Is time universal? NO (and its proof)

[Pete]

I have a fairly tenous grasp on this GPS thing, but I think it's enough to get a grip on basic long-term relativity effects. The short term stuff (changes in signal in times less than a day) is over my head.

So, here's my grasp:

Special Relativity says that due to the velocity of the satellites, clocks on the satellite lose 7.2 microseconds per day compared to ground clocks.

This is easy to calculate for a ground clock on a Pole, but trickier (too tricky for me) for clocks off the pole. This is because the ground clock's speed in the ECI frame is anything up to 460m/s depending on Latitude, and the relative velocity of the ground-clock and satellite clock will change through each orbit and each day. This is where those transient effects come in to confuse me.

So, a ground clock at the South Pole has zero velocity in the ECI (Earth-Centre-Inertial) frame, while the satellite speed in the ECI frame is a pretty constant 3870m/s.

That gives √(1-v²/c&sup2 = 0.9999999999168
Multiply that by 86400 seconds in a day, and we find that SR predicts the satellite clock ticks 86399.9999928 seconds for each 86400 second day on the ground clock, or 7.2 microseconds short.

This is not frame dependent. SR says that it's true in all frames.

BUT... hang on a minute. This is difficult, because if I think of the problem from a GR perspective the "gravitational field" experienced by the satellite constantly changes direction relative to the stationary-satellite clock, and I'm not sure if or how this affects the timing.

The mind-picture that I've constructed incorporates GR only as far as thinking about how deep the clocks are in Earth's gravity well. I wouldn't have thought that the direction of the field would make a difference if the potential stayed the same, but that's not much more than a wild-ass-guess...

So anyway, the easy calculation I make is to use the simple gravitational time dilation formula I found at HyperPhysics...

T = To/√(1-2GM/Rc&sup2

...with the ground clock at R = 6.4x10⁶m, and the satellite clock at R = 2.7x10⁷m

I don't know how that formula is derived, so I can't be sure that it's appropriate, but I do know that it gives a result consistent with what is apparently measured in reality once the velocity dilation is included.

The gravitational time dilation formula says the satellite clock runs 46 microseconds per day faster than the ground clock, ie the satellite clock accumulates 86400.000046 seconds for each 86400 seconds passing on the ground clock (I didn't actually just calculate that, I looked the figure up. But I have calculated it in the past).
Again, this is not frame dependent.

Adding velocity time dilation and gravitational time dilation together, we find that the satellite clock should run 39 microseconds a day faster than the ground clock... a result that I'm led to believe matches the rate that GPS clocks are actually set to run in practice.

Like I said, it's fairly tenuous and ignores transient effects... but it's all I've got.


Edit:
I've just plugged in the numbers and confirmed that the gravitational time dilation formula does indeed predict that the Satellite clock accumulates 86,400.000046 seconds for each 86400 seconds passing on the ground clock.

 

[Pete]

So, where do we go from here?

Hi James,
Ideally we'd show MacM how "reciprocity" is not a prediction of relativity in this situation... but I don't think it's worth trying.

I think that perhaps you do owe Mac an apology, however... after all, he was right when he said that there is no (long term) reciprocity between satellite clocks and ground clocks in the GPS system.

 

[Aer]

That gives √(1-v²/c²) = 0.9999999999168
Multiply that by 86400 seconds in a day, and we find that SR predicts the satellite clock ticks 86399.9999928 seconds for each 86400 second day on the ground clock, or 7.2 microseconds short.

This is not frame dependent. SR says that it's true in all frames.

SR does not say it is true in all frames. It is only true in the frame you calculated it in (ECI). All other frames will give a different value.

Also, postulating that SR effects are different with the presence of gravity is essentially nullifying all of the confirmed SR effects by experiments right here on Earth. Not to mention that gravity essentially exists everywhere even if it is negligible.

 

[MacM]

MacM:

No. What I mean is "Questions I have answered before in considerable detail go away." I have no interest in having a discussion in which I have no chance of learning anything, or of teaching anything.

