Twin Paradox: Einstein's Explanation and Alternative Interpretations

[cos]

You potential explanations here are incomplete and IMO get at the root of your confusion.
First option 2 should be rejected as unrealistic as no change can occur to B since it remains stationary in a single frame. SR certainly will not support the #2 option.

You apparently did not see my original posting in which I argued that option 2 is unrealistic!

But #1 is much to incomplete an explanation;
it has three different possibilities you have not detailed (or considered) since in the travel of A it must use two different Frames; one outbound and one inbound;
this gives four possibilities for the rate of clock A wrt B:

a) both A outbound and A inbound run FAST wrt B
b) both A outbound and A inbound run SLOW wrt B
c) A outbound runs SLOW and A inbound run FAST wrt B
d) A outbound runs FAST and A inbound run SLOW wrt B

SR only rejects option “a”
but based on the given information of the problem “b” “c” or “d” could be true.
Options “c” & “d” work as long as the amount of time A spends at the slow rate is long enough when summed with the time built up by A at the fast rate nets to a total time less than experienced by B for the duration of the round trip.

Perhaps you might care to read my comment to which you refer but to save you having to locate same I herewith reproduce it:-

*********************

"In paragraph 1, chapter 4, OEMB Albert Einstein wrote:-

"If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other one..."

There is, as far as I can tell, only two explanations as to why A is found to lag behind B:

1. During that trip A ticks over at a slower rate than it did before it started moving OR -

2. The rate of operation of clock B increases whilst A is moving.

*********************

Where, in that comment is there any application that 'in the travel of A it must use two different Frames; one outbound and one inbound.'?

You wrote:-

c) A outbound runs SLOW and A inbound run FAST wrt B
d) A outbound runs FAST and A inbound run SLOW wrt B

Neither of these ideas comply with Einstein's chapter 4 depiction which indicates that A inbound (as per your reference c)) runs slow wrt to B and if A had initially been at rest alongside and synchronous with B and had moved the same distance as Einstein's A to B and at the same velocity it, too, would lag behind B by the same amount as does A's clock in Einstein's depiction.

It would not, as you suggest in d), run FAST!

One of the three conditions will be observed by any random observer C moving at any fixed speed wrt B and will always give the same net change from start to finish for both A & B (B always less than A by the same amount) no matter what speed you use for observer C.

There is no random observer C either in Einstein's chapter 4 depictions or in the astronaut's out-and-return journey but even if there was then his observations have absolutely nothing whatsoever to do with the observations made by A or B.

His observations will have no physical affect whatsoever on A's or B's clocks!

His observations are of no interest whatsoever to A or B.

 

[Dale]

Are you suggesting that in Einstein's chapter 4 depiction an observer alongside clock B determines that the distance traveled by A is less than the distance determined by A on the basis that the distance A to B contracts from A's point of view?

No, this is not what I said. Please note very carefully my choice of words above. When I use the word "distance" I am referring to the Euclidean metric s²=x²+y²+z² in space. When I use the word "interval" I am referring to the Minkowski metric s²=-t²+x²+y²+z² in spacetime. The spacetime interval does not contract, it is absolute or frame invariant, not relative, and all observers agree on it.

The frame-invariant spacetime interval from a1 to a2 is less than the frame-invariant spacetime interval from b1 to b2. All observers will agree on this regardless of any relative motion.

This is important, please respond: Do you see how the diagram I drew represents the scenario described by Einstein in section 4? (even if you don't understand or accept the Minkowski spacetime interval)

My extremely limited understanding of Minkowski spacetime is that it is based on mathematical propositions which, according to Einstein, do not refer to reality.

Minkowski geometry is simply a mathematical framework for SR so it "refers to reality" as much as any other formulation of SR (which is very well).

I am not suggesting that mathematics is not of extreme importance but that it does not, by itself, provide proof of a concept or of a theory as some people insist.

Agreed. Validation of a scientific theory is only by experiment.

 

[cos]

Good. We're making some progress. You agree the astronaut in a closed ship cannot know anything about his state of motion before the acceleration. Not his speed nor his direction. Keep this in mind.

His speed and his direction - relatively to what?

As far as he is concerned his internal dynamic experiments indicate that he is at rest!

He could also assume that his ship is moving with uniform velocity but relatively to what is it moving? To the universe? There could be numerous observers all in different reference frames who would totally disagree with each other about how fast his ship is moving and in which direction it is traveling. Which of those numerous observations is the 'correct' one?

Relatively to one of those observers he could be accelerating but relatively to another observer he could be decelerating! Which opinion is the 'correct' one?

According to various observers his ship could be moving up or down; left or right; backwards or forwards. Which 'observation' of the direction in which the ship is moving is 'real'?

[QUOTE cos]Having taken his foot off the gas pedal he knows that he is still moving away from his original location and because he knows that he accelerated at 1g for 1 second he knows his rate of travel away from that original location.

He does not! He only knows his state of motion is different than it was before. It could be faster, slower or even the same depending on whether the acceleration was in the direction of his (unknown) motion, against it or at an angle. Only an outside observer could make the distinction.

Knowing that his state of motion 'is different than it was before' he knows that he is now moving at a different speed than he was before. On that basis he cannot then be traveling at the same speed as he was before he accelerated!

As pointed out above, a number outside observers could be of the opinion that the ship was initially moving in the opposite direction to its exhausts (i.e. that it is moving forward) whilst just as many could argue that from their point of view it was moving in the opposite direction. Whose opinion is the 'correct' one?

[QUOTE cos]The point of view of an outside observer has absolutely nothing whatsoever to do with what the person you described determines!

It most certainly does. Only an outside observer can make a meaningful statement about the astronauts state of motion before or after the acceleration.

Which of the potential dozens of outside observers all located in different reference frames 'can make a meaningful statement about the astronaut's state of motion before or after the acceleration.'? All of them? Are all of those totally opposing points of view meaningful statements? Or are they only 'meaningful statements' as far as the person that made them is concerned?

Your 'observer accompanying Einstein's paragraph 1, chapter 4, OEMB presentation' IS an outside observer. It is this outside observer, who you won't allow me or anyone else to reference, that is fooling you into believing the astronaut can make a meaningful statement of his inertial motion. If you remove this outside observe, as I have done above, you'll realize that the astronaut cannot make a prediction about how his clock is ticking in any absolute sense.

My mistake - I meant to write "an observer accompanying [clock A in] Einstein's paragraph 1, chapter 4, OEMB presentation" however my then reference to "an astronaut who has returned to the planet following an out-and-return voyage.." should have provided a clue that I was talking about an observer accompanying Einstein's clock A.

Neither the observer accompanying Einstein's clock A nor the returning astronaut are outside observers.

Good. We're making some progress. You agree the astronaut in a closed ship cannot know anything about his state of motion before the acceleration. Not his speed nor his direction. Keep this in mind.

His speed and his direction - relatively to what?

As far as he is concerned his internal dynamic experiments indicate that he is at rest!

He could also assume that his ship is moving with uniform velocity but relatively to what is it moving? To the universe? There could be numerous observers all in different reference frames who would totally disagree with each other about how fast his ship is moving and in which direction it is traveling. Which of those numerous observations is the 'correct' one?

Relatively to one of those observers he could be accelerating but relatively to another observer he could be decelerating! Which opinion is the 'correct' one?

According to various observers his ship could be moving up or down; left or right; backwards or forwards. Which 'observation' of the direction in which the ship is moving is 'real'?

[QUOTE cos]Having taken his foot off the gas pedal he knows that he is still moving away from his original location and because he knows that he accelerated at 1g for 1 second he knows his rate of travel away from that original location.

