Simultaneity and the Twin Paradox

[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.

 

[Mister T]

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.

 

[PeterDonis]

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

No! You've forgotten relativity of simultaneity again. "At the same time" is frame-dependent. So is "when" in reference to events that are not co-located.

Here is a correct statement of the first part of the above quote: In the A frame, when B1 is at M, 5 years have passed. That is, A's clock reads 5 years at the event on A's worldline which is simultaneous, in the A frame, with B1 passing M. But that does not mean that A's clock reads 5 years at the event on A's worldline which is simultaneous, in the B frame, with B1 passing M. That is a different event, and as I've already shown, A's clock reads 1.8 years at that event (it's the event at which BN passes A).

 

[PeterDonis]

Here is a correct statement of the first part of the above quote: In the A frame, when B1 is at M, 5 years have passed.

And to expand on this further, here is a corrected statement for the B frame: in the B frame, when B1 is at M, 3 years have passed. Note: not 1.8 years! 3 years. Why? Because we are counting "have passed" from the event where B1 meets A, and if there's one thing that we have always agreed on in this whole discussion, it's that B1's clock ticks 3 years from meeting A to meeting M. And since B1 is at rest in B's frame, B1's clock time is the same as time in B's frame.

Then we have a further statement: in the B frame, 1.8 years elapse on clocks A and M between B1 meeting A and B1 meeting M. That's just a simple consequence of time dilation for A in B's frame--but now it is stated precisely enough to see exactly what it does and doesn't mean. We could also restate the pair of meetings involved as: between BN meeting A and BN meeting M. The statement is true either way.

 

[Freixas]

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

The thing that is leading me astray is that I picture a line of B's spaced 4 LY apart and traveling at 0.8c. Conjunctions are events. Events happen in all frames. Objects are in all frames. Clocks (and their readings) can be viewed in all frames.

I think (incorrectly?) that this means that A sees a B pass by every 5 years. So if a B passes by A and observes that A's clock is not a multiple of 5 years, my brain seizes up. Are B's showing up faster and faster? Or do B's pass by every 1.8 of A's years?

This imagery is preventing me from following any of your detailed explanations. I understand that B sees 1.8 years pass by on A's clock, but that's not the same as the reading that A's clock will show. I can't picture that as anything except a multiple of 5.

 

[Freixas]

And to expand on this further, here is a corrected statement for the B frame: in the B frame, when B1 is at M, 3 years have passed. Note: not 1.8 years! 3 years. Why? Because we are counting "have passed" from the event where B1 meets A, and if there's one thing that we have always agreed on in this whole discussion, it's that B1's clock ticks 3 years from meeting A to meeting M. And since B1 is at rest in B's frame, B1's clock time is the same as time in B's frame.

Then we have a further statement: in the B frame, 1.8 years elapse on clocks A and M between B1 meeting A and B1 meeting M. That's just a simple consequence of time dilation for A in B's frame--but now it is stated precisely enough to see exactly what it does and doesn't mean. We could also restate the pair of meetings involved as: between BN meeting A and BN meeting M. The statement is true either way.

I agree with all this. However (relativity of simultaneity), when B and A are together and agree that their clocks are synchronized at 0, BN disagrees and says that A's clock actually reads 3.2. BN then travels to A, where he arrives in 3 years (his time) and 1.8 years (A's time). A flashes a time of 5 years, which is what goes on the video. So the video starts at 0 (B's video frame) and ends at 5 (with BN's video frame)

 

[PeterDonis]

I picture a line of B's spaced 4 LY apart and traveling at 0.8c.

In the A frame, the B's are traveling at 0.8c, yes, but they are not spaced 4 LY apart. The A's are. The B's spacing is shorter--quite a bit shorter. (Note that I am assuming that the B's are all spaced the way BX, B1, and BN are spaced.)

How much shorter? Consider the "meetings": M meets BX when M's clock reads 3.2 years; M meets B1 when M's clock reads 5 years. So the meetings in the A frame are spaced 1.8 years apart. The B's are traveling at 0.8c in the A frame, so in 1.8 years a B will travel 1.8 * 0.8 = 1.44 light years. So that's how far apart the B's are in the A frame.

Another way of figuring this is to observe that, in the B frame, the B's are all at rest and are 2.4 LY apart. (And A and M, and any other "A" type objects we might concoct, are moving at 0.8c in the opposite direction from the direction the B's are moving in the A frame. For concreteness, I have been thinking of the B's moving to the right in the A frame, and the A's moving to the left in the B frame.) But in the A frame, the distance between the B's is length contracted by a factor of 0.6; so in the A frame, the B's are 2.4 * 0.6 = 1.44 LY apart.

Conjunctions are events. Events happen in all frames. Objects are in all frames. Clocks (and their readings) can be viewed in all frames.

All of these statements are true yes. But it makes a big difference which events/conjunctions you look at.

I think (incorrectly?) that this means that A sees a B pass by every 5 years.

Your guess that you were incorrect is correct. If we space the B's the way B1 and BX and BN are spaced, then an A sees a B pass by every 1.8 years by the A's clock. See above.