"I noticed that with just 100 more views this will be the most read blog post in the last 3 years."

Not that it matters, but you and me contributed almost half the views; there are ~2 views for every reply we wrote. And on a technical point, http://cr4.globalspec.com/blogentry/22812/The-One-way-Speed-of-Light-Controversy of June 2013 produced 1200+ views, with just 15 comments.

People like 'controversy' and disagreement, so the stats get a bit skewed.

Oh, and I keep checking for responses so I guess we're almost alone then. Others probably fall in here by accident and leave screaming. I know I've asked that question about the simultaneity "now slices" before but I never got an answer so far.

"I know I've asked that question about the simultaneity "now slices" before but I never got an answer so far."

And I think is has been answered before. In any case, in 2-d spacetime it is just the x-axis of each observer. In 3-d spacetime it obviously a 2-d surface and so on. What more is there that you want to know?

This question,

3. Just so I have my terms correct, I understand the square coordinates are called Cartesian but what are the rhombic coordinates of the moving frame called? Anyway, the slanted x-axes of the rhombic coordinates are lines of simultaneity from the perspective of the moving frame. If each point on that line had a clock, all clocks on that line would have the exact same time from one length of the universe to the other. If there was a person with each clock and they were all telepathically connected, there would be someone back on earth at the 3.2 year mark of our time telepathically connected with Alice at her 4 year turnaround point her time. SR seems to imply Alice would be telepathically aware of a time in our past (since if Alice is at the 4 yr mark, we are at the 5yr mark) as being in her present. However, if we compensate for the delay of telepathic signal and for the relativity of simultaneity we see that our present at our 5yr mark and Alice's present at her 4 yr mark constitute a universal present which SR forbids. Which interpretation of the spacetime diagram is correct?

It is called Minkowski spacetime.

"If each point on that line had a clock, all clocks on that line would have the exact same time from one length of the universe to the other"

One has to to qualify this a bit: provided that those clocks are all synchronized in the applicable inertial frame and there is no gravity (the latter prevents application on on the larger scales).

I do not want to discuss anything 'telepathic', just observables. The only absolute observables are Alice looking out of the window to read any co-located clock in Bob's reference frame directly. And for Bob's observer at that same clock's position to read Alice's clock directly when she flies past.

Now the turnaround (or handoff) is a nice fixed event on which to do some spacetime interval calculations. As an exercise, write down that event's coordinates in both Alice and Bob's reference frames respectively. Then see if it agrees with the invariant nature of the spacetime interval.

Bob's coordinates at the turnaround point are 5,3 and alice's are 4,0. 25-9=16 and so the spacetime interval for both is 4.

I'm trying to understand what you're saying. Maybe I should drop the idea of clocks on her x-axis and just bring Bob's clock to the turnaround point which she can see without any delay or telepathy. We know Bob is currently at his 5yr mark and Alice is at her 4 yr mark. Alice see's Bob's clock at Bob's 3.2 yr mark (Y=.8, .8*4=3.2) due to time dilation. The difference between Bob's clock at home as he sees it and the time dilation Alice sees is 1.8 years which is the relativity of simultaneity vx= .6*3ly=1.8yrs.

So Bob's past is not really in Alice's present, Alice's velocity is distorting spacetime to make it seem to her that Bob's past is happening now in her present. Is that the correct interpretation?

But if I apply the same analysis for just after the turnaround, does Alice suddenly see Bob's clock jump by twice the relativity of simultaneity to a total of 3.2+2*1.8 = 6.8 yrs? She's not really looking 1.8 yrs into Bob's future which hasn't happened yet for him, but how do you interpret what she's seeing? Bob's clock to her just makes a discontinuous jump like this by 3.6 yrs? How does one interpret what's happening so it makes sense?

"Maybe I should drop the idea of clocks on her x-axis and just bring Bob's clock to the turnaround point which she can see without any delay or telepathy."

No, to suddenly bring Bob's clock to the turnaround point is roughly equivalent to telepathy - quite nonphysical. Bob has all the time in the world to set up and synchronize clocks anywhere in his inertial frame. As I said, Alice then just need to look out of the window and will find a 'Bob-synchronized clock" right there, next to her...

