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GrapesOfWrath
2001-Oct-23, 10:50 AM
I promised I'd repost this once the new board was up, so here goes. It's not against the mainstream of physics, but it is definitely not astronomy:

Joe (mcveetj@ece.orst.edu), on the old BABB, wrote on October 01, 2001 at 14:56:17:



Two rockets, aligned head to tail, are connected with a rope from the tail of Rocket 1 to the head of Rocket 2. Both rockets are at rest wrt each other and the rope. At time t = 0 (synchronized clocks) in the rest frame of the rockets, they both fire their thrusters (single thrusters at the tail of each rocket) such that they accelerate at 1g.

To an oberserver in the inertial frame, the rockets will undergo Lorentz contraction. Because they are both contracting, the distance between the rockets will INCREASE and the rope will break. This makes it appear to the oberserver that the rockets are getting farther apart. This does not match what an oberserver on the back rocket sees. To him (the oberserver on the back rocket), the rockets are undergoing the same acceleration starting from the same initial velocity at the same time. How can the front rocket be accelerating away?

Can anyone explain?

If the two rockets are, in fact, getting farther apart, this would seem to imply that putting two sets of thrusters on a rocket would doom the rockets to be torn apart if both thrusters are turned on.


Bell wrote this up in his book, Speakable and Unspeakable in Quantum Mechanics, p.67, in the chapter titled How to Teach Special Relativity. He says the rope breaks. The sci.physics.relativity faq (http://www2.corepower.com:8080/~relfaq/spaceship_puzzle.html) seems to agree.

SeanF
2001-Oct-23, 11:30 AM
I had posted a response to this just before the board was cleared before, but I'll try again.

I think the issue with this question ties into the SR 'paradox' of simultaneity. When we say that "the rockets are undergoing the same acceleration starting from the same initial velocity at the same time," what that really means is that the two rockets will reach any given velocity (say, 0.5c) at the same time.

However, at that time the rockets are both moving relative to an observer in the original frame. This means that those two events (Rocket A reaching 0.5c and Rocket B reaching 0.5c) cannot be simultaneous for both an observer in the original frame and an observer on one of the rockets.

If the observer in the inertial frame sees the rockets reach that speed simultaneously, then an observer on one of the rockets would see the front rocket reaching that speed before the back rocket.

In other words, all observers see the rope break. The 'stationary' observer sees it break due to the Lorentz contraction. The rocket-bound observers see it break because the front rocket is accelerating away from the back rocket.

Now, if the acceleration were done such that the rocket-bound observers see the simultaneity, then the 'stationary' observer sees the back rocket accelerate faster, thus shortening the distance between the rockets sufficiently to prevent the Lorentz contraction from snapping the rope.

The real question is, if the rockets' engine thrust was pre-programmed, who would see the accelerations as consistent, the 'stationary' observer or the rocket-bound observer(s)?

In the case of dual rockets on a single ship, it would have to be done such that the ship-bound observers see simultaneous acceleration and the 'stationary' observer sees the back of the ship accelerating sooner (or "faster," or however you want to term it). Thus all observers would agree that the ship does not get torn apart.

"Don't worry, she'll hold together! You hear me, girl? Hold together."
-- Han Solo, Star Wars


_________________
SeanF

<font size=-1>[ This Message was edited by: SeanF on 2001-10-23 07:31 ]</font>

Diogenes
2001-Oct-23, 01:27 PM
And I ask again, in the context of the puzzle, 'Why does the rope break?'?...

GrapesOfWrath
2001-Oct-23, 01:52 PM
On 2001-10-23 07:30, SeanF wrote:
Now, if the acceleration were done such that the rocket-bound observers see the simultaneity,


I don't think it would be possible for observers on both ships to see that, would it?

(the word "think" added to post 10/23/2001)

Diogenes

In the frame of the observer not moving along with the ships, the ships stay the same distance apart. The rope contracts, and so breaks.
_________________
rocks

<font size=-1>[ This Message was edited by: GrapesOfWrath on 2001-10-23 10:06 ]</font>

SeanF
2001-Oct-23, 02:28 PM
On 2001-10-23 09:27, Diogenes wrote:
And I ask again, in the context of the puzzle, 'Why does the rope break?'?...


To the observer 'left behind,' the rope breaks because the ships and the rope itself experience Lorentz contraction, but the centers of the ships remain the same distance apart. So, the distance between the 'close ends' of the ships increases as the length of the rope is decreasing, and the rope breaks.

To an observer on one of the ships, the rope breaks because the front ship is accelerating at a faster rate than the back ship, so the distance between the ships increases and the rope breaks.

SeanF
2001-Oct-23, 02:31 PM
On 2001-10-23 09:52, GrapesOfWrath wrote:

I don't think it would be possible for observers on both ships to see that, would it?

Diogenes


I thought about this myself, and to be honest I'm not sure. I think it would be possible if the acceleration was done 'just right,' but I haven't actually tried to work out the specific math yet . . . maybe this evening! /phpBB/images/smiles/icon_smile.gif

Iain Lambert
2001-Oct-23, 02:34 PM
My SR is pretty rusty; can I just check that this 'paradox' boils down as so many do to the idea of whether or not two seperated events are considered to actually be simultaneous? I remember taking a rule of thumb where if events are seperated in space, then at least one frame is going to decide that they are seperated in time as well.

Or have I missed something? I remember some of the previous debate on this, but as that was mainly stuff that had me convinced that the rope didn't break I'm really not sure!

SeanF
2001-Oct-23, 02:55 PM
On 2001-10-23 10:34, Iain Lambert wrote:
My SR is pretty rusty; can I just check that this 'paradox' boils down as so many do to the idea of whether or not two seperated events are considered to actually be simultaneous? I remember taking a rule of thumb where if events are seperated in space, then at least one frame is going to decide that they are seperated in time as well.


Iain,

I think you're right that it boils down to simultaneity -- in fact, that was my original response up above! /phpBB/images/smiles/icon_biggrin.gif

BTW, I think it's more correct to say that if two events are separated by space, then there can be no more than one reference frame which considers those events to be simultaneous, and all other frames will consider them to be non-simultaneous.

Three notes:

1. There is presumably some limit to the measurement of time which could cause a range of observers in different inertial frames to all say the events were simultaneous 'within the limits of measurement,' but in that case those inertial frames are probably not measurably distinguishable from each other, either.

2. Sufficiently separated events could be non-simultaneous in all reference frames, with not even one 'simultaneity-seeing' frame.

3. Events occuring at the same point in space will have their simultaneity agreed on by all reference frames. However, since it's impossible for two events to occupy the exact same point in space-time, such events will always be non-simultaneous.

Diogenes
2001-Oct-23, 02:59 PM
Again, as was brought up in the original post, the 2 ships and the rope would be a single object, that would contract uniformly.

I still see no reason for it to break.

If it breaks, and you want to ignore the physics of why and the result of it breaking (for the sake of argument) and you still want to say that each ship is still accelerating equally/uniformly, then each ship would appear (to a detached observer) to contract individually, with the distance between them growing, giving the illusion that the rocket in front is moving away (thus accelerating) away from rocket number 2.


But!! My last statement would not be true, because.. As the 2 rockets accelerate, their mass and gravitational attraction would increase, drawing them together, by the same amount, that the contraction would be increasing the distance between them! So, no observer, in any reference system, would ever see the gap between the 2 rockets increase!

Irishman
2001-Oct-23, 04:17 PM
This still doesn't make sense to me. They start from the same velocity, fire thrusters at the same time, and travel at the exact same acceleration. At any given instant, won't they be in the same ref frame? Which isn't the same as the observer frame.

Unless you're saying that in order for the third party observer to see them launch at the same time, one actually has to launch earlier than the other. But the wording of the original statement is then ambiguous and misleading.

It's not a paradox, it's a poorly worded and thus misleading statement.

Diogenes
2001-Oct-23, 04:51 PM
I agree Irishman,

The whole thing, is more like an obtuse puzzle, where some of the facts are hidden or revealed after the fact.

There is no paradox, because all observations from any frame of reference, can be accounted for, if the principles of physics and the theory of relativity are adheared to.

again from the original problem as stated:
'To an oberserver in the inertial frame, the rockets will undergo Lorentz contraction. Because they are both contracting, the distance between the rockets will INCREASE and the rope will break.'

Why doesn't the rope contract also?

There are not '2 rockets' and a 'rope'.. There is 1 object, whose major components consist of 2 rockets tied together with a rope.



<font size=-1>[ This Message was edited by: Diogenes on 2001-10-23 12:59 ]</font>

SeanF
2001-Oct-23, 05:29 PM
Diogenes,

Consider three rockets with no rope. One rocket is sitting (stationary) one light-hour away from you. Another is sitting two light-hours away from you. The third is sitting three light-hours away from you. (Measurements are made to the centers of the rockets.)

One of the rockets begins accelerating at a constant rate, and after one hour of acceleration has reached a velocity of 0.5c. During that hour, the rocket would have traveled 0.25 light-hours, correct? (Average velocity of (0.5 + 0 ) / 2 = 0.25c for one hour).

If all three rockets experienced this acceleration simultaneously, then the rockets would now be at distances of 0.75 light-hours, 1.75 light-hours, and 2.75 light hours, right? They have each travelled 0.25 light-hours, and they are all still one-light hour apart.

But what happens if you replace the middle rocket with a rope two light-hours long, connected at the centers of the two rockets. After that hour, Lorentz contraction would reduce that rope to only 1.73 light-hours long, but the centers of the ships are still 2 light-hours apart.

