GPS system and general relativity

[cianfa72]

Think of the simplest case: a standard inertial frame in flat spacetime. Consider how such a frame is constructed, using "measuring rods and clocks" as Einstein described it. The measuring rods and clocks define an orthonormal basis at each point of spacetime.

You mean that the timelike worldlines of "measuring clocks" define a timelike direction at any event along them. The worldtubes of "measuring rods" define also 3 spacelike directions at any point/event. Such 1 + 3 spacetime directions are mutually orthonormal at any point.

So an inertial frame/chart in flat spacetime can be physically constructed by mean of "free-falling measuring rods and clocks" at rest each other (as measured by bouncing light beams) with clocks synchronizated according Einstein's synchronization procedure/convention.

(For extra credit you can also consider the Schwarzschild chart in the exterior region of Schwarzschild spacetime.)

Yes, of course.

Having a hypersurface orthogonal congruence and using the orthogonal hypersurfaces as simultaneity surfaces allows a chart to be constructed that works like the above examples. And that is also the way we intuitively expect a chart to work.

So, the actual possibility of consistently Einstein's synchronize clocks that have as worldlines the members of a timelike congruence and physically construct a chart with "measuring rods and clocks" as above (in which the congruence's worldlines are "at rest") is equivalent to the claim that the timelike congruence is non-rotating (i.e. zero vorticity or hypersurface orthogonal).

 

[cianfa72]

A question perhaps off-topic. We said that the condition $\omega \wedge d\omega = 0$ implies that there are smooth functions $h$ and $t$ defined on open neighborhoods of any point such that locally $\omega = -hdt$.

Suppose there are two function pairs $(h_1,t_1)$ and $(h_2,t_2)$ defined on their own neighborhoods such that $\omega = -h_1dt_1$ and $\omega = - h_2dt_2$ there. In their open intersection the two functions take the same values respectively.

My impression is that then one can basically "glue" such neighborhoods to get globally defined functions $\bar h$ and $\bar t$ such that globally $\omega = - \bar h d\bar t$.

Where is the mistake in a such reasoning ?

 

[PeterDonis]

the timelike worldlines of "measuring clocks" define a timelike direction at any event along them.

The tangent vectors to the worldlines do, yes.

The worldtubes of "measuring rods" define also 3 spacelike directions at any point/event.

Not just the world tubes themselves, no. You need the orthogonality condition to pick out a particular spacelike direction for each measuring rod.

an inertial frame/chart in flat spacetime can be physically constructed by mean of "free-falling measuring rods and clocks" at rest each other (as measured by bouncing light beams) with clocks synchronizated according Einstein's synchronization procedure/convention.

Yes, Einstein described this construction, as I already said.

the actual possibility of consistently Einstein's synchronize clocks that have as worldlines the members of a timelike congruence and physically construct a chart with "measuring rods and clocks" as above (in which the congruence's worldlines are "at rest") is equivalent to the claim that the timelike congruence is non-rotating (i.e. zero vorticity or hypersurface orthogonal).

That is the whole point of Sachs & Wu's definition of a congruence as "synchronizable".

 

[PeterDonis]

Where is the mistake in a such reasoning ?

We went over this in a previous thread a while back. Your basic error is here:

In their open intersection the two functions take the same values respectively.

In general, no, they won't take the same values in the overlap region.

 

[pervect]

TL;DR Summary: How GPS system works in the context of general relativity

Hi, we had a thread some time ago about GPS satellite system.

One starts considering the ECI coordinate system in which the Earth's center is at rest with axes pointing towards fixed stars. One may assume it is an inertial frame in which the Earth's surface undergoes circular motion.

Clocks on Earth's surface and on GPS geostationary satellites are at different gravitational potential in the Earth's gravitational field. Hence there is a gravitational time dilation between them.

First question: does GPS system employ Schwarzschild spacetime model of Earth's gravitational field to evaluate the above time dilation?

Second question: what is the role of ECI ?

Thanks.

The earth is not spherical - I've read that GPS uses WGS-84, see https://en.wikipedia.org/wiki/World_Geodetic_System. Therefore I do not believe GPS uses the Schwarzschild metric.

The current version, WGS 84, defines an Earth-centered, Earth-fixed coordinate system and a geodetic datum.

