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# Relativity and Cosmology

This is a Blog on relativity and cosmology for engineers and the like. You are welcome to comment upon or question anything said on my website (relativity-4-engineers), in the eBook or in the snippets I post here.

Comments/questions of a general nature should preferably be posted to the FAQ section of this Blog (http://cr4.globalspec.com/blogentry/316/Relativity-Cosmology-FAQ).

A complete index to the Relativity and Cosmology Blog can be viewed here: http://cr4.globalspec.com/blog/browse/22/Relativity-and-Cosmology"

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# The 'Half-Twin' Puzzle

Posted January 11, 2010 11:00 PM by Jorrie
Pathfinder Tags: special relativity Twin Paradox

The 'Twin Paradox' of relativity is around a century old, but it still gets a lot of attention on many forums. I have also written on it a few times in this Relativity and Cosmology Blog, e.g. here and here. An interesting variant of the classical twin-paradox is to consider the situation at the halfway stage, which may be labeled the 'Half-Twin Puzzle' of relativity.

On the right is a graphic of what I call the 'RGB scenario', an all-inertial variant of the classical twin paradox. It differs only in that the 'away-twin' (Red) does not make a quick turn-around to come home again, but a third inertial observer (Blue) does a 'R/B flyby' in the opposite direction and performs the return leg. Both speeds are 0.866c relative to Green, giving a Lorentz factor γ = 1/√(1-0.8662) = 2 in both directions. This means that the 'moving' clock appears to tick at half the rate of the 'stationary' clock.

At the R/B flyby, Blue sets her clock to read the same as Red's clock (2 units) and since they are both inertial, presumably their clocks will tick at the same rate relative to Green. When Blue later passes Green, they compare clocks first-hand and conclude that Green has aged 8 units, double the sum of Red's and Blue's aging during the two halves of the test (a sum of 4 units). This is standard, verified relativity and not disputable.

The 'half twin puzzle' is where we simply ask the question: at the halfway (R/B) flyby event, which twin (Green or Red) has aged less since the original (G/R) flyby event? Here are some issues, lurking in very subtle, perhaps confusing logic. To understand them, we must look at the full GRB scenario, after which Green, Red and Blue can hold a teleconference and compare results. They will conclude that the proper time that has elapsed for Red until the R/B flyby (from 0 to 2) equals the proper time that has elapsed for Blue from the R/B to the B/G flybys (from 2 to 4). This is inevitable, since Red and Blue flew the same distance at the same speed relative to Green.

Green, Red and Blue will further agree that Green's elapsed proper time between the G/R and G/B flybys was 8 units, as measured by Green's own clock. It seems logical that between the G/R and R/B flybys, Green "really aged" 4 units while Red "really aged" only 2 units (Green's dotted line of simultaneity). After all, Red and Blue each recorded half of the 4 units, so how can Green not record half of the 8 units at the halfway point?

The first apparent paradox is this: since twins Green and Red were both stationary in their own inertial frames for the duration of the test, they should be equivalent and the one cannot age more (or less) rapidly than the other over the duration of this test. To argue that they age differently will mean giving preference to one inertial frame over another. Since the 2 units for Red is a given, it means that Green should also have aged only 2 units.

Secondly, according to special relativity, Red has the 'right' to view Green as moving at v = -0.866c, with a Lorentz factor of γ = 2. So per Red, Green should have aged only 2/γ = 1 unit between the G/R and R/B flybys (Red's dotted line of simultaneity). Thirdly, to make matters even worse, according to Blue's dotted line of simultaneity, Green's clock should have read 7 units when the R/B flyby occurred. After all, Green flew at 0.866c relative to Blue and while Blue aged 2 units, Green should have aged only 2/γ = 1 unit.

How do we reconcile these apparently paradoxical conclusions, i.e., did Green age 1, 2, 4 or 7 units during this 'Half-Twin test'? We will let interested readers puzzle a little, before offering attempted resolutions of the paradoxes.

-J

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#1

### Re: The 'Half-Twin' Puzzle

01/12/2010 2:13 PM

Delicious.

My "knowledge" of Relativity is gleaned from popular books, that is why it is in quotation marks, it could be faulty but I think I understand the principles.

My first observation is that the velocities are as measured by Green, Red and Blue will not measure each other as moving at 1.732c, further, each will see the other with a different clock speed than Green will see.

Lewis Carroll Epstein said we need a myth, the myth is that everybody moves through spacetime at the speed of light. It seemed to work well, particularly if seconds and light seconds are graphed as equal lengths. If drawn that way, the red and blue lines would be at 30o to the horizontal and the relative velocity problem more obvious. I have to play with that to see if I can figure their view of things.

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#3

### Re: The 'Half-Twin' Puzzle

01/13/2010 9:12 AM

The relative speeds Green_Red and Green_Blue are 0.866c and Red_Blue = 2 x 0.866/(1+0.866^2) ~ 0.99c.

In my diagram, the x- and the t-scales are meant to be the same. It is however a Minkowski spacetime diagram, not an Epstein space-propertime diagram, as you suggested (with the 30o to the horizontal). In the Minkowski, the angle to the vertical (time axis) is atan(v/c). The Epstein diagram is not very useful if you have to consider more than two inertial frames.

-J

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#2

### Re: The 'Half-Twin' Puzzle

01/13/2010 7:57 AM

It is true that Red really ages 2 units between the GR and RB coincidences. But beyond that, there's never a coincident event involving Red, and therefore it is incorrect to conclude that he "would have" aged the same way as Blue did. In fact, it is even absurd to say that. For Red to really compare, he himself must've made the return leg, in which case he has suffered an infinite acceleration at the RB point, thereby making any analysis meaningless.

To say Green's clock "must have read" one thing or another "when" RB happened is meaningless unless some standard of simultaneity is set up. Red and Blue can judge the apparent rate of Green's clock by, say, receiving light signals that Green sent every second of his clock, and then figuring out how long ago in their own frame the signal must've been sent (at any point of their journey, they know how far they are from Green). It is from this apparent rate that, say, Red concludes that "I calculate Green's clock to be half as fast as mine, which means that now when I'm meeting Blue, Green must be 1 unit old". It is in a similar manner that Blue concludes "I calculate Green's clock to be half as fast as mine, and I am now meeting him when his clock shows 8, which means that when I met Red 2 units ago, Green must've been 7 units old."

Of course, these "must've been"s and "when"s don't make any real sense, because there is no coincident event to compare.

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#4

### Re: The 'Half-Twin' Puzzle

01/13/2010 9:19 AM

Hi Guest, you wrote: "It is true that Red really ages 2 units between the GR and RB coincidences. But beyond that, there's never a coincident event involving Red, and therefore it is incorrect to conclude that he "would have" aged the same way as Blue did. In fact, it is even absurd to say that."

But, won't you agree that since they are both true inertial observers, we are allowed to say that they should age the same? On what basis shall we give preference to one or the other?

-J

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#5

### Re: The 'Half-Twin' Puzzle

01/13/2010 11:52 AM

Yes, in fact all three are inertial and their proper times move at the same rate. Indeed there is no preference of one over the other. But one cannot take what Blue observes to conclude that Red "would have" aged the same amount in the latter half. In this phase, Red would have moved somewhere further and he no longer shares any spacetime event with either of Blue and Green, so that they never get a chance to compare their time readings. The only way they could compare elapsed proper times in their respective frames is if their trajectories began and ended at the same two spacetime points.

One way is that whatever clock Red himself has on his ship, he instantaneously hands over to Blue at their meeting, and then to say that that clock records the total proper time. But then, the trajectory in spacetime followed by that clock would involve a bit of acceleration, and so, although both Red and Blue are inertial, the clock they exchange would be noninertial while being exchanged. It would then no longer record the proper time along a spacetime geodesic.

The trajectories of Red, Blue and Green are geodesics, but because it's a flat (Minkowskian) world, they don't intersect at more than one point so as to enable comparison of the proper times along different geodesics. We can compare the proper time along the two contours --- one being Green's trajectory and the other the clock's --- both trajectories going between the same two events GR and GB. They should be the same if both trajectories are inertial. But the clock's trajectory is noninertial, and therefore it is no surprise if it records a different time.

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#6

### Re: The 'Half-Twin' Puzzle

01/13/2010 12:48 PM

Hi Guest, you wrote: "Yes, in fact all three are inertial and their proper times move at the same rate. Indeed there is no preference of one over the other. But one cannot take what Blue observes to conclude that Red "would have" aged the same amount in the latter half."

But, we have a clear condition that Blue and Red are moving at the same speed, just in opposite directions relative to Green. We are not really interested in Red's future aging, only that Blue will age at the same amount between the two 'inbound' events than what Red aged between the two 'outbound' events.

