Bell's Spaceship Paradox and Length Contraction

[GrayGhost]

No, even per B the coordinate system is not 1 to 1. It isn't a matter of perspective or opinion, it is an objective fact about the chosen coordinate system.

Hmmm. Well, if heavenly bodies reshift about B's spacetime coordiante system due to a rotation in B's own sense of simultaneity, why must B's own coordinate system not maintain a 1:1 relation between space and time? The way I see it, instead of forcing everything to remain where it is and contorting the B system, why not leave the B system untorted (ie remain 1:1) and allow the heavens to shift? In the end, it's the same thing viewed 2 different ways.

Because if a coordinate system is not 1 to 1 then you cannot do well defined coordinate transforms any more. If you cannot do coordinate transforms then you cannot use coordinate transforms to determine the laws of physics in that coordinate system and to transform results to other coordinate systems. So suddenly the coordinates become physically useless.

Well, I agree they become useless, but that's only assuming twin B continues to define simultaneity per τ1 = ½( τ0+ τ2), and also assuming B runs the LTs as he would if he were purely inertial (even though he properly accelerates). However, that was defined for cases of uniform linear translation alone, which falls down during a POV undergoing an active proper acceleration.

I don't know of a valid coordinate system for a non-inertial POV. I've heard of a couple, but none were sufficient IMO. One cannot use the Einstein sync convention for non-inertial bodies, assuming spacetime coordinates are "always" to match the predictions made by inertial observers. In any valid theory, all POVs must attain the same solns using the same spacetime transformations.

Thanks for the heads up. I fixed it.

LOL, very funny! I kinda liked it better before you tweeked it. You know, the dynamic worldline there always enters the origin at a vertical orientation at the origin, so just as in the case of a Loedel figure, the horizontal space axis and vertical time axis exist whether displayed or not.

The way I see it, B's calculations using the LTs are just less convenient than any purely inertial observer's LT calculations, because B has extra steps he must do that all inertial observers would not. The reason, when B receives light from a remote body, it does not predict how the remote body has moved since the EM left the body prior. However, if the body is purely inertial, then B should be able to determine where the body really exists by accounting for his proper accelerations from his own proper accelerations (using his own accelerometer data) relative to his initial inertial frame. It's an extra step that inertial observers need not be concermed with, but what they hey. While this is a far less convenient way to determine the location of remote bodies, it should be no less accurate assuming good enough technology is being used in the nav system ... and besides, twin B cannot do it in the usual way during periods of his own proper accelerations.

That said, while I don't really disagree with your points, I don't see the problems you see in such an approach as applied to a twin B POV. Maybe you discard it too hastily. Is there a valid method for twin B to determine the location of any inertial body during his own proper accelerations using the LTs, that you are aware of? I mean, one whose solns precisely match that the LT calculations made by any inertial POV.

 

[pervect]

Hello DaleSpam.

Hmmm. Well, I do understand your point wrt the impact of our usual defined rule for simultaneity per Einstein.

I see twin B's coordinate system as "always 1:1", per B himself.

Well, that's where the difficulty comes in.

If you use Einstein's clock synchronization scheme (I think you already agreed to the importance of doing that?, so I won't discuss it more here without prompting), you _cannot_ synchronize all the clocks in a rotating coordinate system.

Because you can't synchronize all the clocks, you can't define a time coordinate by the usual methods. So your perception that you can or should be able to do this is the problem.

The outcome of the Haefele Keating experiment confirms the expected theoretical difficulties experimentally. If you transport a clock slowly around the world, due to the world rotation it won't read the same time as a clock that stayed behind. Or a clock that was transported around the world in the opposite direction.

 

[PAllen]

Well, I agree they become useless, but that's only assuming twin B continues to define simultaneity per τ1 = ½( τ0+ τ2), and also assuming B runs the LTs as he would if he were purely inertial (even though he properly accelerates). However, that was defined for cases of uniform linear translation alone, which falls down during a POV undergoing an active proper acceleration.

Why would you expect to use LT's for a non-inertial observer? Einstein, in contrast, analogized a non-inertial observer to experiencing a pseudo-gravity field, corresponding to a non-Minkowski metric; and the transform from an inertial frame to this would not be an LT

I don't know of a valid coordinate system for a non-inertial POV. I've heard of a couple, but none were sufficient IMO. One cannot use the Einstein sync convention for non-inertial bodies, assuming spacetime coordinates are "always" to match the predictions made by inertial observers. In any valid theory, all POVs must attain the same solns using the same spacetime transformations.

