I Simultaneity and the Twin Paradox

[PeterDonis]

On the mapped video, A's 10-year frame gets mapped to the 6 year position. A's 1-year frame gets mapped to the -9 position. Cut this off at the 3 year mark (where A reads 8.2).

And doing this maps the frame taken when C's clock reads 6 years to the 6-year position, and the frame taken when C's clock reads 3 years to the -9 position. So the frame at the 3-year position will be the frame taken when C's clock reads 5.4 years. In other words, this portion of the video will have C's clock ticking off 0.6 years while A's clock ticks off 1.8 years, for a Doppler factor of 3, as I said. The fact that the video takes 3 years of "play time" to play doesn't change any of that.

I'm confident that if you could adjust for light delay on B's outgoing trip, you could adjust for light delay on C's incoming trip.

You can always calculate coordinate times and positions by adjusting for light delay from the raw data, yes. But you don't do that by taking raw video frames and just picking out some subset of them and adjusting the playback rate. The relationship between the Doppler shift in the raw data and the time dilation factor in the calculated coordinates is not that simple.

since B didn't get enough video to fill out its 3-year segment

I don't understand what you mean by this. B's raw video, which shows A's clock going from 0 to one year, is captured during B's entire 3-year journey--i.e., the frames showing A's clock going from 0 to one year are captured over a range of B clock readings from 0 to 3 years. There is nothing missing.

A receives B's video as delivered by C in a nicely wrapped package. What I meant is that A cannot receive video recorded by B after B's meeting with C

Ah, ok.

 

[Freixas]

And doing this maps the frame taken when C's clock reads 6 years to the 6-year position, and the frame taken when C's clock reads 3 years to the -9 position. So the frame at the 3-year position will be the frame taken when C's clock reads 5.4 years. In other words, this portion of the video will have C's clock ticking off 0.6 years while A's clock ticks off 1.8 years, for a Doppler factor of 3, as I said. The fact that the video takes 3 years of "play time" to play doesn't change any of that.

Did I get the backwards? Sorry, I do that a lot. Let me try again. A's 10 year frame gets mapped to the 6 year position. A's 7.6 year frame gets mapped to the 3 year position. Since the original, long-forgotten intent of the video was to show what happens during C's leg of the trip if C could instantaneously see A's clock, that's as far back as we need to go.

For B's leg, we only get the first 5/3's of the mapped video. If we wanted to do it the easy way, B would continue mapping frames until a mapped frame landed at the 3-year mark and then would then ship the entire mapped 3 years back to A (via UPS Interstellar). If we wanted to do it the hard way, we would use the video intercepted by C (but sent during B's outbound leg) to do the same mapping. You'd know if you did it right (for some meaning of "right") if B's 3-year section ended with A's clock reading of 2.4.

Since both B and C map a frame to the 3-year mark (the way I've defined it), we would have a "double-exposure" with A's clock reading both 2.4 and 7.6 years.

Again, while the recorded video is 6 years and the mapped video is 6 years, the two are different and I get the feeling they are being confused. The mapped video is trying to capture (from B's or C's perspective) what is happening at A at any point during their trip. While there will be disagreement about whether two events are simultaneous or not when viewed by different observers, there are never any disagreements when viewed by just one observer.

By (properly) correcting for light delay, B (or C) can view what happened at A simultaneous with any point on B's (or C's) journey. We already know the answer, but I find the approach of correcting for light delay appealing. Also, if the answer doesn't come out as expected, it can reveal a lack of understanding about how to properly calculate light delay adjustments.

 

[Herman Trivilino]

For instance, John D. Norton has some of the best explanations of relativity I've seen that avoid math. When he explains the simple scenarios, it all makes sense, but when I try to apply them, it's not so easy.

That's because you have to learn the math to be able to do what he's doing.

 

[PeterDonis]

A's 10 year frame gets mapped to the 6 year position. A's 7.6 year frame gets mapped to the 3 year position.

First, by the implicit method you are using here, it's A's 8.2 year frame (1.8 years before the 10 year frame, which happens when C meets A).

Second, if by "the 3 year position" you mean 3 years in "video play time", that's fine. But the video frame in which A's clock reads 8.2 years is not received by C when C's clock reads 3 years. It is received by C when C's clock reads 5.4 years.

Since the original, long-forgotten intent of the video was to show what happens during C's leg of the trip if C could instantaneously see A's clock

Once more: "instantaneously" is frame dependent. There is no absolute meaning to "instantaneously". Even the implicit definition you are using here, which is "instantaneously" in the inertial frame in which C is at rest, does not generalize to any situation where either C has nonzero proper acceleration, or spacetime is curved (i.e., any gravitating masses are present), because in either of those situations, there is no global inertial frame in which C is at rest.

For B's leg, we only get the first 5/3's of the mapped video.

No, you don't. The frame in which A's clock reads 1 year is the frame received by B when B meets M, i.e., when B's clock reads 3 years, not 5/3 years. The fact that you have jiggered the "video play time" of this segment of video so it only takes 5/3 years to play doesn't change that.

If we wanted to do it the easy way, B would continue mapping frames until a mapped frame landed at the 3-year mark

B already has a frame at his 3-year mark: the frame received when he meets M. That frame is included in the video that C brings back with him and gives to A when they meet. What you are saying here makes no sense.

Since both B and C map a frame to the 3-year mark (the way I've defined it), we would have a "double-exposure" with A's clock reading both 2.4 and 7.6 years.

Even with the implicit frames you are using, this is wrong. At the event on A's worldline which is simultaneous, in B's frame, with B meeting M (and C), A's clock reads 1.8 years, not 2.4. And as above, at the event on A's worldline which is simultaneous, in C's frame, with C meeting M (and B), A's clock reads 8.2 years, not 7.6.

The mapped video is trying to capture (from B's or C's perspective) what is happening at A at any point during their trip.

And even by that criterion, it is failing, because B and C have different simultaneity conventions and so they do not agree on what is happening at A "at the same time" as a point on their respective trips.

While there will be disagreement about whether two events are simultaneous or not when viewed by different observers, there are never any disagreements when viewed by just one observer.

So who is the one observer? You have two, B and C, and their views, even when corrected as you describe, are not compatible.

By (properly) correcting for light delay, B (or C) can view what happened at A simultaneous with any point on B's (or C's) journey.

If you really think this is what you are doing, I can only say that I think you are greatly mistaken. What you are trying to construct makes no sense to me, and does not convey any useful information as far as I can see. It is just a way of deepening your confusion. I'm sorry if that sounds harsh, but that's the way I see it.

