Why is the Wikipedia article about Bell's spaceship paradox disputed at all?

[quantum123]

If Lorentz transform does not apply to accelerating observers, will he/she still experience the Doppler effect of SR?

 

[nakurusil]

If Lorentz transform does not apply to accelerating observers, will he/she still experience the Doppler effect of SR?

Yes, there was another thread on this subject. The equations of the relativistic Doppler effect under accelerated motion will be different from the ones for inertial motion.

 

[quantum123]

It would be nice to see more diagrams, mathematical equations, clear definitions and logic in this discussion. Words are very vague and can be misused easily.
I thought Latex is rather easy to use and you can even send in attachment of pictures?

 

[Fredrik]

You insult me because I exposed your erroneous claims and you ran out of logical and pertinent arguments?

No, I have already explained why I'm calling you a troll. Here's a repost of that:

Why don't you answer my question about why you pretended that I had made claims in post #31?

Why don't you say anything about the fact that you claimed to have discussed the details of #1 with me before, when in fact you had not?

Why don't you say anything about the fact that you claimed that I had said that the object in #8 is rigid, when in fact I had never done so?

You have no right to whine about "personal attacks" as long as you talk to me as if I have opinions that you know I don't have. If you stop doing that, I won't call you a troll again.

Your reply to that was pathetic. You just brought up #8 again, even though your objections to it is just a bunch of stuff that I understood better than you do now when I wrote #8.

What I said in post #8 is still possible in principle. It has nothing to do with Born rigidity though, and nothing to do with what this thread is about. Hmm, I have a feeling I've told you that already.

You gave me plenty of new reasons to call you a troll in #64 and #65. Specifically, you keep returning to my #8 over and over again when we're discussing things that we both know have absolutely nothing to do with #8. You also ignored everything I said about what we were really talking about. That's troll behavior.

By the way, I still want answers to those questions above.

 

[Fredrik]

Lorentz transforms do not apply to accelerated motion. Didn't you know that?
Now I understand your insistance in claiming that "The distance between them is always L in the launcher's frame, so it can't be a constant in the mother ship's frame. (This is easy to see in the space-time diagram). It will be "uncontracted" from L/gamma to L."

I don't think you do. The reason I'm so sure about that is that the specifications of the problem imply that both spaceships will have identical world lines in the launcher's frame, except for their starting position.

You can't simply brute force the problem and apply length contraction, you need to use the equations of hyperbolic motion in order to calculate the proper distance. Didn't you know that? Apparently not.

You need those equations in the version of the problem where the acceleration goes on forever (aren't you dismissing that scenario as "unphysical"?). You don't need them if the rockets turn off their engines after a predefined proper time. (OK, you need them if you want to know the final velocity, but what I said is true no matter what the final velocity is). Once the engines have been turned off the rockets are moving at a constant velocity, so now it's only a matter of comparing the spatial distances in two inertial frames. That's Lorentz contraction, no matter what you want to call it.

 

[Fredrik]

It would be nice to see more diagrams, mathematical equations, clear definitions and logic in this discussion. Words are very vague and can be misused easily.
I thought Latex is rather easy to use and you can even send in attachment of pictures?

OK, let me try to define the problem more clearly.


The problem (my version)

Two spaceships are stationary in a certain inertial frame (the launcher's frame), with their engines turned off. The spaceships are identical in every way in this frame, except of course for their position in space. (Note that this includes clock synchronization, and that they are aimed in the same direction).

The distance between them is L. A string of length L has been attached to the spaceships. To avoid irrelevant complications, we assume that both end points of the string are attached to the same point on both spaceships.

The string is very weak and would break if it's stretched to a length longer than L. The force it exerts on the spaceships is assumed to be negligible.

The engines of both spaceships are controlled by computers on the ships. The computers are of course identical and running identical programs. The program was written to make sure that the engines will start at certain time, and shut off after a finite proper time T. When the engines are on, the ships will accelerate in one of the two directions defined by a straight line between the attachment points of the string.

Will the string break?


Other versions

It is usually postulated that the engines will cause a constant acceleration. I've chosen not to postulate that because it restricts our attention to a special case without making the problem easier.

It's sometimes postulated that the acceleration will go on forever.

It's sometimes not postulated that the engines will shut off.

