Room-temperature superconductivity?

[rcain]

[edit]sorry... reinserted again after checking possible error: [/edit]


Johan, for what my own interpretations are worth, I agree with the logic you present.

To understand a little more clearly, I visualised the (such an...) experiment as follows (with nods to the previously cited 'optical clock experiment'):

Consider not two, but three spaceships O,A and B in open space, far away from any massive bodies. Each ship is of 'finite but negligable' mass.

Each spaceship carries identical clocks, and identical chart recorders, all ships are connected together by (long) equal legths of optical communications fibre - O-A, O-B, A-B.

Though not strictly nececcary, ship O simply serves as our default/laboratory FOR (frame of reference), and can be used as a convenient 'basis' for 'datums', for default definitions of 'events' such as 'synchronisation' of clocks, 'synchronicity, 'casuality and any other 'measures' or 'phenomena' we might wish to 'compare' later. It is also intended to ease the task of describing 'relative' perspectives and resolving any 'apparent paradoxes'.

For consistency, each of the clocks are connected to the chart recorders by 'equal lengths of optical fibre in the following way.

Code: Select all

C(A)-|
C(O)-|-R(O)
C(B)-|

C(O)-|
C(A)-|-R(A)
C(B)-|

C(O)-|
C(B)-|-R(B)
C(A)-|

where C(x) is Clock onboard ship X, R(X) is Recorder onboard ship X and '-' and '|' represent lengths of optical fibre. It is a fully symetric arrangement.

Thus, at 'initial (static) syncronisation' all 'tick events' from ALL clocks are received by ALL recorders at the same instant, albeit after a considerable, but equal (communications) 'delay'.

In addition, for the sake of 'practicality/convenience', each of the ships also carries a 'fax machine' and is thus able to fax a copy of its own chart recorder outputs to each of the other ships, at some convenient time after the experiment, (communicatioing over the same optical fibre connections). Thus no ship may be in any 'doubt' about what any of the others claim to 'observe'.

We can immediatly construe several different dynamical scenarios to explore, vis:

1)
a) Ship A remains stationary with respect to O at all times, whilst B is observed (by O, and A similarly) to be travelling at a 'constant speed' (near the speed of light), and in a 'constant direction', from an 'almost infinite' distance away, shooting past O and A, and continuing along on to an 'almost infinite' distance, still at a constant velocity. although we assume/allow that ship B must have provided the 'inertial' acceleration to get to this steady trajectory, we can also declare those periods to be 'temporarily invisible' to O and, so that we assume no 'knowledge' of it at at the outset.

At some convenient time into the experiment, we transmit a sequence of 'marker events' and 'instruction events' from O, to itself, and to A and to B, 'simultaneously' : they are 'mark1', 'mark 2', 'mark3' , to the pen chart recorders, along with a 'continuous sequence' of time clicks (for convenience), and a final request to the fax machine to 'fax back' their respective pen chart recordings to O where they will all be compared, once all have been received.

In this scanario of course, we expect A's 'record of events' to be 'substantially the same' as O's, since they are 'substantaily colocated' (though that is something else to explore!), and 'static'.

We are about 'describing the situation' as apparent from the THREE chart recordings. These are represented in the figure below:

Code: Select all

recorder trace O::
remote O(C(A))			V| . . . . | . . . . |V . . . . | . . . . |V. . . . |F . . . . | . . . . | . . . . | . . . . |
local O(C(O))			 V| . . . . | . . . . |V . . . . | . . . . |V. . . . |F . . . . | . . . . | . . . . | . . . . |Y(O)
remote O(C(B))			V|  .  .  .  .  |  .  .  .  .  |V .  .  .  .  |  .  .  .  .  |V .  .  .  .  |F .  .  .  .  | 

recorder trace A::
remote A(C(O))			V| . . . . | . . . . |V . . . . | . . . . |V. . . . |F . . . . | . . . . | . . . . | . . . . |
local A(C(A))			 V| . . . . | . . . . |V . . . . | . . . . |V. . . . |F . . . . | . . . . | . . . . | . . . . |Y(A)
remote A(C(B))			V|  .  .  .  .  |  .  .  .  .  |V .  .  .  .  |  .  .  .  .  |V .  .  .  .  |F .  .  .  .  | 

recorder trace B::
remote B(C(O))			V|  .  .  .  .  |  .  .  .  .  |V .  .  .  .  |  .  .  .  .  |V .  .  .  .  |F .  .  .  .  | 
local B(C(B))			 V| . . . . | . . . . |V . . . . | . . . . |V. . . . |F . . . . | . . . . | . . . . | . . . . |Y(B)
remote B(C(A))			V|  .  .  .  .  |  .  .  .  .  |V .  .  .  .  |  .  .  .  .  |V .  .  .  .  |F .  .  .  .  | 

('V' represent synch mark events, F represents final 'fax back' events, and 'Y', the point at which all faxes are/can-be compared. (dots and bars represent simple 'time tick event' markers, and are the recordings of the three clocks (1 local and two remote in each case), down the optical comms links.

this is fine and easy to understand: all it says is that there is an apparent 'cumulative delay' on all events received from a remote moving ship which is due solely determined by the Lorentz factor , and thus an apparrent slowing down of recorded remote-time. local time ticks all show against the 'normal' numbers.

b) a much more practical situation where to start, O, A and B are all considered 'substantially colocated' and 'static'/'at rest' relative to each other.

