I GPS system and general relativity

  • #51
PeterDonis said:
The important thing is the physical definition of the simultaneity surfaces.
Sorry, I believe I lost that physical definition of simultaneity surfaces you were talking about.
 
Physics news on Phys.org
  • #52
cianfa72 said:
I believe I lost that physical definition of simultaneity surfaces you were talking about.
I gave it in post #52.
 
  • #53
PeterDonis said:
The simultaneity convention of the ECI frame is that the simultaneity surfaces are orthogonal to the worldline of the Earth's center of mass, and to an imaginary congruence of non-rotating worldlines at rest relative to the Earth's center of mass.
Yes, this is the definition of ECI simultaneity convention let me say in spacetime mathematical model. Now, from a physical point of view, what does it mean?
 
  • #54
cianfa72 said:
this is the definition of ECI simultaneity convention let me say in spacetime mathematical model.
Not only that. Orthogonality of simultaneity surfaces to worldlines has a physical meaning--actually more than one. It means the congruence of worldlines is non-rotating. It also means that the simultaneity surfaces are the ones that would be established by Einstein clock synchronization between the worldlines.
 
  • #55
PeterDonis said:
Equation 13 in the Living Reviews paper I referenced earlier is an approximate expression for the gravitational potential around Earth including its quadrupole moment.

Sorry didn't see this post earlier. Yes I saw the expression and was able to find a NASA article that includes details on k20 as well as J20. Still studying that article.

Quick question isn't ECEF also oft called the geocentric coordinate reference system? If so then I already have the related mathematics for that coordinate system. Either way I was able to find the mathematical details for ECI and ECEF inclusive with the quadrupole moment.

Edit NASA article was specific to STEP program disregard the J20 and K20 above. Article still is useful otherwise.
 
Last edited:
  • #56
PeterDonis said:
Orthogonality of simultaneity surfaces to worldlines has a physical meaning--actually more than one. It means the congruence of worldlines is non-rotating. It also means that the simultaneity surfaces are the ones that would be established by Einstein clock synchronization between the worldlines.
You mean take the timelike geodesic worldline of Earth's center (it is geodesic since the center is in free-fall) and a set of massive objects whose timelike worldlines do not intersect (a congruence), do not rotate (i.e. hypersurface orthogonal) and are at rest w.r.t. the Earth's center worldline (as measured by constant travel time of bouncing light beams exchanged between Earth's center worldline and each of the congruence's worldlines).

What does physically mean the above conditions to be non-rotating/irrotational ? From Sachs and Wu section 5.3 geodesic and irrotational conditions are equivalent (iff) to locally proper synchronizable.

Capture.PNG

Capture-1.PNG

Then two observers in the congruence, communicating via radar (bouncing light beams), can check the above consistency condition that is basically the Einstein's clock synchronization convention.
 
Last edited:
  • #57
Vanadium 50 said:
Given this, and the signal from one satellite, it can determine its position to a circle.
I don't think it's quite that simple - the receiver would need to know the signal flight time for that, for which it would need an atomic clock ticking ECI time. Without such a clock it needs two satellites and the time difference between signal arrivals for any kind of fix, which places it on something like a paraboloid, which maps to a line on the geoid. A third satellite gives you another pair of paraboloids from the extra pairwise differences, the intersection of which should be a point (except in certain special cases).

Either that or I'm misunderstanding how the system works.
 
  • #58
Ibix said:
the receiver would need to know the signal flight time for that, for which it would need an atomic clock ticking ECI time.
If less than four signals are received, the receiver must use its own clock (which of course is not an atomic clock and is nowhere near that accurate, except for some special cases) to determine signal flight times. However, whenever four signals are in view, the receiver can update its clock, since four signals allows determination of all four coordinates of the reception event in the ECI frame. The latter is the normal mode of operation for receivers.
 
  • #59
cianfa72 said:
What does physically mean the above conditions to be non-rotating/irrotational ? From Sachs and Wu section 5.3 geodesic and irrotational conditions are equivalent (iff) to locally proper synchronizable.
Reading again Sachs & Wu section 2.3 I believe the difference between locally synchronizable vs synchronizable is about the domain of definition of ##C^{\infty}## functions ##h## and ##t## such that the differential 1-form ##\omega## (that defines the 3D spacelike distribution orthogonal to the congruence's timelike worldlines at each point) is written as ##\omega = - hdt##.

In the latter both functions are globally defined on spacetime manifold, while in the former only locally (in an open neighborhood of each point).
 