Apparently there are people who postulate that the string is attached to the front of one ship and to the rear of the other. That only adds a complication to the problem that obscures the real issue.

The problem can also be considered in other space-times than Minkowski space. That makes the problem more complicated. Pervect made a post about this.


The solution

The world lines of the attachment points of the string must be identical in the launcher's frame, except for their starting position in space. (If they weren't, there would be something fundamentally different about those two points in space, and that would violate special relativity). This implies that the distance between the attachment points, and hence the length of the string, in the launcher's frame will be constant. However, since the spaceships' velocity has changed, the string must now be Lorentz contracted. It's Lorentz contracted and it's the same length! This means that its proper length must have increased.

So the string must break.

I'm not going to draw a space-time diagram. The one in the Wikipedia article is good enough.

 

[Boustrophedon]

Your description is acceptable up to the point where you just "throw in" the casual "the string must now be Lorentz contracted". I understand that you mean 'from the launcher frame' but the snag is - where does this glib assumption come from ? Certainly not from Einstein's SR: it is, in fact, the central postulate of Lorentz's theory that was supplanted by SR, and the postulate itself supplanted by Einstein's "constant velocity of light for all inertial observers" postulate instead ( i.e. not 'as well as' ).

The reason acceleration was for a long time not dealt with in SR was because all the original papers dealt only with constant uniform motion from which we knew that lengths in A measured shorter in B and vice versa etc. but it was an open question what would happen if A accelerated to B's frame or vice versa.

Although Bell prefers to champion Lorentz's approach by assuming a length here and now becomes shorter from here when accelerated but constant for a co-mover, it is equally possible to consider the co-mover's lengths becoming longer whilst the length from 'here' stays constant.

The latter is preferable for three reasons:
1) It's consistent with SR being "kinematical" and involving no 'physical' effects - everything follows from the constant c postulate.
2)It's consistent with the 'changing' length effect being associated with the accelerated system rather than the system that is undisturbed.
3)It's consistent with the behaviour of two separate, identically accelerated bodies - particularly in view of the fact that the derivation of SR makes no distinction between solid lengths and separated bodies - it deals abstractly with two points a distance apart moving at certain velocity.

 

[nakurusil]

No, I have already explained why I'm calling you a troll.

Every time you respond with unwarranted abuse I will remind you of your "understanding" (better said lack of thereof) of elementary physics:

Just adding to the list...

5. Every point of the rod is instantaneously (or near instantaneously) boosted to a new velocity, all at the same time in the frame where the rod was at rest before the boost. This stretches the rod to a longer proper length.
6. Every point of the rod is instantaneously (or near instantaneously) boosted to a new velocity, all at the same time in the frame where the rod will be at rest after the boost. This compresses the rod to a shorter proper length.

What I said in post #8 is still possible in principle. It has nothing to do with Born rigidity though, and nothing to do with what this thread is about. Hmm, I have a feeling I've told you that already.

Never admit to your errors, even if it means embarassing you furthr.

 

[nakurusil]

I don't think you do. The reason I'm so sure about that is that the specifications of the problem imply that both spaceships will have identical world lines in the launcher's frame, except for their starting position.


You need those equations in the version of the problem where the acceleration goes on forever (aren't you dismissing that scenario as "unphysical"?). You don't need them if the rockets turn off their engines after a predefined proper time. (OK, you need them if you want to know the final velocity, but what I said is true no matter what the final velocity is). Once the engines have been turned off the rockets are moving at a constant velocity, so now it's only a matter of comparing the spatial distances in two inertial frames. That's Lorentz contraction, no matter what you want to call it.

You may continue claiming that you can use Lorentz transformations and , subsequently, Lorentz contraction all you want. As a matter of fact, you can even use binary arithmetic and Fortran programming. The thing is that you will not get the correct answer to the problem. Lorentz transforms cannot be applied to accelerated motion and expect correct results. Do you know why? Didn't they teach you that in your school? Apparently not.

 

[nakurusil]

Although Bell prefers to champion Lorentz's approach by assuming a length here and now becomes shorter from here when accelerated but constant for a co-mover, it is equally possible to consider the co-mover's lengths becoming longer whilst the length from 'here' stays constant.