The recorders on each ship (O,A,B) ) are started. Each of the clocks of A and B are then synchronized to the clock of O. The 'syncronisation event(s)' are recorded on each of the traces.

Ship A stays at rest relative to O thoughout the experiment, whereas ship B accelerates off in a straight line, very fast, reaches a maximimum speed (close to c) which is maintains for some considerable time, then decelerates and comes to rest, relative to O and A. All clocks and recorders are stopped. and the results faxed back to O for comparison.

We would expect pretty much the same result as in a) above - but with additional expansion and contraction phases at the heads and tails (respectively) of the remote traces, as the remote ships (from each perspective) appeare to acelerate and decelerate.

Note: In this experiment, we can also atach an 'inertial accelerometer' to each of the ships, and an extra chanel on each of the chart recorders to accomodate them. We expect only B's chart reponse to show anything (record a 'self-perceived' acceleration) in this case.

Of course, each ship will suffer considerable delay, in being in final receipt of all the necessary faxes, unless we bring them back together again, as they were at the start of the experiment. We cuold do this in at least two ways:

c1) as to and including the scenario 'b' above, then B turns around, accelerates back totards A in an opposit trajactory, then decelerates again to a stop, reassuming the original co-located configuration.

in this case B will record twice the original 'self percieved' inertial acceleratation. Other than that, the basic seqence of scenario 'a' above, will simply repeat on all chart traces.

or

c2) as to and including the scenario 'b' above, but instead of 'B' turning around and heading back, 'A' is sent off after it, following exactly the same dynamics and path.

again, in this case we would expect a twice repeating pattern, identical to 'c1' above, - with the simple exception in this case, that the 'self percieved' inertial acceleratation is recorded by both A and B equally, though at different times.

c3) we can see by extension/ symetry, however, that if both A and B are sent off on there journeys at precisly the same time, in precisely the same direction, with the same dynamics, then they become the original static pair O and A of this experiment, and the results simply shift around, but are otherwise the same; that is, notwithstanding, we are faced with the following (inconvenience):

if both A and B are allowed to 'approach' the speed of light, and their masses are allowed to 'approach' 0, a discontinuity of calculation arises coinciding with the 'identification' ('classification' of a final state of 'relative staticism' between A and B - ie. if they are both photons, moving together in the same frame of reference. we move from a state of 'knowing virtually nothing of the 'observational history' of the other particle, to 'knowing almost everything' in a single cannonical step. there is only one 'mutually static' frame (within any scenario), whereas there are an infinite (/indeterminable) number of 'relatively moving frames', 'around' such a transition of state.

c4) A and B start of at exactly the same time, same dynamics (as in c3 above), but in opposite directions away from each other.

c5) as per c4 above but in opposite directions, towards each other.

in both c4 and c5 above we might expect similar results to scenario c2.

in all (acellerating) cases i have ignored Doppler shift and in particular 'relativistic Doppler shift' and only considered the Lorentz factor.

Therefore, in an extended experiment:

2) as per experiment(s) 1 above, but with the addition of a 'line of site communications chanels/beams' as well as the fibre optic links and, and additional chanels on the chart recorder to accomodate them. hthis is in order to enable the direct measurement of 'Doppler shift' in the experiment

The (classical and relativistic) Doppler effect depends on the component of the emitter's velocity parallel to the light's direction at emission, and the component of the receiver's velocity parallel to the light's direction at absorption, thus it applies down our fibre-optic cable also. thus we can dispense with the extra line of sight 'laser beam' and just re-use the fibre optiic, which has the additional advantage of taking variable communications distances out of the equation (all accepting as it applies equally and relativistically to each of the cables 'observed lengths').

The 'relativistic Doppler shift' is 'writen as' (thus 'recorded as') the classical Doppler effect multiplied by the ratio of the receiver and source Lorentz factors. (http://en.wikipedia.org/wiki/Relativist ... ler_effect). it is 'relative-direction dependent - thus, the scenarios c4 and c5 in particular could produce very different results from each other, since Lorentz factor is not 'direction dependent'. however,

in all the cases above, (excepting where A and B move in the same dicetcion together at exactly the same time), 'classical Doppler applies, since A and B are always moving in opposite directions to each other at some time or another.