If you weren't so cranky and quick to throw down a red flag you might learn something.

I started off knowing little other than that GPS had to make relativistic corrections, and the general nature of the corrections needed. I have learnt a little more about the system - most of it from my own reading on the web, a little from 2inquisitive's posts, and, as far as I can tell, nothing at all from you.

This is most telling since as I recall I opened the more recent series of GPS issue threads. Further other than a couple of misunderstandings 2Iq and myself have been in almost 100% agreement. 2Iq has added a great deal of technical information but it did not alter my intial claims did it?

Probably you did. I'm sure you didn't explain any of it. In fact, I'm sure you can't give a good description of the Sagnac effect now, either. (And I think it's relevant to the discussion I'm having with Pete.)

I take exception to your description of my knowledge but I chose to not debate that here.

I don't need to argue this. The evidence is all over the forum for all to see.

No you don't but the fact is more people are beginning to see the light and ask simular questions.

I'm talking to Pete about a situation with no gravity. A real orbit is inertial. This orbit is not, because gravity is not the cause of the orbit we're discussing. We're talking about a flat-space problem here. Since this is probably beyond you, why don't you stay out of the conversation?

Why don't you stick it in your ear. If you don't have gravity you have no orbit and no acceleration.

The only reason you think I am inconsistent, MacM, is that you don't understand my posts. I am careful to explain clearly what I mean. I sometimes make mistakes, and I'm happy to admit it when I am wrong. But I do not deliberately lie about things, as you claim. And in your particular case, your accusations that I change from one view to another as is convenient stem from your inability to appreciate subtle differences between different situations, or perhaps just a laziness on your part. You simply look for any opportunity to attack me. Instead of reading the entire argument and getting up to speed, you pick on a single point in a single post, make an unwarranted assumption, and fire away. That's why I don't engage with you any more.

Nice dodge but not factual. Again I chose to not argue that here.

 

[Pete]

SR does not say it is true in all frames. It is only true in the frame you calculated it in (ECI). All other frames will give a different value.

So you say... what's your response to [post=885181]this post[/post]?

Also, postulating that SR effects are different with the presence of gravity...

Who's postulating that?

 

[Aer]

SR does not say it is true in all frames. It is only true in the frame you calculated it in (ECI). All other frames will give a different value.

So you say... what's your response to this post?

OK, picture this.
The Satellite clock and Satellite-Stationary clock both have notepads attached.
As they pass each other, each clock stamps its current time onto both notepads.

Do you agree that the writing on the notepads is an unambiguous record of the readings of the clocks each time they pass each other?

It is not unambiguous because of the relativity of simultaneity inherent in special relativity.

You are still advocating the Local Ether model of wave propogation in which:
1) There is no relativity of simultaneity. All events that are simultaneous in one frame are simultanaeous in all other frames.
2) Time dilation is absolute. That it, the dilation is absolute to some "Local Ether" reference frame.
3) Because of 2), there is no reciprocity (i.e., you must choose the Local Ether as your rest frame in order to calculate time dilations).
4) There is no length contraction as advocated by special relativity.
5) There is length contraction, or if you will "decontraction" around massive bodies which allows for time dilation without motion.

#5 I am not completely sure about regarding local ether theory but I believe it something along those lines. But my point is, in order to make your claims about "all frames" you must drop the relativity of simultaneity which leads you to the above rules which are for the most part defined in local ether theory, not special relativity.

Also, postulating that SR effects are different with the presence of gravity...

Who's postulating that?

I thought James R said something along those lines.

 

[Pete]

It is not unambiguous because of the relativity of simultaneity inherent in special relativity.

You really think that SR predicts that the actual marks on each notepad are ambiguous? That if a notepad is retrieved and examined, what you see on it depends on your reference frame? Get real.

This is tedious, Aer. Go away. Review the basics. Look at the twin paradox again. Come back when you know what you're talking about, and we'll talk more.

