1. 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, 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 seems to agree.

2. 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>

3. Member
Join Date
Oct 2001
Posts
34
And I ask again, in the context of the puzzle, 'Why does the rope break?'?...

4. 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>

5. 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.

6. 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! [img]/phpBB/images/smiles/icon_smile.gif[/img]

7. Established Member
Join Date
Oct 2001
Posts
493
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!

8. 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! [img]/phpBB/images/smiles/icon_biggrin.gif[/img]

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.

9. Member
Join Date
Oct 2001
Posts
34
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!

10. Established Member
Join Date
Oct 2001
Posts
1,468
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.

11. Member
Join Date
Oct 2001
Posts
34
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>

12. 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?

13. Member
Join Date
Oct 2001
Posts
34
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>

14. 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.

15. 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.

16. 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.

17. 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.

18. Established Member
Join Date
Oct 2001
Posts
1,468
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.

19. Guest
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!

20. Member
Join Date
Oct 2001
Posts
34
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?

21. Guest

22. Guest
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.

23. Guest
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.

24. 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.

25. Member
Join Date
Oct 2001
Posts
34
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? [img]/phpBB/images/smiles/icon_biggrin.gif[/img]

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

26. 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.

27. Guest
[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>

28. 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? [img]/phpBB/images/smiles/icon_biggrin.gif[/img]

<font size=-1>[ This Message was edited by: Diogenes on 2001-10-24 08:11 ]</font>
[img]/phpBB/images/smiles/icon_biggrin.gif[/img]

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.

29. 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?
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.

30. Guest
[quote]
On 2001-10-24 08:59, GrapesOfWrath wrote:
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.

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•