Classical Doppler effect, should then be something we should need to subtract from the incomming (local and remote) time signals and record separately on the same chart recorders, before attempting to account for any relativastic dilation effect. (as per the published 'optical clock' experiment citted in my previous post).

there is also of course the Harress-Sagnac effect, to account for in the optical fibre, as well as the distinct Doppler affect, since there 'must be' some resultant rotation of the cable 'medium' related to specific translation of the ships at the ends of it.

so, we shall probably find another 'filterred' chanel on on our recoder helpful in oder to reveal 'solely and wholy relatavisic' effects within our experiments.

(thinking about it, '(comparatively) straight line of sight' comms links might have been easier to run the experiments with, than fiber-optic - though the results we are trying to describe do not depend on that method, so the final results should be 'equivalent'.).

as extensions to the experiment, i'd also be quite intrerested to know what happens with the following scenarios:

3) looking to record and compare relativistic 'polarisation' effects, - since there is already a 'transverse relativistic doppler effect', i'm supposing there is something to see. how about 'charality of circularly polarised light?
4) 'spin states' of the observed 'relativisic clock states' - and the the realm of 'quantum computation space'.
5) the relativistic effects of self and mutual 'gravity fields' and 'perturbations' within them. (mutually contra-rotating ('geared') systems and mutually co-rotating ('chained') systems in particular)

for anyone who has had the fortitude to trot through the above, sorry for no the nodoubt obvious'/self-evident details and lack of formulea: i thought i'd try for a picture of the complete landscape. i'm sure there are other aspects to look at, such as 'relativistic temperature' (/'noise') and 'relativistic image rendering' and so on. i certainly dont attempt to explain Lorentz contraction it'self - i just assume the phenomenon occurs, inline with what i understand of the 'actual experimental results' recorded.

[DeltaV]

To continue the off-topic trend, just saw this related item
One clock with two times: When quantum mechanics meets general relativity

and the referenced paper
Quantum interferometric visibility as a witness of general relativistic proper time

[johanfprins]

Johan, for what my own interpretations are worth, I agree with the logic you present.

Thanks!

To understand a little more clearly, I visualised the (such an...) experiment as follows (with nods to the previously cited 'optical clock experiment'):

I have scanned your post and find it interesting: However, since I am visiting friends and family in the USA and Canada at present, I will not be able to give detailed response to your detailed post. I will, however, raise a few points:

Consider not two, but three spaceships O,A and B in open space, far away from any massive bodies. Each ship is of 'finite but negligable' mass.

When considering three bodies moving relative to each other, one must be careful not to fall into one of Zeno’s paradoxes: For example, by arguing that a faster runner can never overtake a slower runner because when the faster runner reaches the position at which the slower runner was, the slower runner has already moved on.

Each spaceship carries identical clocks, and identical chart recorders, all ships are connected together by (long) equal legths of optical communications fibre - O-A, O-B, A-B.

At this point I am already picking up a problem. As Feynman has also pointed out again and again in his lectures there are two situations in Special Relativity: (1) when not looking outside of your own inertial reference frame; and (2) when looking outside of your inertial into a passing inertial reference frame. Just a pity that Feynman could also not really understand what this really implies.

In effect it means that when you are not looking “outside” you observe “real” or “primary physics”: But when you look “outside” your own inertial reference frame the “primary” physics events within the passing inertial reference frame become distorted within your own reference frame. If within each reference frame there is an identical perfect clock, no matter at which position such a clock is stationary within its own reference frame, both clocks will keep identical time, since both clocks are stationary within their respective reference frames: They WILL keep time at the same time rate whether they have been synchronized or not synchronised.

It is when you look “outside” your reference frame at an identical clock keeping time at exactly the same rate relative to its own inertial reference frame than your clock is keeping within your refrence frame, that you see the distorted, slower rate of the moving clock WITHIN YOUR INERTIAL REFERENCE FRAME. This distortion is given by the Lorentz transformation of the time rate within the passing reference frame into your own inertial reference frame. This does not mean that the two clocks are actually keeping different time rates within their own inertial reference frames respectively. Thus, if they were synchronized at any stage, one knows that the elapse of time on your clock will be exactly the same as the elapse of time on the clock moving with your twin brother. You and your brother will NOT age at different rates.

Though not strictly nececcary, ship O simply serves as our default/laboratory FOR (frame of reference), and can be used as a convenient 'basis' for 'datums', for default definitions of 'events' such as 'synchronisation' of clocks, 'synchronicity, 'casuality and any other 'measures' or 'phenomena' we might wish to 'compare' later. It is also intended to ease the task of describing 'relative' perspectives and resolving any 'apparent paradoxes'.

Here you are on your way to a Zeno paradox!

