The Twins Paradox: A Controversial Truth or a Perplexing Paradox?

[gonzo]

I wanted to add few comments. First I was taking SP as the origin in all inertial frames of interest.

Also, I think it is important to point out that if we imagine this latice works of synchronized clocks in one frame (all frames can have their own latice work), these clocks will not appear synchronized in another inertial frame.

In other words, we are imagining a laticework of synchronized clocks in the EP-SP frame to check time. But these clocks do NOT appear synchronized in the RSF frame. That seems an important point to mention.

 

[JesseM]

As far as I understand it, the twins do not have to meet up again, and there is still no paradox. Is this is the debate? Here is how I understand it, and why I don't think it matters whether or not the twin turns around. I've lost all track of who is what letter, so I'll start from the beginning and see if I cover the issues correctly.

Okay, you have two Twins, Travel Twin (TT) and Home Twin (HT) for ease. They start together at START PLACE (SP) and TT then travels to END PLACE (EP).

Now, we assume the SP and EP are in the same inertial frame. We further imagine a latice of synchronized clocks throughout the whole of this inertial frame. This is a common practice when talking about SR.

As TT zooms out from SP to EP in his Rocket Ship Frame (RSF) he looks out his windows at all these clocks in the SP-EP frame and thinks they are all running slow (show less elapsed time). At each of these clocks we have a little green man in the SP-EP frame who looks in the window of the passing ship and thinks that TT's RSF clocks are running slow.

This is the first apparent problem, but this is easily resolved by issues of simultaneity.

Now, then the apparent problem arises when TT reaches EP. Right when he gets there, there seems to be a conflict. TT looks at the clock on EP and sees not much time has elapsed on that clock. While the little green man on EP looks in the window and sees that it is actually TT RSF clock that shows not much time passing.

So, then you ask what happens when TT stops, who's clock is right?

There's no need to assume the TT stops, he can just whiz by EP, and as he passes arbitrarily close to the clock there, they can compare readings--both frames must agree on what each clock reads at the moment they pass, since different frames can't give different predictions about local events (if they could, different frames might disagree on whether two moving objects would collide or miss each other, for example). The key to resolving this is to notice that from TT's point of view, all these clocks at rest in the SP-EP frame are out of sync, so that although he does see each clock individually running slow, he also sees that each successive clock he passes is ahead of the previous one. So, if both his clock and the clock at SP read the same time at the moment he departs, then at the "same moment" in his frame, the clock at EP started out far ahead of the clock at SP, so despite the fact that the clock at EP was running slow throughout the trip in his frame, it still makes sense that the clock at EP is ahead of his own at the moment they pass.

If he stops at EP, then you're right that this will change his definition of simultaneity, but it will only be distant clocks whose time suddenly jumps forwards or backwards, the local clock at EP won't change if his acceleration is instantaneous.

 

[gonzo]

Sure, you are correct, I was a bit unclear. I was thinking of elapsed time. The rocket frame reads less elapsed time on the EP clock, however, as you pointed out, when he starts at SP in the rocket frame the clocks are not synchronized the clock at EP is already in the future, so when he gets there it all works out nicely so that the clock at EP reads the total time as seen from SP to EP in the EP frame.

The thing to keep in mind is that the problem is meaningless unless there is some frame changing going on somewhere. So when TT shifts frame from SP to RSF then all the SP-EP clocks jump out of synch, or anothe way of looking at it is that a lot of clock ticks that were in his future are suddenly in his past.

 

[JesseM]

The thing to keep in mind is that the problem is meaningless unless there is some frame changing going on somewhere. So when TT shifts frame from SP to RSF then all the SP-EP clocks jump out of synch, or anothe way of looking at it is that a lot of clock ticks that were in his future are suddenly in his past.

Why is it meaningless? Even if the TT flies past the EP without changing velocities, so that both twins are moving inertially throughout the entire problem, we can still ask questions like how the different clocks look in each twin's frame, or why each frame gets the same prediction for the time on the TT's clock and the time on the EP-clock at the moment they pass next to each other.