  • #60
cianfa72 said:
You mean take the timelike geodesic worldline of Earth's center (it is geodesic since the center is in free-fall) and a set of massive objects whose timelike worldlines do not intersect (a congruence), do not rotate (i.e. hypersurface orthogonal) and are at rest w.r.t. the Earth's center worldline (as measured by constant travel time of bouncing light beams exchanged between Earth's center worldline and each of the congruence's worldlines).
Yes.

cianfa72 said:
What does physically mean the above conditions to be non-rotating/irrotational ?
Physically it means there are no "fictitious forces" in the frame, in the Newtonian sense, i.e., no centrifugal or Coriolis force. In the absence of gravity we would say that meant the worldlines of free-falling objects are straight lines in the frame, but since gravity is present (due to the Earth), we can't say that. So the ECI frame is really only "inertial" in the Newtonian sense (where gravity is not considered a "fictitious force" and it's OK for the worldlines of objects free-falling under gravity to not be straight lines in the frame). It is not "inertial" in the GR sense (and in the GR sense it is impossible to define a single inertial frame that covers the entire Earth and its neighborhood, as the ECI frame does).

cianfa72 said:
From Sachs and Wu section 5.3 geodesic and irrotational conditions are equivalent (iff) to locally proper synchronizable.
The geodesic condition is not met since only one worldline at rest in the frame (that of the Earth's center) is a geodesic. The congruence that defines the ECI frame is, in Sachs and Wu terminology, locally synchronizable (a common set of simultaneity surfaces can be defined that are everywhere orthogonal to the congruence) but not locally proper time synchronizable (gravitational time dilation is present).
 
  • #61
PeterDonis said:
Physically it means there are no "fictitious forces" in the frame, in the Newtonian sense, i.e., no centrifugal or Coriolis force. In the absence of gravity we would say that meant the worldlines of free-falling objects are straight lines in the frame, but since gravity is present (due to the Earth), we can't say that. So the ECI frame is really only "inertial" in the Newtonian sense (where gravity is not considered a "fictitious force" and it's OK for the worldlines of objects free-falling under gravity to not be straight lines in the frame). It is not "inertial" in the GR sense (and in the GR sense it is impossible to define a single inertial frame that covers the entire Earth and its neighborhood, as the ECI frame does).
You mean non-rotating/zero vorticity timelike congruences have the feature that in the frame/chart in which they are "at rest", worldlines of free-falling objects, in case of flat spacetime, turn out to be straight lines (i.e. their coordinates are linearly dependent from the proper time parameter along each of them).

PeterDonis said:
The geodesic condition is not met since only one worldline at rest in the frame (that of the Earth's center) is a geodesic. The congruence that defines the ECI frame is, in Sachs and Wu terminology, locally synchronizable (a common set of simultaneity surfaces can be defined that are everywhere orthogonal to the congruence) but not locally proper time synchronizable (gravitational time dilation is present).
Ok, so in Sachs & Wu terminology ##\omega \wedge d\omega = 0##, i.e. locally ##\omega = -hdt## for some functions ##h## and ##t## defined both in an open neighborhood of any point.
 
  • #62
cianfa72 said:
You mean non-rotating timelike congruences have the feature that in the frame/chart in which they are "at rest", worldlines of free-falling objects, in case of flat spacetime, turn out to be straight lines (i.e. their coordinates are linearly dependent from the proper time along each of them).
Non-rotating geodesic timelike congruences have this property in flat spacetime.

cianfa72 said:
Ok, so in Sachs & Wu terminology ##\omega \wedge d\omega = 0##, i.e. locally ##\omega = -hdt## for some functions ##h## and ##t## defined both in an open neighborhood of any point.
Yes.
 
  • #63
PeterDonis said:
Non-rotating geodesic timelike congruences have this property in flat spacetime.
And what if one considers just non-rotating timelike congruences without the requirement to be geodesics (in case of either flat or curved spacetime) ?
 
  • #64
cianfa72 said:
what if one considers just non-rotating timelike congruences without the requirement to be geodesics (in case of either flat or curved spacetime) ?
Then you can no longer say that free-falling worldlines will be straight in the chart derived from the congruence. An obvious example is the Rindler congruence in Minkowski spacetime, which is irrotational but not geodesic. Free-falling worldlines are not straight in the Rindler chart derived from this congruence.
 
  • #65
PeterDonis said:
Then you can no longer say that free-falling worldlines will be straight in the chart derived from the congruence.
Therefore, what is the characteristic feature of the chart derivated from a generic non-rotating timelike congruence (in case of either flat or curved spacetime) ?
 
Last edited:
  • #66
cianfa72 said:
Therefore, what is the characteristic feature of the chart derivated from a generic non-rotating timelike congruence (in case of either flat or curved spacetime) ?
What do you think it is? And why do you feel you need to ask?
 
  • #67
PeterDonis said:
What do you think it is?
I know that non-rotating/zero-vorticity congruence is equivalent to say hypersurface orthogonal. I'd better understand what it means from a physical perspective and what features have charts derivated from them (i.e. coordinate charts where they are "at rest").
 