She then sees "Bob's now" and the Bob's observer, manning that clock, sees "Alice's now". When there is distance between them, the one can only see the others past, so there is no way that anyone can 'see' anyone elses future. Projections into the future is an acceptable art, but it is not 'seeing', neither is it reality.

So my advice, take articles that make claims like 'Alice's now is in Bob's future' with a large pinch of salt.

"When there is distance between them, the one can only see the others past, so there is no way that anyone can 'see' anyone elses future."

Finally, these words make everything crystal clear to me, this is the answer I've been waiting for. They can only see each other's past because of the reciprocal time dilation and the distance adds relativity of simultaneity time to that. Maybe you didn't really look closely enough at Greene's analyses of problems in his course but he has absolutely no understanding of this fact. Every one has the approaching clock's now slice pointing into the stationary frame's future, a future that hasn't happened yet. It's because he doesn't visualize the spacetime diagram correctly. Bob's coordinate system is sliding horizontally along with the moving timeline (Bob's clock is right there as you said.). At the turnaround, his 5yr point is directly superimposed on Alice's 4 yr point (forget about how the x-axis for now.). When Alice turns around, Bob's coordinate system follows her back. So when you draw the now slices, they are not pointing to a future in Bob's original coordinate system to the left, they are pointing to Bob's coordinate system which is back at the turnaround point.Hence whether Alice is approaching or going away from Bob, she is always seeing Bob's past. After 2 years of intellectual imprisonment, Free at last!

I do not want to blow the 'freedom' that you struggled for so long, but your descriptions are not quite valid (maybe freedom is just poorly expressed).

"They can only see each other's past because of the reciprocal time dilation and the distance adds relativity of simultaneity time to that."

They see each others past because of the time delay due to the finite speed of light.

"Every one has the approaching clock's now slice pointing into the stationary frame's future, a future that hasn't happened yet."

Yes, this what Greene said, if I remember correctly.

"Bob's coordinate system is sliding horizontally along with the moving timeline (Bob's clock is right there as you said.). At the turnaround, his 5yr point is directly superimposed on Alice's 4 yr point (forget about how the x-axis for now.). When Alice turns around, Bob's coordinate system follows her back."

Huh?

If Bob is the reference frame, it 'slides' nowhere. However, he can have an observer with a clock synchronized to his that sits permanently at the turnaround point. That observer's clock will read 5 years and he can directly read Alice's clock as 4 years.

The issue about "see the others future" is somewhat like timezone issues: if an event happens here with me "now" (09:00 local time), it will be 03:00 in Ottawa, your "now". So if I immediately tell you about the event and its time, did I give you information from your future. Of course not...

Ok maybe I can explain better what I'm saying through an analogy. It's the same one I used before but now I've corrected it.

Bob is a TV station back on earth. He is transmitting his present line up of TV shows. Alice is working out on a stationary bike. She has a TV screen in front of her receiving Bob's signal. But the faster she pedals, the slower the signal she receives and the slower the motion she sees on the screen. This means the longer she pedals (analogous to the farther she goes out into space) the greater the time lag between what she's watching and what Bob is transmitting (this is the relativity of simultaneity which builds up to 1.8 Bob yrs at the turnaround).

If she stops pedaling, she no longer receives the picture in slow motion but even though she's now in the same inertial time frame as Bob and receiving the signal at the same speed as Bob's transmitting and she shares the same present as Bob, she does not suddenly see the exact same signal Bob is transmitting because of the lag she built up. She has to somehow catch up by somehow going into a fast forward mode. This happens after she stops. The show she was watching will look to be in fast forward for a while until it catches up to the present signal Bob is transmitting. Then Alice will be viewing the show Bob is transmitting, minus the signal transmission delay, at normal transmission speed. It's like time is a rubber band wound up that takes time to recover its original shape once the winding stops and the propeller is let go. Is that right and what does that say about the deeper nature of time?

