About global inertial frame in GR - revisited

  • #1
cianfa72
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TL;DR Summary
Fermi Normal hypersurfaces orthogonal to the FLRW congruence of comoving observer's worldlines
Hi, reading this old thread I'd like a clarification about the following:

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.

Fermi Normal hypersurface at an event on a comoving FLRW worldline is defined by the collection of spacetime orthogonal geodesics. Such geodesics should be spacelike since they are orthogonal to the timelike comoving worldlines at each event along them. On the other hand since FLRW comoving congruence's worldlines are hypersurface orthogonal then such spacelike hypersufaces should be the same as the Fermi Normal ones, I believe.

Am I wrong ? Thank you.
 
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  • #2
You are wrong. The Fermi Normal coordinates are guaranteed to be orthogonal to the cosmological time direction only in the point used to define them whereas the hypersurfaces of constant cosmological time are orthogonal to the cosmological time direction everywhere by construction.

Edit: Compare with the following example in Riemannian geometry. Take polar coordinates on ##\mathbb R^2##. The curves of constant ##r## are everywhere orthogonal to the radial direction. The normal coordinates based on a particular point has the circle’s tangent as the coordinate line. This is orthogonal to the radial direction only at the point used to define the normal coordinates and nowhere else.

Edit 2: See also https://www.physicsforums.com/insights/coordinate-dependent-statements-expanding-universe/ In particular section 3 where the coordinate transformation is constructed explicitly to leading non-trivial order.
 
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  • #3
Orodruin said:
The Fermi Normal coordinates are guaranteed to be orthogonal to the cosmological time direction only in the point used to define them whereas the hypersurfaces of constant cosmological time are orthogonal to the cosmological time direction everywhere by construction
Ah ok, so spacelike hypersurfaces of constant cosmological time cannot be made of spacelike geodesics of spacetime.
 
  • #4
cianfa72 said:
Ah ok, so spacelike hypersurfaces of constant cosmological time cannot be made of spacelike geodesics of spacetime.
Well, it depends on the spacetime. Minkowski space is perfectly conforming to RW coordinates with ##a = 1##.
 
  • #5
Orodruin said:
Well, it depends on the spacetime. Minkowski space is perfectly conforming to RW coordinates with ##a = 1##.
Yes, in that thread we were talking about FRW spacetime. Which are the RW coordinates in Minkowski spacetime ?
 
  • #6
cianfa72 said:
Yes, in that thread we were talking about FRW spacetime. Which are the RW coordinates in Minkowski spacetime ?
Minkowski space is a FLRW spacetime (with no energy content).
 
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  • #7
Orodruin said:
Minkowski space is a FLRW spacetime (with no energy content).
Sorry, I have no idea what the RW coordinates are...
 
  • #8
A Robertson-Walker spacetime is a spacetime that is foliated by homogeneous and isotropic spatial hypersurfaces such that you can construct coordinates where the metric takes the form
$$
ds^2 = dt^2 -a(t)^2 d\Sigma^2
$$
with ##d\Sigma^2## being the metric of the homogeneous and isotropic spatial surfaces. These are the Robertson-Walker coordinates.

A RW spacetime that additionally satisfies the Friedmann equations for some energy content would be a FLRW spacetime.
 
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  • #9
Ah ok, so your point is that Minkowski spacetime is a RW spacetime with ##a(t)=1##. Therefore for Minkowski spacetime the hypersurfaces of RW constant cosmological time are actually hypersurfaces made up of spacelike geodesics of spacetime (i.e. spacelike hypersurfaces of constant cosmological time in RW coordinates are exactly the Fermi Normal hypersurfaces built at events along RW coordinate time congruence).

Right ?
 