He does not! He only knows his state of motion is different than it was before. It could be faster, slower or even the same depending on whether the acceleration was in the direction of his (unknown) motion, against it or at an angle. Only an outside observer could make the distinction.

Knowing that his state of motion 'is different than it was before' he knows that he is now moving at a different speed than he was before. On that basis he cannot then be traveling at the same speed as he was before he accelerated!

As pointed out above, a number outside observers could be of the opinion that the ship was initially moving in the opposite direction to its exhausts (i.e. that it is moving forward) whilst just as many could argue that from their point of view it was moving in the opposite direction. Whose opinion is the 'correct' one?

[QUOTE cos]The point of view of an outside observer has absolutely nothing whatsoever to do with what the person you described determines!

It most certainly does. Only an outside observer can make a meaningful statement about the astronauts state of motion before or after the acceleration.

Which of the potential dozens of outside observers all located in different reference frames 'can make a meaningful statement about the astronaut's state of motion before or after the acceleration.'? All of them? Are all of those totally opposing points of view meaningful statements? Or are they only 'meaningful statements' as far as the person that made them is concerned?

Your 'observer accompanying Einstein's paragraph 1, chapter 4, OEMB presentation' IS an outside observer. It is this outside observer, who you won't allow me or anyone else to reference, that is fooling you into believing the astronaut can make a meaningful statement of his inertial motion. If you remove this outside observe, as I have done above, you'll realize that the astronaut cannot make a prediction about how his clock is ticking in any absolute sense.

My mistake - I meant to write "an observer accompanying clock A in Einstein's paragraph 1, chapter 4, OEMB presentation" however my then reference to "an astronaut who has returned to the planet following an out-and-return voyage.." should have provided a clue that I was talking about an observer accompanying Einstein's clock A.

Neither the observer accompanying Einstein's clock A nor the returning astronaut are outside observers and unlike your astronaut in a closed ship my astronaut can look through a window similarly the observer accompanying Einstein's clock A can see clock B and, having experienced a force of acceleration, knows that he is moving toward B in the same way that the astronaut returning to the planet knows that he is doing so.

Whilst you are referring to a purely hypothetical situation where an astronaut cannot see outside his ship (thus is 'without reference to any outside point') my depiction is in relation to more of a reality. Unlike your astronaut in a closed ship my astronaut can look through a window.

It is ludicrous to suggest that I won't allow you or anyone else to reference an outside observer however on the basis that what an outside observer determines has no affect whatsoever on what is taking place in the astronaut's reference frame I see no reason whatsoever for anybody to repeatedly insist that it does.

Those same people, and others, continue to refuse to respond to my question regarding Hafele's or Keating's ability to realize, whilst hypothetically repeating the first leg of their experiment, that their clocks are ticking over at a slower rate than they were before the flight commenced regardless of the fact that they appear to be ticking over at their usual rate.

 

[JesseM]

My mistake - I meant to write "an observer accompanying [clock A in] Einstein's paragraph 1, chapter 4, OEMB presentation" however my then reference to "an astronaut who has returned to the planet following an out-and-return voyage.." should have provided a clue that I was talking about an observer accompanying Einstein's clock A.

Neither the observer accompanying Einstein's clock A nor the returning astronaut are outside observers.

Ah, so this comment finally makes clear that you are talking about a non-inertial observer who is at rest relative to A at all times, rather than an inertial observer who is at rest relative to A only after A finishes accelerating. The problem here is that unlike with inertial observers, there is no "standard" way to construct a coordinate system for a non-inertial observer--you can pick anyone of an infinite number of different coordinate systems in which that observer is at rest at all times, coordinate systems which may have completely different definitions of simultaneity. With inertial observers, there is a clear physical motivation for constructing their coordinate system in the standard way--it makes sense that an inertial observer would want to use rigid rulers at rest relative to themselves to measure distance, and clocks at rest relative to themselves to measure time, and if different clocks in their system are synchronized using the Einstein synchronization convention (in which clocks are synchronized based on the assumption that light moves at the same speed in all directions in this coordinate system--Einstein discusses this synchronization procedure in section 1 of the 1905 paper), then all the different inertial coordinate systems constructed in this way will find that the laws of physics obey the same equations in their respective coordinate systems.

But with non-inertial observers, there is no such "natural" procedure for constructing their coordinate system since the laws of physics will obey different equations from inertial frames regardless of how you do it. So, just because A and B were synchronized in the frame of an inertial observer at rest relative to B, there is no basis for claiming that in the frame of a non-inertial observer who starts out at rest relative to B but then accelerates along with A, we still have to say that A and B were initially synchronized--you could construct a non-inertial coordinate system with a definition of simultaneity such that they'd have been initially synchronized, but you could equally well construct a different non-inertial coordinate system with a definition of simultaneity such that they weren't synchronized from the start, it's purely a matter of aesthetic taste. And the crux of your argument seems to be your claim that A and B were initially synchronized from the perspective of the observer moving along with A, so that if B is ahead when they meet up that must mean A was running slow--well, if the observer's "perspective" is meant to be shorthand for the coordinate system the observer is using (presumably one in which the observer is at rest at all times), it's purely a matter of taste whether this observer uses a coordinate system where A and B were initially synchronized or whether he uses one where they were out-of-sync, neither coordinate system is any more valid physically than the other. On the other hand, if the observer's "perspective" is supposed to mean something other than his coordinate system, then the burden is on you to explain what it does mean, and why you think A and B were initially synchronized from the observer's "perspective" just because they were originally synchronized in the inertial coordinate system where A and B were initially at rest.

 

[matheinste]

Hello cos.

This is purely so that I can see if my interpretation of what you are saying is correct so that we can avoid any misunderstandings. I realize that while there is any possibility of you thinking that I do not understand what you are saying you will, understandably, not take any answers seriously. Ignoring any later complicayions does the summary below describe your thoughts.


We have two clocks A and B at rest relative to each other. These clocks have been synchronized while at rest relative to each other using the Einstein synchronization procedure. Clock B remains in its original state of inertial motion throughout. Clock A moves to clock B. It is found that when A and B meet clock A lags clock B. They are now back at rest relative to each other. The clocks were originally in sync when at rest relative to each other but now are not because B’s clock lags behind A’s. Nothing has happened to B, and he knows this, so his clock’s rate of ticking must have remained the same the whole time. However A has moved, and both he and B know it, and so both conclude that the change in the situation must be due to A’s movement. So they conclude that A’s movement has caused A’s clock rate to slow during its relative motion and not B’s clock rate to speed up.

We assume that there are no direct effects to clock rates due to acceleration itself.

Matheinste

 

[paw]

His speed and his direction - relatively to what?

That's the whole point. Without reference to anything outside the ship he cannot say anything about his state of motion. If he can't say anything about his state of motion he can't say anything about the behaviour of his clocks.

As far as he is concerned his internal dynamic experiments indicate that he is at rest!

Yes. Again that's my point. He has no way to say anything about his state of motion without reference to something outside his ship.

He could also assume that his ship is moving with uniform velocity but relatively to what is it moving? To the universe? There could be numerous observers all in different reference frames who would totally disagree with each other about how fast his ship is moving and in which direction it is traveling. Which of those numerous observations is the 'correct' one?

Exactly the point JesseM was making earlier to which you objected.. All inertial frames are equally valid. No one of them can claim any special 'correctness'.

Relatively to one of those observers he could be accelerating but relatively to another observer he could be decelerating! Which opinion is the 'correct' one?

According to various observers his ship could be moving up or down; left or right; backwards or forwards. Which 'observation' of the direction in which the ship is moving is 'real'?

Neither and neither. No observer in any inertial frame can claim to be correct. Which is again exactly the point I (and others) have been making all along.

Knowing that his state of motion 'is different than it was before' he knows that he is now moving at a different speed than he was before. On that basis he cannot then be traveling at the same speed as he was before he accelerated!