But what happens if Alice doesn't stop but she starts to back pedal instead (analogous to her returning to Bob). Her pedaling is still slowing down the picture reception and she is still falling behind in her viewing with even more time lag that she has to make up for. We know that by the time she stops pedaling backwards and returns to where she started, she will have caught up with the TV signal. She would have watched 10 years of TV in only 8 years while the whole time the signal was coming in slower than it was being transmitted. This means that even though the signal was still coming in slower on the return journey, the TV shows she was watching were going at fast forward. Weird, she forward pedals and they're playing in slow motion and backward pedals and they're in fast forward (after she's pedaled forward for a while of course).

Part of this effect comes from the decreasing signal delay as she approaches Bob but most of the effect comes from dissipating the lag caused by the relativity of simultaneity. But on the return journey, the rubber band of time is still being wound from the other end while the propeller has been released at the turnaround. I know you'd rather avoid going into my brain but is this the right way to view things?

So are we agreeing now that Greene's interpretation of relativity is wrong? Twice in his course he talks about the reality of past present and future all existing concurrently. He even tells the story of Einstein consoling his friend's widow that his friend lives on in someone else's high velocity perspective. No, his friend does not live on, it's just the illusion of perspective. The present is universal, it's velocity that gives it time zones and time zones are just a convention.

When I say Bob's coordinate system slides over, it's just me visualizing Bob's clock being right beside Alice's clock. It also allows me to see that as Alice returns home, her now slice is not pointing at Bob's future in the starting point coordinate system but backwards to a new coordinate system where Bob's y-axis is at the turnaround point. This helps me visualize that reciprocal time dilation always includes a past time due to slower clocks.

Just let me put more detail into my stationary bike TV signal analogy. If Alice stops, there is a time (1.8 years) for Bob's TV signal to be viewed in fast forward mode by Alice and after that Bob's signal will run at normal mode but will be delayed to Alice by the speed of light delay over 3ly.

If Alice immediately turns around, a 2nd 1.8yr lag begins being added to the one she built up in her away journey. By the time she reaches Bob, the picture speed on her TV will be a combination of incrementally overcoming a 3.6yr lag due to relativity of simultaneity, an increasingly less delay of transmission signal minus the slowing of the signal due to reciprocal time dilation. All of this would make for an accelerating picture that suddenly goes into normal mode at the end of the journey.

Actually I can think of so many different interpretations of what happens to the lag (relativity of simultaneity) for Alice stopping or turning around, I don't know which is correct.

If she stops and the 1.8yr lag just disappears then there will be a sudden discontinuity in the picture she's seeing. She'll somehow miss 1.8 yrs of TV shows. This is not likely because if Alice turns around, all 3.6 yrs of lag will dissipate once she reunites with Bob and there will be no discontinuity or missing programming. And it's also unlikely she suddenly adds another 1.8 yrs of lag at the turnaround point just because she turns around. Mathematically, the total relativity of simultaneity at he turnaround point is 3.6yrs not 1.8 yrs if she just stops.

If Alice stops at the turnaround point or turns around and stops at earth, there's no way the lag should influence the TV picture after she's stopped as I reasoned previously. Relativity of simultaneity is an illusion of perception, it's reciprocal just like time dilation and can't be residual once the two participants are in the same inertial frame, that would make it real like relative aging and it isn't.

So how is the built up relativity of simultaneity satisfactorily handled for both stopping or returning?

"So are we agreeing now that Greene's interpretation of relativity is wrong?"

No, I do not agree. He uses valid concepts, all based on the correct physics, but the words that he uses are imprecise and can be interpreted in many ways. This is the case for all verbal interpretations of SR, including my own efforts.

"So how is the built up relativity of simultaneity satisfactorily handled for both stopping or returning?"

A good example of the vagueness of words. Relativity of simultaneity does not "build up" in the 'twin paradox'. It suddenly changes when one of the two participants 'jumps' inertial frames.