We already know the answer

No, we do not, because there is no unique answer. I think you should read that sentence again and again until it sinks in.

 

[PeterDonis]

B (or C) can view what happened at A simultaneous with any point on B's (or C's) journey.

Here is another reason why what you are trying to do here makes no sense. When C meets up with A and delivers the final video, C's clock reads 6 years and A's clock reads 10 years. There's no way around that: that is an invariant fact which is independent of any choice of frames or any simultaneity convention or any jiggering about with videos. So any video that is going to tell a true and useful story about "what is happening at A while B and C are traveling" has to end up showing A's clock ticking off 10 years while the B/C clock (it's effectively one clock since C sets his clock to B's clock reading when they pass) ticks off 6 years. But you are trying to construct a video in which A's clock ticks slower than the B/C clock, as if that somehow "truly" represents "what is happening at A". That can't possibly be correct.

 

[PeroK]

While there will be disagreement about whether two events are simultaneous or not when viewed by different observers, there are never any disagreements when viewed by just one observer.

That's only true if you are careful to define "observer", "simultaneous" and "reference frame". In practical terms, it may not be at all clear to one observer whether two events are simultaneous in their frame. For example:

https://www.physicsforums.com/threa...ological-order-of-events.946250/#post-5988680

 

[MikeGomez]

My favorite version of the twin paradox involves no acceleration.

I've read articles and watched videos that have reinforced that the fact that one twin experiences acceleration is not relevant.

No acceleration would mean no age differential.

The twin paradox is a thought experiment involving two participants. One twin accelerates and the other does not. The traveling twin accelerates away from the stay at home twin, then re accelerates to switch from outbound to inbound direction, then accelerates again (deceleration) to come to a stop upon returning. Trying to change the scenario by removing acceleration of the traveling twin is not only nonphysical, but adds unnecessary confusion instead of helping.

Introducing a third participant on the inbound leg of the journey in order to record the traveling twins clocks is not a valid step in this thought experiment. The paradox is based on one twin returning to Earth to find his/her twin has aged much more, and that can’t happen if the traveling twin perpetually tours the cosmos instead of returning home. Even more problematic is that introducing the non-accelerating inbound traveler is exactly equivalent to the inertial-less drive, and I agree with Dale that this is not physical.

I think it is much simpler to stop adding more than two participants , and stop trying to invent non-physical solutions like removing acceleration.

This link does a pretty good job ( read up until the Asteroids section)

 

[PeroK]

No acceleration would mean no age differential.

The twin paradox is a thought experiment involving two participants. One twin accelerates and the other does not. The traveling twin accelerates away from the stay at home twin, then re accelerates to switch from outbound to inbound direction, then accelerates again (deceleration) to come to a stop upon returning. Trying to change the scenario by removing acceleration of the traveling twin is not only nonphysical, but adds unnecessary confusion instead of helping.

Introducing a third participant on the inbound leg of the journey in order to record the traveling twins clocks is not a valid step in this thought experiment. The paradox is based on one twin returning to Earth to find his/her twin has aged much more, and that can’t happen if the traveling twin perpetually tours the cosmos instead of returning home. Even more problematic is that introducing the non-accelerating inbound traveler is exactly equivalent to the inertial-less drive, and I agree with Dale that this is not physical.

I think it is much simpler to stop adding more than two participants , and stop trying to invent non-physical solutions like removing acceleration.

I don't agree with any of that. Understanding relativity is about understanding the nature of spacetime. And, in particular, spacetime distances (proper time) along any timelike worldline. That understanding allows one to handle problems where one, both or neither worldline involves acceleration. The acceleration/no acceleration view is a black and white view that only covers a few simple cases and cannot be generalised.

Putting acceleration at the heart of spacetime physics is a false step.

 

[Ibix]

Introducing a third participant on the inbound leg of the journey in order to record the traveling twins clocks is not a valid step in this thought experiment.

It's perfectly fine. The outbound leg is done by one person, who reports their clock reading to the (different) inbound traveller when they meet. The inbound traveller starts their clock with that setting, and then compares the final time to the stay-at home when they meet. The point is to compare elapsed times along different paths, not get bogged down in the details of how the paths are defined.

You personally may find this more confusing, but there's nothing wrong with it. A triangle drawn with three straight strokes of a pen is just as much a triangle as one drawn with one straight stroke and one > shaped stroke. The geometry is exactly the same.

stop trying to invent non-physical solutions

Aside from the bit with the inertialess drive, which is indeed non-physical, acceleration-free versions of the twin paradox are fine. They're just slightly mis-named because you need three players.

 

[Dale]

The twin paradox is a thought experiment involving two participants. One twin accelerates and the other does not.

It's perfectly fine.

I have to admit that my feeling is somewhat in between. The triplet version is physically and mathematically fine, but in my opinion is loses all of the paradoxical character.

The “paradox” part is not a logical contradiction, but a common misunderstanding produced by the usual pedagogy. Students learn that all frames are equivalent (instead of learning that all inertial frames are equivalent), so they see the twins as symmetric. I don’t know of anyone who sees one frame as symmetric with two frames.

So the triplet version is mathematically and physically fine, but IMO it misses the point since it loses even the false symmetry of the traditional twin paradox.

 

[PeroK]

I have to admit that my feeling is somewhat in between. The triplet version is physically and mathematically fine, but in my opinion is loses all of the paradoxical character.

The purpose of logical, physical or mathematically analysis is to cut through the false paradoxical character. Losing the paradoxical character is precisely the point.

 

[Ibix]

So the triplet version is mathematically and physically fine, but IMO it misses the point since it loses even the false symmetry of the traditional twin paradox.

I see what you're getting at. The "paradox" is supposed to make the student think about their own (mis)conceptions about time in relativity, and the single traveling twin is harder to dismiss as some physics professor trick question than the two clock relay.

But I think the non-accelerating version is very useful in understanding that it's not the acceleration that's the problem - it's carelessness when stitching reference frames. It shows that you can have exactly the same situation with no acceleration.

 

[Nugatory]

No acceleration would mean no age differential.

It's possible to set up a twin paradox situation with no acceleration: do both clock comparisons as the twins pass one another, use a hyperbolic orbit around a distant star to do the turnaround. It's still a special relativity problem (more precisely, the general relativistic corrections are negligible) if the trip is long enough. This version has the intriguing property that if the two spaceships have no windows or other way of checking one's relative velocity, both twins will have completely symmetrical experiences: they synchronize their wristwatches while floating in freefall in a windowless box; nothing interesting happens for a while; then they compare wristwatches and find less time has passed for one than the other.

