Bell's Spaceships Paradox explained.

[yuiop]

According to the Wikpedia entry
"[URL
and this paper http://www.aapps.org/archive/bulletin/vol14/14_1/14_1_p03p07.pdf
the conclusion reached by Bell in his paradox is still disputed even by some scientists today.

Since the linked articles give plenty of mathematical explanations I thought a visual explanation of the paradox might help. (See attached diagram)

The left frame is the point of view of an inertial observer that remains in the initial reference frame of the green and yellow rockets that accelerate to simultaneously and with constant equal proper acceleration to a new constant velocity.

The centres of the green and yellow rockets are connected by a spring and slung under the green rocket is a rod supporting an unstressed spring that is connected only at one end to the green rocket. Comparison of the length contracted rockets and unstressed rod and and spring should give a visual indication of stress the connecting spring is under. If the spring was a fairly inflexible string, it is reasonable to assume it would snap if the final velocity is sufficient.

The blue rocket has constant velocity relative to the initial reference frame and is already going at the final velocity of the accelerating rockets. The right hand drawing is the rest frame of the blue rocket. In the blue rocket frame it can be seen that the yellow rocket takes off long before the green rocket. It can also be seen that in the blue rockets frame the green and yellow rockets are de-accelerating from a relative velocity to finally come to rest in the blue frame. In this case the with the chosen parameters the yellow rocket completes its de-acceleration phase and comes to rest just before the green rocket even takes off. It can also be seen that in the blue frame the green and yellow rockets appear to be undergoing "length expansion" rather than the normal length contraction. It is much easier to see in the blue frame why the connecting spring is stretched and why a connecting string would eventually snap.

Hope that the "visual explanation" helps :)

 

[peter0302]

To me it's pretty intuitive that the string would not break. When an inertial observer sees two rockets accelerate away at close to 'c', connected by a string, they in effect become a single object, the entire length of which should contract according to SR. So while the rockets themselves are contracting, so is the string, and so is the _separation_ between those two rockets. In the rockets' frame, nothing happens to the string. In the inertial observer's frame, the rockets appear to move closer to one another as their speed approaches 'c'. The string does not break.

Am I missing something?

In the wikipedia article, they seem to take it as a given that the distance between the two rockets remains constant in the inertial observer's frame. I don't see why that would be the case. If a constant distance is measured between the two rockets in the rockets' frame, then by SR an inertial observer WILL see that length as contracted. All length _measurements_ are contracted, not just physical lengths.

 

[yuiop]

To me it's pretty intuitive that the string would not break. When an inertial observer sees two rockets accelerate away at close to 'c', connected by a string, they in effect become a single object, the entire length of which should contract according to SR. So while the rockets themselves are contracting, so is the string, and so is the _separation_ between those two rockets. In the rockets' frame, nothing happens to the string. In the inertial observer's frame, the rockets appear to move closer to one another as their speed approaches 'c'. The string does not break.

Am I missing something?

In the wikipedia article, they seem to take it as a given that the distance between the two rockets remains constant in the inertial observer's frame. I don't see why that would be the case. If a constant distance is measured between the two rockets in the rockets' frame, then by SR an inertial observer WILL see that length as contracted. All length _measurements_ are contracted, not just physical lengths.

Your statement "If a constant distance is measured between the two rockets in the rockets' frame, .." shows that you are missing the fact that the distance between the two accelerating rockets appears to get larger in the reference frame of the accelerated rockets. When they get to their final velocity the distance they measure between the two rockets is larger by a factor of gamma than the distance measured by the unaccelerated observer in the initial reference frame.

Since the "number of views" of the diagram was zero when you posted your response I guess you did not look at the diagram? The right hand diagram shows that to an observer at the final velocity of the two rockets, one take off before the other (because of the lack of simultaneity) and the string must break from his point of view. If he sees the string break, then all observers must see the string break.

I think you are also missing that the two rockets do not accelerate away as a single object. They have constant and equal proper acceleration while the nose and tail of single rocket undergoing natural length contraction does not experience equal acceleration at the nose and tail of the rocket.

 

[peter0302]

the distance between the two accelerating rockets appears to get larger in the reference frame of the accelerated rockets

Why? If they're both accelerating with proper acceleratoin 1g, they should measure their distance to be constant at all times - they're in the same uniform accelerating reference frame - in free fall -as though they had both jumped off balconies (one just above the other) at the same time. I see no reason why they would not measure their separation to be constant. And GR tells us that the laws of physics work the same in free fall just as they do in inertial reference frames.

However, their distance will actually decrease _as measured by the inertial observer_ as they accelerate toward the speed of light, per length contraction.

 

[yuiop]

Why? If they're both accelerating with proper acceleratoin 1g, they should measure their distance to be constant at all times - they're in the same uniform accelerating reference frame - in free fall -as though they had both jumped off balconies (one just above the other) at the same time. I see no reason why they would not measure their separation to be constant. And GR tells us that the laws of physics work the same in free fall just as they do in inertial reference frames.

However, their distance will actually decrease _as measured by the inertial observer_ as they accelerate toward the speed of light, per length contraction.

That is perfectly true for free falling observers that feel no acceleration and are are indistigishuable from an inertial frame. The accelerating rockets in Bell's spaceship paradox do feel acceleration and are not equivalent to an inertial reference frame.

