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Way back in 2010 GrayGhost and I had a long "standup" debate on differential aging in an all-inertial variant of the classical "twins paradox". The all-inertial scenario (then dubbed the ABC scenario) goes like this (as pictured in 2010 by Gray):

Alice, Bob and Charlie (A, B and C) are all purely inertial "movers" in free space, each carrying a perfect clock. Since they are moving relative to each other, their clocks are not synchronized, but as Bob flies past Alice, they both set their clocks to zero. Let their relative speed by 0.8660c, so that relative time dilation is exactly 1:2, i.e. each perceives the others clock to run at half the rate of his/her own clock.

When Bob's clock reads exactly 2.0 years, Charlie happens to fly past Bob at high speed, but he succeeds to set his clock at 2.0 years at the moment they were next to each other. This is the "time handoff event" in the diagram (tB=2, tC = 2). Say Charlie is now approaching Alice at 0.8660c, i.e. the reverse of Bob's speed as seen by Alice.* When Charlie eventually passes Alice, they read each others clocks and find: tA = 8, tC = 4 years.

This is the same as the classical twins paradox solution and there is no disagreement between Gray and myself on the result. What we disagreed on previously is the answer to this question: what was the time on Alice's clock when the "time handoff" between Bob and Charlie took place?** I maintained that it is not known in any absolute sense, but if I had to put a time on it, I would choose 4 years. Gray argued for either 1 year or 7 years, depending on who looked at it from where: 1 year per Bob and 7 years per Charlie.

Now this I could also agree with, kind of, because it is a coordinate dependent observation. The problem came in when I asked Gray by how many years Alice aged between the time that Bob flew past her and the time of the hand-over event. Grey used all sorts of Lorentz transformation arguments to claim that she have aged only 1 year from Bob's perspective. The problem is that from the handover event until Charlie's eventual fly-by, the same argument is that Alice could have aged only one year from Charlie's perspective.

You can see all the elements for a paradox here. Any bets on what will happen this time around?

--
Regards
Jorrie

* Bob and Charlie pass each other at a relative speed of 0.9897c (by the parallel relativistic addition formula).

** This where the title "half-twins paradox" comes from, because the full scenario is essentially stopped at the half-way mark and the situation analyzed.

BurtJordaan
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BurtJordaan wrote:You can see all the elements for a paradox here. Any bets on what will happen this time around?

Nice puzzle, Jorrie, and one I've not seen before. I'm going to give it a think but I'm not sure if I'll be able to contribute much of value. I'll wait and see first what others make of it.

However I'll be the bookmaker and set the odds for your question. A million to one on this paradox will not be resolved in this thread.

Regards Leo
Obvious Leo
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Jorrie,

Very well stated, thank you. I noted a type-O in my old figure there, as is should be t_B = 0 at the origin.

I have an updated JPG graphic, but when I upload my JPG files here, they look rather crappy on the preview screen. But my old one you posted, seems to look rather good. Not sure how you managed that feat? Do they maybe look better after you post them, than when they're previewed before post? I could email you the updated figure if you like, if the PM or private emailing allows figures or attachments. I'll attach it here, just to see if it looks better after I upload it (or not).

so here's a remake of the OP figure with correction ...

I figure, odds are 2:1 in my favor going in !

Thank you,
GrayGhost
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OK then,

My position is that the extent of a clock's aging depends upon how much the clock's reading advanced over the defined interval. Also, that the clock carrier (in this case Alice) ages as much as her clock does. Any clock's reading is always per the Lorentz transformations, and the LT solns depend upon point of view (POV) ... and POVs will differ when relative motion exists. Each POV is equally correct, equally true, and equally real.

For reference, the Half Twins Paradox figure ...

OK, so the questions posed were ...

BurtJordaan wrote:what was the time on Alice's clock when the "time handoff" between Bob and Charlie took place?**

GrayGhost's position is ...

Per Alice ... Alice's clock reads 4 yr upon the handoff event.
Per Bob ... Alice's clock reads 1 yr upon the handoff event.
Per Charlie ... Alice's clock reads 7 yr upon the handoff event.

BurtJordaan wrote:by how many years Alice aged between the time that Bob flew past her (event A/B) and the time of the hand-over event?

GrayGhost's position is ...

Per Alice ... Alice's clock advanced 4 yr (0->4) from A/B thru handoff, so Alice aged 4 yr.
Per Bob ... Alice's clock advanced 1 yr (0 ->1) from A/B thru handoff, so Alice aged 1 yr.
Per Charlie ... Alice's clock advanced 7 yr (0->7) from A/B thru handoff, so Alice aged 7 yr.

Also, if the question were posed in this way ...

by how many years Alice aged between the hand-over event and the time that Charley flew past her (event A/C)?

GrayGhost's position is ...

Per Alice ... Alice's clock advanced 4 yr (4->8) from handoff thru A/C, so Alice aged 4 yr.
Per Bob ... Alice's clock advanced 7 yr (1->7) from handoff thru A/C, so Alice aged 7 yr.
Per Charlie ... Alice's clock advanced 1 yr (7->8) from handoff thru A/C, so Alice aged 1 yr.

Thank you,
GrayGhost
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Thx for the update, Gray. It looks OK on my screen. Jpg's have the tendency to lower the quality of the originals somewhat.

I find the best SCF results when I upload a graphic and then view it by clicking on the graphic or thumbnail in SCF. When the picture comes up on screen, copy that URL from your browser's URL box and paste the URL into Img tags. This is the permanent SCF storage place for the uploaded image, I hope.

