanation.

71. Originally Posted by xyzt
The puzzle has nothing to do with "force traveling up the rod at a certain time" . It has everything to do with different parts of the rod hitting the ground sequentially (in the embankment frame), rather than simultaneously (in the train car frame). So, the rod would bend in one frame (embankment) and not bend in the other frame (train car). Therein lies the puzzle.
Indeed, but the sequential times in the frame of the embankment equate to times that are simultaneous in the frame of the train, of course.

Originally Posted by xyzt
The far end of the rod was not released later than the rear end - it is simply that from the embankment frame it is a different time of day at one end of the carriage than it is at the other end.
Nope, it is a different time of day of the embankment, not of the carriage.
I said from the embankment frame it is a different time of day at one end of the carriage than it is at the other end. This is why in the original gedanken-experiment the lightning flashes are simultaneous in one frame whilst not being simultaneous in the other. If, in the train frame, the ends of the rods are released, or hit the floor, simultaneously, then from the frame of the embankment those events on the train occur at different times of day. So, as I said, from the embankment frame it is a different time of day at one end of the carriage than it is at the other.

72. Originally Posted by SpeedFreek
Originally Posted by xyzt
The puzzle has nothing to do with "force traveling up the rod at a certain time" . It has everything to do with different parts of the rod hitting the ground sequentially (in the embankment frame), rather than simultaneously (in the train car frame). So, the rod would bend in one frame (embankment) and not bend in the other frame (train car). Therein lies the puzzle.
Indeed, but the sequential times in the frame of the embankment equate to times that are simultaneous in the frame of the train, of course.

Originally Posted by xyzt
The far end of the rod was not released later than the rear end - it is simply that from the embankment frame it is a different time of day at one end of the carriage than it is at the other end.
Nope, it is a different time of day of the embankment, not of the carriage.
I said from the embankment frame it is a different time of day at one end of the carriage than it is at the other end. This is why in the original gedanken-experiment the lightning flashes are simultaneous in one frame whilst not being simultaneous in the other. If, in the train frame, the ends of the rods are released, or hit the floor, simultaneously, then from the frame of the embankment those events on the train occur at different times of day. So, as I said, from the embankment frame it is a different time of day at one end of the carriage than it is at the other.
Ok, so now what? You still don't have an explanation for the puzzle.

73. Gentlemen, I think we all agree that there cannot be a paradox, because it can be shown in a very general way that SR does not permit paradoxes between inertial frames. So the question becomes - what is the resolution of the puzzle ?

I think the rod cannot hit the cabin floor in a different way in the two frames. It either falls straight in both frames or it doesn't. Consider for example part of the floor made of calibrated glass plates - if the rod hits parallel to the floor the impact is absorbed, but if it hits at an angle the glass breaks. Both observers must agree on the outcome - so is post #61 really a possible solution ? I'm not so sure, because we know already that in the train frame the rod will always fall parallel to the floor. It should be no different in the embankment frame, or else we might get a different final outcome.
While I agree with SpeedFreek's analysis of the force lines, I don't think it resolves the basic contradiction of the puzzle; the puzzle can be resolved only if the final outcome, at the end of the sequence, is the same for both observers, regardless of the circumstances.

74. I still think that a major part of this problem is the fact that the rod doesn't touch down as a straight line in either frame. In the train frame, the ends are released simultaneously but the inner parts cannot start moving until a signal that an end has started down reaches them, and that signal can move toward the inner parts at the speed of light at best. Hence the ends touch down first. Then the inner parts land sequentially later, and the bar flattens out at the speed of light at best.

To keep the embankment frame as simple as possible, I will assume that the "release signal" and "now landing signal" travel at the speed of light. In this frame the rear end is dropped first, since it catches up with the light signal, and the lead end is dropped a bit later since it is running away from the light signal. The release signal goes forward from the rear end more slowly than it comes back from the front end, for the same reasons. Thus the middle of the rod gets the signal to drop from each end simultaneously [in this special case only?]. The rod drops down in a bent form. This solves, qualitatively, one of the paradoxes: the rod is bent in both frames.

On landing, as seen from the embankment frame, the rear end lands first. The flattening signal propagates forward at the speed of light, but the rod is also moving forward, so the signal travels more slowly than light relative to the rod. The front end lands later, and the flattening signal propagates backward at the speed of light. The rod is still moving forward, to the signal travels more quickly than light relative to the rod [again, only as measured from the embankment frame]. The result is that the two signals reach the center of the rod at the same time, given the assumption that the signals are traveling at the speed of light. Again, the rod's straightening ends at the middle of the rod in both frames.

