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

[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.

 

[yogi]

Ok - I posted my second question at the same time you were answering my first - so you are headed in the direction of resolving the problem using the thing I have from the beginning sought to avoid - namely making apparent measurements in the other frame - I know that relativity will always provide a way to manipulate the symbols to coax a solution that relativists will embrace. But as I have tried to stress, it is SR that is being questioned from the standpoint of how the experiments should be interpreted - so what is involved is using those aspects of the theory that have been verified, namely in the present discussion, time differences of high speed particles and satellite and airplane clocks to see, if a mechanism is revealed. Starting with a particular view on contraction to illustrate the validity of SR is a bootstrap argument.

When Einstein says that two clocks intially in sync and at rest in the same frame separted by a distance d - will read differently when brought together - I want to know what it is that is acting upon the clocks that brings about the result - I know the formulas - I derived them in graduate studies long before most of the persons posting on these boards hatched out of the egg.

Solutions that depend upon appearances in other frames are unsatisfying - like smoking a cigarette and not inhaling. Any interpretation that depends from Lorentz contraction immediately raises the question of the interpretation to be given to contraction. You may find solice in Eddington's statement: "Contraction is true, but its not really true" but I do not. I would say the reciprocal application of the contraction formulas raises questions at the outset. So in summary I would say, we probably won't be able to find a common turf for meaningful dialog. I do not say you are wrong or that any of my reasons are better - I hate to bring up religeon again - but its sort of like the missionaries that come around once and while and try to give me a Watch Tower. I know from experience the philosophy that separates me from these good folks can never find even a starting point of common ground so... I politely hand back the papar and excuse myself - they leave praying for my salvation - so perhaps jesse - you too should pray for this wayword infidel

 

[JesseM]

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

But C won't arrive at B at almost the same time, because of Lorentz contraction. If the CD system is moving at (\sqrt{3}/2)c in the AB frame, then if the distance between A and B is 1 light-year, and the distance between C and D is also 1 light-year in their own rest frame, then in the AB frame the distance between C and D will only be 0.5 light-years. So at the moment D reaches A, C will still be half a light-year away from B's position in the AB frame:

---------C---------D
B------------------A

Also, remember that if C and D were synchronized in their own frame, in the AB frame they will be out of sync--if AB sees the distance between C and D as x (with x=0.5 light years if you use the numbers above), then in the AB frame D will be ahead of C by \gamma (vx/c^2), or \sqrt{3}/2 years using the values of x and v I gave above. So if A and B's clocks read t=0 at the moment things look like the diagram above in the AB frame, then if you set C and D such that D also reads t'=0 at that moment, then C will read t' = -\sqrt{3}/2 at that moment. Since C is traveling at (\sqrt{3}/2)c and it has to travel 0.5 light-years to reach B, it will take an additional 1/\sqrt{3} years to reach B in the AB frame, so the B-clock will read t = 1/\sqrt{3} when C meets it. Meanwhile, the C clock is only ticking at half the rate of B in the AB frame, so it will only have ticked forward by (1/2)*(1/\sqrt{3}) years in this time, so it should read t' = -\sqrt{3}/2 + (1/2)*(1/\sqrt{3}) = -3/(2\sqrt{3}) + 1/(2\sqrt{3}) = -1/\sqrt{3} years when C meets B.

 

[yogi]

gonzo - that was Jesse's example - and I made the same comment. I am fully aware of the shifting hyper planes and a number of other proposed solutions to the twin paradox - these all depend upon observations (in another frame) You can't look at clocks or lines of constant temporal events or whatever and arrive at a physical age difference. That there is a physical age differece we all agree - it comes about because of some physics - remember - according to SR a clock in an inertial frame always runs at the same speed as any other inertial frame - as I previously posted, I believe the difference in the aging lies in the fact that the fame where the distance is measured is not symmetrical with the fame of the clock which is moving. A straight forward application of the principle of the interval invarience shows this to be the case, although relativists balk at this explanation.