Now back to the odds: I'm not as "hubristic" as Leo, so I'll raise you to 10:1 in my favor. We are talking about "real age" here, not what someone in another frame perceives Alice's age to be. IMO, the answer is trivial, because you already have Alice's assistant, say Alex (your A') in her frame, obviously with a clock that is synchronized to Alice's. All that is needed is for Alex to read his clock when the two travelers (Bob and Charlie) simultaneously fly past him. He can also read the travelers' clocks, so he will record: tA' = 4, tB=2, tC = 2, not so?

So, the answer is that Alice aged 4 years from start to the half-way mark, while Bob aged two years. During Charlie's return flight, he aged 2 years and Alice another 4 years. No paradox...
--
Regards
Jorrie

PS: you may notice that I have softened my stand a little since 2010. Then I maintained that we do not know by how much Alice has aged from the start to the halfway mark, because it depends on what "cheat" (to use Dave_O's term) we used with the synchronization of Alex and Alice's clocks. In a sense, this is still true, but nature seems to "prefer" that we "cheat" a-la Einstein with our clocks. I accept this as standard today, because there are other physical and philosophical arguments for it as well.

BurtJordaan
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GrayGhost wrote:My position is that the extent of a clock's aging depends upon how much the clock's reading advanced over the defined interval. Also, that the clock carrier (in this case Alice) ages as much as her clock does. Any clock's reading is always per the Lorentz transformations, and the LT solns depend upon point of view (POV) ... and POVs will differ when relative motion exists. Each POV is equally correct, equally true, and equally real.

OK, let us follow the recipe that you proposed. The defined interval here is the proper time between flyby event A/B and flyby event B/C. Bob is the only guy present at both events, so we know that his clock measured the proper time interval, $dt_B = 2$, with $dx_B = 0$. Since the question is A's age, we can simply LT the interval to A's inertial frame.

$dt_A = \gamma(dt_B + vdx_B) = 2\times(2-0.866\times 0) = 4$

$dx_A = \gamma(dx_B + vdt_B) = 2\times(0+0.866\times 2) = 3.464$

Hence Alice aged 4 years between the two flyby events.

The only calculation that can give the 1 year answer that you desire is not a LT, but reciprocal POVs, where each observer thinks the others clock is running at half the rate of his/her own clock. IMO, these POVs have nothing to do with aging and only serve to confuse people into thinking that SR is paradoxical.

--
Regards
Jorrie

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Let's start here then Jorrie ...

BurtJordaan wrote:The only calculation that can give the 1 year answer that you desire is not a LT, but reciprocal POVs, where each observer thinks the others clock is running at half the rate of his/her own clock. IMO, these POVs have nothing to do with aging and only serve to confuse people into thinking that SR is paradoxical.

Well, let's put that theory to the test then. You see, Alex is not the only one co-located at the handoff event. Bob is there too. Let's ask Bobby then, see what his POV produces for Alice's clock reading and aging ...

Per the figure, when Bob is at the handoff event, Alice is xB = -1.732 ly downrange and moving at v = -0.866c, per Bob. Let's see if it is true that Bob cannot determine Alice's current clock reading, using the LTs.

The Lorentz transformation ...

tA = γ(tB-vxB/c²)
tA = 2(2-(-0.866)*(-1.732)/1²)
tA = 1

xA = γ(xB-vtB)
xA = 2(-1.732-(-0.866)(2))
xA = 0

The LT soln is ... xA, tA = 0,1

Meaning ... that when Bob's clock reads 2yr-B at the handoff event, Alice's clock then reads 1yr-A, per Bob. The meaning of the soln xA, tA = 0,1 is that Alice's clock is always with Alice, Alice always co-located with her own 3-space system origin as time passes her by. As such, Alice aged 1 yr from A/B flyby to the time-handoff event, per Bob. I submit Alice REALLY AGED 1 yr (not 4 yr) per Bob, because her clock tells him so. Alice may well exist at future points along Alice's worldline (beyond 1yr-A), but that is not the question. The question is, does Alice's clock read 1yr-A per Bob, when Bob is at the handoff event, and is that real?

Validation ... Next, we consider Bob having emitted EM as he goes, the EM always reflecting off Alice and her clock, those images later received by Bob for analysis. Let us also presume there happens to be a clock B' (carried by Buck) at rest and synchronised with Bob's clock, that Alice passes when she reads her clock at 1 yr. As such, the B' clock must read 2 yr when Alice's clock reads 1 yr on intersection ... and per Bob this occurs when he is at the handoff event when his clock reads 2 yr. Bob's subsequent analysis of the received light signals confirms the prior. This is just to say that in theory, light signals would validate the above Lorentz transformations executed by Bob.

Realness ... How real is the 1yr-A clock readout, per Bob? Is it completely real, less than real, merely apparent, unreal, an optical illusion of sort? Well, we might imagine that Alice collides with the B' clock at their flyby event, and so Alice ceases to exist from that time forward. This is not to suggest that the remaining atoms of Alice, or the energies released by her desintegration, do not still exist in space "as simultaneous with Alex at the time-handoff event". However, to assume that 4yr-A is the REAL SOLN (wrt handoff) simply because Alex's clock suggests so per he, is to assume that Alice did not REALLY exist at 1yr-A in Bob's spacetime system, or did not REALLY exist at 7yr-A in Charley's spacetime system, at that same handoff event ... which IMO is incorrect.