The reason the accelerated observers see the separation as expanded while the non accelerated observer sees the separation as constant is because when they send light signals to each other the round trip time for the light appears to longer due to the time dilation of their clocks. The separation distance also seems to be longer to them, because their rulers are length contracted so they they require more rulers to measure the separation distance.

 

[Ookke]

If we modify this slightly so that the rockets do not accerate by engine, but an outside inertial observer kicks both rockets on the go simultaneously, it's easy to see that their distance measured from the inertial observer's (= the kicker) frame remains constant.

I'm not sure what rockets think, but probably their distance will grow in their own frame.

 

[peter0302]

That is perfectly true for free falling observers that feel no acceleration and are are indistigishuable from an inertial frame. The accelerating rockets in Bell's spaceship paradox do feel acceleration and are not equivalent to an inertial reference frame.

This is still the part that I'm not buying. This sounds like it would violate the equivalence principle of GR. A rocket accelerating at 1g should be indistinguishable from being in a gravitational field of 1g. We all know that in my example of two people jumping off different stories of the same building at the same time, for example, their distance remains constant.

And another reason Bell cannot be right. If we instead consider an object made of three molecules, a front, middle, and back, corresponding to spaceship 1, rope, and spaceship 2, by Bell's argument, to an inertial observer the whole object would not actually length contract, but, instead, front and back would, themselves, shrink, but the distance between them would not.

Again, we have to treat the two spaceships and the rope as one system.

 

[yuiop]

This is still the part that I'm not buying. This sounds like it would violate the equivalence principle of GR. A rocket accelerating at 1g should be indistinguishable from being in a gravitational field of 1g. We all know that in my example of two people jumping off different stories of the same building at the same time, for example, their distance remains constant.

And another reason Bell cannot be right. If we instead consider an object made of three molecules, a front, middle, and back, corresponding to spaceship 1, rope, and spaceship 2, by Bell's argument, to an inertial observer the whole object would not actually length contract, but, instead, front and back would, themselves, shrink, but the distance between them would not.

Again, we have to treat the two spaceships and the rope as one system.

The point is that if the two ships and connecting rope are treated as one rigid object then yes, the assembly would length contract but then the front and rear of the assembly will not accelerate at the same rate. In the Bell's experiment the front and rear of the assembly are forced to accelerate at the same rate and something has to give/break.

From the point of view of an inertial observer moving relative to the initial frame of the rockets they do not take off simultaneously. Try connecting two cars with a thin thread and driving off in one car and leaving the other car parked with the handbrake on. The thread is going to break.

 

[kahoomann]

Why? If they're both accelerating with proper acceleratoin 1g, they should measure their distance to be constant at all times - they're in the same uniform accelerating reference frame - in free fall -as though they had both jumped off balconies (one just above the other) at the same time. I see no reason why they would not measure their separation to be constant. And GR tells us that the laws of physics work the same in free fall just as they do in inertial reference frames.

However, their distance will actually decrease _as measured by the inertial observer_ as they accelerate toward the speed of light, per length contraction.

If we assume they're both accelerating by ejecting rocks of same size. The ejecting of the 2nd rock will happen at the same time in the observer frame, but not at the same time in the rocket frame. In the rocket frame, the leading rocket will start accelerating early than the other one. That's the reason why the distance get larger in the reference frame of the accelerated rockets

 

[peter0302]

In the Bell's experiment the front and rear of the assembly are forced to accelerate at the same rate and something has to give/break

Forced to accelerate at the same rate in _their_ frame. The rope follows suit, equally balanced by the two opposing rockets (one pulling on it, the other pushing).

If we assume they're both accelerating by ejecting rocks of same size. The ejecting of the 2nd rock will happen at the same time in the observer frame, but not at the same time in the rocket frame.

No, I believe you have it backwards. The ejecting of the rocks would happen in the same time in the rockets' frame. It is the inertial observer who would see them at different times due to L.C. and T.D., and who would likewise see the distnace between them shrink.

Why would they emit rocks at different times in their own frame? That makes no sense. They'd be sychronized with one another and, experiencing precisely the same acceleration from the start of the experiment, their clocks would remain synchronized.

 

[yuiop]

Forced to accelerate at the same rate in _their_ frame. The rope follows suit, equally balanced by the two opposing rockets (one pulling on it, the other pushing).

I am not sure you have the conditions specified by Bell correctly. Imagine two unconnected rockets take off simultaneously and both rockets have identical rockets and accelerate at the same rate. Before they take of a rope that is just long enough to connect the centre of one rocket to the centre of the other rocket is coiled up and stored aboard the front rocket. Bell's experiment basically asks if when they get to their final velocity and if the rope is uncoiled will it be long enough to reach from one rocket to other without undue strectching?

I hope you agree that two rockets with there own engines and connected by a tough titanium rope would accelerate differently to two rockets that are not connected?

The Bell's thought experiment is the same as two unconnected rockets and it is assumed the string connecting them is not tough enough to influence the flight paths of the rockets or pull the two rockets towards each other and snaps under tension as it tries to length contract. .

No, I believe you have it backwards. The ejecting of the rocks would happen in the same time in the rockets' frame. It is the inertial observer who would see them at different times due to L.C. and T.D., and who would likewise see the distnace between them shrink.

Why would they emit rocks at different times in their own frame? That makes no sense. They'd be sychronized with one another and, experiencing precisely the same acceleration from the start of the experiment, their clocks would remain synchronized.

