Clocks Within Each Ship in Bell Spaceship Paradox

[Stalin Beltran]

Yes the initial conditions are as described in Wikipedia.

PS. if you find that the string doesn't break, then it may be a good idea to present your calculation here for debugging. Comparing the situations at t₀=0 and t₁ >>0 suffices for the analysis.

Of course. I have it published it in my blog (It doesn't mean it is correct, it is just my analysis).
Bell’s spaceship thread does not breaks
The analysis basically goes like this:

Step 0: We have 2 spaceships, A and B, in the same frame of reference that a spaceship S. The rocket motors are about to start, but not yet. A, B and S are "simultaneous", so if motors start, they will start simultaneous to S too.

Step 1: A and B starts motors with exactly the same acceleration for a $\Delta t$ as short as we want. (A, B and S share the same frame of reference when motors start). This results in a final velocity $V_1$ for A and B because both the acceleration and the $\Delta t$'s are equal. At the end of this $\Delta t$ both A and B still share the same frame of reference because they both have the same velocity $V_1$. However S remains at rest, so S is now in another frame of reference. The relativity of simultaneity states that, at the end of this step, both A and B are not longer "simultaneous" to S, so saying that the accelerations are equal in the initial rest frame S is not possible at the end of this step. A and B are in the same frame of reference, so if they accelerate, they both will accelerate at the same time and in the same proportion.

This could be extended to N steps, and the results are the same. A and B being in the same frame of reference means they are at rest relative to each other. The distance between them will stay the same, so the string will not break.

 

[Nugatory]

Of course. I have it published it in my blog (It doesn't mean it is correct, it is just my analysis).
Bell’s spaceship thread does not breaks
The analysis basically goes like this:

Step 0: We have 2 spaceships, A and B, in the same frame of reference that a spaceship S. The rocket motors are about to start, but not yet. A, B and S are "simultaneous", so if motors start, they will start simultaneous to S too.

Step 1: A and B starts motors with exactly the same acceleration for a $\Delta t$ as short as we want. (A, B and S share the same frame of reference when motors start). This results in a final velocity $V_1$ for A and B because both the acceleration and the $\Delta t$'s are equal. At the end of this $\Delta t$ both A and B still share the same frame of reference because they both have the same velocity $V_1$. However S remains at rest, so S is now in another frame of reference. The relativity of simultaneity states that, at the end of this step, both A and B are not longer "simultaneous" to S, so saying that the accelerations are equal in the initial rest frame S is not possible at the end of this step. A and B are in the same frame of reference, so if they accelerate, they both will accelerate at the same time and in the same proportion.

This could be extended to N steps, and the results are the same. A and B being in the same frame of reference means they are at rest relative to each other. The distance between them will stay the same, so the string will not break.

There is no such thing as "being in a frame of reference", so any analysis that casually tosses that phrase around is suspect. The reference frame is simply a convention used for assigning x and t coordinates to events, so any reference frame can always be applied to everything in a given problem.

There is one reference frame in which the two spaceships are at rest relative to one another (meaning that $x_1(t)-x_2(t)$ is a constant where $x_1(2)$ and $x_2(t)$ are the x-coordinates of the leading and trailing spaceships at time $t$ using the $x$ and $t$ coordinates assigned by that frame). That is the frame in which the ground-based observer is at rest ($x_g(t)$ is constant for all $t$ where $x_g(t)$ is the x-coordinate of the ground-based observer's location at time t). In that frame the string is never at rest, so it length-contracts and breaks. In any other frame, the distance between the two ships $x_1(t)-x_2(t)$ (using the $x$ and $t$ coordinates of that frame) increases with time so the string breaks.

 

[harrylin]

I like the thought exper of two clocks, one stays, the other acclerated to the front. stays one year and then acclerated to the rear.
But, this non stationary clock would have had it's time "dilated" during two accelerations and would therefore show less time elapse, not more as you
suggested.
Correct?

Perhaps you are referring to post #13 by stevendaryl. He did not discuss a rocket that is in rest for one year, but a rocket that is accelerating for one year!

The two-way clock transport inside the rocket results in a very slight retardation of the transported clock on the rear clock. If that experiment was done in rest, the one year in the front would be totally irrelevant. And it is easy to understand that also in an accelerating rocket whatever the effect of this transport may be (that requires deeper analysis), it does not depend on the time that the clock is staying in the front of the rocket. Now consider what happens during the year of acceleration. Due to length contraction of the rocket, the front clock will at any instant be going at a very slightly lesser speed than the rear clock, so that during that time the rear clock delays on the front clock. The longer the transported clock stays in front, the greater the delay of the rear clock will be.