Why would connecting the two rockets with a rope suddenly cause the Lorentz contraction to be centered between them rather than on each one individually?

Diogenes
2001-Oct-23, 06:08 PM
SeanF,
Once more, there are not 3 objects.. There were three objects, but they are all connected together now.. This one object, made up of 2 rockets and a rope, will contract uniformly in the direction of travel.

If the rope breaks, it will be because of some force, independant of the initial conditions causes it to do so. What is that force? It CANT be the 2 rockets contracting independently and stretching the rope, because of the reason I stated above.

If you want to ignore that, and say O.K. the rope broke, now we have 2 rockets contracting independently, they still will not appear to move apart because:

" As the 2 rockets accelerate, their mass and gravitational attraction would increase, drawing them together by the same amount, that the contraction would be increasing the distance between them!"



<font size=-1>[ This Message was edited by: Diogenes on 2001-10-23 14:12 ]</font>

Tim Thompson
2001-Oct-23, 06:32 PM
The confusion comes from how the paradox is posed. We are told that the "distance" between the rockets must remain constant. But "distance", according to whom? The observer "at rest" in the laboratory, or the observer "at rest" with respect to the rockets? The answer is: the observer "at rest" in the laboratory. This is the constraint that breaks the rope (and would snap a rigid body in two).

Now, suppose you saw a rigid body (space ship, meter stick, whatever; call it a "rod"), whiz by at a really fast speed. What would you see? The rod would Lorentz contract, and appear to you to be shorter. Now consider the space ships in this problem as if they were connected by a rigid rod, instead of a rope, and treat them as one extended body. what should you observe? You should observer Lorentz contraction of the whole shebang. What do you observe? You do not observe Lorentz contraction of the whole shebang, because the paradoz has been posed with a constraint that prevents it: the distance between the ships is constant as seen by the observer at rest in the laboratory. By definition.

So, what condition must prevail in the "rest frame" of the "moving" rocket ships? What condition must prevail, in the case of a "rigid" rod, if it is allowed to whiz by close to the speed of light, but not be observed as Lorentz contracted? The rod must be seen, in its own "rest frame", to be expanding. That's the only way to negate the Lorentz contraction, as the "paradox" has been posed.

Now does a rope, or a rigid rod, between the space ships, have the ability to expand? We presume not, and so the rope, or the rod, must snap. In fact, no non-expandable "rigid" body can behave as required by the paradox (move without showing Lorentz contraction), and so even if it were one solid piece, it would still be required to break, under the conditions of the paradox.

Now try another view of things. If the rocket ships are carrying rocket ship meter sticks, the observer at rest in the laboratory will watch those sticks, see them "Lorentz contract", and correctly deduce that a "meter" as seen by the rocketeers looks smaller than his own meter. The distance between the ships is constrained to remain constant, as measured by laboratory meter sticks. But the laboratory obsever sees constantly "shrinking" rocketeer meter sticks. So, he deduces that the distance between the ships, as measured by rocketeer meter sticks, must be increasing. So you can easily see that, if the rocketeers see an increasing distance, and it's a rope not a spring, it's going to break (as would a rigid rod, so long as the rockets are stronger than the tensile strength of the rod).

The key to all this is that the distance between two points cannot ever be constant in both the "at rest" and the "moving" reference frames. It can be constant in one or the other, but never both. We would normally expect the distance to be constant in the reference frame that is at rest with respect to the rockets (the co-moving frame), as would be the case for the ordinary length of a rigid rod. But the paradox has been posed with the eccentric condition that the distance remain fixed in the laboratory frame (which is the Lorentz transformed frame). That eccentric & unexpected constraint is what causes the confusion.

SeanF
2001-Oct-23, 06:50 PM
On 2001-10-23 14:08, Diogenes wrote:
SeanF,
Once more, there are not 3 objects.. There were three objects, but they are all connected together now.. This one object, made up of 2 rockets and a rope, will contract uniformly in the direction of travel.


I have to respectfully disagree with this. I don't think Lorentz contraction works quite this way.

I simply can't accept that two individual rockets accelerating simultaneously would maintain a constant distance between their centers, but two rockets with a piano wire connecting them would get closer together.



" As the 2 rockets accelerate, their mass and gravitational attraction would increase, drawing them together by the same amount, that the contraction would be increasing the distance between them!"


Again, I don't think so. Once they stopped accelerating, the Lorentz contraction would stop contracting them, but any gravitational pull would still be pulling them together. Lorentz only contracts them while they are accelerating, but gravity pulls on them all the time, so I don't think they can counteract each other like this.

SeanF
2001-Oct-23, 06:57 PM
On 2001-10-23 14:32, Tim Thompson wrote:
The confusion comes from how the paradox is posed. We are told that the "distance" between the rockets must remain constant. But "distance", according to whom? The observer "at rest" in the laboratory, or the observer "at rest" with respect to the rockets? The answer is: the observer "at rest" in the laboratory. This is the constraint that breaks the rope (and would snap a rigid body in two).


This isn't really correct, Tim -- the original post states:



Two rockets, aligned head to tail, are connected with a rope from the tail of Rocket 1 to the head of Rocket 2. Both rockets are at rest wrt each other and the rope. At time t = 0 (synchronized clocks) in the rest frame of the rockets, they both fire their thrusters (single thrusters at the tail of each rocket) such that they accelerate at 1g.


The question that needs to be addressed is:

Given this situation, in which reference frame (if any) would the rockets maintain a constant distance between them?

If it's the 'lab' frame, then the rope breaks. If it's the rockets' frame(s), it doesn't.

GrapesOfWrath
2001-Oct-23, 07:21 PM
On 2001-10-23 12:17, Irishman wrote:
This still doesn't make sense to me. They start from the same velocity, fire thrusters at the same time, and travel at the exact same acceleration. At any given instant, won't they be in the same ref frame? Which isn't the same as the observer frame.
They will not be in the same reference frame. By that, I assume that you mean the same inertial reference frame. One is farther down the "gravity well."


Unless you're saying that in order for the third party observer to see them launch at the same time, one actually has to launch earlier than the other. But the wording of the original statement is then ambiguous and misleading.
No, that is not the intended meaning. The meaning can be taken on the face of it. The confusion arises only because the result is not intuitive.


It's not a paradox, it's a poorly worded and thus misleading statement.

I disagree. The way Bell phrased it in his book (p.66), is "Let ships B and C be identical, and have identical acceleration programmes." I'd say that is equivalent to Joe's interpretation, wherever he got it. Bell had a "fragile thread" connecting the two ships instead of a rope. The analysis has nothing to do with the mass, or "changing" mass, of the ships.

Irishman
2001-Oct-23, 09:36 PM
I'm still having trouble here.

Tim Thompson said:

The confusion comes from how the paradox is posed. We are told that the "distance" between the rockets must remain constant. But "distance", according to whom? The observer "at rest" in the laboratory, or the observer "at rest" with respect to the rockets? The answer is: the observer "at rest" in the laboratory. This is the constraint that breaks the rope (and would snap a rigid body in two).

and SeanF said:

Given this situation, in which reference frame (if any) would the rockets maintain a constant distance between them?

I think this is the confusion: what is the specific statement of the question? Using the version at the OP of this thread, SeanF's question is the relevant one: which reference frame shows them remain the constant distance? If they are accelerating together at 1 g, I think that constrains the ref frame in question to be theirs, not the "stationary" observer's.

GrapesOfWrath said:

They will not be in the same reference frame. By that, I assume that you mean the same inertial reference frame. One is farther down the "gravity well."

What gravity well? (That's a serious question.) I did not see in the OP any description implying they are under different gravity constraints. I read the OP to assume they are in free space, no planets around. They both accelerate under 1 g. If they are taking off from a planet (from different heights?), then yes the gravity is different, but I would think they would not be accelerating under 1 gravity, as the planet would be subtracting from that. No?


Irishman: "Unless you're saying that in order for the third party observer to see them launch at the same time, one actually has to launch earlier than the other. But the wording of the original statement is then ambiguous and misleading."

No, that is not the intended meaning. The meaning can be taken on the face of it. The confusion arises only because the result is not intuitive.

I disagree. Everyone here appears to be citing different versions of the question and interpreting them differently. Confusion is definitely arising from (1) non-consistent source - this OP, or Bell's original statement? (2) the phrasing being confusing; (3) the situation relying on relativity, and thus being non-intuitive.


I disagree. The way Bell phrased it in his book (p.66), is "Let ships B and C be identical, and have identical acceleration programmes." I'd say that is equivalent to Joe's interpretation, wherever he got it. Bell had a "fragile thread" connecting the two ships instead of a rope. The analysis has nothing to do with the mass, or "changing" mass, of the ships.

Again, I say that if Joe's version is an accurate description of Bell's version, then I do not see how the stationary observer's viewpoint is the one where they have a constant distance between them. Either they are accelerating the same or they are not. Bell's wording says "identical acceleration programmes", which allows if there is a planet involved the difference of the gravity well you mention and my earlier point that they would not be accelerating the same, regardless of their engine output being the same.

See, ambiguous wording. Please explain how if there is no planet (as does not appear in this thread's OP) and they are accelerating identically, how are they moving apart from each other from either rocket's viewpoint? Explain that, and maybe the rest of it will make sense.

I do understand how if the rockets are staying the same distance apart from the stationary observer's viewpoint they are not from their own, I just don't see how from the wording of the problem that is the correct condition that applies.

2001-Oct-24, 12:33 AM
On 2001-10-23 09:27, Diogenes wrote:
And I ask again, in the context of the puzzle, 'Why does the rope break?'?...


