About global inertial frames in GR

In summary: Now consider a finite region of spacetime outside the Earth: take a family of free-falling bodies (a geodesic congruence) foliating it: if we choose a frame (coordinate chart) in which those free-falling bodies are at rest (i.e. having constant spatial coordinates in that frame) then it should match the above definition of (global) inertial frame -- global at least for that finite region of spacetime.This is just terminology. Call it what i want.You're using the word global when you mean local. If it is not the whole space-time, but just an open set, which is part of a coordinate chart, then it is local
  • #1
cianfa72
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TL;DR Summary
About the definition to use for a global inertial frame in GR
Hi,

starting from this thread Principle of relativity for proper accelerating frame of reference I'm convincing myself of some misunderstanding about what a global inertial frame should actually be.

In GR we take as definition of inertial frame (aka inertial coordinate system or inertial coordinate chart) the following:
a frame (aka coordinate chart) is inertial -- by definition -- if accelerometers everywhere at rest in it measure zero proper acceleration.

Now consider a finite region of spacetime outside the Earth: take a family of free-falling bodies (a geodesic congruence) foliating it: if we choose a frame (coordinate chart) in which those free-falling bodies are at rest (i.e. having constant spatial coordinates in that frame) then it should match the above definition of (global) inertial frame -- global at least for that finite region of spacetime.

What is wrong in the above claim ? (sure...I'm aware of global inertial frames actually do not exist in GR).

Thank you.
 
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  • #2
You are using the word global when you mean local. If it is not the whole space-time, but just an open set, which is part of a coordinate chart, then it is local.
 
  • #3
martinbn said:
You are using the word global when you mean local. If it is not the whole space-time, but just an open set, which is part of a coordinate chart, then it is local.
So even if the coordinate chart of post #1 is inertial in a finite region of spacetime (an open set) we cannot claim of it as global, right ?
 
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  • #4
cianfa72 said:
So even if the coordinate chart of post #1 is inertial in a finite region of spacetime (an open set) we cannot claim of it as global, right ?
This is just terminology. Call it what i want.
 
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  • #5
cianfa72 said:
What is wrong in the above claim ? (sure...I'm aware of global inertial frames actually do not exist in GR).
You're aware that global inertial frames do not exist in GR, yet you're asking about a proposed definition of a global inertial frame in GR. That's what's wrong with your claim.
 
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  • #6
cianfa72 said:
even if the coordinate chart of post #1 is inertial in a finite region of spacetime (an open set)
Any coordinate chart covers an open set. Saying the chart "is inertial in an open set" is redundant. The question is whether the chart is inertial.

Strictly speaking, no chart in a curved spacetime can be inertial at more than a single point, since nonzero spacetime curvature will cause second order errors at every other point besides the single point you choose to center the chart on. In practice, we call a chart "locally inertial" over a finite-sized patch of spacetime if the second order errors are smaller than the accuracy of our measurements on the patch.
 
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  • #7
So the definition quoted in post #1 actually make no sense in GR (or simply it does not apply in GR) ?
 
  • #8
cianfa72 said:
So the definition quoted in post #1 actually make no sense in GR (or simply it does not apply in GR) ?
No, the definition itself is fine (with, if you want to be precise, a qualifier that "measures zero proper acceleration" should really be "measures zero proper acceleration to within the measurement accuracy of the accelerometer"). It just is impossible for any global coordinate chart in a curved spacetime to meet it.
 
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  • #9
PeterDonis said:
No, the definition itself is fine (with, if you want to be precise, a qualifier that "measures zero proper acceleration" should really be "measures zero proper acceleration to within the measurement accuracy of the accelerometer"). It just is impossible for any global coordinate chart in a curved spacetime to meet it.
Perhaps I'm missing something, but it seems to me that the definition as written isn't complete. I think you can produce a set of non-crossing inertial timelike worldlines that fill spacetime, at least in some cases - FLRW co-moving observers are an example. The definition in the OP is a frame (aka coordinate chart) is inertial -- by definition -- if accelerometers everywhere at rest in it measure zero proper acceleration. Co-moving accelerometers read zero and are at rest in the usual coordinates, so meet this definition. I think the definition needs something like "...and the distance between those inertial accelerometers doesn't change".
 
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  • #10
Ibix said:
I think the definition needs something like "...and the distance between those inertial accelerometers doesn't change".
Ah, yes, you're right, this needs to be an additional qualifier, otherwise, as you note, standard FRW coordinates would qualify as an inertial frame.
 