 

[harrylin]

[..] A and B are in the same frame of reference, so if they accelerate, they both will accelerate at the same time and in the same proportion. [..] A and B being in the same frame of reference means they are at rest relative to each other.

Coincidentally, I had just before your post identified the interpretation error behind those phrases!
See my post #87.

Proportional mapping between inertial reference frames was assumed for deriving the Lorentz transformations, and inertial A and B in rest relative to each other can according to SR be used to set up an inertial frame for describing natural phenomena. It is the misapplication of such assumptions to mapping of an inertial frame with an accelerating frame that leads to paradoxes of this kind.

 

[Stalin Beltran]

In any other frame, the distance between the two ships x1(t)−x2(t)x_1(t)-x_2(t) (using the xx and tt coordinates of that frame) increases with time so the string breaks.

I can see the frame where the spaceships are at rest. I can see the frame where the string contract. But I just don't see the frame where the distances between spaceships increases. Could you elaborate this please?

 

[Dale]

although time dilatation in accelerated clocks is well accepted, no one has shown length contraction in real life.
Correct?

Lorentz seemed to think that Michelson and Morely showed length contraction in real life. Personally, I think there is a lot of evidence showing length contraction in real life, from MM, to muons, to particle bunches in accelerators, to magnetism.

 

[1977ub]

PS. question to 1977ub: was something like that also the reason for your confusion, or was there a different cause?

Not exactly. But the confusion re laws-of-physics vs lorentz contraction is interesting part of this.

 

[Stalin Beltran]

Lorentz seemed to think that Michelson and Morely showed length contraction in real life. Personally, I think there is a lot of evidence showing length contraction in real life, from MM, to muons, to particle bunches in accelerators, to magnetism.

As far as I understand, length contraction is the way an observer traveling with speed V sees the world in the direction of movement. For example, a muon doesn't notice its clock had slow down. Instead it see its distance to Earth has contracted, and that is the way it reaches the Earth's surface.

Am I right?

 

[Nugatory]

I can see the frame where the spaceships are at rest. I can see the frame where the string contract. But I just don't see the frame where the distances between spaceships increases. Could you elaborate this please?

There is no frame in which the spaceships are both at rest (except the ground frame before and up to the moment that they light off their engines). They remain at rest relative to another (that is, moving in the same direction at the same speed) in the ground frame but of course they are not at rest in that frame once the engines have fired to start them moving. In all other frames, the ships are moving relative to one another.

 

[Dale]

As far as I understand, length contraction is the way an observer traveling with speed V sees the world in the direction of movement. For example, a muon doesn't notice its clock had slow down. Instead it see its distance to Earth has contracted, and that is the way it reaches the Earth's surface.

Am I right?

Yes. Although I am not a fan of the "observer" phrasing. I prefer to simply speak of reference frames and not insist on attaching them to observers.

 

[Stalin Beltran]

To complete the description about length contraction, we must say the length of an object is not contracted (as we usually tend to think) if you take a photograph of it. As stated in wikipedia:

"It was shown by several authors such as Roger Penrose and James Terrell that moving objects generally do not appear length contracted on a photograph. For instance, for a small angular diameter, a moving sphere remains circular and is rotated. This kind of visual rotation effect is called Penrose-Terrell rotation."

 

[Dale]

Really? That is the big "proof" you think is missing? What then, photos of different size objects and different shape objects?

If you are determined to reject SR regardless of the evidence then you can always find some experiment that was not performed and claim incomplete evidence. However, whatever missing experiments you choose to complain about, you are still left with the challenge of explaining all of the previous experiments that have been performed:
http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html

 

[1977ub]

To complete the description about length contraction, we must say the length of an object is not contracted (as we usually tend to think) if you take a photograph of it. As stated in wikipedia:

"It was shown by several authors such as Roger Penrose and James Terrell that moving objects generally do not appear length contracted on a photograph. For instance, for a small angular diameter, a moving sphere remains circular and is rotated. This kind of visual rotation effect is called Penrose-Terrell rotation."

Length contraction and time dilation are not directly perceived but are reconstructed ala forensic analysis and require that you are satisfied that there are clocks remote from the observer which are synchronized with his clock. A photograph does not presume to take measurements of different parts of an object simultaneously in one's own frame - and so doesn't presume to make any statement about actual lengths - contracted or uncontracted.

