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

[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

 

[JesseM]

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?

Because moving on a closed curve is by definition moving non-inertially. If an inertial observer sees an object moving inertially, with its velocity as a function of time given by v(t), then if he wants to know how much time will elapse on the object's clock between $$t_0$$ and $$t_1$$ (as measured in his own inertial frame), he must evaluate the integral $$\int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dt$$. Obviously, if the velocity is constant this integral is simply equal to $$(t_1 - t_0)\sqrt{1 - v^2/c^2}$$, even if the direction of the velocity is changing. On the other hand, if the object moving non-inertially sees the velocity of the inertial observer relative to himself at a given time t' as given by v'(t'), he cannot simply integrate $$\sqrt{1 - v'(t')^2/c^2}$$ to see how much time passes on the inertial observer's clock--if he starts out at the same position as the inertial observer and then later meets up with him again, he would get the wrong prediction for the amount of time elapsed on the inertial observer's clock if he evaluated this integral.

One example of this--suppose I am whirling a clock around my head at some constant velocity v, how slowly will I see the clock ticking, and how slowly will an observer sitting on the clock see my clock ticking? The answer is that I will see the whirling clock running slow relative to my own by a factor of $$\sqrt{1 - v^2/c^2}$$, but unlike with inertial motion there is no disagreement here, an observer sitting on the whirling clock will agree that my clock is running faster than his own by $$1/\sqrt{1 - v^2/c^2}$$. You can prove this by looking at how fast his clock is ticking in my frame, then considering what time light rays from each tick of my clock reach his position, and what time his own clock will read at each moment he receives light signals from a tick of my clock--again, this analysis would all be done from within my own inertial frame, no need to consider any accelerated frames to solve this problem.

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.

Well, the only thing we ever mean by "reciprocal" is that the situation in one frame is symmetrical with the situation in another frame--of course it's true that once you make an arbitrary choice of which reference frame you want to work in, then there will be a single truth about who is aging slower, but this seems pretty trivial. Is that really all you're arguing here, that within the context of a single frame there is only a single truth about which twin is aging more slowly?

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).

In an inertial frame where the center of the circle is at rest, the twin moving in a circle will be moving at the same speed at all times, so of course he'll be aging at an equally slow rate throughout the trip. But in an inertial frame where the center of the circle is moving, the twin will be moving faster at some times than others, so his rate of aging will vary depending on what part of the circle he's on. Either way, all frames will agree on how much time he's lost relative to another person who he departs from and then returns to after making one complete revolution.

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.

All frames will agree that if a twin takes an accelerated path away from a twin moving inertially, and later returns to meet up with the inertially-moving twin again, the accelerated twin will have aged less. But this is fundamentally different from the case where both twins are moving inertially, because in that case different frames cannot agree on who's aging less.

 

[JesseM]

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.

You are correct, circular motion is always non-inertial.

 

[yogi]

Quite right Gonzo and Garth - in actuality we probably cannot find a perfect inertial frame in which to conduct such experiments - at least not on earth, because we are always in elliptical motion about the Sun - but Einstein's formula for the time difference and his statements discounting the significance of acceleration lead us to conclude that A's accceleration wrt to the frame in which both A and B were initially at rest is circumstantial - it is what allows A to acquire motion with respect to B, but it does not impact the aging process nor does it affect the numerical difference. And since A according to Einstein can follow any polygonal path, the journey could be comprised of abrupt saw tooth oscillations that involve many hi accelerations and decelerations - but as long as the velocity is constant along the path (as it is in the one way trip), we can expect the age difference to depend only upon the velocity and total time. This is the basis of the path integral approach to the Twins problem. It is simple, direct and consistent with all experiments.

 

[yogi]

In the last line of part 4 of the 1905 paper - there is no reference to whether the clocks were synced together at one of the poles and then separated or if they were brought into sync at the equator and then separated - Einstein says: "...a balance-clock at the equator must go more slowly..than a precisely similar clock situated at one of the poles.." If the Earth is perfectly spherical this will be the case - so what does acceleration have to do with it? It is true that the clock at the equator will undergo continuous acceleration in the form of directional change - but as I said above this is only circumstantial - there is no acceleration term involved. Same thing is true for GPS - we Sync the satellite clocks in the frame of the rotating Earth - transform to the pole, and preset the rate to account for the orbit height and velocity before launch - lots of accelerations in the form of directional change every second thereafter but the satellite clocks behave just as though they are moving in an inertial frame

 

[yogi]

Jesse - With regard to your last statement in post 91 - in the example of A and B, after the brief impulse in getting A up to speed to move toward B, both A and B are in inertial frames - yet one continues to age more than the other. As I previously stressed, The acceleration can be discounted because If there were another clock Q in the B frame that lies on the AB line but is twice as far, when A arrives at Q, the Q clock would read twice what the B clock read at the time A passed by B. What I am saying is that A must be viewed as being in an inertial frame once he gets up to crusing speed, and the time differential between A as he arrives at each further clock will be proportional to the distance of the successive clocks in the B frame. According to SR, A's frame is equivalent to B's once A gets to up to speed - it is no longer distinguishable - and therein lies the heart of the dilemma because there is no physical justification for mechanical clocks to run at different rates in identical inertial frames. If the two frames are indentical - the A clock and the B clock and the Q clock should all run at the same rate - but experiments tell us otherwise.