I do not think you can dispute that Blue will age 2 units during this interval. It is just a "Minkowski triangle" with two equal, time-like sides. AFAIK, one can analyze it in any inertial frame and get the same numerical result.

"One way is that whatever clock Red himself has on his ship, he instantaneously hands over to Blue at their meeting, and then to say that that clock records the total proper time."

Why can't B just "hand over" the time, not the physical clock, as stated in the puzzle? Blue then sets her own clock according to this time and carries on, with no acceleration of any clock.

-J

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#7

### Re: The 'Half-Twin' Puzzle

01/13/2010 12:59 PM

Hi again Guest, just to get the 'solution' closer to back on track, I fully agree with what you wrote before (I gave a GA for it):

"Red and Blue can judge the apparent rate of Green's clock by, say, receiving light signals that Green sent every second of his clock, and then figuring out how long ago in their own frame the signal must've been sent (at any point of their journey, they know how far they are from Green). It is from this apparent rate that, say, Red concludes that "I calculate Green's clock to be half as fast as mine, which means that now when I'm meeting Blue, Green must be 1 unit old". It is in a similar manner that Blue concludes "I calculate Green's clock to be half as fast as mine, and I am now meeting him when his clock shows 8, which means that when I met Red 2 units ago, Green must've been 7 units old.""

That said, what do you reckon is the age of A at the time of the R/B flyby?

-J

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#8

### Re: The 'Half-Twin' Puzzle

01/13/2010 2:06 PM

Hi again Jorrie,

The thing with this situation is that the first leg of Red's journey and the second leg of Blue's journey are sections of flat-space geodesics, but gluing them together at RB produces a broken line which is NOT a geodesic, and therefore time measured along it does not yield the proper time for inertial observers. You're right, they needn't exchange the clock itself, but the fact remains that while Red ages by 2 between GR and RB, and Blue ages by the same between RB and GB, we have no specific event for Blue before and for Red after until when to consider as the "equivalent" leg to the inner journey.

As for the question "what do you reckon is the age of A (Green) at the time of the R/B flyby?", it is nothing absolutely defined, because simultaneity is relative. So while for Red (from his back-calculations from the light signals etc.), that moment seems simultaneous with the 1-unit point for Green, for Green it appears simultaneous with the 4-unit point, and for Blue, with Green's 7-unit point. There is no universal standard to it.

I think what takes a little time to sink in is that although Red and Blue are at the same event, their current estimates of Green's time are different, because they aren't moving together.

We can take another example in terms of flat-space geodesics. If the times along all piecewise-geodesic contours between the same two end points were the same, then it would mean that time for even noninertial observers, who can be thought of as moving along paths whose infinitesimal sections are geodesics, must experience the same time. But this is not true. We cannot just add the times along the outward and inward sections (which are sections of different geodesics) and get any meaningful measure of time.

Had the metric been curved, say like that of a sphere, then two geodesics could have the same end points and still be distinct, and could have the same length. But in the Minkowski case, this never happens. So the only way two observers meeting at two different events can agree on the time in between is if they move together all along. Otherwise, they can't both be inertial and also meet at two points.

Does this convince you?

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#9

### Re: The 'Half-Twin' Puzzle

01/14/2010 12:26 AM

Hi varun_achar, if these are really your first posts on CR4, then very welcome here!

You are making very good points (worth your first official GA) and I'm glad to have you onboard. I don't necessarily need convincing, but I'm trying to poke holes in whatever resolutions others come up with to see how robust they are. I also throw arguments on the table and see if/how they can be shot down. In the end we should be able to agree on a 'best-buy' solution...

What if we say that since Green and Red are the main actors in the 'half-twin test', the most democratic solution is to assume they are completely equivalent. For convenience, we set up an inertial frame (A) where the situation is drawn symmetrical. Green and Red now moves at 0.577c in opposite directions relative to frame A, still giving their relative speed of 0.866c, as before.

In the A-frame, both Green and Red have aged 2 units at the time of the R/B flyby (Blue is left out for less clutter). Assume Green and Red have agreed to each send the other a time-stamped radio signal when their respective clocks read 2 units (the black arrows). They will each receive the others time signal at 7.46 units on their respective clocks.[1] Doesn't this 'prove' that Green actually aged the same 2 units than what Red aged at the R/B flyby?

I've had the counter-argument that I 'cheated' by selecting a sort-off 'preferred frame' and hence get this symmetrical result.The answer is that for the given inputs, we can choose any frame as reference and we will get exactly the same 7.46 units for Red and Green. This is a frame independent result...

-J

[1] For the general readership, I've used the relativistic Doppler shift to calculate this time, because for me it's the easiest: T2 = (1+0.866)/√[1-0.8662] T1. T1 = 2, the period at the transmitter, giving T2 = 7.46, the period at the receiver.

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#10

### Re: The 'Half-Twin' Puzzle

01/14/2010 1:09 AM

Thank you Jorrie for the warm welcome and for your reception of my answers. Indeed, these have been my first comments here on CR4. I am no engineer, therefore I haven't been a very frequent visitor. My father, who is one, occasionally shares with me posts that interest him. That is how I came upon this your page.

I agree that your suggested frame makes things much more evidently understandable. Just a small remark on my second comment: While there is no requirement for R and B to actually exchange an object, envisaging them doing so serves to highlight the fact that the broken line GR-RB-GB, which in this imagination would be the clock's world line, is not an inertial line, making it analogous to the "Twin Paradox" in the way we usually hear of it.

Thank you for the diagrams, which are extremely helpful in understanding such situations.

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#11

### Re: The 'Half-Twin' Puzzle

01/14/2010 1:34 AM

Hi Jorrie,

A very interesting variant of the Twins paradox, you have there! Takes away all the unnecessary acceleration business, that does nothing but cloud the real solution.

The problem lies in asking questions that do not have any frame invariant meaning. Like "Doesn't this 'prove' that Green actually aged the same 2 units than what Red aged at the R/B flyby?"

The notion of simultaneity is relative and depends on the frame. So which birthday of G was "at with the R/B flyby" event, depends on the frame. In the A-frame it is 2, in the G-frame it is 4, in the R-frame it is 1 and in the B-frame it is 7. This can be seen from your diagrams clearly.

There is no meaning to the question, what is his "real" age when the R/B flyby takes place. It depends on the frame of reference.

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#12

### Re: The 'Half-Twin' Puzzle

01/14/2010 8:50 AM

Hi guest, you wrote: "There is no meaning to the question, what is his "real" age when the R/B flyby takes place. It depends on the frame of reference."

I agree that this is the 'standard view'. But, as I have shown in reply #9, there are some frame-independent quantities available in the 'half-twin' case. My quest is to find out if one of these quantities cannot perhaps be converted to a frame-independent age of Green when some distant event happens. After all, since they are both inertial, one would suspect Green and Red to age at exactly the same rate - the only problem seems to be one of measurement...

I know it's perhaps a futile quest, but I have one more 'ace up the sleeve' for a little later. :)

-J

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#13

### Re: The 'Half-Twin' Puzzle

01/14/2010 12:59 PM

How about some fourth guy, say Yellow, does with Green what Blue does with Red? Then there'd be a corresponding YG crossing to compare with the RB, and that would give a definite meaning to Green's age "at" the RB crossing....

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#14

### Re: The 'Half-Twin' Puzzle

01/14/2010 2:30 PM

Hi varun_achar, "How about some fourth guy, ..."

As a matter of fact, I've already done such a drawing - it is the "ace up the sleeve" that I was talking about earlier. I'm not 100% sure how to interpret it, but I'll post it anyway for all to think along these lines.

I used Lilac for the "fourth guy", L, who first has a flyby with G at 2 units, which time he transfers to his clock. Then he has a flyby with B when they each record 2.4 units on their own clocks. I think this means that we can say with some certainty:

1. B retrospectively knows that G's clock read 2 units when R's clock (and his own) was at 2 units. Or not?

2. This result is coordinate independent, because we get the same values irrespective of what reference frame we use (the A-frame is just a very convenient one).

3. This does not negate the coordinate dependent results, like "per G's definition of simultaneity, G's clock read 4 units when R's clock read 2 units".

4. It is compatible with the notion that all inertial frames should be treated equal, including ticking at the same clock rates.

5. We have replaced the "Minkowski triangle" with a "Minkowski kite". I do not fully understand the implications if this, but maybe it defines a kind of "absolute simultaneity"?

-J

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#15

### Re: The 'Half-Twin' Puzzle

01/14/2010 3:28 PM

Hi Jorrie,

Nice, so my hunch was right. Maybe we don't really need a very profound interpretation of this; to me it seems like merely setting more constraints and thereby defining more features. What do you think?