Well, there are many, not none. You can use Einstein sync convention along with non-minkowski metric (you must use a non-minkowski metric to describe the continuous experience of non-inertial observer - just as you have to incorporate 'inertial forces' if you set up coordinates for a non-inertial observer in classical mechanics). You can also use simultaneity of an MCIF to build valid coordinates locally for the non-inertial observer (these coordinates often have less coverage than rader, but are fine within the region they are valid); however if you build this into a coordinate system covering some spacetime region, you get exactly Fermi-Normal coordinates. The transform from any inertial frame to these is not an LT, and the metric is not a Minkowski metric.

 

[GrayGhost]

Actually, Einstein never used this definition of simultaneity (simultaneity of a momentarily comoving inertial frame) for non-inertial observers. He analyzed non-inertial observers in SR using a single inertial frame; or using GR techniques with a well defined coordinate system.

Hello PAllen. Indeed, what you stated here is of course true. Agreed.

Applying the simultaneity procedure he actually used for inertial frames (radar simultaneity), to non-inertial observers produces a possible well defined set of coordinates; the simultaneity is quite different from that of momentarily com-moving inertial frames. This well defined set of coordinates (radar coordinates) does not have the anomalies of MCIF (or it wouldn't be a well defined coordinate system). However, locally (close to the non-inertial observer's world line) it approaches MCIF; thus it equally well describes local physics as MCIF. It disagrees more and more with MCIF simultaneity for non-inertial motion, the further away you get from the observer's world line.

I agree. And the problem I have had with radar coordinates, is that it is inaccurate overall. I'd like to know long in advance during my proper acceleration that I must make a course correction to avoid a collision with an inbound inertial body, versus awaiting until my nav data is actually accurate when the inbound is at close proximity. In any good theory, all observers should be able to make LT calculations that they all also agree upon. Barring seeing the future of course, one should be able to make LT calculations (maybe with extra procedures, in B's case) that result in its true relative location, which would precisely match any inertial observer's LT calculations.

 

[PAllen]

I agree. And the problem I have had with radar coordinates, is that it is inaccurate overall.

What do you mean by inaccurate? The objective advantage of inertial coordinates is that the laws of physics take the simplest form; and that multiple physical procedures for setting them up yield the same result. This simply cannot be achieved for a non-inertial observer, because you have to associate a proper acceleration with fixed spatial coordinates; and procedures which produce equivalent results for inertial observers produce different results for non inertial observers; each leading to a different coordinates system (and metric), each no more preferred than any other. Any of the coordinates you might pick for this are accurate for calculation when used with the corresponding metric.

I'd like to know long in advance during my proper acceleration that I must make a course correction to avoid a collision with an inbound inertial body, versus awaiting until my nav data is actually accurate when the inbound is at close proximity. In any good theory, all observers should be able to make LT calculations that they all also agree upon. Barring seeing the future of course, one should be able to make LT calculations (maybe with extra procedures, in B's case) that result in its true relative location, which would precisely match any inertial observer's LT calculations.

Why LT calculations? The LT is a transform between inertial coordinate systems. It shouldn't and doesn't apply to non-inertial coordinates. As a practical matter, it would be easier, even for a non-inertial observer to pick a convenient single inertial frame and express their observations in said coordinates. When analyzing the solar system, we do not try extend the coordinates of our local city; we pick convenient global coordinates. A sane rocket pilot, operating in the solar system, would use heliocentric coordinates and (if they needed it) linearized GR metric.

 

[nosepot]

I gave a start in post #99. Did you read it? Did it make sense? I realize it's only a start but how was it for a start?

It's a good start.

Good, that's what we want, understanding.

I'll assume that's not condescension. :)

What does "blew my mind" mean?

Means the latter part of his explanation was hard to follow (or wrong?). By the way, I was trying to find the 48 light-months (4 ly) that you mentioned for planet X in your diagram in #105. I saw approx 92 months. Did I miss something?

 

[Dale]

why must B's own coordinate system not maintain a 1:1 relation between space and time? ...

I don't know of a valid coordinate system for a non-inertial POV.

Here is a reference that answers both of those questions: http://arxiv.org/abs/gr-qc/0104077

Figure 1 shows why the naive synchronization convention is not 1 to 1, and then page 3 provides a valid coordinate system for non-inertial observers.