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  • #11
And we probably need to define "distance" formally in terms of radar pulses, since one could argue that co-moving distance is a distance and doesn't change between co-moving observers.
 
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  • #12
Ibix said:
And we probably need to define "distance" formally in terms of radar pulses
Yes, or use the more specific term "proper distance".
 
  • #13
PeterDonis said:
Ah, yes, you're right, this needs to be an additional qualifier, otherwise, as you note, standard FRW coordinates would qualify as an inertial frame.
A said in post #1 I think the same argument applies as well for any timelike geodesic congruence that foliate a finite region of spacetime (choosing a coordinate chart accordingly).
 
  • #14
cianfa72 said:
A said in post #1 I think the same argument applies as well for any timelike geodesic congruence that foliate a finite region of spacetime
No, it doesn't, because not every timelike geodesic congruence has zero expansion, i.e., not every timelike geodesic congruence has the proper distance between worldlines remaining constant.

In fact, examples of timelike geodesic congruences in a curved spacetime in which the proper distance between worldlines is constant are very rare; the only one I can think of off the top of my head is the congruence of worldlines at rest in the standard coordinate chart on Godel spacetime. Certainly there is no such congruence in FRW spacetime (except the edge case of the "empty" FRW universe, which is just Minkowski spacetime in disguise), or in Schwarzschild or Kerr spacetimes.
 
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  • #15
PeterDonis said:
In fact, examples of timelike geodesic congruences in a curved spacetime in which the proper distance between worldlines is constant are very rare;

Certainly there is no such congruence in FRW spacetime (except the edge case of the "empty" FRW universe, which is just Minkowski spacetime in disguise), or in Schwarzschild or Kerr spacetimes.
Sorry, maybe I was unclear: that was actually my point; so even in Schwarzschild spacetime does not exist a timelike geodesic congruence such that the proper distance between geodesics is constant.
 
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  • #16
cianfa72 said:
so even in Schwarzschild spacetime does not exist a timelike geodesic congruence such that the proper distance between geodesics is constant
Yes.
 
  • #17
The "constant distance" requirement is Euclid's parallel postulate translated for Minkowski geometry. That it doesn't hold is one way to define curved-but-locally-Minkowski space.
 
  • #18
Ibix said:
The "constant distance" requirement is Euclid's parallel postulate translated for Minkowski geometry. That it doesn't hold is one way to define curved-but-locally-Minkowski space.
Is the concept of costant proper distance well defined for a pair of geodesics ?
 
  • #19
cianfa72 said:
Is the concept of costant proper distance well defined for a pair of geodesics ?
Nearby ones, yes. You construct a perpendicular to one and measure the distance along that to reach the other. I prefer using radar because it fits better with my physical intuition, but it amounts to the same thing.
 
  • #20
Ibix said:
Nearby ones, yes. You construct a perpendicular to one and measure the distance along that to reach the other. I prefer using radar because it fits better with my physical intuition, but it amounts to the same thing.
From a spacetime point of view, to construct a perpendicular from an event that belong on the first geodesic, we're actually considering the tangent space at that event -- AFAIK any metric notion (including orthogonality) makes sense just in the tangent space.

Then, from a physical operational point of view, we can use radar distance, I believe.
 
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  • #21
Ibix said:
The "constant distance" requirement is Euclid's parallel postulate translated for Minkowski geometry.
No, it isn't. The parallel postulate is about how many geodesics through a point not on a given geodesic will never intersect the given geodesic (one in Euclidean geometry; zero in spherical geometry; more than one in hyperbolic geometry). Requiring all geodesics in a timelike geodesic congruence to maintain constant distance from neighboring geodesics is a different thing.

Ibix said:
That it doesn't hold is one way to define curved-but-locally-Minkowski space.
If we restrict attention to spacetimes with zero cosmological constant, I think it is correct that the parallel postulate not holding is equivalent to the manifold being curved.

However, as the example of Godel spacetime shows, if there is a nonzero cosmological constant, one can have a timelike geodesic congruence that meets the "constant distance" requirement (and therefore satisfies the parallel postulate), even though the spacetime is curved.
 
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  • #22
Ibix said:
Nearby ones, yes.
What if two timelike geodesics have 'finite' separation (not nearby) ? Can we define as well a proper distance between them ?
 