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  • #10
Orodruin said:
Minkowski space is perfectly conforming to RW coordinates with ##a = 1##.
By this I assume you are referring to the trivial case where ##a = 1## for all ##t##, meaning that both Friedmann equations are simply ##0 = 0## and there is no actual expansion, correct? For this case "RW coordinates" are the same as ordinary inertial coordinates, whose spacelike surfaces of constant ##t## are of course just Euclidean 3-spaces spanned by geodesics of the spacetime.

I bring this up because there is another FRW solution on Minkowski spacetime, or more precisely on the future light cone of the origin in Minkowski spacetime, the Milne model:

https://en.wikipedia.org/wiki/Milne_model

This solution also has zero energy content, but it is expanding: the comoving worldlines are all possible timelike geodesics coming from the origin. The surfaces of constant ##t## in this model are hyperbolas of constant timelike interval from the origin, which are not geodesics of the spacetime.
 
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  • #11
PeterDonis said:
By this I assume you are referring to the trivial case where a=1 for all t, meaning that both Friedmann equations are simply 0=0 and there is no actual expansion, correct? For this case "RW coordinates" are the same as ordinary inertial coordinates, whose spacelike surfaces of constant t are of course just Euclidean 3-spaces spanned by geodesics of the spacetime.
Yes.

PeterDonis said:
I bring this up because there is another FRW solution on Minkowski spacetime, or more precisely on the future light cone of the origin in Minkowski spacetime, the Milne model:

https://en.wikipedia.org/wiki/Milne_model

This solution also has zero energy content, but it is expanding: the comoving worldlines are all possible timelike geodesics coming from the origin. The surfaces of constant t in this model are hyperbolas of constant timelike interval from the origin, which are not geodesics of the spacetime.
Indeed. I also brought up the Milne model in the Insight I linked earlier. It is a great example of what appears as cosmological redshift in one coordinate system is just your regular Doppler shift in another.
 
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  • #12
BTW, the take-home of that thread was that even in some curved spacetime we can define a global inertial coordinate chart. E.g. in Godel spacetime there is a timelike geodesic congruence that has zero expansion and shear. Therefore we can assign fixed spatial coordinates to each worldline in the congruence defining a global chart in which the congruence's timelike geodesics are "at rest". Morever the condition of zero expansion and shear for the congruence does mean "zero geodesic deviation for those geodesics".

The above conditions seem to met the requirements of a global inertial coordinate chart (see the definition in post#9 there).
 
  • #13
cianfa72 said:
BTW, the take-home of that thread was that even in some curved spacetime we can define a global inertial coordinate chart.
No, it wasn't. See below.

cianfa72 said:
E.g. in Godel spacetime there is a timelike geodesic congruence that has zero expansion and shear.
But not zero vorticity, so the chart based on this congruence is not inertial. This was brought up in the previous thread, but you apparently have forgotten it. I suggest re-reading the caution I gave in post #79 of the previous thread.
 
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  • #14
PeterDonis said:
But not zero vorticity, so the chart based on this congruence is not inertial. This was brought up in the previous thread, but you apparently have forgotten it. I suggest re-reading the caution I gave in post #79 of the previous thread.
The consequence of that being that planes constructed from infinitesimal planes locally orthogonal to the congruence don't close, similar to the issues with naively constructing a simultaneity convention from local MCIFs on a rotating disc in flat spacetime?
 
  • #15
Ibix said:
The consequence of that being that planes constructed from infinitesimal planes locally orthogonal to the congruence don't close, similar to the issues with naively constructing a simultaneity convention from local MCIFs on a rotating disc in flat spacetime?
I believe so, yes.
 
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  • #16
Sorry, maybe I missed the point. We said that a global inertial chart requires that accelerometers at rest in it measure zero proper acceleration (i.e. their paths are timelike geodesics). Then we added in post#9 the constraint that the distance between those inertial accelerometers doesn't change (i.e. zero expansion and shear for those accelerometer's geodesic timelike paths).

What is the purpose of the further constraint of zero vorticity for the above congruence (i.e. hypersurface orthogonal condition for that congruence) ? Thanks in advance.
 