However, if the body is purely inertial, then B should be able to determine where the body really exists by accounting for his proper accelerations from his own proper accelerations (using his own accelerometer data) relative to his initial inertial frame.

If you assume that you know the as-yet-undetectable motion then you can solve for its current location in any coordinate system. There is nothing special about an inertial frame in that respect.

That said, while I don't really disagree with your points, I don't see the problems you see in such an approach as applied to a twin B POV.

Look, it simply isn't a valid coordinate system. If the Dolby and Gull explanation didn't do it for you then try chapter 2 (especially p. 33) of Carroll's lecture notes: http://arxiv.org/abs/gr-qc/9712019

 

[nosepot]

That's fine but what was wrong with wikipedia's similar statement that you linked to in post #69 where you summarized it with the word "argh!":

The asymmetry between the Earth and the spaceship is manifested in this diagram by the fact that more blue-shifted (fast aging) images are received by the ship. Put another way, the spaceship sees the image change from a red-shift (slower aging of the image) to a blue-shift (faster aging of the image) at the midpoint of its trip (at the turnaround, 5 years after departure); the Earth sees the image of the ship change from red-shift to blue shift after 9 years (almost at the end of the period that the ship is absent).

The problem with the Wiki page is that that paragraph is embedded among a mess of poorly structured information. That paragraph occurs half way down the page. There is also a large emphasis on the acceleration and deceleration of the traveller in the material which comes further up the page. The next section starts to talk about a general relativity explanation for what happens at the turnaround.

As for the paragraph itself, the inclusion of the actual times would initial distract from the concept you are trying to convey, but the explanation is essentially the same. My general criticism remains for the Doppler explanation, it is challenging to reconcile what we are told should be an observed slowing of the other twin's clock, when the frequencies of the signals which are meant to indicate their aging is Doppler shifted.

That said, I'm still unhappy with this and my similar lay explantion, without digging deeper it doesn't give a sense of why the same time discrepancy wouldn't arise if Earth was the preferred frame and there was no such thing as relativity, as it would still take time for the Earth twin to learn about the turnaround. This is the obvious rebuttal my explanation would meet. The answer is obviously disguised in the fact that the speed of light in each frame is the same, but this doesn't jump out of a lay explanation.

 

[PeterDonis]

it is challenging to reconcile what we are told should be an observed slowing of the other twin's clock

Who told you that? Time dilation is not directly observed. Doppler is. That's the point.

when the frequencies of the signals which are meant to indicate their aging is Doppler shifted.

Again, you've got it backwards. The Doppler shift is the direct observable. You can "explain" the entire twin scenario just by talking about what each twin actually sees, i.e., the Doppler shifted signals they actually observe. That's what the Usenet Physics FAQ page I linked to earlier does: it shows how each twin actually *sees* the other's clock readings change. The asymmetry in *when* the change from Doppler redshift to Doppler blueshift is seen (right at the turnaround for the traveling twin, much later for the stay-at-home twin) explains why the traveling twin's clock has less elapsed time:

* The stay-at-home twin actually sees (meaning, receives light signals, Doppler shift and all, that show) the traveling twin's clock running slow for most of the trip, with a burst of running fast at the end; but that burst isn't enough for the traveling twin's clock to catch up, so it still shows less elapsed time when the two meet.

* The traveling twin actually sees (meaning, receives light signals, Doppler shift and all, that show) the stay-at-home twin's clock running slow for half his trip, then running fast for the other half. The period when it's running fast more than makes up for the period when it's running slow, so the end result is that the stay-at-home twin's clock shows more elapsed time when the two meet.

In other words, as ghwellsjr has already explained, you can account for all the actual observations in the twin scenario without ever talking about time dilation or length contraction or relativity of simultaneity. The only reason time dilation comes into it at all is that you insist on bringing it in, because you haven't yet grasped that you don't need to.

why the same time discrepancy wouldn't arise if Earth was the preferred frame

Frames aren't needed either. Look at my explanation above; did I mention frames? You don't need frames.

Anyway, what does "Earth is the preferred frame" mean, physically? Frames are abstractions that we use to try to help us compute answers. Nature doesn't need frames to compute answers.

it would still take time for the Earth twin to learn about the turnaround.