I'm a bit unclear as to what metric WGS-84 uses (which I think would defined the coordinate part), or the significance and implementation of the geodetic datum.

For general information on GPS, I would consult Ashby's paper, https://link.springer.com/content/pdf/10.12942/lrr-2003-1.pdf, and Misner's paper, "Precis of General Relativity", https://arxiv.org/abs/gr-qc/9508043. The later seems to have a low citation count, but I've always found it helpful and I believe it was written at least in part a response to Ashby's paper. Both papers have metrics, as i recall, I haven't compared the three sources (Misner, WGS-84, and Ashby) to see if they're all the same or other little details I might be missing.

 

[PeterDonis]

I'm a bit unclear as to what metric WGS-84 uses

The Ashby paper, which has been referenced several times now in this thread, gives an approximate line element for GPS, which would also be an approximate line element for WGS-84. I believe a full expression would involve higher order multipole moments, which, as the Ashby paper notes, are not really necessary for GPS (but might be for other applications).

 

[cianfa72]

Not just the world tubes themselves, no. You need the orthogonality condition to pick out a particular spacelike direction for each measuring rod.

Ok, from a physical perspective how does one pick out for each measuring rod such a particular spacelike direction orthogonal to the tangent vector to the timelike worldline passing there ?

That is the whole point of Sachs & Wu's definition of a congruence as "synchronizable".

That is equivalent to the condition to be hypersurface orthogonal.

 

[cianfa72]

In general, no, they won't take the same values in the overlap region.

Not sure to grasp it: basically you are saying that there is not a unique pair of functions $(\bar h, \bar t)$ such that in the overlapping open region $\omega =- \bar h d\bar t$: each of pairs $(h_1,t_1)$ and $(h_2,t_2)$ will do there.

 

[PeterDonis]

from a physical perspective how does one pick out for each measuring rod such a particular spacelike direction orthogonal to the tangent vector to the timelike worldline passing there ?

By reading what the measuring rod says.

basically you are saying that there is not a unique pair of functions $(\bar h, \bar t)$ such that in the overlapping open region $\omega =- \bar h d\bar t$: each of pairs $(h_1,t_1)$ and $(h_2,t_2)$ will do there.

I mean that in the general case you cannot assume that $h_1 = h_2$ and $t_1 = t_2$.

 

[cianfa72]

By reading what the measuring rod says.

Sorry, could you please be more specific ?

I mean that in the general case you cannot assume that $h_1 = h_2$ and $t_1 = t_2$.

Ok, so in the general case there is no way to extend such pairs of locally defined functions into a globally defined pair.

 

[PeterDonis]

Sorry, could you please be more specific ?

What is vague about "read what the measuring rod says"? You asked how one physically picks out a spacelike direction orthogonal to the observer's worldline. Reading what the measuring rod says is a physical action. What more do you want?

Ok, so in the general case there is no way to extend such pairs of locally defined functions into a globally defined pair.

Yes.

 

[cianfa72]

What is vague about "read what the measuring rod says"? You asked how one physically picks out a spacelike direction orthogonal to the observer's worldline. Reading what the measuring rod says is a physical action. What more do you want?

Ah ok, in spacetime the action of reading the "value" stamped on the specific measuring rod, takes place along a null path from the event, though.

 

[PeterDonis]

in spacetime the action of reading the "value" stamped on the specific measuring rod, takes place along a null path from the event, though.

Yes, if you want to be precise, you read the measuring rod by means of light rays that travel a null path from an event on the worldline of the far end of the rod; but a null path from which event?

 

[cianfa72]

but a null path from which event?

Yes, that's my point of confusion. Sorry I'm going in circle.

We said that reading the "advancing in elapsed time" on a clock at a given location picks a timelike direction in spacetime at any given event A along the timelike path followed from that clock in spacetime.

Then regarding spacelike directions, for example, the reading of "advancing in x direction along the measurement rod laid down in x direction starting from event A" should mean pick a spacelike direction orthogonal to the timelike direction at event A.

Edit: for spacelike directions my doubt is that there is not any physical device that can "move/advance" in a such direction from an event (as opposed to timelike directions).