Unless you settle down and start to define and solve the things that you asked algebraically, it is a waste of time and I'm not available to keep on correction verbal interpretations,

I've figured it all out algebraically from the spacetime diagram for the Bob/Alice example at .6c: Yv is the total velocity vector through spacetime of which we can only see a shadow of it through space (x-axis) as velocity v. The time-axis component,applying Pythagoras, is the velocity through time v2Y/c which we can't see (derivation below). v2Y/c = vxY/ct = Vx/(ct') where t'=t/Y. The formula for relativity of simultaneity is Vx/c2 so c* the relativity of simutaneity = the velocity through time * time dilation. So since we have a direct correlation between time dilation and relativity of simultaneity, we no longer need bother with the mathematical involvement of simultaneity since it's the same thing as time dilation multiplied by a constant.

1. Derivation of velocity through time w:

Y2v2 = v2 + w2

w2 = Y2v2 - v2

w2 = v2(Y2-1) where Y2 = c2/(c2-v2)

w2 = v2((c2-c2+v2)/(c2-v2))

w2 = v2(v2/(c2-v2))

w = v2c/(c*sqrt(c2-v2))

w = v2Y/c

If w is the velocity through time what do the time t" and distance x" components mean where w = x"/t"? T" is the time spent traveling through time. It appears to us as the difference in time between the stationary frame's proper time and the moving frame's proper time. In the example of Alice leaving bob at .6c, traveling out 3 ly and returning, Alice ages 4 yrs her proper time at the turnaround point while Bob ages 5 yrs in his proper time. Alice has aged 1 year less than Bob to get to the same universal present and this is the time spent traveling through time. The formula for this is t" = t'(Y-1) and the corresponding formula for x"=vx(Y-1)/c since w=x"/t" where x" is the distance traveled through time.

At this point we need some clarification of the terms. The universal present is the time all clocks are sync'd to. Earth would have sent out a timing signal to a planet 3ly away at least 3 yrs before Alice would go to that planet. Say that signal was sent out in the year 3000, the far planet would have been sync'd to the earth's years starting at 3003. Alice would arrive in the year 3005 and her clock would register 3004. She has not gone 1 yr into the future as relativists would say, she has merely took less of her proper time to reach the same universal present. She took 4 years (plus 1 year of invisible time traveling through the time dimension) while Bob took 5 to reach the same universal present. There is no time travel for one to a future that hasn't happened yet to the other as most people believe.

In the Bob and Alice example where both are running TV stations, Alice would watch a year of Bob's programming at half speed during her movement away from Bob at .6c. If Alice was also transmitting a TV signal, Bob would also be watching her programming in slow motion. (This sounds like time dilation but you'll see later that it is not.)The farther they move apart, the greater the lag of the TV signal reaching them. This is due to the time delay of the speed of light. Alice would receive the first year of programming (3000-3001) between 3000 and 3002 her years. The planet Alice is traveling to would receive the 3000-3001 schedule in 3003-3004. The next schedule from 3001-3002 would be received by Alice in 3002-3004 her years while the planet would receive it from 3004-3005 universal time.

If Alice stops at this point, she starts receiving the same programming at the same speed as the planet does and at the same speed Bob transmits, so she is fully in the present, but her clock reads 1 year less. She is not 1 year in the past, she has just aged 1 year less to get to a common present. Her missing year also doesn't change the fact that the signal is delayed by 3 years. So just as one can't access another's future, there's also no access to or simultaneity with the past. Relativity has no outside the ordinary time travel at all except for a path with less aging to the present.

On her way back, she watches 2 years of Bob's programming in only 1 of her years (which pass at the exact same rate for her as Bob's years pass for him). It looks as if the outside world is running twice as fast as her time but in actuality Bob's time is dilating as slow as it did in the outbound journey. Bob also watches her programming at twice the speed but she'll get his 3002-3004 schedule between her 3004-3005 timeframe (which equals 3005-3006.25 universal time) but he'll have to wait until 3008 to stop viewing Alice's programming at half speed. So he watches her slow motion TV shows for 8 years while she watches his for 4 of her years. We don't need to differentiate between his and her years because both run at the same proper rate.

Even though it looks to her like she's catching up to Bob's programming because it's coming in at twice the rate and the closer she moves to Bob there's less of a time delay due to the signal not having to travel as far, Alice is still losing a quarter year, with respect to Bob, for every year she's speeding back. No matter whether Alice moves away or towards Bob, she ages less than Bob and the time dilation is the same no matter how fast or how slow she's observing time outside of her own.