While the traditional version of the twin paradox works by losing the distinction between "frame" and "inertial frame", this one works by losing the distinction between "inertial frame" and "local inertial frame".

The main pedagogical value is that it shows that the acceleration in the traditional version is something of a red herring. The essential thing is the proper time along a path through spacetime; acceleration is just an inevitable side effect of trying to construct a situation in which objects take different paths between the same two points in a flat spacetime.

 

[Dale]

The purpose of logical, physical or mathematically analysis is to cut through the false paradoxical character. Losing the paradoxical character is precisely the point.

I guess that I see the purpose as slightly different. To me, the student has internalized an incorrect rule "all reference frames are equivalent", and the purpose of the twin paradox is to replace that incorrect rule with the correct rule "all inertial reference frames are equivalent". Particularly with the usual observer-centric terminology, I don't think that the triplet paradox confronts the incorrect rule in most student's minds and I think that directly confronting the incorrect rule is essential for replacing it. I believe that a student who has internalized the incorrect rule will still retain that rule even after analyzing the triplet paradox.

 

[PeroK]

I guess that I see the purpose as slightly different. To me, the student has internalized an incorrect rule "all reference frames are equivalent", and the purpose of the twin paradox is to replace that incorrect rule with the correct rule "all inertial reference frames are equivalent". Particularly with the usual observer-centric terminology, I don't think that the triplet paradox confronts the incorrect rule in most student's minds and I think that directly confronting the incorrect rule is essential for replacing it. I believe that a student who has internalized the incorrect rule will still retain that rule even after analyzing the triplet paradox.

Yes, that's a good point. On the other hand, the issue with not confronting the acceleration is that:

The student has internalised the rule: "the twin who accelerated is really time dilated". And, therefore, "not all inertial reference frames are equivalent": in the sense that an inertial reference frame that results from a period of acceleration is fundamentally different from an inertial frame that (during the course of an experiment) is not the result of an acceleration phase.

If anything, I'd say this latter misconception is more difficult to shake.

 

[Nugatory]

If anything, I'd say this latter misconception is more difficult to shake.

Anecdotally, that's been my experience as well.

 

[Dale]

And, therefore, "not all inertial reference frames are equivalent": in the sense that an inertial reference frame that results from a period of acceleration is fundamentally different from an inertial frame that (during the course of an experiment) is not the result of an acceleration phase.

Yes, that is a good point. That is a particularly difficult one because it goes along with their earthbound preconceived notion that the reference frame of the ground is special, which is probably a notion that they have held most of their life so it is very "sticky".

I think that can be avoided by a careful presentation. The teacher needs to be explicit about the incorrect rule and the correct rule, using the twins paradox as an example. Also, the teacher should teach the difference between inertial and non-inertial frames before teaching relativity.

 

[PeterDonis]

No acceleration would mean no age differential.

More precisely, if both observers start out at rest with respect to each other, and neither one accelerates, there will be no age differential.

In flat spacetime, if the observers are in relative motion and neither one accelerates, "age differential" is not well-defined because they will only meet once.

In curved spacetime, it is possible for two observers in relative motion to both be in free fall and meet multiple times. And in general, in such cases, there will be an age differential.

 

[MikeGomez]

More precisely, if both observers start out at rest with respect to each other, and neither one accelerates, there will be no age differential.

In flat spacetime, if the observers are in relative motion and neither one accelerates, "age differential" is not well-defined because they will only meet once.

In curved spacetime, it is possible for two observers in relative motion to both be in free fall and meet multiple times. And in general, in such cases, there will be an age differential.

Yes.

It's possible to set up a twin paradox situation with no acceleration: do both clock comparisons as the twins pass one another, use a hyperbolic orbit around a distant star to do the turnaround. It's still a special relativity problem (more precisely, the general relativistic corrections are negligible) if the trip is long enough. This version has the intriguing property that if the two spaceships have no windows or other way of checking one's relative velocity, both twins will have completely symmetrical experiences: they synchronize their wristwatches while floating in freefall in a windowless box; nothing interesting happens for a while; then they compare wristwatches and find less time has passed for one than the other.

While the traditional version of the twin paradox works by losing the distinction between "frame" and "inertial frame", this one works by losing the distinction between "inertial frame" and "local inertial frame".

Just curious. Does this also lose the distinction between proper acceleration and coordinate acceleration?

The main pedagogical value is that it shows that the acceleration in the traditional version is something of a red herring. The essential thing is the proper time along a path through spacetime; acceleration is just an inevitable side effect of trying to construct a situation in which objects take different paths between the same two points in a flat spacetime.

I think of it in a contrary manner to that, i.e. the fact that two clocks in flat-space initially at rest with respect to each in an inertial frame requires acceleration (in one form or another) in order to achieve age differential, is a profound concept worthy of careful consideration, and not a red herring or just an inevitable side effect.

This seems like a bias for explaining the twin paradox purely in terms of time dilation due to uniform velocity. The 3-clock scenario accomplishes this. The stay at home twin synchronizes his clock with a passing outbound clock, an inbound clock is synchronized with the outbound clock some distance away, and the inbound clock is compared to the stay at home clock which shows more time has passed, i.e. age differential with no acceleration. Then piggybacking this 3-clock scenario to the twin paradox where the two twins start at rest relative to each other, it is said that the time dilation due to acceleration (gravitational time dilation) is negligible.

Although that may be true, I could have the opposite bias and instead of negligible time dilation due to acceleration, I could specify in the thought experiment negligible time dilation due to uniform relative velocity. The traveling twin could travel with great acceleration the entire trip, completely absent of any periods of coasting.

 

[PeroK]

Although that may be true, I could have the opposite bias and instead of negligible time dilation due to acceleration, I could specify in the thought experiment negligible time dilation due to uniform relative velocity. The traveling twin could travel with great acceleration the entire trip, completely absent of any periods of coasting.

In that case, all the time dilation is due to the relative velocity; it is not directly related to the magnitude of the acceleration.

You can have a scenario where the stay-at-home twin moves in a circle with the same magnitude of proper acceleration as the traveling twin throughout. If the circle is small and the velocity relative to the original inertial frame is small, then there would still be approximately the same differential ageing.

