10-11-05, 01:45 AM #361
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/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.
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.
So anyway, the easy calculation I make is to use the simple gravitational time dilation formula I found at HyperPhysics...
T = To/√(1-2GM/Rc²)
...with the ground clock at R = 6.4x10<sup>6</sup>m, and the satellite clock at R = 2.7x10<sup>7</sup>m
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.
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.
Last edited by Pete; 10-11-05 at 08:09 PM.
10-11-05, 01:50 AM #362Originally Posted by JamesR
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.
Last edited by Pete; 10-11-05 at 08:11 PM.
10-11-05, 02:07 AM #363
Originally Posted by Pete
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.
10-11-05, 02:48 AM #364Originally Posted by James R
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.
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 don't need to argue this. The evidence is all over the forum for all to see.
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?
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.
10-11-05, 03:07 AM #365
10-11-05, 11:37 AM #366
Originally Posted by PeteOriginally Posted by Pete in "this post"
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.
Originally Posted by Pete
10-11-05, 05:05 PM #367Originally Posted by Aer
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.
Prehaps it would be appropriate to reply to the relevant post of James's, and ask him if that is what he is postulating.
Last edited by Pete; 10-11-05 at 05:14 PM.
10-11-05, 05:37 PM #368
Originally Posted by Pete
Originally Posted by Pete
Originally Posted by Pete
10-11-05, 06:54 PM #369Originally Posted by Pete
10-11-05, 07:02 PM #370Originally Posted by Aer
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/c<sup>2</sup>(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
Originally Posted by Aer
10-11-05, 07:20 PM #371Originally Posted by MacM
(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
10-11-05, 08:08 PM #372Originally Posted by 1100f
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.
10-11-05, 08:16 PM #373
Originally Posted by 1100f
Originally Posted by 1100f
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?
Last edited by Aer; 10-11-05 at 08:23 PM.
10-11-05, 09:21 PM #374Originally Posted by MacM
If you are unable to solve this trivial exercise, you are not in a position to claim that tthere are conflicts in SR
10-11-05, 09:30 PM #375Originally Posted by Aer
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.
10-11-05, 09:50 PM #376Originally Posted by 1100f
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.
10-11-05, 09:59 PM #377
10-11-05, 10:29 PM #378Originally Posted by 1100f
10-11-05, 10:34 PM #379Originally Posted by MacM
10-11-05, 10:38 PM #380Originally Posted by 1100f
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.