The latter is preferable for three reasons:
1) It's consistent with SR being "kinematical" and involving no 'physical' effects - everything follows from the constant c postulate.
2)It's consistent with the 'changing' length effect being associated with the accelerated system rather than the system that is undisturbed.
3)It's consistent with the behaviour of two separate, identically accelerated bodies - particularly in view of the fact that the derivation of SR makes no distinction between solid lengths and separated bodies - it deals abstractly with two points a distance apart moving at certain velocity.

Yep, this is the mess of contradictions that one gets into when one insists on using a mathematical formalism (Length contraction/Lorentz transforms) that should NOT be used because it does NOT apply to accelerated motion.

 

[disregardthat]

What applies to the accelerated motion?

Do you deny that an object has a certain velocity that is higher than at the start at one point during the acceleration?

 

[nakurusil]

What applies to the accelerated motion?

Hyperbolic motion. Was designed specifically to deal with accelerated motion in SR.

Do you deny that an object has a certain velocity that is higher than at the start at one point during the acceleration?

What makes you think that I did that? I am just saying that the equations of hyperbolic motion should be used in order to accurately describe accelerated motion in SR.Do you understand why?

 

[disregardthat]

Don't you think it is a little weird that you are the only one that think that lorentz length contraction does not apply?

You are saying that lorentz contraction does not apply here, the only reason for that must be that you are denying that the velocities of the spaceships are the same as the rest-frame... I cannot see any other reason.

 

[quantum123]

Fred:
I have just drawn some spacetime pictures and I think I got you mean - simple enough.
One suggestion - redraw the picture from the frame of one of the rockets when it is very close to the speed of light. Post it as attachment if possible. The picture should highlight the different rates of acceleration of the rockets at the same time, and also the increase in separation of the rocket. Wiki didn't do that.

 

[nakurusil]

Don't you think it is a little weird that you are the only one that think that lorentz length contraction does not apply?

Physics is not a popularity contest.

You are saying that lorentz contraction does not apply here, the only reason for that must be that you are denying that the velocities of the spaceships are the same as the rest-frame... I cannot see any other reason.

I think that you need to take some classes, your post above makes no sense.

 

[disregardthat]

Ok...

You are saying that the lorentz contraction does not apply in this "paradox"
The only reason that you would say that is to me because:
You are denying that the object moving actually is changing velocities when it is accelerating.
Because any object in a different velocity thant in another frame, is contracted by the factor of root'(1+v^/c^2))

If a lorentz contraction should not apply if the object is not in constant motion, please tell me why.

 

[nakurusil]

Ok...

You are saying that the lorentz contraction does not apply in this "paradox"
The only reason that you would say that is to me because:
You are denying that the object moving actually is changing velocities when it is accelerating.

Umm, no.I am not denying that an object is changing speed. I am telling you that you cannot apply length contraction in order to solve this problem. You need to take some classes before attempting to discuss this subject.

Because any object in a different velocity thant in another frame, is contracted by the factor of root'(1+v^/c^2))

You got the factor wrong and you got the idea of applying length contraction wrong. Everything else is right

If a lorentz contraction should not apply if the object is not in constant motion, please tell me why.

Based on your prior posts, this subject is far too advanced for you to grasp. Your buddy Fredrik might be able to explain this to you once he gets it.

 

[Fredrik]

Your description is acceptable up to the point where you just "throw in" the casual "the string must now be Lorentz contracted". I understand that you mean 'from the launcher frame' but the snag is - where does this glib assumption come from ? Certainly not from Einstein's SR

I'm not using anything except the fact that the space-time we're talking about is Minkowski space, and that the two world lines are identical in the launcher's frame except for their starting position.

I will try to explain it in more detail. Look at the space-time diagram in the Wikipedia article. The distance between A and B is the same as the distance between A' and B'. This illustrates how the distance between the two attachment points remains the same in the launcher's frame.

When I said that "the string must now be Lorentz contracted", the "now" I had in mind is the time of event A'. (It would work just as well if I had chosen another event, but then I would have had to draw a new space-time diagram). It was also kind of misleading to say "the string". What I really had in mind was the distance between the two attachment points. That's what's getting Lorentz contracted.

My statement that the distance between the attachment points is Lorentz contracted to length L in the launcher's frame, means that its proper length, i.e. it's length as measured by a co-moving inertial observer is gamma*L.