(Or, back and forward, mimicking the traveling twin's magnitude of acceleration, but never getting to a significant velocity relative to the Earth.)

Proper time is the length of a spacetime path. Period.

 

[Ibix]

Just curious. Does this also lose the distinction between proper acceleration and coordinate acceleration?

Of course not. It's just that there's more than one way to change direction, some involving proper acceleration and some not.

Edit: In fact, I'd go as far as to say that even asking this question is a hint that your "acceleration matters" approach is back-to-front.

two clocks in flat-space initially at rest with respect to each in an inertial frame requires acceleration (in one form or another) in order to achieve age differential

No, two clocks initially at rest require acceleration to achieve non-zero relative velocity. They require more acceleration to meet up again. But the age differential is a result of their paths, not their acceleration. Focussing on the acceleration is like focussing on the details of why I drew a kink in a line (which therefore crosses a straight line twice, forming a triangle) and trying to understand the triangle inequality from that.

Acceleration is important. Lines without corners do not cross more than once in flat spaces. But focusing on why they have a corner isn't really relevant to understanding why two sides together are longer than one.

This seems like a bias for explaining the twin paradox purely in terms of time dilation due to uniform velocity

This cannot be correct since you cannot explain the twin paradox in those terms. You need the relativity of simultaneity as well to resolve the paradox.

Although that may be true, I could have the opposite bias and instead of negligible time dilation due to acceleration, I could specify in the thought experiment negligible time dilation due to uniform relative velocity

How? The time dilation formula depends on velocity. How are you going to remove this dependence? You can stop having uniform relative velocity, sure, but that doesn't change anything about the argument.

By switching to constant acceleration you simply switch from a triangle to a smoother shape. You don't remove the differential ageing (due to the different intervals along the paths), you don't remove its dependence on relative velocity, and you don't remove the problems with the relativity of simultaneity. You just smear the latter two out over the whole path, making the problem intractable without calculus.

 

[MikeGomez]

Apparently I have confused being in different gravitational potentials with change of gravitational potential. Thanks for clearing that up.

 

[Freixas]

Wow, I leave for a while and a lot of interesting discussion has filled my thread!

I disagree with PeterDonis about the video mapping and the results I expect. I suspect Peter has some other mapping in mind, and we are talking at cross-purposes, but really, it’s not worth pursuing. Let me re-focus on a simpler thought experiment.

There are two very precise clocks: A and B.

Clock B is accelerated (relative to A) until it is moving at constant speed v toward A. As B passes by A, both clocks synchronize to 0 and begin counting. In terms of the Lorentz transform, this is where reference frames coincide exactly and (x, y, z, t) = (x’, y’, z’, t’) = (0, 0, 0, 0).

At time t on clock A, A fires a pulse of light at B. When B receives the pulse, it stops counting. We can now decelerate B to inspect the time on the clock.

I have two questions:

• If I run the experiment a hundred times using the same procedure, same t and same v, will B always display the same number? If yes, proceed to the next question. If no, stop here. A hint at why would be interesting.
• Given the parameters of the experiment as described here, the value of v, and the reading on clock B, would a competent physicist be able to calculate the time t at which clock A fired its pulse? This is not a theoretical number; it is the time clock A used to trigger the firing of the pulse. If yes, we’re done. If no, a hint as to why would again be nice.

The reverse of the second question might also be interesting. Given t and v, could one calculate the end time on B? Note: the difficulty of the calculation is immaterial, but it needs to be possible.

 

[PeterDonis]

If I run the experiment a hundred times using the same procedure, same t and same v, will B always display the same number?

Yes. (I'm not sure why a "no" answer would even be considered possible, since relativity is a deterministic theory.)

Given the parameters of the experiment as described here, the value of v, and the reading on clock B, would a competent physicist be able to calculate the time t at which clock A fired its pulse?

Yes.

Given t and v, could one calculate the end time on B?

Yes.

the difficulty of the calculation is immaterial, but it needs to be possible.

None of these calculations would be at all difficult for a competent physicist with knowledge of relativity. (I am assuming that the experiments are all to be done far out in empty space, far away from all gravitating bodies, so that spacetime can be assumed to be flat in the region used for the experiments.)

 

[PeterDonis]

I suspect Peter has some other mapping in mind, and we are talking at cross-purposes

It's possible that I'm misunderstanding the mapping you have in mind or what you are trying to do with it. My biggest concern with what I understand you to be trying to do with the video is what I said in post #55.

 

[Freixas]

It's possible that I'm misunderstanding the mapping you have in mind or what you are trying to do with it. My biggest concern with what I understand you to be trying to do with the video is what I said in post #55.

OK, from post #55:

But you are trying to construct a video in which A's clock ticks slower than the B/C clock, as if that somehow "truly" represents "what is happening at A". That can't possibly be correct.

And yet, that's exactly the bill of goods a lot of sites seem to be selling, including Michael Weiss on the web page that Nugatory pointed me to way back in post #2. Michael Weiss says: "Now different inertial reference frames have different notions of simultaneity. The Outbound reference frame says: "At the same time that Stella makes her turnaround, Terence's clock reads about two months." The Inbound reference frame says: "At the same time that Stella makes her turnaround, Terence's clock reads about 13 years and 10 months." The apparent "gap" is just an accounting error, caused by switching from one frame to another."

So Michael Weiss makes claims of simultaneity for each reference frame and he claims a very specific correspondence (from the viewpoint of that frame, of course). When I say the same thing, I get all sorts of crap about there being alternate ways to draw simultaneity planes or that simultaneity doesn't actually mean anything. Well, perhaps at some rarefied physics level, that's true, but at the level of the tutorials I'm reading, people seem totally willing to assign a precise meaning (and value) to simultaneity (again from a single reference frame).

The thought experiment I posted pokes at exactly that concept: if the clock experiment always gives the same answer and if t is calculable, then it implies a precise and real-world meaning of simultaneity that matches real-world numbers. There may be other curiosities for advanced physicists, but I don't need to go there right now.

 

[PeterDonis]

Michael Weiss makes claims of simultaneity for each reference frame

For each inertial frame, meaning the one in which the outbound traveler (B in your scenario) and the inbound traveler (C in your scenario) are at rest. In other words, he is using a particular specific definition of simultaneity, the one that goes with global inertial frames in flat spacetime.

But you'll notice what he doesn't say: he doesn't say that there is a single frame in which the traveler is always at rest. Nor does he say that any inertial frame's notion of simultaneity is absolute. Nor does he say that the "at the same time" calculations according to any inertial frame are the "true" representation of what is happening to the stay at home twin (Terence in his version, A in yours).