 

[JesseM]

In the last line of part 4 of the 1905 paper - there is no reference to whether the clocks were synced together at one of the poles and then separated or if they were brought into sync at the equator and then separated - Einstein says: "...a balance-clock at the equator must go more slowly..than a precisely similar clock situated at one of the poles.." If the Earth is perfectly spherical this will be the case - so what does acceleration have to do with it? It is true that the clock at the equator will undergo continuous acceleration in the form of directional change - but as I said above this is only circumstantial - there is no acceleration term involved. Same thing is true for GPS - we Sync the satellite clocks in the frame of the rotating Earth - transform to the pole, and preset the rate to account for the orbit height and velocity before launch - lots of accelerations in the form of directional change every second thereafter but the satellite clocks behave just as though they are moving in an inertial frame

The situation with one clock on the equator and one clock on the pole is exactly like the situation where I am whirling a clock in a circle around my head while carrying a clock of my own. In this case, all inertial frames will agree that the clock whirling around ticks less time after a full revolution than the clock I am holding (which is at rest in an inertial frame), so we can simply say that the whirling clock "must go more slowly" without referring to which inertial frame we're talking about, just as Einstein said the same thing about the clock at the equator. But the fact that the whirling clock is moving in a circle is critical here--if instead we were comparing a clock held by me with a clock moving in a straight line at constant velocity relative to me, then different frames would disagree about which clock is going more slowly, so we couldn't make a statement like that without specifying which frame we're talking about. Do you agree that the two situations are different in this sense?

As for GPS, no, the satellite clocks will not behave like they are in an inertial frame--for example, if a GPS clock is moving at velocity v(t) in the instantaneous rest frame of a clock on the Earth's surface at time t as measured by that clock, you can't find the elapsed time on the GPS clock by integrating $$\sqrt{1 - v(t)^2/c^2}$$.

Jesse - With regard to your last statement in post 91 - in the example of A and B, after the brief impulse in getting A up to speed to move toward B, both A and B are in inertial frames - yet one continues to age more than the other.

Not in any frame-independent sense, no. B ages more than A in B's rest frame, and A ages more than B in A's rest frame. The situation is completely symmetrical, as I tried to show with the addition of objects C and D to this example at the end of post #87.

 

[yogi]

Jesse - it can't be symmetrical. In the linear straight path case - Its only symmetrical in the sense that either A could have taken off toward B (in which case A will be younger when they are compared at arrival) or B takes off toward A in which case B is younger when the clocks meet. Once you decide upon which one takes off, things are not symmetrical - even though A, once up to speed, travels in an inertial frame. This is the unresolved mystery of SR...but Einstein clearly tells you that the two clocks will be out of sync when the meet - and yet he disregards any influence of acceleration. If it is A that moves toward B, all during that journey the A clock will running at a reduced rate - because when they meet A will have accumulated less time as mesured by the B clock - and the longer the distance betwen A and B at the outset, the greater the age difference. And that is the puzzle of the Special Theory - it obviously isn't a puzzle to You just just like the Trinity isn't a worry to Catholics - but it is to me and many others ---this aspect of relativity is at best - unexplained. You can't have it both ways...as we have both said, once A is up to speed, A's inertial frame should be as good as B's (there is no apparent reason why A's clock should continue to run slow after the acceleration period has ended, and there is no reason why it should run slow during the initial takeoff).

Given that all inertial frames are guaranteed by the constitition to be created equal, maybe, like people, some are more equal than others

 

[gonzo]

These seem strange things to assert, Yogi. I don't really understand what you mean in a few places.

For example, what do you mean by "takes off"? Is this another way of saying "changes inertial frames"?

You seem to be trying to make it more complicated than it is, and it is very odd to invoke religion here. There is nothing unresolved or confusing here as far as I understand it.

Are you talking about A and B starting in sync and then one of them changing inertial frames? Have you tried drawing the world lines? That often clears these things up. The other thing to look at is assumptions you make about simultaneity, that is often another point of artificial confusion.

Time is measured differently in different inertial frames. Simultaneity is different. I don't really see the issue or the need to compare it to religous belief.

 

[JesseM]

Jesse - it can't be symmetrical.

What can't be symmetrical? Look over the example I gave in post #87 with the addition of D (who moves at constant velocity towards B, and who is traveling alongside A after A accelerates) and C (who is a constant distance d from D in D's rest frame, until C passes B, at which point it changes velocity so it is at rest relative to B):

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).

Do you agree with all the statements in my last paragraph? If not, please specify which ones you think are wrong.

Once you decide upon which one takes off, things are not symmetrical - even though A, once up to speed, travels in an inertial frame.