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  • #17
cianfa72 said:
We said that a global inertial chart requires that accelerometers at rest in it measure zero proper acceleration (i.e. their paths are timelike geodesics). Then we added in post#9 the constraint that the distance between those inertial accelerometers doesn't change (i.e. zero expansion and shear for those accelerometer's geodesic timelike paths).
Yes, and in post #40 I added the zero vorticity requirement.

cianfa72 said:
What is the purpose of the further constraint of zero vorticity
Nonzero vorticity means that there will be "fictitious forces" present--the worldlines of at least some freely falling objects will not be straight lines in the chart. In the vast majority of cases, this also means that the chart itself will not be geodesic--worldlines at rest in the chart will have nonzero proper acceleration, so they won't meet the geodesic requirement anyway. But in the case of Godel spacetime, there is a geodesic congruence that still has nonzero vorticity and for which the spatial paths of geodesics not in the congruence will not be straight lines in the corresponding chart. So to be sure we are specifying an inertial chart, we need to also specify zero vorticity for the congruence the chart is based on.
 
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  • #18
cianfa72 said:
i.e. hypersurface orthogonal condition for that congruence
@Ibix in post #14 brought up another relevant point, which is that if the congruence is not hypersurface orthogonal, it will be impossible to construct a global chart using it, because the surfaces of "constant time" will not close. The example he was implicitly referring to was the Born chart on Minkowski spacetime.
 
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  • #19
PeterDonis said:
But in the case of Godel spacetime, there is a geodesic congruence that still has nonzero vorticity and for which the spatial paths of geodesics not in the congruence will not be straight lines in the corresponding chart.
You mean the timelike geodesic congruence of worldlines at rest in standard coordinate chart in Godel spacetime. In this coordinate chart the spatial paths of timelike geodesics not in the congruence are not straight lines, however.
 
  • #20
cianfa72 said:
You mean the timelike geodesic congruence of worldlines at rest in standard coordinate chart in Godel spacetime.
Yes.

cianfa72 said:
In this coordinate chart the spatial paths of timelike geodesics not in the congruence are not straight lines, however.
Yes.
 
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  • #21
PeterDonis said:
So to be sure we are specifying an inertial chart, we need to also specify zero vorticity for the congruence the chart is based on.
So this further constraint make sure that every timelike geodesic either in the timelike geodesic congruence the coordinate chart is based on or not in it, will always have straight spatial path in that coordinate chart.
 
  • #22
cianfa72 said:
So this further constraint make sure that every timelike geodesic either in the geodesic congruence the coordinate chart is based on or not in it, will always have straight spatial path in that coordinate chart.
A straight worldline, not just a straight spatial path.
 
  • #23
PeterDonis said:
A straight worldline, not just a straight spatial path.
Sorry, in post#17 you talked about spatial paths of timelike geodesics not in the congruence in that coordinate chart.
 
  • #24
cianfa72 said:
Sorry, in post#17 you talked about spatial paths of timelike geodesics in that coordinate chart.
I said both "worldlines" and "spatial paths" in that post. But actually only "worldlines" is necessary, because if they are straight lines, the spatial paths (the projection of the worldline into a spacelike surface of constant time) will also be straight lines.
 
  • #25
PeterDonis said:
But actually only "worldlines" is necessary, because if they are straight lines, the spatial paths
So this basically amounts to check if the spatial coordinates of the timelike geodesics are linearly dependent with constant coefficients on the coordinate time along the path in spacetime.
 
  • #26
cianfa72 said:
So this basically amounts to check if the spatial coordinates of the timelike geodesics are linearly dependent with constant coefficients on the coordinate time along the path in spacetime.
Yes.
 
  • #27
I've some difficulty to grasp the concept of coordinate time. From a physical point of view, since it corresponds to timelike paths, it should be related to the proper time along those timelike paths "at rest" in the coordinate chart.