There are other absolutes besides a universal present that relativity doesn't openly accept but secretly does. There's also a universal rate of aging within each inertial frame no matter what its constant velocity. Distance is also an absolute and so is who is moving. At the beginning of any journey, the two participants decide who is staying and how far the other is going, there is no need to consider both points of view as the stationary frame. The one that's moving is the one that goes from point A to point B with minimal distance separation from both points. With x=0, the spacetime interval is minimized to time only (distance adds to a longer spacetime interval). The proper distance for both is judged from the stationary frame and length contraction during motion need not be considered.

The stationary frame is a common ground at the start of the journey. Either participant can be moving relative to the stationary frame and that relative velocity can be used to determine which participant has a greater share of their mutual relative velocity. This common patch of ground relative to both allows one to understand why there are cases where a high relative velocity between participants yields no relative time dilation between them.

Other points of interest on the cartesian coordinates (stationary frame) of the spacetime diagram of the Bob/Alice example:

Start: t=0, x=0.

1. Velocity through spacetime: Yv

2. Velocity through time: v2Y/c or vx/ct' or t'(Y-1/Y)

3. Time spent travelling through the time dimension: t" = t'(Y-1)

4. Slope of Alice's timeline = t/x or 1/v.

5. X-axis coordinates (t=0) corresponding to t': x = t'(Y-1/Y)/v

6. Time-axis coordinates (x=0) when Bob starts transmitting signal which will be received at t': t = t'Y(1-v/c2)

7. Number of years each takes to watch a year of transmitted programming: outbound = 1/Y(1-v/c2), inbound is inverse

8. Time-axis coordinate for Alice's lines of simultaneity: t = t'/Y

9. The universal present time: t= Yt'

10. The relativity of simultaneity (diff between the universal present and #8): Vx/c2 = (Y - 1/Y)t'

11. Coordinate translation from (t',0) to (t,x) is (Yt', (Yt'-t'/Y)/v)

Here are some sample calculatons for v=.6c, Y=10/8

1. Yv = .75c

2. w = .45c

3. t" = .25t'

4. 1/v = 5/3

5. x = .75t'

6. t = .5t'

7. outbound is 2 yrs to watch 1 yr in slow motion. inbound 1/2 yr to watch 1 yr at fast forward

8. t = .8t'

9. t = 1.25t'

10. Vx/c2 = .45t' aka relativity of simultaneity = velocity through time * time dilation

11. (t',0) = (1.25t', .75t')

Here are some sample calculatons for v=.8c, Y=5/3

1. Yv = 1.333c

2. w = 1.0666c

3. t" = .666t'

4. 1/v = 1.25

5. x = 1.333t'

6. t = .333t'

7. outbound is 3 yrs to watch 1 yr in slow motion. inbound 1/3 yr to watch 1 yr at fast forward

8. t = .6t'

9. t = 1.666t'

10. Vx/c2 = 1.0666t' aka relativity of simultaneity = velocity through time * time dilation

11. (t',0) = (1.666t', 1.333t')

I'm stuck on how to interpret parts of the spacetime diagram in the .6c example. Specifically, I don't know how relativity interprets past, present and future. I know how Greene defines them and he clearly articulates it very well but I can't agree with his conclusion that all 3 exist concurrently. To me, if you're passing a clock and you get a signal from it that differs from the signal a stationary person at the clock is getting, then you're either in the past or in the future of that stationary person which is impossible. (However, I've since figured out how 2 functioning clocks at the same spot can have different readings of the same present moment without implying either is in the future or past of the other; read on.)

When people say relativity allows a short cut to the future, it doesn't mean you're arriving at the future and waiting for the people trudging the regular way through time to catch up. It's more like you arrive at the same present moment but you've aged relatively slower to get there. That's a big difference in interpretation.

Let's look at the spacetime diagram where t'=1. By the formula t=Yt', t=1.25. Since t' and t are on the same horizontal x-axis, it means that t' is in our present even though the two clocks differ. A ship at t' would be passing a clock synchronized to earth time but it would not see a signal from it saying t=1.25. By reciprocity of time dilation, that clock would be moving relatively to the ship and dilating. If the ship's time was 1, the time on the earth clock would read 1/Y or .8 to the guy in the moving ship.

