This article with drawings is in these links

https://www.academia.edu/40876793/Synchronizing_moving_GPS_clocks Word, PDF https://pengkuanonphysics.blogspot.com/2019/11/synchronizing-moving-gps-clocks.html PDF

1. Light pulse synchronization

Can we synchronize clocks of a moving frame? Let us see Figure 1 where we have stationary frame F1 and moving frame F’2. The 2 clocks in the frame F1 are synchronized with the master clock through a light pulse.

The 2 clocks of the frame F’2 are moving. At the time t0 in the frame F1, the 2 clocks in the moving frame F’2 coincide with that of the frame F1, but due to relativity of simultaneity the time of the moving clocks are different,

dt’=t’2-t’1=/=0. So, the light pulse does not synchronize the clocks in the moving frame.

2. Fixed clocks in orbit

GPS satellites are moving, so they suffer from relativity of simultaneity. Let us imagine a disk centered at the center of the Earth whose rim is the GPS orbit and on the rim are equally spaced clocks, see Figure 2. The disk does not rotate with the Earth and the frame of the disk is the frame F1. These clocks can be synchronized because they are fixed and they are synchronized with the master clock at the North Pole of the Earth through a light pulse. They all show the time t0 when each GPS satellite pass in front of each corresponding fixed clock. But the clocks of the GPS satellites are not synchronized with the master clock because they are moving.

3. Time in GPS satellites

The event the satellite 1 coincides with a fixed clock and the satellite 2 coincides with the next fixed clock are simultaneous in the frame F1. But these 2 events are not simultaneous in the frame F’2, which is the frame moving with the satellite 1 and containing the satellite 2, see Figure 3.

In the frame F’2, the clock of the satellite 1 reads t’1 and the clock of the satellite 2 reads t’2 because of non-simultaneity. So, we have the gap of time dt’=t’2-t’1=/=0.

In the same way, the event the satellite 2 coincides with a fixed clock and the satellite 3 coincides with the next fixed clock are simultaneous in the frame F1, but not in the frame F’3, which is the frame moving with the satellite 2 and containing the satellite 3, see Figure 3. Also for the same reason, the gap of time in the frame F’3 is dt’=t’3-t”2=/=0.

Note that here in the frame F’3, the time of the satellite 2 is denoted by t”2 with double prime to distinguish it from the time of the satellite 2 in the frame F’2.

4. Time of the satellite 2

We have 2 different notations of the time of the satellite 2: t’2 with single prime in the frame F’2 and t”2 with double prime in the frame F’3. The satellite 2 has only one time which is shown by its clock. For the event the satellite 2 coincides with the fixed clock in Figure 3, its value in the frame F’2 is t’2. Now we ask: does the reading of the clock of the satellite 2 change to a different value t”2 when seen in the frame F’3? Logically no, because t’2 is the time of the satellite 2 while t”2 is the time of the same satellite at the same event. So, we must have t’2 = t”2. In this case, we have

t’3 = t”2+dt’=t’2+dt’=t’1+dt’+dt’=t’1+2dt’

5. Time of the satellite n

For the same reason, the double-primed t”3 equals the single-primed t’3 and each double-primed t”i equals each corresponding single-primed t’i. Then, the n+1th satellite has the time: t'(n+1) =t’1+ndt’.

Because ndt’ is not zero, t'(n+1) =/=t’1 while the satellite n+1 being the satellite 1, hence contradiction.

6. If t'(n+1) =t’1?

We can assert that the satellite n+1 being the satellite 1 and having the same time, that is,

t'(n+1) =t’1. In this case, we first suppose that the single-primed t’2 does not equal the double-primed t”2, but instead their different cancels dt’, that is, t”2=t’2-dt’. So, t’3=t”2+dt’=t’2-dt’+dt’=t’1+dt’ .

In this case, we will have for the n+1th satellites t'(n+1) =t’1+dt’ .

However, we still do not have t'(n+1) =t’1. For achieving this equality, we have to suppose the difference of time between single-primed and double-primed time to make t”2 = t’2 -dt'(1+1/n). This way, we will have at the n+1th satellite: t'(n+1) =t’1+dt’-dt’*n/n=t’1

But this assumption is absurd and cannot be true. So, due to relativity of simultaneity, the satellite n+1 and the satellite 1 are the same satellite but cannot have the same time.

The only way for the satellite n+1 and the satellite 1 to have the same time is that relativity of simultaneity is zero, that is, dt’=0.