How does GPS correct for time dilation?

1. Mar 3, 2015

doaaron

Hi all,

as I understand it, GPS satellites are offset prior to launch in order to correct for the time dilation that they experience in orbit. I am very confused about how this works in practice. I think the velocity of a GPS satellite wrt a position on earth would depend on the position on earth where the GPS data is to be analyzed.

I understand that all GPS satellites have the same orbital inclination, so wrt a fixed position on earth, the average velocity over time of the GPS satellites is the same for all of the satellites. So I can safely ignore the effect of the orbit on the amount of time dilation.

But what about the position on Earth? For example, the velocity of an observer at the north pole is very different from the velocity of an observer at the equator. I believe that the relative velocities of GPS satellites wrt the two positions would also be different. So how would this affect the accuracy of GPS?

thanks,
Aaron

2. Mar 3, 2015

Staff: Mentor

That is true due to the rotation of earth, but it does not influence the relative synchronization of the satellites with respect to a fixed observer (resting at the poles, or some arbitrary other point in space). The motion of the receiver does not matter as (typical) receivers do not use a clock on their own, and "what happens at the same time at the same place?" is independent of the reference frame.

3. Mar 3, 2015

Staff: Mentor

Relative to an inertial frame centered on the Earth, yes. But the observer at the equator is also further from the center of the Earth. The two effects on time dilation (slower due to speed, faster due to higher altitude) cancel. An observer at rest at sea level anywhere on the rotating Earth has the same clock rate. This clock rate is the reference clock rate that the GPS satellite clocks are adjusted to.

4. Mar 3, 2015

doaaron

Thanks for the replies.

I will need to think on this. I understand that GPS receivers do not use atomic clocks, but I don't know that that is important. I think that the clock signal that the reference frame receives from the GPS satellite will still be different depending on where he is on Earth due to the difference in relative velocities.

I'm also not quite convinced by this argument. I have briefly read of the argument you are making before, but I think in this case, the situation is a little different. Just taking a hypothetical example, suppose we have 3 clocks, and A is taken as the rest frame, B is moving in the -x direction with speed v, and C is moving in the +x direction with speed v. Relative to A, both B and C would have the same amount of time dilation and A would say that B and C show the same time. However, B thinks that C's time is different from his and vice versa. My point is that even if two clocks appear to be synchronized (the one at the equator and the one at the pole), it doesn't mean that a third clock moving in a different frame will show the same time wrt those two clocks.

Please feel free to correct me if I'm wrong.

best regards,
Aaron

5. Mar 3, 2015

Staff: Mentor

The frequency of the radio signal received from the GPS satellite by the Earth receiver will be different, yes. (Note that this frequency is not just affected by relative speed--it's also affected by the passage of the signal through the Earth's ionosphere, which contains charged particles that do various things to radio signals passing through.) But that frequency is not the "clock signal". The "clock signal" is a digital message encoded in the radio signal; it is the same message regardless of the frequency of the signal. The only thing the frequency change affects is what bandwidth the antenna has to have in order to be sure of receiving the signal.

Then I strongly suggest that you actually look at the math instead of trying to make arguments based on your intuitive understanding.

Speed is only one thing that affects time dilation in the gravitational field of a static, massive body like the Earth; the other is altitude. You need to take that into account.

Also, you need to be careful to distinguish "clock rate" and "clock synchronization". They are not necessarily the same. See below.

This is true in general; but the specific application I think you intend here is not correct.

Consider three clocks: clock A at the North Pole, clock B at the equator at 0 degrees longitude, and clock C at the equator at 180 degrees longitude. Relative to an inertial frame centered on the Earth, these three clocks all have different speeds: clock A is at rest, clock B is moving at about +450 m/s (call this speed $v$), and clock C is moving at about -450 m/s (i.e., $-v$). Also, clock A is at altitude $R$ (the polar radius of the Earth), while clocks B and C are at altitudes $R + h$ (where $h$ is the height of the Earth's equatorial bulge).

Let me first give the correct GR formula for the rates of these clocks. In an inertial frame centered on the Earth (i.e., not rotating), the rate of a clock at altitude $r$ (distance from the center of the Earth) and with speed $v$ is

$$\frac{d\tau}{dt} = \sqrt{ 1 - \frac{2 G M}{c^2 r} - \frac{v^2}{c^2}}$$

where a clock rate of $1$ corresponds to an observer at rest at infinity (i.e., very far away from the Earth, so the effect of the Earth's gravity is negligible). If you plug in values for the three clocks, A, B, and C, you will see that they all give the same answer for $d\tau / dt$ to a good approximation--the effect of the increase in $r$ for clocks B and C is compensated for by the effect of the nonzero $v$. (A better approximation would apply a correction for the Earth's quadrupole moment, but we won't go into that here.) So all three clocks have the same clock rate, relative to an inertial frame centered on the Earth.