For consistency, each of the clocks are connected to the chart recorders by 'equal lengths of optical fibre in the following way.

Thus each clock now “looks outside” of its own inertial reference frame and you will thus compare distorted time rates. The optical fibres act as Lorentz transformations from one spaceship to the other. Thus you will be comparing “transformed distorted” time rates, not the actual time rates of the clocks within each space ship.

In addition you will find that although all clocks within a long spaceship will keep exactly the same time rate within the spaceship, independent of where they are on the spaceship, the distorted time-rates of these clocks as seen from another passing spaceship will be such that a clock in the tail of the spaceship will have a distorted time that is ahead of the distorted time in the nose of the spaceship. This then seems to give a “time-coordinate” which changes with position: But within an inertial reference frame the time is actually exactly the same at each and every position.

[johanfprins]

To continue the off-topic trend, just saw this related item
One clock with two times: When quantum mechanics meets general relativity

and the referenced paper
Quantum interferometric visibility as a witness of general relativistic proper time

This is further proof that NATURE PUBLISHING HOUSE should be closed down and permanently liquidated in disgust. When you send them a model which accurately explains why an amorphous superconductor can form a superinsulating phase; you cannot get it past their second rate editors. But the above article, which is based on pure paranormal speculation is prominently published.

The fact is that the Aharanov-Bohm effect is caused exactly because one KNOWS where the electron-wave's centre-of charge (and thus centtre-of mass) is. The centre-of-charge moves SMACK-BANG through the magnetic field AND NOT past the outside of the solenoid. But, of course, it is much more fun to believe that quantum mechanics is paranormal metaphysics.

Furthermore, in all quantum mechanical experiments using the two beam Mach-Zehnder interferometer, the interpretation is done in terms of metaphysics instead of actual cause-and-effect physics! That is why they come up with utter nonsense like: "What we do at present affects the past".

[deane]

To understand a little more clearly, I visualised the (such an...) experiment as follows (with nods to the previously cited 'optical clock experiment'):

Note that while your ship is moving at constant speed, all or parts of the optical fiber connecting it to the others is experiencing acceleration.

The ship must also expend energy to compensate for that acceleration, which means that the ship loses mass as it travels.

Finally, since your ship starts at "almost infinite" distance, the optical fibre must be almost inifinitely long and therefore of almost infinite mass, which starts pushing you up against various boundary conditions.

Alas, I don't have the chops to determine how much all of that would affect your results, but it does complicate the experiment.

EDIT: Immediately after posting this I realized that I was assuming a physical connection between the ship and the optic fibre, which isn't necessary. The ship could fly along the length of a stationary fibre and communicate with it without a physical connection, thereby eliminating all of the complications I list above.

[rcain]

Johan, for what my own interpretations are worth, I agree with the logic you present.

Thanks!

...and thank you for spending the time to read mine.

To understand a little more clearly, I visualised the (such an...) experiment as follows (with nods to the previously cited 'optical clock experiment'):

I have scanned your post and find it interesting: However, since I am visiting friends and family in the USA and Canada at present, I will not be able to give detailed response to your detailed post. I will, however, raise a few points:

Consider not two, but three spaceships O,A and B in open space, far away from any massive bodies. Each ship is of 'finite but negligable' mass.

When considering three bodies moving relative to each other, one must be careful not to fall into one of Zeno’s paradoxes: For example, by arguing that a faster runner can never overtake a slower runner because when the faster runner reaches the position at which the slower runner was, the slower runner has already moved on.

Each spaceship carries identical clocks, and identical chart recorders, all ships are connected together by (long) equal legths of optical communications fibre - O-A, O-B, A-B.

At this point I am already picking up a problem. As Feynman has also pointed out again and again in his lectures there are two situations in Special Relativity: (1) when not looking outside of your own inertial reference frame; and (2) when looking outside of your inertial into a passing inertial reference frame. Just a pity that Feynman could also not really understand what this really implies.

In effect it means that when you are not looking “outside” you observe “real” or “primary physics”: But when you look “outside” your own inertial reference frame the “primary” physics events within the passing inertial reference frame become distorted within your own reference frame. If within each reference frame there is an identical perfect clock, no matter at which position such a clock is stationary within its own reference frame, both clocks will keep identical time, since both clocks are stationary within their respective reference frames: They WILL keep time at the same time rate whether they have been synchronized or not synchronised.

It is when you look “outside” your reference frame at an identical clock keeping time at exactly the same rate relative to its own inertial reference frame than your clock is keeping within your refrence frame, that you see the distorted, slower rate of the moving clock WITHIN YOUR INERTIAL REFERENCE FRAME. This distortion is given by the Lorentz transformation of the time rate within the passing reference frame into your own inertial reference frame. This does not mean that the two clocks are actually keeping different time rates within their own inertial reference frames respectively. Thus, if they were synchronized at any stage, one knows that the elapse of time on your clock will be exactly the same as the elapse of time on the clock moving with your twin brother. You and your brother will NOT age at different rates.

