I GPS system and general relativity

[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.