The GPS system actually applies a correction to the Earth-centered inertial frame I described just now. The correction is to rescale all clock rates so that a clock rate of $1$ corresponds to the clock rate of an observer at rest on Earth at sea level. That is, clock rates are rescaled so that clocks A, B, and C all have clock rates of $1$, and thus the clock rate of an observer at rest at infinity is now greater than $1$--it is the reciprocal of the (less than $1$) rate for clocks A, B, and C calculated from the above formula. So the "GPS clock rate" of a given clock is $d\tau / dt$ for that clock, calculated from the above formula, divided by $d\tau / dt$ for the reference clock (clock A, B, or C--they're all the same). The clocks on all the GPS satellites have their rates adjusted so that they effectively tick at the same rate as the reference clock; i.e., their adjusted GPS clock rates are all $1$.

However, having the same clock rate is not sufficient for clock synchronization. Consider clocks A, B, and C again. Each of these clocks is at rest in a different inertial frame (and for clocks B and C, which inertial frame they are at rest in changes as the Earth rotates). So even if they are all running at the same rate, they don't all have the same "natural" simultaneity convention--that is, the inertial frames in which they are at rest do not agree on which events happen "at the same time". So in order to fully synchronize these clocks, we have to agree on a single simultaneity convention that they all will share (even if it is not the "natural" simultaneity convention for the inertial frame in which they are at rest). The convention that used for GPS is that of the Earth-centered inertial frame (which is the same as the "natural" simultaneity convention of clock A). All GPS clocks have their definition of what events happen "at the same time" (which basically means what time counts as "time zero" for the clock) adjusted to be the same as that of the reference clock.

So the sense in which the GPS clocks are "synchronized" is, if you want to look at it that way, "artificial"--both the rates and the simultaneity conventions of GPS clocks are changed from their "natural" ones, in order to impose a common convention on all of them. There is nothing wrong with this: it has to be done as a practical matter in order for GPS to work. But it does mean that you can't just apply your intuitions about the "natural" rates of clocks in various states of motion (or even at various altitudes) if you want to understand how timing in the GPS system works.

6. Mar 3, 2015

Filip Larsen

I think it would be more accurate to say that the two relativistic effects are opposite, but do not exactly cancel for GPS satellites. According to [1], the L1 frequency is reduced by 4.57 mHz (milli Hertz) in the space segment so that the nominal frequency of L1 at sea-level is 10.23 MHz.

Edit: just missed Peters own post where he elaborates on this.

[1] GPS, Theory and Practice, Hofmann-Wllenhof et al., Springer 1997.

Last edited: Mar 3, 2015
7. Mar 3, 2015

Staff: Mentor

Just to clarify, the cancellation I was referring to is for clocks at sea level on the rotating Earth.

8. Mar 3, 2015

doaaron

First off, thanks for taking the time to reply.

I wasn't referring to frequency......

Unfortunately, I think the math involved would be far too cumbersome. I think that reasoning out the problem is a more efficient approach in this case. Or maybe I'm wrong since you seem to have done the math :P

I'll need some more time to digest the rest of your post. It could take some time, so thanks for now.

Aaron

9. Mar 3, 2015

Staff: Mentor

Not really; I wrote down the key equation in my last post. You don't need anything more complicated than algebra.

(Of course, to derive the equation I wrote down from first principles, you need a lot more, yes; but for this discussion we can just assume it, we don't need to derive it.)

10. Mar 3, 2015

Staff: Mentor

Maybe that approach helps:
The satellites know their own position and the time of the observer N at the north pole (both as seen in the frame of N) with a high precision. They send it to Earth.

Observer E at the equator has a setup with two GPS devices: one is at rest for him (so it rotates with earth), one zips past with an arbitrary velocity. Exactly at the moment they meet each other, both check the satellite signals. They will get the same signals because they measure at the same point in spacetime. The velocity does not matter - it changes the frequency of the signals, but not the synchronization of them.

11. Mar 4, 2015

doaaron

Hi PeterDonis,

I've been mulling over your reply. However, it still seems that you are simply taking the clock rate at different points (e.g. north pole, equator, GPS satellite), and showing that the difference in clock rates (tGPS - tNorthPole) = (tGPS - tEquator). Have I misunderstood you?

I'm not sure that this approach is valid as (tGPS wrt tEquator) ≠ (tGPS - tEquator). You need to calculate the GPS satellite speed wrt your position on Earth. This is where the mathematical difficulty comes in (for me anyway).

Anyway I think I understand the solution based on the example below.

Hi mfb,

This is a good example to work with since we don't need to consider the odd shape of the Earth and its effect on clock rates. I think I finally understand it based on a modified version of this example. I explained it below for anybody interested...

So to make it even simpler, lets assume that we have five GPS satellites in view and their positions are fixed. Device A is at a fixed point on Earth which we can call the rest frame. Device B is whizzing around the Earth at velocity v. Since we have assumed that the GPS satellites are fixed in position, they are also in the rest frame. This is equivalent to the argument that the GPS satellites are calibrated before launch to tick at the same rate as a pre-defined rest frame on Earth.