Yes, this would still be true in a classical Newtonian model with a finite speed of light. But this time delay alone is not the full explanation; see above. The relativistic Doppler shift formula is quite different from the non-relativistic Doppler shift formula, so the explanation I gave above would make a different prediction if non-relativistic formulas were used.

 

[PAllen]

Coordinates based on MCIF simultaneity for uniform acceleration

To make some of the discussions of non-inertial coordinates concrete, I thought it might be instructive to give one way of constructing a coordinate patch for a the trivial case of uniform acceleration. I emphasize patch, because neither radar simultaneity nor MCIF simultaneity can construct valid coordinates for all of spacetime for an eternally accelerating observer. In fact, in this particular case (uniform acceleration in x direction), their coverage is identical. Also, in this particular case, they arrive at the same surfaces of simultaneity. The only difference between them is scaling of coordinate distance in radar coordinates to force coordinate speed light to be constant (along null rays through the origin). [For any more complex non-inertial motion, radar and MCIF simultaneity differ.]

Thus, if you use MCIF simultaneity from each event on a uniformly accelerating world line (proper acceleration of g), and measure coordinate distances as proper distance along these (Euclidean flat) surfaces, and use the accelerated observer's clock to define a t coordinate, then for the x,t plane you get the metric:

ds^2 = dx^2 - (1 + gx)^2 dt^2

Note, it is not Minkowski metric, except at x=0. Note that for x=-1/g, the gtt term becomes zero and the metric is no longer valid (this is the Rindler horizon). The region x≤ -1/g is simply not covered by any valid coordinates based on MCIF. If one simply translates the horizon to be the origin via x' = x + 1/g, you get the Rindler metric as commonly given. These coordinates are the simplest case of Fermi-Normal coordinates (actually, Fermi-Normal coordinates for an inertial observer are just standard Minkowski coordinates). Note that the coordinate speed of a null ray along x is 1+gx. Again, this becomes degenerate for x ≤ -1/g.

In addition to the obvious non-Minkowski metric applicable to the accelerated observer, it is also true that LTs are inapplicable. The transform from inertial coordinates (X,T) to these is given by the following, not by an LT:

t = (1/g) arctanh(T/X)
x = √(X^2 - T^2) - 1/g

 

[PAllen]

Yes, this would still be true in a classical Newtonian model with a finite speed of light. But this time delay alone is not the full explanation; see above. The relativistic Doppler shift formula is quite different from the non-relativistic Doppler shift formula, so the explanation I gave above would make a different prediction if non-relativistic formulas were used.

Adding to what Peter noted, the key difference is symmetry. If you adopt a classical aether theory (e.g. light is similar to sound in air), two observers in relative motion, one stationary relative to the aether, the Doppler they observe will asymmetric - one will see more red/blue shift than the other. If you adopt, instead, a Newtonian corpuscular model (which has the feature that emitter velocity affects observed speed of light, as for bullets shot from a standard gun), you have symmetric Doppler, but also you predict double images to result from a sudden turnaround of an emitter observed by an inertial observer. What is known is that Doppler depends only on relative motion and light speed is unaffected by emitter speed (no double images). These two features are sufficient to derive differential aging.

 

[ghwellsjr]

Oh, I thought you meant that he was going to measure the passing starship with just his ruler like he did when they were at rest together.

Well, the only problem with your assumption is that eyeballing a passing starship traveling at 86.6% light speed on passing with a stationary meter stick would be rather difficult. Yes?

No, you're talking about the fact that our eyeballs, neurons, and brains are not fast enough to process the light signals that are available to us but we can solve that problem with high speed cameras and computers (in principle).

I have, from the very beginning of this exchange, said that you cannot use just a ruler to measure the length of a moving object like you can when they are at rest.

Instead, you have applied a "measurement" that involves the prior adjustment of two clocks.

Agreed, however it also makes use of the original meter stick.

Then you agree with what I have been saying.

And how do you do that? Well, one way is to just start with the clocks unadjusted and then do your measurement of the starship. Chances are that it comes out to be some value other than 1/2 meter. So you tweak one of the clocks in the correct direction to get it closer to 1/2 meter. Keep repeating until it comes out exactly 1/2 meter. Do you call this a measurement?

All one does is synchronise the 2 clocks using the usual Einstein clock synchronisation method, before takeoff. The test only need be run once. Nothin to it.