The tension on the rope is what breaks the rope. Only a force or some quantity that has information on the force can be a cause that is invariant to observer. Write the tension in covariant form, any covariant form (generalized force 4-vector, covariant Hamiltonian, etc.) and you will have a description of what happens to the rope.

You guys are discussing the kinematics, which is just mathemtics. The physical use of SR is in its dynamics. SR does allow forces. Use them!

Diogenes
2001-Oct-24, 12:38 AM
SeanF,You seem to be ignoring the theory, that, the objects only appear to contract to a detached observer.

That is one of the problems with the breaking rope. Since nothing 'really' contracts, the rope will not break, so how can the observer see it break?

2001-Oct-24, 11:21 AM

2001-Oct-24, 11:34 AM
The tension breaks the rope, fellas.

You guys are talking about the kinematics.
By definition, the kinematics do not contain forces. However, a rope is physical. It holds things together by forces. The force exists
irregardless of kinematics.

The force is the only thing that is
physical here. A measurement can't be made
without forces.

Another point. Absolutely rigid bodies are forbidden in relativity. The tension moves away from the rockets at a speed less
than the speed of light. In fact, the tension will move at the speed of sound
in the rope, as seen in the pseudo-frame of the rockets. The rope has to stretch a little before breaking. It needs time for the tension to distribute itself.

The rope is itself a type of field. By
that I mean "a region of space where
a body experiences a force due to
other bodies." The strength of the field is the force divided by a physical property of the body that is affected by the field.

So in this case, I would define the "field"
as the rope and "strength of the field" as the tension divided by the mass density of the rope. One can't use a rigid body approximation here, but one can use a force
law that is covariant. However, there must
be an appropriate "field equation" for the rope. The tension at every point on the rope at every time can be calculated using the field equation.

The circular logic of kinematics is always broken by the dynamics. In any theory. Forces are important.

2001-Oct-24, 11:48 AM
That is one of the problems with the breaking rope. Since nothing 'really' contracts, the rope will not break, so how can the observer see it break?
[/quote]

Some observers see the two spaceships
part. Some observers see the rope contract.
There is always a "reason," appropriate to each observer. Unstated is the fact that the observers can only make measurements by perturbing forces or (equivalently) fields. If they look at it using some type of light, the light exerts radiation pressure. It also
pushes electrons around which is why the rope reflects light. But we ignore the forces in the measurement by calling the device "an observer." "An observer" is
a device that exerts forces that result in a measurement.

All observers agree that the rockets are emitting a plume. The rocket engine has a thrust measured in Newtons or pounds. The
rope has a tension measured in Newtons or pounds. All the observers may see the rope start to stretch, the rope start to fray, and rope breaking. I

An observer in the time-like region will see the rockets turn on and then the rope breaks. If the observer is far enough away, in the space-like region, he may see the rope break and the rockets turn off later.
Even the sequence of events is ambiguous.
One can't assign a "cause" that is observer independent if the sequence of events is observer dependent.

The only thing both observers will agree on is that the rope showed signs of tension. The breaking is a sign of tension. It was the tension that did it.

SeanF
2001-Oct-24, 11:50 AM
On 2001-10-23 20:38, Diogenes wrote:
SeanF,You seem to be ignoring the theory, that, the objects only appear to contract to a detached observer.

That is one of the problems with the breaking rope. Since nothing 'really' contracts, the rope will not break, so how can the observer see it break?


Well, the observer sees it break because it does break. If the rocket-bound observers perceive that the front rocket is accelerating faster than the back rocket, they would expect to see the rope pulled apart -- the rope breaks.

However, because of the simultaneity of SR, this can happen in such a way that the stationary observer sees the rockets maintaining the exact same distance between them. Yet, the rope breaks, so this observer must see the rope break. To what cause does he ascribe the rope breaking even though the distance between the rockets remained constant? Lorentz contraction.

SR predicts that the stationary observer will measure a shorter distance between the rockets than the rocket-bound observers. The only way to answer the question, "Does the rope break?" is to first answer the question, "Which observer sees the distance remain the same?"

Scenario One:
If the rocket-bound observers sees the distance remain the same, the stationary observer would then see the distance contract. The rocket-bound observers see no reason for the rope to break, and the stationary observer sees the distance contract at the same ratio as the rope itself is Lorentz-contracted, so he also sees no reason for the rope to break. It doesn't break.

Scenario Two:
If the stationary observer sees the distance remain the same, then the rocket-bound observers would see the distance increase. The rocket-bound observers therefore see a force being applied to the rope, which must break. The stationary observer also sees the rope break, but the distance between the rockets remained constant. Therefore, it must have been the Lorentz contraction of the rope which caused it to break.

My personal opinion? Given preprogrammed identical acceleration rates on the two rockets, the rope breaks. Once the rockets start accelerating, they would no longer see each other as accelerating equally due to the simultaneity problem with separated events.

Diogenes
2001-Oct-24, 12:00 PM
Rosen,
Why does the tension on the rope increase if both rockets are accelerating with uniform velocity/accelleration?

If you say, because of the contracting rockets, you are not addressing the fact that the two rockets connected by a rope is one (1) object!


What if you had 2 rockets with a length of rebar welded to the front of one and the rear of another?

What if you had a silo with one rocket engine near the front and one near the end, and a mural painted on the side depicting two rockets connected with a piece of rope? Would the rope break? /phpBB/images/smiles/icon_biggrin.gif

<font size=-1>[ This Message was edited by: Diogenes on 2001-10-24 08:11 ]</font>

GrapesOfWrath
2001-Oct-24, 12:30 PM
On 2001-10-23 17:36, Irishman wrote:
What gravity well? (That's a serious question.) I did not see in the OP any description implying they are under different gravity constraints. I read the OP to assume they are in free space, no planets around. They both accelerate under 1 g.

Correct, there are no planets involved. Or the gravity of the mass of the ships. When I mentioned "gravity well" I deliberately used the quotation marks. I thought that it might be easier to visualize the scene in the context of a gravity well, but I should have included more detail. The two ships are both experiencing one g (in Joe's OP; in Bell's formulation, he is more general and says they have "identical acceleration programmes") and so can be considered to be in a large gravity well, produced by a fictitious mass. That is one way of looking at it, but it is consistent with the other answers--that the two ships cannot remain in a single inertial reference frame.



On 2001-10-23 17:36, Irishman wrote:
Everyone here appears to be citing different versions of the question and interpreting them differently. Confusion is definitely arising from (1) non-consistent source - this OP, or Bell's original statement? (2) the phrasing being confusing; (3) the situation relying on relativity, and thus being non-intuitive.

(1) I don't see the two versions (Joe's and Bell's) as inconsistent. Bell's is just a generalization of Joe's. Joe's has the "acceleration programme" at a constant one g. (2) I personally don't find the phrasing confusing, in the problems themselves. I do see a lot of confusion arising from interpretation of the answers, though. (3) That the result is non-intuitive was just what I said. Bell even mentions that he presented the problem to the physicists at the CERN canteen (they took a poll!), and almost everyone got it wrong. So, we're in good company.

2001-Oct-24, 12:46 PM
[quote]
On 2001-10-24 08:00, Diogenes wrote:
Rosen,
Why does the tension on the rope increase if both rockets are accelerating with uniform velocity/accelleration?

The rope broke because the rockets
each had a certain amount of thrust.
Thrust is also a force. You can discuss forces without worrying about simultaneity.
You can't discuss shrinking or moving without worrying about simultaneity.

The rockets turned on, the thrust passed into the rope becoming tension, it
traveled through the rope on sound waves, and the tension reached a maximum when it broke. Foward, backward, stationary frame, pseudo-frame. The event that the rope broke occurs simultaneous and at the same position as the tension reaching a maximum. To everybody. No question of simultaneity. There is more than one way to measure
tension and, if done right, they will all agree.

In fact, it is only because of the thrust that the rockets are not in an inertial frame. Remember, an stationary frame is one where both Newton's laws and Maxwells equations apply (Einstein 1905). Inertial frames are defined relative to the stationary frame (i.e., they go at a constant velocity relative to the stationary frame). The definition of inertial frame, even the definition of straight line in four dimensional space, requires that there be no force on the body. So Einstein managed to slip forces into the definition of the kinematics. This stuff about "space-time
warping" is merely a statement that "there are certain forces that I am ignoring because they confuse the issue."

For an observer in the pseudo-frame of the rocket, the third law of Newton doesn't apply. His g-force has no reaction in his pseudo-frame, so it is not a true inertial frame. However, it is the thrust of the rockets that made the observer go into a noninertial frame.

My experience with certain people has taught me that one should not ignore the dynamics entirely when explaining relativity. The kinematics is not sufficient to describe any real system, and even a layman can sense it. Some of them start ranting and raving against scientists, but that is another issue. Some of the forces, those necessary for measurement, are hidden in a broad category called "the observer." But forces are there too.

<font size=-1>[ This Message was edited by: Rosen1 on 2001-10-28 20:29 ]</font>

SeanF
2001-Oct-24, 12:49 PM
On 2001-10-24 08:00, Diogenes wrote:
Rosen,
Why does the tension on the rope increase if both rockets are accelerating with uniform velocity/accelleration?

If you say, because of the contracting rockets, you are not addressing the fact that the two rockets connected by a rope is one (1) object!


What if you had 2 rockets with a length of rebar welded to the front of one and the rear of another?