At t = 0, lets say that device B passes device A. Now lets say that you are right, and B receives the same signals from the GPS satellites as A and therefore concludes that it is in the same position as A. B then continues on its journey, and at t = 1, it completes one revolution of the Earth and passes A again. Since B has been moving while A and all GPS satellites have been at rest, A will observe that the time passed by B is less than that passed by A. So B will observe that more time has passed for the GPS satellites than for B. However, none of this matters, as the way that B finds its position is to take the distance information from four of the satellites, and the time information from the last satellite. Therefore, although his own time will not be the same as that of A, his own time is not needed to calculate his position. Furthermore, as PeterDonis pointed out, it the real case, device B will not even be out of sync due to the cancellation of GR and SR time dilations.

Aaron

12. Mar 4, 2015

A.T.

The simple reasoning is that the rotating Earth forms an equipotential surface in it's rest frame. Since the Earth is at rest in that frame, there is no kinetic time dilation between clocks resting w.r.t to the Earth. The only time dilation can come from positions at different potentials, which is not the case for clocks on the equipotential surface.

13. Mar 4, 2015

jbriggs444

The "GR" and "SR" time effects cancel for an object at rest on the surface of the rotating earth. In your scenario, B is in motion and loops around the earth before coming back again to A's position. B's internal clock (if any) will not read the same as A's internal clock.

But that is irrelevant. Both A and B again receive identical satellite signals, perform identical GPS calculations and compute identical positions. Their internal clocks are not needed.

14. Mar 4, 2015

doaaron

When I said the real case, I was referring to the case where A is at the equator, B is at a pole, and both are resting on the surface of the Earth......

I also already mentioned that the internal clocks are not used in practice.

thanks anyway,
Aaron

15. Mar 4, 2015

Staff: Mentor

I have actually been trying to say two things, and that is only one of them. The proof of it is simple, if we are working in an inertial frame centered on the Earth. In that frame, we know that, for clocks at sea level and at rest on the (rotating) Earth, we have tNorthPole = tEquator. Therefore, for any given GPS satellite, the equality in the above quote must hold (since tGPS is obviously the same in both cases). Obviously, this proof is only valid in an inertial frame centered on the Earth, though; in some other inertial frame, we will not, in general, have tNorthPole = tEquator.

All of the above talks only about "natural" clock rates, however. The other thing I was trying to do is to draw an important distinction between the "natural" clock rate of a given clock (the rate at which it ticks if no corrections are applied, just due to the basic physics of the clock's path through spacetime and the choice of inertial frame in which to describe the clock) and the "GPS time" clock rate, which is artificially imposed on all the clocks in the system, since almost none of them are clocks at rest at sea level on the rotating Earth (the satellite clocks are obviously not, and even most of the ground station clocks are not at sea level), by applying corrections based on their altitude and state of motion relative to an idealized reference clock that is at rest at sea level on the rotating Earth. It looks like you now understand that GPS uses this artificial clock rate, so that the variation in "natural" clock rates does not affect its operation.

16. Mar 5, 2015

doaaron

Hi PeterDonis,

So the first point is that (tGPS - tNorthPole) = (tGPS - tEquator).

But as I argued earlier,
Actually I understood this from the beginning. From my original question...

I was also vaguely aware that clocks on the surface of the Earth run at the same rate. My big problem was that even though they run at the same rate, they have different relative velocities wrt the GPS satellites. So I was concerned that that would affect their reading of GPS time.

As a last point, what I've understood is that the whole system does end up working out because a person on the ground takes his time measurement from the GPS clocks. i.e. tEquator and tNorthPole never enter the equations. Although I was aware of this, I wasn't sure whether an observers velocity wrt a GPS satellite would affect his reading of tGPS. Now it is clear to me that it won't.

It seems that there was some misunderstanding about what each of us were saying, but anyway your information on natural clock rates on Earth was very helpful.

thanks,
Aaron

17. Mar 5, 2015

Staff: Mentor

Yes. But note that in order to do this, the ground observer must be able to receive signals from four GPS satellites, whereas if only position was being determined, three would be enough.

18. Mar 5, 2015

doaaron

Are you sure about that? I remember reading the math before. For triangulation using radar, 4 signals would be needed in a 3D coordinate system. However, GPS doesn't work by triangulation, and the time information from a 5th satellite is necessary even if you are only calculating your position. If you consider the surface of the Earth a 2D coordinate system, then 3 Satellites would be needed for triangulation, but 4 are needed using GPS. If I can find the link I will post it.

regards,
Aaron

19. Mar 5, 2015

A.T.

Why not only 2, if each measures the direction to the object (if that’s what you mean by "triangulation")

I would say 2 for triangulation (direction info), 4 for timing (distance info) and 3 for timing + constraint to surface of the Earth.

20. Mar 5, 2015

jbriggs444

If I remember correctly, the additional satellite is needed is to disambiguate between multiple possible solutions. Take the simple case of two satellites and a receiver that already has a properly synchronized time source. With a signal from one satellite you can solve for distance from the satellite. The solution set is a spherical shell. With two satellites you can solve for distance from both. The solution set is a circle. With the additional constraint that you are on the surface of the earth, the solution set is reduced to two points. So you need a third satellite to figure out which is the correct point.

If you receiver has no clock source, you need an additional satellite for that. If you are not figuring for a receiver tied for the surface of the earth you need a satellite for that. Total of five satellites.