Is GPS consistent with relativity? (Split from is Relativity 100% proven)

[Bart]

The relativity bit comes from the fact that the GPS Satellites have an atomic clock on board - this enables the timings to be good and the receivers to make an accurate trigonometrical calculation of its separation from a number of satellites and thus its position on the earth. However it was predicted in relativity that clocks in frames with a relative velocity will be seen from the observers rest frame as running slowly. It is also predicted by general relativity that clocks at a higher gravitational potential will run faster.

 

If we just ignored these predictions from relativity our GPS would not work - even the most accurate system would give incorrect answers. So we must adjust for two influences of the fact that the satellite is both in relative velocity with respect to the GPS receiver, and is at a higher gravitational potential. Special Relativity tells us, through simple sums, that the clocks on the GPS-Sats will run slower than earth bound by 7200 nano-seconds per day due to relative velocity . General Relativity tells us that the clocks on the GPS-Sats will run faster than earth bound by 45,900 nano-seconds per day. The net result is that we need to correct by 38,700 nano-seconds per day. We do this and the system works.

 

This is what we mean by saying that GPS provides evidence to support relativity.

In the GPS satellites detuning frequency of the signal generators is 10 229 999.99543 Hz, while the frequency of such generators in ground receivers is equal 10 230 000.00000 Hz.

From this it follows that the detuning of generators frequency, gives the time deviation equal 38597.0653 nanoseconds / day.

This is not consistent with the theory of relativity, as the required sum of relativistic deviations, resulting from the velocity of satellites and gravity, should be only 38419.1878 nanoseconds / day ( 45629.9702-7210.7824).

As you see the difference to the relativity requirements is significant.

[swansont]

Cite your source(s) for the values you've presented. Where do you get e.g. 45629.9702 ns?

[imatfaal]

I am curious how your figures have such accuracy.

 

Also it might be worth looking at XYZT's excellent post concerning the GR based derivation of the overall time difference.

 

[latex] \displaystyle \frac{f_1}{f_2}= \sqrt{\frac{1-r_s/r_2-\frac{(v_2/c)^2}{1-r_s/r_2}-(\frac{r_2 \omega_2}{c})^2}{1-r_s/r_1-\frac{(v_1/c)^2}{1-r_s/r_1}-(\frac{r_1 \omega_1}{c})^2}} [/latex]

 

if you plug average earth values (one of the reasons I am a little suspect with your accuracy is earth is not that round) into the above formula you get 0.9999999995705. If I multiply through by that ratio I get agreement with the claimed figure to 4 decimal places.

====

Do we use a sidereal day or the normal sort for these forms of calculation? I would have thought sidereal but you have used normal solar I believe. This would wipe out half your anomaly in a flash

[Strange]

The clocks on the satellites run at a different speed just to do a very rough compensation for relativistic effects. The detailed correction is done at the receiver where it takes into account the exact difference in gravitational potential (I can't remember now if this includes the altitude of the receiver, but it might), relative velocity, etc. The trigonometry used to calculate position by triangulating from 4 (or more) satellites (so maybe that should be "quadrangulating") has to take relativity into account in correcting distances to get anywhere near correct results. (I used to work with someone who wrote the software for GPS receivers, and he talked me through all this at the time, even though I have forgotten most of the details now.)

[xyzt]

I am curious how your figures have such accuracy.

 

Also it might be worth looking at XYZT's excellent post concerning the GR based derivation of the overall time difference.

 

[latex] \displaystyle \frac{f_1}{f_2}= \sqrt{\frac{1-r_s/r_2-\frac{(v_2/c)^2}{1-r_s/r_2}-(\frac{r_2 \omega_2}{c})^2}{1-r_s/r_1-\frac{(v_1/c)^2}{1-r_s/r_1}-(\frac{r_1 \omega_1}{c})^2}} [/latex]

 

if you plug average earth values (one of the reasons I am a little suspect with your accuracy is earth is not that round) into the above formula you get 0.9999999995705. If I multiply through by that ratio I get agreement with the claimed figure to 4 decimal places.

====

Do we use a sidereal day or the normal sort for these forms of calculation? I would have thought sidereal but you have used normal solar I believe. This would wipe out half your anomaly in a flash

Sidereal.

 

This is just another Bart "I can prove relativity wrong" thread that belongs in the Trash Can. Does not belong in "Science".

Edited June 28, 2014 by xyzt

[swansont]

The trigonometry used to calculate position by triangulating from 4 (or more) satellites (so maybe that should be "quadrangulating")

 

I've learned that the term is trilateration. You aren't using angle measurements to determine the position.

[Strange]

 

I've learned that the term is trilateration. You aren't using angle measurements to determine the position.