Only if you know the answer before you make the measurement. How did you know the answer was going to be 1/2 meter? Unless you know that, you can't synchronize two clocks that are the correct distance apart for the amount of Length Contraction for the particular speed of the starship relative to your rest frame. You left out how you measure the speed and how you determine the correct length for the measured speed.

But we're nitpicking. We agree that a ruler by itself is insufficient to measure the length of a moving object and that some reference to the one-way speed of light in a particular coordinate system is required which is why we say that Length Contraction is a coordinate effect.

Lest you think that I'm playing foul, realize that any other way of adjusting those two clocks is exactly equivalent to the method I just described.

But you said to take the measurement multiple times with unadjusted clocks and reteeking the clocks until one force fits a 1/2 meter passing starship to be recorded, and I said to take the measurement (only once) with clocks adjusted prior per a pre-takeoff clock sync procedure and the recorded startship length should precisely match the LT predicted contracted-length. How then is your method equivalent to mine?

I was talking about the standard method of synchronizing two remote clocks using the one-way speed of light in your rest frame versus synchronizing them by a more complicated method knowing the speed of the starship and what its Length Contraction would be in your rest frame.

 

[ghwellsjr]

You're right. You need some spacetime figures. Here's one for your scenario in the earth/planet's mutual rest frame. Earth is in red, planet X is in black and twin B is in blue. The dots represent one-month intervals of Proper Time for each observer/object. I have drawn in some thin black Doppler signal lines:

Your figure here looks OK ghwellsjr. However, it seems to focus only on the doppler and only during the inertial phases of twin B's flight, and does not present what happens during the virtually instant twin B acceleration at beginning, nor deceleration at the end. Those are more important to understanding my prior post.

I was trying to help you. If you want me to make one that is different, tell me what it is.

I only included those Doppler signals because you talked about it at the end of your post. I said I didn't know why you brought up Doppler but I included it in both diagrams to show you that it doesn't change in different IRF's. I also don't understand why you are concerned about Doppler during acceleration. What is the issue?

 

[GrayGhost]

No, you're talking about the fact that our eyeballs, neurons, and brains are not fast enough to process the light signals that are available to us but we can solve that problem with high speed cameras and computers (in principle).

I have, from the very beginning of this exchange, said that you cannot use just a ruler to measure the length of a moving object like you can when they are at rest.

Then we have always been in agreement.

Only if you know the answer before you make the measurement. How did you know the answer was going to be 1/2 meter?

Because, the special theory is an accepted theory, and I therefore assume the LTs a valid model of reality, or at least our best non-quantum model of space and time. The LT's require the passing inertial starship to be length-contracted by 1/γ per the inertial space station POV, per my scenario definition.

Unless you know that, you can't synchronize two clocks that are the correct distance apart for the amount of Length Contraction for the particular speed of the starship relative to your rest frame.

Well, the clocks may be synchronised anywhere in the space station frame at any time prior to the commencement of the starship's flight test. We can move those clocks about w/o effecting their sync, because it's equivalent to a slow clock transport. I figured the clocks would be synchronised at the ends of the meter stick pre-positioned for test. So while I assumed the starship would be length-contracted by 1/γ per the LTs, I positioned the lasers accordingly. However, I did not need to know the starship's expected contracted-length to synchronise the clocks in the station frame.

You left out how you measure the speed and how you determine the correct length for the measured speed.

That is not required, since I stated the speed in my scenario definition. This is why I love thought experiments so much. I can just say "this is how it is", and so it was written.

But we're nitpicking. We agree that a ruler by itself is insufficient to measure the length of a moving object and that some reference to the one-way speed of light in a particular coordinate system is required which is why we say that Length Contraction is a coordinate effect.

I have always agreed that the length of a body is per POV, and as such a coordinate effect. My posted thought experiment does not dispute that. So I'm not sure why you raise that matter.

I was talking about the standard method of synchronizing two remote clocks using the one-way speed of light in your rest frame versus synchronizing them by a more complicated method knowing the speed of the starship and what its Length Contraction would be in your rest frame.

Hmm. Well, I assumed the standard method of clock sync (Einstein's), as executed in the space station's frame. So, it would seem neither of us were talking about "other more complicated methods of clock sync". I will state it again ... the proper length of the starship was stated by scenario defintion as 1m. The starship passes the station at 0.866c inertial. The station MUST record the passing starship as length-contracted to ... 1m/γ. The LTs require it, per an accepted theory in physics. I just tried to keep it as simple as possible. I could easily have assumed an infinite number of virtual micro-laser systems, and just considered the break in continuities after the fact. Either way, the moving length would be found to be 1m/γ.

