# One Way Twin Revisited

• I

## Summary:

New mode of travel in a one-way 'twin's paradox. Is this method valid?

## Main Question or Discussion Point

I have always had an interest in relativity theory, though have struggled to wrap my head around it. But, now that I have retired, I thought I should try once again to understand its intricacies.

Following the discussions in 'One Way Twin', initiated by Buckethead on 16th June last year, has recalled a problem I had in previous attempts at comprehension.

My enquiry starts off like the 'One Way Twin', but instead of travelling to a distant plant (Mars), my 'twins' live on a very long, flat planet, called 'Home', adjacent to which is another long, flat planet, known as 'Northbound, which whizzes past at 3c/5 - I chose that value as it makes the maths easy.

Twins Tim and Tom have top quality watches, which tick in synchronisation with each other, and display the same time. The watches are also extremely shockproof.

My understanding is that if one of them leaves Home at 3c/5, travelling for four minutes on his watch, then returns at the same speed, he will 'lose' two minutes when they compare watches, that is when the traveller's watch displays three o'clock, the stay-at-home's will display two minutes past three.

Tim steps onto Northbound, and waits for four minutes to elapse on his watch, then leaps back onto Home.

What I should like to know is: what is the difference between their watches? As suggested in the 'One Way Twin' discussion, this can be discovered. Tom could use triangulation to measure how far away Tim is, if Tim provides two signalling devices at a known separation.

Once Tom knows how far away Tim is, Tim sends a light signal when his watch is at exactly three o'clock. Tom will see the signal, note the time on his watch and as he knows haw far away Tim is, he can calculate how long it took the light to come from Tim. Thus he can say "the time on my watch when Tim's watch was at 3:00 was... and that should allow Tom to work out how Tim's time varied with respect to his on his one-way trip.

Can you tell me what the answer will be please, and obviously more importantly, why?

Of course, that's the easy bit; my real puzzle comes once I get this information.

My apologies if this seems very simple stuff.

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jbriggs444
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Tim steps onto Northbound, and waits for four minutes to elapse on his watch, then leaps back onto Home.
[...]
what is the difference between their watches?
Since $v=\frac{3}{5}c$, relativistic gamma is given by $\gamma=\frac{1}{\sqrt{1-v^2/c^2}}=1.25$

Calculation is simple in the home frame. We can use the time dilation formula instead of the full Lorentz transform. The moving watch (running at 0.8 speed) has advanced by 4 minutes. So the home watch must have advanced by $\frac{4}{0.8}=4 \times 1.25$ = 5 minutes.

Five minutes on the home watch. Four minutes on the moving watch. Delta = 1 minute.

The watches are now separated by a distance (as reckoned in the home frame) of 5 minutes times 3/5c = 3 light minutes. A meaningful remote comparison between the displayed times on the two watches is possible because we have decided to use the home coordinate system as the standard of synchronization. [You have chosen to employ home rulers and home clocks as the physical realization of the watch comparison].

Ibix
So Tom remains inertial. Tim synchronises his watch, accelerates instantaneously to 0.6c for four minutes by his watch, then decelerates instantaneously again.

According to both Tom and Tim, Tim's watch will show four minutes to have passed and Tom's will show five minutes. Other observers will agree about Tim's watch (because there is a well-defined answer to "what time did Tim's watch read when he started and stopped moving?"). They will disagree about Tom's, because the relativity of simultaneity means that there is not a unique answer to the question "what time did Tom's watch read when Tim stopped?", because Tom's watch is not in the same place as Tim.

You may wish to look at the spacetime diagrams I drew in the last post in the thread you reference. They apply equally to this, except that the blue lines represent Tom and a clock that just happens to be where Tim steps back onto Home.

Since $v=\frac{3}{5}c$, relativistic gamma is given by $\gamma=\frac{1}{\sqrt{1-v^2/c^2}}=1.25$

Calculation is simple in the home frame. We can use the time dilation formula instead of the full Lorentz transform. The moving watch (running at 0.8 speed) has advanced by 4 minutes. So the home watch must have advanced by $\frac{4}{0.8}=4 \times 1.25$ = 5 minutes.

Five minutes on the home watch. Four minutes on the moving watch. Delta = 1 minute.