Precisely, that is my view also, although in truth I believe it is a little more 'subtle' than simply 'looking' inside or outside. Looking 'over' or 'through' 'what' is key in my mind. (a third invariant, or relative variant, in this case - the (fiber optic) 'channel' or specific 'time-line recording'.)

Thus in the (second) 'faxed chart recording' example above, we can both 'define' and 'ensure' that, at time 'F+1bar', say (relative to any of the observers local clocks/charts), ship B completes its round trip and comes to rest stationary again beside ship A (and O).

At this time, the pilots pop out of their space capsules, shake each other by the hand and compare results.

According to both of their (local) chart recordings (the middle trace in each chart), this reunion indeed takes place at F+1bar.

However, at this time, they are both 'waiting' to be in receipt of the 'record' of the this 'present event' in the now 'permanently delayed' signals down the comms channel between them. However, any new comms channel established between them at this (new stationary) time, suffers no such (additional, cumulative) delay.

They (each) wait until point Y (approximately) against their own chart time lines (which are the same), for final receipt of the last faxed message and the full record of A and B comming to rest wrt each other.

At that time, according to the 'remote trace' on each recording, the 'rate or remote time' when viewed over the cable, resumed it's 'normal rate' wrt the 'local time line'. However, there is just the matter of this big fat delay on the remote trace, that simply cannot be 'unwound' (unlike conventinal Doppler effect which can be 'unwound' by moving in the opposit direction, Lorrentz distortion is 'insensitive' to direction, merely to 'relative (light) speed'.) It can be wound up, but not down (by any practical method we know of), thus appeares to permanently deform the recorded timeline over moving channel, rather like 'stringing' out a peice of chewing gum beyond its elastic limit.

If they now wish to 'resynchronize' their respective chart recordings, thay can only do so from the time the remote trace in each case, resumed its 'normal rate' relative to the local trace (at time F+1bar on ALL traces). Thus, in each case, they each end up with a 'floppy' loop of longer chart paper, for the remote trace, who's data points can ony be 'lined up' with the constant local trace above it, if it hangs in a curve underneath it. Also, any such 're-lining-up' of the most recent event on the (come to a stop) remote trace, with the corrsponding event on the local trace, has the unfortunate effect that they have both foreshortend the useful range of their local trace line (discarding to the right).

(This assumes of course they use a pair of scisors, with which to cut up the chart recording paper into its respective strips, and provide it with an extra dimension (the 'hanging', curved one) so that that the very ends can be glued back together again, from the time/event when 'proper time' between them can be reestablished again. (or more precisely, when the delayed and stretched recording of the remote system ALSO shows this - point Y in the chart, rather than F+1bar).

Importantly, no matter how long, they both wait there, (after stop), starring down their optical fiber's (via the chart traces), they will never again be able to view 'the present of the remote system' (or anything near to it), on the remote trace, over the 'original channel'. (Obviously here, one possible Zeno's paradox to beware, as you point out, but only if we (mistakenly) take 'apparent' things too literally).

They can however ask each other 'what did you do during your trip away' - 'oh, nothing much, I built a book case' replies the other, 'just finished it in fact as I arrived back here. Here it is.'. 'Funny that', says the first, 'I built one too, an identical one in fact, finished it just as you were arriving. Here it is.'. 'Although, whilst I was travelling, you seemed to be taking a lot longer to build yours, from what I could see. If I did't have a record of 'your' clock ticks also. In fact up to few moment ago, I could see you were still building it (down the channel) '. 'Strange, I have precisely the same view of your history', says the first. But, here both we are, with our two identical book cases. (Attempt at work-done/energy comparison implicit here).

Two further questions arise:
- at what 'point' is the fibre optic comms line 'usable' again, for real time/undilated use. Although it resumes conveyance of real time 'rate' as soon as everything comes to a stop, the Lorentz induced delay is still present, at least for all communication/recording of 'connected/sequential events' established at the outset of experiment. If we put it away and then reuse it again for some other experiment later, it behaves as normal!
- at what 'point' (or over what 'space') can the Lorentz expansion of sequential events apparent down the fiber, be said to contract again (or collapse), to the present rate of time (FOR). In the example above, we can state without doubt (it seems) that, it happens over some period between F and F+bar (the return journey/deceleration). Or, alternatively, over the entirety of the seqential record (contained along it).

From the conundrums above, it becomes clear to me that FOR's are not as simple as simply being 'stationary' wrt eah other or not. In fact they relate to the 'total accumulated' time lines (or histories) of each of the participants (observers) in experiment together with their respective views of their 'local' universe. Further, that, the 'channel' cannot be simply thought of as the 'fiber optic cable, in this case. Apparently, there is some 'extended connection' between the physicality of the cable and the events of the whole of space-time at each end of it, and along/within it, for so long as they are all 'connected', that is, the cable records a 'continuous' record of events at both ends and along its own length.