 

Thank you!

[studiot]

There is more to it than trilateration v triangulation.

 

In order to obtain a 'position fix' you require the intersection of a minimum of two intersecting position lines, in 2D, and 3 lines in 3D.

 

A position line is provided either as a distance (circle) or direction from a known point, thus 2 or 3 known points are required.

 

The known points may be either actual points in 2D or 3D space or they may be projections of actual points onto a standard surface, eg a sphere.

 

Calculation is then performed on the triangles, rather than quadrilaterals or other figures to establish coordinate differences.

 

These triangles may be spherical triangles as in the case of astro work or they may be euclidian, corrected for relativity.

 

The use of additional known reference points and position lines provides multiple estimates of the fix, which may be weighted and adjusted statistically to improve accuracy.

[Bart]

Cite your source(s) for the values you've presented. Where do you get e.g. 45629.9702 ns?

The calculation of the above values can be checked on the basis of formulas:

Gravitational time dilation in the orbit Delta t1 = to / (1-2GM / (Rsat * c ^ 2)) ^ 0.5

Gravitational time dilation on the surface of the Earth Delta t2 = to / (1-2GM / (R * c ^ 2)) ^ 0.5

Daily gravitational time dilation of satellites = (Delta t2 - Delta t1) * 24 * 3600 * 10 ^ 9 ns

 

where:

 

Earth's mass M = 5.97 E+24 kg

G = 6.6726 E-11 m3/kgs ^ 2

The radius of the Earth R = 6 378 000 m

The average radius of the orbit Rsat = 26 571 000 m

The average speed of the satellite Vsat = 3 873.2 m / s

to = 1 second

Edited June 28, 2014 by Bart

[imatfaal]

The calculation of the above values can be checked on the basis of formulas:

Gravitational time dilation in the orbit Delta t1 = to / (1-2GM / (Rsat * c ^ 2)) ^ 0.5

Gravitational time dilation on the surface of the Earth Delta t2 = to / (1-2GM / (R * c ^ 2)) ^ 0.5

Daily gravitational time dilation of satellites = (Delta t2 - Delta t1) * 24 * 3600 * 10 ^ 9 ns

 

where:

 

Earth's mass M = 5.97 E+24 kg

G = 6.6726 E-11 m3/kgs ^ 2

The radius of the Earth R = 6 378 000 m

The average radius of the orbit Rsat = 26 571 000 m

The average speed of the satellite Vsat = 3 873.2 m / s

to = 1 second

1.

Your figure for G strikes me as wrong. I have Newtons Universal Gravitation Constant as 6.67344e-11 with uncertainty of .00080e-11.

 

http://physics.nist.gov/cgi-bin/cuu/Value?bg|search_for=universal_in!

 

2.

Your figure for average radius of the earth is not the usual one - although I understand this is a matter of taste rather than strictly arguable. 6371000m is the more usual

[Sensei]

Bart, while satellite is orbiting around Earth, it's once closer to Sun, once further from Sun.

You have not addressed issues caused by Sun gravitation though.

The same with Moon and other solar system objects.

Moon can rise water for a couple meters, so it's not so small influence.

 

1.
Your figure for G strikes me as wrong. I have Newtons Universal Gravitation Constant as 6.67344e-11 with uncertainty of .00080e-11.

http://physics.nist.gov/cgi-bin/cuu/Value?bg|search_for=universal_in!

2.
Your figure for average radius of the earth is not the usual one - although I understand this is a matter of taste rather than strictly arguable. 6371000m is the more usual

6378100 is radius at equator. After using average 6371000 and more correct constant values, result is higher than Bart's initial 45629.9

 

I attached file, so anyone can load and modify.

satellite calcs.zip

Bart, you should t₀ multiply by 60*60*24*10^9 prior dividing. This way you will have higher precision.

IEEE floats have limited precision.

Edited June 28, 2014 by Sensei

[imatfaal]

Sensei - I have just finished doing exactly the same calculations. I am very happy that I have the same answers as you

 

[swansont]

The calculation of the above values can be checked on the basis of formulas:

 

 

Which require the proper values be used, and also don't carry with them nine significant digits.