Perhaps a good place to start is to think about two events that happen in the same place at the same time.
Try transforming those events to any other frame. Let me know if you find a case where those events transform to different times in the other frame.

It can't be done, of course - in SR, events which are simultaneous and in the same place in one frame are simultaneous and in the same place in all frames. That's so obvious that I'm surprised we're even talking about it.

I thought James R said something along those lines.

So why'd you put it in a reply to my post? :bugeye:
Prehaps it would be appropriate to reply to the relevant post of James's, and ask him if that is what he is postulating.

 

[Aer]

You really think that SR predicts that the actual marks on each notepad are ambiguous? That if a notepad is retrieved and examined, what you see on it depends on your reference frame? Get real.

It appears you do not understand the relativity of simultaneity once again. Relativity states that making of these "notepad markings" of yours is not simultaneous in another frame.

This is tedious, Aer. Go away.

Is this you last stand of defense? Do you agree that relativity predicts the relativity of simultaneity or not?

Review the basics. Look at the twin paradox again. Come back when you know what you're talking about, and we'll talk more.

The only one that needs to review the basics is you. You still do not know how the relativity of simultaneity applies to the twins paradox.

 

[MacM]

You really think that SR predicts that the actual marks on each notepad are ambiguous? That if a notepad is retrieved and examined, what you see on it depends on your reference frame? Get real.

I want to simply point out that this is my same arguement I've been making regarding the accumulted time on physical clocks.

 

[1100f]

Relativity states that making of these "notepad markings" of yours is not simultaneous in another frame.

Relativity states thet if you have 2 events (let us call them 1 and 2) that occur in one frame of reference at point x1 and x2, at time t1 and t2 in one reference frame, then in another frame they will occur at positions x1' and x2' at times t1' and t2'. All these quantities are related by the Lorentz transformation. Then t2' - t1' != t1 - t2. (I use != as the symbol for differnt)
So that if the events are simultaneous in one frame (say t2-t1 = 0), they will not be simultaneous in another frame (t2'-t1' !=0)., in fact the difference in the primed frame will be given by t2' - t1' = γ((t2-t1) -v/c²(x2-x1))
However you can see that if the two events are simulaneous in the unprimed frame (t2 = t1) and occur at the same position (x2 = x1) then you get that t2'-t1' = 0. They will be simultaneous in all reference frame

Do you agree that relativity predicts the relativity of simultaneity or not?

Yes SR predicts the relativity of simultaneity but in the case of simultaneous events at the same location, the simultaneity is absolute. This is of course in contrast with Galilean relativity that predicts that all events simultaneous in one frame will be simultaneous in all other frames.

 

[1100f]

I want to simply point out that this is my same arguement I've been making regarding the accumulted time on physical clocks.

To understand SR I suggest that you do the following exercice:
(For simplicity of notation I suggest that you take c = 1)
In reference frame unprimed, you have two identical clocks. One clock lacated at x = 0 at t = 0, and one clock having velocity -v, located at x = L at t = 0. Both clocks show t = 0. At this instant of time each one send a light signal in the y direction (this represent if you want a tick of the clock)
Calculate the time it takes to the moving clock (in the unprimed reference frame) to reach the clock located at the origin. and what will be the reading of each clock when they meet.

Now do a Lorentz transformation to the rest frame of the second clock (the one that was moving in the unprimed frame), this will be the primed reference frame. And perform the same calculations.

I think it will help you to understand SR

 

[MacM]

Relativity states thet if you have 2 events (let us call them 1 and 2) that occur in one frame of reference at point x1 and x2, at time t1 and t2 in one reference frame, then in another frame they will occur at positions x1' and x2' at times t1' and t2'. All these quantities are related by the Lorentz transformation. Then t2' - t1' != t1 - t2. (I use != as the symbol for differnt)
So that if the events are simultaneous in one frame (say t2-t1 = 0), they will not be simultaneous in another frame (t2'-t1' !=0)., in fact the difference in the primed frame will be given by t2' - t1' = γ((t2-t1) -v/c²(x2-x1))
However you can see that if the two events are simulaneous in the unprimed frame (t2 = t1) and occur at the same position (x2 = x1) then you get that t2'-t1' = 0. They will be simultaneous in all reference frame


Yes SR predicts the relativity of simultaneity but in the case of simultaneous events at the same location, the simultaneity is absolute. This is of course in contrast with Galilean relativity that predicts that all events simultaneous in one frame will be simultaneous in all other frames.