So I see no qualitative paradox in the simplified case. Since I believe special relativity, I expect that the more complicated analysis when the rod adjustments propagate more slowly than light will also work out. I am not anxious to tackle the math.

75. Originally Posted by Markus Hanke
I think the rod cannot hit the cabin floor in a different way in the two frames. It either falls straight in both frames or it doesn't.
I'm not quite so sure of that fundamental principle, Markus.

I believe this might be related to the puzzle at hand and might be worth looking into in order to resolve this puzzle:

Imagine two long rods, at rest. One rod is sitting at an angle in relation to the other, such that the rods are neither parallel or perpendicular. See the diagram below. The blue and red lines represent the rods, whereas the grey boxes are there to aid visualisation of length contraction along the axis of motion.

If the distance between the rods was decreasing, due to there being relative motion between them, then:

From the frame of A (blue), the length contraction of B (red) causes both ends of B to intersect with A simultaneously, as the length contraction of B makes it parallel with A.

But from the frame of B, the length contraction of A causes both ends of B to intersect with A at different times. The two rods are not parallel.

Special Relativity/Simultaneity, time dilation and length contraction

If a rod makes an angle with its direction of motion toward or away from an observer the component of its length in the direction of motion will be contracted. This means that observed angles are also transformed during changes of frames of reference.
By the way, I am not suggesting that the above example alone will solve the puzzle, as I feel the answer to this lies purely in the relativity of simultaneity, but I just wondered whether the rod on the train had to fall parallel to the floor from the frame of the embankment just because it falls parallel in the frame of the train.

This particular line of thought stems from a discussion we were having over at BAUT a couple of years ago, where the above example was raised (that diagram was my final contribution to that thread, but the diagrams have since disappeared due to a forum rebuild). It seems that the ends of one rod intersect with the other rod simultaneously in the frame of one of the rods, but not in the frame of the other, which raises the same question as we are discussing here - imagine one rod was impacting the other...

76. Originally Posted by SpeedFreek

Special Relativity/Simultaneity, time dilation and length contraction

If a rod makes an angle with its direction of motion toward or away from an observer the component of its length in the direction of motion will be contracted. This means that observed angles are also transformed during changes of frames of reference.
The above settles it, since it confirms the same exact formula, I derived two years ago in analyzing this paradox. At the time I solved the puzzle, I was not happy with the math, I thought it was wrong but, on further reflection, it is correct. So Markus is right, the rod hits the axis y=0, the same exact way in all frames.

I just wondered whether the rod on the train had to fall parallel to the floor from the frame of the embankment just because it falls parallel in the frame of the train.
Yes, it does.

77. Originally Posted by Markus Hanke

I think the rod cannot hit the cabin floor in a different way in the two frames. It either falls straight in both frames or it doesn't.

78. that's what i said all along.

79. I'm not quite so sure of that fundamental principle, Markus.
Ha ! This forum would be boring if we'd all agreed all the time, wouldn't it.

Anyways, the fundamental principle I was referring to is the fact that the physical outcome must be indistinguishable in both frames at the end of the sequence. And I think we can agree on that, because otherwise we'd have a true paradox at our hands.

All I was saying is that, if the rod falls differently in different frames, you could conceivably obtain different outcomes given appropriate experimental setups, e.g. the simple glass plate I was mentioning. Obviously this cannot be.

By the way, I am not suggesting that the above example alone will solve the puzzle, as I feel the answer to this lies purely in the relativity of simultaneity,
I am now almost sure that the answer lies in a combination of RoS, SpeedFreek's rod diagrams, and xyzt's maths, all of which is correct and valid. Just to summarize this quickly :

1. The train frame is trivial - both wires are cut simultaneously, and the rod falls straight down
2. In the embankment frame, one of the wires is cut before the other, due to RoS. This causes a torque around the endpoint which is still attached, so the rod starts to rotate around that point. The rod is also skewed along its cross-section.
3. The relative motion between embankment and train also leads to a phenomenon called Thomas precession. Once the second wire is cut and the rod starts falling, the precession causes another torque which is equal in magnitude but opposite in sign to the one above ( see xyzt's maths ).
4. Both rotations cancel, and, despite the relative motion and RoS, the rod still falls straight down, just like in the train frame.
5. It remains to be noted that the observers may not agree on when exactly the rod hits the floor.

I must say this is a very interesting puzzle, a welcome change from the usual time dilation / length contraction scenarios.