 

There's a nuance here that one might miss in the quote in the OP: if the GPS satellites were not corrected the system would still nominally work. Since all of the satellites' clocks would be off, the differences would still be valid as long as the clocks were not drifting off from each other, and it's timing differences that give the position. But they would drift, so they need to be synched up, and the choice was to use a ground control station to do that, hence the decision to shift the clocks to reflect time on earth (specifically UTC(USNO)). There are some claims out there that positioning errors of ~10 km per day would get into the system if the clocks weren't adjusted for GR, but that assumes all the other clocks in the system were adjusted. The current clocks are good enough that you only get a few ns of timing error when synching twice a day, good for 1.5 m or less of positioning error. (With GPSIII that should be reduced)

[xyzt]

The calculation of the above values can be checked on the basis of formulas:

Gravitational time dilation in the orbit Delta t1 = to / (1-2GM / (Rsat * c ^ 2)) ^ 0.5

Gravitational time dilation on the surface of the Earth Delta t2 = to / (1-2GM / (R * c ^ 2)) ^ 0.5

Daily gravitational time dilation of satellites = (Delta t2 - Delta t1) * 24 * 3600 * 10 ^ 9 ns

 

where:

 

Earth's mass M = 5.97 E+24 kg

G = 6.6726 E-11 m3/kgs ^ 2

The radius of the Earth R = 6 378 000 m

The average radius of the orbit Rsat = 26 571 000 m

The average speed of the satellite Vsat = 3 873.2 m / s

to = 1 second

Yeah, exactly as expected, you are using an incomplete formula (and your numerical accuracy is wrong, to boot). The oversimplified formula you are using does not account for the difference in angular speed (the satellites are NOT geostationary, they MOVE wrt. the Earth surface) and the trajectories are NOT circular, they are elliptical. Yet another attempt by Bart to "I can prove relativity wrong" that falls flat on its face.

Edited June 28, 2014 by xyzt

[imatfaal]

Bart

And if I do the sums properly with the formula provided by XYZT - I get agreement to the change in frequency required to the 5th decimal place 10229999.99542880 Hz

Can you do the decent thing now and concede that GPS is a very consistent with relativity?

 

=====

by the way XYZT - I dropped the radial movement section as i) I had no idea what it was ii) it seemed very small (abs and rel) as the orbits are desribed as very near circular iii) I felt it would average out during a day

[xyzt]

Bart

 

And if I do the sums properly with the formula provided by XYZT - I get agreement to the change in frequency required to the 5th decimal place 10229999.99542880 Hz

 

Can you do the decent thing now and concede that GPS is a very consistent with relativity?

 

=====

by the way XYZT - I dropped the radial movement section as i) I had no idea what it was ii) it seemed very small (abs and rel) as the orbits are desribed as very near circular iii) I felt it would average out during a day

Yes, the radial motion is very small but it is still accounted for by the wonderful people who designed GPS. The adjustment is table driven based on the ephemeris.

Edited June 28, 2014 by xyzt

[Bart]

Sensei - I have just finished doing exactly the same calculations. I am very happy that I have the same answers as you

 

calc sheet.jpg

 

Explanation and some questions:

1. The GPS system has been introduced for commercial use in the 90s, hence in my calculation was use value of G such that applied to 2002.
2. Orbits of GPS satellites are strongly close to circular, so the average radius of the orbit 26 571 000 m is required in order to obtain an average orbital speed of the satellite 3 873.2 m/s. Such a speed is required in order to provide the GPS satellite orbital period equal to half of sidereal day i.e.11h58m2s .
3. Adopted for calculation the radius of the Earth was as at the equator , because on the equator GPS also needs to work properly.
4. Even the gravitational deviation of 45 721.203ns , as by your actual calculations, still does not ensure compliance with the detuning frequency used in satellites (45 721.203 – 7 210.7824 = 38 510.4206ns).
5. Satellite orbits are inclined at an angle of 55 degrees to the equator. The Earth rotates at a speed of 463.8 m/s at the equator and 0 m/s at the poles. The question then arises of what the actual speed of the satellite is in relation to GPS receivers, anywhere on the Earth? How does this apply to the calculation of time dilation?
6. Apart from the refraction of radio waves in the stratosphere and the troposphere of the Earth, the GPS receiver can see moving satellites at different and changing angles, so there is inevitably the Doppler effect, which changes the frequency of the received radio signals from the satellite.
7. It should be noted that the GPS receiver does not have its own time, as the time in the GPS receiver is digitally synchronized with the satellite clock. There is no room, therefore, for any time dilation between satellite and GPS reciver, because the numerical values transmitted from the satellites do not change.

[imatfaal]

I will address your 7 points when I have a bit more time. Would you comment on the fact that my use of the single GR calculation provides an answer that is correct?

 

We all realise that there will never be a single solution that will fit all parameters - there are too many changes over the earth, during the orbit, and other variations; but if you use the average parameters you get the figures used. Your claim way back at the top was the the alteration did not concur with the calculations under special relativity and general relativity - we have now shown that the alteration is exactly what GR would calculate on average parameters.

 

You said "This is not consistent with the theory of relativity,..." do you know acknowledge that it is completely consistent?