This is all true and correct and factual. HOwever, it only tells half of the story. While accumulated time on clocks in motion or timing of events by a remote or moving observer one must have a simultaneous start and stop from that frame of referance to make any useful statements.

BUT accumulated time is a direct function of tick rate and tick rates between frames require no such synchronous timing. That is the number of tick in each frame over a fixed and equal test time period.

Taking that into consideration the accumulated times can be easily calculated and accurate predictions of such time per each clock made without the clocks having to be synchronized.

That fact clearly shows that the Theory of Simultaneity does not alter conclusions about conflicts in SRT predictions. Most obvious is an actual analysis of data and acceptable logic with regard to the assumptions made in SRT that only relative velocity counts and that each clock ticks slower than the other.

Simultaneity does not resolve the time dilation nor spatial contraction emperically derived data conflicts.

That has not once been demonstrated in 100 years of relativity.

 

[Aer]

Relativity states thet if you have 2 events (let us call them 1 and 2) that occur in one frame of reference at point x1 and x2, at time t1 and t2 in one reference frame, then in another frame they will occur at positions x1' and x2' at times t1' and t2'. All these quantities are related by the Lorentz transformation. Then t2' - t1' != t1 - t2. (I use != as the symbol for differnt)
So that if the events are simultaneous in one frame (say t2-t1 = 0), they will not be simultaneous in another frame (t2'-t1' !=0)., in fact the difference in the primed frame will be given by t2' - t1' = γ((t2-t1) -v/c²(x2-x1))
However you can see that if the two events are simulaneous in the unprimed frame (t2 = t1) and occur at the same position (x2 = x1) then you get that t2'-t1' = 0. They will be simultaneous in all reference frame

Are you defending the claim that time dilation is absolute according to special relativity? Because it certainly is not. The round trip of the twin paradox is explained in terms of a shift in simultaneity of events. That is, from the return frame of reference, the clocks onboard each satellite did not start ticking simultaneously.

Yes SR predicts the relativity of simultaneity but in the case of simultaneous events at the same location, the simultaneity is absolute. This is of course in contrast with Galilean relativity that predicts that all events simultaneous in one frame will be simultaneous in all other frames.

You fail to realize that elapsed time requires two points in spacetime, not just one.

If you are willing to admit that time dilation is absolute for the satellite frame as seen by the satellite stationary frame (same frame as the so-called local ether frame by the way), then consider for a moment the following example:

Our satellite stationary frame is no longer the frame of the local ether, but rather, it is an inertial frame that moves between two points on the satellite orbit path going directly through the center of the Earth. Also, set the timing up such that the satellite and the satellite-through-earth meet each other at the two points. What is the time dilation of the satellite compared to the satellite-through-earth frame?

 

[1100f]

That fact clearly shows that the Theory of Simultaneity does not alter conclusions about conflicts in SRT predictions.

Please solve this little exercise, and show me according to the results of this exercise the conflicts.

If you are unable to solve this trivial exercise, you are not in a position to claim that tthere are conflicts in SR

 

[1100f]

Are you defending the claim that time dilation is absolute according to special relativity? Because it certainly is not. The round trip of the twin paradox is explained in terms of a shift in simultaneity of events. That is, from the return frame of reference, the clocks onboard each satellite did not start ticking simultaneously.


You fail to realize that elapsed time requires two points in spacetime, not just one.