What we need to understand to see this is to a co-moving observer, the distance between the attachment points is the space-time distance along the dotted line between A' and B'', because that line is the subset of Minkowski space that the co-moving observer considers "space" at the time of event A'.

It is straightforward to calculate what distance the co-moving observer would measure. The math is the same as for any other Lorentz contraction and the result is gamma*L.

This is one way to do it explicitly.

The slope of the solid lines are 1/v. The slope of the dotted line is v. We are looking for the space-time distance between A' and B''. Let's call this quantity M(A',B''). By definition (of special relativity), this quantity is frame-independent, so we can choose any frame for the calculation. I choose one that's co-moving with the launcher and has it's origin at A'.

$$M(A',B'')^2=-t(B'')^2+x(B'')^2=x(B'')^2 \big(1-\frac{t(B'')^2}{x(B'')^2}\big)$$
$$=x(B'')^2 (1-v^2)=\gamma^{-2} X(B'')^2$$

I used the fact that the slope of the dotted line is v. Now we have to calculate x(B'').

$$x(B'')=x(B')+K=L+K$$

where K satisfies

$$\frac{t(B'')}{K}=\frac{1}{v}$$

I used the fact that the slope of the solid line is 1/v. Now we see that

$$K=vt(B'')=v^2x(B'')$$

I used the fact that the slope of the dotted line is v. Insert this into the equation where we first used the variable K=x(B'')-x(B').

$$x(B'')=L+K=L+v^2x(B'')$$

and solve for x(B'').

$$x(B'')=\frac{L}{1-v^2}=\gamma^2L$$

Now use this in the calculation of the invariant distance.

$$M(A',B'')^2=\gamma^{-2} X(B'')^2=\gamma^{-2} \gamma^4 L^2=\gamma^2 L^2$$

$$M(A',B'')=\gamma L$$

...and we're done.

 

[nakurusil]

I'm still editing this post. Don't bother replying to it yet.I'm not using anything except the fact that the space-time we're talking about is Minkowski space, and that the two world lines are identical in the launcher's frame except for their starting position.

I will try to explain it in more detail. Look at the space-time diagram in the Wikipedia article. The distance between A and B is the same as the distance between A' and B'. This illustrates how the distance between the two attachment points remains the same in the launcher's frame.

When I said that "the string must now be Lorentz contracted", the "now" I had in mind is the time of event A'. (It would work just as well if I had chosen another event, but then I would have had to draw a new space-time diagram). It was also kind of misleading to say "the string". What I really had in mind was the distance between the two attachment points. That's what's getting Lorentz contracted.

My statement that the distance between the attachment points is Lorentz contracted to length L in the launcher's frame, means that its proper length, i.e. it's length as measured by a co-moving inertial observer is gamma*L.

What we need to understand to see this is to a co-moving observer, the distance between the attachment points is the space-time distance along the dotted line between A' and B'', because that line is the subset of Minkowski space that the co-moving observer considers "space" at the time of event A'.

It is straightforward to calculate what distance the co-moving observer would measure. The math is the same as for any other Lorentz contraction and the result is gamma*L.

This is one way to do it explicitly.

The slope of the solid lines are 1/v. The slope of the dotted line is v. We are looking for the space-time distance between A' and B''. Let's call this quantity M(A',B''). By definition (of special relativity), this quantity is frame-independent, so we can choose any frame for the calculation. I choose one that's co-moving with the launcher and has it's origin at A'.

$$M(A',B'')^2=-t(B'')^2+x(B'')^2=x(B'')^2 (1-\frac{t(B'')^2}{x(B'')^2})=x(B'')^2 (1-v^2)=\gamma^{-2} X(B'')^2$$

I used the fact that the slope of the dotted line is v. Now we have to calculate x(B'').

$$x(B'')=x(B')+K=L+K$$

where K satisfies

$$\frac{t(B'')}{K}=\frac{1}{v}$$

Here I'm using that the slope of the solid line is 1/v. So we have that

$$K=vt(B'')=v^2x(B'')$$

Here I'm using that the slope of the dotted line is v. Insert this into the equation that defined our K

[tex]x(B'')=L+v^2x(B'')[tex]

and solve for x(B'').