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#16

### Re: The 'Half-Twin' Puzzle

01/14/2010 9:51 PM

Hi Varun,

I'm still tying to work out what we can and cannot say about such a scenario. Returning to my prior 5-pointer:

"1. B retrospectively knows that G's clock read 2 units when R's clock (and his own) was at 2 units. Or not?"

I think it is clear that the L/G flyby had to be at tG=tL=2, otherwise the identical readings tG=tL=2.4 at the L/B flyby could not be true.

"3. This does not negate the coordinate dependent results, like "per G's definition of simultaneity, G's clock read 4 units when R's clock read 2 units"."

The definition of simultaneity is conventional (Einstein's), which is a good convention, because it simplify physics considerably. It is however not the only convention that will satisfy the Lorentz transformations (LTs) and so predict the same outcomes for experiments.

"5. We have replaced the "Minkowski triangle" with a "Minkowski kite". I do not fully understand the implications if this, but maybe it defines a kind of "absolute simultaneity"?"

It looks like for every event in Minkowski spacetime, there is at least one other 'complimentary event' that can be viewed as simultaneous without using Einstein's simultaneity convention. By this I mean doing the experiment 'BGRL' above, which relies on Minkowski space (and hence on the LTs), but not on Einstein's simultaneity convention.

IMO, this is the same as saying that for every space-like[1] pair of events in Minkowski spacetime, there exists one inertial frame in which the events are simultaneous as per Einstein's simultaneity convention.

There may also be time-like complimentary events, where there is one inertial frame in which the events are co-located.

Any more features?

-J

[1] Space-like means nothing, not even light, can travel between the events, while time-like means material objects can.

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#17

### Re: The 'Half-Twin' Puzzle

01/14/2010 10:05 PM

You are right. The terms "spacelike" and "timelike" are in fact motivated by the fact that spacelike (timelike) -separated events always have some inertial frame in which their separation is purely in space (time), and their time (space) coordinate(s) is (are) the same. In other words, spacelike-separated events don't have a "proper time interval" between them but what I'd call a "proper space interval", whereas timelike-separated events have a "proper time interval" between them, which is the time-separation between them in a frame where they have the same spatial coordinates.

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#18

### Re: The 'Half-Twin' Puzzle

01/15/2010 1:55 AM

Yup!

I suppose we can now take the terminology further by saying that like we have 'proper distance' versus 'coordinate distance' and 'proper time' versus 'coordinate time', we can have 'proper simultaneity' versus 'coordinate simultaneity'.

Interestingly, the lines of 'proper simultaneity' is nothing more than the hyperbolas of constant time-like interval in Minkowski spacetime, as shown on the right.[1] Each event on the ct-axis has only on hyperbola (i.e., one definition of 'proper simultaneity'), irrespective of who observes the event. Contrast this to 'coordinate simultaneity', a straight line with a different slope for each inertial observer of the event. Galilean simultaneity obviously has a single, zero-slope line of simultaneity for each event, so 'proper simultaneity' must not be confused with 'absolute simultaneity'.

A single time-like hyperbola represents the set of events with the same proper time separation from the origin, i.e. as observed by various observers co-located with the origin and with those events. This is what I call a 'line of proper simultaneity'.

A single space-like hyperbola represents the set of events with the same proper spatial separation from the origin, i.e. as observed by various observers for whom those events are apparently simultaneous with the origin. May we call this a 'line of proper separation', or is there a better term?

-J

[1] For interested parties, the figure depicted is Fig. 2.7 of the eBook Relativity-4-Engineers. More discussion there...

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#19

### Re: The 'Half-Twin' Puzzle

01/15/2010 7:17 PM

This is a nice construction. I have one question to ask, whose answer doesn't seem trivial and I must think about: What is the extent of "objectiveness" of these lines with respect to a change of the origin (translation), and to a boost? For now we can ignore spatial rotations since we're visualizing it in a plane.

My hunch is that the whole scene would diffeomorphically deform into one which looks identical. For instance, if we wanted to draw the same diagram with the ct'-axis as the chosen time axis (which amounts to a boost) and ct'=2 as the origin (translation), then the proper-time-1 hyperbola would deform into proper-time-zero asymptotes (+ and - pi/4 lines), while the rest of them would all deform in accordance with these new asymptotes. The boost part would deform them to a new parametrization, but would keep them congruent to themselves before the deformation. The interesting part is the translation, which would deform each hyperbola to the shape of its immediate neighbour. If, for instance, we shifted the origin to the ct=1 point, then as mentioned before, the corresponding hyperbola would deform into the V-shaped asymptotes. At this point, if we apply a small translation further (to some ct=1+delta), the V would suddenly get reflected into the inverted V and thereafter become a hyperbola on the bottom half-plane.

Similar things would happen to the spacelike hyperbolas on applying spatial translations.

It would be great if we had some sort of a tool where we could graphically choose the origin and axes and then see the diffeomorphism evolve the diagram into the new frame.

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#20

### Re: The 'Half-Twin' Puzzle

01/15/2010 7:25 PM

I forgot to mention: Since the points sharing a hyperbola continue to do so under these diffeomorphisms, and ones on different hyperbolas remain on different ones, this is a Lorentz-invariant definition of simultaneity.

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#22

### Re: The 'Half-Twin' Puzzle

01/15/2010 7:58 PM

Good point about the Lorentz invariance.

BTW, I did not notice your #19 and #20 before I posted my #21 (they did a 'flyby' in cyberspace :)

I'll also have to think a bit about translations and boosts.

-J

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#21

### Re: The 'Half-Twin' Puzzle

01/15/2010 7:49 PM

Hi Varun,

Before I get too carried away and I take others with me, it may be wise to point out that AFAIK, "proper simultaneity" is not an accepted mainstream term and there may not even exist such a concept in modern science. It may well be that it has been treated in some paper that I do not know of - the closest I've got to a good summary of the Conventionality of Simultaneity is in a document by Prof. Allen I. Janis, formerly of the University of Pittsburgh. He referenced a large number of papers that I have not worked through yet. Einstein's simultaneity is usually called "standard synchrony", while non-Einstein definitions of simultaneity are usually called "nonstandard synchrony".

Be that as it may, it is scientifically valid to define a 'new' simultaneity, different from Einstein's and then analyze the effects of such a definition, whatever its official terminology may be. Prof. Janis mentioned that "all of special relativity has been reformulated (in an unfamiliar form) in terms of nonstandard synchronies (Winnie, 1970a and 1970b)".

Returning to my definition of "proper simultaneity" (or maybe I should have called it "hyperbolic simultaneity"). In order to make it science and not philosophy, we should answer some technical questions:

1. How would clocks be synchronized in such a definition of simultaneity?
2. What would the observable effects be, if any?
3. What will the "unfamiliar laws" of physics look like?

The answers are not all clear to me, but for 1 and 2, I can mention the following:

A1. 'Fast move' a number of clocks, all set to zero as they pass the origin simultaneously, at different constant speeds until their co-located observers read time t on the clock faces. Do not resynchronize them (the Einstein way) and they will sit on a "hyperbola of (proper) simultaneity". This is somewhat loosely described, but I think you get the gist. Any recommendations?

A2. The full 'twin-paradox scenario' will yield the same result as Einstein's, because no clocks are ever resynchronized there. The Michelson-Morley experiment would not be influenced, because it too does not use clock synchronization. Any one-way result will however differ from Einstein's, including the one-way speed of light. Light propagation will not be isotropic - the one-way speed of light depends on the simultaneity convention used.

A3. Other than the anisotropy of light propagation, I have no idea. I can however imagine that things may be rather ugly. Clever Einstein...

-J

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#23

### Re: The 'Half-Twin' Puzzle

01/15/2010 8:24 PM

Hi Jorrie,

These are issues to be addressed indeed. My suggestions are as follows: a light pulse will travel along the hyperbola containing the source event. So for synchronized clocks, we could just use a source of light stationary in some accepted frame which emits a spherically-symmetric (in the 3D case) pulse every second, say. The zeroth pulse would propagate along the rectangular asymptotes, the first one would propagate along the ct=1 hyperbola, and so on. Observers in other frames would measure time based on the count of pulses they have received. This way, when they compare time readings of events with other observers, they would all agree on your definition, i.e. the "proper simultaneity". One could perhaps also call it "null simultaneity"?

The only problem is that such a system could be established only between nearby locations, since the light intensity would wane with distance. A possible solution is to have relay points, where the light pulse is detected and a new pulse with appropriate calculations is broadcast, with prior understanding between stations about the calculations on how to get their readings to match.