He knows he is moving at a different velocity than he was before. His speed could be faster, slower or even the same as it was before (in the frame where he was at rest before accelerating). You do understand the difference between speed and velocity don't you? If the acceleration was applied correctly he could be going the same speed but in a different direction (wrt the initial rest frame).

As pointed out above, a number outside observers could be of the opinion that the ship was initially moving in the opposite direction to its exhausts (i.e. that it is moving forward) whilst just as many could argue that from their point of view it was moving in the opposite direction. Whose opinion is the 'correct' one?

Exactly. Only an outside observer could make this distinction as I aready stated. The astronaut cannot make this distinction without reference to something outside the ship. He cannot know his state of motion before the acceleration so he cannot know it after the acceleration. Therefore he cannot claim his clocks are running slow in any meaningful sense.

Which of the potential dozens of outside observers all located in different reference frames 'can make a meaningful statement about the astronaut's state of motion before or after the acceleration.'? All of them? Are all of those totally opposing points of view meaningful statements? Or are they only 'meaningful statements' as far as the person that made them is concerned?

All inertial frames can make meaningful statements about the astronauts state of motion both before and after the acceleration. All the apparently opposing pov are meaninful to the observer who makes them. Further, they could all make calculations that will agree with each other after a suitable transform.

It is ludicrous to suggest that I won't allow you or anyone else to reference an outside observer however on the basis that what an outside observer determines has no affect whatsoever on what is taking place in the astronaut's reference frame I see no reason whatsoever for anybody to repeatedly insist that it does.

This is getting to the heart of why you can't seem to properly interpert Einsteins scenario. You are analyzing it from the pov of a non inertial observer, the observer accompanying clock A (the astronaut). I have been attempting to remove that outside observer to show you how the astronaut cannot make any valid claim about the rate his clock is ticking at (beyond the fact it appears to him to be ticking normally). If you still can't see it, after basically agreeing with my statements, then I guess you'll have to stick with your incorrect analysis and continue to believe SR is somehow flawed.

 

[RandallB]

I heard it in a movie – “How can you be so obtuse? – Is it intentional?

Where, in that comment is there any application that 'in the travel of A it must use two different Frames; one outbound and one inbound.'?

There is no “Twins Paradox” without one twin returning to the other. The only way A can get back to B is no make two trips one outbound the inbound. Period.
If that is not clear to you – then you have no chance of understanding this.

If you think you can turn the twin problem into a one way trip – that does not belong in a thread titled “twin paradox”!
You cannot do twins with a oneway trip, without taking Einstein quotes completely out of context - most importantly ignoring his point that simultaneity can only exist for two things that are collocated next to each other; as in when Clock A and B are next to each other on the outbound trip and when they pass next to each other again during the inbound trip by B.

You wrote:-

c) A outbound runs SLOW and A inbound run FAST wrt B
d) A outbound runs FAST and A inbound run SLOW wrt B

Neither of these ideas comply with Einstein's chapter 4 depiction which indicates that A inbound (as per your reference c)) runs slow wrt to B and if A had initially been at rest alongside and synchronous with B and had moved the same distance as Einstein's A to B and at the same velocity it, too, would lag behind B by the same amount as does A's clock in Einstein's depiction.

It would not, as you suggest in d), run FAST!

There is no random observer C either in Einstein's chapter 4 depictions or in the astronaut's out-and-return journey but even if there was then his observations have absolutely nothing whatsoever to do with the observations made by A or B.

His observations will have no physical affect whatsoever on A's or B's clocks!

His observations are of no interest whatsoever to A or B.

If you have not already done the math to see this is true you are not even trying.
Just pick 5 different “C” observers to calculate and record all the times and locations for Clocks A & B wrt to the stationary observer C.
C1 anywhere on the B ref frame
C2 on a frame moving 0.5c wrt B in the outbound direction.
C3 on the 0.8c outbound frame used by A for part of the time.
C4 on a frame moving 0.5c wrt B in the inbound direction
C5 on the 0.8c inbound frame used by A for part of the time.

Just remember the turnaround point is defined in just one frame so C must calculate where that is in the stationary frame it is using. None of the 5 “C” observers will agree with each other on almost anything concerning where when and how fast except for two measurements! They will all agree and predict the same times on clocks A & B when they pass each other at the start and the time on each clock when A returns to meet B again. All showing the traveler A to have aged less even though one of the observers clearly saw the A clock running FAST wrt B for one leg of the trip.
You don’t need to take my word for it – just do the math for yourself.

 

[phyti]

JesseM;
This is the most ridiculous posting I've encountered yet with regard to confusing the issue, and lack of comprehension of posts. It reminds me of the psychiatrist who after passing someone who says "good morning", thinks, "I wonder what he meant by that?".
To make the point as simple as possible, if A is at rest relative to this distance of 20 ls, which is marked off with some type of markers, then when he is moving at .8c, you say the distance is shrunk to 12 ls. If he decelerates to 0, then he is no longer moving relative to the markers, and the distance is now 20 ls. It can't be 12 ls when moving
at .8c and 12 ls when moving at 0, but that is what you said in post 31!

 

[phyti]

Hello cos.

This is purely so that I can see if my interpretation of what you are saying is correct so that we can avoid any misunderstandings. I realize that while there is any possibility of you thinking that I do not understand what you are saying you will, understandably, not take any answers seriously. Ignoring any later complicayions does the summary below describe your thoughts.


We have two clocks A and B at rest relative to each other. These clocks have been synchronized while at rest relative to each other using the Einstein synchronization procedure. Clock B remains in its original state of inertial motion throughout. Clock A moves to clock B. It is found that when A and B meet clock A lags clock B. They are now back at rest relative to each other. The clocks were originally in sync when at rest relative to each other but now are not because B’s clock lags behind A’s. Nothing has happened to B, and he knows this, so his clock’s rate of ticking must have remained the same the whole time. However A has moved, and both he and B know it, and so both conclude that the change in the situation must be due to A’s movement. So they conclude that A’s movement has caused A’s clock rate to slow during its relative motion and not B’s clock rate to speed up.

We assume that there are no direct effects to clock rates due to acceleration itself.

Matheinste

You have my vote, that is the context of the case described in the book. The point A.E. is making is, when the clocks are compared, there is a difference, and that decides the behavior of the clocks. He is not discussing equivalent frames, or how many ways to skin a cat. The relativity police however are looking for someone they think is trying to overthrow the status quo. As I said before, history is full of it.

 

[cos]

This is important, please respond: Do you see how the diagram I drew represents the scenario described by Einstein in section 4? (even if you don't understand or accept the Minkowski spacetime interval)
Minkowski geometry is simply a mathematical framework for SR so it "refers to reality" as much as any other formulation of SR (which is very well).

Does the diagram you drew show that clock A does not lag behind clock B? Does it show that clock A does not 'go more slowly' (i.e. tick over at a slower rate) than clock B?

 

[cos]

We have two clocks A and B at rest relative to each other. These clocks have been synchronized while at rest relative to each other using the Einstein synchronization procedure. Clock B remains in its original state of inertial motion throughout. Clock A moves to clock B. It is found that when A and B meet clock A lags clock B. They are now back at rest relative to each other. The clocks were originally in sync when at rest relative to each other but now are not because B’s clock lags behind A’s. Nothing has happened to B, and he knows this, so his clock’s rate of ticking must have remained the same the whole time. However A has moved, and both he and B know it, and so both conclude that the change in the situation must be due to A’s movement. So they conclude that A’s movement has caused A’s clock rate to slow during its relative motion and not B’s clock rate to speed up.

The underlined section of your comment is incorrect in that, according to Einstein, A's clock lags behind B's clock.