$$x(B'')=\frac{L}{1-v^2}=\gamma^2L$$

Now use this in the calculation of the invariant distance.

$$M(A',B'')^2=\gamma^{-2} X(B'')^2=\gamma^{-2} \gamma^4 L^2=\gamma^2 L^2$$

$$M(A',B'')=\gamma L$$

You are still using Lorentz transforms.
Try using the appropriate instruments, you might be able to get the correct result. Here:
http://www.ph.utexas.edu/~gleeson/NotesChapter13.pdf

 

[Fredrik]

You are still using Lorentz transforms.
Try using the appropriate instruments, you might be able to get the correct result. Here:
http://www.ph.utexas.edu/~gleeson/NotesChapter13.pdf

Wow, it must have been like your birthday when you got a chance to quote one of my posts in its entirety before it was finished.

You still got it wrong. My solution is more general than the one you linked to. There's no need to postulate that the acceleration is constant, so I haven't used that detail.

 

[Fredrik]

If a lorentz contraction should not apply if the object is not in constant motion, please tell me why.

Lorentz contraction does apply to accelerated objects, but the Lorentz contraction formula won't help you calculate the length of an object in a certain frame if the object is also being stretched or squeezed while it's being accelerated. That's why the concept of "Born rigid acceleration" was invented.

An object is said to be doing Born rigid acceleration if the Lorentz contraction formulas can be used at any time during the acceleration to calculate the distance, in any frame, between any two points on the object.

Edit: [ The last sentence above should have said "any two points that are infinitesimally close to each other". This condition actually implies that e.g. the end points won't be moving at the same velocity, so we can't use the Lorentz contraction formula to calculate the distance between them. ]

That's what's so funny about Nakurusil's claims. He keeps saying that it's "Born rigidity" that shrinks an accelerating rocket, rather than Lorentz contraction, when in fact Born rigidity is just a name for a mathematical idealization of how solid objects get Lorentz contracted. It's extra funny since he's also rejecting mathematical idealizations.

 

[nakurusil]

Wow, it must have been like your birthday when you got a chance to quote one of my posts in its entirety before it was finished.

You still got it wrong. My solution is more general than the one you linked to. There's no need to postulate that the acceleration is constant, so I haven't used that detail.

Umm, no. You insist on using Lorentz transforms and length contraction.The Lorentz transforms are linear, the transforms that you need to use (if you read the reference I gave you) are hyperbolic (on linear). If you apply the correct mathyou will get the correct results.If you insist in applying the incorrect math , you will continue to get the incorrect results. The Lorentz transforms don't work with accelerated motion (contrary to your insitance), don't they teach that in Sweden? I think they do teach it but you must have missed the class.

 

[MeJennifer]

They don't work with accelerated motion, don't they teach that in Sweden? I think they do teach it but you must have missed the class.

This "Lorentz transformations don't work with accelerated motion" matra that you are proclaiming here is getting a bit old.

Perhaps it is time for you to explain what exactly does not work and why instead of insulting everybody here.

 

[nakurusil]

Lorentz contraction does apply to accelerated objects, but the Lorentz contraction formula won't help you calculate the length of an object in a certain frame if the object is also being stretched or squeezed while it's being accelerated. That's why the concept of "Born rigid acceleration" was invented.

An object is said to be doing Born rigid acceleration if the Lorentz contraction formulas can be used at any time during the acceleration to calculate the distance, in any frame, between any two points on the object.

That's what's so funny about Nakurusil's claims. He keeps saying that it's "Born rigidity" that shrinks an accelerating rocket, rather than Lorentz contraction, when in fact Born rigidity is just a name for a mathematical idealization of how solid objects get Lorentz contracted. It's extra funny since he's also rejecting mathematical idealizations.

You are mixing up two of my criticisms of your approach:

1. Your claims 5,6 that are in violation of Born rigid motion (looks like you read a lot on it in the last two days, this is good) as applied to calculating the the SPEED of individual parts as the rear and the front of a SINGLE rocket.

2. The fact that you cannot use the Lorentz transforms in order to calculate the SEPARATION distance between TWO rockets in ACCELERATED motion.

Try getting your facts straight, will you?