In fact, what Weiss is doing on that page is not telling you "how it really is". He is explaining the limitations of any viewpoint that treats the simultaneity of any specific inertial frame as absolute or "real". For example, that switching frames in mid-trip makes it seem like there's a "gap" in what is happening at A, when in fact the gap is just an artifact of switching frames.

When I say the same thing

But you're not saying the same thing as Weiss. That's the whole point. You are trying, or at least you certainly seem to be trying, to treat the simultaneity convention of a particular frame as absolute or "real". And it isn't. It's just a convention. If it were absolute or "real", then Weiss's dismissal of the apparent gap as "just an accounting error" would make no sense.

if the clock experiment always gives the same answer and if t is calculable, then it implies a precise and real-world meaning of simultaneity that matches real-world numbers.

It implies no such thing. Every quantity you talk about in that experiment is an invariant, and can be calculated without even using frames of reference or assuming any simultaneity convention at all. It says nothing at all about simultaneity and certainly does not give any "real-world meaning" of it.

 

[Ibix]

Given the parameters of the experiment as described here, the value of v, and the reading on clock B, would a competent physicist be able to calculate the time t at which clock A fired its pulse? This is not a theoretical number; it is the time clock A used to trigger the firing of the pulse.

There's a trap for the unwary here.

A does not do anything at time t. He can't, because coordinate time isn't a real thing, just a convention. What he actually does is send a signal when his ship's clock shows a time he interprets as meaning he's reached the event we've agreed to call t. We deliberately picked the coordinate time in A's rest frame to be the same as A's clock readings, but that doesn't mean that the coordinates and the clock readings are the same thing: A can always choose to accelerate, or maybe never zeroed his clock and just adds the appropriate offset whenever he has to talk to anyone else about time coordinates.

I can calculate the coordinate time, t, at which the light pulse was emitted. I can also calculate the coordinate time t' at which the pulse was emitted. Neither of these are what A uses to trigger his pulse. He uses his ship's clock, which measures his time but does not imply any simultaneity criterion anywhere else.

 

[Dale]

When I say the same thing, I get all sorts of crap about there being alternate ways to draw simultaneity planes or that simultaneity doesn't actually mean anything

If he posted here we would probably mention that simultaneity doesn’t mean anything, but the reason that we use him as a reference and you get all sorts of crap is that you didn’t say the same thing. He very carefully specified the frame and the quantity using standard language in a way to convey a unique meaning.

What happens with you and most relativity initiates is actually quite unavoidable. You don’t yet know enough to properly frame your question, so you include lots of irrelevant details and exclude some key relevant details. The result is that the question as written is ambiguous or unanswerable.

Respondents have two choices, either we can answer a different question, or we can try to get you to improve your question. If we answer a different question then there is the impression that we are being evasive. If we try to get you to improve your question then there is the impression that we are giving you crap.

I don’t know a way to avoid one of those outcomes.

 

[m4r35n357]

Maybe should have posted these videos earlier. Watch in HD, read the explanatory text, and watch all the clocks all the time. BTW Mr Tomkins is an early and misguided attempt to describe what you see, the linked videos include light delay.

 

[Freixas]

Respondents have two choices, either we can answer a different question, or we can try to get you to improve your question. If we answer a different question then there is the impression that we are being evasive. If we try to get you to improve your question then there is the impression that we are giving you crap.

I don’t know a way to avoid one of those outcomes.

Hey, Dale, my phrasing was poor and I actually didn't mean it in a mean way (to clarify: I didn't believe you guys were being mean). Don't worry about it—you can continue to give me crap. :-)

He very carefully specified the frame and the quantity using standard language in a way to convey a unique meaning.

Well, I tried to make sure my frames (inertial frames at that) were carefully specified. I'll admit to using simultaneous where Weiss uses at the same time. These sound equivalent to my ear.

The puzzle is that I can create what looks like an at-the-same-time correspondence between an event in A's and B's frame. B receives a video from A and calculates when the video was sent. The time the video was received and the calculated time it was sent are all from B's inertial frame of reference. The contents of the video establish what looks to me like a non-theoretical correspondence of B to A. I have never claimed that A would agree with any of this only that this is B's view of an at-the-same-time mapping.

Nor am I worried about a gap in the A, B, C scenario. B and C could both record the entire 10 year transmission from A. They would each have a mapping of the entire sequence and their mappings would be different, agreeing only when they meet at M. Splicing the videos together would clearly be artificial and only to show the point that we are changing reference frames.

Again, there might be some hair-splitting distinction as to whether a correspondence established by this method has any "real" meaning. If it is reproducible and repeatable, I can live with it. I am perfectly fine with the idea that someone in a different frame might completely disagree; I tried to make sure I always said that the correspondence was in one inertial frame and went one way only.

A does not do anything at time t. He can't, because coordinate time isn't a real thing, just a convention. What he actually does is send a signal when his ship's clock shows a time he interprets as meaning he's reached the event we've agreed to call t. We deliberately picked the coordinate time in A's rest frame to be the same as A's clock readings, but that doesn't mean that the coordinates and the clock readings are the same thing: A can always choose to accelerate, or maybe never zeroed his clock and just adds the appropriate offset whenever he has to talk to anyone else about time coordinates.

Interesting. Now my unwary response would that if A accelerates, we have stepped out of my experiment into a different one. And if clock time and coordinate time can't be said to have any real-world correspondence, then I would hate to planning the next space mission to some distant rock. I'm OK with the clock being "close enough" to the perfection of coordinate time. If the answer is within so many significant digits, I can live with it. Same for the at-the-same-time correspondence. This all seems like nit-picking: even if we can't build a perfect clock, it seems like a big jump to claim that there is no good-enough correspondence. And, in a thought-experiment, I can have a clock that perfectly matches coordinate time.

 

[Freixas]

Oh, I meant to add that , if it seems we are in an infinite-loop here, it's OK to close this thread. My feelings won't be hurt. There are some concepts that are best left for a face-to-face discussion and I do have that neighbor who is a retired physics instructor and a pretty nice guy (he's been out of town for a while, which is why I haven't brought him into the picture just yet).

I didn't want to end the discussion if someone else felt strongly about responding. I'll leave that up to you guys.