80. Originally Posted by Markus Hanke
5. It remains to be noted that the observers may not agree on when exactly the rod hits the floor.
Yes, the observers disagree where the rod lands on the floor. they also disagree on the markings left on the floor (hey, you can't escape length contraction! :-) )

I must say this is a very interesting puzzle, a welcome change from the usual time dilation / length contraction scenarios.
Yes, it was a fun thread! Thank you all for your contributions.

81. Originally Posted by xyzt
Originally Posted by curious mind
but that is for the length contraction and time dilation, and (v+c)/(1+v*c/c^2) = c" will show that the light will travel at c in both frames. so what both would disagree on is to when the body hit the ground due to time dilation and length contraction, and not which part of the body hit the ground first.
I don't think you understood the exercise. The exercise has nothing to do with "time dilation, length contraction". Nor does it have anything to do with c being frame invariant.
hmmm.....

82. Originally Posted by curious mind
Originally Posted by xyzt
Originally Posted by curious mind
but that is for the length contraction and time dilation, and (v+c)/(1+v*c/c^2) = c" will show that the light will travel at c in both frames. so what both would disagree on is to when the body hit the ground due to time dilation and length contraction, and not which part of the body hit the ground first.
I don't think you understood the exercise. The exercise has nothing to do with "time dilation, length contraction". Nor does it have anything to do with c being frame invariant.
hmmm.....

83. Originally Posted by xyzt
Originally Posted by curious mind
Originally Posted by xyzt
Originally Posted by curious mind
but that is for the length contraction and time dilation, and (v+c)/(1+v*c/c^2) = c" will show that the light will travel at c in both frames. so what both would disagree on is to when the body hit the ground due to time dilation and length contraction, and not which part of the body hit the ground first.
I don't think you understood the exercise. The exercise has nothing to do with "time dilation, length contraction". Nor does it have anything to do with c being frame invariant.
hmmm.....
nothing, since i had it right.

84. Originally Posted by curious mind
Originally Posted by xyzt
Originally Posted by curious mind
Originally Posted by xyzt
Originally Posted by curious mind
but that is for the length contraction and time dilation, and (v+c)/(1+v*c/c^2) = c" will show that the light will travel at c in both frames. so what both would disagree on is to when the body hit the ground due to time dilation and length contraction, and not which part of the body hit the ground first.
I don't think you understood the exercise. The exercise has nothing to do with "time dilation, length contraction". Nor does it have anything to do with c being frame invariant.
hmmm.....
nothing, since i had it right.
Good for you, when you posted something totally incoherent you can always claim that you "had it right" afterwards.

85. The x,z plane is shown below and aligned with the x axis.
Blue line is light path, cyan line is for measuring or alignment.
Ignoring lc & td, Δt is approx. 2.5 ns for a 2m rod moving at .6c.
The back end of the rod will fall a distance h during Δt.
P will perceive the cuts simultaneously (assuming a reflection from each).
Since the lag Δt sets the simultaneity of perception for P for all events, he sees the fall of the front end and back end at the same time, i.e. the rod falls horizontally for P.
falling rod.jpg

86. Originally Posted by Markus Hanke
Ha ! This forum would be boring if we'd all agreed all the time, wouldn't it.

Anyways, the fundamental principle I was referring to is the fact that the physical outcome must be indistinguishable in both frames at the end of the sequence. And I think we can agree on that, because otherwise we'd have a true paradox at our hands.

All I was saying is that, if the rod falls differently in different frames, you could conceivably obtain different outcomes given appropriate experimental setups, e.g. the simple glass plate I was mentioning. Obviously this cannot be.
So, how do we deal with the rods I described in post #74, where length contraction causes the rods to be parallel when they meet as judged by an observer in the frame of the blue rod, whilst the rods are not parallel as judged from the frame of the red rod?

87. Originally Posted by SpeedFreek
Originally Posted by Markus Hanke
Ha ! This forum would be boring if we'd all agreed all the time, wouldn't it.

Anyways, the fundamental principle I was referring to is the fact that the physical outcome must be indistinguishable in both frames at the end of the sequence. And I think we can agree on that, because otherwise we'd have a true paradox at our hands.

All I was saying is that, if the rod falls differently in different frames, you could conceivably obtain different outcomes given appropriate experimental setups, e.g. the simple glass plate I was mentioning. Obviously this cannot be.
So, how do we deal with the rods I described in post #74, where length contraction causes the rods to be parallel when they meet as judged by an observer in the frame of the blue rod, whilst the rods are not parallel as judged from the frame of the red rod?
Yes, now that you are bringing it back up, I like this way at looking at the phenomenon. It takes the internal response of the rod out of the problem entirely since you can set up the problem at the observer's leisure in the absence of any external forces. I wish I had seen this when you posted this before.