Other than that - I am of the opinion that, apart from your comment "..both conclude that the change in the situation must be due to A’s movement. So they conclude that A’s movement has caused A’s clock rate to slow during its relative motion and not B’s clock rate to speed up" this is precisely what Einstein stated in paragraph 1, chapter 4 of OEMB.

Your comment applies to my interpretation of why A lags behind B.

 

[JesseM]

JesseM;
This is the most ridiculous posting I've encountered yet with regard to confusing the issue, and lack of comprehension of posts.

We are discussing the example I gave in post #31, and the statements I made about this example--are you really so presumptuous that you would accuse me of failing to comprehend my own posts? Are you willing to consider even for a second that maybe there was a misinterpretation on your end?

To make the point as simple as possible, if A is at rest relative to this distance of 20 ls, which is marked off with some type of markers, then when he is moving at .8c, you say the distance is shrunk to 12 ls. If he decelerates to 0, then he is no longer moving relative to the markers, and the distance is now 20 ls. It can't be 12 ls when moving
at .8c and 12 ls when moving at 0, but that is what you said in post 31!

I can only repeat that in post #31 I was not talking about what things looked like from A's perspective, I was talking about what things look like in a frame where B is moving at 0.8c at all times (I think I already made this pretty obvious in my previous post to you when I said 'I'm just talking about what is happening in a particular inertial frame, not what the non-inertial clock A "perceives"'). A is not an inertial observer, there is no single "correct" way to construct the coordinate system for non-inertial objects, and thus no single "correct" answer to how things look from their perspective, you could construct a coordinate system where the non-inertial observer A is always at rest and the distance from A to B is 50 trillion light years if you want, this distance would be no more or less correct as a statement of how things look from A's "perspective" than any other distance because all non-inertial coordinate systems are equally valid, it's only for inertial objects that there is a standard "correct" way to construct their rest frame and thus a single "correct" answer to what distances look like in their rest frame (I'm trying to avoid any presumptions about what you do or don't understand about SR, but if you have any objection to this sentence, you are indeed failing to understand something important about the difference between inertial and non-inertial frames in SR). Once again, I am only talking about how things look from the perspective of the inertial frame where B is always moving at 0.8c, and I made that perfectly clear in post #31:

So, in the frame where A and B are initially moving at 0.8c, they will be out-of-sync by (0.8c)*(20 light-seconds)/c^2 = 16 seconds. Since we are picking a frame where B is moving in the direction of A, B is the trailing clock here, so its time is the one that's ahead by 16 seconds. So, if A suddenly decelerates and comes to rest in this frame when it reads 0 seconds, B will already read 16 seconds at the "same moment" in this frame. From then on B will be moving towards A at 0.8c, and hence slowed down by a factor of 0.6 in this frame while A now ticks at the normal rate in this frame since it's at rest. Since the initial distance between them is 12 light-seconds in this frame, it will take 12/0.8c = 15 seconds for B to catch up with A. During this time A will advance forward by 15 seconds but B will only advance forward by 15*0.6 = 9 seconds. Since A started out reading 0 seconds at the moment it came to rest, and B started out reading 16 seconds "at the same moment" in this frame, then when B catches up with A, A will read 0 + 15 = 15 seconds, while B will read 16 + 9 = 25 seconds. So, in this frame we get the exact same prediction that A is behind B by 10 seconds when they meet, in spite of the fact that in this frame A was ticking faster than B after A accelerated, not slower.

Can you not see that my repeated uses of the phrase "in this frame" were referring to the inertial frame where B was moving at 0.8c at all times, not any kind of non-inertial frame belonging to A where the equations of SR wouldn't apply? Do you not understand that if B has a rod attached to it which is 20 ls long in B's inertial rest frame, so that A was next to the other end of the rod before it accelerated, then in this inertial frame where B (and the rod) is moving at 0.8c at all times, the rod will be 12 ls long at all times, regardless of how A changes speed? There is never a moment--before, during, or after A's acceleration--where the rod is 20 ls long in this frame, it's always 12 ls long in this frame!

 

[cos]

That's the whole point. Without reference to anything outside the ship he cannot say anything about his state of motion. If he can't say anything about his state of motion he can't say anything about the behaviour of his clocks.

On the basis that “he cannot say anything about his state of motion” his speed and motion can only be determined by an outside observer and due to the fact that numerous outside observers can determine numerous different speeds and directions of travel for his ship none of them has the right to state that their interpretation is the correct one.

Ergo anyone of those observers who insists that the ship was originally moving relatively to him at a certain speed and in a certain direction thus that when he fires his engine he is decelerating is insisting that the ship was moving at that speed without taking into account the fact that it could, from the point of view of another outside observer be stationary. That first observer insists that the ship does not accelerate but that it decelerates and that this is in his opinion, for the astronaut in your ship, reality.

Yes. Again that's my point. He has no way to say anything about his state of motion without reference to something outside his ship.

And that’s my point! Your depiction has absolutely nothing whatsoever to do with my depiction of an astronaut (or Einstein's observer A) who have reference to an outside point.

Exactly the point JesseM was making earlier to which you objected. All inertial frames are equally valid. No one of them can claim any special 'correctness'.

My point precisely - an outside observer cannot claim any special correctness if he ‘determines’ that from his point of view the ship was moving in reverse (i.e. in the direction of its main engine’s exhaust pipe) thus that it decelerates when that astronaut steps on the gas pedal and comes to a stop relatively to the outside observer.

Your outside observer could be of the opinion that ‘all inertial frames are equally valid’ thus that before the astronaut gunned his engines he could in his inertial frame have been ‘at rest’ thus that in his (the astronaut’s) reference frame he is accelerating.

I believe that at this point it is well worth repeating your comment that “All inertial frames are equally valid. No one of them can claim any special 'correctness.'”

Therefore for anyone to suggest that your astronaut could be moving backwards and decelerates coming to a stop this is only the opinion of one single outside observer whose determinations have absolutely no affect whatsoever on what is taking place in your astronaut’s reference frame.

Neither and neither. No observer in any inertial frame can claim to be correct. Which is again exactly the point I (and others) have been making all along.

So the observer in a reference frame who determines that the ship was already moving and that it decelerated to a stop is not entitled to claim that his observation is the correct one and that determinations arrived at by the astronaut in that ship (that he has accelerated) are wrong!

So your suggestion that, having gunned his engines, the astronaut’s impression that he is moving is negated by an outside observer’s opinion to the contrary does not comply with your suggestion that “No observer in any inertial frame can claim to be correct.”

He knows he is moving at a different velocity than he was before. His speed could be faster, slower or even the same as it was before (in the frame where he was at rest before accelerating).

QUOTE: “in the frame where he was at rest[.B] before accelerating.”

You do understand the difference between speed and velocity don't you? If the acceleration was applied correctly he could be going the same speed but in a different direction (wrt the initial rest frame).

QUOTE: “wrt the initial rest frame.”

Exactly. Only an outside observer could make this distinction as I already stated. The astronaut cannot make this distinction without reference to something outside the ship. He cannot know his state of motion before the acceleration so he cannot know it after the acceleration. Therefore he cannot claim his clocks are running slow in any meaningful sense.

All inertial frames can make meaningful statements about the astronauts state of motion both before and after the acceleration. All the apparently opposing pov are meaninful to the observer who makes them. Further, they could all make calculations that will agree with each other after a suitable transform.

QUOTE: “All the apparently opposing pov are meaningful to the observer who makes them.”

They are meaningful to the observer who makes them but they have no effect whatsoever on what actually takes place in the astronauts reference frame.

Although to perhaps one of those potentially numerous outside observer’s he could appear to have been moving thus decelerated when he guns his engines this is just one pov and does NOT mean that it is reality merely that, in his opinion, it appears to be reality.