Summary ... GrayGhost submits that each POV is as real as the next, and POVs may differ. All are equally TRUE, REAL, and CORRECT. That fact that no disagreement exists for co-located clocks, does not lead that their readings are MORE TRUE. It means only that there is no disagreement. If Alice had scratched her left ear only when her clock read 1yr-A, the reflected light signals analyzed later by Bob would show she REALLY did so when Bob's own clock read 2yr-B at handoff.

It is TRUE that ...

(1) Alice's clock reads 1 yr per Bob, at handoff.
(2) Alice's clock reads 4 yr per Alex, at handoff.
(3) Alice's clock reads 7 yr per Charley, at handoff.

All 3 are equally REAL. What makes (2) appealing, resides only in the fact that Alice is not then effected by relativistic effects per Alex, being of the same frame. It's a natural desire to prefer that situ, because it's consistent with everyday experience. However ...

To suggest that (2) is the only TRUE answer, is IMO to adopt the PRESENTISM ontological philosophy of time. To suggest that (all 3) are TRUE, is IMO to adopt the ETERNALISM ontological philosophy of time. And, this does seem to come down to just that.

Thank You,
GrayGhost
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Good thoughts Gray. I like this one in particular.

GrayGhost wrote:Summary ... GrayGhost submits that each POV is as real as the next, and POVs may differ. All are equally TRUE, REAL, and CORRECT.

Regards Leo
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Thanx Leo,

I do fully understand Jorrie's line of reasoning here, and I am rather certain he understands mine as well. Technically speaking, we have no disagreement on the LTs. I think there may be a difference of opinion as to the meaning of the LTs, far as what is real versus lesser-real. I see it more as a philosophical argument, although I for one would consider it an important one. Let's face it, we all want to know exactly what the theory means, and we want to know deep down that we are correct in our understanding. I'm looking forward to seeing how our interpretations stand up to scrutiny. And besides, Jorrie gave me 10:1 odds :-)

GrayGhost
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BurtJordaan wrote:PS: you may notice that I have softened my stand a little since 2010. Then I maintained that we do not know by how much Alice has aged from the start to the halfway mark, because it depends on what "cheat" (to use Dave_O's term) we used with the synchronization of Alex and Alice's clocks. In a sense, this is still true, but nature seems to "prefer" that we "cheat" a-la Einstein with our clocks. I accept this as standard today, because there are other physical and philosophical arguments for it as well.

Yes, Dave Oblad's cheat, indeed. As Poincare pointed out long ago, we elect SR's convention of simultaneity because of convenience. Minkowski had mentioned that LET's upholding of the principle-of-relativity was "a gift from above", because it was argued to be upheld by coincidence. One can argue that SR's upholding of the principle-of-relativity was "a gift from Occam".
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GrayGhost wrote:And besides, Jorrie gave me 10:1 odds :-)

My facetious exercise in bookmaking was directed at the odds of a reasonable number of players being able to arrive at a consensus. This is doubly true if Dave decides to join the game.

Regards Leo
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And I absolutely agree that this is as much a philosophical question as it is a physical one.

Regards Leo
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GrayGhost wrote:Per the figure, when Bob is at the handoff event, Alice is xB = -1.732 ly downrange and moving at v = -0.866c, per Bob. Let's see if it is true that Bob cannot determine Alice's current clock reading, using the LTs.

Gray, what you are doing here is to define a new coordinate origin at the event B/C flyby and using B as the reference frame. Yes, you can do that, but then your diagram and the x and t coordinates are no longer what you have pictured.

The LTs are valid for two inertial frames in "standard configuration", with origins coinciding, so in your diagram, xB is not = "-1.732 ly downrange" for the LTs. For your diagram, event B/C is at xB = 0, tB = 2. This Lorentz transforms to my original tA = 4, etc.

From: http://en.wikipedia.org/wiki/Lorentz_transformation#Boost_in_the_x-direction

These are the correct expressions to use in the LTs for your diagram (you want to transform the values of the primed "moving frame" to the reference frame). So: gamma = 2, t'=2, v= 0.866, x'=0. You are using x' = 1.732, which is incorrect (mixing of frames). B is present at both events, so how can his spatial coordinate be anything other than x'=0? The LT answers for A are x=3.464, t=4.

As far as I'm concerned, the LTs do not give the POVs that you talk about; they give transformations of time and space coordinates from one inertial frame to another that are in relative movement, with the two frames in the standard configuration (0,0 coinciding in spacetime). If coordinate origins are not coinciding, we are forced to use the invariance of the spacetime interval, ds2 = dt2 - dx2 = dt'2 - dx'2 to transform space and time intervals. It incidentally always gives identical results to the LT's, when the latter is correctly used.

The coordinate position that you are referring to, xA=0, tA = 1 on you diagram, is a different event, which represents neither of the two flybys in question. Yes it has the same time coordinate in B's frame, i.e. 2.0, but its xB coordinate is not 1.732. Like in 2010, it seems to me that you are "freely" mixing coordinate systems in the LTs and that is unfortunately guaranteed to give incorrect results.

This misuse (or not?) of the LTs must first be cleared up - otherwise we may again go into 15 pages of fruitless argumentation. I will invite Lincoln and Marshall to help with "arbitration" here. Please also nominate someone if you like.

--
Regards
Jorrie

BurtJordaan
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Jorrie,

Lincoln and Marshall are excellent, so I would very much invite their opinions. I don't think arbitration is required, however I am fine with them arbitrating since you are.

We'll discuss convention first then, at your request. You said I misapplied the LTs in my calculations, that what I did was not a Lorentz transformation at all, and suggested that I did not follow convention properly. Also, you posted the Wiki reference for velocity boost, showing the meaning of the inverse LTs ...

http://en.wikipedia.org/wiki/Lorentz_transformation#Boost_in_the_x-direction.