The watches are now separated by a distance (as reckoned in the home frame) of 5 minutes times 3/5c = 3 light minutes. A meaningful remote comparison between the displayed times on the two watches is possible because we have decided to use the home coordinate system as the standard of synchronization. [You have chosen to employ home rulers and home clocks as the physical realization of the watch comparison].
Thanks for that, however, I said that was the easy bit. Where I get a problem is as follows.
Applying your logic to another situation: Bill and Bob live on a long, flat planet called 'Normal', adjacent to which is another long, flat planet called 'High-speed', passing Normal at 3c/5. They have watches just like Tim and Tom in my first query.
Bill steps onto High-speed, and waits four minutes on his watch before stepping back onto Normal. Going through the triangulation process again, Bob should find that Bill's watch is one minute behind his - five minutes for Bob, four minutes for Bill. Alternatively, four minutes for Bob is three minutes, twelve seconds for Bill.
Next, their entire world gets bumped onto High-speed, which gets renamed 'Home', and Normal is renamed 'Northbound'.
Fast forward to Tim and Tom performing their experiment. If Tim steps onto Northbound for four minutes, surely only three minutes twelve seconds will have elapsed on Tom's watch? If not, where is the mistake in my logic?
Rather unfairly of course, I can invent history where they or their ancestors, occupied another planet, from which they transferred, which leads us round in a circle.
Please note, I am not trying to be difficult, I am just trying to get my head around this, and am happy if I am merely thick, and don't understand the details.
Thanks again.

So Tom remains inertial. Tim synchronises his watch, accelerates instantaneously to 0.6c for four minutes by his watch, then decelerates instantaneously again.

According to both Tom and Tim, Tim's watch will show four minutes to have passed and Tom's will show five minutes. Other observers will agree about Tim's watch (because there is a well-defined answer to "what time did Tim's watch read when he started and stopped moving?"). They will disagree about Tom's, because the relativity of simultaneity means that there is not a unique answer to the question "what time did Tom's watch read when Tim stopped?", because Tom's watch is not in the same place as Tim.

You may wish to look at the spacetime diagrams I drew in the last post in the thread you reference. They apply equally to this, except that the blue lines represent Tom and a clock that just happens to be where Tim steps back onto Home.
Thanks Ibix. I replied to jbriggs444, explaining my real problem, so if you don't mind, could you please look there? If the real problem is just that I am dense and shall never understand relativity, I can accept it. But if you can point out the flaws in my logic, I shall be grateful.

jbriggs444
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Next, their entire world gets bumped onto High-speed
When?

That question has teeth.

When?

That question has teeth.
Some time after Bill and Bob, and before Tim and Tom do their experiment.
Is that important, and is that what I am failing to grasp?
Thanks again.

Nugatory
Mentor
Some time after Bill and Bob, and before Tim and Tom do their experiment.
Is that important, and is that what I am failing to grasp?
You have forgotten the relativity of simultaneity, and that's why the "when?" question has teeth. If they both step from one world to the other at the same time (using one world's definition of "the same time") then it's not at the same time using the other world's definition of "at the same time".

OK,I am being thick here.
Some time after Bob has found out that Bill has 'lost' a minute, the relevant population on 'Normal' are scooped up and land on 'High-speed'. Bill is no longer of any concern.
Some time after this event, hours, days, weeks possibly, Tim and Tom perform their experiment. The only relevance I can see from Bill and Bob's experiment is that four minutes on High-speed will occupy five minutes on Normal.
What is it that I am missing?

Mister T
Gold Member
Next, their entire world gets bumped onto High-speed, which gets renamed 'Home', and Normal is renamed 'Northbound'.
You imagine, I suppose, that every piece of their entire world gets bumped onto High-speed at the same time? But at the same time according to which world? It can't be both! Simultaneity is relative.

Still struggling a bit here. But, if a large arm comes out of High-speed, and catches everyone required, especially Tim and Tom, and takes them onto High-speed.
They will then be together on High-speed, in a fairly small space. Their watches should still be ticking at the same rate, and they could reset the time to be twelve o'clock, though I don't see why that would be necessary.
Some time after that, Tim can step back onto Normal, which has been renamed Northbound.
What am I failing to grasp?

It sounds to me like you see "high speed" as being absolute, as in, it's always high speed compared to the other.

But for anyone on "high speed" (later re-named "Home"), it is "normal" (later re-named "Northbound") that is "moving fast" compared to them.