If you are willing to admit that time dilation is absolute for the satellite frame as seen by the satellite stationary frame (same frame as the so-called local ether frame by the way), then consider for a moment the following example:

Our satellite stationary frame is no longer the frame of the local ether, but rather, it is an inertial frame that moves between two points on the satellite orbit path going directly through the center of the Earth. Also, set the timing up such that the satellite and the satellite-through-earth meet each other at the two points. What is the time dilation of the satellite compared to the satellite-through-earth frame?

I was just saying that although simultaneity is relative, in some cases it is absolute.
For example if in some frame you have two particlesthat move and two points A and B. If the two following events, particle 1 reaching point A and particle 2 reaching point B, are simultaneous, they will not be necessarily simultaneous in another reference frame.
However, if the two following events, particle 1 reching point A and particle B reaching the same point A are simultaneous, i.e. the two particles collide, these two events will be simultaneous in all other frames. In all other frames the particles will collide.

 

[MacM]

Please solve this little exercise, and show me according to the results of this exercise the conflicts.

If you are unable to solve this trivial exercise, you are not in a position to claim that tthere are conflicts in SR

My pleasure. It has been done numerous times before. But everytime a poster raises the issue of time dilation and reciprocity relativist try to interject (with negative innuendo) the ignorance of the poster as to The Theory of Relativity of Simultaneity.

I then point out that simultaneity is only affective at causing differentials in accumulated times based on methods of synchronous start/stop times.

That if you realize that accumulated time is a direct and unavoidable result of tick rates that the tick rate of a moving clock can be independantly (not synchronized) recorded by any number of frames and those results will not vary without direct comparison being possible.

For example if a stationary clock records the (transmitted) tick rate of a moving clock for an hour according to his clock with its proper time tick rate, he will record (i.e. - 1,800 from the clock in motion) ticks vs 3600 ticks (a tick being the standard 1 second) of his clock.

The moving clock will obviously record 3600 ticks according to his clock with its dilated tick rate such that his hour is longer than the stationary clock.

But since accumulated time is a direct function of tick rate, the stationary clock knows the dilated tick rate of the moving clock and can calculate that the moving clocks hour is actually longer than his 3,600 ticks and is actually 2 hours according to his time base.

Simultaneity does not destroy the capacity to compare clocks directly without synchronization.

 

[1100f]

My pleasure. It has been done numerous times before. But everytime a poster raises the issue of time dilation and reciprocity relativist try to interject (with negative innuendo) the ignorance of the poster as to The Theory of Relativity of Simultaneity.

I then point out that simultaneity is only affective at causing differentials in accumulated times based on methods of synchronous start/stop times.

That if you realize that accumulated time is a direct and unavoidable result of tick rates that the tick rate of a moving clock can be independantly (not synchronized) recorded by any number of frames and those results will not vary without direct comparison being possible.

For example if a stationary clock records the (transmitted) tick rate of a moving clock for an hour according to his clock with its proper time tick rate, he will record (i.e. - 1,800 from the clock in motion) ticks vs 3600 ticks (a tick being the standard 1 second) of his clock.

The moving clock will obviously record 3600 ticks according to his clock with its dilated tick rate such that his hour is longer than the stationary clock.

But since accumulated time is a direct function of tick rate, the stationary clock knows the dilated tick rate of the moving clock and can calculate that the moving clocks hour is actually longer than his 3,600 ticks and is actually 2 hours according to his time base.

Simultaneity does not destroy the capacity to compare clocks directly without synchronization.

Please solve this problem

 

[MacM]

Please solve this problem

Before I waste time, do you agree that simultaneity does not prevent direct comaprison of clock tick rate (time dilation).

 

[1100f]

Before I waste time, do you agree that simultaneity does not prevent direct comaprison of clock tick rate (time dilation).

OK, I understand that you are unable to solve this exercise
thank you

 

[MacM]

OK, I understand that you are unable to solve this exercise
thank you

That doesn't cut it. You merely chose to dodge the issue.

Do you agree that the claim that simultaneity precludes direct clock comparisons is false or not. Your failure to respond is selfevident. I said only after you state your position would I waste my time doing the exercise.

You seem to be wasting ours since you have no rebuttal of my position.