To keep the internal mechanics of the rod out of the problem as much as possible, I'm going to coat the rods with super-instant glue, so that once any corresponding parts of the rods come into contact they stick to each other. Now, off to considering more of what you said about this....

88. Originally Posted by SpeedFreek
So, how do we deal with the rods I described in post #74, where length contraction causes the rods to be parallel when they meet as judged by an observer in the frame of the blue rod, whilst the rods are not parallel as judged from the frame of the red rod?
Yeah, good point. My feeling would be that once again Thomas precession might play a role here.
Just to make this clear, I am not saying that I have all the answers to the original puzzle. Post #78 is my take on things, but of course that doesn't necessarily mean it is physically correct. This entire scenario is quite non-intuitive.

89.

90. Originally Posted by Markus Hanke
I'm not quite so sure of that fundamental principle, Markus.
Ha ! This forum would be boring if we'd all agreed all the time, wouldn't it.

Anyways, the fundamental principle I was referring to is the fact that the physical outcome must be indistinguishable in both frames at the end of the sequence. And I think we can agree on that, because otherwise we'd have a true paradox at our hands.

All I was saying is that, if the rod falls differently in different frames, you could conceivably obtain different outcomes given appropriate experimental setups, e.g. the simple glass plate I was mentioning. Obviously this cannot be.

By the way, I am not suggesting that the above example alone will solve the puzzle, as I feel the answer to this lies purely in the relativity of simultaneity,
I am now almost sure that the answer lies in a combination of RoS, SpeedFreek's rod diagrams, and xyzt's maths, all of which is correct and valid. Just to summarize this quickly :

1. The train frame is trivial - both wires are cut simultaneously, and the rod falls straight down
2. In the embankment frame, one of the wires is cut before the other, due to RoS. This causes a torque around the endpoint which is still attached, so the rod starts to rotate around that point. The rod is also skewed along its cross-section.
I don't believe that there will be any rotation or torque. Let's assume that the rod is extremely rigid so that the speed at which the impulse of the end of the rod being released travels at c through the rod, If you compare the delay between the two ends being released to the time it would take the impulse to travel between ends, it turns out that the second end releases before the information that the other end released ever gets to it. If fact, it releases before the information has even reached the mid point of the rod. These impulses will meet at the midpoint of the rod, (just like they do in the train frame, as pointed out by MVB.[/tex]
3. The relative motion between embankment and train also leads to a phenomenon called Thomas precession. Once the second wire is cut and the rod starts falling, the precession causes another torque which is equal in magnitude but opposite in sign to the one above ( see xyzt's maths ).
4. Both rotations cancel, and, despite the relative motion and RoS, the rod still falls straight down, just like in the train frame.
5. It remains to be noted that the observers may not agree on when exactly the rod hits the floor.
The rod cannot fall "flat" (so that both ends land at the same time) in the embankment frame. We Know that the ends land simultaneously in the train frame. If we put clocks at the landing point of the ends, they will read the same at impact. However, we also know that these two clocks cannot be in sync in the embankment frame. The reading on each clock when its respective end impacts must be the same in both frames. Thus one end must hit before the other.

I still think that the answer is in how the force of impact propagates through the rod. As in the release, one end will hit, then the other and finally the midpoint. the forces propagate through the rod just as they did during the release with the impulses from the ends impacting meeting in the middle of the rod. The end result will be a final deformation of the rod that is the same as that seen in the train frame.

I must say this is a very interesting puzzle, a welcome change from the usual time dilation / length contraction scenarios.

91. The rod cannot fall "flat" (so that both ends land at the same time) in the embankment frame. We Know that the ends land simultaneously in the train frame. If we put clocks at the landing point of the ends, they will read the same at impact. However, we also know that these two clocks cannot be in sync in the embankment frame. The reading on each clock when its respective end impacts must be the same in both frames. Thus one end must hit before the other.
That is a very good point. I must admit that I overlooked this.