Though not strictly nececcary, ship O simply serves as our default/laboratory FOR (frame of reference), and can be used as a convenient 'basis' for 'datums', for default definitions of 'events' such as 'synchronisation' of clocks, 'synchronicity, 'casuality and any other 'measures' or 'phenomena' we might wish to 'compare' later. It is also intended to ease the task of describing 'relative' perspectives and resolving any 'apparent paradoxes'.

Here you are on your way to a Zeno paradox!

I hope not. In this example i Iargely ignore the third ship A (or alternatively O) anyway.Paradox seems just as likely to encounter with two as opposed to three entities. And Zeno's in particular, is caused by (incorrectly) taking ever decreasing 'limits'. But forsure, I am forwarned. In the example, I avoid it by concluding (correctly I hope), that however 'predictable' the history of remote events appears down the 'delayed' cable, we will 'never again' be able to 'confirm the present (near time)', down it; such a position will forever reside ahead in 'our future' - ie. we will wait forever+some. So I don't go there. I simply draw a close to the experiment at time Y once all important patterns are available for all to see.

For consistency, each of the clocks are connected to the chart recorders by 'equal lengths of optical fibre in the following way.

Thus each clock now “looks outside” of its own inertial reference frame and you will thus compare distorted time rates. The optical fibres act as Lorentz transformations from one spaceship to the other. Thus you will be comparing “transformed distorted” time rates, not the actual time rates of the clocks within each space ship.

Exactly. Although see my refiinement to the concept of such a 'channel', alluded to above.

In addition you will find that although all clocks within a long spaceship will keep exactly the same time rate within the spaceship, independent of where they are on the spaceship, the distorted time-rates of these clocks as seen from another passing spaceship will be such that a clock in the tail of the spaceship will have a distorted time that is ahead of the distorted time in the nose of the spaceship. This then seems to give a “time-coordinate” which changes with position: But within an inertial reference frame the time is actually exactly the same at each and every position.

Absolutely. That is my understanding of the phenomena also. (though when you say the tail appeares 'ahead', I simply say that the remote recording of the nose is 'lagged' and 'stretched' more than the remote record of the tail). Since the (finitely separated) nose and tail of each ship describe separate 'integral' paths through space time, thus the Lorentz factor at the nose will be advanced relative to that at the tail, from the point of view of stationary external observer, and once established, this phase difference can never be recovered/unwound (over the same, singular channel observing it from a different FOR). Rotating bodies, however, cause some further refinement to be considered.

Hope you enjoyed your trip to the US/canada.

[rcain]

To understand a little more clearly, I visualised the (such an...) experiment as follows (with nods to the previously cited 'optical clock experiment'):

Note that while your ship is moving at constant speed, all or parts of the optical fiber connecting it to the others is experiencing acceleration.

very true, and very important to recognize i think. an aspect we might explore further.

The ship must also expend energy to compensate for that acceleration, which means that the ship loses mass as it travels.

true in a practical sense, although we can easily 'discount' the factor (relatively speaking), by considering both ships A and B sharing the fuel/energy bill. (my second or third example scenario above). You could even 'propel' them using long elastic bands if you really wanted to (so no mass loss, or even energy loss). The chart results will be the same, so far as apparent dilation is concerned.

Finally, since your ship starts at "almost infinite" distance, the optical fibre must be almost inifinitely long and therefore of almost infinite mass, which starts pushing you up against various boundary conditions.

yes, although the distance is arbitrary. (and this is a thought experiment after all so we could consider, the mass-less interior of a vacuum light-pipe if we preferred, just the same, for our purposes here).

boundary conditions in general are critical, as you point out. the very reason i chose my first example (or ship B, suddenly 'come-upon', already travelling at speed, was to 'obviate' that very factor. in particular, how did the universe 'arrive' at this state of affairs to begin with. eg: some necessary energy expenditure at some point to establish a relative velocity in the first place - and whether or not we have 'properly accounted' for it in experiment; also to what 'extent' is any account ever fully 'complete' or fully 'consistent'.

questions here for sure.

Alas, I don't have the chops to determine how much all of that would affect your results, but it does complicate the experiment.

i'm sure you do, certainly no worse than I.

with all the various different scenario's i've presented above, i hoped to cover pretty much all the important cases (and even described the difference in recordings which should be apparent in each case, compared with the first example).