 

[nakurusil]

This "Lorentz transformations don't work with accelerated motion" matra that you are proclaiming here is getting a bit old.

Perhaps it is time for you to explain what exactly does not work and why instead of insulting everybody here.

I'm not insulting anybody. I explained this issue several times, go back and re-read my posts. You can also read the refence to the appropriate treatment of accelerated motion I gave Fredrick a few posts back.Here is another one, specially for you:
http://www.arxiv.org/PS_cache/physics/pdf/0405/0405038.pdf

 

[Fredrik]

You are mixing up two of my criticisms of your approach:

1. Your claims 5,6 that are in violation of Born rigid motion (looks like you read a lot on it in the last two days, this is good) as applied to calculating the the SPEED of individual parts as the rear and the front of a SINGLE rocket.

2. The fact that you cannot use the Lorentz transforms in order to calculate the SEPARATION distance between TWO rockets in ACCELERATED motion.

Try getting your facts straight, will you?

1. That's still not relevant, because 5 and 6 have nothing to do with the subject of this thread. And as I said before, I understood Born rigidity very well before I started this thread.

2. I'm not calculating any distances in an accelerated frame. I'm using inertial frames. (In my version of the problem the rockets have turned off their engines and are moving at constant velocity at the events where I do the calculation, but that's actually irrelevant. If they had been accelerating, I could have used a co-moving inertial frame).

I will explain to you one last time why my 5 and 6 (i.e. post #8) are valid. That post was a digression from the main topic of this thread. We were talking about SR in general, and not specifically about the spaceship problem. 5 and 6 deserve to be on that list because they are verbal descriptions of a set of curves in Minkowski space. That makes them valid. SR is the claim that space and time can be represented by Minkowski space, so 5 and 6 are definitely allowed by the rules of SR. They also have a pedagogical value, as I have explained before. They do not however have anything at all to do with the subject of this thread, so you can't "prove me wrong" by explaining that they are impossible in the real world, or that they don't help us solve the spaceship problem. We weren't talking about the real world, or about the spaceship problem. We were talking about SR.

 

[nakurusil]

1. That's still not relevant, because 5 and 6 have nothing to do with the subject of this thread. And as I said before, I understood Born rigidity very well before I started this thread.

2. I'm not calculating any distances in an accelerated frame. I'm using inertial frames. (In my version of the problem the rockets have turned off their engines and are moving at constant velocity at the events where I do the calculation, but that's actually irrelevant. If they had been accelerating, I could have used a co-moving inertial frame).

The string(rod) gets stretched starting from from time 0, doesn't it? Not only after the acceleration has stopped. If you want to compute the string
stretching during the acceleration phase you cannot use Lorentz transforms.OK?

If you are simply trying to calculate the rocket separation AFTER the engines have shut off, then you are making up your own, simplified problem. The string may have long snapped during the acceleration phase. Read the reference I gave you, would you?

I will explain to you one last time why my 5 and 6 (i.e. post #8) are valid. That post was a digression from the main topic of this thread. We were talking about SR in general, and not specifically about the spaceship problem. 5 and 6 deserve to be on that list because they are verbal descriptions of a set of curves in Minkowski space. That makes them valid. SR is the claim that space and time can be represented by Minkowski space, so 5 and 6 are definitely allowed by the rules of SR. They also have a pedagogical value, as I have explained before. They do not however have anything at all to do with the subject of this thread, so you can't "prove me wrong" by explaining that they are impossible in the real world, or that they don't help us solve the spaceship problem. We weren't talking about the real world, or about the spaceship problem. We were talking about SR.

5. Every point of the rod is instantaneously (or near instantaneously) boosted to a new velocity, all at the same time in the frame where the rod was at rest before the boost. This stretches the rod to a longer proper length.
6. Every point of the rod is instantaneously (or near instantaneously) boosted to a new velocity, all at the same time in the frame where the rod will be at rest after the boost. This compresses the rod to a shorter proper length.

To which I answered that you can't do that to realistic objects (unless you attach a little motor to each atom in the object :-) ) Why do you persist? Why not admit that you were wrong. Especially since you are claiming that you understand Born rigidity?