Ibix
I'm a bit confused by your description @Gordie19, and I think others may be too. If I understand right, you are simply running your experiment twice, once with everybody starting at 0.6c and one dropping to zero for four minutes, and once with everybody starting ar zero and ine accelerating to 0.6c for four minutes. Is that correct?

If so, you are correct that the exact mechanics of how everyone gets from one planet to another don't matter. If you attempt to check the difference between Bill and Tim's watches then it does matter, but you don't seem to be doing that.

The answer is that the outcome of the experiments is the same. The twin who jumped to the other world (whichever world you are calling "other" at the time) shows four minutes and the one who didn't shows five. This is easy to see: remember that the theory is called relativity. Either world can regard itself as at rest, so the experiments must come out the same since their descriptions are identical - in each case one twin accelerates to 0.6c as measured by the stay-at-home, stays at that speed for four minutes by his own watch, then decelerates to rest again.

Where you are getting confused is that you are forgetting the mechanics of comparing watches with someone a long way away. They have to exchange light signals and correct for the travel time. But different frames will disagree on the travel time, because they will disagree about how fast the twins are moving and in which direction - and this disagreement added to disagreement about whose watch was ticking slowly leads to agreement about what the twins determine their offset to be.

As I noted in my last post, other frames will not agree the offset. In particular, the frame of the "other" world (whichever is the other at the time) will measure an offset of 48 s, as you appear to be thinking.

Thanks again Ibix.
Please let me sleep on this (in Australia, we sleep while everyone else is up and active).
I think you have almost understood me, but I still am confused (don't worry, by the time I am finished, you will be tearing your hair out) by your reply, as indeed you are by mine.
I hope I may address my ongoing queries to you - everyone else should be grateful you are the only one who will be a gibbering wreck when I am finished.

PeroK
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OK,I am being thick here.
Some time after Bob has found out that Bill has 'lost' a minute, the relevant population on 'Normal' are scooped up and land on 'High-speed'. Bill is no longer of any concern.
Some time after this event, hours, days, weeks possibly, Tim and Tom perform their experiment. The only relevance I can see from Bill and Bob's experiment is that four minutes on High-speed will occupy five minutes on Normal.
What is it that I am missing?
There have been many people like you on this site who try to use the twin paradox to understand SR. What you're missing (IMO) is that you must learn SR before you can understand the twin paradox.

If you want to learn SR you can safely leave the twin paradox until you have the conceptual and mathematical basis nailed down. And, then, the twin paradox won't look much like a paradox at all.

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I'm a bit confused by your description @Gordie19, and I think others may be too. If I understand right, you are simply running your experiment twice, once with everybody starting at 0.6c and one dropping to zero for four minutes, and once with everybody starting ar zero and ine accelerating to 0.6c for four minutes. Is that correct?

If so, you are correct that the exact mechanics of how everyone gets from one planet to another don't matter. If you attempt to check the difference between Bill and Tim's watches then it does matter, but you don't seem to be doing that.

The answer is that the outcome of the experiments is the same. The twin who jumped to the other world (whichever world you are calling "other" at the time) shows four minutes and the one who didn't shows five. This is easy to see: remember that the theory is called relativity. Either world can regard itself as at rest, so the experiments must come out the same since their descriptions are identical - in each case one twin accelerates to 0.6c as measured by the stay-at-home, stays at that speed for four minutes by his own watch, then decelerates to rest again.

Where you are getting confused is that you are forgetting the mechanics of comparing watches with someone a long way away. They have to exchange light signals and correct for the travel time. But different frames will disagree on the travel time, because they will disagree about how fast the twins are moving and in which direction - and this disagreement added to disagreement about whose watch was ticking slowly leads to agreement about what the twins determine their offset to be.

As I noted in my last post, other frames will not agree the offset. In particular, the frame of the "other" world (whichever is the other at the time) will measure an offset of 48 s, as you appear to be thinking.
You have been most helpful again Ibix, actually I think I am running the experiment three times; the first part establishes the ratio of lapsing time for those involved, with one of them going from zero to 0.6. The second part involves a number of them, going from zero to 0.6, and third one going from 0.6 to zero.