The clocks onboard two unconnected rockets accelerating at exactly the same rate will be seen to slow down by the unaccelerated observer but will appear to remain synchronised to the observer in the initial unaccelerated frame. Relativity says that if the clocks appear to be synchronised to an observer in an inertial frame then they will not appear to be synchronised to observer that is not at rest with that inertial frame. The accelerating rocket observers see their own clocks as getting progressively more out of sync.

Generally speaking when you you accelerate to a new velocity your clocks do not resychronise themselves to the new velocity and you have to go around your ship adjusting all the clocks manually to get them back in sync again. It is very difficult to come up with an acceleration scheme that keeps clocks self synchronised after acceleration. Constant and equal acceleration won't do the trick and nor will Born-rigid acceleration which is an acceleration scheme designed to keep the apparent length of the rocket/assembly constant as far as the onboard accelerated observers are concerned. An external unaccelerated observer watching a rocket or assembly of rockets undergoing Born-rigid acceleration will measure the rocket/assembly as length contracting. The acceleration method used in the Bell experiment is explicity not Born-rigid acceleration.

 

[kahoomann]

Forced to accelerate at the same rate in _their_ frame. The rope follows suit, equally balanced by the two opposing rockets (one pulling on it, the other pushing).No, I believe you have it backwards. The ejecting of the rocks would happen in the same time in the rockets' frame. It is the inertial observer who would see them at different times due to L.C. and T.D., and who would likewise see the distnace between them shrink.

Why would they emit rocks at different times in their own frame? That makes no sense. They'd be sychronized with one another and, experiencing precisely the same acceleration from the start of the experiment, their clocks would remain synchronized.

If you draw a space-time diagram, you will see clearly in which reference frame the ejecting of the rocks would happen at the same time. I must admit that I did not find out what's wrong with the arguments until I drew the space-time diagram.

 

[peter0302]

The clocks onboard two unconnected rockets accelerating at exactly the same rate will be seen to slow down by the unaccelerated observer but will appear to remain synchronised to the observer in the initial unaccelerated frame. Relativity says that if the clocks appear to be synchronised to an observer in an inertial frame then they will not appear to be synchronised to observer that is not at rest with that inertial frame. The accelerating rocket observers see their own clocks as getting progressively more out of sync.

And I'm saying that the clocks onboard two uniformly accelerating ships will remain synchronized in _their_ frame, so it's the inertial observer who sees them lose synchronization as they approach the speed of light, and who sees their spatial separation shrink.

Neither of you has persuaded me as to why the two ships' clocks would not be in sync the entire time. You say "It is very difficult to come up with an acceleration scheme that keeps clocks self synchronised after acceleration. Constant and equal acceleration won't do the trick" but I am not understanding why that would be the case. Clocks have no problem staying synchronized on Earth in a uniform 1g field - why would it matter whether the clocks were on two separate spaceships or two separate locations on earth?

 

[kahoomann]

And I'm saying that the clocks onboard two uniformly accelerating ships will remain synchronized in _their_ frame, so it's the inertial observer who sees them lose synchronization as they approach the speed of light, and who sees their spatial separation shrink.

Neither of you has persuaded me as to why the two ships' clocks would not be in sync the entire time. You say "It is very difficult to come up with an acceleration scheme that keeps clocks self synchronised after acceleration. Constant and equal acceleration won't do the trick" but I am not understanding why that would be the case. Clocks have no problem staying synchronized on Earth in a uniform 1g field - why would it matter whether the clocks were on two separate spaceships or two separate locations on earth?

I don't need to convince you of anything. In his book "Speakable and unspeakable in quantum mechanics", John Bell said: "A distinguished experimental physicist refused to accept that the thread would break, and regarded my assertion, that indeed it would, as personal misinterpretation of special relativity. We decided to appeal to the CERN Theory Division for arbitration, and made a canvas of opinion in it. There emerged a clear consensus that the thread would Not break! Of course many people who give this wrong answer at first get the right answer on further reflection. Usually they feel obliged to work out how things look to observers B or C. they find that B, for example, see C drifting further and further behind, so that a given piece of thread can no longer span the distance"

 

[granpa]

if you add a long line of synchronized (with the stationary observer) clocks to the thought experiment then it makes it clear what is happening. after the acceleration the rockets clocks will indeed be ticking at the same rate. but the line of clocks will be out of sych (from the point of view of the rockets) . if one rocket is behind the other (and it has to be for the thought experiment to work) then obviously it will be affected. during acceleration objects before and behind an object are affected by the increasing loss of simultaneity.

so the clocks on the rocket will indeed be out of synch as measured by the rockets themselves. but as measured by the stationary observer they should not be.

 

[Hurkyl]

Peter0302: don't 'think it out' -- try actually performing the calculation.

 

[Hurkyl]

Clocks have no problem staying synchronized on Earth in a uniform 1g field

Actually, they do. If you take two clocks and place them at the top and bottom of a tall tower, the one at the top will run (slightly) faster.

 

[MeJennifer]

The Earth has no uniform gravitational field.

 

[Hurkyl]

The Earth has no uniform gravitational field.

Yes, but the effects of that fact are negligible compared to gravitational time dilation.

 

[peter0302]

I guess I'm in good company then, but I am convinced there is some aspect of GR that is not being taken into account here.

I hope you agree that two rockets with there own engines and connected by a tough titanium rope would accelerate differently to two rockets that are not connected?

Maybe here's the problem. No, I do not agree.

Actually, they do. If you take two clocks and place them at the top and bottom of a tall tower, the one at the top will run (slightly) faster.