As per WIKI ...

Convention, for Lorentz transformations ...

t’ = γ(t-vx/c²)
x’ = γ(x-vt)

Convention, for inverse Lorentz transformations ...

t = γ(t'+vx'/c²)
x = γ(x'+vt')

These conventions, require the input of positive velocities exclusively. Only difference ...

Normal convention ... positive velocity for motion in the direction of increasing space.
Inverse convention ... positive velocity for motion in the direction of decreasing space.

If the Inverse LT convention was never created, one could use only the normal LT convention and just plug in the actual velocity. A positive velocity for motion in the direction of increasing x, and a negative velocity for motion in the direction of decreasing x. For a velocity in the direction of decreasing x, we'd then have ...

t’ = γ(t-vx/c²)
t’ = γ(t-(-v)x/c²)
t’ = γ(t+vx/c²)

This would be prefectly fine, if folks could ever keep straight which frame is the primed and which is the unprimed. Unfortunately, so many folks have trouble keeping frames straight in SR analyses, the inverse Lorentz transform was invented. As such, the convention was invented, to help keep the frames straight for those who had trouble.

OK, so in my LT computation, we were determining the ALICE POV per BOB. We named the frames after their first initials, so xA,tA and xB,tB. As such, we have no primed and unprimed frames, and therefore the convention is not so important ... ie, it's pretty difficult to mix up Alice and Bob. All that's required, is that the LTs be appropriate for the polarity of the velocity used, plain and simple. OK, so ...

Regarding whether I have improperly used the LTs, or produced something that was somehow not really an LT transformation ...

tA = γ(tB-vxB/c²)
tA = 2(2-(-0.866)*(-1.732)/1²)
tA = 1

xA = γ(xB-vtB)
xA = 2(-1.732-(-0.866)(2))
xA = 0

The LT soln is ... xA, tA = 0,1

Note that while I began with the substitution of a negative velocity for v, the result is as follows ...

tA = γ(tB-vxB/c²)
tA = 2(2-(-0.866)*(-1.732)/1²)
tA = 2(2+(0.866)*(-1.732)/1²)

which is identical to the inverse transform of ...

tA = γ(tB+vxB/c²)
tA = 2(2+(0.866)*(-1.732)/1²)

whereby only positive values of v may be input in the latter.

Look at the figure, and note the following ...

(1) Alice moves in the direction of decreasing xB, per Bob.
(2) Alice is not at the Bob worldline at tB=2, per anyone.
(3) Alice is -1.732 ly wrt Bob's 3-space origin at tB=2, per Bob.

In conclusion, I submit that ...

1) my LT computation above is in fact, correct, even though I did not begin the computation with the standardized inverse LT convention.

2) my LT soln is completely consistent with the referenced Minkowski figure, and had my LT calculation been wrong, the error would be visible on the figure (it is not).

3) Alice is in fact -1.732 ly from Bob (per Bob) when Bob is at the handoff event, and xB,tB = -1.732, 2 is a valid input for LT transformation.

4) Any coordinate may be input into the LTs for transformation, and not just those that reside on Bob's own worldline. Ie, they are spacetime coordinate transformations for any and all points, all points unique.

5) Any point in Bob's space that defines the current location of a body (eg Alice) must transform to a point in the other spacetime system where the body (eg Alice) resides.

OK Jorrie, I hope this helps here. While I agree that I did not follow standard convention, I think I have shown that my calculations and solns are correct. Please point out where you disagree.

Thank You,
GrayGhost
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GrayGhost wrote:Convention, for inverse Lorentz transformations ...

t = γ(t'+vx'/c²)
x = γ(x'+vt')

These conventions, require the input of positive velocities exclusively. Only difference ...

Normal convention ... positive velocity for motion in the direction of increasing space.
Inverse convention ... positive velocity for motion in the direction of decreasing space.

Actually, since the primed frame is by convention the "relatively moving" one, the normal equation is for conversion of unprimed to primed vales. The inverse equation is for converting the primed ("moving") to unprimed (Reference frame) values. Positive x (and speed) is always taken to the right. In the question stated: "how much have Alice aged between the two flyby's", it seems clear to me that we need to convert Bob's primed time to Alex's unprimed time. My problem is with the fact that you mix coordinates, not which way you choose speed.

In Bob's frame the time between the two relevant events is 2y and the distance between the two events is zero. We are trying to LT from Bob's (x', t') to Alice's, (x,t). You work with Bob's t = 2 yr, but then you go and use x = -1.732 lyr, which is not the distance between the relevant events in Bob's frame. It is the distance between B/C flyby and another event, xA=0, tA = 1* as observed from Bob's frame. Where you labeled the 1.732 lyr is slightly confusing, because it insinuates that it is the distance between the A/B and B/C events per B, which it is not. I think we should not carry on until we get this cleared up. Here is your chart again for ease of reference for everyone.

Despite what I said on carrying on, I do want to ask you for a reality check on Alice's aging process. You are surely not holding that between the A/B flyby and B/C flyby, Alice aged 1 yr and then "instantly" aged 6 more years during the time hand-off, so that Charlie can observe her as aged by 7 yrs at that time? This would represent a serious discontinuity for Alice...

I hold that if Bob and Charlie wants to know how much Alice has aged between the first two flybys, they should simply read tA' at their flyby (on a clock in sync with Alice's), i.e. 4 ly, which is also what a correctly used LT tells us.

