Is GPS consistent with relativity? (Split from is Relativity 100% proven)

[xyzt]

Yes, I can follow the maths. So what you say is that we have to change the clocks to account for gravitational effect prior to their launch because it would be impossible to do so once in orbit,

It is impossible to do it the way you want to do it,

 

 

 

and my question is, can we change them for the radial motion effect too or do we have to program the receivers to account for that?

 

The math that I posted shows the adjustment for the radial motion. See the term in [math]\frac{dr}{dt}[/math] in my post?

 

 

Also, what about the transverse effect, does it affect the system?

 

See the term in [math]r\frac{d\theta}{dt}[/math] in my post?

[Strange]

So what you say is that we have to change the clocks to account for gravitational effect prior to their launch because it would be impossible to do so once in orbit

 

There is a "gross" adjustment made in the design of the satellites by setting the clocks to run slower than an Earth-bound reference. This means that the regular adjustments made from the ground are much smaller.

 

do we have to program the receivers to account for that

 

The satellites are adjusted to stay synchronised to the Earth's coordinate system. The receivers then have to adjust for any further differences between the satellites (the receiver has to use at least 4 satellites to calculate its position) and the receiver's frame of reference (which may not be the same as the Earth reference).

[Le Repteux]

It is impossible to do it the way you want to do it,

The math that I posted shows the adjustment for the radial motion. See the term in [math]\frac{dr}{dt}[/math] in my post?

See the term in [math]r\frac{d\theta}{dt}[/math] in my post?

The formula that you referred me to is:

 

 

I understand that the "r" distances are those to the center of the earth, and that the "v" speeds are radial to that center, so I conclude that the omega term refers to the tangential speeds to the same center. Am I right?

[xyzt]

The formula that you referred me to is:

 

 

I understand that the "r" distances are those to the center of the earth, and that the "v" speeds are radial to that center, so I conclude that the omega term refers to the tangential speeds to the same center. Am I right?

Yes. So, I hope that you now understand that GR takes care of both radial and tangential motion, as I explained to you earlier. Yes?

Incidentally, the above is the formula used for correcting the atomic clock frequency prior to launch. It is confirmed in practice with an amazing precision.

Edited September 4, 2014 by xyzt

[Le Repteux]

Yes I understand how it works, and I thus understand why it is simpler to adjust the clocks prior to launch, but since the tangential speed of the receivers vary with latitude, I suppose that it is the equator speed that is used the equations. Does it mean that the receivers are programmed to correct this factor?

[xyzt]

Yes I understand how it works, and I thus understand why it is simpler to adjust the clocks prior to launch, but since the tangential speed of the receivers vary with latitude, I suppose that it is the equator speed that is used the equations.

Once in orbit the tangential speed is constant, does not vary.

 

 

 

Does it mean that the receivers are programmed to correct this factor?

 

No, see above.

[Le Repteux]

I was talking of the tangential speed at the earth surface, which vary with latitude. I suppose that one of the omega terms represents the tangential speed of the satellite, and the other the equator one. Am I right?

[xyzt]

I was talking of the tangential speed at the earth surface, which vary with latitude. I suppose that one of the omega terms represents the tangential speed of the satellite, and the other the equator one. Am I right?

The angular speed is wrt to ECIF (Earth Centered Inertial Frame) ,

The tangential speed is the product between the Schwarzschild radial coordinate , [math]r[/math] and the angular speed.

So, it has nothing to do with any latitude. May I suggest a few very good references on GPS? Because reading them would clear a lot of your misconceptions.

[swansont]

I was talking of the tangential speed at the earth surface, which vary with latitude. I suppose that one of the omega terms represents the tangential speed of the satellite, and the other the equator one. Am I right?

The speed of the earth doesn't matter. Clocks on the geoid run at the same rate, regardless of latitude.

[xyzt]

The speed of the earth doesn't matter. Clocks on the geoid run at the same rate, regardless of latitude.

Yes, there is a very good paper on this subject. Published in American Journal of Physics before the editor (Tobochnick and his followers) turned the journal into the garbage it is today.

[Le Repteux]

I suppose that the clocks on earth run at the same rate because the difference in frequency shift due to their rotational speed is compensated by the difference in frequency shift due to the deformation of the rotating earth, whose diameter is larger at the equator than at the poles. Is that so?

[imatfaal]

Have a read of this article

 

http://www.aapt.org/doorway/tgru/articles/ashbyarticle.pdf

 

Whilst it does talk of separate rel velocity correction and grav pot correction - it is easy to follow and will answer your questions

[Strange]

I suppose that the clocks on earth run at the same rate because the difference in frequency shift due to their rotational speed is compensated by the difference in frequency shift due to the deformation of the rotating earth, whose diameter is larger at the equator than at the poles. Is that so?

 

Run at the same rate as what? Each other? That is because they are not moving relative to one another (*). I'm not sure where deformation of the Earth comes into it.

 

(*) Well, unless they are, of course. And the altitude will have an effect.

[xyzt]

I suppose that the clocks on earth run at the same rate because the difference in frequency shift due to their rotational speed is compensated by the difference in frequency shift due to the deformation of the rotating earth, whose diameter is larger at the equator than at the poles. Is that so?

Yes, you got this right. It is somewhat more complicated than that but you got the basic idea: the difference in tangential speed is compensated by the difference in gravitational potential.Here is the AJP paper I was talking about.

Edited September 5, 2014 by xyzt

[Le Repteux]

I began reading Imatfaal's link, and at page 14, I was glad to see that it confirmed my supposition. I'll go on with this article until I understand everything, and I'll have a look on yours too xyzt.