Heh, ok sure. I think we can agree that that's rather negligible in the context of this problem.

Peter0302: don't 'think it out' -- try actually performing the calculation.

I was initially looking for bad assumptions in Bell's proof, which I still believe I found. When I have nore time I will try to do as you suggest, though as I said I suspect the answer is found in GR.

 

[Fredrik]

The answer is found in SR, and it's pretty simple. The distance between the rockets is constant in the the frame where they were stationary before the engines were switched on. At any time after they started, the distance between the rockets in a co-moving inertial frame is different from the original distance by a factor of gamma.

In other words, it's a simple Lorentz contraction that breaks the string. To remain the same distance away from each other in the original rest frame, they must move apart in a sequence of co-moving inertial frames.

E.g. consider the frame co-moving with the rear endpoint of the string at proper time t₁, and then the frame co-moving with the rear endpoint of the string at proper time t₂ > t₁. The distance is greater in the latter than in the former, and it's easy to see this in a space-time diagram.

 

[Fredrik]

In the inertial observer's frame, the rockets appear to move closer to one another as their speed approaches 'c'.
...
In the wikipedia article, they seem to take it as a given that the distance between the two rockets remains constant in the inertial observer's frame. I don't see why that would be the case.

I haven't read every post, so I don't know if this has already been answered. The distance must remain constant in the inertial frame where they were at rest before the acceleration began, because of translation invariance. The rockets are assumed to be identical, except for their starting positions, so if their world lines turn out to be different, then there must be something different about their starting positions.

 

[kahoomann]

I haven't read every post, so I don't know if this has already been answered. The distance must remain constant in the inertial frame where they were at rest before the acceleration began, because of translation invariance. The rockets are assumed to be identical, except for their starting positions, so if their world lines turn out to be different, then there must be something different about their starting positions.

If we assume they're both accelerating by ejecting rocks of same size. The ejecting of the 2nd rock will happen at the same time in the observer frame, but not at the same time in the rocket frame. In the rocket frame, the leading rocket will start accelerating early than the other one. That's the reason why the distance get larger in the reference frame of the accelerated rockets

In post #10, peter0302 disagree in which frame the 2nd ejection of rocks happen at the same time
Do you agree that the 2nd ejection of rocks happen at the same time in the observer's frame, not in the rocket frame?

No, I believe you have it backwards. The ejecting of the rocks would happen in the same time in the rockets' frame. It is the inertial observer who would see them at different times due to L.C. and T.D., and who would likewise see the distnace between them shrink.

Why would they emit rocks at different times in their own frame? That makes no sense. They'd be sychronized with one another and, experiencing precisely the same acceleration from the start of the experiment, their clocks would remain synchronized.

 

[Wizardsblade]

If we assume they're both accelerating by ejecting rocks of same size. The ejecting of the 2nd rock will happen at the same time in the observer frame, but not at the same time in the rocket frame. In the rocket frame, the leading rocket will start accelerating early than the other one. That's the reason why the distance get larger in the reference frame of the accelerated rockets

In post #10, peter0302 disagree in which frame the 2nd ejection of rocks happen at the same time
Do you agree that the 2nd ejection of rocks happen at the same time in the observer's frame, not in the rocket frame?

I think about it like this..
What if there is a flashing light inbetween the two ships on the rope that flashs every second. When it flashs the ships toss a rock and wait for another flash.

If we use this don't both ships accelerate together, so that peter0302 is correct?

 

[kahoomann]

I think about it like this..
What if there is a flashing light inbetween the two ships on the rope that flashs every second. When it flashs the ships toss a rock and wait for another flash.

If we use this don't both ships accelerate together, so that peter0302 is correct?

No, both rockets suppose to reset their clocks at launch and eject one rock every time the clocks tick

 

[Wizardsblade]

No, both rockets suppose to reset their clocks at launch and eject one rock every time the clocks tick

1 clock or 2 in the same frame does not matter. Think this through...
First everyone/thing is at rest. Then both rocks are tossed at the same time.
Now we have 2 frames. Why would the 2 ship's clock read differently?

 

[peter0302]

The biggest problem I have with the conventional explanation is that it screams preferred reference frame. We're giving the inertial observer on Earth preferred status and forcing the rockets to do what the earthling thinks they should do. But the laws of physics are the same in free fall as they are in zero acceleration. So the two ships should also see one another maintain a constant distance the entire time. No one so far has addressed that problem. Everyone seems to be ok with the fact that the laws of physics are different for the rockets than for the earthling, and GR tells us you cannot do this.

 

[yuiop]

The biggest problem I have with the conventional explanation is that it screams preferred reference frame. We're giving the inertial observer on Earth preferred status and forcing the rockets to do what the earthling thinks they should do. But the laws of physics are the same in free fall as they are in zero acceleration. So the two ships should also see one another maintain a constant distance the entire time. No one so far has addressed that problem. Everyone seems to be ok with the fact that the laws of physics are different for the rockets than for the earthling, and GR tells us you cannot do this.

The free falling observers experience everything the same as the observers with no acceleration including the fact that they do not feel acceleration. The free falling observers are effectively inertial observer's (over a small scale).

The observers on the accelerating rockets are not equivalent to free falling observers because they do feel acceleration. These observers are equivalent to observers in a gravitational field and they do not share an inertial frame.

 

[Hurkyl]

We're giving the inertial observer on Earth preferred status

Wrong.

and forcing the rockets to do what the earthling thinks they should do.