Feynman once posed the question of why so many of our physical formulas have the same form - he queried further by asking what one thing they all have in common. He concluded: "It is the space - the framework into which the physics is put." And that is the difference between the two frames - in one frame the interval is compsed only of time - in the other there is a physical change in the spatial coordinates betwen the beginning and ending events - and that in my opinion is where we should look for the mechanism

 

[JesseM]

Ok - I posted my second question at the same time you were answering my first - so you are headed in the direction of resolving the problem using the thing I have from the beginning sought to avoid - namely making apparent measurements in the other frame - I know that relativity will always provide a way to manipulate the symbols to coax a solution that relativists will embrace.

yogi, forget about frames, and think only about physical facts that will be agreed upon by all frames. For example, in the example I gave, it is a physical fact that at the time D and A meet, both their clocks read a time of zero years. (On the other hand, the question of what B and C read at the 'same time' as this event is not a frame-independent physical fact, so let's not even think about it.) Likewise, it is a physical fact that when C and B meet, B reads 1/\sqrt{3} years and C reads -1/\sqrt{3} years. Likewise, we could calculate the time read on all four clocks at the moment they all meet. "Reference frames" help you calculate these numbers, but just think of them as scaffolding, once we have the physical facts of the situation, we can get rid of the scaffolding and just list these facts without any adornment. You do agree that these statements don't depend on reference frames, and that the numbers I've given will be correct, right?

If so, go back to the question at the end of post #87, and note that the parts in bold are only saying things about such frame-independent physical facts:

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).

So, do you agree with this last statement in bold, which has nothing to do with "reference frames" at all?

But as I have tried to stress, it is SR that is being questioned from the standpoint of how the experiments should be interpreted

But there is no "interpretation" in physical facts like the time that two clocks read at the moment they meet at a single point in space, agreed? "Interpretation" only arises when we start making statements that depend on reference frame, like what C read "at the same time" that A and D were meeting.

 

[yogi]

Jesse - My view of the ABCD problem is this - First - I do not believe in actual space contraction - While it is true that each frame measures space as contracted in the other frame - there is not a real contraction in the sense envisioned by Lorentz - this notion appears to be a carryover from Lorentz Ether theory... and while measurments made in the other frame are real in the same sense of all measurements, they do not correctly depict things in the other frame. So from my point of view, ABC and D can all be brought into sync - the distance between A and B is x and the distance between C and D is x - so even though A would calculate the distance x in the CD frame is different from the x measured in his own frame - A and D could be coincident and brought into sync and C and B can be brought into sync when they are coincident and all 4 clocks will read the same number (if they could instantantly communicate with one another) .. But since they can't, each clock could record the time at which they are synced on a permanent memory to be later compared).

In your last post - why is it a physical fact that when C meets B the time shift is
1/(3)^1/2. Doesn't this follow again from your premise re the reality of contraction.



Permit me to go back to the two clock set up - A and B are two clocks brought to sync in a common frame in which they are both at rest - A accelerates to v and travels at v until he meets A. Any path will do. A and B could initailly have been side by side and A buys a round trip ticket - or they could have been originally separated - the time differential doesn't distinguish between the two situations. Do you think it is possible to have the same result if A were already in motion relative to B? For example let's assume A is an incoming rocket with velocity v that passes near a clock E that is in sync with B and at rest in the frame of B - the two times are compared as they pass (e.g., A resets his clock on the fly to correspond with that E -- A continues on toward B w/o changing his velocity v - when A arrives at B what will he report as to his trip time from E to B?

 

[JesseM]

Jesse - My view of the ABCD problem is this - First - I do not believe in actual space contraction - While it is true that each frame measures space as contracted in the other frame - there is not a real contraction in the sense envisioned by Lorentz - this notion appears to be a carryover from Lorentz Ether theory... and while measurments made in the other frame are real in the same sense of all measurements, they do not correctly depict things in the other frame. So from my point of view, ABC and D can all be brought into sync - the distance between A and B is x and the distance between C and D is x - so even though A would calculate the distance x in the CD frame is different from the x measured in his own frame - A and D could be coincident and brought into sync and C and B can be brought into sync when they are coincident and all 4 clocks will read the same number (if they could instantantly communicate with one another) .. But since they can't, each clock could record the time at which they are synced on a permanent memory to be later compared).