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  • #23
cianfa72 said:
What if two timelike geodesics have 'finite' separation (not nearby) ? Can we define as well a proper distance between them ?
If there's a single spacelike geodesic connecting them, yes, but if there are multiple ways to connect them with spacelike geodesics that are initially perpendicular to one worldline then which one should you pick?
 
  • #24
cianfa72 said:
What if two timelike geodesics have 'finite' separation (not nearby) ? Can we define as well a proper distance between them ?
As @Ibix said, in general there will not be a unique spacelike geodesic between the worldlines. However, in one very specific case, there will be: if the timelike geodesic congruence is hypersurface orthogonal, and if geodesics of any hypersurface of the orthogonal foliation to the congruence are also geodesics of the spacetime. The only case I'm aware of that meets all these requirements is FRW spacetime.
 
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  • #25
PeterDonis said:
As @Ibix said, in general there will not be a unique spacelike geodesic between the worldlines. However, in one very specific case, there will be: if the timelike geodesic congruence is hypersurface orthogonal, and if geodesics of any hypersurface of the orthogonal foliation to the congruence are also geodesics of the spacetime. The only case I'm aware of that meets all these requirements is FRW spacetime.
Actually, the geodesics of the indicated hypersurface are not geodesics of the spacetime. This is why Fermi Normal hypersurface to a comoving observer is different from the hypersurface orthogonal slice. The former has the property that geodesics of the surface are geodesics of the spacetime. The latter does not.
 
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  • #26
PAllen said:
This is why Fermi Normal hypersurface to a comoving observer is different from the hypersurface orthogonal slice.
Ah, yes, I had forgotten about that
 
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  • #27
PAllen said:
Actually, the geodesics of the indicated hypersurface are not geodesics of the spacetime.
Does your claim actually refer specifically to FRW spacetime? In other words there exists such a curved spacetime that what @PeterDonis said is true, I believe -- namely there is an hypersurface orthogonal timelike geodesic congruence having the property that geodesics of any hypersurface of the orthogonal foliation to the congruence are also geodesics of the spacetime.
 
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  • #28
I'd say the only way to define such a distance is the standard way given in Landau Lifshitz. You start with infinitesimally close points ##A## (coordinates ##q^{\mu}##) and ##B## (coordinates ##q^{\mu}+\mathrm{d} q^{\mu}##) you can measure a spatial distances by sending a lightsignal from ##A## to ##B## being reflected back to ##A## and measure the time this needs. Then the distances is half this time times ##c##.

The light signal follows a null geodesics, i.e.,
$$g_{\mu \nu} \mathrm{d} q^{\mu} \mathrm{d} q^{\nu}=g_00 \mathrm{d} q^{02} + 2 g_{0j} \mathrm{d} q^0 \mathrm{d} q^j +g_{jk} \mathrm{d} q^j \mathrm{d} q^k =0.$$
From this you get the coordinate-time interval
$$\mathrm{d} q^0_{12}=\frac{1}{g_{00}}\left (-g_{0j} \mathrm{d} q^j \pm \sqrt{(g_{0j} g_{0k}-g_{jk} g_{00})\mathrm{d} q^j \mathrm{d} q^k} \right),$$
referring to the path from A to B and back from B to A. So the coordinate-time interval of the round-trip light signal is
$$\mathrm{d} q_1^0-\mathrm{d} q_2^0=\frac{2}{g_{00}} \sqrt{(g_{0j} g_{0k}-g_{jk} g_{00})\mathrm{d} q^j \mathrm{d} q^k}.$$
The proper-time interval is this times ##\sqrt{g_{00}}## and thus the spatial distance between the points this times ##1/(2c)##. So we have
$$\mathrm{d} l^2 = \left (\frac{g_{0j} g_{0k}}{g_{00}}-g_{jk} \right) \mathrm{d} q^j \mathrm{d} q^k.$$
This gives a local spatial metric, but if ##g_{\mu \nu}## depends on time, you cannot integrate these infinitesimal lengths uniquely to give a definite length between two points separated by a finite distance, as discussed above.
 
