# Global Positioning System and Relativity

1. Dec 3, 2004

### Thomas2

The Global Positioning System is nowadays considered as the prime example for the everyday importance of Relativity. It is claimed that without the relativistic corrections (which amount to 38 microseconds/day) the error in the determination of the position would accumulate quickly to values much larger then the observed accuracy.
However, in reality the positions are actually not obtained by comparing the time signal received from the satellite with the receiver time, but by observing the difference between the time signals obtained from a number of different satellites (see http://en.wikipedia.org/wiki/Global_positioning_system for details).

Consider for simplicity a one dimensional problem where the receiver is located somewhere on the line connecting the two transmitters. In this case the signal from transmitter 1 reaches the reveiver at time

t1=t0+x1/c

and the signal from transmitter 2 reaches the receiver at time

t2=t0+x2/c ,

where t0 is the time the signal is being sent out (assuming both transmitter clocks are synchronized), x1 is the distance of the receiver from transmitter 1, x2 the distance of the receiver from transmitter2, and c the speed of light.
Now if one subtracts the two equations one gets

x1-x2=c*[t1-t2].

One knows therefore the position of the receiver just by comparing the time signals from the two transmitters (the receiver clock is completely irrelevant).
If one assumes now that the transmitter clocks are running fast or slow by a relative factor (1+e), one has instead:

x1-x2=c*[(1+e)*t1 -(1+e)*t2] = c*(1+e)*(t1-t2)

which means that the position will simply be wrong by a relative factor e, but there is obviously no accumulation as the transmitter clocks run at the same rate relatively to each other.

Now the quoted value of 38 microseconds/day due to relativity (allegedly) corresponds to e=4.4*10^-10. As the satellites are at a distance of around 20000 km (=2*10^9 cm), the positional error due to relativity should actually only be 4.4*10^-10 * 2*10^9 cm = 0.8cm! This is even much less than the presently claimed accuracy of the GPS of a few meters, so the Relativity effect should actually not be relevant at all!

Last edited: Dec 4, 2004
2. Dec 3, 2004

### Staff: Mentor

Question: How do you know the clocks are synchronized?

3. Dec 3, 2004

### Nereid

Staff Emeritus
Q2: Is the '38 microseconds/day' correction the only place where Relativity enters the whole GPS system? IOW, if the GPS system were designed using only Newtonian physics, would all aspects be the same as the GPS system we actually have, except for that one correction?

4. Dec 3, 2004

### yogi

In GPS all clocks are preset relative to the non-rotating earth centered reference system. We use the transforms that were derived by Lorentz and later by Einstein to preset the clock rate due to the relative velocity between the non rotating earth centered system and the satellite velocity, and we use GR to calculate the rate of offset due to the difference in gravitational potential between the earth clock and the clock aboard the satellite(s) prior to launch. Once these factors are determined there are only minor corrections that are transmitted to update the sateelite clocks to account for slight differences in altitude etc. Although GPS relies upon the clock rate transforms, it is non-the-less a preferred reference system. This does not mean that the offsets cannot be calculated via traditional SR, but the synchronization is conceptually simplier if the earth is taken as a preferred reference frame and all offsets are referenced thereto.

5. Dec 4, 2004

### Thomas2

a) you make sure that you use identical clocks (i.e. clocks running at exactly the same rate)
b) you start both clocks at the same moment

As shown above, condition b) is actually not necessary for determining the position as only time differences are important.

6. Dec 4, 2004

### Thomas2

This would be a further question of mine. Why is the basic relativistic correction actually physically applied by adjusting the satellite clocks pre-launch? As shown above, the correction could also simply be done by applying a constant factor to the positions when calculating the latter in the receiver. The amount of code that needs to be added to the already existing receiver software should therefore be completely negligible because of the multitude of corrections applied already. This wouldn't even have to be a fixed correction but could be made variable according to the relevant parameters which could be transmitted from the satellites together with the time signal.

7. Dec 4, 2004

### Alkatran

The satellites don't tick at the same rate, depending on their orbit.

8. Dec 4, 2004

### Thomas2

The GPS satellites are all in similar orbits, i.e. they are at the same height (same GR correction) and have identical speeds (same SR correction). There are only small corrections due to small differences of the orbits. These are being corrected for when the positions are calculated in the receiver and these corrections should be even much smaller than the 0.8cm due to the generally assumed 38 microseconds/day mismatch (see my opening post above).