In other words we can calculate the chart's coordinate time of an event that occurs along the timelike path at rest in the chart from the reading of the clock "carried" by its implied observer.
 
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  • #28
cianfa72 said:
I've some difficulty to grasp the concept of coordinate time. From a physical point of view, since it corresponds to timelike paths, it should be related to the proper time along those timelike paths "at rest" in the coordinate chart.

In other words we can calculate the chart's coordinate time of an event that occurs along the timelike path at rest in the chart from the reading of the clock "carried" by its implied observer.
Generally, coordinate time ##t## is only related to the proper time ##\tau## of an observer with fixed spatial coordinates as ##\tau = f(t)## where ##f## is some monotonically increasing function with ##f’(t)## is the square root of the factor in front of ##dt^2## in the line element. However, in RW coordinates that factor is one and so ##\tau = t##, ie, the coordinate time is the proper time of the comoving observers. This is by construction.

At the same time, for example, this is not the case for stationary observers in a Schwarzschild spacetime.
 
  • #29
Orodruin said:
Generally, coordinate time ##t## is only related to the proper time ##\tau## of an observer with fixed spatial coordinates as ##\tau = f(t)## where ##f## is some monotonically increasing function with ##f’(t)## is the square root of the factor in front of ##dt^2## in the line element.
So, how do other observers at rest in the coordinate chart with fixed spatial coordinates "assign" coordinate time to events occurring along their worldlines ?
 
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  • #30
cianfa72 said:
In other words we can calculate the chart's coordinate time of an event that occurs along the timelike path at rest in the chart from the reading of the clock "carried" by its implied observer.
You can do, but the calculation may depend on your position and time.

A simple analogy on the surface of the Earth: if you start on the equator and move north, your latitude coordinate is roughly your distance travelled divided by ##111\ \mathrm{km/°}##. On the other hand, if you start on the Greenwich meridian and move only East then your longitude coordinate is roughly ##111\sec(\mathrm{latitude})\ \mathrm{km/°}## (using the geographer's definition of latitude that is zero at the equator).

Just as these coordinates are related to ##\int\sqrt{g_{ab}u^au^b}dt##, but not necessarily trivially, so is coordinate time.
 
  • #31
cianfa72 said:
So, how do other observers at rest in the coordinate chart with fixed spatial coordinates "assign" coordinate time to events occurring along their worldlines ?
That would depend on the coordinates being used. There is nothing actually physical about the concept of a coordinate time. It only holds physical significance in relation to whatever process is used to assign it.
 
  • #32
Ibix said:
Just as these coordinates are related to ##\int\sqrt{g_{ab}u^au^b}dt##, but not necessarily trivially, so is coordinate time.
ok, you're saying that from the spatial position (spatial coordinates) of an event in the chart and the elapsed proper time along a timelike path joining the (0,0,0,0) event with the given event we can calculate back the coordinate time ##t## for it.
 
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  • #33
Orodruin said:
It only holds physical significance in relation to whatever process is used to assign it.
That was my point. A coordinate chart for spacetime (or for the Earth surface) is defined from the process employed to assign those coordinates to events (points on the Earth surface as in the example from @Ibix ).
 
  • #34
cianfa72 said:
ok, you're saying that from the spatial position (spatial coordinates) of an event in the chart and the elapsed proper time along the timelike path with fixed spatial coordinates joining the t=0 hypersurface with the given event we can calculate back the coordinate time ##t## for it.
With the bolded changes above, yes.
 
  • #35
Ibix said:
ok, you're saying that from the spatial position (spatial coordinates) of an event in the chart and the elapsed proper time along the timelike path with fixed spatial coordinates joining the t=0 hypersurface with the given event we can calculate back the coordinate time for it.
ok, that's basically like moving on the Earth only East along a parallel with a fixed latitude of an amount of longitude starting from Greenwich meridian (all points on Greenwich meridian have longitude 0 like ##t=0## hypersuface events).
 
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