In your last post - why is it a physical fact that when C meets B the time shift is
1/(3)^1/2.

Actually the time shift is 2/(3)^1/2, since B reads 1/(3)^1/2 and A reads -1/(3)^1/2.

Doesn't this follow again from your premise re the reality of contraction.

It follows from the fact that the equations of the laws of physics have the mathematical property of Lorentz-invariance, and that both observers "synchronize" their two clocks by turning on a light at the midpoint between the two clocks and setting the clocks to read the same time at the moment the light reaches them. Are you saying you're not just disagreeing with the "interpretation" of relativity, but with the actual physical predictions of relativity? You think that even if they both use the "synchronization" procedure described above, it is possible for A and D to read the same time when they meet and for C and B to read the same time when they meet? Do you think the equations that correspond to the most fundamental known laws actually give incorrect predictions about physical events, and that the "true" fundamental laws would not be Lorentz-invariant?

Permit me to go back to the two clock set up - A and B are two clocks brought to sync in a common frame in which they are both at rest - A accelerates to v and travels at v until he meets A. Any path will do. A and B could initailly have been side by side and A buys a round trip ticket - or they could have been originally separated - the time differential doesn't distinguish between the two situations. Do you think it is possible to have the same result if A were already in motion relative to B? For example let's assume A is an incoming rocket with velocity v that passes near a clock E that is in sync with B and at rest in the frame of B - the two times are compared as they pass (e.g., A resets his clock on the fly to correspond with that E -- A continues on toward B w/o changing his velocity v - when A arrives at B what will he report as to his trip time from E to B?

If A resets his clock to match that of E at the moment he passes it, and E is in sync with B in the BE rest frame (which means they aren't in sync in A's frame) then of course when A reaches B his clock will read a smaller time than B's. On the other hand, if there is a clock F at a constant distance from A and in sync with it in the AF rest frame, and B resets his clock to read the same time as F when he passes it, then B will read a smaller time than A when A and B meet. Once again, the situation is completely symmetrical.

 

[CronoSpark]

Light travels at a constant speed.
We percieve light at a constant rate.
Light rebounds from the clocks and reaches the reciever at a certain time.
When both clocks are going away from each other, we percieve a slower FPS.
When both clocks are traveling towards each other, we percieve a higher FPS.
I do not think there is a paradox. Both twins should be the same age in their own frames of reference for the whole trip. When B is x distance away and is stationary, then A will see B being x/c time younger. When B reaches Earth again, both should be the same age.

 

[JesseM]

Light travels at a constant speed.
We percieve light at a constant rate.
Light rebounds from the clocks and reaches the reciever at a certain time.
When both clocks are going away from each other, we percieve a slower FPS.
When both clocks are traveling towards each other, we percieve a higher FPS.

What's "FPS"?

I do not think there is a paradox. Both twins should be the same age in their own frames of reference for the whole trip. When B is x distance away and is stationary, then A will see B being x/c time younger. When B reaches Earth again, both should be the same age.

Not according to relativity, no. And everything about relativity follows from the assumptions that the laws of physics should work the same in every inertial reference frame, and that the speed of light should travel at the same speed in every inertial reference frame. So if you disagree with relativity's prediction that the twin who departs and then returns will be younger than the twin whose velocity never changed, then you must disagree with one of these two assumptions.

 

[yogi]

Hi Jesse - In the ABCD problem, the issue for me is not the validity of the LT - it is the interpretation - in each frame the separation between the end points is x - if relative motion means actual contraction then there is no coincidence between B and C when A and B are concurrent, and vice versa. That is to me an interpretional issue - not a rebuke of Einstein's theory. Let me again regress to the simplier question of whether you find a difference in the situation where two separated objects are in at rest in the same frame and in sync and a third clock enters the picture flying by one of the at rest clock to copy its reading. I see no way the transforms can be interpreted to predict a different age for the newcomer clock when it continues on to meet the other at rest clock.