 

[yogi]

Perhaps the disagreement between our viewpoints is at least partially semantic rather than substantive - I brought up this caper with the idea of taking it a step further - As Jesse correctly surmises, and as I have more than hinted at, I feel that Einstein's great contribution was not the rederivation of the LT, but his interpretation of the meaning to be given to time as belonging to a particular frame. For those who have read different texts on the subject of the twin (or clock paradox) it is apparent there are different views - basically these views are divided into two camps - the dynamic and the kinematic. So to try to make sense out of things in the light of the fact that experimental evidence supports Einstein's physical argument, I used his example - but what is not obvious from his physical example is how he arrived at it from the concept that two passing clock frames will each measure time in the other frame as running slow. In other words, there is a jump from observation to the reality of a physical age difference between the two frames in the situation set forth...For example, we could install two clocks separated by a distance L on A's frame and measure the rate of a single clock at rest in the D-B frame (what I called the stationary frame) - and the single passing clock will always be measured to be running slow. More later -

 

[yogi]

So how does Einstein transition from observational appearances (which is what is required for reciprocity) to a statement about real age difference between two frames? - it seems that when the problem is set with Einsteins's initial conditions imposed in a frame in which the two events are separated by proper length of space, then the reality of age difference must follow irrespective of the fact that in the moving frame clocks and lengths can be constructed to show (measure) that a single passing clock in the stationary frame will appear to be running slower than the clocks in the moving A frame. It would seem then that the real age difference between the stationary frame and the moving frame is not dependent upon turn around but upon the fact that the experiment is non-symmetrical at the outset - that is, the interval as measured in the frame of A during motion comprises only time, whereas the interval as measured in the stationary frame comprises both length and time components. And while the interval is invarient vis a vis the two frames, the components of the interval in each frame are not equal.

 

[JesseM]

For those who have read different texts on the subject of the twin (or clock paradox) it is apparent there are different views - basically these views are divided into two camps - the dynamic and the kinematic.

I have never heard of a dynamic vs. kinematic view of the twin paradox--can you explain what you mean by these terms?

So to try to make sense out of things in the light of the fact that experimental evidence supports Einstein's physical argument, I used his example - but what is not obvious from his physical example is how he arrived at it from the concept that two passing clock frames will each measure time in the other frame as running slow. In other words, there is a jump from observation to the reality of a physical age difference between the two frames in the situation set forth

What is the difference between an "observation" that one twin is aging slowly vs. "the reality of a physical age difference"? Does "the reality of a physical age difference" refer to the two twins meeting up again and comparing ages in a single location, in which case every frame will agree on what their respective ages are?

...For example, we could install two clocks separated by a distance L on A's frame and measure the rate of a single clock at rest in the D-B frame (what I called the stationary frame) - and the single passing clock will always be measured to be running slow.

Sure. Would you call this a mere "observation", or would you say that the slowness of the passing clock as measured by these two clocks in A's frame is a physical reality?

So how does Einstein transition from observational appearances (which is what is required for reciprocity) to a statement about real age difference between two frames?

Again, I don't understand what distinction you're making here. If I have two clocks A and B which I have synchronized using Einstein's method, and a clock C is passing by them both, then the statements "at the moment A and C passed next to each other, A read 12:00 and C read 12:00" and "at the moment B and C passed next to each other, B read 1:00 and C read 12:30" are both statements about objective physical events which are true in every frame...but to jump from that to "C was ticking at half the rate of A and B", you have to assume that A and B were synchronized, which is only true in the AB rest frame. So would you say the first two statements are physical truths while the third statement is just about observational appearances? It's nevertheless true that if you impose the condition that every frame must see moving clocks slow down according to measurements of the times they passed clocks at rest in that frame which were synchronized according to Einstein's rule, it logically follows that if a clock travels away and then returns to another clock which doesn't accelerate, then the clock that changed velocities must have accumulated less time when they reunite. There is no way for the first idea to be true in every frame while the second idea is false.

it seems that when the problem is set with Einsteins's initial conditions imposed in a frame in which the two events are separated by proper length of space, then the reality of age difference must follow irrespective of the fact that in the moving frame clocks and lengths can be constructed to show (measure) that a single passing clock in the stationary frame will appear to be running slower than the clocks in the moving A frame.

I don't understand what "reality of age difference" is supposed to mean, if you don't actually reunite the clocks so that every frame must agree on what each one reads when they reunite. As long as they are at different locations, there can be no single objective reality about which clock has elapsed less time, since there is no objective reality about simultaneity.

It would seem then that the real age difference between the stationary frame and the moving frame is not dependent upon turn around but upon the fact that the experiment is non-symmetrical at the outset - that is, the interval as measured in the frame of A during motion comprises only time, whereas the interval as measured in the stationary frame comprises both length and time components.