[xyzt]

 

 

1. Satellite orbits are inclined at an angle of 55 degrees to the equator. The Earth rotates at a speed of 463.8 m/s at the equator and 0 m/s at the poles. The question then arises of what the actual speed of the satellite is in relation to GPS receivers, anywhere on the Earth? How does this apply to the calculation of time dilation?

 

Easy:

 

[latex] \displaystyle d \tau_2= d \tau_1 \sqrt{\frac{1-r_s/r_2-\frac{(v_2/c)^2}{1-r_s/r_2}-(\frac{r_2 \omega_2}{c})^2}{1-r_s/r_1-\frac{(v_1/c)^2}{1-r_s/r_1}-(\frac{r_1 \omega_1}{c})^2}} [/latex]

 

Integrate over one sidereal day the RHS and you get [math] \tau_2[/math].

 

 

 

Even the gravitational deviation of 45 721.203ns

 

The "gravitational deviation" is not 45 721.203ns. Find another hobby, "proving relativity wrong" has been a failure.

Edited June 29, 2014 by xyzt

[swansont]

Satellite orbits are inclined at an angle of 55 degrees to the equator. The Earth rotates at a speed of 463.8 m/s at the equator and 0 m/s at the poles. The question then arises of what the actual speed of the satellite is in relation to GPS receivers, anywhere on the Earth? How does this apply to the calculation of time dilation?

 

GPS uses an earth-centered earth-fixed (ECEF) reference frame.

http://relativity.livingreviews.org/open?pubNo=lrr-2003-1&page=articlese2.html

http://books.google.com/books?id=DH7jEf48KjgC&pg=PA11#v=onepage&q&f=false

 

 

Apart from the refraction of radio waves in the stratosphere and the troposphere of the Earth, the GPS receiver can see moving satellites at different and changing angles, so there is inevitably the Doppler effect, which changes the frequency of the received radio signals from the satellite.

 

 

Yes, and that's why it sometimes takes a while for a receiver to find satellites if it doesn't know where to look and what Doppler shift to apply, so that it has to scan over all of the possibilities. But this has nothing to do with whether GPS is consistent with relativity, as the Doppler shift only affects the carrier.

[Strange]

6. ...the GPS receiver can see moving satellites at different and changing angles, so there is inevitably the Doppler effect, which changes the frequency of the received radio signals from the satellite.

 

Obviously the receiver has to adjust for this. It cannot just have a radio tuned to a fixed frequency, it has to scan the range of possible Doppler-shifted frequencies to find each satellite (initially it doesn't know whether each satellite is moving towards, away or tangentially).

 

 

7. It should be noted that the GPS receiver does not have its own time, as the time in the GPS receiver is digitally synchronized with the satellite clock. There is no room, therefore, for any time dilation between satellite and GPS reciver, because the numerical values transmitted from the satellites do not change.

But the receiver uses GR to correct the time signals from the satellite (as well as correcting the distances to do the correct triangulation). Hence the time determined by the receiver is not just a copy of the time received from the satellite. After all, it is getting data from at least 4 satellites.

[Endy0816]

I'm going to have to try and remember to take a look at the DAGR :/

 

I know it receives a fair amount of information and typically we barely look at it.

[swansont]

 

But the receiver uses GR to correct the time signals from the satellite (as well as correcting the distances to do the correct triangulation). Hence the time determined by the receiver is not just a copy of the time received from the satellite. After all, it is getting data from at least 4 satellites.

 

And it's only determining timing differences from those satellites over a short interval, which is why you can get away with having a relatively cheap quartz oscillator in the GPS receiver and have it work.

[Le Repteux]

Swantson,

 

If the GPS system can get away with SR while taking the center of the earth as a common reference frame, then it appears to me that we can do the same with any inertial motion since any of them is subjected to gravitation. For instance, the twin's paradox disappears if we take the center of the sun as a reference frame for both twins, and if one of them goes around it at high speed, because from the sun's point of view, none of them would have aged more than the other. Is that what you meant?

[swansont]

Swantson,

 

If the GPS system can get away with SR while taking the center of the earth as a common reference frame, then it appears to me that we can do the same with any inertial motion since any of them is subjected to gravitation. For instance, the twin's paradox disappears if we take the center of the sun as a reference frame for both twins, and if one of them goes around it at high speed, because from the sun's point of view, none of them would have aged more than the other. Is that what you meant?

 

No. I can't find a post of mine that would be a reference for this.

 

The common frame for GPS (the ECEF frame) is the surface of the earth, not the center; since clocks on the geoid all run at the same rate that's a fairly obvious choice to make. And your statement of the twin paradox is wrong; at the end of the experiment everyone agrees that the twin who traveled is younger. Everyone has to agree.