 

[PeterDonis]

that's my point of confusion

Because you're not answering the specific question I asked you. I'll ask the question again with more supporting detail: you carry along with you a measuring rod with two ends. One end is co-located with you. The other end isn't. You want to read off a "distance" to the other end of the measuring rod. You do that by receiving a light signal from the rod. That light signal leaves the other end of the measuring rod at an event we'll call A, and intersects your worldline at an event we'll call B. The question is, given that you know event B, which event will A be? Which specific event on the worldline of the other end of the measuring rod will the light you are seeing at event B be emitted from?

Remember we are assuming an irrotational (and therefore hypersurface orthogonal) congruence of worldlines, and both your worldline and the worldline of the other end of the measuring rod are members of the congruence. Remember also that we have three special cases where we know of a coordinate chart that is adapted to these geometric facts: an inertial chart in Minkowski spacetime, the Rindler chart in Minkowski spacetime, and the Schwarzschild chart in Schwarzschild spacetime. For all three of these cases, you should be able to calculate directly the answer to the question above, and that should help you with understanding the general case.

 

[PeterDonis]

Which specific event on the worldline of the other end of the measuring rod will the light you are seeing at event B be emitted from?

Btw, if you want to make it easier to see intuitively what is going on, imagine that instead of a measuring rod you have a mirror, whose worldline is the same as what the worldline of the other end of the measuring rod would have been, and that you send a round-trip light signal that bounces off the mirror and returns to you. The light signal is emitted by you at an event we'll call C, reflected off the mirror at event A, and received by you at event B. Then you can calculate what specific event A is, and verify that that event has a particular useful property for the cases we have discussed.

 

[cianfa72]

you carry along with you a measuring rod with two ends. One end is co-located with you. The other end isn't. You want to read off a "distance" to the other end of the measuring rod. You do that by receiving a light signal from the rod. That light signal leaves the other end of the measuring rod at an event we'll call A, and intersects your worldline at an event we'll call B. The question is, given that you know event B, which event will A be? Which specific event on the worldline of the other end of the measuring rod will the light you are seeing at event B be emitted from?

Remember we are assuming an irrotational (and therefore hypersurface orthogonal) congruence of worldlines, and both your worldline and the worldline of the other end of the measuring rod are members of the congruence.

Since the congruence is irrotational it is hypersurface orthogonal, then there exists always a chart where the metric mixed terms $g_{t,\alpha}=0, \alpha =1,2,3$. In that chart the timelike worldlines of both ends of measuring rod have fixed spatial coordinates (assumed known) and varying $t$. Therefore from $ds^2 =0$ (null path equation), knowing the coordinates of event B along my worldline in that chart, one can solve for the coordinate time of event A.

The above equation is 2nd order in $(dt)^2$ and one gets two opposite solutions for the coordinate time $dt$ of event A (one for the ingoing and the other for the outgoing light signal).

 

[cianfa72]

The light signal is emitted by you at an event we'll call C, reflected off the mirror at event A, and received by you at event B. Then you can calculate what specific event A is, and verify that that event has a particular useful property for the cases we have discussed.

Yes, as in post #90 one gets two equal values for $dt$, therefore the coordinate time $t_A$ of event A is $t_A = (t_B + t_C)/2$.

 

[cianfa72]

Coming back to last posts.

For example in Schwarzschild spacetime pick the timelike congruence of observers hovering at constant Schwarzschild radial-coordinate $r$. Such observers carry with them 3 orthogonal measurement rods and a wristwatch.

Now if the rate of their wristwatches is "adjusted" according to the factor $1 -r_s/r$ (i.e. their rate is the same as that of Schwarzschild coordinate time $t$ at radial-coordinate $r$) then they successfully verify using light signals exchanged between them that their "rate adjusted wristwatches" are actually synthonized (no time dilation). Furthermore, starting from observer/my "rate adjusted wristwatch" then all other observers in the congruence can synchronize their wristwatches in order to measure the Schwarzschild coordinate time $t$.

Now, the spacetime direction from an event that takes place on my co-located end of the measurement rod and the far end at the same coordinate time $t$, actually defines a spacelike direction orthogonal to the timelike direction along my worldline at that event.

