- #1

francisco

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how is relativity theory applied to gps?

thanks

thanks

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- Thread starter francisco
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- #1

francisco

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how is relativity theory applied to gps?

thanks

thanks

- #2

mathman

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- #3

russ_watters

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- #4

mikeu

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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.

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.

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- #5

George Jones

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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 [itex]m[/itex] be the mass of the satellite and [itex]M[/itex] be the mass of the Earth. For [itex]\theta=\pi/2[/itex], the Schwarzschild metric is

[tex]

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}.

[/tex]

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

[tex]

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

[/tex]

where [itex]v=rd\phi/dt[/itex] 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 [itex]v[/itex] 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

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

[/tex]

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

[tex]

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

[/tex]

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

Now,

[tex]

\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}}}.

[/tex]

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

[tex]

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

[/tex].

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

Regards,

George

- #6

TheAntiRelative

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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

russ_watters

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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 theyTheAntiRelative said:

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.

- #8

TheAntiRelative

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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

surd100

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