IMO, what you have done is just and exercise in reciprocal time dilation between two purely inertial frames, which does not represent aging. Lincoln has written an Expert Note on this Forum where he warns against wrong interpretations of the LTs, and their applicability. I'm not sure if your case is covered there, but I think the general principle is applicable.

--
Regards
Jorrie

* The best operational meaning that I can think of for the event (say E) at xA=0, tA = 1 is: it happened when Alice's clock read 1 yr, 3 yrs before the B/C flyby. According to Bob, the event was simultaneous with his B/C flyby, separated by the -1.732 lyr. The events were obviously not simultaneous according to Alice, because they were co-located, but happened at 1 yr and 4 yrs respectively.

Hope this helps in clearing up the LT confusion.
Last edited by BurtJordaan on October 18th, 2013, 11:19 am, edited 5 times in total.

BurtJordaan
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BurtJordaan wrote:* The best operational meaning that I can think of for the event (say E) at xA=0, tA = 1 is: it happened when Alice's clock read 1 yr, 3 yrs before the B/C flyby. According to Bob, the event was simultaneous with his B/C flyby, separated by the -1.732 lyr. The events were obviously not simultaneous according to Alice, because they were co-located, but happened at 1 yr and 4 yrs respectively.

Hope this helps in clearing up the LT confusion.

I would state the above "meaning" ...

xA,tA = 0,1 is Alice's own coordinate per herself when Bob holds Alice simultaneous wrt the handoff, Alice then being -1.732 ly downrange from Bob per Bob. Alice disagrees with Bob that she is then simultaneous with the handoff event, and in fact won't happen for 3 more years per Alice.

***************

At any rate, it sounds like you now agree in that the readout of Alice's clock, and hence its extent of aging wrt the interval, is dependent strictly upon the POV of the observer. As such, upon the handoff ...

Bob holds Alice's clock at 1y-A, and hence Alice has aged 1y post A/B flyby, per Bob.
Alex holds Alice's clock at 4y-A, and hence Alice has aged 4y post A/B flyby, per Alex.
Charley holds Alice's clock at 7y-A, and hence Alice has aged 7y post A/B flyby, per Charley.

All are equally correct, as no frame is preferred.

***************

Next, let's discuss "LT confusion" ...

You have suggested that I may be using the LT improperly, seemingly because I did not use a coordinate of xB = 0 in the LT. Let's explore this ...

Jorrie's LT transformation ...

xA = γ(xB+vtB)
xA = 2(0+0.866*2)
xA = 3.464

tA = γ(tB+vxB/c²)
]tA = 2(2+0.866*0/1²)
tA = 4

The soln being ... xA,tA = 3.464, 4

GrayGhost's LT transformation ...

xA = γ(xB+vtB)
xA = 2(-1.732+0.866*2)
xA = 0

tA = γ(tB+vxB/c²)
]tA = 2(2+0.866*-1.732/1²)
tA = 1

The soln being ... xA,tA = 0, 1

OK, so Jorrie used a point on Bob's worldline for transformation, while GrayGhost used a point on Alice's worldline for transformation. So, what then is the difference between these 2 methods ...

(1) Jorrie's method ... Determines the coordinates of the handoff per Alice, when Alice holds the event as simultaneous per herself. This answers the question ... what does Alice's clock read upon handoff, per Alice?

(2) GrayGhost's method ... Determines the coordinates of Alice per herself, when Alice is simultaneous wrt the handoff event per Bob. This answers the question ... what does Alice's clock read upon handoff, per Bob?

Both sets of coordinate inputs are valid for LT transformation. One means one thing, the other means another thing.

This takes us back to the original question, which was ...

BurtJordaan wrote:question: what was the time on Alice's clock when the "time handoff" between Bob and Charlie took place?**

The question is asked without any specification of an observing POV. As stated, the answer depends upon the POV one considers, and each POV will input different coordinates into the LTs to determine the soln, and the solns for each differing POV will differ.

Jorrie, I am in agreement that there is no absolute single answer to the question, as stated. I think you mentioned that early on here. My position wrt the LTs however, is that my transformations were appropriate for the POV of Bob. Do you still disagree?

Thank You,
GrayGhost
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BurtJordaan wrote:* The best operational meaning that I can think of for the event (say E) at xA=0, tA = 1 is: it happened when Alice's clock read 1 yr, 3 yrs before the B/C flyby. According to Bob, the event was simultaneous with his B/C flyby, separated by the -1.732 lyr. The events were obviously not simultaneous according to Alice, because they were co-located, but happened at 1 yr and 4 yrs respectively.

OK, herein lies the resolution of debate. This is essentially what GrayGhost was saying all the time... :-) He actually hinted above at the proper operational meaning of the event, which I will paraphrase thus to couple Alice's aging to it.

When Alex was at +1 yr, event (E) happened co-located to Alice, which she recorded. Bob, flying away did not know about event (E) until much later (at tB=3.732 yrs) when the light message caught up with him. In the meantime, at tB=2 yrs, the Bob/Charlie flyby event happened. Using the isotropy of the speed of light in his frame, Bob then calculated that the two events (E and B/C flyby) happened simultaneously at 2 yrs. But we know that event E happened when Alice aged 1 year.

Alice's assistant (Alex) observes the B/C flyby on location at tA'=4 yrs and let Alice know this value (the signal takes 3.464 years), so Alex can deduce that the B/C flyby event happened at tA=4 yrs. This is then the full resolution of the "half-twins paradox".

So Gray was entitled to write:

It is TRUE that ...

(1) Alice's clock reads 1 yr per Bob, at handoff.
(2) Alice's clock reads 4 yr per Alex, at handoff.
(3) Alice's clock reads 7 yr per Charley, at handoff.