 

[PeterDonis]

if the rate of their wristwatches is "adjusted" according to the factor $1 -r_s/r$ (i.e. their rate is the same as that of Schwarzschild coordinate time $t$ at radial-coordinate $r$)

Your factor here is wrong. Check your math. (Two hints: first, clocks hovering at finite $r$ run slow relative to Schwarzschild coordinate time; second, the time dilation factor is not $g_{tt}$ itself, remember that the line element is a formula for $ds^2$.)

they successfully verify using light signals exchanged between them that their "rate adjusted wristwatches" are actually synthonized (no time dilation).

With the correct adjustment factor applied to the clocks, yes.

Furthermore, starting from observer/my "rate adjusted wristwatch" then all other observers in the congruence can synchronize their wristwatches in order to measure the Schwarzschild coordinate time $t$.

With the correct adjustment factor applied to the clocks, yes.

the spacetime direction from an event that takes place on my co-located end of the measurement rod and the far end at the same coordinate time $t$, actually defines a spacelike direction orthogonal to the timelike direction along my worldline at that event.

Yes.

 

[cianfa72]

Your factor here is wrong. Check your math. (Two hints: first, clocks hovering at finite $r$ run slow relative to Schwarzschild coordinate time; second, the time dilation factor is not $g_{tt}$ itself, remember that the line element is a formula for $ds^2$.)

Yes sorry, the conversion factor is actually $1/\sqrt{g_{tt}}$ hence for the specific case it is $$\frac {1} {\sqrt {(1 - r_s/r)}}$$
From my understanding, in principle, the "construction" in the previous post can be done locally in any spacetime (in other words there is always a spacetime transformation such that locally $g_{0\alpha} = 0$ and using the timelike congruence "at rest/adapted" to such a local chart the above construction can be applied). In a sense it defines 4 spacetime directions at any point/event such that the 3 spacelike directions are orthogonal to the timelike one.

My question is: in the general case does always exist a transformation that brings the metric components locally in the form $g_{00}=1, g_{0\alpha}=0$ leaving "at rest" the "old" timelike coordinate lines in the new local chart being defined (i.e. leaving at rest in the new chart the timelike curves described by $\{x_\alpha = c_\alpha, \alpha =1,2,3 \}$ in the old chart one started with) ?

 

[PeterDonis]

the conversion factor is actually $1/\sqrt{g_{tt}}$ hence for the specific case it is $$\frac {1} {\sqrt {(1 - r_s/r)}}$$

Yes.

From my understanding, in principle, the "construction" in the previous post can be done locally in any spacetime (in other words there is always a spacetime transformation such that locally $g_{0\alpha} = 0$ and using the timelike congruence "at rest/adapted" to such a local chart the above construction can be applied). In a sense it defines 4 spacetime directions at any point/event such that the 3 spacelike directions are orthogonal to the timelike one.

Yes, this is just a version of constructing a local inertial frame centered on a point.

in the general case does always exist a transformation that brings the metric components locally in the form $g_{00}=1, g_{0\alpha}=0$ leaving "at rest" the "old" timelike coordinate lines in the new local chart being defined (i.e. leaving at rest in the new chart the timelike curves described by $\{x_\alpha = c_\alpha, \alpha =1,2,3 \}$ in the old chart one started with) ?

You can always construct Fermi normal coordinates on an open region centered on a chosen timelike worldline. You might have to make some additional adjustments to enforce $g_{00} =1$ and $g_{0 \alpha} = 0$ on the chosen worldline. If you want those conditions to hold in an open region centered on the worldline, the congruence of timelike worldlines you choose must be irrotational.

 

[cianfa72]

Yes, this is just a version of constructing a local inertial frame centered on a point.

Sorry, to get a local inertial frame centered on a point, the metric components in that local chart should be exactly $(1,-1,-1,-1)$ with vanish derivatives on that point.

You can always construct Fermi normal coordinates on an open region centered on a chosen timelike worldline. You might have to make some additional adjustments to enforce $g_{00} =1$ and $g_{0 \alpha} = 0$ on the chosen worldline. If you want those conditions to hold in an open region centered on the worldline, the congruence of timelike worldlines you choose must be irrotational.

You mean that locally (i.e. in an open neighborhood of any point) in any spacetime there is always an irrotational timelike congruence.