For you to suggest that he could have been moving and decelerated to a stop you are allocating the responsibility of the determination of reality to one out of possibly dozens of outside observers all with varying points of view.

Reality as far as I am concerned is what takes place in the astronaut’s reference frame and on the basis that he determines that he is initially at rest then the pov of one outside observer out of dozens has no effect whatsoever on what the astronaut determines.

This is getting to the heart of why you can't seem to properly interpret Einstein's scenario. You are analyzing it from the pov of a non inertial observer, the observer accompanying clock A (the astronaut). I have been attempting to remove that outside observer to show you how the astronaut cannot make any valid claim about the rate his clock is ticking at (beyond the fact it appears to him to be ticking normally). If you still can't see it, after basically agreeing with my statements, then I guess you'll have to stick with your incorrect analysis and continue to believe SR is somehow flawed.

The observer accompanying clock A is NOT an outside observer who you have apparently been attempting to remove.

The observer accompanying clock A is the only observer who can claim to be correct but only when he arrives at B’s location and - ‘looking out of his window’ - finds that his clock lags behind clock B.

Whilst you have steered the discussion away from this fact - he can either conclude that B ticked over at a faster rate than his clock as per the nonsensical claim to which my op referred OR that his clock ticked over at a slower rate than B.

Returning to his original location and repeating the experiment he would then be fully justified in believing that although the rate of operation of his clock appears to be unchanged it is, most likely, as it did during his first trip, ticking over at a slower rate than it was before he commenced that second journey in the same way that Hafele and Keating making a hypothetical repeat of the first leg of their experiment would have been fully justified in assuming that their clocks are ticking over at a slower rate than they were before the flight commenced in precisely the same way that they did during the first trip.

 

[cos]

There is no “Twins Paradox” without one twin returning to the other. The only way A can get back to B is no make two trips one outbound the inbound. Period.
If that is not clear to you – then you have no chance of understanding this.

My reference to ‘that comment’ was in relation to Einstein’s comment regarding clock A moving to B’s location which does not allow for two trips.

If you think you can turn the twin problem into a one way trip – that does not belong in a thread titled “twin paradox”!

Having taken my comment out of context you are side-stepping my argument that there would be no difference to Einstein’s conclusion that A lags behind B if A moved to B’s location (i.e. the twin’s return journey) or if A was initially alongside and synchronous with B and traveled away from B for the same distance, and at the same velocity (i.e. the twin’s outward-bound journey), as A’s trip to B’s location

You cannot do twins with a oneway trip, without taking Einstein quotes completely out of context

I didn’t; you took my comments out of context.

According to Einstein’s paragraph 1, chapter 4, OEMB depiction when the astronaut makes his outward bound trip his clock will lag behind his twin’s clock by precisely the same amount as it additionally lags behind the twin’s clock on his return journey by .5tv²/c².

most importantly ignoring his point that simultaneity can only exist for two things that are collocated next to each other;

Where, in any of my messages did I ignore that point?

as in when Clock A and B are next to each other on the outbound trip and when they pass next to each other again during the inbound trip by B.

What are you talking about? A and B are not, in Einstein’s chapter 4 depiction, “next to each other on the outbound trip.” nor are they “next to each other on the outbound trip.” in the twin paradox!

A and B do not “pass next to each other again during the inbound trip by B.” In the twin paradox. A (the astronaut’s clock) arrives at B’s location (his twin’s clock) it does not pass B.

If you have not already done the math to see this is true you are not even trying.

I have not already done the math due to the fact that I cannot do the math however I agree with Einstein that “As far as the propositions of mathematics are certain; they do not refer to reality.” I am of the opinion that physics should be a subject that deals with reality.

Just pick 5 different “C” observers to calculate and record all the times and locations for Clocks A & B wrt to the stationary observer C.

There are no ‘C’ observers either in Einstein’s depiction or the twin paradox. Determinations arrived at, or predictions made, by purely hypothetical observers C have absolutely no affect whatsoever on what physically takes place in A’s or B’s reference frames. I am of the opinion that physics should be a subject that deals with reality.

 

[Dale]

Does the diagram you drew show that clock A does not lag behind clock B? Does it show that clock A does not 'go more slowly' (i.e. tick over at a slower rate) than clock B?

Forget for a moment about what the clocks show. Do you see how the diagram represents the kinematics, the motion, of the clocks? In other words, do you see that a vertical line represents a clock at rest and a diagonal line represents a moving clock? Can you piece those rules together to see that the diagram correctly represents the motion of the clocks in Einstein's chapter 4?

 

[matheinste]

Hello cos.

In reply to your post #186. Thanks for pointing out my error. It is of course not what i meant. You say that you agree with my interpretation of waht you are saying. Thats helpful. I am currently busy with less interesting things but will get back as soon as.

Just to recap without comment from me. -- You are saying that Einstein says in the place quoted that the difference in clock readings is due to A's movement. ---That A's clock slowed down during the journey is your interpretation of why there is a difference.

Matheinste.

 

[RandallB]

There are no ‘C’ observers either in Einstein’s depiction or the twin paradox. Determinations arrived at, or predictions made, by purely hypothetical observers C have absolutely no affect whatsoever on what physically takes place in A’s or B’s reference frames. I am of the opinion that physics should be a subject that deals with reality.

Nobody said that the selection of C will change the reality of anything, only what the observer thinks might be real.
But you are not using reality when you insist on arguing the twins as a one way trip, because you can not justify which POV give the correct and “real” view of reality; the view from Clock B or the view from moving clock A.

I know enough about the many folks that have tried to help you in this thread to know they all know much more than you about the Twins than you. And yet in over 180 posts you don’t seem to have learned a single think since your OP. That’s because you are arguing instead of listening and learning.

I’m all for reality so you tell use which view is the correct view of reality A or B when the only thing they agree on is the start time when A passes B and both set there clocks to O (zero). All the other C frames I mentioned (that you refuse to do the math on) also agree that A & B both simultaneously read “0” at that defined starting point. But never again will any two reference frames agree on what both A & B read simultaneously in your one way problem – never!
So if you big on reality; which frame is giving you the correct version of reality and just how do you justify your choice of frame as correct?

However if you do the twins correctly and have A turnaround at a well defined time or place and eventually pass the ‘stationary’ B again all frames will continue to disagree about almost every thing except one and only one thing. Each and every frame will show the same times for both A & B (with A less than B) at the moment A & B pass each other again.
After you do the math how can you every create a reality where the returning A twin was anything but younger than B. Not only do all frames agree A will be younger they all agree on the same ages and differences.
That all frames agree to the same reality at the point the two come back together – is the only point that can be taken as being “real” in the Twins Paradox.
But as to the other predictions made by the various ref. frames; nothing (at least within SR rules) can rule out any of as incorrect or not “real”.
And if you cannot rule out any frame as incorrect you cannot assume anyone frame is in reality logging time simultaneously including your own.
Meaning no matter how well you use light or radio signals to synchronize clocks from coast to coast on zulu time – you cannot depend on them all in reality hitting 12:00:00 simultaneously only that in your POV in this frame of reference they appear to.

You will not understand SR or Twins until you understand that,
and you best first step to actually learning something about it is DO THE DARN MATH.
It really is not that hard.

 

[cos]

Forget for a moment about what the clocks show. Do you see how the diagram represents the kinematics, the motion, of the clocks? In other words, do you see that a vertical line represents a clock at rest and a diagonal line represents a moving clock? Can you piece those rules together to see that the diagram correctly represents the motion of the clocks in Einstein's chapter 4?

On the basis that 'what the clocks show' is the central issue of my argument I see no reason whatsoever for me to forget that phenomenon even for just a moment.