The interval of what? You mean the path from taken by A? If so, that's not what I meant by "symmetrical"--it's symmetrical in the sense that the path taken by B as seen in frame A looks exactly like the path taken by A in frame B. In frame A, A's path comprises only time while B's path comproses both length and time components, while in frame B, B's path comprises only time while A's path comprises both length and time components. The symmetry here refers to the fact that if you exchange the labels of A and B, and you exchange +x for -x, then the situation is precisely identical in both frames.

 

[yogi]

No Jesse - that is point - the situation cannot be identical in both frames because of the initial conditions - if it were the comparison of clocks - as in the Einstein experiment, would never yield a different age for one twin than the other (clocks A and B read differently when A arrives at B). Einstein imposed an initial condition in which A and B were separated by a distance d in the stationary frame - and then A moves along the line AB - so the spatial interval is measured in the stationary frame of B. Since the spacetime interval is invariant, clocks in the stationary frame must intrinsically accumulate more time in order to offset the spatial distance. What I am saying, the age difference between the clocks when they are compared is not due to acceleration or turn around, or changing frames per se, but rather the reality of the time difference as between A and B is consequent to A's motion along a spatial path defined in B's inertial frame. Of course, we could have B move toward A instead of vice versa, in which case the spatial interval would be measured in A's frame, and the A clock would register more time when B arrived (if that is what you mean by symmetry - then yes, there is no difference since neither frame has any property which would render it preferred). But once you decide upon which one moves, you immediately create the asymmetry that leads to differential aging

 

[JesseM]

No Jesse - that is point - the situation cannot be identical in both frames because of the initial conditions

What initial conditions? What situation are we talking about? Isn't it the situation where the two twins are moving apart at constant velocity, without either one turning around? If so, this situation is indeed completely symmetrical.

- if it were the comparison of clocks - as in the Einstein experiment, would never yield a different age for one twin than the other (clocks A and B read differently when A arrives at B).

So now you are talking about a situation where A turns around and returns to B so they can compare clocks at a single point in space? You have to be specific about what scenario you're talking about when you switch from one to another. I agree that if one turns around, he will have aged less; if the traveling twin turns around in the Earth's frame, the traveling twin is younger when they meet, and if the earth-twin turns around in the traveling twin's frame, the earth-twin is younger when they meet.

Einstein imposed an initial condition in which A and B were separated by a distance d in the stationary frame

What does "the" stationary frame mean? Einstein said the choice of which frame you label the stationary one is arbitrary. What are the velocities of A and B in the frame you are labelling as stationary?

- and then A moves along the line AB - so the spatial interval is measured in the stationary frame of B.

So you're using "stationary frame" to mean the frame where B is at rest? That's not how Einstein used the term, as I said above. Please just say something like "B's rest frame" if that's what you mean, it'll be less confusing for both of us.

And what do you mean "so the spatial interval is measured in the stationary frame of B"--why "so"? Is there something about this problem that obligates us to use the spatial interval in B's rest frame, or do you agree that the choice of which frame's spatial interval we choose to use is a completely arbitrary one?

Since the spacetime interval is invariant, clocks in the stationary frame must intrinsically accumulate more time in order to offset the spatial distance.

What do you mean by "intrinsically?" Do you agree that we could just as easily analyze this problem in a frame where clocks in B's rest frame tick more slowly than A's clock, and we'd get exactly the same answer for the spacetime interval?

What I am saying, the age difference between the clocks when they are compared is not due to acceleration or turn around, or changing frames per se, but rather the reality of the time difference as between A and B is consequent to A's motion along a spatial path defined in B's inertial frame.

In the problem you have described, there was no initial synchronization of A's clock with B's at a single spatial location like in the twin paradox--A started out at a distance from B and moved towards it. Therefore, which clock reads a greater time when they meet depends entirely on how they were synchronized at the beginning when they were a distance d apart in B's frame. If they were synchronized in A's frame at that moment, then when A and B meet, B's clock will have accumulated less time.

Of course, we could have B move toward A instead of vice versa

How do you tell the difference? Are you assuming A and B were initially at rest at a distance d apart in their rest frame, and then A accelerated towards B? If so, you should have specified. However, this still does not mean that "A was moving towards B"--after all, there will be a frame where A and B were initially moving at velocity v, then when A accelerated its velocity dropped to zero in this frame, so it was now at rest while B moved towards it.