 

[PeterDonis]

to get a local inertial frame centered on a point, the metric components in that local chart should be exactly $(1,-1,-1,-1)$ with vanish derivatives on that point.

Yes.

Your mean that locally (i.e. in an open neighborhood of any point) in any spacetime there is always an irrotational timelike congruence.

That's not what I said. Go read what I said again, carefully.

 

[cianfa72]

That's not what I said. Go read what I said again, carefully.

From Synchronous frame in any spacetime in any open neighborhood there is a (synchronous) coordinate chart such that $g_{00}=1, g_{0\alpha}=0$. I believe the timelike curves (actually geodesics) at rest in it form an irrotational congruence.

 

[PeterDonis]

From Synchronous frame in any spacetime in any open neighborhood there is a (synchronous) coordinate chart such that $g_{00}=1, g_{0\alpha}=0$. I believe the timelike curves (actually geodesics) at rest in it form an irrotational congruence.

You already had a previous thread on this topic. Go read it.

https://www.physicsforums.com/threads/synchronous-reference-frame-definition-and-usage.1007040/

 

[cianfa72]

You already had a previous thread on this topic. Go read it.

https://www.physicsforums.com/threads/synchronous-reference-frame-definition-and-usage.1007040/

Sure, I remember that thread. The take-home message was that in a finite open patch of any spacetime one can always build a synchronous reference frame/chart (such a chart may not extend globally since sooner or later the timelike geodesics starting orthogonal to the initially chosen spacelike hypersurface will intersect).

 

[PeterDonis]

Sure, I remember that thread. The take-home message was that in a finite open patch of any spacetime one can always build a synchronous reference frame/chart (such a chart may not extend globally since sooner or later the timelike geodesics starting orthogonal to the initially chosen spacelike hypersurface will intersect).

Yes, and what I said in post #55 did not contradict any of that. But it did not just repeat it either.

 

[cianfa72]

Ok, so the fact that in any open patch of spacetime one can always build a synchronous coordinate chart implies that any spacetime admits a locally proper time synchronizable congruence/frame using the terminology of Sachs and Wu section 2.3 (i.e. $d\omega = 0$ and by Poincaré lemma $\omega = dt$ for some smooth function $t$ in that open region).

 

[PeterDonis]

in any open patch of spacetime one can always build a synchronous coordinate chart

Actually, as you state this, it's too strong. The correct statement is that, given a spacelike hypersurface, one can always find some open neighborhood of that hypersurface in which Gaussian normal coordinates, i.e., a "synchronous coordinate chart", can be constructed. But one cannot guarantee that such coordinates will be valid for any open neighborhood, of any size whatever.

any spacetime admits a locally proper time synchronizable congruence/frame using the terminology of Sachs and Wu section 2.3 (i.e. $d\omega = 0$ and by Poincaré lemma $\omega = dt$ for some smooth function $t$ in that open region).

With the qualifications given above, yes.

 

[cianfa72]

Actually, as you state this, it's too strong. The correct statement is that, given a spacelike hypersurface, one can always find some open neighborhood of that hypersurface in which Gaussian normal coordinates, i.e., a "synchronous coordinate chart", can be constructed. But one cannot guarantee that such coordinates will be valid for any open neighborhood, of any size whatever.

Ok, let me say the point is that, given an open patch in spacetime, it might be so much larger that timelike geodesics starting orthogonal from a spacelike hypersurface within it will intersect inside that region, though.

 

[PeterDonis]

the point is that, given an open patch in spacetime, it might be so much larger that timelike geodesics starting orthogonal from a spacelike hypersurface within it will intersect inside that region, though.

It doesn't have to be "so much" larger, just large enough for geodesics to intersect.

This is getting pretty far off the original topic of this thread, btw. The frames used in GPS are not examples of synchronous coordinates.

 

[binis]

Hi, we had a thread some time ago about GPS satellite system.
One starts considering the ECI coordinate system in which the Earth's center is at rest with axes pointing towards fixed stars. One may assume it is an inertial frame in which the Earth's surface undergoes circular motion.

Do you mean this ? I would thank you for your contribution.

 

[cianfa72]

Do you mean this ? I would thank you for your contribution.

No, the thread I was referring to is this.