I am of the opinion that the diagram is an interpretation of what the math shows and as far as I am concerned the reality of 'what the clocks show' takes precedence.

I can only repeat my question "Does the diagram you drew show that clock A does not lag behind clock B? Does it show that clock A does not 'go more slowly' (i.e. tick over at a slower rate) than clock B?"

Unless you can answer this important and overriding question we should forget the whole thing.

If the diagram shows that A does not lag behind B thus that A does not 'go more slowly' than B then as far as I am concerned the diagram contradicts Einstein's chapter 4 regardless of the fact that, as you point out, it "correctly represents the motion of the clocks in Einstein's chapter 4?." It is not their motion with which I am concerned; my interest lies specifically in what happens to clock A.

 

[cos]

You are saying that Einstein says in the place quoted that the difference in clock readings is due to A's movement. ---That A's clock slowed down during the journey is your interpretation of why there is a difference.

That's what I'm saying.

In his 1918 article Einstein said the same thing - that it was the clock that changed reference frames that physically 'goes more slowly than' (my interpretation being 'ticks over at a a slower rate than' or 'incurs time dilation relatively to') the clock (B) that does not change reference frames (i.e. does not accelerate).

 

[Dale]

as you point out, it "correctly represents the motion of the clocks in Einstein's chapter 4?."

I am going to take this as a "yes" that you do understand how vertical lines represent objects at rest in the given reference frame and how slanted lines represent objects moving in that frame and therefore you agree that the diagram correctly represents the motion of the clocks in section 4.

I can only repeat my question "Does the diagram you drew show that clock A does not lag behind clock B? Does it show that clock A does not 'go more slowly' (i.e. tick over at a slower rate) than clock B?"

Unless you can answer this important and overriding question we should forget the whole thing.

The diagram shows that the interval between a1 and a2 is shorter than the interval between b1 and b2. Therefore A, having measured a shorter interval than B, will read less when they meet. Again, this is seen to be a property of the path rather than a property of the clocks. Both clock A and B correctly measured the interval of their respective paths through spacetime, but clock A took a "shortcut" so it reads less than clock B.

 

[cos]

Nobody said that the selection of C will change the reality of anything, only what the observer thinks might be real.

The argument was that A and B may not initially have been at rest as Einstein depicted but that from (a purely imaginary) C's point of view A and B may have initially been moving at v relatively to him thus that A did not accelerate then move toward B at v but that A decelerated (and came to a stop in C's reference frame) and that B moved toward A at v (i.e. at v relatively to C) thus that it was B that was moving thus that it was B that 'went more slowly' than A.

On the basis of C's point of view that B 'goes more slowly' (i.e. ticks over at a slower rate) than A then, when the clocks are bought together, B should lag behind A but that's not what Einstein said! He specifically stated the complete opposite - that A lags behind B!

But you are not using reality when you insist on arguing the twins as a one way trip, because you can not justify which POV give the correct and “real” view of reality; the view from Clock B or the view from moving clock A.

Would you please explain the difference between A moving the distance A to B at v (the astronaut's return trip)and A moving the distance B to A at v (the astronaut's outward bound trip)? Einstein's equation applies equally to both trips!

Could you please explain why you are of the opinion that Einstein's equation .5tv²/c² applies to a journey in one direction but not to a journey over an identical distance in the opposite direction?

On the basis that an observer located alongside clock B and an observer accompanying clock A have both read and agree with Einstein's chapter 4 depiction as well as his 1918 article they can both be of the opinion that A is 'going more slowly' (i.e. ticking over at a slower rate) than B.

In paragraph 1, chapter 4, Einstein refers to a one way trip however in paragraph 2 he points out that the same results will be arrived at "if the clock moves from A to B in any polygonal line." An astronaut's out-and-return trip IS a polygonal line!

The astronaut travels at the same velocity (v) for the same period of time (t) and covers the same distance ergo both trips comply with Einstein's one way depiction and his polygonal line depiction.

I know enough about the many folks that have tried to help you in this thread to know they all know much more than you about the Twins than you. And yet in over 180 posts you don’t seem to have learned a single think since your OP. That’s because you are arguing instead of listening and learning.

I have no doubt whatsoever that back in 1905 Einstein was told that the many folks that had tried to explain to him that Newton's theory invalidated his theory also knew more about the subject of physics than he did.

You are obviously laboring under the misapprehension that my OP was an attempt to learn something by listening to the opinions of others - it was NOT!

It was a statement in relation to which numerous responses have tried to convince me that the pov of an external observer has validity - it does NOT!

I’m all for reality so you tell use which view is the correct view of reality A or B when the only thing they agree on is the start time when A passes B and both set there clocks to O (zero).

That situation NEVER ARISES neither in Einstein's chapter 4 NOR his 1918 article!

It is similarly inapplicable comments such as yours that create confusion!

All the other C frames I mentioned (that you refuse to do the math on) also agree that A & B both simultaneously read “0” at that defined starting point.

Even if I were capable of 'doing the math on' all the other C frames you mentioned I would refuse to apply that math to the situation referred to in Einstein's chapter 4 or his 1918 article on the basis that the opinion expressed by (i.e. the mathematical determinations arrived at by) numerous C frames has absolutely no affect whatsoever on what A and B determine!

On the basis of the 'Relativity of Simultaneity' NONE of the C frames "agree that A & B both simultaneously read “0” at that defined starting point." i.e. the starting point depicted by Einstein in paragraph 1, chapter 4 where A is at a fixed distance from B. Try to get your facts straight!

But never again will any two reference frames agree on what both A & B read simultaneously in your one way problem – never!

I never, in any of my postings, made any comment in relation to an agreement or otherwise as to the simultaneity of those clocks ergo your comment is totally unwarranted and inapplicable!

So if you big on reality; which frame is giving you the correct version of reality and just how do you justify your choice of frame as correct?

On the basis of my assumption that observers located in both frames (A and B) have read and fully agree with Einstein's chapter 4 as well as his 1918 article then both frames will realize that reality is that clock A (the astronaut's clock) 'goes more slowly' (i.e. ticks over at a slower rate than) B (the Earth clock).

However if you do the twins correctly and have A turnaround at a well defined time or place and eventually pass the ‘stationary’ B again all frames will continue to disagree about almost every thing except one and only one thing. Each and every frame will show the same times for both A & B (with A less than B) at the moment A & B pass each other again.

After you do the math how can you every create a reality where the returning A twin was anything but younger than B. Not only do all frames agree A will be younger they all agree on the same ages and differences.

Has anything I have written given you the impression that I was saying that A will not lag behind B?

That all frames agree to the same reality at the point the two come back together – is the only point that can be taken as being “real” in the Twins Paradox.

I believe that when Einstein wrote in his chapter 4 that a clock at the equator "must go more slowly...than a precisely similar clock at one of the poles under otherwise identical conditions." he was of the opinion that this is real!

His paragraph 1 depiction of clock A moving to B's location effectively incorporates the same factor - clock A will 'go more slowly' than B thus arrives at B's location lagging behind B.

But as to the other predictions made by the various ref. frames; nothing (at least within SR rules) can rule out any of as incorrect or not “real”.

I don't know if it would come under the heading of an SR 'rule' however on the basis that it is part of SR Einstein's paragraph 3, chapter 4 indicates to me that he was suggesting that although an observer located at the equator (on a hypothetical massless transparent sphere the size of the Earth) might 'see' the 'stationary' polar clock ticking over at a faster rate than his own clock thus insist that this is reality Einstein suggested that it is not reality; he suggested that the equatorial clock physically ticks over at a slower rate than the polar clock.