Again, you seem to have this crazy idea that the details of who accelerated obligates us to consider things from the point of view of one inertial reference frame rather than another. But you never answered my questions about this idea before, like what if we see two asteroids in deep space moving towards each other at constant velocity, if we want to analyze which is "really" aging more slowly do we have to know which was the last one to accelerate, even if neither has accelerated for millions of years?

in which case the spatial interval would be measured in A's frame, and the A clock would register more time when B arrived (if that is what you mean by symmetry - then yes, there is no difference since neither frame has any property which would render it preferred).

No, that's not what I mean by symmetry, I meant that if you look at some region of spacetime where two objects are both moving at constant velocity, the situation looks totally symmetrical in each frame. It's irrelevant that at an earlier point in time outside the region you're considering, the symmetry may have broken by one accelerating, that doesn't change the fact that it's symmetrical in that region of spacetime, unless you think that in order to analyze a problem taking place in some limited region of spacetime we are obligated to know the entire history of the universe leading up to the events in that region to decide which object was the last to accelerate, and we are then obligated to use the frame of the object that has been moving inertially for the longest time.

But once you decide upon which one moves, you immediately create the asymmetry that leads to differential aging

If you're using "moves" as a synonym for "accelerates", that's bogus, both for the reasons I mentioned above about needing to know the whole history of the universe to know who's "really" moving, and also for the reason I mentioned earlier, that there will always be a frame where the object was moving before the acceleration but came to rest afterwards.

 

[gonzo]

Jesse, I meant it is a meaningless comparison unless someone changes frames because otherwise you are comparing two different inertial frames, and while possibly interesting, the measurements of time and distance won't agree.

Your counter example shows that you didn't understand what I meant. If you are talking about twins, you are talking about two people starting in the same inertial frame, and then one of them shifting to a nother inertial frame, thus someone shifted frames, one of the twins. This has no bearing on what that twin later does, whether he stops or keeps moving with the new frame, he has shifted frames.

Otherwise you are just talking about measuring time and distance in two different inertial frames, and there is no even seeming paradox. They measure time differently, they measure space differently, they measure simultaneity differently. Measurements made from any two inertial frames are equally valid, it's just a different coordinate system for looking at the invariant spacetime interval between two events.

This thread I assumed by about the apparent paradox of two twins aging at different rates depending on their motion, and that whole question only has meaning if one of them moves with regard to the other.

Is that more clear?

 

[Hurkyl]

For fun, let's consider a completely symmetrical problem.


There are two pairs of clocks, A and B, and C and D.

A and B are stationary with respect to each other, and are synchronized in their rest frame, and separated by a distance L.
C and D are stationary with respect to each other, and are synchronized in their rest frame, and are separated by a distance L.

However, the two pairs are not stationary with respect to each other: there is a relative velocity between them.

The clocks travel so that:
B and C meet before B and D meet.
B and C meet before A and C meet.

(So, one might diagram it as saying they started out arranged like this:
A...B..C...D and ended like C...D..A...B)

For convenience, let us also say that B and C both happen to read zero when they meet.

---------------------------

So, what can we say about this problem?

First, we can consider the times on clocks when they meet. They can be ordered as follows:

(1) The time on B and C when they meet.
(2) The time on C when A and C meet. The time on B when B and D meet.
(3) The time on A when A and C meet. The time on D when B and D meet.
(4) The time on A and D when they meet.

In particular, when A and C meet, the time on A is greater than the time on C. And, when B and D meet, the time on D is greater than the time on B.

Now, some frame specific statements.

In the frame where A and B are stationary (and in sync):
D meets B before C meets A.
The time on D is always greater than the time on C.

In the frame where C and D are stationary (and in sync):
A meets C before B meets D.
The time on A is always greater than the time on B.


Hrm, I can't think of anything else interesting to say for this example.

(P.S. you notice that if you ignore, say, clock A, the problem is essentially the one you are analyzing?)

 

[JesseM]

Your counter example shows that you didn't understand what I meant. If you are talking about twins, you are talking about two people starting in the same inertial frame, and then one of them shifting to a nother inertial frame, thus someone shifted frames, one of the twins.

But you are free to only start considering the problem at the instant where the traveling twin has finished accelerating. Like I said to yogi, you don't have to consider the entire history of the system back to the the beginning of the universe.