92. Originally Posted by Janus
I don't believe that there will be any rotation or torque. Let's assume that the rod is extremely rigid so that the speed at which the impulse of the end of the rod being released travels at c through the rod, If you compare the delay between the two ends being released to the time it would take the impulse to travel between ends, it turns out that the second end releases before the information that the other end released ever gets to it. If fact, it releases before the information has even reached the mid point of the rod. These impulses will meet at the midpoint of the rod, (just like they do in the train frame, as pointed out by MVB.
Interesting argument but you need to consider the complete argument:

1. The endpoint farthest to the platform observer releases first (as viewed by the platform observer)
2. The endpoint closest to the platform observer releases next.
3. Various points release at some later time, allowing for the information to propagate from the endpoints (at the speed of sound).

The effect is that :

4. The endpoint farthest to the platform observer hits the ground first (as viewed by the platform observer)
5. The endpoint closest to the platform observer hits next but LATER.
6. Various points hit at some later time than the two ends.

In the frame of the train car things are different:

1a. The endpoints release simultaneously
2a. Various points release at some later time, allowing for the information to propagate from the endpoints (at the speed of sound). The center point falls out last.

The effect is that :

3a. The endpoints hit the ground simultaneously
4a. Various points hit at some later time than the two ends. The center point hits the ground last.

I still think that the answer is in how the force of impact propagates through the rod. As in the release, one end will hit, then the other and finally the midpoint.
Like I said, I would love to see the math supporting your assertion. You would have to show, through math, that there is no difference in the distortion measured by the two observers.

93. The thought experiment with the blue and red rods that I laid out in post #74 is asking the same question about the impact of the rod (if the red rod has the same angle as the blue rod at impact, in the frame of the blue rod, how does the impact play out from the frame of the red rod, where one end of the red rod impacts before the other end), but it does away with all the complications that arise due to the information of the drop propagating along the rod, so we can do this with perfectly straight rods until impact. mvb alluded to this in post #86.

The math might be easier. ()

94. Originally Posted by SpeedFreek
The thought experiment with the blue and red rods that I laid out in post #74 is asking the same question about the impact of the rod (if the red rod has the same angle as the blue rod at impact, in the frame of the blue rod, how does the impact play out from the frame of the red rod, where one end of the red rod impacts before the other end), but it does away with all the complications that arise due to the information of the drop propagating along the rod, so we can do this with perfectly straight rods until impact. mvb alluded to this in post #86.

The math might be easier. ()
I very much doubt it, once one abandons infinite rigidity, one "buys" the formulation of of the solution in terms of relativistic elasticity, something that makes the solution infinitely harder.

95. I wasn't really talking about infinite rigidity, I was simply doing away with any initial forces and assuming inertial motion. If one wants to, one could imagine one rod was pushed towards the other by a force that was equally distributed along its length. Whatever, we can then simply look at two straight rods approaching each other parallel in one frame whilst they are not approaching parallel in the other. The rods could be made from wood, ice, rubber or bread - it doesn't matter until the impact, where that impact is simultaneous along the length of the red rod in the blue frame, but not simultaneous along the length of the red rod from the red frame.

Perhaps I misunderstood you but isn't it an explanation of the end result from different frames you are interested in?

96. Originally Posted by SpeedFreek
I wasn't really talking about infinite rigidity, I was simply doing away with any initial forces and assuming inertial motion. If one wants to, one could imagine one rod was pushed towards the other by a force that was equally distributed along its length. Whatever, we can then simply look at two straight rods approaching each other parallel in one frame whilst they are not approaching parallel in the other. The rods could be made from wood, ice, rubber or bread - it doesn't matter until the impact, where that impact is simultaneous along the length of the red rod in the blue frame, but not simultaneous along the length of the red rod from the red frame.
The very link you posted earlier shows that is not the case:

In the car frame . What does this tell you about ?

97. Sorry, you will have to explain what is not the case. Are you saying it is impossible for one rod to approach another parallel in one frame but not parallel in the other?

98. Originally Posted by SpeedFreek
Sorry, you will have to explain what is not the case. Are you saying it is impossible for one rod to approach another parallel in one frame but not parallel in the other?
What does the link say? what does the math say?

99. Above, I was talking about my example in post #74. What is the "car frame", the blue rod or the red rod?

100. Originally Posted by SpeedFreek
What is the "car frame", the blue rod or the red rod?
Doesn't matter. In BOTH frames, the situation is the same, the rod falls parallel to the y=0 axis, this is what the math in your OWN link teaches you.If you think otherwise, you are welcome to produce your own math. So far, you haven't produced any.

101. Originally Posted by SpeedFreek

So, how do we deal with the rods I described in post #74, where length contraction causes the rods to be parallel when they meet as judged by an observer in the frame of the blue rod, whilst the rods are not parallel as judged from the frame of the red rod?