  • #29
cianfa72 said:
Does your claim actually refer specifically to FRW spacetime? In other words there exists such a curved spacetime that what @PeterDonis said is true, I believe -- namely there is an hypersurface orthogonal timelike geodesic congruence having the property that geodesics of any hypersurface of the orthogonal foliation to the congruence are also geodesics of the spacetime.
No, for FLRW solutions, the hypersurfaces orthogonal to the congruence of comoving observers have the property that geodesics of the surface are not geodesics of the spacetime. This follows immediately from the fact that the Fermi Normal surface at an event on a comoving world line is defined by the collection of spacetime orthogonal geodesics, and that this is not the same as a slice of constant cosmological time. But the cosmological slice is tangent to the Fermi slice at each event on the origin comoving observer for the comoving slice. Since there can be exactly one geodesic through a given point with given tangent (trivially follows from parallel transport definition of geodesic), the cosmological slice cannot be made up of geodesics of the spacetime, since these define the Fermi Normal slice.
 
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  • #30
I suspect, but have not tried to prove, that:

If a spacetime region supports a timelike geodesic congruence such that there is a hypersurface orthogonal foliation that has the property that geodesics of the surface are geodesics of the spacetime, then the spacetime is flat in that region.

Note, this does not contradict what I said about FermiNormal slices. The timelike congruence of constant FermiNormal position is not a geodesic congruence. Only the comoving origin would be a geodesic, except in the edge case of the flat Milne cosmology, in which case the FermiNormal coordinates are ordinary Minkowski coordinates centered on the chosen comoving geodesic.
 
  • #31
cianfa72 said:
Summary:: About the definition to use for a global inertial frame in GR

take a family of free-falling bodies (a geodesic congruence) foliating it: if we choose a frame (coordinate chart) in which those free-falling bodies are at rest
You can take such a family of bodies and define a basis (tetrad) based on that, but it will be a non-coordinate basis so you cannot chose a coordinate chart based on that. See e.g. the last appendix in the Carroll's book.
 
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  • #32
Demystifier said:
You can take such a family of bodies and define a basis (tetrad) based on that, but it will be a non-coordinate basis so you cannot chose a coordinate chart based on that. See e.g. the last appendix in the Carroll's book.
I’m confused at what you are claiming. Lemaitre coordinates for Schwarzschild geometry are synchronous coordinates in which the congruence of radial infalling test bodies from infinity are at coordinate rest (similar to comoving observers in FLRW standard coordinates). The Lemaitre foliation of the infalling congruence is also hypersurface orthogonal.
 
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  • #33
PAllen said:
I’m confused at what you are claiming. Lemaitre coordinates for Schwarzschild geometry are synchronous coordinates in which the congruence of radial infalling test bodies from infinity are at coordinate rest (similar to comoving observers in FLRW standard coordinates). The Lemaitre foliation of the infalling congruence is also hypersurface orthogonal.
I think he means in general. In a spcific spacetime and a specific congruence it is possible.
 
  • #34
martinbn said:
I think he means in general. In a spcific spacetime and a specific congruence it is possible.
Ok, but the OP specifically asked about the region around earth.

Also, you can construct Gaussian Normal (also called synchronous) coodinates in a finite region in any spacetime. As to lack of global coverage, I don't see how that is a differentiator, because in the general case, there is unlikely to be be any global coordinate charts.

Note the features of Gaussian Normal coordinates:

1) Lines of constant position are timelike geodesics
2) This timelike congruence admits foliation by spacelike surfaces everywhere orthogonal to the congruence
3) The proper time along any of the defining congruence lines between foliation surfaces is the same (this is why such coordinates are also called synchronous).

[add: Just to be clear, as noted by @PeterDonis and @Ibix much earlier in this thread, the conditions satisfied by a Gaussian normal chart do not define an inertial frame because the distances between the congruence lines change (in almost all cases). In curved spacetime it is impossible to add the constant distance requirement - there will be no solution to the combined constraints. As an aside, in Godel spacetime, maximum coverage of a Gaussian patch is small, being inversely proportional to the vorticity of the Godel spacetime. ]
 
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  • #35
PAllen said:
I’m confused at what you are claiming. Lemaitre coordinates for Schwarzschild geometry are synchronous coordinates in which the congruence of radial infalling test bodies from infinity are at coordinate rest (similar to comoving observers in FLRW standard coordinates). The Lemaitre foliation of the infalling congruence is also hypersurface orthogonal.
So the geodesic congruence of radial infalling test bodies from infinity in Schwarzschild geometry basically define the Lemaitre coordinate chart: assign different spatial coordinate value to each infalling test body and take the proper time along any of the congruence worldline.
 
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