What if you had a silo with one rocket engine near the front and one near the end, and a mural painted on the side depicting two rockets connected with a piece of rope? Would the rope break? /phpBB/images/smiles/icon_biggrin.gif

<font size=-1>[ This Message was edited by: Diogenes on 2001-10-24 08:11 ]</font>


/phpBB/images/smiles/icon_biggrin.gif

Actually, Diogenes, these are all the same things. Even in the last case, with the silo with two rocket engines, if the front engine accelerates faster than the back, it would tear the silo apart (yup, that painted-on rope would break). The stationary observer could see the two ends of the silo stay the same distance apart, but he would see Lorentz contraction of the whole shebang -- essentially contracting towards the ends and pulling the middle apart.

Whether this occurs in the exact middle of the silo or not would depend on exactly where the engines are located and how they're firing . . . each molecule of the silo is accelerated differently and may be in different inertial frames at different times.

GrapesOfWrath
2001-Oct-24, 12:59 PM
On 2001-10-24 08:00, Diogenes wrote:
What if you had a silo with one rocket engine near the front and one near the end, and a mural painted on the side depicting two rockets connected with a piece of rope? Would the rope break? :D

Sorry, I missed this one. This seems to be just a reconstruction of the original problem with a more "substantial" rope. I've mentioned that Bell used a "fragile thread" instead of Joe's rope. The key observation is that the engines are producing the same accleration for observers with the engines. So, the silo breaks.

In a real silo launch, the silo captain makes subtle control tweaks to make sure that the silo reaches its destination intact.

2001-Oct-24, 02:24 PM
On 2001-10-24 08:59, GrapesOfWrath wrote:
[quote]
Sorry, I missed this one. This seems to be just a reconstruction of the original problem with a more "substantial" rope. I've mentioned that Bell used a "fragile thread" instead of Joe's rope. The key observation is that the engines are producing the same accleration for observers with the engines. So, the silo breaks.



May I point out that in Newtonian mechanics, without Lorentz contraction, the rope may break also? The rope has inertia. So the rocket in the back pushes on the portion of rope nearest it and creates a region of compression. The rope in front pulls the portion nearest it and stretches it into a region of rarification. A sound wave which is Galilean invariant may have enough tension to break the rope. Even though the distance between rockets is constant in ALL reference frames.

Now, this is slightly different from
the SR case. The rope ALWAYS breaks because the sound wave has to be Lorentz invariant, and is not Galilean invariant. The average tension on the rope in SR is positive, not zero as in the case Newtonian case. So I am not saying that the picture of a rocket pulling on a rope is perfectly clear. However, due to SR, the equation of force on the rope can't be a simple Hookes Law type of force, it has to take a more complex form that fits the constraints of the theory. So
the real question is: what type of force law is both covariant and consistent in the "v<<c" region with simple Hooke's Law type ropes.

The rope wouldn't break unless there was thrust on the rockets. In any inertial frame. Therefore, I vote for the thrust, and the resulting tension, as being the "cause" of the rope breaking.

robert_d
2001-Oct-24, 03:35 PM
Hang in there Diogenes and Irishman, I still think you are right. Although others talk about "forces" - if the forces are applied equally the rope can't break. The point of strongest persuasion in the origianl thread was that the distances are not "really" changing it is just a measurement issue. And the Time of occurance MUST change in other reference frames.



<font size=-1>[ This Message was edited by: robert_d on 2001-10-24 11:37 ]</font>

GrapesOfWrath
2001-Oct-24, 04:00 PM
robert_d

You mean, if the forces of each ship's engines are applied equally?

In that case, it seems it is clear that the rope does break, no? This is strictly a thought experiment dealing with relativity. In a real experiment, slight errors would be very important. In this experiment, we are free to assume that each ship follows the acceleration programme exactly.

SeanF
2001-Oct-24, 04:00 PM
On 2001-10-24 11:35, robert_d wrote:
Hang in there Diogenes and Irishman, I still think you are right. Although others talk about "forces" - if the forces are applied equally the rope can't break. The point of strongest persuasion in the origianl thread was that the distances are not "really" changing it is just a measurement issue. And the Time of occurance MUST change in other reference frames.

<font size=-1>[ This Message was edited by: robert_d on 2001-10-24 11:37 ]</font>


Robert,

You said "if the forces are applied equally, the rope can't break." That's not entirely true.

In order to apply the forces "equally" they have to be applied at the same time. That's an issue in SR. If the forces are applied equally in the 'stationary' observer's reference frame, then they are not applied equally in the rocket-bound observers' reference frames.

In fact, it is impossible for the forces to be applied equally in all reference frames.

If, in any reference frame, the front rocket accelerates faster than the back rocket, the rope will break.

There can be an observer, moving relative to the rockets, who sees the two rockets accelerating exactly the same as each other. However, the rope breaks, and this observer has to see the rope break (SR says different observers can disagree on when an event occurs, but they will agree that the event did, in fact, occur). Therefore, it is possible for an observer to see the forces being applied to both rockets equally and simultaneously, but still see the rope break.

And as far as the distances not really changing, just measurements, that may be true -- but the bottom line is that while the observers disagree on the distance between the rockets, and the observers disagree on the length of the rope, they all agree that the distance between the rockets ends up greater than the length of the rope and the rope breaks. All they disagree on is whether the rope got shorter or the rockets got farther apart.

2001-Oct-24, 04:08 PM
[quote]
<Although others talk about "forces" - if the forces are applied equally the rope can't break.>

Why does it break in the steady state Newtonian case? The tension is defined as the force between two adjacent line segments. The tension is constant throughout the rope. However, consider two small segments of rope right next to each other, right in the middle. One segment has a net force from the right side the rope pointed to the right, and the other has a net force from the left side of the rope that is the pointed to the left. Why don't the two segments fly apart? Because the force between the two segments exactly balances the forces from each side of the rope. This force has a maximum. When the maximum is exceeded, the two segments fly apart. According to Newton and according to Einstein. Einstein merely chose his coordinate system to simplify the forces
(i.e., his "field equations")

>The point of strongest persuasion in the origianl thread was that the distances are not "really" changing it is just a measurement issue. <
No, but the rope really breaks. That is not a measurement issue. This is an event.
An event is a a feature that occurs at a point in 4-D space-time, or at least that is one way to analyze it.

>And the Time of occurance MUST change in other reference frames.
However, the event in 4-D space-time doesn't change. The rope parts at the same 4-D point where the tension reaches a maximum.
In other words, one feature occurs at the same 4-D point as another feature. In all reference frames.


I think the other points of view are valid. Maybe even ALL of them. That's the problem. One usually thinks of a "cause" as unique. As in "the real cause."





<font size=-1>[ This Message was edited by: Rosen1 on 2001-10-28 20:07 ]</font>

2001-Oct-24, 04:33 PM
There is another way to look at it another way to look at it. In Newtonian mechanics, it is permissable for the force of tension to be the same in every inertial frame. However, this is not allowed in SR. There has to be a velocity dependent force for every static force. When the velocity of the particle is zero, the velocity dependent part is zero.

An example is the magnetic field at zero velocity that appears as a electric field for nonzero velocities.

Think of this in terms of ropes. The tension is not velocity dependent. According to SR, there has to be another type of force, between adjacent parts of the rope, that is velocity dependent. It is zero if the rope isn't moving but is nonzero when the rope is moving. This force would resemble tension for a moving rope. Together with "tension," we have "ropinism."


So for the observer in the stationary frame, where the distance between space ships is constant, the rope is moving so fast that it breaks because of the force of "ropinism," not tension.

2001-Oct-25, 01:59 AM
On 2001-10-23 09:27, Diogenes wrote:
And I ask again, in the context of the puzzle, 'Why does the rope break?'?...


I'll make it clear why the concept of
"shrinking" alone won't help. Suppose that
I program the computer of each rocket ship to move so that the distance between the front of the rocket in the back is a constant distance from the back of the rocket in the front as measured in the stationary plane. This means that the acceleration of the center of mass of one
rocket is very slightly different than that of the other rocket, but the difference is irelevent. The rope stretches from the front of the back to the back of the front. In
the stationary frame, its measured length is constant.

The rocket engines rev up so the two
rockets eventually move at a speed close to that of light. The zero tension length of the
rope shrinks in the stationary frame, However, the two rockets keep
the same distance. So the two rockets end up
stretching the rope. The tension rises above
zero. When the tension is high enough, the rope breaks. Even though the measured length
of rope was constant in the stationary frame. The two rockets were moving away from each other in the pseudo-frame of
the rocket, but as far as the stationary plane goes there was no change in length. Only in tension.

Since in our experience tension is
independent of velocity, we usually don't think of a rope as a dynamic object. That is allowed in Newtonian mechanics. However, a velocity dependent tension is also allowed in Newtonian theory. Newtonian theory doesn't restrict the forms of the forces in any way.

SR does not allow forces that are strictly independent of velocity. If a velocity independent force existed, and Newton's Laws held true, and Maxwell's equations held true (where there are velocity dependent forces), one could make an experiment to figure out how one is moving. Even in a closed system. Therefore, SR does not allow a force law totally independent of velocity.

Let us restrict our POV to the stationary frame. If the rope is pulled taunt by a velocity dependent tension, there must also be a force exerted by the rope which is velocity dependent. It is zero at zero speed, and increases with velocity divided by the speed of light. This force is to tension like the magnetic field is to the electric field.

The rope isn't changing length in the stationary frame. However, the force pulling apart the rope increases due to this velocity dependent force. If the observer is travelling in one of the space ships, the measured velocity of the rope is nearly zero so this force is nearly zero. However, due to effects on the internal forces holding together the meter stick that is in the rocket, the observer in the rocket thinks that the distance between rockets is growing. The forces holding the meter stick together also has a velocity dependent part. SR doesn't allow any old force, it has to satisfy all the conditions of SR.