 

[yogi]

jesse - to recap: A and B are two clocks brought to sync in a common frame in which they are both at rest - A accelerates to v and travels at v until he meets B. Any path will do. A and B could initailly have been side by side and A buys a round trip ticket - or they could have been originally separated - the time differential doesn't distinguish between the two situations. Do you think it is possible to have the same result if A were already in motion relative to B? For example let's assume A is an incoming rocket with velocity v that passes near a clock E that is in sync with B and at rest in the frame of B but separated from B by a distance d- the two times (A and E) are compared as they pass (e.g., A resets his clock on the fly to correspond with that E -- A continues on toward B w/o changing his velocity v - when A arrives at B what will he report as to his trip time from E to B?

If A arrives at B having aged less than B after he has set his clock to correspond with E, how could that happen since in this situation we have no way of knowing whether E and B are moving relative to a space defined by a distance d in the A frame or a distance defined in the EB frame. There is symmetry in this case. Contrary wise if E takes off and flies parallel to A to arrive at B coincident with A. In this case the symmetry is broken because E is traveling in a space defined by distance measured in the B frame - so E's clock will read less than B's when E meets B.

 

[JesseM]

Hi Jesse - In the ABCD problem, the issue for me is not the validity of the LT - it is the interpretation - in each frame the separation between the end points is x

Ok, let's not say anything about the distance between them, since "distance" is not an objective physical concept. Let's just state the following objective facts: if a light is turned on at the midpoint of B and A before A accelerates, then A's clock and B's clock will read the same time at the moment the light hits them; likewise for C and D before C accelerates. And before any acceleration happens, if A aims his telescope at B, he will see B's clock 1 year behind the clock he is carrying, and vice versa; same goes for what happens when C aims his telescope at D or vice versa. Given these conditions, do you agree with the numbers I gave for what times C and B will read when they meet, given that A and B both read a time of 0 years when they meet? Do you agree that the situation is "symmetrical" in the sense I described at the end of post #87, so if you instead set things up so C and B read 0 years when they meet, D will read 1/\sqrt{3} and A will read -1/\sqrt{3} when A and D meet, and likewise the situation is also symmetrical when all four meet?

 

[JesseM]

jesse - to recap: A and B are two clocks brought to sync in a common frame in which they are both at rest - A accelerates to v and travels at v until he meets B. Any path will do. A and B could initailly have been side by side and A buys a round trip ticket - or they could have been originally separated - the time differential doesn't distinguish between the two situations. Do you think it is possible to have the same result if A were already in motion relative to B? For example let's assume A is an incoming rocket with velocity v that passes near a clock E that is in sync with B and at rest in the frame of B but separated from B by a distance d- the two times (A and E) are compared as they pass (e.g., A resets his clock on the fly to correspond with that E -- A continues on toward B w/o changing his velocity v - when A arrives at B what will he report as to his trip time from E to B?

If t_E is the time that E read as A passed it, and t_B is the time that B read when A met it, then the time that A measures between passing E and meeting B will be \sqrt{1 - v^2/c^2}(t_B - t_E).

If A arrives at B having aged less than B after he has set his clock to correspond with E, how could that happen since in this situation we have no way of knowing whether E and B are moving relative to a space defined by a distance d in the A frame or a distance defined in the EB frame.

A hasn't "aged less", his clock just reads a lesser time than B's. From A's point of view, the reason for this is that B's clock is always ahead of E's clock. (If I put a clock in my hallway that's one hour behind the clock in my study, and it takes you only 15 seconds to walk through my hallway and into my study, am I justified in saying that you've aged 59 minutes and 45 seconds less then me when we meet?) If you want to stick to objective physical facts that don't depend on your choice of reference frame, the question of "who has aged less" is meaningless for two observers moving at constant velocity, just like the question of which of two meter sticks in relative motion "really shorter". The only situation where you can talk objectively about which of two observers has aged less is where they first compare clocks at a single location, then move apart, then come back together to compare clocks again.