An observer accompanying Einstein's paragraph 1, chapter 4 clock A in accordance with SR's chapters 1 through 3 'rules' would determine or predict that clock B, moving toward him at v, is ticking over at a slower rate than his own clock thus that when he arrives at B's location it should, according to his predictions in accordance with SR rules, lag behind his own clock however he finds, much to his surprise, that his clock lags behind B as Einstein suggested it would.

And if you cannot rule out any frame as incorrect you cannot assume anyone frame is in reality logging time simultaneously including your own.

What do you mean by 'logging time simultaneously'?

An astronaut, having read and fully accepted Einstein's chapter 4 depiction as well as his 1918 article, can come to the conclusion that although his clock appears to be ticking over at its normal rate it is, in reality, ticking over at a slower rate than it was before he started moving.

Having carried out the first leg of their experiment and discovering that their clocks lagged behind the laboratory clocks Hafele and Keating would have been fully justified in concluding that, during their flight, their clocks were physically ticking over at a slower rate than they were before the flight commenced analogous to the fact that a clock at the equator will, according to Einstein, tick over at a slower rate (i.e. 'go more slowly') than a clock at one the poles.

In his book 'Was Einstein Right? Clifford M Will points out that Einstein's polar clock is analogous to a hypothetical master clock at the center of the planet.

In paragraph 3, chapter 4, Einstein points out that if a clock is made to move in a closed curve around an 'at rest' clock the moving clock (as will his equatorial clock) 'go more slowly' than the at rest clock (i.e. the master clock, above) in accordance with the equation .5tv²/c².

Therefore the laboratory clocks in Washington tick over at a slower rate than a master clock and because of their faster rate of travel (v in Einstein's equation) the clocks in the first leg of the HKX went more slowly than the laboratory clocks.

If they had repeated that same trip Hafele and Keating would have been fully justified in realizing, all appearances to the contrary, that their clocks were, in reality, ticking over at a slower rate than they were before the flight commenced.

Meaning no matter how well you use light or radio signals to synchronize clocks from coast to coast on zulu time – you cannot depend on them all in reality hitting 12:00:00 simultaneously only that in your POV in this frame of reference they appear to.[/QUOTE

Irrelevant! Nothing whatsoever to do with what the astronaut determines is taking place in his own reference frame.

You will not understand SR or Twins until you understand that,

Irrelevant! I have no interest in 'understanding SR' per se but only in what Einstein wrote in chapter 4 and its application to an astronaut's out-and return journey.

and you best first step to actually learning something about it is DO THE DARN MATH.
It really is not that hard.

On the basis of my agreement with Einstein that as far as the propositions of mathematics are certain they do not refer to reality I see absolutely no reason whatsoever to 'DO the darn math' on the basis that my interest is IN reality!

 

[cos]

The diagram shows that the interval between a1 and a2 is shorter than the interval between b1 and b2. Therefore A, having measured a shorter interval than B, will read less when they meet. Again, this is seen to be a property of the path rather than a property of the clocks. Both clock A and B correctly measured the interval of their respective paths through spacetime, but clock A took a "shortcut" so it reads less than clock B.

I'm going to take this (that clock A reads less than clock B) as a "no" in response to my question ""Does the diagram you drew show that clock A does not lag behind clock B?"

Would you please respond to my second, and much important question "Does [the diagram you drew] show that clock A does not 'go more slowly' (i.e. tick over at a slower rate) than clock B?"

 

[Dale]

I'm going to take this (that clock A reads less than clock B) as a "no" in response to my question ""Does the diagram you drew show that clock A does not lag behind clock B?"

Correct, that is what I said. The diagram shows that clock A does lag behind clock B on their reunion since the interval it travels is shorter.

Would you please respond to my second, and much important question "Does [the diagram you drew] show that clock A does not 'go more slowly' (i.e. tick over at a slower rate) than clock B?"

Correct. It shows that clock A does not physically "go more slowly" than clock B, it just takes a shorter path through spacetime. But along this path clock A still ticks over at the same rate as clock B does along its path (i.e. both tick at 1 second/light-second).

 

[cos]

Correct. It shows that clock A does not physically "go more slowly" than clock B, it just takes a shorter path through spacetime. But along this path clock A still ticks over at the same rate as clock B does along its path (i.e. both tick at 1 second/light-second).

So would a similar diagram show that the equatorial clock to which Einstein referred in chapter 4 does not, as he suggested 'go more slowly than a clock at one of the poles'?

 

[Dale]

So would a similar diagram show that the equatorial clock to which Einstein referred in chapter 4 does not, as he suggested 'go more slowly than a clock at one of the poles'?

Unfortunately, Einstein was wrong about that, a clock at the equator does not "go more slowly" than one at the poles*. Here is http://arxiv.org/PS_cache/gr-qc/pdf/0501/0501034v2.pdf" you might like on the subject, it is very light on math. Basically, the SR and GR effects cancel each other out.

*I.e. no net transverse/gravitational Doppler effect

 

[cos]

Unfortunately, Einstein was wrong about that, a clock at the equator does not "go more slowly" than one at the poles*. Here is http://arxiv.org/PS_cache/gr-qc/pdf/0501/0501034v2.pdf" you might like on the subject, it is very light on math. Basically, the SR and GR effects cancel each other out.

*I.e. no net transverse/gravitational Doppler effect

OK I read that paper. I had assumed that Einstein's comment "under otherwise identical conditions." would have (albeit, perhaps, inadvertently) allowed for the effects of gravity (i.e. altitude and the Earth's greater diameter at the equator) but what about Einstein's similar comment in relation to a clock that moves in a closed curve around an 'at rest' clock?

Does the traveled clock end up, as Einstein suggested, lagging behind the 'stationary' clock?

Does that traveled clock 'go more slowly' than the stationary clock in order to end up lagging behind same?

Did the Hafele-Keating clocks 'go more slowly' than the laboratory clocks? i.e. did they tick over at a slower rate than the laboratory clocks after gravitational time variation effects were taken into account and removed from the equations as Will's did in 'Was Einstein Right?'?

I'm specifically talking about what physically happened to those clocks not what a Minkowski spacetime diagram 'shows'.

You wrote - "...the SR and GR effects cancel each other out." however was Einstein's comment regarding clocks at the equator and at one of the poles applicable to SR?

Was the paper to which you refer published in a peer-reviewed science journal? Has it been accepted by the physics community?

 

[atyy]

Audoin and Guinot, p9, 10:

"The postulate leaves no room for differential ageing of the various natural phenomena. ... Although it has never yet been found to fail, it may one day be brought into question by experimental progress."

"...Einstein's general relativity has appeared on the scene. According to this theory, only local time can be directly measured with a clock. In other words, it is the proper time of this clock or an observer in the immediate vicinity that is measured. The time ...over an extended region of space including, for example, the Solar System, is just a coordinate time freely chosen ... and to which no physical reality is attributed."

http://books.google.com/books?hl=en...=X&oi=book_result&resnum=1&ct=result#PPA10,M1

 

[Idjot]

Shouldn't clocks on opposite sides of the planet serve to test this theory? They are facing away from one another and both under 1G. According to Relativity, what's the difference?
Gravity is acceleration. The real challenge is how to actually observe one from the perspective of the other. The "paradox" is really only an observatory phenomena. Only if one of them accelerates more than the other will there be any real dilation and that will take place on the clock under more acceleration (resulting in a higher velocity relative to 'Big Ben'). When the faster clock returns to 1G all observers will notice the dilation effect on the faster clock, even relative to the paradox. Wrap your head around that :)

 

[cos]

Audoin and Guinot, p9, 10:

"The postulate leaves no room for differential ageing of the various natural phenomena. ... Although it has never yet been found to fail, it may one day be brought into question by experimental progress."