I thought time dilation reciprocity would yield the same clock value for each; but instead we see his clock at 1.0 and he sees ours at .8 while this is all happening at 1.25 our time. But if he is reading our clock as .8 then that implies he is presently .45 years in our past. He has supposedly traveled .45 years into the past plus we see him simultaneously share our present (although his clock reads differently than ours) while he's in our past? Not only that, but at his 1 yr mark his most recent TV signal from us is from 3/4 yr ago even though his present time points to being .45 yr in our past. This would suggest he can view our TV sooner than it could arrive by light speed which is impossible. There must be another interpretation even though this is what the spacetime diagram is suggesting. The answer must be that the readings themselves mean nothing because they are between two relatively unsynchronized clocks from both perspectives.

We need to introduce the fudge factor of relativity of simultaneity to read the clocks as if they were synchronized so we don't come to the erroneous conclusion that the moving guy is simultaneously existing in our past or traveling in any time but our present.

The relativity of simultaneity fudge factor = .45 from the moving guy's perspective. Add that to the .8 time dilation he sees of our time and the total is 1.25 which agrees with our present time. So when you correct for the lack of sync between our clock and his, they both agree on what the present time is. But from our perspective there is no relativity of simultaneity fudge factor. Instead there is the moving guy's .25 yr invisible time spent in the time dimension. Add that to the 1.0 time dilation we see of his clock and there is again agreement that the present time is 1.25.

But this is an unsatisfying interpretation for me. Why would it be that from one perspective there is this invisible time spent in the 4th dimension that clocks can't record and from the other perspective there is this invisible fudge factor of relativity of simultaneity that clocks also can't record. This led me to look for a mathematical relationship of how to express relativity of simultaneity also as invisible time spent in the 4th dimension. Here it is:

We know that the velocity component through time w = c * time for the relativity of simultaneity(trs)/ t' which can be expressed also as (Y-1) t'/Y and t'(Y-1)(1+1/Y). t'(Y-1) = t" which is the invisible time spent in the 4th dimension by the moving frame. So trs = (1+1/Y)t".

So now we can express the fudge factor of how to correct the clock readings of both perspectives using only the concept of invisible time spent in the 4th dimension. The time on our universal stationary clock (no need to read me the riot act on the unorthodoxy of that) is 1.25. In order to reconcile our reading of the moving clock which we see as 1.0, we need to add t"= .25. In order for the .8 reading of our clock from the perspective of the moving clock, they need to add the fudge factor of (1+1/Y)t" = .45. No need to ever mention relativity of simultaneity again except as a relic of the old interpretation just like lorentz contraction and relativistic mass are.

So what have we learned here:

1. There is a universal present time that both (not all) frames can agree upon. The Earth could seed the universe with clocks that are all synchronized in the Einstein way meaning they take into account the speed of light delay from the source to each of their proper distances from the source. There is no ability to take a snapshot of the universe in the present moment because of the information delay in creating that snapshot. However, if each clock took timestamped photos, the present moment could be post-processed sometime later of what everything looked like at a universal present. Otherwise everything around each clock is a view of the past from that clock; the farther away, the more into the past. This is how the fabric of the present moment is distorted.

2. Nothing exists in the past because the past does not exist. We can see the past due to the speed of light delay but things from 2 different times cannot occupy the same point in space. There is no time travel in the movie sense (or Brian Greene sense) to either the past or an already written future.

3. The only reason clocks at the same point in space can differ because one is moving is that clocks cannot register time movement through the 4th dimension. Just as we have no instruments to register the velocity through the time dimension of the total spacetime velocity (Yv), we can't register the time spent traveling through the time dimension. That time manifests itself as lost time and slower aging for moving frames.