The programming of the rockets are to continually apply a thrust with a constant magnitude, and stop thrusting when its internal clock reads a certain time¹.

The experiment is designed so that the rockets launch simultaneously as measured by the Earthling; that is certainly something we can do physically; e.g. we use Einstein's convention to synchronize the clocks on the rockets, and have them both launch at a specific reading of their clock.

None of this depends on what the 'Earthling thinks they should do'.


1: We are assuming the rockets are small enough that we can ignore the fact clocks at the nose and tail of the rocket will get out of sync, as well as other ways it may deform.

But the laws of physics are the same in free fall as they are in zero acceleration. So the two ships should also see one another maintain a constant distance the entire time.

By "distance" I assume you mean the coordinate distance as measured in a frame where gravity is expressed by a uniform (coordinate) force field, and the rockets are in freefall? Then yes, the ships maintain a constant distance, and the thread snaps². And in terms of coordinates, this is exactly the same as the original problem, as measured in an inertial frame centered on the Earthling.

2: Naïvely applying Hooke's law, the tension in the string would be inversely related to \sqrt{c^2 - v^2}, where v is the coordinate velocity of the string. I'm not familiar with the right way to deal with elasticity relativistically.

 

[Fredrik]

If we assume they're both accelerating by ejecting rocks of same size. The ejecting of the 2nd rock will happen at the same time in the observer frame, but not at the same time in the rocket frame. In the rocket frame, the leading rocket will start accelerating early than the other one. That's the reason why the distance get larger in the reference frame of the accelerated rockets

In post #10, peter0302 disagree in which frame the 2nd ejection of rocks happen at the same time
Do you agree that the 2nd ejection of rocks happen at the same time in the observer's frame, not in the rocket frame?

Yes. This is implied by the fact that they rockets are assumed to be identical.

Let's say that rocket A is in front of rocket B, just so we can refer to them by names.

If the clock on rocket A shows time T when rocket A ejected the second rock, then the clock on rocket B will show time T when rocket B ejects the second rock. Again, this is implied by the two rockets being identical.

Now consider a frame co-moving with rocket B as it ejects the second rock. It's easy to see that in this frame, rocket A has not ejected the second rock yet.

1 clock or 2 in the same frame does not matter. Think this through...
First everyone/thing is at rest. Then both rocks are tossed at the same time.
Now we have 2 frames. Why would the 2 ship's clock read differently?

They don't. Not if you look at them at the two events where the ships eject their second rocks, but if you consider one of those two events, it's not simultaneous with the other in a frame that's co-moving with that rocket. If you would read the two clocks at the same time in a co-moving frame, they show different times.

The biggest problem I have with the conventional explanation is that it screams preferred reference frame. We're giving the inertial observer on Earth preferred status and forcing the rockets to do what the earthling thinks they should do.

That's not true. It's convenient to draw a space-time diagram of the events as seen from the inertial frame where the rockets are initially at rest. That frame has no special significance apart from that.

But the laws of physics are the same in free fall as they are in zero acceleration. So the two ships should also see one another maintain a constant distance the entire time. No one so far has addressed that problem. Everyone seems to be ok with the fact that the laws of physics are different for the rockets than for the earthling, and GR tells us you cannot do this.

I saw that you talked about two people jumping from different floors of the same building earlier in this thread. Your argument seems to be that since the distance between those guys (as measured by one of them) won't increase, even if we pretend that the pull of gravity doesn't vary with altitude, the distance between the rockets (as measured by one of the rockets) won't either. I would reverse that argument and say that since my posts above prove in a very simple way that the string must break, we are forced to conclude that the distance between the two guys in free-fall must also increase.

 

[Hurkyl]

I would reverse that argument and say that since my posts above prove in a very simple way that the string must break, we are forced to conclude that the distance between the two guys in free-fall must also increase.

What distance are you referring to?

 

[Fredrik]

I would reverse that argument and say that since my posts above prove in a very simple way that the string must break, we are forced to conclude that the distance between the two guys in free-fall must also increase.

It's really annoying that we can't edit posts after the first 30 minutes. I just wanted to clarify by saying that it's the distance between the two guys as measured by one of them (either of them) that must increase. It won't increase in the building's rest frame, because of translation invariance (which we only have in the idealized situation, where the gravitational field really is homogeneous). Another way of looking at it is that it won't increase in the building's frame because Lorentz contraction exactly cancels the stretching of the distance that can be seen from a co-moving frame.

Edit:

What distance are you referring to?

I didn't see your question until after I answered it. You asked while I was writing the answer.

 

[peter0302]

I would reverse that argument and say that since my posts above prove in a very simple way that the string must break, we are forced to conclude that the distance between the two guys in free-fall must also increase.

Except then the laws of physics are being violated in the freefaller's frame, which is a problem.

The rockets shouldn't care that they're accelerating away from earth. All they know is that their velocity w/r/t one another, as well as the string, is zero at all times. Ergo they should see no time dilation effects between the two of them, and no ill effect on the string. If they do, then the Earth is right, and they are wrong, and the equivalence principle is thrown out the window.

What if the Earth accelerated right along with them as soon as they launched? Would the string still break?

 

[Fredrik]

Except then the laws of physics are being violated in the freefaller's frame, which is a problem.

The only real problem is that the gravitational field isn't homogeneous in the "two guys and a building" scenario, so we don't really have translation invariance.