 

[Ibix]

And if clock time and coordinate time can't be said to have any real-world correspondence,

Coordinate time is the time on a set of imaginary clocks. You can easily work out what one of those imaginary clocks would read, but they're imaginary. You can't use them for anything.

It's like latitude and longitude coordinates on Earth. There are no actual lines you can use to navigate. But you can use your compass and your sextant and what have you to take measurements and work out which latitude and longitude line would be beneath your feet if there were any.

Coordinates are incredibly useful to plan and communicate. But they're not real things, even when the imaginary clocks we imagine specifying coordinate time happen to tick (or we imagine them ticking) at the same rate as your clock.

It's that distinction between the imaginary coordinate clocks and the real clocks we use to keep track of how we imagine the imaginary clocks to work that is the trap I was trying to warn of. The Lorentz transforms relate one set of imaginary clocks and rulers at rest with respect to each other to another set of imaginary clocks and rulers all moving at constant velocity with respect to the first lot. The relationship between those imaginary clocks and rulers and your own measurement apparatus is well defined, so you can work out the readings they give when a particular coordinate time is T. But you aren't reacting to the coordinates - you are reacting to your measurements.

So sure I can calculate t and t'. But neither of those things is (directly) what A used to time his signal.

 

[Freixas]

But you aren't reacting to the coordinates - you are reacting to your measurements.

Let me see if I can paraphrase some useful content out of this. I send a real person on a real spaceship to a real location, transmitting video the whole way. All of this is real, but when we take the video and adjust frames for light delay, we have shifted out of the real and into the imaginary. While I can appreciate that the resultant calculations might not be perfect, they won't exactly be random numbers nor will they lack some connection to the real world.

So your point is correct, but the distinction is not that interesting to an engineer. Close enough is good enough for me. :-)

 

[Ibix]

So your point is correct, but the distinction is not that interesting to an engineer. Close enough is good enough for me. :-)

You are completely missing the point. It is not about precision. I can build a clock that keeps time to whatever degree of precision I want. However, when I try to subtract out the light speed delay from looking at that clock I have to make a decision about which definition of distance and time I use - the one associated with this set of imaginary clocks, or the one associated with that set of imaginary clocks. Once I've made that decision I can make calculations with any degree of precision I like. That does not invalidate someone choosing the other set of clocks. There is no "correct choice". I can switch between choices using Lorentz transforms.

Choosing coordinates is like choosing how to draw a map. I can choose to have north pointing up the page, or off at 45°. That will change my opinion about what "at the same height on the page" means in practical terms. It has absolutely nothing to do with the precision of my measurements. Similarly, I can choose to have one set of clocks at rest and decide "simultaneous" means one thing, or choose to have a different set of clocks at rest and decide "simultaneous" means something else. It has nothing to do with the precision of my measurements.

There is no randomness here (or in practical terms, experimental error can be tightly controlled). There is an arbitrary decision about which plane in spacetime you are calling "now".

 

[Dale]

Hey, Dale, my phrasing was poor and I actually didn't mean it in a mean way (to clarify: I didn't believe you guys were being mean). Don't worry about it—you can continue to give me crap. :-)

I understand, I am just saying that your feeling is justified. We are in fact giving you crap, but I don’t know anything else we could do other than be evasive.

I'll admit to using simultaneous where Weiss uses at the same time. These sound equivalent to my ear.

Me too. Also, “instantaneous” would be the same.

You did describe A, B, and C’s frames well, and I don’t think that caused any crap to be given. Where we ran into ambiguities is in your description of the correction process. For instance, you said A would do the correction so I assumed he would use his frame, but you had intended him to use another frame. And in the description of the correction process you used a simultaneous word without specifying which frame.

The puzzle is that I can create what looks like an at-the-same-time correspondence between an event in A's and B's frame.

Events are not in a specific frame. They are in all frames. They merely have different coordinates in different times.

B receives a video from A and calculates when the video was sent. The time the video was received and the calculated time it was sent are all from B's inertial frame of reference. The contents of the video establish what looks to me like a non-theoretical correspondence of B to A.

Sure. The whole point of the Lorentz transform is that it describes that correspondence quantitatively.

Again, there might be some hair-splitting distinction as to whether a correspondence established by this method has any "real" meaning. If it is reproducible and repeatable, I can live with it. I am perfectly fine with the idea that someone in a different frame might completely disagree; I tried to make sure I always said that the correspondence was in one inertial frame and went one way only.

I am with you on this. I never worry about the “real” or not distinction. I simply identify it as frame-variant. If it is measurable, then I am fine with using it even if it is frame variant or require using some other convention.

 

[Freixas]

Events are not in a specific frame. They are in all frames. They merely have different coordinates in different times.

Sorry, more imprecise language. I agree with you. I'm afraid to re-phrase it, but may something like "an at-the-same-time correspondence between a time in which an event occurred in A's and occurred in B's frame and from B's point of view." This might still sound like the event occurred in A--I do mean that the event occurred, period. The time is the only thing that I am establishing a correspondence with.

 

[Dale]

The time is the only thing that I am establishing a correspondence with.

Yes, that is the purpose of the Lorentz transform.

 

[PeterDonis]

A does not do anything at time t. He can't, because coordinate time isn't a real thing, just a convention.

To be clear about the answers I gave, I interpreted $t$ to be the time on A's clock, not a coordinate time. The time on A's clock when he emits the light pulse towards B is the necessary input to the problem.

 

[PeterDonis]

I interpreted $t$ to be the time on A's clock, not a coordinate time.

And this appears to be what @Freixas intended, since his own problem statement says:

At time t on clock A, A fires a pulse of light at B.

 

[Herman Trivilino]

Let me see if I can paraphrase some useful content out of this. I send a real person on a real spaceship to a real location, transmitting video the whole way. All of this is real, but when we take the video and adjust frames for light delay, we have shifted out of the real and into the imaginary.

When you adjust for light travel time you are assuming that a clock located there (where the movie was made) is synchronized with a clock located here (where the movie is viewed). Others will simply not agree that the clocks are synchronized. It doesn't make anything less real, it just means that simultaneity is not absolute.

Consider the location of an accident involving the collision of two cars. That location can be described as, say, two blocks east of the courthouse, but it can also be described as two blocks west of Bob's Diner. Each description is valid, the fact that they differ doesn't make the collision any less real. Everybody who is familiar with this scheme of locating events will find nothing peculiar about it.

Likewise, everybody who is familiar with Einstein's relativity will find nothing peculiar about simultaneity being relative.