EDIT: Immediately after posting this I realized that I was assuming a physical connection between the ship and the optic fibre, which isn't necessary. The ship could fly along the length of a stationary fibre and communicate with it without a physical connection, thereby eliminating all of the complications I list above.

precisely. though it helps getting my head around it to consider a 'physical manifestation' of 'channel' as in the fiber. (also to explore/cancel-out the effects of 'classical' path length, etc, in this thought experiment).

the phrase (situation?) 'without a physical connection' i think bears some further reflection as to meaning here. (see the points made to Johan above on the idea of an 'extended' channel space, and his points about Feynman's interpretation of 'continuity' of observations).

[Teemu]

So what do you think about these tests where atomic clocks are synchronized and then one of them is flown around?
http://en.wikipedia.org/wiki/Hafele%E2% ... experiment
Based on the result table in the wikipedia, 1971 Hafele-Keating experiment result seems to match predicted values for both eastward and westward around the world flight.

The results were verified in an improved experiment in 1976 by the University of Maryland, this time verifying the relativistic predictions to a precision of about 1%

Additional reenactments in 1996 and 2010

http://resource.npl.co.uk/docs/publicat ... nstein.pdf

On return to NPL the travelling clock was predicted to have gained 39.8 ns, including an additional geometric factor. This compared remarkably well with a measured gain of 39.0 ns.

http://www.npl.co.uk/news/time-flies

When the two clocks are reunited, according to Einstein's theories, they should no longer be in sync - the clock that travelled around the world should be ahead of the stationary clock by a significant amount. In this case, the time-shift was predicted to be 246 ± 3 nanoseconds after taking account of the aircraft speed and height for the different flights taken in the journey. The actual measured result using the clocks was 230 ± 20 nanoseconds, which agrees (to within the measurement error) with the prediction.

[johanfprins]

...and thank you for spending the time to read mine.

I posted that I mostly scanned it. It will take a lot of time to go step-by-step through all your arguments and I am a bit tested for time at present. I believe that there are easier experiments to prove that two clocks moving with a speed v relative to one another outside a gravitational field do actually keep time at exactly the same rate within their respective intertial refrence frames. Furthermore, it is not even necessary to do an experiment, since this synchronous time-keeping is directly mandated by the Lorentz transformation, and can thus be proved by invoking the latter equations. Thus it is really a no-brainer.

[sparkyy0007]

Johan

Fascinating book Johan!
Just a suggestion but you might want to include a reference to this thread in future book deliveries. I find your clarifications a great Q&A to supplement the book.

Thanks for the quick delivery.

[johanfprins]

So what do you think about these tests where atomic clocks are synchronized and then one of them is flown around?
http://en.wikipedia.org/wiki/Hafele%E2% ... experiment

The results were verified in an improved experiment in 1976 by the University of Maryland, this time verifying the relativistic predictions to a precision of about 1%
Additional reenactments in 1996 and 2010

http://resource.npl.co.uk/docs/publicat ... nstein.pdf

There is no direct proof that a clock which is stationary within its own inertial reference frame and moving with a speed v relative to a second clock which is stationary within its own inertial reference frame, is keeping time at a slower rate than the second clock when both clocks find themselves within gravity free space. All that has been measured is that the travelling clock has kept faster time when the gravitational field-strength decreased: And that this must be so is not in dispute since it follows directly from the mechanics of matter waves.

They have not directly measured a decrease caused by the speed of the travelling clock but only deduced indirectly that it is happening. To do this they used an "average height" for the aeroplane. This will end up giving an experimentally mesured lower time increase than the calculated time increase; since aeroplanes in general follow a curved path between taking off and landing.

In order to prove that time goes slower within a clock's own inertial reference frame, when this clock's reference frame moves with a speed v relative to another "stationary" clock, the journey of the "moving clock" must occur through gravity-free space. If it would then be found that the clock does actually keep slower time (which I sincerely doubt), this result will invalidate the Lorentz transformation and thus also Einstein's Special Theory of Relativity.

[johanfprins]

Johan

Fascinating book Johan!
Just a suggestion but you might want to include a reference to this thread in future book deliveries. I find your clarifications a great Q&A to supplement the book.

Thanks for the quick delivery.

Thank you for your response and your advice. I will consider it seriously.

[Teemu]

http://en.wikipedia.org/wiki/Hafele%E2% ... experiment

eastward flight
gravitational (general relativity): 144±14 ns
kinematic (special relativity): -184 ± 18 ns
total prediction: -40 ± 23 ns
measured: -59 ± 10 ns

westward flight
gravitational (general relativity): 179±18 ns
kinematic (special relativity): 96±10 ns
total prediction: 275±21 ns
measured: 273±7 ns

so
|eastward gravitational - westward gravitational | = 35 ns
|eastward kinematic - westward kinematic | = 280 ns

According to that the difference between westward and eastward kinematic component is much more dramatic than the westward and eastward gravitational component, because

According to special relativity, the rate of a clock is greatest according to an observer who is at rest with respect to the clock. In a frame of reference in which the clock is not at rest, the clock runs slower, and the effect is proportional to the square of the velocity. In a frame of reference at rest with respect to the center of the earth, the clock aboard the plane moving eastward, in the direction of the Earth's rotation, has a greater velocity (resulting in a relative time loss) than a clock that remains on the ground, while the clock aboard the plane moving westward, against the Earth's rotation, has a lower velocity than the one on the ground, resulting in a relative time gain.