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#24

### Re: The 'Half-Twin' Puzzle

01/15/2010 10:38 PM

Nope, I was wrong about all that. I treated it as though the hyperbola defined by a constant distance from the origin contains points which are null-separated, which is not true! Now my description of the diffeomorphism and all that falls apart. Light obviously travels along null geodesics and not along hyperbolas.

Now it is a matter to consider what happens to the diagram under Lorentz transformations. By your definition, points having the same Minkowski separation from the origin are simultaneous. However, with a translation, the origin would change and so would the sets of points which have the same separation from it. The hyperbola containing the new origin wouldn't "turn into" the light cone, alas!

It seems to me that the only definition of simultaneity which would result in the same sets of simultaneous points in all inertial frames is to say that null-separated points are simultaneous. What do you think?

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#25

### Re: The 'Half-Twin' Puzzle

01/16/2010 4:48 AM

Because of the 'standard configuration' Minkowski, with the various frame clocks synchronized at the origin, I think we need not consider translations, but only spacetime rotations (Lorentz transformation). I think my definition does result in one set of simultaneous points for all inertial frames in the standard configuration (figure on the right).

In any case, my interest in Lorentz invariant simultaneity stems from the quest to determine if there are any technical grounds for a view that Green and Red have the same proper age at the R/B flyby (figure below). It seems to be so from this discussion, but I'm sure there will be a lot of resistance as well. Anyway, there may still be a fatal flaw in the whole argument.

-J

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#26

### Re: The 'Half-Twin' Puzzle

01/17/2010 4:03 PM

A further thought on the problem: you wrote:

"It seems to me that the only definition of simultaneity which would result in the same sets of simultaneous points in all inertial frames is to say that null-separated points are simultaneous. What do you think?"

The null-separated events do indeed form the 'x-axis' for all frames in this scheme, but then there are hyperbolas above it that form the lines of simultaneity for time-like separated events.

One problem is that with a null x-axis, distance coordinates are undetermined and so are all one-way speeds, including the one-way speed of light. However, two-way speed is determined and hence we can have a conventional one-way distance, I think. Does that mean we can still have a conventional one-way speed determination?

Any ideas?

-J

PS: I've found this interesting article on The Conventionality of Simultaneity by Vesselin Petkov.

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#27

### Re: The 'Half-Twin' Puzzle

01/18/2010 4:48 AM

Further to my prior post, where I wrote: "However, two-way speed is determined and hence we can have a conventional one-way distance, I think. Does that mean we can still have a conventional one-way speed determination?".

A light signal (dotted red) from the origin will propagate up the light cone and if reflected back by a mirror colocated with the ct' = 2 event, the return signal will arrive at ct=1.74. This can be used to fix a coordinate distance of x=0.87 for the event, irrespective of which synchrony convention is followed (only one clock involved). This gives the two-way speed of light at c.

One can also postulate that the one-way speed of light must then also be c, but it is not directly measurable. The fact that the light-cone is defined as the 'real x-axis' means that the one-way light travel time was zero - hence infinite one-way speed for light. This is a consequence of the 'new' clock synchrony model - a clock present at the reflection event must read zero.

Interestingly, while the relative speed of the two frames (ct and ct') was set as v = 0.4c in standard Einstein synchrony, their relative speed in the new synchrony is v' = 0.87/2=0.435c. This does mean that we have a 'mixed' method of distance measurement and clock synchrony. For distance we still use half the two-way light travel time, but for clock synchrony we don't.

Ouch... This is becoming very complex now! The question is: since clock synchrony is a convention, is this type of clock synchrony valid?

-J

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#28

### Re: The 'Half-Twin' Puzzle

01/18/2010 9:37 AM

Interesting questions raised. I will have to look through in detail.....

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#29

### Re: The 'Half-Twin' Puzzle

01/19/2010 2:42 PM

I wrote: "... their relative speed in the new synchrony is v' = 0.87/2=0.435c. This does mean that we have a 'mixed' method of distance measurement and clock synchrony. "

This bears some resemblance to the "Epstein space-propertime" diagram[1], where we plot proper time against coordinate space (or 'your time' against 'my space'), as pictured on the right. The diagram is drawn for v=0.8c, giving the relative time dilation factor as 1/γ = √(1-0.82) = 0.6.

The Epstein diagram shows the light-cone (the "real" x-axis) at an angle perpendicular to the time axis. In essence, the hyperbolas are just replaced by circles. The angle of the ct' axis is now v/c = sin(-Φ), where in the Minkowski diagram it is v/c = tan(-Φ). Here the scale of the ct' axis is the same as that of the ct axis, directly indicating that there is no desynchronization of clocks (the lines of simultaneity are circles).

An interesting connection, maybe, but not quite the same thing, I think...

-J

[1] Relativity Visualized, Lewis Carrol Epstein, 1997

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#30

### Re: The 'Half-Twin' Puzzle

01/19/2010 3:25 PM

This is another interesting construction. I read the paper on the conventionality of simultaneity. It seems to me that a definition which would conform to our intuition would suffice for everyday (nonrelativistic) life, while the proper time definition works very well in the context of, say, cosmology, where the universe we see "now" is more or less a light cone (except that the spacetime isn't globally Minkowski).

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#31

### Re: The 'Half-Twin' Puzzle - Answer

01/20/2010 3:07 PM

After all these 'diversions', it may be time to try and answer the final question of the OP: "How do we reconcile these apparently paradoxical conclusions, i.e., did Green age 1, 2, 4 or 7 units during this 'Half-Twin test'?"

The modern answer is that because the definition of simultaneity is a convention, devoid of any "real physical meaning", we cannot really tell. Einstein adopted a convention for his Special Relativity that makes the speed of light isotropic, i.e., the one-way speed of light will be observed as the same in all directions. This gives every inertial observer a different view of what is simultaneous and what not.

However, there is nothing that prevent us from adopting a different method (convention) for synchronizing clocks, e.g., like the "proper simultaneity" that I suggested in #18 above. The price that we will pay for such a 'non-standard' synchronization procedure is that the laws of physics will be much more difficult to express mathematically. Sonego & Pin wrote in Foundations of anisotropic relativistic mechanics, 2009: Misner, Thorne and Wheeler concisely and effectively wrote: "Time is defined so that motion looks simple". We could paraphrase them saying: "Clocks are synchronized so that physics looks simple."

-J

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#32

### Re: The 'Half-Twin' Puzzle - Answer

01/21/2010 2:10 PM

I wrote: "Sonego & Pin wrote in Foundations of anisotropic relativistic mechanics, 2009: Misner, Thorne and Wheeler concisely and effectively wrote: "Time is defined so that motion looks simple". We could paraphrase them saying: "Clocks are synchronized so that physics looks simple.""

Just to illustrate how things can become more complex than necessary when we deviate from Einstein's convention for clock synchronization, here is a (more or less) complete graphical representation of "proper simultaneity" (as roughly suggested in #18 above).

According to the literature that I've referenced so far, this is a perfectly valid convention for defining simultaneity - in fact it even has some technical appeal to it as well.[1] However, having a non-linear coordinate system will never 'look simple' and it may be difficult to calculate (or even plot) certain characteristics.

In a nutshell, the bottom part (below the blue/black/red ±X,X,X axes, the light-cone), may be ignored completely, because it is the non-causal area of Minkowski spacetime, where neither material objects, nor electromagnetic effects can go. The light cone is the new x-axis for all inertial frames. The time- and spatial axes are still linear, but the spacetime 'slices of constant time' are hyperbolic and all inertial observers share the same definition of simultaneity. Hence, all inertial frames agree on the x, cT coordinates of any event that happens inside the light-cone.

Distances are still measured by the Einstein (or radar) method - time the two-way propagation of light, using only one clock. Take half the time and divide it by the speed of light and you have the spatial separation. We set up a grid of distances in this way, with non-synchronized clocks, static in the inertial frame, at every grid position of interest. To achieve 'proper simultaneity' (in principle), we 'shoot' a set of identical clocks, synchronized at the origin, in every direction in such a way that speeds (or magnitude of momenta) in the original coordinates are all identical. A perfectly symmetrical 'explosion' should be able to achieve this.

As such clocks pass the static clocks of the grid, each grid clock is set in accordance with the passing clock's reading, with no correction for travel time or other offsets. The moving clocks all register proper time and hence the synchronization is according to proper time. While such clocks are traveling, they are assumed to be perfectly inertial, with the initial acceleration either happening before they pass the origin, or alternatively, is of negligible duration.

If this does not make any sense, don't worry - we will not use such a scheme - it is just to illustrate an alternative definition of simultaneity, however impractical. If it makes sense and you spot a flaw in the argument (in principle, not in practice), please let me know.