So I say that only the forces can be described as "the real cause" of anything. However, in addition to twisting space and time, SR restricts the possible forces in the universe. Every velocity dependent force field, in the stationary frame, has associated with it another force field.

In analogy with electromagnetic theory, I am developing a tensoropetic theory /phpBB/images/smiles/icon_smile.gif
There are two fields represented by the rope. One with a velocity independent tension, and one with a velocity dependent force. There are field equations connecting them, a rule that the change in tension can't move faster than the speed of light in a vacuum, and in fact analogs to all the dynamics described by Einstein.

With regard to forces, what is the same for each inertial frame is the restrictions on forces as defined by SR. SR is useless without forces. Without forces, the twin paradox IS a paradox. The dynamics of SR are always very important, even when they are refered to indirectly.

GrapesOfWrath
2001-Oct-25, 03:04 AM
On 2001-10-24 21:59, Rosen1 wrote:
With regard to forces, what is the same for each inertial frame is the restrictions on forces as defined by SR. SR is useless without forces. Without forces, the twin paradox IS a paradox.


I guess we'll have to ask Bill Clinton just exactly what a paradox is, but, barring that, where do you see the force in this description (http://mentock.home.mindspring.com/twins.htm) of the twin paradox?

Nevermind. If you want to pursue that line, let's do it in the twins paradox very own thread (http://www.badastronomy.com/phpBB/viewtopic.php?topic=38&forum=1&0).

Wally
2001-Oct-25, 11:05 AM
In SR, you must have inertial frames of reference. Acceleration takes you into the realm of GR. Now, lorentz contraction is an SR fact, is it not? Let's ignore acceleration and just look at the situation where the ships (tied by the UNBROKEN rope) are travelling at some fraction of c. That WHOLE frame of reference undergoes L.C. from the aspect of the "stationary" observer. Thus, the entire ship/rope/ship thing contracts. The rope does not break. in SR terms, we can just as easily say it's the "stationary" observer who is moving (it's RELATIVE, remember!!!). Obviously, then, the rope between the 2 "stationary" ships DOES NOT BREAK! It sounds to me like many of you are overthinking this thing, or are getting too involved with trying to combine SR with GR.

GrapesOfWrath
2001-Oct-25, 11:43 AM
On 2001-10-25 07:05, Wally wrote:
In SR, you must have inertial frames of reference. Acceleration takes you into the realm of GR.

That is a misapprehension. SR is quite capable of analyzing accelerated systems. In fact, it was such analysis that led Einstein to GR. The key step was to equate such analysis of acceleration with gravitation--the equivalence principle.


Now, lorentz contraction is an SR fact, is it not? Let's ignore acceleration and just look at the situation where the ships (tied by the UNBROKEN rope) are travelling at some fraction of c. That WHOLE frame of reference undergoes L.C. from the aspect of the "stationary" observer. Thus, the entire ship/rope/ship thing contracts. The rope does not break. in SR terms, we can just as easily say it's the "stationary" observer who is moving (it's RELATIVE, remember!!!). Obviously, then, the rope between the 2 "stationary" ships DOES NOT BREAK!

No need to shout. /phpBB/images/smiles/icon_smile.gif

It's not quite as obvious as that. The conditions of the problem have to be satisfied, and you can't satisfy them if you take that approach.


It sounds to me like many of you are overthinking this thing, or are getting too involved with trying to combine SR with GR.


Well, I didn't start it (Joe neither), Bell did (http://math.ucr.edu/home/baez/physics/Relativity/SR/spaceship_puzzle.html).

SeanF
2001-Oct-25, 11:49 AM
On 2001-10-25 07:05, Wally wrote:
In SR, you must have inertial frames of reference. Acceleration takes you into the realm of GR. Now, lorentz contraction is an SR fact, is it not? Let's ignore acceleration and just look at the situation where the ships (tied by the UNBROKEN rope) are travelling at some fraction of c. That WHOLE frame of reference undergoes L.C. from the aspect of the "stationary" observer. Thus, the entire ship/rope/ship thing contracts. The rope does not break. in SR terms, we can just as easily say it's the "stationary" observer who is moving (it's RELATIVE, remember!!!). Obviously, then, the rope between the 2 "stationary" ships DOES NOT BREAK! It sounds to me like many of you are overthinking this thing, or are getting too involved with trying to combine SR with GR.


Wally,

There are really two different situations here; one in which the rope breaks and one in which it doesn't. The rope breaks if the rockets get farther apart in the rockets frame of reference. It doesn't break if they don't.

The question is, if the two rockets are preprogrammed with identical acceleration programs and begin running those programs simultaneously, will they get farther apart in their own frame of reference?

My thoughts on it go like this: if you take just one spaceship with a preprogrammed acceleration program and situate it distance d from the observer and start the program at t=0 and it accelerates such that it reaches velocity v at time t1, then at that time it will be a distance from the observer of d + ( v * t1 ) / 2 (assuming steady acceleration).

If you take that same rocket with the same accleration program but start it at a distance of d + x, then at time t1 it will be at a distance of d + x + ( v * t1 ) / 2.

In other words, the rocket started out a distance of x from where the first rocket started and it ended at a distance of x from where the first rocket ended.

If we do both rockets at the same time, they will maintain a distance of x from each other throughout the entire procedure from the observer's POV.

Therefore, from any POV moving in the same direction as the rockets, they will get farther apart. Don't worry about the rockets' POV at this point, because they're accelerating. However, if you have an observer already moving at velocity v, that observer would see the rockets moving farther apart.

Therefore, if you consider simply one of the rocket's POV at the start (same as the first observer's) and that rocket's POV at the end (same as the moving observer's), the rockets must also see themselves getting farther apart.

Therefore, the rope breaks.

QED.

/phpBB/images/smiles/icon_smile.gif
_________________
SeanF

<font size=-1>[ This Message was edited by: SeanF on 2001-10-25 07:54 ]</font>

Tim Thompson
2001-Oct-25, 07:54 PM
Tim Thompson: "The confusion comes from how the paradox is posed. We are told that the "distance" between the rockets must remain constant. But "distance", according to whom? The observer "at rest" in the laboratory, or the observer "at rest" with respect to the rockets? The answer is: the observer "at rest" in the laboratory. This is the constraint that breaks the rope (and would snap a rigid body in two)."

SeanF: "This isn't really correct, Tim -- the original post states: ... If it's the 'lab' frame, then the rope breaks. If it's the rockets' frame(s), it doesn't."

I agree with the stated "if" conditions. My interpretation that the paradox requires the distance in the lab frame to remain constant, and therefore break the rope, comes from the fact that Baez explicitly says so in his "Bell's Spaceship Paradox (http://www2.corepower.com:8080/~relfaq/spaceship_puzzle.html)", linked in one of the earlier posts.

Quote 2nd paragraph of Baez: "To begin, a statement of the paradox--- and if you notice some ambiguities in my formulation, that's the point! (That's always the point in SR paradoxes.) Bell asks us to consider two rocket ships, each accelerating at the same constant rate, one chasing the other. The ships start out at rest in some coordinate system (the "lab-frame"). Since they have the same acceleration, their speeds should be equal at all times (relative to the lab-frame) and so they should stay a constant distance apart (in the lab-frame). But after a time they will acquire a large velocity, and so the distance between them should suffer Lorentz contraction. Which is it?" End quote.

I have never seen Bell's original, so I am going on Baez's formulation. If baez is wrong, then so am I, but the persistent parenthetical qualifier "in the lab-frame" seems explicit enough.

SeanF
2001-Oct-25, 08:00 PM
On 2001-10-25 15:54, Tim Thompson wrote:

Quote 2nd paragraph of Baez: "To begin, a statement of the paradox--- and if you notice some ambiguities in my formulation, that's the point! (That's always the point in SR paradoxes.) Bell asks us to consider two rocket ships, each accelerating at the same constant rate, one chasing the other. The ships start out at rest in some coordinate system (the "lab-frame"). Since they have the same acceleration, their speeds should be equal at all times (relative to the lab-frame) and so they should stay a constant distance apart (in the lab-frame). But after a time they will acquire a large velocity, and so the distance between them should suffer Lorentz contraction. Which is it?" End quote.



Hmm, that's a rather different set-up and a completely different question than the whole "rockets and ropes" thing!

If the rockets, with their equal acceleration, are maintaining the same distance in the "lab frame", then they are moving apart in the "rockets frame". The fact that they stay the same distance apart in the lab frame is the Lorentz contraction!

2001-Oct-25, 11:29 PM
If the rockets, with their equal acceleration, are maintaining the same distance in the "lab frame", then they are moving apart in the "rockets frame". The fact that they stay the same distance apart in the lab frame is the Lorentz contraction!
[/quote]

In the lab frame, the rope has not contracted. You just said so. The rope has gotten tenser for some reason not covered in pure Newtonian physics.