 

[Stalin Beltran]

Really? That is the big "proof" you think is missing? What then, photos of different size objects and different shape objects?

If you are determined to reject SR regardless of the evidence then you can always find some experiment that was not performed and claim incomplete evidence. However, whatever missing experiments you choose to complain about, you are still left with the challenge of explaining all of the previous experiments that have been performed:
http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html

Proof against Bell Paradox? Not at all. If I try to reject Bell Paradox saying that length contraction is not real that will be circular reasoning. The last post was to complete the idea of length contraction asked by bligh in post 89.

What my analysis reveals about Bell Paradox is that A and B could not accelerate simultaneously to frame S. Please note that if we found (like Bell did) that A and B are accelerating simultaneously in the S frame, then Bell is right and the string breaks.

 

[Stalin Beltran]

Length contraction and time dilation are not directly perceived but are reconstructed ala forensic analysis and require that you are satisfied that there are clocks remote from the observer which are synchronized with his clock. A photograph does not presume to take measurements of different parts of an object simultaneously in one's own frame - and so doesn't presume to make any statement about actual lengths - contracted or uncontracted.

To be honest I have not checked the Roger Penrose paper (I am looking for other facts of relativity), but It sounds good. While studying the relativity of simultaneity I found that length contraction could be an "artifact" of the relativity of simultaneity. That is, the object could actually no contract, but look contracted when we measure it.

 

[stevendaryl]

To be honest I have not checked the Roger Penrose paper (I am looking for other facts of relativity), but It sounds good. While studying the relativity of simultaneity I found that length contraction could be an "artifact" of the relativity of simultaneity. That is, the object could actually no contract, but look contracted when we measure it.

You have to be careful about what it means to "actually" contract, as opposed to "looking contracted when we measure it".

 

[Dale]

What my analysis reveals about Bell Paradox is that A and B could not accelerate simultaneously to frame S.

Sure they can. And if they don't then you are not analyzing Bells Paradox but instead analyzing some other scenario.

 

[A.T.]

...look contracted when we measure it.

A fast moving sphere doesn't look contracted visually, despite actually being contracted as a measurement shows.

 

[1977ub]

A fast moving sphere doesn't look contracted visually, despite actually being contracted as a measurement shows.

Such a measurement will also show that the back of the sphere is older than the front FWIW.

 

[A.T.]

Such a measurement will also show that the back of the sphere is older than the front FWIW.

If the sphere was linearly accelerated from rest, then the back has aged less, not more, than the front in that process.

 

[1977ub]

If the sphere was linearly accelerated from rest, then the back has aged less, not more, than the front in that process.

Interesting. Is this frame dependent? This effect must battle with the frame-dependent effect whereby the back clock of relativistic-velocity non-accelerating train has rec'd its ticks sooner than the train's front clock as determined by observer "on the platform." If the train is long enough & moving fast enough, back clock may have ticked quite a bit sooner than front...

 

[Nugatory]

Interesting. Is this frame dependent? This effect must battle with the frame-dependent effect whereby the back clock of relativistic-velocity non-accelerating train has rec'd its ticks sooner than the train's front clock as determined by observer "on the platform." If the train is long enough & moving fast enough, back clock may have ticked quite a bit sooner than front...

It is not frame dependent. Light reflected from different parts of the moving object reaches the photographic film at the moment that the shutter of the camera is open, and all observers in all frames must agree that it's the same light that hits the film.

There's no great mystery here. All photographs always show different things as they appeared at different times. Right now I'm looking at a photo of a bird in a tree in the front yard of my house, maybe ten meters from the camera, and my neighbor's house about 100 meters farther from the camera is in the background. The photograph has accurately captured the light that reached the film, but in the image the bird is about 300 nanoseconds younger than the house because that's how much longer it took the light reflected from my neighbor's house to travel to the camera.

Usually we can ignore this effect and act as if the photograph has captured an "at the same time" picture of what's going on (try googling around to find the origin of the term "photo finish") because the differences are not noticeable - it's not as if my neighbor's house changes much in 300 nanoseconds. If something is moving at a significant fraction of the speed of light, so that one end moves a fair amount during the time that it takes for light from the other end to make it to the camera, then the differences start to matter.

 

[A.T.]

Is this frame dependent?

I don't think so. During the acceleration of a rigid object the back ages slower than the front. The offset in age remains after the acceleration ended, so all frames agree there is an offset compared to the initial synchronization.

