# How is relativity theory applied to gps?

1. Aug 31, 2005

### francisco

how is relativity theory applied to gps?

thanks

2. Aug 31, 2005

### mathman

The clocks in the satellites have to be very precise. General relativity is needed to correct the clocks compared to what they would do if they were sitting on the ground.

3. Aug 31, 2005

### Staff: Mentor

The clock tick rates are actually adjusted for SR and GR prior to launch, meaning while sitting on the ground, they don't stay synchronized with earth-based clocks, but once in orbit, they do.

4. Aug 31, 2005

### mikeu

Yes, time dilation causes the satellites' clocks to lose about 7us per day, and gravitational redshift causes them to gain about 45us. The net effect is a gain of 38us per day which, as Russ said, is accounted for by adjusting the frequency of the clocks when they are on the ground.

Check out http://relativity.livingreviews.org/Articles/lrr-2003-1 [Broken] (especially section 5) for information on the subject from Neil Ashby, one of the leading experts on the physics of GPS.

Last edited by a moderator: May 2, 2017
5. Sep 1, 2005

### George Jones

Staff Emeritus
As mikeu has said, if gravitational time dilation is not taken into account, a time error between the ground and the satellites accumulates at the rate of about 39 millionths of a second per day. Distance is determined by timing signals that pass between the ground and the satellites. A rough estimate on the error in position that this causes can be obtained by calculating the distance that a signal moving at the speed of light travels in 39 millionths of a second. This gives 11.6 kilometres.

To show this, I'm going to play fast and lose with differentials, but I am not going to split things up into different effects. The analysis that I give below isn't as accurate or as detailed as Ashby's, but it's still pretty good.

The Schwarzschild metric reveals (almost) all!. Let $m$ be the mass of the satellite and $M$ be the mass of the Earth. For $\theta=\pi/2$, the Schwarzschild metric is

$$d\tau^{2}=\left( 1-\frac{2M}{r}\right) dt^{2}-\left( 1-\frac{2M}{r}\right) ^{-1}dr^{2}-r^{2}d\phi^{2}.$$

Consider 2 clocks, 1 rotating along with the Earth on the Earth's surface and one in an orbiting satellite. Boths clocks have constant $r$ values, so $dr=0$ for both clocks, and, after factoring out a $dt^{2}$ the above equation becomes

$$\left( \frac{dt}{d\tau}\right) ^{2}=1-\frac{2M}{r}-v^{2},$$

where $v=rd\phi/dt$ is, approximately, the speed of something moving along a circular path. Use this equation twice - once for the clock on the Earth and once for the clock on the satellite.

How is $v$ found? By using Newtonian gravity in flat space(time)! I will proceed without justifying this approximation. A clock at the equator on the Earth moves 1 Earth circumference in 1 day, so

$$v_{Earth}=\left( 2\pi r_{Earth}\right) / \left( 1 day\right) = 1.544\times10^{-6}$$

in relativistic units. For the satellite, setting centripetal force equal to Newtonian gravitational force results in

$$m\frac{v_{sat}^{2}}{r_{sat}}=\frac{GmM}{r_{sat}^{2}}.$$

Using this with $v_{sat}=\left( 2\pi r_{sat}\right) /T$, where $T=12 hours$ is the period of the satellite's orbit, gives $r_{sat}=2.611\times10^{7}$ and $v_{sat}=1.2910\times10^{-5}$.

Now,

$$\frac{d\tau_{sat}}{d\tau_{Earth}}=\left( \frac{d\tau_{sat}}{dt}\right) \left( \frac{d\tau_{Earth}}{dt}\right) ^{-1}=\sqrt{\frac{1-\frac{2M}{r_{sat}}-v_{sat}^{2}}{1-\frac{2M}{r_{Earth}}-v_{Earth}^{2}}}.$$

Plugging values into some calculators won't work because the result is

$$\frac{d\tau_{sat}}{d\tau_{Earth}}=1.0000000004479=1+4.479\times10^{-10}$$.

The error accumulated over the course of one day is $4.479\times10^{-10}\times 1 day=38.7\times10^{-6}$ seconds.

Regards,
George

6. Sep 1, 2005

### TheAntiRelative

One other note of interest is that while the large GR effects are very pronounced and GPS would nearly immediately be off by a mile (no pun intended) if those considerations were not in the timing, the fact that the satellites are in elliptical orbits, combined with the random drift of the clocks themselves (3 us/day I think it was?)along and other small effects (sagnac and perhaps lense-thirring) requires that they be re-synchronized with earth clocks on a regular basis.

So while it is a beautiful GR test, I believe the level of error actually exceeds the kinematic effects over time and so it is not the best SR lab from what I hear.

Does anyone know how frequently they are reset? I thought I heard daily.

7. Sep 1, 2005

### Staff: Mentor

They are reset every time they pass over a ground station. But the drift rate due to SR and GR is substantially higher than the error in the clocks themselves, so they do make an excellent test-bed for Relativity.

8. Sep 2, 2005

### TheAntiRelative

You sure?

I know the GR effects are significantly higher but I'm pretty sure the kinematic (SR) effects are not higher than all the errors. When you add clock drift, eliptical orbit and all the other effects, I think the total cumulative errors are roughly the same size as the kinematic shift.

I'll have to see if I can't find a few good links to see if I'm wrong about that.

I like to be precise...

9. Nov 27, 2011

### surd100

George's workthrough (#5) to 38.7us/day is impressive. It is sometimes said that if that figure were not taken into account, the position indicated by a GPS user device would drift by about 11km/day but I've never seen an accompanying explanation. Multiplying 38us/day by the speed of light yields 11.6km/day but I don't see that explaining an about 11km/day indicated position drift. Anyone got any ideas?