The ongoing synchronisation discussion arose later because nosepot asked about my position regarding the classic twins scenario. That always opens up cans of worms, eg as conventions for clock sync, and whether Einstein's convention of simultaneity is correct wrt nature ... versus be elected only for its convenience.

I tend to believe that the 1-way speed of light really does equal the 2 way speed of light, per inertial observers, versus it only seeming as such by coincidence. Mainly, per Occam's Razer. Just don't ask me to prove that, although a few seem to think they already have.

 

[GrayGhost]

Here is a reference that answers both of those questions: http://arxiv.org/abs/gr-qc/0104077

Figure 1 shows why the naive synchronization convention is not 1 to 1, and then page 3 provides a valid coordinate system for non-inertial observers.

I will take a look at that. Thanx.

If you assume that you know the as-yet-undetectable motion then you can solve for its current location in any coordinate system. There is nothing special about an inertial frame in that respect.

Agreed. However, if one assumes the LTs are a correct model of reality, then any method used by the non-inertial observer should produce coordinates for remote events that precisely match the LT solns of the all inertial observer, assuming we were to compare their nav and track data "after the fact". Yes?

Look, it simply isn't a valid coordinate system. If the Dolby and Gull explanation didn't do it for you then try chapter 2 (especially p. 33) of Carroll's lecture notes: http://arxiv.org/abs/gr-qc/9712019

I'll look at that as well, after I go thru the Dolby and Gull paper carefully first. I'm not sure there can be a valid coordinate system, if a valid coordinate system must meet the usual 1:1 requirement as in the Minkowski or Euclid models. It seems to me that 1:1 is maintained, but not in the usual sense. Usually, there is a 1:1 mapping between the systems that never changes over time. Per the non-inertial POV, it seems that a 1:1 mapping exists in every moment, but each moment maps 1:1 in a different way (all points still covered). One's sense-of-NOW dynamically rotates when one undergoes proper acceleration, and so it would seem that an unwaivering mapping between the 2 systems is impossible. Why not allow for a dynamic 1:1 mapping for non-inertial motion?

I agree that the radar coordinates are maybe the easiest to use for non-inertial POVs. However, we do assume that the LTs are accurate. If one can use the MCRF approach, the non-inertial POV solns match the inertial POV solns, which one would think is a good thing. I just don't see how radar coordinates can achieve such, and it seems to me one would want to achieve such. I'm all for simplicity, so long as the solns are both consistent and accurate. If I may ask ...

During a proper acceleration, ranges to remote events should dynamically change since the dilation should dynamically change. We know this per SR. I mean, we assume the LTs are accurate per all inertial POVs, yes? Why should one assume that simultaneity can be based upon the bisection of the ray's round trip, per he who undergoes proper acceleration? Assuming such, should lead to incorrect results of remote events, if the LTs are assumed an accurate model of nature ... and SR is an accepted theory in physics. Far as covariance is concerned, I would wish radar coordinates would work out perfectly. But wishing is good for little to nothing, and I doubt it can work out as such. Maybe when I disect the Gull paper, it'll make more sense. Thanx.

Regards,
Grayghost

 

[GrayGhost]

I was trying to help you. If you want me to make one that is different, tell me what it is.

I only included those Doppler signals because you talked about it at the end of your post. I said I didn't know why you brought up Doppler but I included it in both diagrams to show you that it doesn't change in different IRF's. I also don't understand why you are concerned about Doppler during acceleration. What is the issue?

I have a spacetime diagram, however I think it's too large to be uploaded as a pic. It takes a full page. If I can convert it to PDF, I should be good. Need to buy some software to convert to PDF.

Regarding doppler, it's not that I am concerned about it. My point was only that the non-inertial observer would not be aware that remote heavenly bodies where wildly flying about his spacetime system due to his own proper accelerations, or that their clocks might be spinning wildly in his spacetime system, except by the consideration and analysis of the change in relativistic doppler per EM received.

 

[Dale]

Agreed. However, if one assumes the LTs are a correct model of reality, then any method used by the non-inertial observer should produce coordinates for remote events that precisely match the LT solns of the all inertial observer, assuming we were to compare their nav and track data "after the fact". Yes?