Also, you can consider two clocks which are moving at constant velocity through space, and at the instant they pass infinitesimally close to each other, they both read the same time. Then as they move apart, this is just like the twins moving apart.

Otherwise you are just talking about measuring time and distance in two different inertial frames, and there is no even seeming paradox. They measure time differently, they measure space differently, they measure simultaneity differently. Measurements made from any two inertial frames are equally valid, it's just a different coordinate system for looking at the invariant spacetime interval between two events.

Well, tell that to yogi! He doesn't seem to agree with this, and that's one of the main reasons our debate has dragged on for so long.

 

[yogi]

No - I do not agree that measurements made in all frames stand on the same footing - those made in the rest frame by an observer in that same frame are proper distances and proper times - measurments made to assess the length of a rod in a frame in relative motion is not a proper distance - it will have different values depending upon the relative velocity. Measurements of lengthts and times made in the moving frame by an observer in the moving frame are proper measurements in the moving frame - they don't vary with velocity. Measurements of lengths in another frame are apparent - Jesse and Hyrkyl - if you don't like that, again that is your right - but I am also entitled to adopt the interpretation of Eddington and Resnick and other recogonized experts on SR.

Moreover Jesse - you continue to distort what I have said - I post a few lines to clarify something and you want to write a book about it. If I ask you what time it is, your going to tell me how to build a clock. What I have said is consistent with the few paragraphs that Einstein wrote in connection with the physical meaning to be accorded the transforms that were derived by making observations of how things looked in a relatively moving frame. Moreover you have got your clocks mixed up - it is A that moves to B, and there is no turn around - A and B are together at the completion of the one way trip.

 

[Hurkyl]

those made in the rest frame

What is "the rest frame"? A universal frame of reference?

 

[yogi]

Hurkyl - Of course not - when you do a relativistic analysis, you are always free to select one frame and consider it at rest. Then by definition the other is considered moving. Does that mean that there may not be a tie between SR and the rest of the universe - no again - but since we don't have a good theory of why SR works so well as a stand alone .. we go along using it as though it is a final resolution - maybe it is !

 

[Hurkyl]

Of course not - when you do a relativistic analysis, you are always free to select one frame and consider it at rest.

But that is precisely what we mean by "equal footing". The measurements in that frame aren't special: they're simply relative to the frame you selected. If I chose to make measurements according to a different reference frame, you have no grounds to claim that your measurements are any more "proper" than my measurements.


You're entirely right that, in SR, you are free to pick a single reference frame and use that as a coordinate chart for doing the analysis. In fact, I would usually suggest it as a good practice to work in a single coordinate chart unless there's good reason to do otherwise.

It doesn't follow that doing so is any more "proper" than anything else.

but since we don't have a good theory of why SR works so well as a stand alone

What do you mean by that?

 

[JesseM]

No - I do not agree that measurements made in all frames stand on the same footing - those made in the rest frame by an observer in that same frame are proper distances and proper times - measurments made to assess the length of a rod in a frame in relative motion is not a proper distance - it will have different values depending upon the relative velocity. Measurements of lengthts and times made in the moving frame by an observer in the moving frame are proper measurements in the moving frame - they don't vary with velocity.

The concept of "proper length" as a quantity which all frames can agree on only makes sense if you're talking about the length of physical objects like rulers, it doesn't apply to the distance between two objects in space, since a distance in space doesn't have a rest frame like a ruler does.

Moreover Jesse - you continue to distort what I have said

How? Why don't you clarify what you meant, if I'm misunderstanding?

What I have said is consistent with the few paragraphs that Einstein wrote in connection with the physical meaning to be accorded the transforms that were derived by making observations of how things looked in a relatively moving frame.

Please quote those paragraphs, I don't know what you're talking about here. You are not correct that Einstein used the phrase "stationary frame" to mean the rest frame of the object being analyzed, for example.

Moreover you have got your clocks mixed up - it is A that moves to B, and there is no turn around - A and B are together at the completion of the one way trip.

I understood that A is the one who accelerates (the phrase 'A moves to B' is meaningless if you don't specify what frame you're referring to), and that there is no turnaround--where did I say otherwise? It would really help if you would quote my post when responding, rather than just making general comments that don't address most of the points I brought up or the questions I asked you.