4. Sometimes what's on a clock is not the correct time even though the clock is running correctly.

a) See #3 above. The reading on the clock is a measure of time spent in the space dimension but is missing the time spent in the time dimension.

b) Reciprocal time dilation due to relative velocity. Both are illusions of perspective so far as I'm concerned because if their effects leave no lasting impression , they are illusions. For example, 2 ships racing together, towards each other or away from each other at the same speed relative to a common stationary frame will register no time dilation on each other's clocks even though their relative velocity can be any value. For ships moving at different speeds relative to a common frame, their relative velocity gives no indication of who is aging slower until one or the other returns to the same constant velocity as the other. The reciprocal time dilation becomes real at that point for only one of them, manifested as slower relative aging, while it evaporates for the other as a mirage.

c) A biological clock's read out, as in suspended animation, can be slowed by cold temperature slowing just the clock itself without any slowing of time. Apparently whether time slows or just the clock slows, one can arrive at a subsequent present having aged less in either case.

5. The past can be seen by the speed of light delay. This does not mean you have traveled into the past or that past and present co-exist simultaneously.

6. The passage of time on the stationary frame can look slowed down or sped up when viewed by the moving frame. This is a combination of the speed of light delay, the speed and direction of the ship. It is independent of relative clocks as a TV show from earth can look to be in fast forward when viewed by an incoming ship even though time is slowed from both perspectives.

Should I submit this as a paper to Airvx? Is that the right name?

I now have only 1 mathematical proof to work on. I don't accept the interpretation that time does not pass for a photon. I need to therefore prove that massless particles have no velocity or time component in the time dimension, their total time and velocity is spent only in the space dimension. I have to somewhere prove as Y -> infinity, t" collapses suddenly to zero. But it may be deeper than this since a change in inertia is required to determine which frame is relatively aging which is somehow related to mass which massless particles don't have. It could take me a while.

Maybe the answer is as simple as that since light is always at a constant velocity and can never experience inertia because it has no mass, that it must always be considered the stationary frame. As the stationary frame , it can have no velocity component or time component through time; only the moving frame can have those. Well I guess I've reached the end of special relativity.

Hmmm no response. I guess I've committed blasphemy again even with math. I suppose I'll go on the science chat forum http://www.sciencechatforum.com/index.php now and see if anyone will bite there.

Sorry Ralf, I'm in the process of moving house and have not looked at the forums much lately. In any case, what you have suggested as a 'solution' is so ridiculously wrong that I do not even want to respond to it. This thread is considered closed now.

Just make sure you post on "personal theories" on other forums, otherwise you may meet with some grief there...

-J

Whew! I was worried you were going to say this is so painfully obvious that there's absolutely no new idea or interpretation here. Now I'm elated except for the coming months of being set upon by the ravenous wolves that inhabit physics forums. Someone's going to eventually see the elegance of what I've stated here. I assume you don't want me bringing up your name in any future discussions?

Yes, you've assumed right. Past experience has not been good when you have quoted me, because you tend to color it with your own 'strange' relativistic thinking.

All I can say is good luck!

-J

OOps I found 2 mistakes. w= v2Y/c = vx/(ct') = ctrs/t' = Y-1/Y not t'(Y-1/Y).

Also trs= t"(1+1/Y)/c.

Now I can answer your question:

If Alice only did a flyby with Bob and then a flyby of the distant clock, she and Bob would not have agreed that the remote clock measures Bob's aging correctly. She was not present at the synchronization and although she has read 12.5 years on the clock's display, she is hence not forced to conclude that Bob has aged that much. In fact she may be inclined to think that maybe Bob has aged only 4.5 years! Can you think why I'm entitled to say this?

There has been no inertial frame jump so reciprocal time dilation is in effect. From Alice's view, Bob could be moving and she's stationary. His clock could be dilating in relation to hers so 7.5 * .6 = 4.5. The rulebook rules!

x

The answer to your twin paradox variant question is in your book. It is the twin paradox done where each participant is put in the "stationary" frame. Alice will age less in both scenarios. When they open their envelopes is critical. The answer is from your post in the conventionality of relativity:

Point 3: If the probe only whizzed by us and also whizzed by Pluto, we would have no idea about elapsed time difference (ignoring the obvious gravity influence in this hypothetical scenario). If it 'stopped' at Pluto there had to be a change of inertial frames and we would have known the absolute time aging difference, although we could not have directly measured it.