I don't see how that follows. The LT claim to be the correct transformation between inertial frames. So, if you assume them to be a "correct model of reality" then that means you assume that they are the correct transformation between inertial frames. That assumption says nothing about the form of non-inertial transforms. There is therefore no requirement that non-inertial transforms must even approximately match the Lorentz transform, even assuming the Lorentz transform to be a "correct model of reality".

Per the non-inertial POV, it seems that a 1:1 mapping exists in every moment, but each moment maps 1:1 in a different way (all points still covered).

This is fine. This is what is called a momentarily co-moving inertial frame (MCIF). You can take any of these momentarily co-moving frames and describe all of the physics in terms of that frame.

One's sense-of-NOW dynamically rotates

Here is where you get into trouble. You can use any of the MCIF's and do all of the physics for the entire scenario. However, as soon as you try to take a "sense-of-NOW" slice (plane of simultaneity) from each MCIF and stitch those together into a single non-inertial "POV" (coordinate system) then you run into the problem that Dolby and Gull discuss.

I mean, we assume the LTs are accurate per all inertial POVs, yes?

Yes.

Why should one assume that simultaneity can be based upon the bisection of the ray's round trip, per he who undergoes proper acceleration?

We don't have to assume it. We can prove it. Simply apply the process and see that it produces a valid coordinate patch which is everywhere 1-to-1 and meets the other mathematical requirements of coordinates.

Assuming such, should lead to incorrect results of remote events, if the LTs are assumed an accurate model of nature ... and SR is an accepted theory in physics.

Again. The LT's are "an accurate model of nature" for transforming between inertial frames, which says nothing about non-inertial transforms.

Maybe when I disect the Gull paper, it'll make more sense.=

I hope so. If not, please ask here or in a new thread. I will be glad to explain as best I can.

 

[ghwellsjr]

By the way, I was trying to find the 48 light-months (4 ly) that you mentioned for planet X in your diagram in #105. I saw approx 92 months. Did I miss something?

It isn't clear from the diagrams in #105 which were customized to show Doppler.

Here is a diagram that shows the Proper Time dots for planet X (in black) aligned with those for Earth (in red):

The two events we are concerned with are the bottom red and black dots which are simultaneous in the mutual earth/planet rest frame at the Coordinate Time of 0 months and a Coordinate Distance between them of 24 light-months.

Now if we transform this diagram into one moving at 0.866c, we get:

Now you can see that the separation between those same two first events for the Earth and planet x is 48 light-months. I thought that was the 48 light-months that GrayGhost was talking about when he said:

although planet X is then 1 ly distant per B, the separation between Earth and planet X (relative to takeoff) is not 1 ly, but rather 4 ly...

It dilates to 4 ly because the location of planet X relative to B's takeoff exists in B's past, not B's NOW. His NOW said it was 2 ly distant before takeoff at v=0, and his NOW says it's 1 ly distant after takeoff at steady v=0.866c. However, after takeoff, and after rapidly attaining 0.866c, planet X had to exist far in B's own past (not his NOW) at a range of 4 ly ... for X to then be at 1 ly range NOW (per B). It's as though the rapid acceleration of B caused the location of X to fast forward 75% along its own worldline, from 4 ly to 1 ly range. In fact, as though is not accurate. That's what Einstein's theory requires, per B. The location of planet X "at takeoff" goes from 2 ly just before the virtually instant acceleration, to 4 ly just after the acceleration, because planet X's position at takeoff shifts from B's NOW to a point in B's PAST. That shifting into the past makes the interval longer, and a longer duration requires a longer earth-planet X separation.

However, he said I got it wrong:

Nope. I said 2 ly, not 4 ly. Before twin B's takeoff, the Earth and planet X are separated by 2 ly proper. That be per the earth/planetX rest frame. The dilated 4 ly sep is per twin B alone, and only after he completes his virtually instant proper acceleration from v=0 to v=0.866c ... and that's not the current separation between Earth and planet X per B (which is 1 ly), that's the separation between Earth and planet X wrt the 2 defined events of takeoff and turnabout (which is 4 ly).

However, I still don't have any idea where this 4 ly exists. He has promised to produce a diagram that shows what he's talking about:

I have a spacetime diagram, however I think it's too large to be uploaded as a pic. It takes a full page. If I can convert it to PDF, I should be good. Need to buy some software to convert to PDF.