Please allow me to summarise (again), as you may be able to straighten me out. I understand that different frames will disagree about travel time, so I am trying to limit it to just the two reference frames, simulated by the planets (H & N).
Bill and Bob perform their experiment, and establish that four minutes lapse on High-speed/Home for Bill while five minutes lapse for Bob. This implies that if Bill had spent three minutes twelve seconds on H, four minutes would have elapsed for Bob on N.
Bob can inform Tim and Tom what he has found out, then Tim and Tom step onto H together.
Tim and Tom are then on H and ready to perform their experiment, in which Tim steps onto N for four minutes on his watch, then steps back onto H.
What I am unable to grasp is why five minutes have elapsed on Tom's watch rather than three minutes twelve seconds. Am I wrong in thinking that while Tim is on N, his watch ticking rate will be the same as Bob's? That is while the time displayed on their watches will be wildly different, won't ten seconds on Bob's watch be ten seconds on Tim's? They are on the same planet/reference frame.

P2 - once Bill has played his part in the first experiment, he has no subsequent role
P3 - that is my puzzle, hopefully explained better above
P4 - I understand that different observers disagree about travel time, I have also done some maths
which indicates all observers and participants agree, if the traveller does an out-and-back trip
P5 - not sure I quite understand that one.

Once again, thank you for your patience and taking the time to try to explain for my old and weary brain.

Nugatory
Mentor
Bill and Bob perform their experiment, and establish that four minutes lapse on High-speed/Home for Bill while five minutes lapse for Bob. This implies that if Bill had spent three minutes twelve seconds on H, four minutes would have elapsed for Bob on N.
That’s where you’re going wrong. Time dilation is symmetrical, so that if four minutes pass on Highspeed while five minutes pass on Home then it is also true that five minutes pass on Highspeed while four minutes pass on Home.

This sounds completely paradoxical and impossible, but when you consider the relativity of simultaneity it turns out to make sense. Look at https://www.physicsforums.com/threads/special-relativity-time-issues.846242/post-5307390 and https://www.physicsforums.com/threads/weird-time-dilation-question.883303/post-5552719 for an explanation.

The Bill/Bob measurement is establishing something else, namely the distance along the two different paths to two different points that the two twins follow through spacetime. These are paths through spacetime, so we measure their lengths with a clock instead of a ruler, and when we do the math (drawing a spacetime diagram is very helpful here) we find that one path is four minutes long and the other is five minutes long.

Ibix
P5 - not sure I quite understand that one.
And that's your problem (along with 99% of newcomers to relativity). Time dilation and length contraction are fairly well publicised effects of relativity, but there's a third (arguably more important) effect called the relativity of simultaneity. What it means is that different frames do not agree on what "at the same time" means except for things that happen in the same place as well as at the same time. That bites you here, because after travelling, the twins aren't in the same place, and what they want to ask is "at the same time as his watch reads 4 minutes, what does mine read?" If you use the definition of "at the same time" that is natural for the stay-at-home's planet (whichever that is) then the answer will be five minutes. If you use the other planet's definition, the answer will be three minutes twelve seconds. In either case, you can work out what the other planet's definition would tell them, even if you don't agree with it.

That's the resolution to your problem. If you study both of the experiments using the frame of one planet, then in one case you will say that five minutes elapsed and in the other that three minutes twelve second elapsed (but you can deduce that the twins would say five minutes).

Einstein's train thought experiment is the usual way to introduce the concept of relativity of simultaneity. Einstein imagines two lightning flashes striking opposite ends of a train and deduces that if a passenger on the train would say that the strikes occurred simultaneously then another standing on a station platform cannot say the same. I'd suggest looking up that thought experiment.

And that's your problem (along with 99% of newcomers to relativity). Time dilation and length contraction are fairly well publicised effects of relativity, but there's a third (arguably more important) effect called the relativity of simultaneity. What it means is that different frames do not agree on what "at the same time" means except for things that happen in the same place as well as at the same time. That bites you here, because after travelling, the twins aren't in the same place, and what they want to ask is "at the same time as his watch reads 4 minutes, what does mine read?" If you use the definition of "at the same time" that is natural for the stay-at-home's planet (whichever that is) then the answer will be five minutes. If you use the other planet's definition, the answer will be three minutes twelve seconds. In either case, you can work out what the other planet's definition would tell them, even if you don't agree with it.

That's the resolution to your problem. If you study both of the experiments using the frame of one planet, then in one case you will say that five minutes elapsed and in the other that three minutes twelve second elapsed (but you can deduce that the twins would say five minutes).