[Strange]

 

Run at the same rate as what? Each other? That is because they are not moving relative to one another (*). I'm not sure where deformation of the Earth comes into it.

 

(*) Well, unless they are, of course. And the altitude will have an effect.

 

Clearly, I was wrong. Again!

I also need to read those articles...

[Le Repteux]

I'm through with the articles, and I fell on that consideration at page 15 in Imatfaal's one:

 

"In the older satellites these terms were compensated by setting the atomic clock frequencies down by this amount before launch–the so-called “factory frequency offset.” Atomic clocks that have been recently launched are based on Rubidium atoms. These clock frequencies may be bumped during launch so they are measured after orbit insertion and the necessary frequency corrections are transmitted to the receivers in the navigation message."

If we can completely correct the clocks from the ground, why couldn't we correct them completely by essay and error, I mean without using the relativity formulas? Of course it would be longer, but it should be possible, no? It is only an hypothetical question, it is just for fun. I am not trying to say that relativity is useless, I am just trying to see if we could operate the GPS system without knowing about the relativity effects. In the old days, we could predict the position of the planets without knowing that the sun was at the center of the system: isn't that comparable?

[xyzt]

If we can completely correct the clocks from the ground, why couldn't we correct them completely by essay and error, I mean without using the relativity formulas? Of course it would be longer, but it should be possible, no? It is only an hypothetical question, it is just for fun. I am not trying to say that relativity is useless, I am just trying to see if we could operate the GPS system without knowing about the relativity effects.

 

At this point, I am not sure that you want to learn, I am becoming more and more convinced that you are intent on pushing your fringe ideas. You can't correct the clocks if you have no means to know what the correction should be.

 

 

 

In the old days, we could predict the position of the planets without knowing that the sun was at the center of the system: isn't that comparable?

 

You mean, you want us to go back to epicycles?

[Le Repteux]

At this point, I am not sure that you want to learn, I am becoming more and more convinced that you are intent on pushing your fringe ideas. You can't correct the clocks if you have no means to know what the correction should be.

If you could put a clock in a perfect circular orbit, and if you could correct its rate from the ground as it seems to be for Rubidium clocks, it seems to me that, with time, you could find the correct rate so that it stays synchronized with the ground clock. Can you give me the reason why you say it is impossible?

You mean, you want us to go back to epicycles?

No, I don't think that we can play with clocks that way! But we can play with our minds that way though! Analogies are a way for a better understanding of phenomenon.

Edited September 5, 2014 by Le Repteux

[xyzt]

If you could put a clock in a perfect circular orbit, and if you could correct its rate from the ground as it seems to be for Rubidium clocks, it seems to me that, with time, you could find the correct rate so that it stays synchronized with the ground clock. Can you give me the reason why you say it is impossible?

 

I already did, earlier in the thread. Please go back and read, understand and stop trolling.

[Le Repteux]

You only said that it was impossible xyzt, without explaining why. Synchronizing two clocks at a distance without knowing why they are out of sync is not an important matter, and if you think it is impossible I'll take note of that, but before I quit, give me a last chance to illustrate what I mean. If an alien spaceship is going strait to yours, and you have the possibility to change the rate of your clock to synchronize it with the alien's one, could you nullify the doppler effect? And if so, couldn't you do the same with the alien ship's clock if you could adjust it from your own ship?

[xyzt]

You only said that it was impossible xyzt, without explaining why.

Yes, I explained why. So, you either don't read or you don't understand what I write, or you simply pretend that it isn't there.

 

 

 

If an alien spaceship is going strait to yours, and you have the possibility to change the rate of your clock to synchronize it with the alien's one, could you nullify the doppler effect?

 

 

[math]f_{observed}=f_0 \sqrt{\frac{1+v/c}{1-v/c}}[/math]

 

1. You don't know their speed, [math]v[/math]

2. You don't know the frequency of their clock,[math]f_0[/math]

4. You do not know their offset from zero

5. Your example is not relevant for the case is being discussed

 

 

I have to conclude that you are not here with the intent of learning, you are here with an intent of pushing your fringe agenda.

Edited September 5, 2014 by xyzt

[swansont]

 

If we can completely correct the clocks from the ground, why couldn't we correct them completely by essay and error, I mean without using the relativity formulas?

 

Hypothetically yes, but since a knowledge of relativity is required for building the clocks, one has to suspend their disbelief at the technology being developed in the first place.

[Le Repteux]

[math]f_{observed}=f_0 \sqrt{\frac{1+v/c}{1-v/c}}[/math]

 

1. You don't know their speed, [math]v[/math]

2. You don't know the frequency of their clock,[math]f_0[/math]

4. You do not know their offset from zero

5. Your example is not relevant for the case is being discussed

All we know is the frequency that we observe, and we want to sync our clock with it.

I have to conclude that you are not here with the intent of learning, you are here with an intent of pushing your fringe agenda.

You conclude too fast.

Hypothetically yes, but since a knowledge of relativity is required for building the clocks, one has to suspend their disbelief at the technology being developed in the first place.

This is a better argument. Since it is local time that we want them to keep with precision, what would be the use to know about relativity to build atomic clocks? We do not use the formulas to build the clocks, do we?

Edited September 6, 2014 by Le Repteux

[xyzt]

All we know is the frequency that we observe, and we want to sync our clock with it.

If you do that , you are NOT synching at all. Your clock has [math]f_{observed}[/math]. The other clock has [math]f_0[/math]. You are rapidly devolving into pure crackpottery.