 

[Freixas]

When you adjust for light travel time you are assuming that a clock located there (where the movie was made) is synchronized with a clock located here (where the movie is viewed).

This is not my assumption at all; my problem statement long ago implies the opposite. The assembly of the final video is reference-frame-independent, in any case. As for viewing a video, as long as the viewer and the video player are in the same reference frame, you are fine. If the viewer and the player are not, the viewer won't see the show for very long, anyway.

 

[Freixas]

OK, I believe I figured out my error by working it through to a logical contradiction.

Let's review the A, B, C, M scenario: for A and M are 4 LY apart and in the same inertial frame. B travels from A to M at a constant 0.8c. A and B synchronize clocks at 0 when they pass. C also travels at a constant 0.8c, reaching M at the same time as B and continuing on to A. For B and C, the distance between A and M is 2.4 LY and the travel time is 3 years. For A (and M) B and C take 5 years to cover 4 LY. A sends a video of his clock to B and C.

What I added was B' who is in the same inertial frame as B but trails by 2.4 LY (B's view) or 4 LY (A's view). There is also a C' in the same inertial frame as C, but leading by 2.4 LY (or 4 LY).

When B reaches M, B' reaches A. My incorrect thinking was in believing the following were both true:

• Two observers are co-located but in different inertial frames. An event occurs at that location at time t for the first observer and t' for the second. They now know when the event occurred in their own inertial frame and the other's inertial frame.
• Two observers are in the same inertial frame. They consider two moments simultaneous if each of their synchronized clocks displays the same time.

The incorrect conclusion, then, is that if the co-located observer's t maps to t', then t maps to t' for all observers in the same inertial frame. Again, bitten by the relativity of simultaneity.

Let's examine this using B and B'. B reaches M, receives A's video clock showing 1yr. Correcting for light delay, B would say "I was 1 2/3 years into my journey when A sent this video frame." Continuing to gather video after passing point M, B eventually receives a video clock of 2.4 years. Correcting for light delay would put B at M when the signal was sent. B' would have been at A at the same time as B was at M, so would have seen the signal as it was sent. So B' would directly see A's clock showing 2.4 years.

The idea I had was that a light-delay correction was equivalent to a direct observation at the event's location. Running the same logic for C and C', one would conclude that C' saw A's clock showing 7.6 years at the same time and location as B' saw 2.4 years. This is impossible.

When B is at A, their clocks both read 0. For A, that means M's clock is also 0, but for B (if I did the math correctly) it reads 2.6 years. A thinks his clocks are synchronized, but B doesn't. B', 2.4 LY back, has the same problem with A's clock: it doesn't read 0, it reads 2.6 years. When B' reaches A, A's clock reads 5 years. When B reaches M, M's clock reads 5 years. Neither is surprised since they considered the clock ahead of them off by 2.6 years. They both think that A's clocks are running slow by a factor of 0.6.

I could run the numbers for A's view and C's view, but it's not important.

Sometimes things are clearer when one can figure out the problem one's own.

For what it's worth, a light-delay-corrected video will show an effect that appears to reflect the planes of simultaneity. In other words, A's clock will be seen to run slow by a factor of 0.6 and there will be a gap at turn-around. People seemed to have trouble with the whole video mapping concept, so let me see if I can clarify:

• Record the video as received.
• For B's recorded video, map each frame to a new time on the mapped video. Say frame n occurs at B's time t. We then place n at time f(t), where f() is the light-delay correction formula for B. We completely ignore the recorded value of A's clock on frame n when we do this mapping.
  
• When B is at A, f(0) = 0, so the frame (showing 0) is placed at time 0.
• When B is at M, f(3) = 1 2/3, so the frame (showing 1) is placed at time 1 2/3.
• B's entire 3 years of video is mapped to a new video 1 2/3 years long.

B transfers the video to C, who continues the recording.

• We need two mapping functions for C's leg: g() and h().
  
• g() maps a frame to the light-delay-corrected position of B. This is a tricky, of course. It fills in the mapped video from 1 2/3 to 3 years.
  
• h() maps a frame to the light-delay-corrected position of C.
  
• We use g() until g(t) = 3 (when B reached M).
  
• We use h() after g(t) = 3. However, when g(t) = 3, h(t) < 3. In other words, C will be processing frames it received before it reached M. It throws these away. When h(t) = 3, he starts filling the incoming leg of the video. The frame will show A's clock at 7.6. h(6) = 6, so that frame (showing 10) is placed at time 6.

I'll leave determining f(), g() and h() as an exercise for the reader. :-)

Let's say we have an infinite line of B' ships, so there's always one adjacent to A. Let's say the video frame rate is 30 frames/sec and that, on each 1/30th second mark, the B' closest to A captured one frame showing A's clock. This repeats until the lead B reaches M. Assemble the frames in sequence (remember that all B's clocks are synchronized). The video produced this way will show A's clock going from 0 to 5 years over a 3 year period. It is not equivalent to the light-delay-corrected video, as I originally thought.

 

[PeterDonis]

When B reaches M, B' reaches A.

In whose frame? You can't use the word "when" in this way without specifying a frame. It looks like you mean in the common rest frame of B and B', but you should make this explicit.

The incorrect conclusion, then, is that if the co-located observer's t maps to t', then t maps to t' for all observers in the same inertial frame. Again, bitten by the relativity of simultaneity.

Yes, you've got it.

The idea I had was that a light-delay correction was equivalent to a direct observation at the event's location. Running the same logic for C and C', one would conclude that C' saw A's clock showing 7.6 years at the same time and location as B' saw 2.4 years. This is impossible.

Yes, you've got it.

When B is at A, their clocks both read 0. For A, that means M's clock is also 0, but for B (if I did the math correctly) it reads 2.6 years.

You have the right idea here, but you need to check your math. I get that M's clock reads 3.2 years at the event on M's worldline which is simultaneous, in B's frame, with the event of B meeting A. (And M is 2.4 light-years from B, in B's frame, at this event, and is moving towards B at speed 0.8, so M will meet B in 3 years, in this frame. Btw, knowing that M's clock reads 5 years when M meets B, and that M's clock runs slow by the time dilation factor of 0.6 in B's frame, you can check that the 3.2 years that I just quoted is correct.)

 

[PeterDonis]

The video produced this way will show A's clock going from 0 to 5 years over a 3 year period.

No, it won't. It will show A's clock going from 0 to 1.8 years over a 3 year period. All of the B clocks are sychronized in the B rest frame, not the A rest frame, but you are thinking of simultaneity in the A rest frame.