 

[yogi]

Jesse - read my post 55. I think it will answer most of your questions

 

[JesseM]

yogi, the quote by Einstein you provided in post #55 was:

If at points A and B there are stationary clocks which viewed in the stationary system, are synchronous, and if clock A is moved with the velocity v along the line AB to B then on its arrival the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by (1/2)t(v/c)^2 ...

In what way do you think this supports your argument? Do you agree that the following quote by Einstein shows that he considered it an arbitrary choice which inertial frame you denote the "stationary system" in a given problem?

Let us take a system of co-ordinates in which the equations of Newtonian mechanics hold good. In order to render our presentation more precise and to distinguish this system of co-ordinates verbally from others which will be introduced hereafter, we call it the "stationary system."

In particular, do you agree that he would have been perfectly happy to analyze this same physical experiment in terms of a different "stationary system" in which the clocks are not initially synchronized, and in which the clock which was moving in the other frame is now at rest, and the clock which was at rest in the other frame is now moving?

 

[yogi]

Jesse - of course - where did you get the idea that I considered the stationary frame as preferred - I called it stationary for the same reason as Einstein did - not because it is preferred, but because we can always pick one fame and consider it as stationary for the purpose of making measurments in a relatively moving frame. I think if you read my other posts, in particular # 75, I have made it clear. What Einstein denominated stationary has evolutionized to proper in modern parliance.

What I really wanted to talk about in connection with this interesting subject is how Einstien's physical explanation is arrived at from observational assumptions - going back to a point you raised in a previous post as to how would a person know - once the A clock is accelerated into motion along the AB line - what distinguishes the two frames if you don't know the history (the initial conditions). For example, let us assume a fourth clock D aboard an inbound rocket ship is also moving toward B on a line coextensive with AB and as A gets up to speed it is adjacent to D and traveling at the same speed. Since A and D are now in the same inertial frame, would you agree that D's clock and A's clock should run at the same rate. If so, then would the D clock lag behind the B clock at the time D arrived at B (we know the A clock will read less because Einstein told us). But if that is the case, how does this square with the proposition that D (who is only aware of his relative motion wrt to B) can consider himself at rest and B in motion, expect to measure a lesser time on B's clock when they meet? Jesse - try to keep your answer under 4 chapters, thanks

 

[JesseM]

What I really wanted to talk about in connection with this interesting subject is how Einstien's physical explanation is arrived at from observational assumptions - going back to a point you raised in a previous post as to how would a person know - once the A clock is accelerated into motion along the AB line - what distinguishes the two frames if you don't know the history (the initial conditions). For example, let us assume a fourth clock D aboard an inbound rocket ship is also moving toward B on a line coextensive with AB and as A gets up to speed it is adjacent to D and traveling at the same speed. Since A and D are now in the same inertial frame, would you agree that D's clock and A's clock should run at the same rate. If so, then would the D clock lag behind the B clock at the time D arrived at B (we know the A clock will read less because Einstein told us). But if that is the case, how does this square with the proposition that D (who is only aware of his relative motion wrt to B) can consider himself at rest and B in motion, expect to measure a lesser time on B's clock when they meet? Jesse - try to keep your answer under 4 chapters, thanks

It's all about whose frame the clocks are synchronized in. In your example, at the moment before A accelerates towards B, you're probably assuming that A and B are synchronized in their mutual rest frame. On the other hand, say that at the moment D catches up to A (after A has finished its acceleration), D's clock reads the same time as B in D's own rest frame. In this case, the D clock will be ahead of the A clock, so even though they both tick at the same rate, at the moment they pass B, the A clock will be behind B while the D clock will be ahead of B.

 

[yogi]

The same question arises in connection with Einstein's statement that a clock at the Earth's' equator (assume Earth is spherical so we can ignor GR, Earth's oblateness etc) will run a little bit slower than one at the pole. But does this follow from the fact that the two were initially in sync at the north pole and one is carried to the equator so that the equator clock (call it E) has changed frames - or is it a consequence of E's motion relative to the rest frame of the N (the north pole clock). What if both clocks are brought to sync at the equator and N is carried to the pole. Why can't E consider itself at rest and deduce that it is N that is in motion around him and conclude that it is N that is running slower. Einstein gave no credence to acceleration as having anything to due with why one clock runs faster - albeit E will experience some acceleration, Einstein discounted it (reaffirmed in his 1912 manuscript).

 

[yogi]

With regard to your post 81, what I had in mind was that D starts counting time on his clock when he sees A pull along side. If D logs 4 hours from this point until his arrival at B, then A's clock should also log 4 hours?

