GPS and relativity.

  • Thread starter OS Richert
  • Start date
  • #1
35
0
I'm working through "Exploring Black Holes" by Taylor and Wheeler. Their first project is on the GPS system and how GR must be accounted for. At the end you arrive at the conclusion that the satellite atomic clock will advance approximately 38,700 nanoseconds a day with respect to an atomic clock on earth. This part I understand, what I don't understand is why this matters. And I think my problem is that I don't understand how the GPS really works.

I understand that three satellites send out their signals and I take the intersection of those three spheres as my location. I also understand that to figure out what the radius of the sphere is, I would calculate the radius as c*(t_receive - t_send) where t_send was encoded into the signal.

Now, let's assume I had an atomic clock in my receiver (which I realize I don't have), that was synchronized with the satellite clock before it went into space. Furthermore, lets assume I ignore GR. Here I agree that every day that the satellite atomic clock advanced, my answer will get further and further skewed (since t_receive - t_send would get smaller and smaller and eventually turn negative). But, I don't have an atomic clock in my receiver, and the cheap clock I do have is not accurate enough for this calculation. The clock in my receiver probably skews more then 38,700 ns all on its own. Is this not why we add a fourth satellite to send us a time that we use to "correct" our own time? I'm guessing the algorithm for this correction will answer my question, but let me continue. Since I am somehow setting my receive time off this satellite clock which experiences the same effects as the other three satellite clocks, I would think that this daily advance of the satellite clocks would be canceled out since my reference (the 4th satellite) is also advancing. Now there would still be an error of the time dilation difference during the propagation of the signal, but I would think it would only be the difference experienced during that time, which would be on the order of .06 seconds. The time error effect do to GR then we only be on order tens of picoseconds (nothing at all to fuss about).

So I guess my question is, how does the 4th satellite "correct" my clock in my hand? Do I really care about the daily advance of the satellite clocks, doesn't the 4th satellite cancel this out since it is also advanced?
 

Answers and Replies

  • #2
mgb_phys
Science Advisor
Homework Helper
7,774
13
The clocks in the satelites are corrected for relativity effects - actualy the GR effect of being in a weaker gravitational field and so time going faster is greater then the SR effect of them moving and time going slower.

The receiver uses the time sent by the satelites, this avoids having to carry your own synchronised atomic lcock with you, but does mean a receiving an extra satelite - 4 rather than 3.
In simple terms:
Each satelite sends it's own time (clocks are corrected for relativity)
The receiver works out the time from the redundant 4th satelite.
It knows the position from the empheris broadcast by the satelite
Since it knows the local time and the time sent y each sat it can work out the distance.
It iterates to get a better solution for the local time and hence the position.
 
  • #3
35
0
I'm afraid this didn't answer my question as I pretty much know already everything you wrote.

I see it all the time that people claim if we didn't correct for GR, then the distance would be off by several kilometers in a single day. Taking Taylor's number of 39k nano seconds clock advance, and the speed of light c, I agree that we would be over 10k different if we calculated our receive time from an earth based atomic clock. But as you said in your reply and as I made clear in my original post, we don't get our receive time from a local atomic clock, we get it from this fourth redundant satellite. My question is probably along the lines of what is the exact algorithm for this, because it seems to me that since our receive clock (derived from the reference clock of this 4th redundant satellite) is also advanced by 39k nano second s a day, it will simply cancel out.

Let me illustrate with some equations.

distance = c * (t_recieve - t_send).

First: assume we are using are own ground based atomic clock, have ignored GR, and one day has passed since the satellites where in orbit.

distance = c * (t_recieve - (t_send + 39000nS) does indeed create distance error on order of 10km.

Second assume we are using the 4th satellite as our reference clock, have ignored GR, and one day has passed since the satellites where in orbit.

distance = c * [(t_receive + 39000ns) - (t_send + 39000ns)] = c * (t_receive - t_send)

the 39000 ns advance in time is canceled out sine our receive clock, based on the 4th satellite clock, has experienced the same GR effects. So now our error is no longer dependent one how far it might skew in one day, but only how far it might skew during the actual time of transmission, a very short time indeed which would have a very small error due to GR.

What am I missing?
 
  • #4
mgb_phys
Science Advisor
Homework Helper
7,774
13
The cummulative effect wouldn't matter because as you said, you are only interested in the time difference. But since the clocks are running slower the ticks aren't being sent out 1second apart (from the receivers point of view) and so the distance = (speed of light * time diff) would be off although only by a much smaller amount.

Plus you do need to correct for relativity if you want to use the GPS as an absolute time reference.
 
  • #5
35
0
The cummulative effect wouldn't matter because as you said, you are only interested in the time difference.
So does this mean that anytime someone claims that if we didn't correct for GR we would be off by several kilometers in a day is wrong?

distance = (speed of light * time diff)
time difference between how they view 1 second ticks?

For every second on earth, satellite clock experiences 1 second + .44 nano seconds. Time difference is .44 nano seconds. That is about 13 centimeters distance difference. Is this really what is being corrected for?
 
  • #6
1,997
5
For an interesting technical overview on GPS see: http://tycho.usno.navy.mil/ptti/1996/Vol%2028_16.pdf [Broken]
 
Last edited by a moderator:
  • #7
mgb_phys
Science Advisor
Homework Helper
7,774
13
According to the paper posted by MeJennifer the effect is negligble for a single reading but does add up when positions are averaged over a longer period.

This doesn't seem to be a big deal - you generally use a differential GPS to get an averaged position.
 
  • #8
35
0
So then I want to restate my question. Does this mean that anytime someone claims that if we didn't correct for GR we would be off by several kilometers after one day is wrong? As long we we take one reading, actual GR effects on position are negligible, even if the clocks are running 39000 nanoseconds per day faster on the satellites?
 
Last edited:

Related Threads on GPS and relativity.

  • Last Post
Replies
2
Views
1K
  • Last Post
3
Replies
60
Views
10K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
4K
  • Last Post
Replies
1
Views
2K
  • Last Post
2
Replies
45
Views
8K
  • Last Post
2
Replies
31
Views
2K
  • Last Post
Replies
8
Views
3K
Replies
8
Views
4K
Replies
1
Views
4K
Top