xyzt

Here

No, it is not, you understood absolutely nothing, the clock on the ground and the clock in the satellite end up ticking at the same exact frequency. They also show the same exact elapsed time due to having the same exact offset from zero. You are not here to learn as you profess, you are here to push some fringe ideas. Not how to build them but how they function.

Classical.

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.

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. [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.

This is, like all your other claims, false. There are several tests that falsify your above claims. You only need to learn, instead of making wild and incorrect claims.

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

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

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.

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.

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.

Once in orbit the tangential speed is constant, does not vary. No, see above.

Your posts reflects misunderstanding about relativity: -it is the physical quantities (length, mass, time, etc) that change under the Lorentz transforms -it is NOT their units of measurement ([m],[kg],) that change under the Lorentz transforms

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.

Generally I post a lot of math backing up my claims, so I take exception to your dig.. Secondly, if you do not understand why RoS "conventionality" precludes testability, I could explain it to you. Just ask, don't throw digs.

Here is a very good paper from the Stanford library that explains why RoS is untestable.

Actually, it is true.

False, GR subsumes SR, it does not "rest on SR" False. SR rests on a set of axioms. Relativity of simultaneity is a CONSEQUENCE of those axioms. False. You simply misunderstand the thought experiment. In addition, thought experiments cannot be used to disprove (falsify) a theory, only real experiments can do that. As an aside, RoS is not testable. There is some debate about this issue in the philosophical (not physical circles) but the bottom line is that it isn't testable. False. It is your understanding of physics that needs to be revised. This is why your post got deleted.

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?

No. I explained why. False, GR handles BOTH the effects of gravitational potential and of motion. You claimed that you read my linked post, it is explained there. Can you follow the math?

The problem with your reasoning is that the clocks do not "drift the same amount", the amount of drift accumulates over time. So, measuring the amount of drift is useless, you need to adjust the frequency (as it is being done, at launch). You do not know by how much to adjust the frequency UNLESS you apply the predictions of GR. You did not read my post I linked in, the one that explains all this, did you?

This is false, the corrections ARE mandated by GR and are much more complex than Doppler effects. I explained the corrections here. Yes, you are wrong. Very wrong.

He doesn't mean that, you misunderstand.

If you don't "know about GR", you don't have a means of synchronizing the clocks.

The theoretical foundation of GPS is entirely GR. With one exception: there is a Sagnac effect that needs to be accounted for in the communication between satellites. The Sagnac effect is a classical effect, not a relativistic one, so GR (nor SR) are necessary in explaining it.