"...Einstein's general relativity has appeared on the scene. According to this theory, only local time can be directly measured with a clock. In other words, it is the proper time of this clock or an observer in the immediate vicinity that is measured. The time ...over an extended region of space including, for example, the Solar System, is just a coordinate time freely chosen ... and to which no physical reality is attributed."

On the basis that your first paragraph relates to Poincare's postulate and the second paragraph relates to general theory I assume that this message is not in relation to my argument which is specifically in relation to special theory.

 

[Dale]

what about Einstein's similar comment in relation to a clock that moves in a closed curve around an 'at rest' clock?

Does the traveled clock end up, as Einstein suggested, lagging behind the 'stationary' clock?

Does that traveled clock 'go more slowly' than the stationary clock in order to end up lagging behind same?

Such a scenario would be much better as it deals only with SR effects and does not add GR effects into the mix. For convenience let us speak of clock A on the rim of a rotating "wheel type" space station, and clock B in the hub. If we were to draw the spacetime diagram we would get http://upload.wikimedia.org/wikipedia/commons/thumb/8/87/060322_helix.svg/250px-060322_helix.svg.png" like the one posted by JesseM in post 133 as the worldline of clock A. The worldline of clock B would simply be the axis of the helix.

Note, that clocks A and B never meet so you have to define the endpoints of each worldline completely separately. One typical choice would be to choose the intersection of each worldline with a "beginning" and an "ending" hypersurface of simultaneity, usually defined using Einstein synchronization in the rest frame of the hub.

Now, if you do that you find that the interval along worldline A is shorter than the interval along worldline B. So if clock A and B are set to zero at the beginning then clock A will read less than clock B at the ending. Each clock still measures the same 1 second/light-second along their respective paths, but clock A just travels a shorter path.

In case you missed them in the paragraph above that is a yes for your "lagging" question and a no for your "go more slowly" question. In (Euclidean) geometrical terms this scenario is analogous to the fact that the distance from the Atlantic coast to the Pacific coast is shorter when measured from Veracruz to Acapulco than when measured from New York to Los Angeles.

 

[matheinste]

Hello cos.

I feel that eventually this frame will slowly come to an end because people will realize that you cannot be convinced by logical reasoning. You will feel able to claim you are right by default because people have given up, not because they think you are right but through sheer frustration. I hereby claim the dubious honour of being the first to give up, unless someone in some other frame has already done so.

Matheinste

 

[RandallB]

On the basis of C's point of view that B 'goes more slowly' (i.e. ticks over at a slower rate) than A then, when the clocks are bought together, B should lag behind A but that's not what Einstein said! He specifically stated the complete opposite - that A lags behind B!

Would you please explain the difference between A moving the distance A to B at v (the astronaut's return trip)and A moving the distance B to A at v (the astronaut's outward bound trip)? Einstein's equation applies equally to both trips!

Could you please explain why you are of the opinion that Einstein's equation .5tv²/c² applies to a journey in one direction but not to a journey over an identical distance in the opposite direction?

I have no interest in 'understanding SR' per se but only in what Einstein wrote in chapter 4 and its application to an astronaut's out-and return journey.

All of that is made clear when you just do the simple math from all 5 “C” observer views. That will help you understand SR!
But you make it clear you do not want to understand SR
--- I can only assume you intentionally just want to be argumentative and I do see why you came to these forums at all.
Waste others time but not mine - I’ll unsubscribe from this thread.
IMO a mentor should lock it simply as a lost cause; you are not listening to anyone.

 

[cos]

Such a scenario would be much better as it deals only with SR effects and does not add GR effects into the mix. For convenience let us speak of clock A on the rim of a rotating "wheel type" space station, and clock B in the hub. If we were to draw the spacetime diagram we would get http://upload.wikimedia.org/wikipedia/commons/thumb/8/87/060322_helix.svg/250px-060322_helix.svg.png" like the one posted by JesseM in post 133 as the worldline of clock A. The worldline of clock B would simply be the axis of the helix.

Note, that clocks A and B never meet so you have to define the endpoints of each worldline completely separately. One typical choice would be to choose the intersection of each worldline with a "beginning" and an "ending" hypersurface of simultaneity, usually defined using Einstein synchronization in the rest frame of the hub.

Now, if you do that you find that the interval along worldline A is shorter than the interval along worldline B. So if clock A and B are set to zero at the beginning then clock A will read less than clock B at the ending. Each clock still measures the same 1 second/light-second along their respective paths, but clock A just travels a shorter path.

In case you missed them in the paragraph above that is a yes for your "lagging" question and a no for your "go more slowly" question. In (Euclidean) geometrical terms this scenario is analogous to the fact that the distance from the Atlantic coast to the Pacific coast is shorter when measured from Veracruz to Acapulco than when measured from New York to Los Angeles.

Whilst you point out that clocks A and B never meet this does not comply with Einstein's chapter 4 depiction which starts off with two synchronous clocks alongside each other. One of them moves in a closed curve until it returns to its original location and is once again alongside the other clock where it is found that the traveled clock will lag behind the clock that has remained at rest.

On the basis that they do meet we, presumably, do not "have to define the endpoints of each worldline completely separately."

The rest of your post applies to the mathematically determined Minkowski spacetime concept which, as I have pointed out on several occasions, is not - according to Einstein - reality.

I note that you declined to respond to my question regarding the HKX and other salient points so I will repeat same:-

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Did the Hafele-Keating clocks 'go more slowly' than the laboratory clocks? i.e. did they tick over at a slower rate than the laboratory clocks after gravitational time variation effects were taken into account and removed from the equations as Will's did in 'Was Einstein Right?'?

I'm specifically talking about what physically happened to those clocks not what a Minkowski spacetime diagram 'shows'.

Was the paper to which you refer published in a peer-reviewed science journal? Has it been accepted by the physics community?

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Here is another question which although applicable to GR also applies to Einstein's chapter 4 SR depiction specifically a polygonal line clock A relocation but which has similarly been ignored by others in this thread - an observer is located on top of a mountain; he notes that a clock at that location ticks over at the same rate as his own clock which is obviously ticking over at it's 'normal' rate. He moves to sea-level and again notes that a clock at that location ticks over at the same rate as his own clock - which is still ticking over at it's 'normal' rate.

Does he insist that the clock at the top of the mountain and the clock at sea-level are ticking over at the same rate as each other as determined by his observations or does he apply his knowledge of the Wallops Island experiment and general theory and realize that although the sea-level clock appears to be ticking over at the same rate as the mountain top clock it is physically ticking over at a slower rate?

An astronaut comes to a stop at the end of his outward-bound journey and notes the rate of operation of his clock. He then accelerates and again looks at his clock which, although appearing to be ticking over at a normal rate, is physically ticking over at a slower rate than it was before he started accelerating in the same way that the above mentioned mountain-descending observer's clock ticks over at a slower rate than it did before he started moving.

My specific interest is in relation to what is physically happening to the clocks!

Although I am of the opinion that this analogy is highly relevant it will most likely be emitted from your response as were the above-referred to salient points.

 

[Idjot]

Although I am of the opinion that this analogy is highly relevant it will most likely be emitted from your response as were the above-referred to salient points.

That's "omitted" not "emitted". You must have a non-qwerty board.

Sorry, I couldn't help myself.

 

[phyti]

JesseM post 187;

Since the initial distance between them is 12 light-seconds in this frame, it will take 12/0.8c = 15 seconds for B to catch up with A.

The shrinking distance is the alternate explanation by B instead of his time dilation. [.6*(20/.8)=15]

During this time A will advance forward by 15 seconds but B will only advance forward by 15*0.6 = 9 seconds.

Here you are applying time dilation twice! You have done this before on previous posts.

A is not moving at .8c, therefore his clock will not experience B's dilation, and B cannot apply his dilation to A's clock.