 

[1977ub]

It is not frame dependent. Light reflected from different parts of the moving object reaches the photographic film at the moment that the shutter of the camera is open, and all observers in all frames must agree that it's the same light that hits the film.

There's no great mystery here. All photographs always show different things as they appeared at different times. Right now I'm looking at a photo of a bird in a tree in the front yard of my house, maybe ten meters from the camera, and my neighbor's house about 100 meters farther from the camera is in the background. The photograph has accurately captured the light that reached the film, but in the image the bird is about 300 nanoseconds younger than the house because that's how much longer it took the light reflected from my neighbor's house to travel to the camera.

Usually we can ignore this effect and act as if the photograph has captured an "at the same time" picture of what's going on (try googling around to find the origin of the term "photo finish") because the differences are not noticeable - it's not as if my neighbor's house changes much in 300 nanoseconds. If something is moving at a significant fraction of the speed of light, so that one end moves a fair amount during the time that it takes for light from the other end to make it to the camera, then the differences start to matter.

I was responding to the issue

I don't think so. During the acceleration of a rigid object the back ages slower than the front. The offset in age remains after the acceleration ended, so all frames agree there is an offset compared to the initial synchronization.

Hmm. In Bell scenario, clocks by engines of both ships continue to tick at the same rate for the initial frame observer.
of the back clock in an accelerating ship being younger due to unequal clock rates vs being older in constant velocity ship due to RoS.

 

[A.T.]

Hmm. In Bell scenario...

I wasn't talking about Bell scenario. I was talking about accelerating a rigid object that keeps its proper length, so it gets shorter during the acceleration in the initial inertial rest frame.

 

[1977ub]

I wasn't talking about Bell scenario. I was talking about accelerating a rigid object that keeps its proper length, so it gets shorter during the acceleration in the initial inertial rest frame.

Taking a rigid object that accelerates to velocity V and stays there, there is the time difference that the clock in back of rigid object is younger than the front as determined in the original frame - due to acceleration, and there is the time difference that a clock in the moving vehicle is measured older for the original observer due to RoS. I would imagine there are cases where one effect predominates, and other cases where the other one does?

 

[A.T.]

I would imagine there are cases where one effect predominates, and other cases where the other one does?

I have no idea what two effects you talk about. Both clocks are subject to time dilation in the initial inertial rest frame. But during the acceleration phase their time dilation is different.

 

[1977ub]

I have no idea what two effects you talk about. Both clocks are subject to time dilation in the initial inertial rest frame. But during the acceleration phase the time dilation is different.

A) During acceleration of a rigid object, the clocks at both ends tick at different rates. (back clock is younger)
(Then let's say the object stabilizes at a particular velocity.)
B) A rigid object measured by observer in the initial frame has clocks at both ends that an observer moving with the object believes to be ticking at the same rate as one another. The observer in the initial frame agrees that they are ticking at the same rate as one another, but finds that the clock at the back end ticks first - before the one at the front end. This is the familiar connection between Lorentz contraction and RoS. (back clock is older)

 

[A.T.]

A) During acceleration of a rigid object, the clocks at both ends tick at different rates. (back clock is younger)
(Then let's say the object stabilizes at a particular velocity.)
B) A rigid object measured by observer in the initial frame has clocks at both ends that an observer moving with the object believes to be ticking at the same rate as one another. The observer in the initial frame agrees that they are ticking at the same rate as one another, but finds that the clock at the back end ticks first - before the one at the front end. This is the familiar connection between Lorentz contraction and RoS. (back clock is older)

No. Both frames will agree that the back clock lags behind the front clock, after the acceleration stopped.

 

[1977ub]

No. Both frames will agree that the back clock lags behind the front clock, after the acceleration stopped.

Let us make clear that I refer to a specific frame - let's say S, where the rigid object begins its acceleration.
There is a fixed quantity of difference which builds up between the clocks over the period of the acceleration, as measured by S, and makes the back clock appear to be younger. Ok so far?
variant A: after reaching target velocity V, observers in the object synchronize their clocks. Thereafter, everyone in S (or any other frame moving wrt the object) finds that the clock at the back reads a later time than the clock at the front.
variant B: but what if the synchronization never takes place. Since the acceleration difference puts the back clock ahead and the velocity difference puts the back clock behind, there might in fact be a perfect arrangement whereby the front and back clocks could start out reading the same time as one another before the acceleration begins, and then also end up reading the same time as one another during the velocity phase according to observers in S - right?