-J

[1] One of the problems with most alternative, non-Einstein conventions of simultaneity is that they tend to make the speed of light (and hence the Newtonian laws of motion) anisotropic for moving frames. The described convention seems to retain the isotropy of light propagation and hence the same laws of motions in different directions for all inertial frames.

I could not find a mention of this specific convention in the literature, but I will keep looking...

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#33

### Re: The 'Half-Twin' Puzzle - Answer (Correction)

01/22/2010 10:31 AM

Oops! In the footnote of my prior post, there is a subtle problem. I wrote: "The described convention seems to retain the isotropy of light propagation and hence the same laws of motions in different directions for all inertial frames."

This "proper simultaneity" convention has the rather nasty characteristic that the measured one-way speed of light, although isotropic, will be infinite! The reason is that all clocks 'crossing' the x-axis read zero at that time. If the light-sphere is emitted when the clock at the origin reads zero, it will arrive at the reflection event just as the clock there also reads zero. The light travel time will hence appear to be zero...

-J

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#34

### Re: The 'Half-Twin' Puzzle - Answer

02/01/2010 2:47 AM

I kept looking and I found this type of synchrony inside the paper that I referenced earlier: "Sonego & Pin wrote in Foundations of anisotropic relativistic mechanics, 2009"

It is in fact a peer reviewed paper, with full citation and (partial) synopsis:

"Foundations of anisotropic relativistic mechanics, Sonego, Sebastiano; Pin, Massimo (Universit`a di Udine, Via delle Scienze 208, 33100 Udine, Italy); Journal of Mathematical Physics, Volume 50, Issue 4, pp. 042902-042902-28 (2009).

"We lay down the foundations of particle dynamics in mechanical theories that satisfy the relativity principle and whose kinematics can be formulated employing reference frames of the type usually adopted in special relativity. Such mechanics allow for the presence of anisotropy, both conventional (due to non-standard synchronisation proto-cols) and real (leading to detectable chronogeometrical effects, independent of the choice of synchronisation)."

The relevant equations are in section 2.1.1, page 7 of the pdf. If I understand the paper correctly, the non-standard synchrony described in this Blog is valid as one of the limit-cases.

-J

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#35

### Re: The 'Half-Twin' Puzzle

03/02/2010 4:59 PM

G aged 4 units and R aged 2 units. In R's frame (and B's), the distance between G and the point where R and B cross is half the distance as measured by G and his clock ticks half as many times as G until they cross. This remains true, and becomes more apparent, in the diagram where R considers his own frame as stationary and G as receding, because the relative velocity of R and B to each other is less than twice their velocities with respect to G.

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#36

### Re: The 'Half-Twin' Puzzle

03/02/2010 11:09 PM

Hi Neil,

You are right, but only by the convention you adopt for simultaneity, as I explained in my Answer above. One has to complete the other half of the 'half-twin paradox' before you can really tell. One-way aging remains a thing defined by convention, not physics...

-J

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#37

### Re: The 'Half-Twin' Puzzle

03/03/2010 12:01 PM

Ah, this was a puzzle on synchronizing distant clocks. I didn't pick up on that.

I've ordered your book. I'm no engineer but my math isn't too bad, so hopefully I can follow some of it. OT, but I am rereading Epstein's Relativity Visualized. Whatever happened to him? His geometric explanations in this book are truly ingenious.

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#38

### Re: The 'Half-Twin' Puzzle

04/30/2010 7:54 AM

Hi Neil,

Sorry, I missed this reply of yours.

You can ask questions on Relativity 4 Engineers here, if you like (or perhaps in the FAQ section that I've opened in this Blog - see header above).

Yes, I also enjoyed Epstein's visualizations of relativity very much. His work did not get much credit in the scientific circles, because it is a little narrow-minded, although very useful for the novice. One cannot do much more than that with most of it - e.g., his 'Epstein space-propertime diagram' has very limited utility, but it is a very nice visualization...

-J

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#39

### Re: The 'Half-Twin' Puzzle

08/05/2015 11:38 AM

Hi Jorrie,

It's been about a month since you re-directed me here to continue our discussion from the CR4 general discussion forum on relativity and specifically the twin paradox. I've been re-reading your book and going over Brian Greene's worldscienceu relativity course intently in that time.

To recap, we differed on 2 major points of relativity concerning the twin paradox and although I still differ, my extra studying has hopefully improved my arguments.

1. I know the most basic tenant of relativity is that there is no preferred frame of reference and that I'm crazy to even propose a counter-argument but here it is anyway:

If you get 2 transparencies and draw 2 planets, A and B and a guy on planet A on one transparency. On the 2nd draw a spaceship. I get it that there are infinite combinations of moving these transparencies relative to each other at different speeds that will yield the same relative velocity between the guy fixed on planet A and the guy in the spaceship. But there's one inescapable difference, only the guy in the spaceship can move from planet A to planet B. It doesn't matter that you can fix him on your slide and move the planets behind him and say it's equivalent to fixing the planets and moving him between them because in either case he is the only one who can move from point A to point B so he is the only one who is actually moving.

Even without planets or points in space, both twins can leave breadcrumbs where one will be left sitting in a pile of breadcrumbs while the moving one does not. Hence, there is a preferred frame of reference and only 1 of the twins is actually moving.

Of course they can both be moving either parallel or toward or away from each other at the same speed (which yields 3 different relative speeds) but this movement yields the same result as if both were stationary relative to each other. Neither ages slower than the other unless one is moving faster than the other. I'm not sure what to call this relativistic relative velocity that yields absolute differences is aging even though both are moving through points in space.

It's not just his positions in space that confirm this, it's his relative aging. At some point in the past, one of the twins will have accelerated to a speed greater than the other twin. The relative speed between them will not reveal to either of them which twin is actually going faster in an absolute sense unless they share a log between them of their travels and accelerations. The other way to tell is that they can communicate their ages between them as in your example which leads to the 2nd point on which we differ:

2. Relativity says there is no way to tell who is actually aging slower unless one of the twins switches inertial frames AND (a very important AND) returns to meet back with his twin. If these two things don't happen there is supposedly no way to tell who has aged slower. Here's my counter-argument:

Let's use your example of the half twin paradox but instead of going back, the twin in the spaceship just stops and both twins are moving through time separated by space and still sending out beacons to each other on a yearly basis. They will be able to ascertain their relative ages and that the twin in the spaceship remains two years younger. This is from the spacetime diagram and the final value is not up in the air until the stopped spaceship returns to earth. The analysis holds true even if the twin never stops, it's possible from the beacons to determine how much slower the moving twin ages. The age difference will accumulate over time. Again, he is the only one who's moving because he's the only one moving from point to point in space.

I just don't believe in the "magic" of relativity where both twins remain in a state of superposition where the wave function collapses to a final value of their age difference only if one accelerates AND meets back up to the point of origin (which is somehow the only point where their data is valid and can be compared on a magical level playing field). I believe more in the mathematical reality of the spacetime diagram which does not need special rules to determine who ages slower; it's an ongoing process of the twin who's actually moving according to my definition of movement above.

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#40

### Re: The 'Half-Twin' Puzzle

08/05/2015 2:55 PM

"Even without planets or points in space, both twins can leave breadcrumbs where one will be left sitting in a pile of breadcrumbs while the moving one does not. Hence, there is a preferred frame of reference and only 1 of the twins is actually moving."

Well, then the one has accelerated (changed inertial frames) and then the reciprocal nature of SR has been violated.

"Of course they can both be moving either parallel or toward or away from each other at the same speed (which yields 3 different relative speeds) but this movement yields the same result as if both were stationary relative to each other. Neither ages slower than the other unless one is moving faster than the other."

Not true, unless they have made identical inertial frame changes.

"Relativity says there is no way to tell who is actually aging slower unless one of the twins switches inertial frames AND (a very important AND) returns to meet back with his twin."

True, because when you compare two clocks (or aging) at a distance, you are depending on a synchronization (or simultaneity) convention. Hence, the comparison is theory dependent.

"The analysis holds true even if the twin never stops, it's possible from the beacons to determine how much slower the moving twin ages. The age difference will accumulate over time. Again, he is the only one who's moving because he's the only one moving from point to point in space."

Only if the twin is the one that changed inertial frame. "Point to point in space" is meaningless, because it is observer dependent. What has got meaning is "point to point in spacetime", because it is the invariant spacetime interval.

It seems that you still have a lot of relativity to learn. ;)

Finally, 'superposition of states' is not relativity theory, so it should not enter the arguments. Quantum physics does not enter the twin 'paradox' in any way.

