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From the official GPS website:

How accurate is GPS for timing?

GPS time transfer is a common method for synchronizing clocks and networks to Coordinated Universal Time (UTC). The government distributes UTC as maintained by the U.S. Naval Observatory (USNO) via the GPS signal in space with a time transfer accuracy relative to UTC(USNO) of ≤40 nanoseconds (billionths of a second), 95% of the time. This performance standard assumes the use of a specialized time transfer receiver at a fixed location.

How is this possible, though, if relativistic time dilation effects (both from SR and GR) are already in the order of 38µs/day which the satellite's time signal clearly has to correct for? (In contrast, such a correction is not needed for mere position fixing, see e.g. this answer.)

I mean, the GR effects assume that we live in a more or less perfect Schwarzschild spacetime (which is clearly not the case) and so the time dilation calculations are merely an approximation. Now I haven't done any calculations but to me it seems very unlikely that the time dilation of 38µs/day is accurate to within 0.1% (= 40ns / 40µs) 95% of the time. (For instance, the gravitational acceleration $g$ at ground level already varies from place to place by 0.5%.)

Now that I'm thinking about it: Could my initial assumption be wrong in that the time in the GPS time signal does not actually come from the atomic clocks on board the satellites but from a ground station (whose signal the satellites merely reflect / amplify)?

But even then I'm still astounded by the accuracy of 40ns, given that there's also a significant delay (far greater than a few nanoseconds) between the time the satellite emits the signal and the time the receiver on the ground gets it. But maybe I'm not understanding the term "time transfer accuracy" correctly?

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  • $\begingroup$ (1) relativistic corrections are applied to the satellite clocks. (2) the GPS receiver knows where the satellites are and how far away they are. (3) Averaging across multiple satellites helps. $\endgroup$ – Jon Custer Jun 18 at 14:49
  • $\begingroup$ Thanks for your comment! (1) As stated above I'm aware of the relativistic corrections. But these assume a Schwarzschild spacetime and don't correct for Earth's specific geometry. (2) What does this have to do with time signals? (3) True. But can an accuracy of <= 40ns really be guaranteed that way? $\endgroup$ – balu Jun 18 at 14:53
  • $\begingroup$ (1) The corrections are more direct - how far off is the satellite's current time relative to the controlling ground station with the system master clock. (2) If you know how far away the satellite is, you can get the delay time roughly, which (3) allows the averaging across multiple satellites seen in the sky to do its job. $\endgroup$ – Jon Custer Jun 18 at 15:02
  • $\begingroup$ (1) Could you provide a source for that? I read somewhere that the clocks on board the satellites are slowed down a bit by a fixed rate to counter relativistic effects but I haven't heard of any adjustments beyond that. $\endgroup$ – balu Jun 18 at 15:04
  • $\begingroup$ en.wikipedia.org/wiki/… may help. $\endgroup$ – Jon Custer Jun 18 at 15:05
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Your link to the official GPS website has a lot of information about how the system work, including the fact that ground stations can and do check clock accuracy and update satellite clock time as needed. The difference in clock rate built into the system is not the final answer - satellite clock times are tweaked should they get out of spec. So, we can assume that the clocks on the satellites are kept current, including all relativistic effects.

Then one can read through the Wiki article on Error analysis of the GPS system. One thing to remember is that for accurate position fix (which is equivalent to an accurate time fix) the receiver needs to see more than one satellite. That way the path length differences can be (mostly) eliminated - the clocks on each satellite are all the same time and the receiver is told where they are and the rest is math.

On my old consumer GPS, I routinely see position fixes with estimated errors of under 10 meters, or less than 30 nanoseconds. That fits well with the Wiki article above.

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  • $\begingroup$ I don't see how accurate position fixes are equivalent to accurate time fixes. For the former, only the time difference (between the arrivals of satellite signals) plays a role. In particular, assuming that the satellites are all affected by relativistic effects in the same way and that their atomic clocks all show the same time (after the occasional synchronization), relativistic affects don't play a role for position fixing of a ground receiver at all. See e.g. this answer. $\endgroup$ – balu Jun 18 at 16:19
  • $\begingroup$ Regarding "So, we can assume that the clocks on the satellites are kept current, including all relativistic effects." – I don't see how we can. My question was specifically about general-relativistic corrections due to the fact that spacetime near Earth is not perfectly Schwarzschildian. E.g. Earth is not a perfect ball (but rather an ellipsoid) and, moreover, the gravitational field above mountains is different compared to the gravitational field above the ocean (though the latter effect is certainly smaller than the overall deviation of the geoid from a ball). … $\endgroup$ – balu Jun 18 at 16:28
  • $\begingroup$ … All in all, the necessary relativistic corrections to the satellites' clocks would depend on where exactly each satellite currently is and I don't see how this is accounted for (if possible at all) or how an accuracy of 40ns can be achieved in light of this. $\endgroup$ – balu Jun 18 at 16:29
  • $\begingroup$ So, first, you need to understand that time and position are intimately related here - a time uncertainty is a position uncertainty. If the time is not right, the position of the satellites is not right, and your position on the ground is not right. Second, you seem to conflate the small changes in time experienced by the satellite because of small differences in gravity at a given point in the orbit with large overall time changes at the 40 ns or greater level. That just isn't right. Again, the Wiki goes into reasonable depth with plenty of references. $\endgroup$ – Jon Custer Jun 18 at 17:54

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