Small difference in gravitational components can be mostly explained apparently by the westward flight taking some hours more time than eastward flight.

How do you explain that dramatic difference only with gravitational component, and without kinetic component (special theory)?

[Teemu]

http://en.wikipedia.org/wiki/Twin_paradox

This result appears puzzling because each twin sees the other twin as traveling, and so, according to a naive application of time dilation, each should paradoxically find the other to have aged more slowly. In fact, the result is not a paradox in the true sense, since it can be resolved within the standard framework of special relativity.

Resolution of the paradox in special relativity

The standard textbook approach treats the twin paradox as a straightforward application of special relativity. Here the Earth and the ship are not in a symmetrical relationship: the ship has a turnaround in which it undergoes non-inertial motion, while the Earth has no such turnaround. Since there is no symmetry, it is not paradoxical if one twin is younger than the other. Nevertheless it is still useful to show that special relativity is self-consistent, and how the calculation is done from the standpoint of the traveling twin.
Special relativity does not claim that all observers are equivalent, only that all observers at rest in inertial reference frames are equivalent. But the space ship jumps frames (accelerates) when it performs a U-turn. In contrast, the twin who stays home remains in the same inertial frame for the whole duration of his brother's flight. No accelerating or decelerating forces apply to the homebound twin.
There are indeed not two but three relevant inertial frames: the one in which the stay-at-home twin remains at rest, the one in which the traveling twin is at rest on his outward trip, and the one in which he is at rest on his way home. It is during the acceleration at the U-turn that the traveling twin switches frames. That is when he must adjust his calculated age of the twin at rest.

The asymmetry in the Doppler shifted images

Light paths for images exchanged during trip
Left: Earth to ship. Right: Ship to Earth.
Red lines indicate low frequency images are received
Blue lines indicate high frequency images are received

More details from wikipedia for those interested. So why is that asymmetry good enough explanation for twin paradox?

[johanfprins]

http://en.wikipedia.org/wiki/Hafele%E2% ... experiment

eastward flight
gravitational (general relativity): 144±14 ns
kinematic (special relativity): -184 ± 18 ns
total prediction: -40 ± 23 ns
measured: -59 ± 10 ns

westward flight
gravitational (general relativity): 179±18 ns
kinematic (special relativity): 96±10 ns
total prediction: 275±21 ns
measured: 273±7 ns

The kinematic change in time must ALWAYS be a decrease in the time of the "moving" clock within the reference frame of the stationary clock; and must thus be negative for both the eastward and westward flights. During the westward flight the experimenters are claiming that the kinematic factor actually increased the clock rate of the flying clock within the inertial reference frame of the clock on earth. Even this is not possible according to the time-dilation formula of Einstein. These researchers must have butchered Special Relativity to obtain reach the conclusion that this is possible.

According to that the difference between westward and eastward kinematic component is much more dramatic than the westward and eastward gravitational component, because

According to special relativity, the rate of a clock is greatest according to an observer who is at rest with respect to the clock. In a frame of reference in which the clock is not at rest, the clock runs slower, and the effect is proportional to the square of the velocity. In a frame of reference at rest with respect to the center of the earth, the clock aboard the plane moving eastward, in the direction of the Earth's rotation, has a greater velocity (resulting in a relative time loss) than a clock that remains on the ground, while the clock aboard the plane moving westward, against the Earth's rotation, has a lower velocity than the one on the ground, resulting in a relative time gain.

Exactly here, this reasoning violates Einstein’s Special theory of Relativity. The only speed that must be taken into account is the relative speed measured as if one of the clocks is uniquely stationary. To use a third reference frame (the centre of the earth) as if it is the uniquely stationary reference frame relative to which both clocks are moving with different speeds, and then to calculate the relative speeds between the clocks by subtracting their speeds relative to the centre of the earth, one violates the relativistic rule for the addition of velocities.

How do you explain that dramatic difference only with gravitational component, and without kinetic component (special theory)?

They probably got what they wanted to get; since it can definitely not be caused by the moving clock slowing down within its own inertial refrence frame by purely having a speed v relative to the other clock. To prove so-called time-dilation according to Einstein's Special Theory of Relativity, for a clock within its own inertial refrence frame, the two clocks MUST move relative to one another WITHIN GRAVITY-FREE SPACE. In these experiments the travelling clock did NOT move through gravity free space; and the anlysis was obviously flawed.