Wally
2001-Oct-26, 10:32 AM
So, if everything's from the perspective of the lab's frame of reference (as stated a few posts up), and the ships remained a constant (apparent) distance apart from the lab's frame, then in the rocket's frame, they would have in fact moved farther apart, and did not accelerate simutaneously and at the same rate, so the rope breaks. I guess I'm still confused. The original question seems to introduce a paradox which in fact really isn't there, or can't be there, if the facts are taken verbatum, right??? PS. Sorry for shouting in my last post! I'm still somewhat new to SR, but find it incredibly fascinating! I'm even trying to get some of my buddies to learn about it, so I have others (on my same limited level of knowledge!) that I can chat with! /phpBB/images/smiles/icon_smile.gif

SeanF
2001-Oct-26, 11:42 AM
On 2001-10-25 19:29, Rosen1 wrote:



If the rockets, with their equal acceleration, are maintaining the same distance in the "lab frame", then they are moving apart in the "rockets frame". The fact that they stay the same distance apart in the lab frame is the Lorentz contraction!


In the lab frame, the rope has not contracted. You just said so. The rope has gotten tenser for some reason not covered in pure Newtonian physics.


When did I say the rope has not contracted? In fact, I was responding to a posted thought experiment which didn't even include a rope.

Even in the "original" (such as it is) question, "the rope" and "the distance between the rockets" are not the same thing. The rope is a physical object which only experiences measured changes in length due to Lorentz contraction. The distance between the rockets is a relative measurement which can experience absolute changes simply by moving one of the rockets!


_________________
SeanF

<font size=-1>[ This Message was edited by: SeanF on 2001-10-26 09:14 ]</font>

2001-Oct-28, 01:23 AM
When did I say the rope has not contracted? In fact, I was responding to a posted thought experiment which didn't even include a rope.

SeanF

<font size=-1>[ This Message was edited by: SeanF on 2001-10-26 09:14 ]</font>
[/quote]

I said that the the rope didn't contract in the stationary frame. It didn't contract because its actual length is the distance between the two rockets, where the ropes are attached. The length of the rope hasn't really contracted. The hypothetical "normal" length of rope has contracted, but that is a purely mathematical concept. The question is "what broke the rope in the stationary frame." My answer: The tension increased until it reached a breaking threshold. In the rocket frame, the rope broke when the rockets moved further apart. However, why didn't the rope simply stretch to accomodate the distance between rockets? A really elastic rope wouldn't break. My answer, the tension increased until it reached a tension threshold.

Now, you guys are arguing over what caused the tension to increase. The answer will vary with the observer. That is an amazing result from SR. The tension increases for a different reason in the rocket frame then in the stationary frame. I interprete this plurality of explanations as saying that those "reasons" are nice, but not fundamental. What seems to be fundamental in breaking the rope is the forces, not the abstractions of time and space.

Lorentz said that one can only make measurements using forces. He said that explicitly in "Theory of the Electron." He also said that measuring devices have to be in a "quasi-stationary state." I think that Einstein makes equivalent assumptions, but he folded them into his formal definitions. Maybe they are camouflaged assumptions, but they are very important.

I vaguely remember that you did something to remove the physical rope from the problem. However, this does not address the question of "what broke the rope." Is that really what everyone wants to know?

Kaptain K
2001-Oct-28, 10:41 AM
Two pages of posts on this board (and who knows how many posts on the old board) and, as far as I can tell, nobody has changed their mind.

GrapesOfWrath
2001-Oct-28, 11:00 PM
How would you be able to tell? /phpBB/images/smiles/icon_smile.gif

SeanF
2001-Oct-29, 01:34 PM
On 2001-10-27 21:23, Rosen1 wrote:


When did I say the rope has not contracted? In fact, I was responding to a posted thought experiment which didn't even include a rope.

SeanF


I said that the the rope didn't contract in the stationary frame. It didn't contract because its actual length is the distance between the two rockets, where the ropes are attached. The length of the rope hasn't really contracted. The hypothetical "normal" length of rope has contracted, but that is a purely mathematical concept. The question is "what broke the rope in the stationary frame." My answer: The tension increased until it reached a breaking threshold. In the rocket frame, the rope broke when the rockets moved further apart. However, why didn't the rope simply stretch to accomodate the distance between rockets? A really elastic rope wouldn't break. My answer, the tension increased until it reached a tension threshold.

Now, you guys are arguing over what caused the tension to increase. The answer will vary with the observer. That is an amazing result from SR. The tension increases for a different reason in the rocket frame then in the stationary frame. I interprete this plurality of explanations as saying that those "reasons" are nice, but not fundamental. What seems to be fundamental in breaking the rope is the forces, not the abstractions of time and space.

Lorentz said that one can only make measurements using forces. He said that explicitly in "Theory of the Electron." He also said that measuring devices have to be in a "quasi-stationary state." I think that Einstein makes equivalent assumptions, but he folded them into his formal definitions. Maybe they are camouflaged assumptions, but they are very important.

I vaguely remember that you did something to remove the physical rope from the problem. However, this does not address the question of "what broke the rope." Is that really what everyone wants to know?


Well, in the simplest terms, the rope breaks in any reference frame because the distance between the ships becomes greater than the length of the rope, so the rope can no longer reach.

I think the point of the original post was that there was some confusion as to what SR actually predicted. It seems like what the original experiment posed was that the rope would have to break in the 'stationary' frame because of the Lorentz contraction, but why would it break in the 'moving' frame, since there should be no Lorentz contraction in that frame? My original post was to postulate the answer that in the 'moving' frame the distance between the rockets would increase.

I don't think it was ever intended to be a question of what force or tension actually causes the rope to break (as opposed to stretching or whatever), just the mis-estimation that in the 'moving' frame the rockets would maintain the same distance and thus the rope should continue to 'reach' between them.

The whole thread just went on from there.

Ducost
2001-Nov-04, 09:35 AM
What if we shortened the rope to length 0m? in other words if we stuck the nose of rocket a into the rear of rocket b, would they separate?

_________________
Don't ask me, I don't know what I'm talking about

<font size=-1>[ This Message was edited by: Ducost on 2001-11-04 04:37 ]</font>

Kaptain K
2001-Nov-04, 12:32 PM
Don't ask me, I don't know what I'm talking about


That does not seem to be a handicap on this thread. /phpBB/images/smiles/icon_biggrin.gif

_________________
All else (is never) being equal.

<font size=-1>[ This Message was edited by: Kaptain K on 2001-11-04 07:35 ]</font>

GrapesOfWrath
2001-Nov-04, 11:50 PM
On 2001-11-04 04:35, Ducost wrote:
What if we shortened the rope to length 0m? in other words if we stuck the nose of rocket a into the rear of rocket b, would they separate?

Yes, if you satisfy the other conditions.

SeanF
2001-Nov-05, 12:39 PM
On 2001-11-04 04:35, Ducost wrote:
What if we shortened the rope to length 0m? in other words if we stuck the nose of rocket a into the rear of rocket b, would they separate?

_________________
Don't ask me, I don't know what I'm talking about



This is still basically the same thing, Ducost.

If the two rockets are accelerating equally for the stationary observer, they will maintain the same distance (center to center). Since the rockets themselves will be Lorentz contracted, the distance between them (nose to tail) will increase.

To observers on the rockets, the front rocket will be accelerating faster; for these observers, both distances will increase.

Now, if the rockets are accelerating equally for the rocket-bound observers, then the rockets do not move apart. In this case, the stationary observer would say the back rocket was accelerating faster, but not enough to overtake the Lorentz contraction and "crash into" the front rocket! In this case, the rockets remain touching nose-to-tail.

It's actually kind of beautiful the way the different predictions of SR (time dilation, Lorentz contraction, and simultaneity) all take care of each other and prevent these things from becoming paradoxes! /phpBB/images/smiles/icon_biggrin.gif


_________________
SeanF

<font size=-1>[ This Message was edited by: SeanF on 2001-11-05 07:40 ]</font>

GrapesOfWrath
2001-Nov-05, 12:50 PM
On 2001-11-05 07:39, SeanF wrote:
It's actually kind of beautiful the way the different predictions of SR (time dilation, Lorentz contraction, and simultaneity) all take care of each other and prevent these things from becoming paradoxes!

That's the mathematics of the relativity. Once you've satisfied the conditions to apply special or general relativity, you can use the appropriate arbitrary reference frame to do your analysis of the scenario, and the calculations "work out".

If you screw up somewhere along the way, and make an invalid assumption, or skip a step, and you seem to get two different answers depending on which reference frame you use, then the tendency is to call that a "paradox."

AJ
2001-Nov-05, 05:03 PM
Just a Thought. If the initial reference frame viewer (the one not on either rocket) knew that the rockets would be accelarting to a high percentage of v wouldn't they look at the relative distance between the rockets and not the absolute distance. For example the rockets are 500m long and the rope is 100m long. After x amount of time the stationary observer sees the rockets 400m long and the rope at 80m long. Thus the relative distance between the rockets hasn't changed. Needless to say from the rockets point of view it the rope would still be 100m from nose to tail. Just my two cents

-AJ

GrapesOfWrath
2001-Nov-11, 06:26 PM
On 2001-11-05 12:03, AJ wrote:
After x amount of time the stationary observer sees the rockets 400m long and the rope at 80m long. Thus the relative distance between the rockets hasn't changed.

No, stationary observer sees that the distance hasn't changed, and the rope shortened. So, the rope breaks.

DoctorDick
2001-Nov-30, 03:25 PM
Ok guys, there are three observers here who we can talk about: the rest observer, the observer on the rear rocket and the observer on the front rocket.

The rule was that the rockets accelerations are identical. It makes absolutely no difference whether or not that acceleration is constant or varies with time: so long as all three observers agree to preprogram that acceleration in some fixed way.

The rest frame observer will see the rockets starting from some fixed separation and accelerating away with identical accelerations (this you can conclude from the simple symmetry of the events). The separation of the rockets will not change and the rope must break (as it will be Lorentz contracted).