--

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Jorrie

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#41

### Re: The 'Half-Twin' Puzzle

08/06/2015 11:28 AM

Ok I tend to put too many issues in one post. So in this example, two ships start out at the same point and separate using identical flight paths. When they're far enough apart, one at planet A and the other at planet B, they zoom back towards each other straight on through and past each other at the point where they started.

Now, the relative velocity of them coming together follows the relativistic addition rule so if they're each going .6c, their relative velocity is .84c. Because they know their past flight path, they know they're both aging at the same rate, there is no time dilation between them that will cause one to age slower and this can be verified continuously as they send out time beacons between them. My analogy of using two transparencies will show that there's no way one can be deemed stationary and the other moving because in no way can one be stuck at his planet while the other passes overhead. There is a true symmetry between them but only the point between them can be deemed stationary and they're moving relative to it and not really relative to each other in a relativistic sense. What I mean by relativistic sense is that without a time dilation that causes one or the other to age slower, their apparent relative velocity is the same are if they were moving in the same inertial frame with no relative velocity between them.

Now let's say the two pilots didn't know their past flight paths. The one on planet B flies to planet A at .84c where the other pilot is parked. Relativity says neither knows who is actually moving and to each of the pilots, ignoring the planets and the background space, their approach looks identical to their former approach at .84c relative velocity. But it's not identical, there is an asymmetry. You can consider each of the pilots as stationary but whichever one you choose, only pilot B will visit both planets. As such, he is the only one who goes from point B to point A and hence he is the only one who is actually moving. This is further proven by the fact that if you go through all the relativistic calculations for the twin paradox, considering each of the pilots as stationary, the one that ages slower is indeed pilot B.

Now you'll say there's no way to determine this for sure unless both start and end at the same point in space. This conclusion must be backed up mathematically somehow and not just because Einstein said so.

P.S. A consequence of what I'm saying is that muons coming from space and the GPS satellites in space do not see our time as dilated from their perspective even though relativity allows us to say they're stationary and we're moving. I'm saying the basic asymmetry of only 1 participant visiting both events means that only that participant will have a slower clock running at proper time (which runs at the same universally slowest rate for every inertial frame). But ignore this for now, I'm interested if you can poke any holes in what I've said above apart from the fact that I'm not presenting my arguments in a relativistic light.

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#42

### Re: The 'Half-Twin' Puzzle

08/06/2015 12:15 PM

"Now let's say the two pilots didn't know their past flight paths. The one on planet B flies to planet A at .84c where the other pilot is parked. Relativity says neither knows who is actually moving and to each of the pilots, ignoring the planets and the background space, their approach looks identical to their former approach at .84c relative velocity. But it's not identical, there is an asymmetry. You can consider each of the pilots as stationary but whichever one you choose, only pilot B will visit both planets. As such, he is the only one who goes from point B to point A and hence he is the only one who is actually moving."

Who is to say that the two planets and pilot A were not the ones moving and just happen to each pass pilot B?

That is unless pilot B accelerates/decelerates heavily, changing his inertial frame and lands on the planet of B. Then the situation is asymmetrical in an absolute sense and we all know the result.

"Now you'll say there's no way to determine this for sure unless both start and end at the same point in space. This conclusion must be backed up mathematically somehow and not just because Einstein said so."

The math backing it up is actually quite trivial - the calculations of simultaneity. I've shown it in the eBook and many times on this forum; do you really want me to repeat it again?

--

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Jorrie

PS: I must give it to you, you are quite persistent and I have high hopes that you will eventually find SR to be a valid theory. It is not the only one, Lorentz ether theory (LET) in its modern form is also valid - just very, very cumbersome and lacking the generality of SR.

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#44

### Re: The 'Half-Twin' Puzzle

08/10/2015 9:17 AM

I finally understand the source of our disagreement. Einstein was wrong, beating a dead horse over and over and getting different results leads to discovery, not insanity.

My example of the twins racing head-on at at .6c to yield a relative velocity of .84c and showing that neither one ages slower than the other while different spacetime diagrams that involve an imbalance of non-inertial motion either future or past (one of the twins accelerates more than the other) could still have the same relative velocity of .84c but have a whole gamut of relative aging. This is what you've been saying.

In relativity, relative velocity alone is insufficient to tell who is aging slower, one must not only have a non-inertial asymmetry, but also one must not carry around stored time from the relativity of simultaneity bank. Here's what I mean by this:

The formula for relativity of simultaneity is Vx/c2. That time gets stored the farther things separate from each other by a factor of the velocity of that separation. Using v=.6c, Y=1.25. From the twin on planet earth's inertial frame, for every year the away twin ages, the earth twin potentially ages 1.25 years due to time dilation. From algebra one can see that the now slice for each year of travel increases by .8 yrs so the time banked into relativity of simultaneity per year is 1.25-.8 = .45 years.

So let's say the twin decides to stop at 4 yrs. His now slice would suddenly swing from 3.2 years, burning up the simultaneity of 1.8 years ( time burnt by deceleration?) for a total of 5 yrs for the earth twin. Even though the twins share the same now slice in the same inertial frame and can send signals to each other confirming the earth twin has aged 1 more year than the away twin, relativity prevents us from concluding that this is true because the twins are separated and hence still have a hidden time bank of relativity of simultaneity to burn off.

One way is to have the away twin turn around. His bank is suddenly at +1.8 yrs which he will burn off at .45 yrs per year of his travel back. By then his bank will be zero and relativity allows us to now conclude that the earth twin has aged 2 extra years.

I don't believe relativity needs to constrain itself to a universe with nothing in it except the two pilots because that universe doesn't exist. The relative position of the two pilots is not the two events. As you said, if there are only 2 pilots, then the two events are symmetrical from either perspective, they meet and separate. There's no way under the constraints of relativity to determine who's actually stationary and who's actually moving in order to definitively say who's aging slower.

But, in my example, I've added the two planets and now the perspectives are not symmetrical. The two events are both pilots on planet earth and only 1 is ever on planet B and he is the only only one from either perspective that has visited both planets. I'm not constrained by the no absolute motion assumption and can definitively say that the only twin moving through 2 points in space is the one that ages slower. Similarly, in a universe with points in space, I can say that a twin leaving the earth twin at .6c will age .8 times slower than the earth twin regardless of any inertial frame asymmetry, past or present, because the travelling twin is the only one travelling through points in space that the other does not. If the earth twin also takes off, his motion relative to those points in space must also be taken into consideration. For example, both twins heading towards each other at equal speeds to a midway point between planets, may have a higher relative velocity but it will be equivalent to zero relative velocity w.r.t. relative aging.

The above method will agree with relativity under its constrained assumption but will also allow definitive results where relativity can't venture. The aging between the twins is continuous and is not dependent for its determination on some arbitrary past or future conditions.

I hope I have convinced you that I understand what you're telling me, I just disagree with it.

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#45

### Re: The 'Half-Twin' Puzzle

08/10/2015 11:07 AM

"The above method will agree with relativity under its constrained assumption but will also allow definitive results where relativity can't venture. The aging between the twins is continuous and is not dependent for its determination on some arbitrary past or future conditions.

I hope I have convinced you that I understand what you're telling me, I just disagree with it."

Not quite, but we all know that special relativity is constrained as a special case of general relativity, i.e. it is only applicable when one can ignore spacetime curvature. The mere presence of 'points in space' does not mean that there is spacetime curvature and hence SR is still applicable. Even in GR there is no notion of 'absolute space' or 'absolute movement' and all movement is still relative.

But then, you do not agree with SR or GR in this respect, so there is little that I can do to help you. What you support seems to be akin to Lorentz ether theory, which has been abandoned long ago as unnecessarily complicated for no benefit at all.

As a final piece of advice: remember that the one-way speed of light being 'c' in all inertial frames is based on convention, not observable fact. The one-way speed has never been measured without there being a circularity in the measurement, more specifically a clock sync convention. In other words, 'using the speed of light to measure the speed of light'.

The only value that is indisputably 'c', irrespective of how the apparatus moves, is the two-way speed of light. In a sense, this is why the 'inertially moving' twin must make a two-way journey before there is an indisputable age difference. Communicating afterwards (over a distance) with one another about their respective ages does not give that 'absolute answer' that you are after, because it depends on a clock sync convention.

That said, Einstein's choice of convention was still a brilliant one...

--

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Jorrie

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#46

### Re: The 'Half-Twin' Puzzle

08/10/2015 12:01 PM

Thanks for all your help, I guess this is the end of the road for me.

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#47

### Re: The 'Half-Twin' Puzzle

08/10/2015 5:08 PM

Sorry I just had an idea for a proof. In my example of the away twin going away at .6c and after he travels 4 years the home twin has aged 5. Now you're saying there's definitely no way to set this in stone until they meet up again. Hence there must be some sort of change in either's motion after the away twin hits the 4 year mark that the home twin sees less than 5 years aging. I doubt you can produce a spacetime diagram that could show this because I think it would involve backwards time travel. So if nothing can change this milestone, it is set in stone and relativity is wrong in stating it isn't.