You can't have it both ways, sometimes rejecting ideas which depend on reference frames like Lorentz contraction, at other times wanting to talk in terms of language that has no frame-independent meaning, like which of two observers moving at constant velocity has "aged less".

There is symmetry in this case.

No, this is not symmetrical, because you have two separated clocks at rest and synchronized in B's frame, while A has only one clock. To make it symmetrical, you'd have to add another clock F which is at rest relative to A, and synchronized with A in A's rest frame. Then it's clear that you can either arrange things so that A synchronizes his clock with E as he passes it, or that B synchronizes his clock with F as he passes it--and whichever arbitrary choice you make, the one that synchronized his clock with the clock at rest in the other's frame will read a smaller time when A and B meet. Do you agree?

 

[gonzo]

Yogi, it might help if you didn't think about inertial frames as having any existence of their own. As I understand it, they are just artificial constructs for humans to look at things

Frames don't exist, only events in spacetime exist, whether that event is a tick of a clock, or two particles being in the same place. We humans can then try and measure things about these events, using various metrics. It turns out that these metrics differ in different inertial frames, though the spacetime interval is always the same. The events are still real, still happen, and aren't affected by our way of measuring them.

Taylor and Wheeler's book has a good analogy of two people making a map of a city, using slightly different directions for North. Using the city center as the origin, they will assign different East-West and North-South distances to the same point in the city. That point still exists in the same place, and that point doesn't really care about how it is being measured, but they will still have different numbers for it. However, straight distance from the city center will be the same for both cartographers.

In your example in #115 of A flying between E and B which are stationary and in sync, assuming no acceleration, you are looking at it in a funny way as I see it. There are simply two events. Event one is when A arrives at E, and event two is when A arrives at B. You can even take event one to be the agreed upon origin.

These are two events that have nothing to do with any inertial frames. However, we can if we want measure the time and distance between these events. If this measurement is done in different inertial frames, then the interval between these events will be same.

Luckily, this is a simple example since if we look at the inertial frame following A, the distance is 0 between the two events, so the the Interval is equal to the Time.

Assuming everyone sets their origin to when A passes E, then you can compare measurements when A passes B. And yes, A's clock and B's clock will show different elapsed times, just like they will show different distances. For A, there was zero distance between the two events. They both happened at A. For B there was the distance BE between the two events. Different distance means different times. So what?

I don't really see what you mean by "symetrical" in this case.

 

[yogi]

Jesse: From your post 117: "The only situation where you can talk objectively about which of two observers has aged less is where they first compare clocks at a single location, then move apart, then come back together to compare clocks again."

Simply not true - look again at what I have quoted from the 1905 paper - A and B are not together - they are at rest separated by a distance d, brought to sync - and then A moves to meet B. There is now an age difference. A's clock will have recorded less time than B's. Seems we have already established this about 10 times.

 

[yogi]

Let us assume Yogi and Jesse are on orbiting spacecraft - I am flying around the equator east to west and you are flying west to east. Each time we pass each other we check the other guys clock using two clocks in our own spacecraft - You will say poor Yogi - not only is he slow to understand SR - his clock is running slow also. And I will look out my window and measure the rate of your clock as it flies by and say - Jesse is to busy posting on the forums to set his clock rate up to speed. Gonzo is at the North pole - at the same height above the Earth as you and I - he will see both our clocks running slow. Now each time we pass I will see your clock running slow and you will see mine running slow - but a funny thing happens -neither of us accumulates any age difference relative to the other - our situation is entirely reciprocal. So after many orbits we decide to land and we do so bu detouring to the North Pole - we compare clocks - Jesse and yogi's logged times will be the same - but Gonso's clock will read more than Jesse's and Yogi's Or do you have a different conclusion?