 

[yogi]

The interesting question to the above is - even though A and D now travel together in the quote moving frame - they got into the situation because of entirely different histories. If both clocks are now running at the same rate - is there an intrinsic rate of time passage assocated with every inertial frame?

 

[JesseM]

The same question arises in connection with Einstein's statement that a clock at the Earth's' equator (assume Earth is spherical so we can ignor GR, Earth's oblateness etc) will run a little bit slower than one at the pole. But does this follow from the fact that the two were initially in sync at the north pole and one is carried to the equator so that the equator clock (call it E) has changed frames - or is it a consequence of E's motion relative to the rest frame of the N (the north pole clock). What if both clocks are brought to sync at the equator and N is carried to the pole. Why can't E consider itself at rest and deduce that it is N that is in motion around him and conclude that it is N that is running slower. Einstein gave no credence to acceleration as having anything to due with why one clock runs faster - albeit E will experience some acceleration, Einstein discounted it (reaffirmed in his 1912 manuscript).

E is not in an inertial reference frame, the situation is not symmetrical, E cannot assume that N is ticking slower--in this case every possible inertial reference frame will agree that E's clock ticks less time than N's clock after a full revolution of the Earth (although there may be times when N is ticking slower at some point in the day from the perspective of a given inertial reference frame). If you're saying the fact that E is accelerating (moving in a circle) is not relevant to this problem, then I'm sure you're misunderstanding what Einstein was saying about accelerating--could you provide the quote?

With regard to your post 81, what I had in mind was that D starts counting time on his clock when he sees A pull along side. If D logs 4 hours from this point until his arrival at B, then A's clock should also log 4 hours?

They will both log an interval of 4 hours between the time A and D start traveling alongside each other and the time they meet B, regardless of whether D's clock is in sync with A. If you mean that D synchronizes his clock with A at the moment they start traveling together, then both clocks will read the same time when they meet B.

The interesting question to the above is - even though A and D now travel together in the quote moving frame - they got into the situation because of entirely different histories. If both clocks are now running at the same rate - is there an intrinsic rate of time passage assocated with every inertial frame?

I'm not sure what you mean by "intrinsic", but any two clocks which are at rest wrt each other will tick at the same rate, regardless of which frame is observing them.

 

[yogi]

If you look at the 1905 paper - Einstein simply gives the formula for the difference between the clocks - whether the path followed by A is straight or polygonal - no time is added or subtracted because A is not following a straight trajectory - See also his 1912 paper. We know there is a real age difference between A and B with A accumulating less time at the event of arrival - this is not a reciprocal situation - A accumulates less time as per B's clock - but why is D not entitled to conclude that B's clock is running slow ... he can consider himself at rest at the time A pulls along side and measure B's motion toward him. Will he not believe that B's clock is slow compared to his own.

What I meant by intrinsic - it is an issue that has been much discussed in the literature - but here we have a non-reciprocal situation where clocks in the A-D frame accumulate less time than the clock in the B frame - so either they don't really run at different rates and we have to explain the age difference at the end some other way such as 1) there is a specific rate of time passage (intrinsic) associated with every inertial frame that depends upon its motion with respect to every other intertial frame or 2) the two frames are not equal in all respects or 3) the spatial part of the interval is different for the two frames and this inequality leads to the different ages or.. (there are other ideas also). Einstein didn't profer an answer - now 100 years later, the subject is still debated and unresolved - it came up frequently in Einstein's own life, but he didn't shed any light upon it. My own view is that of 3) which I have previously stated.

 

[JesseM]

If you look at the 1905 paper - Einstein simply gives the formula for the difference between the clocks - whether the path followed by A is straight or polygonal - no time is added or subtracted because A is not following a straight trajectory - See also his 1912 paper.

I have looked at the 1905 paper, which is online here:

http://www.fourmilab.ch/etexts/einstein/specrel/www/

If you look at his derivation of the Lorentz transform in section 3, he specifies that he is talking about a system which is in a state of "uniform motion of translation" relative to the stationary system--ie an inertial frame. If you think this derivation would apply to non-inertial frames, then you aren't understanding the derivation. In section 4 he does talk about motion in a polygonal line, but he's only talking about looking at the motion of a non-inertial object from the perspective of an inertial frame, he's not saying the non-inertial object has its own frame which works just like an inertial one.