All 3 are equally REAL. What makes (2) appealing, resides only in the fact that Alice is not then effected by relativistic effects per Alex, being of the same frame. It's a natural desire to prefer that situ, because it's consistent with everyday experience.

There is no discontinuity in Alice's aging, as I incorrect stated. She ages normally and experience (and perceive) different events at different times than what Bob experience them.

So 10 kudus to Gray for finally 'winning' the consensus on this one, congrats and many thanks to him for his persistence. Sad for LEO who lost a thousand (I think), for betting that this one would not be resolved.

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Jorrie

PS: I wrote this before your last post, so I will look at the LT issues a little later on Saturday. Maybe you are winning that as well. ;-) We'll see...

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BurtJordaan wrote:So 10 kudus to Gray for finally 'winning' the consensus on this one, congrats and many thanks to him for his persistence. Sad for LEO who lost a thousand (I think), for betting that this one would not be resolved.

Thanx Jorrie. 10 kudus feels better than 1, or even 2. Between us, I think I may be 1 for 10 now :-)

Anyone can toss about the LTs and play with them, but to disect the deeper meaning of the theory is the challenging part. It's also to best part, and while it can be hard, the hard it was makes it great. You generally probe deep into the theory, and diving in can be somewhat addicting.

In your OP, when you said ...

BurtJordaan wrote:This is the same as the classical twins paradox solution and there is no disagreement between Gray and myself on the result. What we disagreed on previously is the answer to this question: what was the time on Alice's clock when the "time handoff" between Bob and Charlie took place?** I maintained that it is not known in any absolute sense, but if I had to put a time on it, I would choose 4 years. Gray argued for either 1 year or 7 years, depending on who looked at it from where: 1 year per Bob and 7 years per Charlie.

I must say, I was compelled on a few occasions to consider 4 yr "a better answer of sort", and deep down I just wanted to prove to myself why that happens. In the end, (for me) I think it comes down to the fact that we live in a single POV, and so we gravitate toward that environment as "more desirable". In the case of Alice and Alex, they were (originally) the only 2 of a single frame, and so we may assign that frame as more desirable even if we don't realize it.

In your case, you recognize the importance of invariants. In fact, you were likely the first to make me more fully aware of their importance. In this discussion here, I think you may have combined the absoluteness of simultaniety within a single inertial frame, with the invariance of a defined spacetime interval. As such, you connected Alice-to-Alex to AB-to-BC. While all that is fine, the rest of the overall picture may have seemed a bit less significant, yet all aspects of the picture weigh in the same. I mean, it is rather difficult to grasp the concept of "all NOWs coexist on equal footing in the spacetime continuum", and that's what the 3 POVs at handoff tell us. It's contrary to everyday experience, yet the math requires it and the evidence supports it.

Thank You,
GrayGhost
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Congratulations Jorrie and Gray. I'm happy to concede that I was wrong in assuming that a consensus would be impossible, although you'll have a hard job extracting any dough out of me. It may have been a different story with more players but as far as I can judge the matter you two blokes are the ones that understand the relativity theories the best within these pages. Once again, I capitulate, hopefully with good grace.

I actually learnt a lot from this little exercise because I was unaware that SR could produce such a simple common sense conclusion.

GrayGhost wrote: I mean, it is rather difficult to grasp the concept of "all NOWs coexist on equal footing in the spacetime continuum", and that's what the 3 POVs at handoff tell us. It's contrary to everyday experience, yet the math requires it and the evidence supports it.

Personally I've always found it difficult to grasp how all NOWs could not co-exist on an equal footing in a common sense reality and my everyday experience tells me exactly this, not the contrary, as Gray suggests above. I remain convinced that this little exercise is an instructive demonstration in the artificiality of simultaneity and that the same conclusion could have been reached far more easily without spacetime at all.

I never questioned the ingenuity of the SR model and I still don't, merely its metaphysical validity. Once again, thanks to you both.

Regards Leo
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GrayGhost wrote:In this discussion here, I think you may have combined the absoluteness of simultaniety within a single inertial frame, with the invariance of a defined spacetime interval.

I guess you meant the "relativeness" of simultaneity (not the absoluteness")?

I still have a problem with the way that you have used the LT to arrive at the "per Bob" answer. The LTs ensure the invariance of the spacetime interval between two events. You have used it on 3 events, with two different timelike intervals: (i) the interval A/B flyby to the B/C flyby (2 years) and (ii) the interval between the A/B flyby and the event when A's clock read 1 yr. In order to do that in one step, you had to mix frames inside the LT, which is not something I would encourage.

IMO, the proper use is to convert a set of coordinates (a single x,t) in one inertial frame to values in another inertial frame (a single x',t'), both in "standard configuration). In the "half-twins" scenario, the single (x,t) is point (0,2), the B/C flyby in Bob's frame (the other event being 0,0) in both frames. This transforms to coordinates (0,4) in Alice's reference frame.

As I see it, you could have used time dilation to find the point (0,1) on Alice's worldline and then calculated the spacelike interval between the points (0,1) and (3.464, 4) in Alice's frame. This gives 1.732 lyrs, which holds for both frames (invariance of interval). Then you could have used that value to LT (-1.732,2) per Bob back to the starting point (0,1) per Alice, which is circular.