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#48

### Re: The 'Half-Twin' Puzzle

08/11/2015 12:32 AM

Remember, we are talking about 2 clocks in purely inertial motion, no acceleration whatsoever, especially not at the 4 year mark.

The 'home twin' can be any one of the two and neither can "see" the other one aging less (or more), unless one of then makes a u-turn and they meet again.

Yes, they can each have a setup of clocks synchronized in their own respective frames and that will show a perfectly reciprocal "I'm aging faster than my twin" from either side, due to different clock syncs. That's all there is to it.

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#49

### Re: The 'Half-Twin' Puzzle

08/11/2015 4:50 AM

Yes that's the relativistic interpretation but at the 4 year mark, both twins look out their window and 1 sees earth and the other sees planet B. Now in order for both to compare their ages, one of the twins must exit his inertial frame through acceleration in order that both meet at their point of origin. Only the twin at planet B can do that. There's nothing the earthbound twin can do. Even if the traveling twin stops at planet B and the earth twin fires up his engines to meet him there (which relativity says is out of bounds?), the outcome will not change: the twin at planet B will have been the one who aged 4 years while the earth twin will have aged 5 by the time he fires up his engines.

Even if relativity allows them to compare ages at a point other than the origin, it uses the trick of unleashing time from the relativity of simultaneity time bank after one or the other twin falls out of his inertial frame. So I guess relativity CAN say both twins had aged the same amount and it was the change in inertial frame that added a sudden block of time from the relativity of simultaneity time bank. The only problem is, that if either twin falls out of his inertial frame, it's only the one on planet B which will have aged 4 years and the one on earth 5. So I guess the 2nd trick relativity uses to resolve this problem is the caveat that both must compare ages at the origin. Hence the only result relativity will allow to be true is if the planet B twin turns around.

Sorry these caveats of relativity look like lame, fudge factor, band-aids to me to support relativity's main assumption that only relative motion can be used as a reference when relative position arrives at the same answer much more straightforwardly. I know, I know, go read about Lorentz Ether theory.

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#50

### Re: The 'Half-Twin' Puzzle

08/11/2015 7:02 AM

"Yes that's the relativistic interpretation but at the 4 year mark, both twins look out their window and 1 sees earth and the other sees planet B. "

Remembering that the "half-twin-paradox" is an all-inertial scenario, I can just as well say: twin 2 is the reference, minding his own business, static as far as he is concerned. Suddenly Earth (with the earth-twin) flies past twin 2 at high speed. Four years later, planet B flies past him at the same speed. Both twins look out of their windows and '1 sees earth and the other sees planet B', just like you have said. Who has done the 'moving'? Who must have aged more in the 4 years? You cannot tell.

Contrary to what you have said, if any one of the twins would start his engine and join the other one, he would demonstrably have aged less than the other one during the duration of the total test. That's all there is to it. No experiment in a century of trying has ever proven otherwise.

The sooner you drop the quest for 'absolute motion', the sooner you will have some relative peace...

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#51

### Re: The 'Half-Twin' Puzzle

08/11/2015 11:50 AM

Well, you blew my mind, I must now accept Einstein was right. If the home twin turns on his burners and catches up with the away twin in less than a year, he can end up younger than the away twin. What's also surprising is that you're allowing them to meet up at a place other than the origin where they separated. I'm going to haveto look at this closer.

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#52

### Re: The 'Half-Twin' Puzzle

08/11/2015 12:39 PM

"If the home twin turns on his burners and catches up with the away twin in less than a year, he can end up younger than the away twin. What's also surprising is that you're allowing them to meet up at a place other than the origin where they separated."

The 'home twin' can take as long as he wishes, as long as he catches up. He will be the younger one, since he was the only one to change inertial frames (or more correctly, he has changed inertial frames more severely, because the other twin did nothing).

On the second point, yes it does not matter where they meet up again. As long as they meet more than once, they can compare clocks directly. Who ages more will depend on the spacetime geometry of their respective journeys. It is just for simplicity that we normally use the 'no change twin' as reference. Should both change inertial frames, it can become a little more complex.

BTW, in the scenario where "the home twin turns on his burners and catches up with the away twin", the 'away twin' obviously becomes the 'home twin' in the classical 'twins paradox' scenario.

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#53

### Re: The 'Half-Twin' Puzzle

08/13/2015 8:45 AM

I've been very depressed these past 2 days trying to reconcile my intuitive view of reality to relativity. I mean, how can time be so arbitrary and undefined until a particular set of circumstances and comparative measurements occur to bring it into reality (hence my analogy to quantum physics which seems way more straightforward than relativity). I also lost touch with what the X-axis meant on spacetime diagrams if it was no longer attached to absolute space coordinates (it never was, only in my mind). So I began to draw spacetime diagrams and here's the simplest one I came up with:

The away twin leaves earth at .6c and stops at planet B 3 lt yrs away (so he travels 4 years in (his) proper time). (I could have left him keep going after 4 years but that just complicated the math with the relativistic combo law to allow the earth twin to catch up). Now after 4 yrs (the earth twin's) proper time, the earth twin goes off at .6c to meet his twin on planet B. You'd expect that when they meet, they'll both be the same age as the time dilation of the earth twin will cancel out the time dilation of the away twin. The spacetime diagram confirms this. If the earth twin had traveled faster than .6c, he would have arrived younger than the away twin.

The Earth twin's journey allowed him to make back the year he "would have" lost to the away twin. For every year of the 4 years he travels at .6c, he makes back .25 yrs. I say "would have" lost because relativity says all that time is indeterminate until they meet. I made the mistake of saying that if the away twin traveled 4 years it meant the earth twin was automatically a year older but he isn't since he can make that year back. In fact if the earth twin goes very close to the speed of light, he can be up to 2 years younger than the away twin when they meet up on planet B.

The key is to use relativity to do the analysis when they're both 4 years into the space time diagram because that's when each can say the other has dilated equally. The fact that the earth twin can repeat the away twin's journey and can make up the year he lost to end up the same age means that the away twin was indeed 1 yr younger and not indeterminate as relativity states. The end result can be altered by the earth twin's motion but that's only if the intermediate result is equally real. This means they don't have to meet up to validate intermediate steps as both can jockey back and forth between them to settle the final result which is the only valid result according to relativity. I say all intermediate results are valid and if the away twin stops on planet B after 4 years, he remains 1 year younger than his earth twin barring whatever the earth twin does to correct this in the future.

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#54

### Re: The 'Half-Twin' Puzzle

08/13/2015 9:21 AM

P.S. In my spacetime diagram where I allow the away twin to keep going without stopping, if the home twin goes very close to c 4 yrs after the away twin left, he will catch up with the away twin at a distance of 6 lt yrs when the away twin has aged 8 yrs and the earth twin has only aged little more than 4 yrs. This is all in keeping with the previous analysis of the home twin going close to c to meet the away twin having aged 6 years while the earth twin has only aged 4. So the earth twin gets 2 yrs younger for every 4 yrs the away twin travels while the earth twin catches up at close to c. I'm just saying it all works out consistently.

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#55

### Re: The 'Half-Twin' Puzzle

08/13/2015 12:01 PM

Ralf, congratulation, you have come a long way!

As long as there is a 'catch-up', there has been some acceleration (different change of inertial frames) and you are perfectly right. What you must be aware of is that in such a case relativity does not say: "all that time is indeterminate until they meet." Whoever 'changes inertial frame most' would have aged less, even irrespective of them being in the same place twice. It is only in the all-inertial scenario where relativity gives a complementary answer, i.e. where it has nothing to say at the interim.

But, and this a big but, relativity has no way of proving the premise unless they pass each other at least twice. The 'distant result' is relativity theory dependent and one cannot use relativity theory to prove relativity theory. You know that relativity's simultaneity definition is a convention and only assumes an absoluteness when there is no distance between observers.

It is just like when relativity says: 'the one-way speed of light is 'c' in all inertial frames, irrespective of relative movement of the frames'. This depends on the assumption of that statement in the first place. It cannot be proven or disproved by observation.

It is only the two-way speed of light that is the same in all directions, irrespective of relative movement. This can and has been proven to be an absolute fact. And we know that this in not true in Newtonian/Galilean theory - this is special relativity's main claim to fame.

Likewise, we can only prove the twin paradox solution by making it a two-way trip. Trying to attach an "absoluteness" to the half-way result is sneaking in some form of absolute frame of reference - one that is preferred by physics above all others, e.g. where clocks will run faster than in all other inertial frames.