 

[Freixas]

You have the right idea here, but you need to check your math. I get that M's clock reads 3.2 years at the event on M's worldline

Ah, yes, I forgot the dilation factor and I might have swapped time and distance. Let's see, for B, A's clocks run slower by 0.6. B travels to M in 3 years, [edit so 5. 3 * 0.6 = 1.8 years pass for M], so for 1.8 years to have passed and A's clock to read 5 at the end of that, the clock must have started at 3.2. Got it now.

 

[Freixas]

No, it won't. It will show A's clock going from 0 to 1.8 years over a 3 year period. All of the B clocks are sychronized in the B rest frame, not the A rest frame, but you are thinking of simultaneity in the A rest frame.

OK, let's say B1 is the lead B ship and BN is the last B ship. When B1 is at A, B1's clock reads 0 and A's clock reads 0, so we know the video starts at 0. When A meets B1, it would seem that B1 would view M's clock in the same way that BN views A's clock. So I thought BN would, on arrival, also view A's clock as reading 5. When at A, BN would not view M's clock as reading 5, but 8.2. In another 3 years, BN would reach M where the clock would read 10.

BN's clock reads 3 years when it reaches A, so the video frame at 3 years should show 5.

Another way to think about it is that when B1 is at M, A thinks 5 years have passed. BN should be at A at the same time that B1 is at M, no matter whose frame I approach it from. So if the video ends with 1.8 at year 3, either I don't understand something or you are assembling the video in a different way than I am. B1 might think that A's clock is 1.8 when B1 is at M, but the video is captured at A by BN, where it should be 5.

 

[Janus]

OK, I believe I figured out my error by working it through to a logical contradiction.

Let's review the A, B, C, M scenario: for A and M are 4 LY apart and in the same inertial frame. B travels from A to M at a constant 0.8c. A and B synchronize clocks at 0 when they pass. C also travels at a constant 0.8c, reaching M at the same time as B and continuing on to A. For B and C, the distance between A and M is 2.4 LY and the travel time is 3 years. For A (and M) B and C take 5 years to cover 4 LY. A sends a video of his clock to B and C.

What I added was B' who is in the same inertial frame as B but trails by 2.4 LY (B's view) or 4 LY (A's view). There is also a C' in the same inertial frame as C, but leading by 2.4 LY (or 4 LY).

When B reaches M, B' reaches A.

Wait. If B' trails B by 2.4 LY in B's rest frame, then it trails B by 1.44 LY in A's rest rest frame.
Or, if B' trails B by 4 LY in A's rest frame, then it trails B by 6 2/3 LY in the rest frame of B and B'.

Only the second choice allows B to reach M as B' reaches A, and then only according to the rest frame of A and M.

Thus in the rest frame of A things start like this:

B'-------------------B
----------------------A--------------------M
------------------------------------------------------------------C--------------------C'
Which will eventually become:

----------------------B'--------------------B
----------------------A--------------------M
--------------------------------------------C--------------------C'
The rest frame of B will start like this:

B'--------------------------------B
-----------------------------------A------------M
----------------------------------------------------C-------C'
The distance between A and M is 2.4 LY
Also, for B, the addition of velocities gives a relative velocity of 0.9756c for C and C', and thus a distance between them of ~1.47 LY

Which as events unfold you will reach this point.

B'--------------------------------B
---------------------A------------M
-----------------------------------C-------C'

note that When B and M meet, B' and A are well short f meeting.

 

[PeterDonis]

When B1 is at A, B1's clock reads 0 and A's clock reads 0, so we know the video starts at 0.

Yes. But this "when" doesn't require any notion of simultaneity, because B1 and A are co-located.

When A meets B1, it would seem that B1 would view M's clock in the same way that BN views A's clock.

I'm not sure what you mean by this, but it seems to be leading you astray, so let's take a step back.

First, we need to be clear about what defines which B clock is clock BN. Clock BN is the clock that reads exactly 3 years at the instant at which it passes clock A. That is the definition in B's frame of "happens at the same time as B1 meets M".

It will also be helpful to define another B clock, which I'll call BX, which is the clock that reads exactly 0 at the instant at which it passes M. The reading on M's clock at this event will be, as we've already seen, 3.2 years

Now, we know that when B1 meets M, M's clock reads 5 years, and B1's clock reads 3 years. And we've just seen that M's clock reads 3.2 years when clock BX passes it--and clock BX reads 0 at that same event. So M's clock ticks 1.8 years between those events. And that corresponds to M's clock ticking 1.8 years while M is traveling from BX to B1, while 3 years elapse in the B frame.

The same reasoning applies to A traveling from B1 to BN: B1's clock reads 0 when A passes B1, and BN's clock reads 3 years when A passes BN. And we know A's clock reads 0 when A passes B1; therefore, A's clock must read 1.8 years when A passes BN. In other words, the difference in A clock times between passing B1 and passing BN, must be the same as the difference in M clock times between passing BX and passing B1.

Another way of putting this: we have that, in B frame, M's clock reads 3.2 years when A's clock reads 0. In other words, that is the clock offset between A and M in the B frame (which is another way of describing the relativity of simultaneity between the two frames). But that clock offset is constant--it's true for any pair of A and M clock readings that are taken at the same time in the B frame. So it must also be true that, when M's clock reads 5 years, A's clock reads 1.8 years in the B frame. And that is the same as saying that A's clock reads 1.8 years when A passes BN.

 

[PeterDonis]

When at A, BN would not view M's clock as reading 5, but 8.2. In another 3 years, BN would reach M where the clock would read 10.

No. Let's continue the analysis from my last post and figure out what M's clock reads when M passes BN.

The simplest way of doing this is to realize that clocks BN, B1, and BX are all equally spaced in the B frame. (Can you see why?) So M's clock must tick the same amount of time between passing BX and B1, as it does between passing B1 and BN. We know the former time is 1.8 years, so the latter must be as well. Hence, M's clock reads 6.8 years when M passes BN.

Another way of doing it is to realize that, on clock B1, 3 years elapse between A passing and M passing. Therefore, 3 years should also elapse on clock BN between A passing and M passing. And the difference between the A and M clock readings that B1 sees on the two passings is 5 years. So the difference between the A and M clock readings that BN sees on its two passings (A and M) must also be 5 years. Since BN sees A's clock reading 1.8 years when A passes, BN must see M's clock reading 1.8 + 5 = 6.8 years when M passes.