When the front rocket observer looks back at the rear rocket (because it takes light a finite time to get him), he sees the rear rocket where it was earlier. When he makes the correction for the finite speed of light (in his instantaneous inertial frame), he will get exactly the same result he would have obtained were he above the rear rocket and at rest in a gravitational field. Thus we know he will see the lower (rear) rocket's clock as running slow! From his perspective, the observer on the rear rocket is accelerating at a rate slower than what was agreed upon and won't keep up -- the rope will break!

The observer on the rear rocket will, in the same vein, see the observer on the front rocket (after making corrections for the speed of light) exactly as if the rear rocket were at rest in a gravitational field: i.e., he will see the front rocket's clock as running fast! From his perspective,the observer on the front rocket is accelerating at a rate faster than what was agreed upon and will outrun him -- the rope will break!

Since this puzzle seemed to create so much interest, I will start a new thread with a new puzzle which yields some interesting insights.

Have fun -- Dick

GrapesOfWrath
2001-Dec-01, 01:26 AM
On 2001-11-30 10:25, DoctorDick wrote:
Since this puzzle seemed to create so much interest, I will start a new thread with a new puzzle which yields some interesting insights.

Waiting with bated keyboard.

DoctorDick
2001-Dec-03, 05:52 PM
Ok guys,let's look at this a little differently! There are three observers here who we can talk about: the rest observer, the observer on the rear rocket and the observer on the front rocket.

The rule was that the rockets accelerations are identical. Suppose we instead make the rule that the rope is strong enough that it won't break (under the forces contemplated here). In that case, the outcome is only slightly different.

The at rest observer will see the acceleration of the front rocket to be slightly below what was expected (as it is retarded by the Lorentz contracting rope) and the acceleration of the rear rocket to be slightly above what was expected (as it is pulled forward by the Lorentz contracting rope).

The observer on the front rocket will find that he is unable to maintain the agreed upon acceleration because the rear rocket is not accelerating at the agreed upon rate and the rope is holding him back.

The observer on the rear rocket will find that he is unable to maintain the agreed upon acceleration because the front rocket is accelerating at more than the agreed rate and the rope is pulling him forward.

Of course, in the above, I have made some very specific assumptions: namely that the observers on the rockets have no ability to change the thrust of their rockets once the problem begins.

If they do have the power to adjust the thrust of their rockets the situation would be slightly altered. The front guy wants to go faster so he will add thrust. The rear guy wants to go slower so he will subtract thrust. If the front rocket is strong enough to accelerate at the agreed upon rate even when pulling the rear rocket with no applied power, the guy on the rear rocket will reverse the direction of thrust of his rocket (he will still want to slow down).

If the two rockets are of equal power, the rear rocket would have enough thrust to bring the tied system to rest but then he would be going slower than the agreed upon acceleration so he wouldn't go that far.

The net result if the observers on the rockets can control the thrust of the rockets is that the rear rocket will accelerate at the agreed upon rate and the front rocket will be held back by the rope.

The rest observer will see essentially the same result (of course the accelerations need to be relativisticly corrected).

Now, if it is given that the intended acceleration aboard the rockets was 1g, the actual acceleration of the front rocket as seen by that observer (presuming the rope is strong enough so as not to break) is a function of the length of the rope. Can anyone here tell me what that function is?

Have fun -- Dick

GrapesOfWrath
2001-Dec-05, 08:01 PM
On 2001-12-03 12:52, DoctorDick wrote:
Now, if it is given that the intended acceleration aboard the rockets was 1g, the actual acceleration of the front rocket as seen by that observer (presuming the rope is strong enough so as not to break) is a function of the length of the rope. Can anyone here tell me what that function is?
After acres of thought, I conclude that it must be a function of the length of the rope plus the length of the space ship, no?

DoctorDick
2001-Dec-07, 01:27 PM
By the way, there exists another scenario: If the rear rocket cannot reverse his thrust and the front rocket has sufficient thrust to reach his desired acceleration even pulling the rear rocket with the rear rocket thruster shut off, then the front rocket is in charge. The rear rocket will be exceeding the agreed upon acceleration even when his motor is off and he has no further options.

With regard to the fact that the acceleration is a function of the length of the rope, the actual fact is that the acceleration is a function of the distance between the particular site which is accelerating at the agreed upon rate and the site which has lost control of its acceleration because of the rope. Actually, since the function of the rope is no more than an excuse to make the total system obey the expected Lorentz contraction, we are essentially talking about the fact that various parts of an accelerating ship are not accelerating at the same rate (not if the ship is to hold together).

Put the scenario above together with the scenario proposed in my previous post and consider three rockets in a line where only the middle ship can reverse his thruster. In that case the middle rocket has complete control over the acceleration of the whole system; the rear rocket is being dragged along by the front pair and the rear rocket of the front pair is in charge! The scenario can be extended to any number of rockets strung out in a line (you have to do a little complex planing as to who can do what but, with sufficient subtle adjustment of the available thrust, it can be done).

This is essentially the opening part of the solution to my puzzle posted on that other thread. Forget the rockets and look at the rope. Pick a point on the rope which is to be accelerating at a known rate and deduce the acceleration at every other point on the rope. (The rope stays straight because every point on the rope is independently accelerated in the mental model.)

If you think about it for a while, you will realize that the front end of the rope (at plus infinity) is standing still no matter what the acceleration of the reference point is and that, somewhere behind the reference point, the rope cannot keep up even if its acceleration is infinite (it has to go faster than light - view the experiment from the rest observers perspective). Since the acceleration is a continuous function of where you are on the rope, all accelerating ropes are identical: i.e., the only difference is, how far is it from the singular point in the rear to the point on the rope which is in control of the acceleration.

Essentially, there is only one possibility here. If you see the rope as a coordinate system, only one possibility exists: stating your acceleration is equivalent to stating where you are in the coordinate system.

So, starting from infinity at the singular point, what is the acceleration as a function of the distance from the singular point?

grewwalk
2001-Dec-17, 08:23 AM
Not only are the bodies in motion contracting, the space around them are also contracting. The distortions would cause optical effects to all observers, yet the true length of rope, and the bodies never change accounting for the bends in space.
Their distances, to all observers would change but how they look would be distorted.
The rope will not break until something interferes outside of this "problem".
As others have stated many interfering factors, none of them are being introduced for this argument.
I wish I could graph this out visually for everybody, because it can be visualized simply.

GrapesOfWrath
2001-Dec-17, 08:15 PM
Welcome to the BABB, grewwalk. I believe that the answer is that the rope does break, given the information in the OP. It breaks from the point of view of both sets of observers.

robert_d
2001-Dec-19, 05:38 PM
Sorry GofW,
I know it seems that those of us that are unconvinced seem dense, but we are standing by the statement that the 'Lorentz' contraction is an observational effect, not an actual change in length. All of the above seem defficient in some meaningful way. For example, an accelleration profile is not neccessary to pose the problem. The two ships with rope (say unattached at first) could be moving at some steady high velocity with the front and back ends just touching the rope. This would remove all the apparent acceleration differences discussed above. Yet you would tell me that the stationary observer would see a space between the rope ends and the rockets that would be different each time they passed by at the current **invariant** velocity. What if they were stationary and the observer went flying by?

SeanF
2001-Dec-19, 06:06 PM
Robert, the acceleration is necessary for the original question.

In your example, an observer moving relative to the rockets would measure a shorter rope length and a shorter rocket-to-rocket distance than would an observer stationary relative to the rockets.

In the original example, we have a single observer who is at rest relative to the rockets to begin with and in motion relative to the rockets later, correct? As in your example, he would measure a shorter rope length and a shorter rocket-to-rocket distance at the later time.

The key is that in the original post, it was stated that the stationary observer sees the rockets accelerating identically. This means that at the later time, the observer would still see the rockets being the same distance apart. After all, if they accelerate identically, they can't get closer together. However, the observer would still measure a shorter rope length than he did originally. Since the rope is now shorter, but the rockets are not closer together . . . **snap**.

As to your statement:



. . . the 'Lorentz' contraction is an observational effect, not an actual change in length.


It may be just semantics, but the Lorentz contraction of distance is "real" -- at least, it is just as "real" as the distance itself is. There is no such thing as the "true" distance between two space-time events, it is dependent on the relative motion of the measuring observer.

robert_d
2001-Dec-19, 08:11 PM
SeanF,
Thanks for the reply. I think I finally may be getting this. If the acceleration is measured as 1 G by the observer for both rockets then it would be something different for the rocket drivers. but I think we interpreted the original note as measuring the 1 G acceleration from the rockets. (somewhat reasonably). We saw the part about the observer seeing the Lorentz contraction, but did not interpret that as necessarily meaning the acceleration was being measured from the same vantage point. It may be that the original problem in the book was more clear as to who was measuring the acceleration.

SeanF
2001-Dec-19, 08:56 PM
Robert, that may be . . .

If the accelerations are equal for the stationary observer, then they're unequal for the rocket-bound observer. Specifically, the rocket-bound observer sees the front rocket accelerating faster, so the distance between the rockets increases, and the rope is pulled apart.

If the accelerations are equal for the rocket-bound observer, then they're unequal for the stationary observer. Specifically, the stationary observer sees the back rocket accelerating faster, so the distance between the rockets decreases to match the Lorentz contraction of the rope, and the rope remains intact.

As I recall the original, original, way-back on the old board post (yikes!), it specified that the rockets were pre-programmed with identical acceleration code. This should cause the accelerations to be identical for the stationary observer (in the reference frame in which all observers started), resulting in the rope breaking.