Einstein's train thought experiment is the usual way to introduce the concept of relativity of simultaneity. Einstein imagines two lightning flashes striking opposite ends of a train and deduces that if a passenger on the train would say that the strikes occurred simultaneously then another standing on a station platform cannot say the same. I'd suggest looking up that thought experiment.
This will be my last epistle on this subject, as I am clearly unable to follow your logic. I can understand that different frames do not agree about what "at the same time" means, however, I was not suggesting any measuring being made across frames. In each case, the traveller had stepped back into the same frame as the 'stay-at-home' before measuring. Then if they can measure how far away they are from each other, and the speed of light is constant, each actor should be able to signal to the other at a specific time, and the other calculate what that difference is. The measurement should be the same for both.
I should actually expect that the difference would be neither 'gain one minute' or 'lose three minutes twelve seconds', but be somewhere in between. That would allow for consistency with all measurements.
Anyhow, thank you for taking the time and trouble to try and explain it, though your explanation has fallen on stony ground.

jbriggs444
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In each case, the traveller had stepped back into the same frame as the 'stay-at-home' before measuring.
A frame is not something you step into. It is scribbles on a piece of paper -- a convention you adopt. It has nothing to do with the ground on which you stand.

PeterDonis
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In each case, the traveller had stepped back into the same frame as the 'stay-at-home' before measuring.
I don't think this is at all clear from your descriptions of the scenarios. It also doesn't help that you piled on two more scenarios before you had resolved your questions with the first one.

The best technique when verbal descriptions of a relativity problem seem to be confusing is to switch to math, meaning writing down explicitly labeled events and their coordinates. I will give such a description of your original scenario (the one in your OP). You should then try to write down similar descriptions of your other scenarios.

You say you want to analyze everything using just one frame, which is apparently the frame you call "Home" (but it seems like you switch which frame that name refers to from scenario to scenario, which also doesn't help--I'll stick to the way you used that term in your OP). So let's write down the key events from the scenario in your OP, with their coordinates in the Home frame.

Event O: Tim steps onto Northbound from Home. Tom stays at Home. Coordinates $(x, t) = (0, 0)$.

Event T: Tim steps back onto Home from Northbound. Coordinates $(x, t) = (3, 5)$. (Tim travels for four minutes by his watch, which with a gamma factor of 1.25 means 5 minutes according to clocks at rest in Home; he travels at speed 3/5 so he travels 3 light minutes in 5 minutes. We are using units where $c = 1$ so the $x$ coordinates are in light-minutes and the $t$ coordinates are in minutes.)

Event L: Tim sends a light signal to Tom. This happens at 3 o'clock by Tim's clock, which is 1 minute behind Home clocks (Tim's clock reads 4 at time $t = 5$, the time of event T), so if we say $t = 0$ is noon by Home clocks, the coordinates of event L will be $(x, t) = (3, 181)$ (3 hours and 1 minute after noon by Home clocks, at Tim's location).

Event R: Tom receives the light signal from Tim. Coordinates $(x, t) = (0, 184)$. (Can you see how I figured that out?) If Tom knows that Tim send the signal at $t = 181$ (presumably the signal contains some kind of data that gives that information), Tom knows that Tim is located at $x = 3$. If he also knows that Tim's watch read 180 minutes (3 hours) when the signal was sent (again, the signal could contain data that gives this information), he knows that Tim's watch is 1 minute behind Home watches (since Tim's watch read 180 at $t = 181$).

PeterDonis
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You should then try to write down similar descriptions of your other scenarios.
I won't write these additional descriptions down explicitly at this point, but here are a few items to ponder:

(1) You renamed "High-speed" to "Home" in your second scenario (the one in post #5). You said this "Home" is moving relative to "Normal" at 3/5 c, but you didn't say in which direction (relative to the direction that Northbound is moving relative to Home). If we keep everything relative to "Home" (renamed from "High-speed"), then "Normal" is moving at minus 3/5 c (opposite direction, same speed), but again, we need to know which direction this is relative to the direction that Northbound is moving. Without that information we can't analyze the scenario since it is not well-defined.

(2) You have Bob and Bill doing their experiment, where (from the point of view of Home/High-speed) they are both moving, then Bill stops for 4 minutes by his watch (which, since during this time he is at rest relative to Home, will be 4 minutes by Home clocks too), then Bill starts moving again. Then, some time later, Bill stops once more (presumably along with Tim and Tom? Or are they supposed to always be at rest relative to Home? this is another thing you left out). Then we are supposed to ignore Bob for the rest of the scenario. But that means everything that Bill and Bob did is completely irrelevant to what Tim and Tom do. You can still analyze the Tim and Tom scenario exactly the same way, because that scenario only depends on the relative motion of Home and Northbound and how Tim and Tom move relative to Home; how something else called "Normal" or whatever is moving doesn't affect Tim and Tom at all. (Note that you don't appear to grasp that in post #5, since you seem to think the previous events somehow change how Tom's clock behaves--which they won't.) So why are Bill and Bob even there? What is the point of this whole scenario?