9. Dec 4, 2004

### yogi

Thomas2 - I would imagine that all the calibration could be carried out in the receiver software - but the philosophy that was adopted is to establish as nearly as possible a correct common rate for all transmitting clocks. The receivers use this to make a first cut at the position, then they use a refining algorithm to compensate for other things like the "one way sagnac effect." To have all the transmitters in sync with the non rotating earth based reference would seem to be easier (at least to my way of thinking). Otherwise, it would seem that the receiver software would have to make iterative adjustments before rendering a position. This would lengthen the process and would probably lead to greater error in assessing position when the receiver is in motion relative to the earth. Another advantage of having the transmitter clocks "right on" is that the clock in the receiver does not need to be very accurate - (cheap receiver clocks lead to cheap GPS receivers).

10. Dec 4, 2004

### Staff: Mentor

There is no such thing as "identical clocks." All clocks have some random error rate - including atomic clocks.
GPS clocks were not all started at the same moment. And even if they were - see "a" above.
Huh? If the clocks aren't synchronized, how do you know whether the measured time difference is due to the distance the signal traveled or clock mis-synchronization?
An atomic clock measures oscillations (sorta) and divides that by the known oscillation rate. Thus the simplest way to ensure the clocks run at the same rate is to adjust the divider.

Thomas2, while it may very well be that the common layman's explanation of the error induced by Relativity is oversimplified. That's pretty common for layman's explanations, but it in no way means that Relativistic corrections aren't still essential to the operation of the GPS system. The system you are describing may work reasonably well: but it would not work as well as the GPS system does now.

Last edited: Dec 4, 2004
11. Apr 5, 2005

### Creator

Actually, that is not correct; each orbital clock has a different speed not only with respect to each other orbital clock but also with respect to different angular positions on the earth's surface. Calculating the resulting and continously varying SR time dilation (from relative velocities) would be practically intractable. Yogi is correct; it is far simpler (in fact, necessary) to calibrate all clocks according to one frame, the non-rotating center of earth frame. From THAT frame all clocks have the same SR & GR time dilation correction.
Thus other error adjustments due to drift, sagnac, or atmospheric diffraction,etc. are determinable.

Creator

Last edited: Apr 5, 2005
12. Apr 5, 2005

### Staff: Mentor

You're correct, but this soooo didn't need to be resurrected.

13. Apr 6, 2005

### Thomas2

All GPS satellites are essentially at the same height (see for instance http://hyperphysics.phy-astr.gsu.edu/hbase/gps.html ; in fact, http://www.losangeles.af.mil/smc/pa/fact_sheets/gps_fs.htm gives this height exactly as 10,988 nautical miles). Satellites at exactly the same height must also have exactly the same orbital speed with respect to the center of the earth (which is about 5 km/sec) and hence also the same time dilation in this frame. The rotational speed of the earth at its surface is at best 0.5 km/sec (at the equator), so the time dilation for the receiver clock should be only 1/100 of that for the satellite clocks as the time dilation factor depends on (v/c)² Even if still significant, the latter, like many other effects, is variable and hence the satellite clocks can not be preset for this and it has to be calculated in the receiver. The point that I made earlier was, why the simple constant correction is not done in the receiver as well when there are already so many complicated variable effects being taken into account. This would hardly increase the computing requirements in the receiver and one would not have to mess around with the satellite clocks, especially as relativistic effects should in fact not be relevant at all if time differences from satellites rather than absolute times are used (as shown in my opening post).

Last edited by a moderator: Apr 21, 2017
14. Apr 6, 2005

### pervect

Staff Emeritus
15. Apr 25, 2005

### mcoy

hey i'm having a homework about GPS, and this thread has proved to be very useful, but i would also like to know how waves are used in the GPS system, the principles, etc... hope you could help :P

16. Apr 25, 2005

### mcoy

please.....

17. Apr 25, 2005

### pervect

Staff Emeritus
You might try

http://www.gpsworld.com/gpsworld/static/staticHtml.jsp?id=7860 [Broken]

or browse around the USNO site

http://tycho.usno.navy.mil/gps.html

Last edited by a moderator: May 2, 2017
18. Mar 6, 2007

### psychedelic

Newtonian Model.

HI there. In all your threads and discussion, reference has been made to the Newtonian model. But where exactly is the Newtonian model? Every discussion pertaining to PGS has necessarily some relativity in it. But can someone not give any explanation of how the system incorporated Newtonian physics in it.Am doing a project on GPS and one section in it is to show the Newtonian aspect of the system. But I have not, up to now, seen a single site that talks of Newtonian aspects of GPS. Should we use the metric with diagonals all being equal to 1s? Thanks for replying.
psychedelic

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