Sure they're symmetrical. Suppose you synchronize A such that, in D's frame, when A and B both read exactly the same time at the moment that A started accelerating and traveling alongside D. In this case, when A and B meet, B will read less time. It's only if you choose to synchronize A and B such that they read the same time in B's frame at the moment A accelerates that A will read less time than B when they meet. Once again, it all comes down to a completely arbitrary choice of synchronization, there is no "real" truth about whether A or B has aged less since the moment A accelerated.

 

[yogi]

gonzo - I was being facetious - doesn't really have anything to do with religeon or the US constitution. When I say takes off - that implies a change from sitting at rest in the AB frame and the start of A's motion toward B. When A takes off, A goes from zero velocity to v (granted an acceleration and a change of frame). A then continues toward B for a long time at v. When he gets to B his clock reads less than B.

The situation is very simple - except it doesn't follow. During the major part of his trip A moves at constant velocity - so he is in an inertial frame. His clock should not run at a different rate than B's clock since two identical mechanical devices cannot run at different speeds w/o a mechanical reason. The reason is absent. It has nothing to do with observations in other frames - there are none - only a beginning event (take-off) and an ending event (arrival) These two events exist at the same spatial point in both frames - but the arrival event corresponds to different time in each frame.

 

[yogi]

Jesse - I am trying to follow your logic. D and C are separated by a distance X as measured in the CD frame and A and B are separated by the same number of meters as measured in the AB frame - do I have that right? C and D are in motion v relative to AB and moving to the left - are you saying that when D is abreast of A, C will be abreast of B. Then A launches into motion so there is now no relative motion between A and D.

And then if C is separated from D by x meters - are you saying C will or will not reach B at a time that is definable in each frame relative to the time(s) that are measured by A and D when they meet. I am having a problem here because we have not synced the AB frame with the moving CD frame initially - so could you clarify this before we go on. Thanks

 

[JesseM]

Jesse - I am trying to follow your logic. D and C are separated by a distance X as measured in the CD frame and A and B are separated by the same number of meters as measured in the AB frame - do I have that right?

Yeah, that's right.

C and D are in motion v relative to AB and moving to the left - are you saying that when D is abreast of A, C will be abreast of B.

No. Each frame sees the distance between the other two as Lorentz-contracted--so in B's frame, D will catch up with A before C reaches B, and in D's frame, B catches up with C before A reaches D.

Then A launches into motion so there is now no relative motion between A and D.

Yes, at the moment A and D are at the same location, A instantaneously accelerates so there is no relative motion between A and D. Likewise, at the moment B and C are at the same location, C instantaneously accelerates so there is no relative motion between C and B.

And then if C is separated from D by x meters - are you saying C will or will not reach B at a time that is definable in each frame relative to the time(s) that are measured by A and D when they meet.

Both frames agree that at the moment C reaches B, C changes velocity so it's now at rest relative to B. But they disagree over whether this happens before or after A reached D and A accelerated so it was at rest relative to D.

I am having a problem here because we have not synced the AB frame with the moving CD frame initially - so could you clarify this before we go on. Thanks

There is no way to synchronize two clocks that are moving relative to each other permanently, but what we can do is set the two pairs of clocks either so that C reads the same time as B at the moment they arrive at the same location (and since C then accelerates so it's at rest wrt B, their times will be the same thereafter), or so that A reads the same time as D when they arrive at the same location (in which case A and D will read the same time thereafter).

 

[yogi]

Anyway - maybe another question will clarify your experiment - Taking premise # 1 you state that if A and B clocks read the same upon meeting, then C's clock cannot read the same as B. If the A and B clocks are initally in sync and the C and D clocks are initally in sync - and there is very little time loss in A's coming up to the speed of D, A and B will not have appreciable time difference at the outset - even though they are now in different inertial frames - as A travels toward B that time differential will grow. But if C arrives at B at the almost same time (which of course is undefined universally and must be subjectively measured in each frame) what is the rationale for the impossibility of C and B not registering the same time when they meet - I am missing something I know - so please clarify. Thanks

 

[gonzo]

gonzo -
The situation is very simple - except it doesn't follow. During the major part of his trip A moves at constant velocity - so he is in an inertial frame. His clock should not run at a different rate than B's clock since two identical mechanical devices cannot run at different speeds w/o a mechanical reason. The reason is absent. It has nothing to do with observations in other frames - there are none - only a beginning event (take-off) and an ending event (arrival) These two events exist at the same spatial point in both frames - but the arrival event corresponds to different time in each frame.

What doesn't it follow? The act of changing frames changes a lot of things, whether or not you do it instantaneously. One of these things is lines of simultaneity, and that means which events are in the future or in the past.

One way to think of it is that when one of them changes frames, a bunch of clock ticks in the other frame shift between the past and future as seen by the one in the new frame.

As for needing a mechanical reason for clocks to run differently -- why is that? In different frames, time and space are measured differntly, but there is no paradox or contradiction when you look at the other issues involved like simultaneity.

 

[gonzo]

Yogi, I was looking at your ABCD example above, and one thing you are missing is a reference event. You need an event in spacetime that both frames use for the origin to start counting from.