I sincerely hope the stress is subsiding now...

-J

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#56

### Re: The 'Half-Twin' Puzzle

08/17/2015 11:52 AM

"Whoever 'changes inertial frame most' would have aged less, even irrespective of them being in the same place twice. It is only in the all-inertial scenario where relativity gives a complementary answer, i.e. where it has nothing to say at the interim."

So relativity forbids using inertial jumps as a tool to measure what the non-inertial scenarios would have said about relative time dilation. How convenient.

Hypothetically, what if Newton's laws were based on a caveat that the theory does not apply to any shells fired at greater than a 45 degree angle; their trajectory would be indeterminate. Yet it was found that their trajectory was determinate and followed from the theory for shells fired at less than a 45 degree angle. A theory isn't fact, it's a story that ties facts together and if the story is incomplete and doesn't tie all the facts together, then the theory is wrong.

There can be no possible answer in the all inertial scenario other than the earth twin has aged 1 more year than the the away twin at the 4 year mark. There is no spacetime diagram you can draw where the away twin has a vertical line at x=0 and the earth twin has to execute a manouver where he can prove he is indeed younger than the away twin at the 4 year mark. Either twin can execute a manouever to prove the earth twin was actually 1 year older during the 4 year inertial phase.

Relativity has made an assumption that there is no absolute space and only considers scenarios that justify this assumption. Yet assuming there is immutable space with no length contraction, not only confirms all of relativity's results in a much simpler and more intuitive way but also includes scenarios for which relativity is not applicable. I think Einstein violated his caveat about making a theory as simple as possible but no simpler. Maybe there's something I don't know about where assuming an absolute space violates relativistic facts.

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#58

### Re: The 'Half-Twin' Puzzle

08/17/2015 12:23 PM

"Either twin can execute a manoeuver to prove the earth twin was actually 1 year older during the 4 year inertial phase."

Wrong. Whoever executes such a manoeuvre would have aged less at the end. So what right do you have to assume that the earth twin was actually the oldest during the half-way stage?

This thread is getting stale again, going round and round in philosophical circles, so I think it is time for closing remarks and then leaving it to the jury.

-J

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#60

### Re: The 'Half-Twin' Puzzle

08/17/2015 2:58 PM

Yup I'm going round and round like a broken record. I can only stay on the right path for a little while before I want to veer off into the ditch. Ok so if I take the mirror image of my example and have earth going .6c on the negative x quadrant while the away twin stays 4 years on the vertical ct axis, then the away twin would make his way back to earth and rendezvous with earth. In your book, the away twin would have to go faster than .6c to catch up with earth. In order not to the violate the twin paradox, the away twin would have to meet up with the earth twin at a speed that would not only make up the earth twin's year younger but also end up an additional 2 years younger. I don't see how this is symmetrical to saying the earth twin doesn't move for 10 years and the away twin moves at .6c on both legs of his journey but I see this must be the answer to understanding how the 4 year mark is really indeterminate. There's the proof but I'll do anything to wiggle out of it if I can but in the meantime I have to accept that the non-inertial phase is indeterminate.

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#61

### Re: The 'Half-Twin' Puzzle

08/18/2015 7:26 PM

I just thought of an experiment that would settle this. Send a spacecraft off into the solar system with an atomic clock on board. It will receive our atomic clock time and send back it's atomic clock time and the time it's receiving from us. According to relativity, it will receive the time signals from us going at a slower rate than the rate on its clock. We will see its clock rate going at a slower rate than ours and we'll also see our clock rate but I'm confused as to whether it'll match our clock rate or be even slower than his clock rate. I think it's the latter but that answer doesn't matter right now. What does matter is that if relativity's wrong, the spaceship won't see our clock slowing relative to his, but we'll see his slowing relative to ours. Again my brain is frozen as to what our clock transmitted back to us will look like so I'll have to think hard on this.

The fact that the ship will accelerate away from us won't matter because as you say in your book, acceleration does not cause time dilation itself, it's the cumulative speeds the acceleration causes. So long as there's no change in direction, the cumulative speeds can be averaged out to be inertial velocity. So the example above is indeed the purely inertial scenario and may indirectly settle the one-way speed of light controversy you mentioned.

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#62

### Re: The 'Half-Twin' Puzzle

08/19/2015 3:21 AM

As I have indicated above, this will be a closing argument on this thread and there will be no further specific replies from my side.

Einstein's special relativity (SR) has been around for 110 years an his general relativity (GR) will be a century old next year. All possible practical tests that the world's most brilliant minds could come up with, have been thrown at it. No violation of its predictions has ever been found, within it's domain of applicability, of course.

So why is it that people like the above respondent are still trying to come up with amateurish efforts to "prove it right or wrong". One problem may be with the many popularisations that are often not quite correct, or at least inadequate in its efforts to demystify it. A second one may be that it is a quite complex theory and without math it is virtually incomprehensible in its total scope.

Like in any trial, if the physical evidence is overwhelming in pointing to the innocence of a suspect, the judge or jury will set him free. We have long since reached that point and the "suspect" is fortunately free to live a productive life, helping to make things like, satellite TV, GPS and CERN possible.

Yes, science is still trying to test GR at the limits of its applicability, because we suspect that it will fail at extreme density and energy levels (and at the quantum particle level). My advice to everybody is to stop wasting energy and learn as much as they can of the present GR and hopefully be ready when the next big breakthrough comes - quantum gravity.

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#63

### Re: The 'Half-Twin' Puzzle

08/19/2015 4:47 AM

I've found strong opinions elicit strong responses and the weakest opinion of all is a question. Yes it's an annoying method to get answers but I've gotten them. Every response was an eye opener. Unfortunately I didn't know enough to assign the correct significance to some of the words in the answers in order to understand the answers.

Science isn't about having faith in the scientists or in the track record of the theory. In fact, the longer the track record, the more likely scientific consensus will be overturned like an over ripe apple falling from a tree and smashing to the ground. How dare I question the theory? because I don't understand it. Have there been adequate explanations of it out there? I'd think they would have surfaced to the top by now. As Einstein said, and I'm paraphrasing here, if you can explain it to your grandmother then you understand it yourself. You've understood it more than anyone else I've come across.

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#57

### Re: The 'Half-Twin' Puzzle

08/17/2015 12:02 PM

"You know that relativity's simultaneity definition is a convention and only assumes an absoluteness when there is no distance between observers"

In your book you stated time dilation and length contraction were also conventions. I'm confused, what is real then? I know length contraction isn't real but the other two are also not real?

"one that is preferred by physics above all others, e.g. where clocks will run faster than in all other inertial frames."

No all clocks run at the same absolute rate in all inertial frames. No clock rate can be found that runs slower but when comparing time between inertial frames one will run relatively faster than the other. There is no frame where clocks run faster than in all other inertial frames just because I'm assuming absolute space.

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#59

### Re: The 'Half-Twin' Puzzle

08/17/2015 2:35 PM

"In your book you stated time dilation and length contraction were also conventions. I'm confused, what is real then? I know length contraction isn't real but the other two
are also not real?"

It is only purely inertial movement where the time dilation is 'conventional', because it depends on the convention of simultaneity. As soon as there is any non-inertial motion (or gravity) involved, time dilation is a real phenomenon, which can be absolutely measured. Just like the two-way speed of light that can be absolutely measured.

This will be my last reply to specific statements/questions. Then I'll deliver a 'closing argument'.

-J

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#43

### Re: The 'Half-Twin' Puzzle

08/07/2015 1:27 AM

"P.S. A consequence of what I'm saying is that muons coming from space and the GPS satellites in space do not see our time as dilated from their perspective even though relativity allows us to say they're stationary and we're moving."

Just one thing about GPS clocks. They are rate-adjusted to run at the same rate as our master clocks. Since they are so high, the main effect is gravitational, which makes them naturally faster than earth clocks by 45 ms/day. Since they are in closed orbits (a-la twins paradox), their speed relative to the ground makes their clocks lose 7 ms/day relative to ours, for a net 38 ms/day time gain per day. This is rate-adjusted out before they are launched and then they run at the same rate as ours, +-10 or so nanosecond per day.

To ensure that this tiny offset does not build up over time and cause inaccuracies, they are generally resynchronized once per day.

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#64

### Re: The 'Half-Twin' Puzzle

06/16/2017 1:30 AM

I have found a solution. In R's frame, when R is 1 B is 1+0.5. And in B's frame, when B is 1+0.5 R is 1+0.5+0.25. And in R's frame, when R is 1+0.5+0.25 B is 1+0.5+0.25+0.125. So both times of R and B converge to 2!

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