 

[MeJennifer]

I'm not insulting anybody. I explained this issue several times, go back and re-read my posts. You can also read the refence to the appropriate treatment of accelerated motion I gave Fredrick a few posts back.Here is another one, specially for you:
http://www.arxiv.org/PS_cache/physics/pdf/0405/0405038.pdf

What has a document on the clock paradox to with your statement that Lorentz transformations do not work with accelerated motion?

This is getting ridiculous and is boosted by the fact that you are claiming expertise and making continious denigrating remarks to several members on this forum here without apparently even understanding the basic scope of the Lorentz transformations with regards to boosts, rotations and reflections in flat space-time.

 

[disregardthat]

We don't need to calculate what length the rope would be at every moment, we are merely trying to find proof that the string actually IS stretched when the spaceships are moving.

If you measured the average velocity of the spaceships in acceleration to a specific moment, you could use the lorentz contraction for velocity. This would only give you the amount of contraction at that specific point. But that's all you need. If the velocity faster than at the beginning, the string WILL get stretched.

Nakurusil, if you want a proper discussion, please answer on all the questions and statements stated here. You are not only ignoring the relevant parts of what Fredrik is saying, you are denying small irrelevant parts. That will not make this discussion advance any further.

And stop saying I need to take some classes, I can't, ok? I am only here to try to learn, and discuss and finding the correct answer. And I believe most of us here are...

 

[Fredrik]

The string(rod) gets stretched starting from from time 0, doesn't it? Not only after the acceleration has stopped.
...
If you are simply trying to calculate the rocket separation AFTER the engines have shut off, then you are making up your own, simplified problem. The string may have long snapped during the acceleration phase.

You're missing the point as usual. If the distance between the attachment points in the new inertial frame turns out to be larger than it was from the beginning, the stretching of that distance must have happened during the acceleration. Nothing interesting happens once the ships have turned off their engines, as we can show explicitly.

If you want to compute the string
stretching during the acceleration phase you cannot use Lorentz transforms.OK?

This is actually correct. The point of postulating constant acceleration (which I didn't) is that it makes it possible to explicitly calculate how much the string has stretched in the accelerating frame, at any time. (Edit: Hmm, is this really true? There's an infinite number of accelerating frames here. I'm not even sure if the proper distance is the same in the two frames defined by the attachment points. I need to think about this some more).

When I said that I could have used a co-moving inertial frame, I was actually making a mistake. (Note that I just refuted your claim that I never admit mistakes). It's actually not even obvious from what I wrote that I was making a mistake, so I didn't even have to admit this, but the mistake was to think that it's possible to calculate the proper distance between the attachment points during the acceleration in a co-moving inertial frame. The result of such a calculation would only have been a lower bound on the proper distance. (That's enough to prove that the string breaks of course, but my thoughts about it were still wrong).

To which I answered that you can't do that to realistic objects (unless you attach a little motor to each atom in the object :-) ) Why do you persist? Why not admit that you were wrong. Especially since you are claiming that you understand Born rigidity?

I wasn't wrong, and you know it. That's why this is trolling, and nothing else.

 

[nakurusil]

What has a document on the clock paradox to with your statement that Lorentz transformations do not work with accelerated motion?

This is getting ridiculous and is boosted by the fact that you are claiming expertise and making continious denigrating remarks to several members on this forum here without apparently even understanding the basic scope of the Lorentz transformations with regards to boosts, rotations and reflections in flat space-time.

It gives you the application and the proper formulas of hyperbolic motion. Try learning how to apply them, it solves the Bell paradox in a few lines.

 

[nakurusil]

You're missing the point as usual. If the distance between the attachment points in the new inertial frame turns out to be larger than it was from the beginning, the stretching of that distance must have happened during the acceleration. Nothing interesting happens once the ships have turned off their engines, as we can show explicitly.

Sure, you are making as if I didn't say the same thing. Problem with your derivation is that you are deriving the distance between the ships, during the acceleration phase incorrectly. Your derivation produces the incorrect result.

This is actually correct. The point of postulating constant acceleration (which I didn't) is that it makes it possible to explicitly calculate how much the string has stretched in the accelerating frame, at any time. (Edit: Hmm, is this really true? There's an infinite number of accelerating frames here. I'm not even sure if the proper distance is the same in the two frames defined by the attachment points. I need to think about this some more).

Well, thank you, you are starting to understand.