We know there is a real age difference between A and B with A accumulating less time at the event of arrival - this is not a reciprocal situation - A accumulates less time as per B's clock

Only because at the moment A finished its instantaneous acceleration, at that moment A and B were synchronized in B's frame--if they had been synchronized in A's frame at that moment, then B would have accumulated less time at the event of arrival.

but why is D not entitled to conclude that B's clock is running slow ... he can consider himself at rest at the time A pulls along side and measure B's motion toward him. Will he not believe that B's clock is slow compared to his own.

Once again, it's simply a question of what D's clock is set to read at the moment it catches up with A. Do you agree that, since A and D have never been at the same point in space before (and thus there was no initial moment when both could agree their clocks were synchronized), the choice of whether to have B and D read the same time in B's frame or D's frame when D catches up with A is a completely arbitrary one?

What I meant by intrinsic - it is an issue that has been much discussed in the literature - but here we have a non-reciprocal situation where clocks in the A-D frame accumulate less time than the clock in the B frame

No, it's completely reciprocal, because the choice of whose frame they should read the same time in at the moment D catches up with A is completely arbitrary, there is no physical reason to prefer one frame's definition of simultaneity.

To show more clearly how reciprocal it is, imagine that when A and B are at rest relative to each other, A is x meters to the right of B in their rest frame. Meanwhile, suppose that as D heads left towards A, D also has another object C which is at rest relative to D, and which is x meters to the left of D in their rest frame. At the moment D reaches A's location, A instantaneously accelerates so it's now at rest relative to D; at the moment C reaches B's location, C instantaneously accelerates so it's now at rest relative to B. Also, until C and A accelerate, A is synchronized with B in the AB rest frame and C is synchronized with D in the CD rest frame. Now, in terms of how D and A's times relate when they meet, and how C and B's times relate when they meet, you have two choices:

1. When D meets A, their clocks read the same time, and read the same time thereafter. This means C's clock will not read the same time as B's when they meet.
2. When C meets B, their clocks read the same time, and read the same time thereafter. This means A's clock will not read the same time as D's when they meet.

You can't have both, but do you agree there's no reason to prefer one over the other, since if you pick #1 the situation in B's frame looks exactly how the situation in D's frame looks if you pick #2, and vice versa? (this also means that if you pick #1, and when all four meet A and D's clocks read time x while C's reads y and B's reads z, then if you pick #2, you know that when all four meet C and B's clocks will read time x while A's reads y and D's reads z--once again, totally reciprocal).

 

[yogi]

I would agree the derivation is based upon inertial frames - but the physical explanation disregards any effect of changing direction Says Einstein: "If one of two synchronous clocks at A is moved in a closed curve with constant velocity until it returns to A, the journey lasting t seconds, then by the clock which remained at rest the traveled clock on its arrival at A will be 1/2t(v/c)^2 second slow." How do you conclude this refers to a non-inertial object?

I would agree as I have previously stated that either frame can be considered at rest, and in that sense the situation is reciprocal - but once the goal posts that define the proper distance in one frame are set, the age difference is one way - the clock that moves between two spatial points separated by a proper distance defined in one frame always reads less than a clock in the frame where the proper distance is measured.

Going back to my quote about the continuous curved path - Einstien has implicitly resolved the twin paradox w/o considering acceleration, changing frames's, shifting hyperplanes ect. Specifically, if B lies to the east of A and both are initially in the same frame, and A starts out headed north, then gradually veers east and then South so as to pass by B and continues by curving west and then North to return to his original starting point so that A's path describes a large circle, A will return having aged less than some sibling that remained at A's starting point. Moreover, because the circle is symmetrical and continuous and because A travels at constant velocity, there can be no logical reason to assume the aging effect comes about at any particular point in the journey (such as turn around). Upon passing by B, A's age as judged by B will be 1/2 that lost for the round trip (no pun intended). In other words, the differential aging between the twins is a continuous linear function of the time spent in moving at a constant velocity relative to the frame in which the twins are at rest.

 

[gonzo]

I must be more confused than I thought. I was under the impression that any time you changed direction you would be required to be changing inertial frames. So that going in circle, no matter how constant your velocity around the circle was, would involve constantly changing inertial frames.

 

[Garth]

yogi Going round in a circle is accelerating, it is not an inertial frame.
Nevertheless the conclusion is valid, the person moving relative to the observer encounters her again after clocking up less proper time.

Garth