  • #68
cianfa72 said:
I know that non-rotating/zero-vorticity congruence is equivalent to say hypersurface orthogonal.
Yes. This is a consequence of the Frobenius theorem.

cianfa72 said:
I'd better understand what it means from a physical perspective
Do you understand what having simultaneity surfaces that are orthogonal to the worldlines in the congruence means physically? More basically, do you understand what an orthonormal basis of vectors means physically?

cianfa72 said:
what features have charts derivated from them (i.e. coordinate charts where they are "at rest").
Do you understand what hypersurface orthogonality implies about a chart that is adapted to that feature?
 
  • #69
PeterDonis said:
Do you understand what having simultaneity surfaces that are orthogonal to the worldlines in the congruence means physically?
No, I don't.

PeterDonis said:
More basically, do you understand what an orthonormal basis of vectors means physically?
Can you kindly clarify it ?

PeterDonis said:
Do you understand what hypersurface orthogonality implies about a chart that is adapted to that feature?
Yes, it implies there exists a chart where the mixed metric components ##g_{t\alpha}, \alpha =1,2,3## vanish.
 
  • #70
cianfa72 said:
No, I don't.
Can you kindly clarify it ?
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.

For a non-geodesic case, consider the Rindler congruence in Minkowski spacetime and the Rindler chart adapted to it. The Rindler chart can also be realized with "measuring rods and clocks" in Einstein fashion, but accelerated ones instead of inertial (freely falling) ones. Again, those measuring rods and clocks define an orthonormal basis at each point of the patch of spacetime covered by the Rindler chart (but now a different orthonormal basis from the inertial one).

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

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.

cianfa72 said:
it implies there exists a chart where the mixed metric components ##g_{t\alpha}, \alpha =1,2,3## vanish.
Yes. Which, again, is the way we intuitively expect a chart to work.
 
  • #71
PeterDonis said:
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.

PeterDonis said:
(For extra credit you can also consider the Schwarzschild chart in the exterior region of Schwarzschild spacetime.)
Yes, of course.

PeterDonis said:
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).
 
Last edited:
  • #72
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 ?
 
  • #73
cianfa72 said:
the timelike worldlines of "measuring clocks" define a timelike direction at any event along them.
The tangent vectors to the worldlines do, yes.

cianfa72 said:
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.

cianfa72 said:
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.

cianfa72 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".
 
  • #74
cianfa72 said:
Where is the mistake in a such reasoning ?
We went over this in a previous thread a while back. Your basic error is here:

cianfa72 said:
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.
 
  • #75
cianfa72 said:
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.

wiki said:
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.
 
  • #76
pervect said:
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).
 
  • #77
PeterDonis said:
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 ?

PeterDonis said:
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.
 
  • #78
PeterDonis said:
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.
 
  • #79
cianfa72 said:
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.

cianfa72 said:
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##.
 
  • #80
PeterDonis said:
By reading what the measuring rod says.
Sorry, could you please be more specific ?

PeterDonis said:
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.
 
  • #81
cianfa72 said:
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?

cianfa72 said:
Ok, so in the general case there is no way to extend such pairs of locally defined functions into a globally defined pair.
Yes.
 
  • #82
PeterDonis said:
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.
 
  • #83
cianfa72 said:
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?
 
  • #84
PeterDonis said:
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).
 
Last edited:
  • #85
cianfa72 said:
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.
 
  • #86
PeterDonis said:
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.
 
  • #87
PeterDonis said:
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).
 
Last edited:
  • #88
PeterDonis said:
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##.
 
  • #89
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.
 
  • #90
cianfa72 said:
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##.)

cianfa72 said:
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.

cianfa72 said:
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.

cianfa72 said:
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.
 
  • #91
PeterDonis said:
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) ?
 
Last edited:
  • #92
cianfa72 said:
the conversion factor is actually ##1/\sqrt{g_{tt}}## hence for the specific case it is $$\frac {1} {\sqrt {(1 - r_s/r)}}$$
Yes.

cianfa72 said:
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.

cianfa72 said:
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.
 
  • #93
PeterDonis said:
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.

PeterDonis said:
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.
 
Last edited:
  • #94
cianfa72 said:
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.

cianfa72 said:
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.
 
  • #95
PeterDonis said:
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.
 
  • #96
  • #97
PeterDonis said:
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).
 
  • #98
cianfa72 said:
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.
 
  • #99
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).
 
  • #100
cianfa72 said:
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.

cianfa72 said:
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.
 

Similar threads

Replies
34
Views
3K
Replies
31
Views
6K
Replies
7
Views
2K
Replies
39
Views
3K
Replies
19
Views
1K
Replies
58
Views
5K
Replies
21
Views
1K
Back
Top