When I said that I could have used a co-moving inertial frame, I was actually making a mistake. (Note that I just refuted your claim that I never admit mistakes).

Indeed. If you managed to see your mistake in deriving the separation between rockets using Lorentz transforms next...

It's actually not even obvious from what I wrote that I was making a mistake, so I didn't even have to admit this, but the mistake was to think that it's possible to calculate the proper distance between the attachment points during the acceleration in a co-moving inertial frame.

Actually you can. But you need to learn how to use the proper formalisms.

I wasn't wrong, and you know it. That's why this is trolling, and nothing else.

Here you go again. I was thinking about showing you how to do things correctly but now I'll just show you again that you don't have a clue:

5. Every point of the rod is instantaneously (or near instantaneously) boosted to a new velocity, all at the same time in the frame where the rod was at rest before the boost. This stretches the rod to a longer proper length.
6. Every point of the rod is instantaneously (or near instantaneously) boosted to a new velocity, all at the same time in the frame where the rod will be at rest after the boost. This compresses the rod to a shorter proper length.

 

[nakurusil]

We don't need to calculate what length the rope would be at every moment, we are merely trying to find proof that the string actually IS stretched when the spaceships are moving.

If you measured the average velocity of the spaceships in acceleration to a specific moment, you could use the lorentz contraction for velocity.

Bad idea. The rope doesn't break due to any "lorentz contraction", it breaks due to increased separation between rockets.

 

[Fredrik]

As far as I can tell, by only looking at a few of the external sources, they all seem to be focused on the detail of constant acceleration. There's nothing wrong with that of course. When the acceleration is constant, it's possible to calculate explicitly how much the has string has stretched at the time of any given event on one of the world lines. (In this post, I will sometimes be talking about the string as if it's able to stretch without breaking, and sometimes as if it breaks at the very first moment of stretching. I hope it's obvious what I mean. If it's not, ask). But it's not necessary to calculate this explicitly. All we need to prove is that the string breaks.

Since we don't need to calculate exactly how much the string breaks, we shouldn't have to postulate that the acceleration is constant. We should be able to show that the string breaks no matter how the spaceships accelerate.

There are many different versions of this problem, for example:

a) constant acceleration for ever (this is Bell's original version, I think)
b) arbitrary acceleration until a certain proper time when the engines shut off (my version)
c) arbitrary acceleration (the most general version)

I want to prove that the string breaks in version c). I have already proven that the string breaks in version b), by explicitly calculating the proper length of the string after the engines have been shut off. (See #76 and #88).

I think it's intuitively obvious that if the string breaks in b) it must also break in c), but I'd like to find a rigorous argument.

I think my solution of b) can be used as a starting point. This is the kind of reasoning I have in mind: The proper length doesn't increase once the engines have been turned off, so it must increase during the acceleration phase. This means that the string would also break in c).

I pretty sure this line of reasoning is valid, but as it stands, I don't think it's rigorous enough to prove that the string breaks in c). I'm kind of busy today, so I'm not going to try to work this out now. Maybe I'll try to fill in the missing details tomorrow. If anyone else feels like taking a shot at it in the mean time, go ahead...

 

[cincirob]

The Rope a Rocket Problem

This is an old problem first posed by JS Bell at a gathering of physicists. A number of physicists didn't get the right answer the first time. Because of this anti-relativists have latched onto it as a proof agianst relativity because relativity physicists sometimes disagree.

In any case, the outcome is a little strange. While most (even antirelativists) will agree the rope breaks, looking at the clocks on both ships is a bit confusing.

1. Before they accelerate the clocks on the ships are synchronized.

2. If the two ships stop accelerating at the same instant in the stationary frame, they will be going the same velocity and therefore it is possible to synchronize their clocks.

3. Since the clocks underwent the same process they will read identical times when compared to clocks in the stationary frame.

4. #3 seems problematical because,if the rocke clocks are synchronized, they shoud read differently for the stationary observers; that is, a stationary observer near the rear rocket will read Tr and because the clocks are moving at some v, Tf shoul read Tr-vd/c^2(1-(v/c)^2^-.5.

So it's a bit confusing that the clocks read the same. Lately I have been working on the math to see if I can predict this outcome but I'm not finished yet. You might try it as a challenge.