If Alice kept going and Bob stayed put we couldn't determine who aged less. If Alice stopped and Bob stayed put we would have known the absolute time aging difference, although we could not have directly measured it.

"The answer to your twin paradox variant question is in your book. It is the twin paradox done where each participant is put in the "stationary" frame. Alice will age less in both scenarios."

All the diagrams in the eBook are for the standard twin paradox, which is case one. In case two, Bob also has to accelerate and he has to make a much larger inertial frame change than Alice (in order to catch up). So please rethink the answer.

I think the situation before the opening of the envelopes is more or less covered in my prior reply on the 'time hand-off' scenario. So consider it in that light.

-J

You may think we're at the end but we're not, I have other questions about relativity that aren't connected to the subject discussed here and are equally as frustrating.

Anyhow, back to this subject matter. You say the idea of the speed of light being constant in all frames is a convention, not directly supported by empirical evidence. The evidence comes indirectly from two-way speed of light analysis supported by a clever clock synchronization scheme and a multitude of experimental evidence that does not contradict the initial assumption hence the initial assumption must be correct.

As a result of that initial assumption, we can't make any determination of who is aging slower when a ship is leaving earth on a one-way trip. The ship is not allowed to get empirical evidence of whether the earth's clock is ticking faster than his because relativity says our clock must be ticking slower than his.

The problem is that if we get the empirical evidence to support this, it contradicts the basic assumption that relative velocity by definition means there is no way to tell what is each individual's share of that relative velocity.

In the analysis where Alice is put in the stationary frame for half the journey, yes, Bob's clock on earth IS ticking slower in the virtual analysis sense but when she fires up her rockets to catch up with Bob, her extra speed erases his slower aging and she ends up aging even slower than Bob overall. If the ship and the earth had been in constant communication with each other, compensating for the light delay of the messages in post-processing, there could not have been two contradictory scenarios of messaging one from assuming Alice was initially moving and another assuming Alice was stationary. In the first assumption, Alice would have always been aging slower throughout the journey. In the second assumption, Bob would have initially been aging slower than Alice but then Alice would have been aging much slower than Bob. There is only going to be one of these scenarios that is real and by the messaging one can actually tell who had the individual share of the relative velocity which relativity forbids.

"As a result of that initial assumption [conventional synchronization], we can't make any determination of who is aging slower when a ship is leaving earth on a one-way trip."

We cannot make a proper elapsed time observation in a one-way situation, but if Alice started in Earth's inertial frame, we know that she had to changed inertial frames to fly away. Doesn't that tell us enough to make an inference on aging?

"There is only going to be one of these scenarios that is real and by the messaging one can actually tell who had the individual share of the relative velocity which relativity forbids."

No, messages can give us only a relative velocity. As above, if starting off in the same frame initially, one can infer something from the magnitudes of frame changes that each suffered.

-J

You wrote: "Anyhow, back to this subject matter. You say the idea of the speed of light being constant in all frames is a convention, not directly supported by empirical evidence."

The correct terminology would have been "the observed propagation speed of light is isotropic in all inertial frames" or simply "the observed one-way speed of light is the same in all inertial frames".

We know that this needs two Einstein-synchronized clocks, some distance apart, but inertially at rest relative to each other. There is in fact direct evidence for Einstein's synchronization method: https://en.wikipedia.org/wiki/One-way_speed_of_light#Slow_clock-transport

"If however one clock is moved away slowly in frame S and returned the two clocks will be very nearly synchronized when they are back together again. The clocks can remain synchronized to an arbitrary accuracy by moving them sufficiently slowly. If it is taken that, if moved slowly, the clocks remain synchronized at all times, even when separated, this method can be used to synchronize two spatially separated clocks. In the limit as the speed of transport tends to zero, this method is experimentally and theoretically equivalent to the Einstein convention."

This has been experimentally verified to great precision. See http://ntrs.nasa.gov/search.jsp?R=19940006498, or more fully described in: http://tycho.usno.navy.mil/ptti/1992papers/Vol%2024_10.pdf. This is good enough for me (and for every scientist/engineer that I know of).

-J

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