All they know is that their velocity w/r/t one another, as well as the string, is zero at all times.

This is definitely incorrect. Identical rockets plus translation invariance means that the world lines of the two rockets are identical (except for starting position) in the original rest frame. That means that the event where the clock on rocket B (which is behind rocket A) shows t₁ is simultaneous, in an inertial frame that's co-moving with rocket B, with an event where the clock on rocket A shows t₂ > t₁. So rocket A has had more (proper) time to accelerate and has a higher velocity relative to the starting position. This implies that the velocity of rocket A in the frame that's co-moving with rocket B is greater than 0.

What if the Earth accelerated right along with them as soon as they launched? Would the string still break?

I don't understand this question. Do you want to consider the effect on space-time geometry created by a mass the size of a planet that's forced to accelerate to near light speed right behind two rockets? That might be a fun problem, but you should probably make sure you understand the original problem first.

I solved the original problem completely in my first two posts in this thread, so I encourage you to take a look at them again and try to find something wrong with my argument. l also recommend that you draw a space-time diagram.

 

[Wizardsblade]

Let's say that rocket A is in front of rocket B, just so we can refer to them by names.

If the clock on rocket A shows time T when rocket A ejected the second rock, then the clock on rocket B will show time T when rocket B ejects the second rock. Again, this is implied by the two rockets being identical.

Now consider a frame co-moving with rocket B as it ejects the second rock. It's easy to see that in this frame, rocket A has not ejected the second rock yet.
...

They don't. Not if you look at them at the two events where the ships eject their second rocks, but if you consider one of those two events, it's not simultaneous with the other in a frame that's co-moving with that rocket. If you would read the two clocks at the same time in a co-moving frame, they show different times.

Would this not also apply to the two rockets when they toss their first rocks in Earths frame?
It is a co-moving frame w/r/t the rockets, but this implies that the two rocket's clocks are not synced.

 

[yuiop]

It seems odd that Fredrik and Pete0302 seem to disagree on this topic and yet they both manage to draw wrong conclusions.

...
I saw that you talked about two people jumping from different floors of the same building earlier in this thread. Your argument seems to be that since the distance between those guys (as measured by one of them) won't increase, even if we pretend that the pull of gravity doesn't vary with altitude, the distance between the rockets (as measured by one of the rockets) won't either. I would reverse that argument and say that since my posts above prove in a very simple way that the string must break, we are forced to conclude that the distance between the two guys in free-fall must also increase.

The distance between the two guys in free-fall in a uniform gravitational field will not increase.

The reason is this:

2 observers (A and B) that are in free fall in a uniform free fall (after jumping off a tall building is equivalent to 2 inertial observers (C and D) far out in space far away from any large gravitational body that are at rest with respect to each other and watching a tall but massless building being accelerated past them by a rocket.

Observers (C and D) do not see the distance between them as changing with time and any string joining them will not be stretched and the same is true for observers A and B. It is is also true that observers A and B and observers C and D do not feel any acceleration. They only see it when they look at the building.

2 observers (E and F) that are accelerating with constant equal acceleration according to an inertial observer are not equivalent to the situation of observers A and B and nor are they equivalent to the situation of observers C and D and this is made clear when it is noted that observers E and F feel acceleration and would know the difference even with their eyes closed.

2 observers (G and H) that are on separate floors of a tall building located in a gravitational field (and at rest with their respective floors) are in an equivalent situation to observers (J and K) that are in a massless building far away from any significant massive body that is being accelerated artificially. An inertial observer watching the building being accelerated in space sees the building as length contracting as it accelerates. Observers G and H in the gravitational field obviously do not see the distance between floors of their building as increasing over time. Because the situations of observers G and H and J and K are equivalent it is also obvious that observers J and K do not not see their separation as increasing over time. Neither the situation of observers G and H and J and K are are equivalent to that of observers E and F even though all thee pairs of observers actually feel acceleration.

The nearest equivalent to the situation of observers E and F is that of two observers in a building being artificially accelerated and being artificially expanded at the same time, or two observers on different floors of a tall building in a gravitational field where the building is getting rapidly taller over time. Observers E and F are equivalent to the classic situation described in Bell's paradox and Bell's paradox can not be compared to the situation of observers (A and B) or (C and D) or (G and H) or (J and K) as described above because none of those situations are equivalent.

...

I solved the original problem completely in my first two posts in this thread, so I encourage you to take a look at them again and try to find something wrong with my argument. l also recommend that you draw a space-time diagram.

You could always refer to the diagram I posted in post#1 of this thread which has the paths and points in spacetime accurately drawn using geometrical software with coordinates tranformed using the the Lorentz transformations.

What if the Earth accelerated right along with them as soon as they launched? Would the string still break?

The original Bell's paradox does not include the Earth as a gravitational body but just as a point of reference. As in the twins paradox the Earth is not meant to represent a source of acceleration and is loosely used as inertial reference frame even though it is not in reality. In the though classic thought experiments, the Earth is imagined to be an ideal massless point of reference with no significant gravitational field. As such it would make no difference if the Earth accelerated right along with accelerating rockets. The rockets are only required to maintain constant proper acceleration which they can measure without even looking out of a window by using onboard accelerometers. If the Earth is replaced by a small spacestation it should be clear that the spacestation accelerating after the rockets have accelerated would make little difference to the proper acceleration measured by the onboard rocket accelerometers.