OK, you may have argued differently (e.g. by using Minkowski chart principles), but I can't help seeing a circularity in the process. This is where we could use Lincoln's insight on proper use of the LTs in this scenario. He is presently very busy, so he said he will have a look when time allows (he's not retired like me! :-)

This does not change the results we agreed upon, though...
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Jorrie

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BurtJordaan wrote:I still have a problem with the way that you have used the LT to arrive at the "per Bob" answer. The LTs ensure the invariance of the spacetime interval between two events. You have used it on 3 events, with two different timelike intervals: (i) the interval A/B flyby to the B/C flyby (2 years) and (ii) the interval between the A/B flyby and the event when A's clock read 1 yr. In order to do that in one step, you had to mix frames inside the LT, which is not something I would encourage.

OK, so let's discuss FRAME MIXING next ...

So, I inputted coordinates of Bob's system, and transformed them to coordinates of Alice's system, similarly as you did. So, there was no "mixing of frames" inside the LTs in that respect.

Technically, the only DIFFERENCE between the BurtJordaan and GrayGhost transformation methods was ...

BurtJordaan inputted Bob-frame-coordinates that represented a point on Bob's own worldline, simultaneous with the handoff event, per Bob.

GrayGhost inputted Bob-frame-coordinates that represented a point on Alice's own worldline, simultaneous with the handoff event, per Bob.

Regarding valid LT inputs ... note that one could input any point, including a point on no worldline. Let's see what happens ...

Consider and EVENT M marked by the coordinate of Bob's system that's (say) midway between Bob and Alice upon handoff, per Bob (xB = -1.732/2 = -0.861) ...

xA = γ(xB+vtB)
xA = 2(-0.861 + 0.866*2)
xA = 1.732

tA = γ(tB+vxB/c²)
tA = 2(2 + 0.866*(-0.861)/1²)
tA = 2.5

The soln being ... xA,tA = 1.732, 2.5

So Bob's coordinate of xB,tB = -0.861, 2 transforms to a point in Alice's system at xA,tA = 1.732, 2.5.

Per Bob ... EVENT M occured remotely at -0.861 ly downrange at time 2y, the same time as the handoff event. A clock synchronised in Bob's frame at EVENT M would read 2y.

Per Alice ... EVENT M occured remotely at 1.732 ly uprange at time 2.5y, however Alice disagrees the handoff event occured at that time. A clock synchronised in Alice's frame at EVENT M would read 2.5y.

However, inputting those coordinates do not answer the question ... what time does Alice's clock read upon handoff, per Bob? Instead, it answers the question ... what time does a clock synchronised in Alice's frame read that's only half of Alice's distance upon handoff, per Bob?

The following spacetime diagram applies (EVENT M infos, in purple) ...

Jorrie, I see no improper use of the LTs, or mixing of frames in what I present here. Can you articulate further as to what I may have done inappropriately?

Thank You,
GrayGhost
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Thanks Leo, graceful capitulation accepted with glee... ;-)

Obvious Leo wrote: I remain convinced that this little exercise is an instructive demonstration in the artificiality of simultaneity and that the same conclusion could have been reached far more easily without spacetime at all.

Not easily provable, as you well know. To take a leaf out of the canonical Minkowski spacetime tree:

(sorry for the extra large picture - don't know how to set size in this editor when using a Wikipedia url)

For any two separate events, we can randomly pick a Cartesian coordinate system and then anchor it to one of the events (observer in the picture) and to the lightcone (of course), which is by convention shown upright for flat space. The second event (say E, but not shown) may lie inside, on, or outside the light cone. This leads to a number of observations.

1. If E lies on the lightcone, it is separated by a null (or lightlike) interval from the origin. No particle with mass can travel between the events (be present at both events), only a massless particle like a photon can.

2. If E lies inside the light cone, it is separated by a timelike interval and there exist one other inertial frame (also centered on the origin) in which the two events are co-located, i.e. separated in time only. Only a particle with mass can be present at both events, but obviously at different times.

3. If E lies outside the light cone, it is separated by a spacelike interval and there exist one other inertial frame (centered on the origin) in which the two events are simultaneous, separated in space only. No known particle, be it massless or massive, can be present at both events. Hypothetical "tachyons" could perhaps, but their existence is uncertain, if not impossible.

4. Roughly half of all possible such event pairs can be simultaneous in some or other reference frame and the other half cannot.

Are these observations also true in your view of reality, Leo?

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Jorrie

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If E lies on the same horizontal plane as the origin, what does that mean in physical terms? What is the special significance of the "hypersurface of the present" in the diagram? What particular property makes some spacelike intervals lie on that plane, and others not?
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BurtJordaan wrote:Are these observations also true in your view of reality, Leo?

Yes, more or less. Of course with gravity missing the time assumptions can only be approximations, as can the flat space, but other than that there doesn't seem to be much wrong with the logic.

Tachyons make no sense to me whatsoever and I regard them as nothing more than a mathematical oddity but I like this conclusion.

BurtJordaan wrote: Only a particle with mass can be present at both events, but obviously at different times.

Nice to see particle superposition disposed of in such a simple and efficient manner.

Regards Leo
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BurtJordaan wrote:3. If E lies outside the light cone, it is separated by a spacelike interval and there exist one other inertial frame (centered on the origin) in which the two events are simultaneous, separated in space only. No known particle, be it massless or massive, can be present at both events. Hypothetical "tachyons" could perhaps, but their existence is uncertain, if not impossible.

Leo, I am surprised that you agree with the above, as it seems to contradict your no-space and no-simultaneity model. I thought you disagreed with the standard concept of an inertial frame containing two distinct but simultaneous events.
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Positor wrote:Leo, I am surprised that you agree with the above, as it seems to contradict your no-space and no-simultaneity model. I thought you disagreed with the standard concept of an inertial frame containing two distinct but simultaneous events.