(3) Later on you talk about other people being "scooped up" from Normal to Home/High-speed, but again nothing that happens to them affects Tim and Tom at all, so I don't see what the point is of including them.

I won't write these additional descriptions down explicitly at this point, but here are a few items to ponder:

(1) You renamed "High-speed" to "Home" in your second scenario (the one in post #5). You said this "Home" is moving relative to "Normal" at 3/5 c, but you didn't say in which direction (relative to the direction that Northbound is moving relative to Home). If we keep everything relative to "Home" (renamed from "High-speed"), then "Normal" is moving at minus 3/5 c (opposite direction, same speed), but again, we need to know which direction this is relative to the direction that Northbound is moving. Without that information we can't analyze the scenario since it is not well-defined.

(2) You have Bob and Bill doing their experiment, where (from the point of view of Home/High-speed) they are both moving, then Bill stops for 4 minutes by his watch (which, since during this time he is at rest relative to Home, will be 4 minutes by Home clocks too), then Bill starts moving again. Then, some time later, Bill stops once more (presumably along with Tim and Tom? Or are they supposed to always be at rest relative to Home? this is another thing you left out). Then we are supposed to ignore Bob for the rest of the scenario. But that means everything that Bill and Bob did is completely irrelevant to what Tim and Tom do. You can still analyze the Tim and Tom scenario exactly the same way, because that scenario only depends on the relative motion of Home and Northbound and how Tim and Tom move relative to Home; how something else called "Normal" or whatever is moving doesn't affect Tim and Tom at all. (Note that you don't appear to grasp that in post #5, since you seem to think the previous events somehow change how Tom's clock behaves--which they won't.) So why are Bill and Bob even there? What is the point of this whole scenario?

(3) Later on you talk about other people being "scooped up" from Normal to Home/High-speed, but again nothing that happens to them affects Tim and Tom at all, so I don't see what the point is of including them.
Thank you Peter, I did say to Ibix earlier that I was giving up, as I clearly do not and cannot understand the concept of relativity. Also, I am clearly not expressing myself very well, but since you have been kind enough to try to help, I shall make this my final attempt, and if it makes you think about my mental weakness, that will do nicely. If some or all of it is irrelevant, please accept my apologies.

To summarise the entire procedure:
1. A group of 'actors' occupy part of a long, flat planet, 'N'. They each have a watch and all watches are synchronised by time displayed and 'tick rate' 10 seconds on one = 10 seconds on all
2. One of the actors, Bill steps onto the adjacent planet 'H', which is passing at 3c/5 = can't see why the direction matters
3. After four minutes on his watch, Bill steps back onto N, and sets up signals to allow the other actors to measure how far away Bill is (note they are all in the same reference frame)
4. Actor Bob measures how far away Bill is, reads the time displayed on Bill's watch (say 15:00) and calculates the time on his own watch when the light left Bill's watch showing 15:00. I know this involves simultaneity at a distance, but cannot see a fault in the logic. According to others on PhysicsForums, this should be 15:01, four minutes, five minutes.
5. Bob tells Tim and Tom to hold hands and step onto H.
6. Some time after they have stopped wobbling (hours/days) Time steps onto N and waits for four minutes to pass on his watch.
7. Tim steps onto H, and Tim and Tom perform the same calculation process as Bill and Bob did earlier. Tom does his calculation based on Tim's watch showing 16:00.
I am unable to see how Tom's calculation will show anything other than 15:59:12. Otherwise, consistency is gone.

I am happy if you want to let me know where my stupidity lies. I struggled through three years of maths at university, but that was 50 years ago, I was never much good, but I am excellent at logic, so if you choose to reply, please use logic, not mathematics.
Thanks again

PeterDonis
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To summarise the entire procedure
You don't need to summarize the scenario again. You need to do what I suggested: write down actual coordinates, in whatever frame you choose, for the events. Verbal descriptions are not helping, so you need to use actual math.