Why identical pre-programming would cause non-identical acclerations from the rocket-bound observer's point-of-view is certainly not inherently obvious, but I think it does follow from the simultaneity aspects of Relativity . . .


_________________
SeanF

<font size=-1>[ This Message was edited by: SeanF on 2001-12-19 15:57 ]</font>

grewwalk
2001-Dec-20, 08:51 AM
I went back and thought through the problem again. Almost convinced myself that the rope would break. I researched deeper into Lorentz contraction and Lorentz transformation. Here is some information that I searched for.

***

The equations used in relativity theory to change from a coordinate system, or frame of reference, in which the observer is at rest to a second system that is moving at constant velocity with respect to the first system are known as the Lorentz transformation. The Lorentz transformation will result in a stationary observer recording an effect equivalent to the Lorentz contraction when observing an object in uniform motion relative to his system of coordinates. Einstein showed, however, that this effect is due not to the actual deformation of the body in question, as Lorentz had originally supposed, but to a change in the way space and time are measured.

***

In the end, I think we may never truly know ... or go insane.

SeanF
2001-Dec-20, 01:36 PM
Grewwalk,

Mind if I ask where you got that quote from? I might be interested in reading the entire work. /phpBB/images/smiles/icon_smile.gif

Even within that framework, though, the thought experiment still works. Consider it this way:

Before the rockets begin moving, both observers ("stationary" and "rocket-bound") would agree on the measurements; the distance between the rockets is measured at 100m, and the rope length is measured at 100m (I'm just picking numbers here).

Once the rockets are moving, the measurements disagree. The rocket-bound observer will measure a greater distance between the rockets than the stationary observer measures, as well as a greater rope length than the stationary observer measures.

Since the rocket-bound observer is still not moving relative to the rope, he should not see a change in the length of the rope (ignoring stretching, etc.). Therefore, he will measure the rope as still being 100m. The "stationary" observer, who is now moving relative to the rope, would measure a Lorentz contraction, and would measure the rope as, say, 80m.

If the rocket-bound observer measured the rockets' accelerations as identical, than he would still measure the rockets as being 100m apart as well, so the rope still reaches. In this scenario, the stationary observer (moving relative to the rockets) would measure a Lorentz contraction -- the rockets are only measured as 80m apart. The rope still reaches for him, as well. However, since the back rocket started out 100m behind the front and is now only 80m behind the front, it must have accelerated faster.

Now, let's look at another scenario, where the rocket-bound observer sees unequal acceleration. In this scenario, he sees the front rocket accelerate faster, so that once moving the rockets are 125m apart. Since he still measures the rope as 100m long, the rope cannot reach and must snap (note that the rope length is 80% of the rockets' distance).

In this scenario, the stationary observer (moving relative to the rockets) would measure the Lorentz contraction -- he would measure that 125m distance between the rockets as only 100m. Since he is measuring the rope as 80m long, the rope cannot reach and must snap. Note that the rope length is 80% of the rockets' distance for this observer as well. But also note that since the rockets started out 100m apart and are still 100m apart, their accelerations were identical.

So, the rope breaks for both observers in the second scenario because the rope is only 80% long enough and can't reach. The Lorentz contraction causes the observers to measure different distances for the rope length (80m vs. 100m) as well as the rockets' distance (100m vs. 125m), but the fact that the rope isn't long enough to reach is absolutely real either way.

GrapesOfWrath
2001-Dec-20, 02:21 PM
On 2001-12-19 15:11, robert_d wrote:
If the acceleration is measured as 1 G by the observer for both rockets then it would be something different for the rocket drivers. but I think we interpreted the original note as measuring the 1 G acceleration from the rockets. (somewhat reasonably). We saw the part about the observer seeing the Lorentz contraction, but did not interpret that as necessarily meaning the acceleration was being measured from the same vantage point. It may be that the original problem in the book was more clear as to who was measuring the acceleration.
Still, we are discussing the OP, not the book.

It doesn't matter who is measuring the acceleration, or whether the 1 g is measured from the intertial frame or the rockets, actually.

If the folks on the rockets each see themselves as accelerating at 1 g (which I agree is a reasonable interpretation), then the inertial observer will see them as accelerating the same--though maybe not g. But the same is true in reverse if we assume the intertial observer sees them as accelerating at 1 g--they'll still observe a similar acceleration. In neither case, though, will they observe the other ship as accelerating at the same rate, which is why the rope breaks.

grewwalk
2004-Aug-10, 07:45 AM
I know this reply is waaaay late, but ...

I Think this is the place where I got it from.

http://www.bartleby.com/65/lo/Lorentzc.html

The article seems to be in about 30 different sites though.


Just having fun with the old stuff ...
:wink: [/url]

Wally
2004-Aug-10, 01:09 PM
Holy Cow Grewwalk! Almost 3 whole years to dig up that site!?!? And no posting in between, by the looks of things.

Glad you finally got this off your chest!!!! :lol:

fyi though. I think folks here would prefer you start a new thread for something like this. In the new thread, you can provide a link to the original thread. Kinda keeps things neat and tidy that way. Plus, you won't get confused with certain trolls who revel in digging up old threads for no particular reason.

ToSeek
2004-Aug-10, 01:23 PM
fyi though. I think folks here would prefer you start a new thread for something like this. In the new thread, you can provide a link to the original thread. Kinda keeps things neat and tidy that way. Plus, you won't get confused with certain trolls who revel in digging up old threads for no particular reason.

That seems needlessly complex. Why not add to this thread if what you have to say is relevant to it?

Wally
2004-Aug-10, 03:01 PM
To each his own. . . I really don't have a preference. I just figured with the recent Jim Crow event, I'd mention the alternative method. . .

amateurphysicist
2004-Aug-11, 02:11 AM
Truncated for the sake of brevity...



Both rockets are at rest wrt each other and the rope. At time t = 0 (synchronized clocks) in the rest frame of the rockets, they both fire their thrusters (single thrusters at the tail of each rocket) such that they accelerate at 1g.


If an object starts from rest and accelerates at 1g, wouldn't it take an awfully long time to reach relativistic speeds for this to even matter?

grewwalk
2004-Aug-11, 06:34 AM
It's been 3 years? Must have been in a time warp! :wink:

No, actually superbusy life kept me ... well, busy. I recapped some old stuff, then searched it and added. It was nice to have that closure. I think I'm done twisting my brain on this subject, though.

What do you think? A record for least posts in all these years? :D

milli360
2004-Aug-11, 09:45 AM
amateurphysicist:
Truncated for the sake of brevity...



Both rockets are at rest wrt each other and the rope. At time t = 0 (synchronized clocks) in the rest frame of the rockets, they both fire their thrusters (single thrusters at the tail of each rocket) such that they accelerate at 1g.


If an object starts from rest and accelerates at 1g, wouldn't it take an awfully long time to reach relativistic speeds for this to even matter?
One g is about 10/m/s/s, so at that acceleration, it would take about one year to reach the speed of light--much shorter than the length of this thread. :)

However, someone who perceives that they are accelerating at one g, would never reach the speed of light.

Wally
2004-Aug-11, 01:00 PM
What do you think? A record for least posts in all these years? :D

Gotta be pretty darn close anyways! =D>

Welcome back, by the way!

ToSeek
2004-Aug-11, 01:06 PM
Truncated for the sake of brevity...



Both rockets are at rest wrt each other and the rope. At time t = 0 (synchronized clocks) in the rest frame of the rockets, they both fire their thrusters (single thrusters at the tail of each rocket) such that they accelerate at 1g.


If an object starts from rest and accelerates at 1g, wouldn't it take an awfully long time to reach relativistic speeds for this to even matter?

I figure 354 days to reach lightspeed at 1g acceleration (disregarding GR effects in order to provide an approximate value).

milli360
2004-Aug-11, 02:58 PM
What do you think? A record for least posts in all these years? :D

Gotta be pretty darn close anyways!
Now, if only Hook would post again. :)

Lorentz
2004-Aug-16, 11:59 PM
And I ask again, in the context of the puzzle, 'Why does the rope break?'?...

The disturbing thing about relativity is that the explanation of "why" changes with the observer, even though the physical rope breaking does not.
The when and where the rope breaks changes with the observer, the rope breaking does not. That the rope breaks at some point everyone agrees to.
Answers to questions that are the same for all observers are called covariant properties, answers to questions that differ with observers are called noncovariant.

Covariant: At some place, at some time, the rope got so tense it broke. Everyone agrees that the rope got tense, and at some point broke.


Noncovariant:
Observer drifting in free space who was at the scene of the break when it happened: I saw the two spaceships contract relativistically, making the rope so tense it broke at 12:00, noon. Both rockets were going at the same speed at every moment of time, the rope was between them.
Observer on rocket in back: The guy ahead of me accelerated faster than me, he moved away from me, and because he moved away from me, the rope got so tense, it broke at 11:50 AM. The rope was between us at every point of time.

Observer of rocket in front: The guy behind me couldn't accelerate as fast as me, he fell behind, and because he couldn't keep up, the rope got tense and broke at 12:10 PM. The rope was between us at every point of time.

Observer in inertial frame, 1 light year away and moving close to the speed of light relative to the rope in a direction opposite the direction of acceleration:
I wasn't there, but I used a telescopic video camera and recorded the movie a year later, when the light reached me. The camera shows what happened exactly.
The rope was tense but strong up to 11:50. It got soft at 11:50 AM, due to some chemical reaction, the rope broke at 12:00 noon, the front rocket to start accelerating at 12:10 noon and then the back rocket started moving at 12:20.