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38 comments

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
In reply to #1

Re: The 'Half-Twin' Puzzle

01/13/2010 9:12 AM

Hi passingtongreen,

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
In reply to #2

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
In reply to #4

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
In reply to #5

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
In reply to #5

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
In reply to #7

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
In reply to #8

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
In reply to #9

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
In reply to #9

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
In reply to #11

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
In reply to #12

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
In reply to #13

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
In reply to #14

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
In reply to #15

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
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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
In reply to #17

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
In reply to #18

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
In reply to #19

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
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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
In reply to #17

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
In reply to #21

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
In reply to #23

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
In reply to #24

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
In reply to #24

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
In reply to #24

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
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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
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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
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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
In reply to #30

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
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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
In reply to #32

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
In reply to #32

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
In reply to #35

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
In reply to #37

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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