 

[Fredrik]

Would this not also apply to the two rockets when they toss their first rocks in Earths frame?
It is a co-moving frame w/r/t the rockets, but this implies that the two rocket's clocks are not synced.

You can consider four different frames here. The frames can be co-moving with a rocket either before or after the toss, and they can be co-moving with either rocket A or rocket B.

The two frames that are co-moving with the rockets before the toss have the same velocity as the original rest frame. (I prefer not to think of it as the Earth frame, because that suggests gravitational effects that we aren't going to consider anyway). The only difference between them is a translation. At this time, in these frames, the clocks are both showing 0.

If we consider the frame that's co-moving with rocket B (which is behind rocket A) after the toss, then the clock on B still shows 0 immediately after the toss (assuming that the toss was instantaneous). The event on rocket A that's simultaneous with the event where rocket B tossed its first rock, occurs some time after rocket A tossed its first rock. So at this time, in this frame, clock A shows 0 and clock B shows a positive time.

If we consider the frame that's co-moving with rocket A after the toss, then the clock on A still shows 0 immediately after the toss. The event on rocket B that's simultaneous with the event where rocket A tossed its first rock, occurs some time before rocket B tossed its first rock. So at this time, in this frame, clock B shows 0 and clock A shows a negative time.

 

[yuiop]

Let Z be a spacestation of isgnificant mass far out in space.

A and B are 2 rockets initially at rest with Z and spatially separated along the x-axis on which observers A,B and C all lie on.

Rockets A and B accelerate simultaneously according to observer Z to reach a new velocity v relative to Z.

Rockets A and B accelerate simultaneously again as observered by observer Z.

Since observer Z sees the the second burst of acceleration happen simultaneously, it follows that the the rocket observers in the frame with velocity v relative to Z would not see the the second burst of acceleration as happening simultaneously.

This is because if two spatially separated events are simultaneous in one reference frame (Z) it is impossible to find another inertial reference frame (Z') that has velocity relative to Z such that frame Z' will also consider the events to be simultaneous.

Therefore the assumption the two rockets share an (instantaneous) inertial reference frame is incorrect or the assumption that they also see the acceleration events as simultaneous is incorrect.

 

[Fredrik]

It seems odd that Fredrik and Pete0302 seem to disagree on this topic and yet they both manage to draw wrong conclusions.

The distance between the two guys in free-fall in a uniform gravitational field will not increase.

It's not that strange, considering that we're talking about two different problems. It's certainly possible to be right about one and wrong about the other.

I know I understand the spaceship problem, because I've been thinking about that a lot, but it seems I didn't give the other problem enough thought.

You could always refer to the diagram I posted in post#1 of this thread which has the paths and points in spacetime accurately drawn using geometrical software with coordinates tranformed using the the Lorentz transformations.

Oooh, nice. I actually didn't read #1 until now.

One thing that would be nice to see in the diagram on the left is this: Two dots, somewhere in the middle of each world line, marking two events where the two clocks show the same time. (They would be on the same horizontal line in the diagram). Also, simultaneity lines, showing what events co-moving observers consider simultaneous with those two events.

 

[Wizardsblade]

2 observers (E and F) that are accelerating with constant equal acceleration according to an inertial observer are not equivalent to the situation of observers A and B and nor are they equivalent to the situation of observers C and D and this is made clear when it is noted that observers E and F feel acceleration and would know the difference even with their eyes closed.

2 observers (G and H) that are on separate floors of a tall building located in a gravitational field (and at rest with their respective floors) are in an equivalent situation to observers (J and K) that are in a massless building far away from any significant massive body that is being accelerated artificially. An inertial observer watching the building being accelerated in space sees the building as length contracting as it accelerates. Observers G and H in the gravitational field obviously do not see the distance between floors of their building as increasing over time. Because the situations of observers G and H and J and K are equivalent it is also obvious that observers J and K do not not see their separation as increasing over time. Neither the situation of observers G and H and J and K are are equivalent to that of observers E and F even though all thee pairs of observers actually feel acceleration.

This seems to disagree with...

Acceleration does not cause time dilation. This is known as the http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html#Clock_Hypothesis" and has been experimentally verified up to about 10^18 g. Consider also muons created from cosmic rays in the upper atmosphere. They do not accelerate but instead are created at their high relative velocity. They are a textbook example of time dilation without acceleration.

You can either say that velocity causes time dilation or that time dilation is just what happens when a clock takes a shorter path through spacetime. I prefer the second approach, which is the spacetime geometric explanation.

I take this to mean that E and F is equivalent to G and H and J and K.

 

[Wizardsblade]

Sorry I just realized my mistake with what Kev was saying. But I tend to think that this paradox is talking about the G,H / J,K pairs not the E,F pair. What makes you think its the E,f pair and not the others?

 

[yuiop]

...

Oooh, nice. I actually didn't read #1 until now.

One thing that would be nice to see in the diagram on the left is this: Two dots, somewhere in the middle of each world line, marking two events where the two clocks show the same time. (They would be on the same horizontal line in the diagram). Also, simultaneity lines, showing what events co-moving observers consider simultaneous with those two events.

I have uploaded two more diagrams.

The first has solid magenta lines added that show the lines of simultaneity of an inertial observer that remains in the initial reference frame of the green and yellow Bell's rockets.

The second has solid blue lines added that shows the lines of simultaneity of the blue rocket that represents an inertial observer that stays in the final reference frame of the Bells rockets.

Not exactly what you asked for but the rest can easily be extrapolated.

