# Why do GPS satellites care about Earth clock at all? (relativity)

Many online resources make grandiose claims about how GPS would be useless without relativistic corrections.

Specifically, they outline that satellite clocks runs slower by 7μs/day due to special relativistic effects as the satellites are moving at ~ 14,000km/h. And also time runs faster by 45μs/day due to general relativity effects coming from the higher orbit.

This yields a 38μs/day net drift, which we're told would cause errors in the order of 11km/day if left uncorrected.

But if all satellites are orbiting at the same altitudes and speed, why do we care at all about time dilation between the satellites and the ground? My GPS receiver on the ground only cares about the differences between the timestamps it receives from the various satellites, right?

Specifically, receiving $[ (t_0, p_0) @ u_0, ..., (t_n, p_n) @ u_n ]$ gives me no additional information vs. receiving $[ (t_0-K, p_0) @ u_0, ..., (t_n-K, p_n) @ u_n ]$ for some unknown $K$ (where $t_i$ is the transmission time of the signal from the $i$-th satellite and $p_i$ is its position at that point and $u_i$ is the receiver stopwatch time (not an absolute clock)).

(There is one obvious way in which satellites care about time dilation, and it is that they would overestimate the earth's rotation and therefore all readings would drift west, but only by 37μs/day $\times$ 1700km/h ≈ 2cm/day! And this could be easily corrected by periodically having satellites verify their positions vs ground objects)

I'm not well versed in the theory of relativity, so surely I must be missing something. Please enlighten me!

(Other related questions:

• What do you find inadequate about the highest-voted answer to the second question you linked? – Chris Apr 28 '18 at 23:39
• It's the only online post I've found that agrees with me, but it doesn't have any kind of sources backing it, so I can't be sure that it's any more reliable than my own (untrustworthy) reasoning. – obadz Apr 29 '18 at 0:21

My GPS receiver on the ground only cares about the differences between the timestamps it receives from the various satellites, right?

If you start with the correct constellation geometry, and all the vehicles have synchronized clocks, you can use the timestamp differences to determine your position within the constellation. The absolute time doesn't matter.

But information about the geometry of the constellation is calculated from their orbital ephemerides and the current time. You can't take the pseudoranges and construct a unique geometry directly.

• A side issue is that GPS doesn't just tell you where you are, it tells you when you are. That is, a GPS unit functions as a clock that is as accurate as the atomic clocks aboard the satellites, and this is helpful for some applications. – user4552 May 3 '18 at 0:51

The essential point is: we care about relativistic corrections because the time measured on the satellite is used to determine its position.

Roughly speaking, the GPS satellites continously broadcast their position $\mathbf r_i$ and the time $t_i$ at which they've done so. The receiver then solves the following system of equations for its coordinates ($t,\mathbf r$): $$c|t-t_i|=||\mathbf r-\mathbf r_i ||\tag{*}$$ From this equation, if the clock on each satellite is shifted by the same amount $\Delta t$, the measured time $t$ would be wrong by $\Delta t$, but the position $\mathbf r$ would be unaffected. No need of relativistic corrections here.

The time $t_i$ is time as measured by the atomic clock on satellite $i$. But how does a satellite know the value of $\mathbf r_i$? What the satellite actually broadcasts is not its actual position, but something called the ephemeris, see here and here. The data contained in the ephemeris is listed here. An example of how you can compute the position of the satellite based on its ephemeris $E_i$ and its measured time $t_i$ can be found here, so actually $\mathbf r_i = \mathbf r_i(E_i,t_i)$ in a complicated way.

The coordinates used are called earth-centered earth-fixed (ECEF). If everyone measured the same time, we'd have no problem. Othwerwise, since we're interested in the ECEF coordinates, you first need to convert satellite coordinates to ECEF coordinates and then do the calculations (the orbital parameters of the satellites are determined from Earth and uploaded). But it's just easier to make the satellite clock tick slower and pretend everyone measures the same time.

• But why special relativity applies to satelites and gps if they undergo accelaration and therefore they are non-inertial frames? – Antonios Sarikas Sep 17 '19 at 15:50
• @ado sar Good question! You can do the full calculation in GR and it turns out the result is the same as naively applying the SR speed contribution plus the GR height contribution – John Donne Sep 24 '19 at 19:28

I agree with you entirely that the case of GPS satellites time-keeping usually presented in a form that is dramatized to an extent that it doesn't make any sense.

It is like an article about enforcement of car speed limits, dramatically stating that 'if engineers would not understand doppler effect police officers would not have working radar guns.'

Even if you would assemble a team of engineers who have never heard of relativistic phyiscs and have them design and deploy a GPS system you would eventually end up with a working GPS system. The team would notice the anomaly, they would identify every type of error accumulation, and they would compensate for each. In general, any error that accumulates consistently (hence predictably) can be compensated for in software.

The objective way of stating it:
GPS requires such precise time-keeping that the error in the time-keeping is hundreds of times smaller than the relativistic time effect of Earth orbit.

The dramatization is just out of whack.
Why do so many authors use it?
My best guess is that they want to convey it's not that the time-keeping on the satellites is just barely able to show the relativistic effect, no: the error in the time-keeping is hundreds of times smaller than the relativistic time effect of Earth orbit. I guess the authors are kind of saying: if random drift in the time-keeping would be as large as the deviation arising from relativistic time effect then the position error would be huge.