That's why I qualified my statement with "more or less" and pointed to the absence of gravity as the problem with this representation. In a sense I was agreeing with Jorrie in terms of the narrow parameters of his question. In other words "if" we accept space-time itself as a valid construct then this is the conclusion we can draw from it. I didn't think it was appropriate to start my argument all over again in this thread by arguing over the foundational assumptions. As I've repeatedly said elsewhere, in a spaceless model the word simultaneous can never be used without the qualifying adjective "approximately".

I most certainly do disagree with the standard concept of an inertial frame containing two distinct but absolutely simultaneous events. If you and I are standing side by side on my verandah and watching the world go by it is utterly impossible for us both to be seeing the same thing at the same time. That's pretty much all that the bloody obvious asserts. In my ontology to suggest otherwise is nothing more than a mathematical contrivance and I reckon Gray has shown us this in his slightly different take on the LTs. However my understanding of the SR model is not actually advanced enough to judge whether what Gray has done is kosher or not.

Regards Leo
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GrayGhost wrote:Jorrie, I see no improper use of the LTs, or mixing of frames in what I present here. Can you articulate further as to what I may have done inappropriately?

Gray, unfortunately there is an ongoing trend in your replies of applying the LTs in an unconventional way, which may give you the right answers (by using some other "crutches" as I will attempt to show), but your method may be very confusing to the uninitiated.

The first time you used the LT in this thread was in this reply:
GrayGhos wrote:Per the figure, when Bob is at the handoff event, Alice is xB = -1.732 ly downrange and moving at v = -0.866c, per Bob. Let's see if it is true that Bob cannot determine Alice's current clock reading, using the LTs.

The Lorentz transformation ...

tA = γ(tB-vxB/c²)
tA = 2(2-(-0.866)*(-1.732)/1²)
tA = 1

xA = γ(xB-vtB)
xA = 2(-1.732-(-0.866)(2))
xA = 0

The LT soln is ... xA, tA = 0,1

You are giving a great Minkowski with frames in standard configuration. When you do the calculation above, you suddenly shifted B's origin to xB=0, tB=2. Then you presumably used $v\Delta t = -0.866 \times 2 = -1.732$ to find the distance to the new event, per the "new B origin, (i.e. the value $\Delta x_B$). Only now could you use LTs to find the coordinates of point (0,1) in the old reference frame (A's). I think you used this roundabout method not to find the point, but to show that the LTs can produce the point if we stretch the definition somewhat.

I'm sure you know that it is a straight time dilation calculation and the roundabout way above may be simply confusing to some. E.g. it originally struck me as simply wrong, but once I figured out the roundabout, it was OK-ish.

OK, so what would the most straight forward LT method (if we really wanted to use one) have been?

Firstly, I would have kept the origin in its rightful place, 0,0 for both. Then use the given event coordinates per B (0, 2), find the handoff point per A, by straight forward LT, getting (3.464, 4), because we needed that for the primary answer anyway.

To answer the secondary question: "what would Alice's clock read at handoff according to B?", I would have considered an observer B' (with clock synchronized to B) to the left of the origin, reaching A when the B' clock reads 2 yrs. At 0.866c, B's clock must have read 0 when he was at $x_B = 0.866\times (-2 ) = -1.732$. To answer the question, the input values are $\Delta t_B = +2$ yrs and $\Delta x_B =+1.732$.

Now the LT is similar to Gray's original, but conforms to LT standard configuration principles:

$\Delta t_A = \gamma(\Delta t_B - v \Delta x_B) = 2(2-0.866\times1.732) = 1$ yr.

Modified ABC scenario

Ultimately, this is still a round-about way for a very simple problem: we could have just used reciprocal time dilation straight away and came to the same answers in a jiffy. But the argument was around LTs.

The issue may seem trivial an just nitpicking, but I have seen many a confusion in time dilation and/or LTs through non-standard use. You can handle it, but not many novices could, IMO.
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Regards
Jorrie

PS: Readers are advised to also check Lincoln's http://www.sciencechatforum.com/viewtopic.php?f=84&t=8856 for deeper discussion of LTs and reciprocal time dilation.
Last edited by BurtJordaan on October 21st, 2013, 4:09 am, edited 1 time in total.
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Positor wrote:If E lies on the same horizontal plane as the origin, what does that mean in physical terms? What is the special significance of the "hypersurface of the present" in the diagram? What particular property makes some spacelike intervals lie on that plane, and others not?

All clocks on that plane read zero simultaneously in the reference frame, as per Einstein's definition of simultaneity (not necessarily as per Leo's definition ;) Since the reference frame can be arbitrarily chosen, there is nothing really special about that plane - it is just special for the chosen frame. But so have all other inertial frames in motion relative to whatever reference their own planes.

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Jorrie

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Obvious Leo wrote:If you and I are standing side by side on my verandah and watching the world go by it is utterly impossible for us both to be seeing the same thing at the same time. That's pretty much all that the bloody obvious asserts.

Einstein's simultaneity definition says nothing about "seeing the same thing at the same time". Neither does Minkowski spacetime.

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Jorrie

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BurtJordaan wrote:Einstein's simultaneity definition says nothing about "seeing the same thing at the same time". Neither does Minkowski spacetime.

I realise that, Jorrie, the observer dependence is still preserved. I'm just drawing a simpler conclusion ( I hope)

BurtJordaan wrote:as per Einstein's definition of simultaneity (not necessarily as per Leo's definition ;)

Just to clarify this point. I have no definition of simultaneity at all since I deny that such a thing exists.

Regards Leo
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