 

[yuiop]

Sorry I just realized my mistake with what Kev was saying. But I tend to think that this paradox is talking about the G,H / J,K pairs not the E,F pair. What makes you think its the E,f pair and not the others?

Have a look at this link to a drawing of the classic Minkoski diagram of an accelerating rocket. http://www.mathpages.com/home/kmath422/Image5456.gif it has the worldlines converging and represents Born rigid acceleration ,where the accelerating observers measure their separation to be constant while the inertial observer measures the separation of the accelerating observers to be length contracting. This represents the classic equivalence of an artificaly accelerated rocket to gravitational acceleration. The equivalence princple is only valid over a short range so if you wanted to nit pick you could point out that this represents acceleration proportional to GM/R rather the normal Newtonian GM/R^2. You could also point out that this classic equivalence diagram shows the event horizon being at the origin rather than at 2GM/Rc^2 but we won't go there for now. However, the point is that the acceleration is inversely proportional to the distance from the origin as is normal for gravitational type acceleration. This represents the G/H and J/K pairs.

The rockets in Bell's paradox experience constant and equal acceleration. Their wordlines in the Minkowski diagram would be parallel (translated) rather than converging as shown in the Born rigid case. The bell's rockets have acceleration that is independent of the displacement along the x axis. That is the E/F type acceleration described earlier.

 

[peter0302]

I believe I finally found the flaw in the logic and, I believe this proves the string will not break.

Let's start with the wikipedia article as a reference:
http://en.wikipedia.org/wiki/Bell's_spaceship_paradox

The very first assumption they make is that the following equations are true for all times "t":
x_a(t) = a0 + f(t)
x_b(t) = b0 + f(t)

This assumes that the observer's coordinate (let's call this O) is zero. But in reality the equations are:
x_a(t) = (a0 - O) + f(t)
x_b(t) = (b0 - O) + f(t)

Now here's where I believe the error is. I believe (a0 - O) and (b0 - O) need to be divided by gamma because those lengths, corresponding to the respective starting points of the rockets, should be length contracted just as surely as the length of the rockets themselves, no different than if the rockets had tails that extended all the way back to the Observer at the start of the experiment. In other words, once the rockets are moving at relativistic velocities, you can no longer simply add their original starting points relative to the inertial observer when the rockets were at rest without accounting for relativistic effects - length contraction - on that displacement caused by high velocity of the rocket. Otherwise you're adding apples and oranges.

If that's right, then the equation becomes:

x_a(t) = (a0 - O)/gamma + f(t)
x_b(t) = (b0 - O)/gamma + f(t)

x_a(t) - x_b(t) = (a0-O-b0+O)/gamma

x_a(t) - x_b(t) = (a0-b0)/gamma

So the apparent distance between the two rockets DOES length contract, exactly as we would expect it to if they were treated as one object, they do not appear to accelerate at the same rate in the inertial observer's frame, although they do in their own frame, and therefore the string does not break.

 

[Fredrik]

This assumes that the observer's coordinate (let's call this O) is zero. But in reality the equations are:
x_a(t) = (a0 - O) + f(t)
x_b(t) = (b0 - O) + f(t)

Now here's where I believe the error is. I believe (a0 - O) and (b0 - O) need to be divided by gamma because those lengths, corresponding to the respective starting points of the rockets, should be length contracted just as surely as the length of the rockets themselves,

Those lengths are Lorentz contracted in a frame that's co-moving with one of the rockets, but the equations above are in the original rest frame.

 

[Hurkyl]

I believe I finally found the flaw in the logic and, I believe this proves the string will not break.

Let's start with the wikipedia article as a reference:
http://en.wikipedia.org/wiki/Bell's_spaceship_paradox

The very first assumption they make is that the following equations are true for all times "t":
x_a(t) = a0 + f(t)
x_b(t) = b0 + f(t)

This assumes that the observer's coordinate (let's call this O) is zero. But in reality the equations are:

x_a(t) = (a0 - O) + f(t)
x_b(t) = (b0 - O) + f(t)

There is no third observer in Wikipedia's presentation of the experiment. And if there was a third observer stationary in the (t, x)-coordinate chart, a₀ was defined to be the initial x coordinate of spaceship A, not the initial coordinate distance between the third observer and the third observer, so you'd still be wrong.

Edit:Corrected a typo[/color]

 

[Fredrik]

There is no third observer in Wikipedia's presentation of the experiment. And if there was a third observer stationary in the (t, x)-coordinate chart, a₀ was defined to be the initial x coordinate of spaceship A, not the initial coordinate distance between the third observer and the third observer, so you'd still be wrong.

FYP

 

[peter0302]

Neither of you addressed my point, that you cannot use a measurement taken in an inertial frame and add it to a distance calculated using the proper acceleration of a relativistically moving object. You've got to include the gamme factor in the original distance measurement (a0, b0), OR adjust the displacement function for one of the two ships, which means f(t) for the first ship is not the same as f(t) for the second ship in the inertial observer's frame. Either way, you get the same result: x(a)-x(b) length contracts as the ships move faster.

 

[Hurkyl]

Neither of you addressed my point,

Your argument contains errors, and is therefore invalid.

And both of our responses touch upon what I think is your underlying problem -- in your (misguided) attempts to apply relativistic ideas, you've apparently forgotten how to use coordinates to analyze a problem.

 

[peter0302]

Since you can't give me any specific criticisms, your critiques are not helpful.