# Reference frame vs coordinate system

• I
• lriuui0x0
I'm not sure what you mean by "basis". A basis for a set of vectors in a vector space is a set of vectors such that every vector in the set can be written in terms of the members of the basis. In the case of the frame field, the basis would consist of the 4 vectors in the tangent space at that point. However, you don't need a basis for the frame field, since it is a mapping from a spacetime to 4 vectors.Is this concept related to inertial/non-inertial frame?No, the concept of the frame field is unrelated to the concepts of inertial/non-inertial frames.

#### lriuui0x0

Just want to clarify some concepts.

There seems to be difference between reference frame and coordinate system. See https://en.wikipedia.org/wiki/Frame_of_reference#Definition . A reference frame is something has physical meaning and is related to physical laws, whereas coordinate system (chart) is purely mathematical tools that convert reference frame to numbers. The difference between reference frame and coordinate system is analogous to the difference between vector and number tuple, or linear map and matrix. Is this understanding correct?

I also heard that a reference frame is an ordered set of basis vector (on the tangent space if it's GR). Is this correct? If that's the case, is a non-inertial frame also a set of basis vector?

lriuui0x0 said:
There seems to be difference between reference frame and coordinate system.
There is a lot of variation in terminology in the literature. Some sources use "reference frame" to mean "coordinate system". Others don't.

lriuui0x0 said:
Wikipedia should definitely not be your go-to source for something like this. You really need to look at actual textbooks or peer-reviewed papers.

Basically, there are three distinct concepts involved here:

(1) A mapping of points in some open set (which might be an entire spacetime or just an open region of one) to 4-tuples of real numbers. The most common term for this is "coordinate chart", but "reference frame" is also sometimes used to mean this, as noted above.

(2) A mapping of points in some open set (which might be an entire spacetime or just an open region of one) to tetrads--i.e., sets of 4 vectors, one timelike and three spacelike, all of unit length and orthogonal to each other. The most precise technical terms for this in the literature are "frame field" or "tetrad field", but "reference frame" is also sometimes used to mean this.

(3) An actual physical apparatus for making measurements, which will typically occupy a small finite region of spacetime. There does not seem to be a single technical term for this, although "measurement apparatus" or "laboratory apparatus" are sometimes used; but "reference frame" is also sometimes used to mean this.

So, unfortunately, the term that invites the most ambiguity and misunderstanding is "reference frame", since it is used for all three of the above concepts. But unfortunately it is also the most commonly used term, and the one most likely to provoke acrimonious discussions where participants are talking past each other because they are using it to mean different things.

• dextercioby, cianfa72, martinbn and 1 other person
PeterDonis said:
Wikipedia should definitely not be your go-to source for something like this. You really need to look at actual textbooks or peer-reviewed papers.
Oops, you caught me again ;)

Ok I think for my purpose, we don't need to consider (3), so the difference I'm considering is "frame field" vs "coordinate chart". For the moment let's consider flat spacetime only (classical or SR). Based on your definition, "frame field" is a mapping from spacetime to sets of 4 vectors. How does this concept degenerate in classical mechanics and SR? Does it mean we only need a single set of 4 vectors (a basis)? Is this concept related to inertial/non-inertial frame? What happens in non-inertial frame?

• dextercioby
lriuui0x0 said:
For the moment let's consider flat spacetime only (classical or SR).
There is no such thing as "flat spacetime" in classical mechanics, if by that you mean Newtonian mechanics. One can formulate Newtonian mechanics in terms of a spacetime (the Cartan formulation), but the spacetime does not have a metric (there is a spatial metric and a time function but no spacetime metric), so it makes no sense to say that spacetime is flat (it only makes sense to say that space is flat).

lriuui0x0 said:
Based on your definition, "frame field" is a mapping from spacetime to sets of 4 vectors.
More precisely, it is a mapping from each point of the spacetime to a set of 4 orthonormal vectors in the tangent space at that point.

lriuui0x0 said:
How does this concept degenerate in classical mechanics and SR?
I'm not sure what you mean by "degenerate". The frame field concept applies in SR just as it is. In classical mechanics there is no such thing as an orthonormal tetrad because there is no spacetime metric, as noted above. So there is no way to define a frame field at all in the SR/GR sense.

lriuui0x0 said:
Does it mean we only need a single set of 4 vectors
No. You need a set of 4 vectors at each point, since the vectors are vectors in the tangent space at that point. (You can sometimes get away with ignoring this in flat spacetime and treating the vectors as though they "lived" in the spacetime itself instead of the tangent space at each point, but this will come back to bite you as soon as you deal with either non-inertial frames or curved spacetime, so it's best to avoid it from the start.)

lriuui0x0 said:
(a basis)?
You can treat a tetrad at a given point as an orthonormal basis for the tangent space at that point, yes.

Whether or not you can use a tetrad field as the basis for a coordinate chart on whatever open set it covers depends on whether the tetrad vectors commute. In general they won't, which means they can't serve as a coordinate basis. They can always serve as a non-coordinate basis; whether or not that helps depends on what computational methods you are using (some work better in a coordinate basis, some work better in an orthonormal basis).

lriuui0x0 said:
Is this concept related to inertial/non-inertial frame?
A tetrad field can be considered inertial or non-inertial depending on whether the worldlines defined by the integral curves of its timelike vectors are geodesics (inertial) or not (non-inertial). The general tetrad field method works fine in both cases.

• dextercioby, cianfa72 and BvU
Just to be clear: a frame field that is defined on a open spacetime region (or on the entire spacetime since itself is open) does not assign coordinates to events in that open set. However it can be useful to represent an observer along its timelike worldline carring a wristwatch and 3 mutually orthogonal spatial axes.

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cianfa72 said:
a frame field that is defined on a open spacetime region (or on the entire spacetime since itself is open) does not assign coordinates to events in that open set.
This is true, but it's also true that, in order to have useful expressions for the frame field vectors, one generally has to already have a coordinate chart defined on the open region. It's just that the frame field vectors won't in general be the same as the coordinate basis vectors.

• cianfa72
I'm trying to bridge my understanding from classical mechanics to relativity by thinking about these problems more carefully. Is frame field also a concept in classical mechanics? Is there any resources explaining things from this perspective? I feel I might ask better questions once I read a bit more on this.

• dextercioby
lriuui0x0 said:
Is frame field also a concept in classical mechanics?
Not that I'm aware of. I'm not sure how you would define one in Newtonian mechanics since there is no spacetime metric and therefore no way to form an orthonormal set of one timelike and 3 spacelike vectors. (Even the concepts of "timelike" and "spacelike" vectors are not really well-defined in Newtonian mechanics, since there is not one single vector space that contains both ordinary spatial vectors and "vectors" pointing along the worldlines of objects.)

• vanhees71
PeterDonis said:
Not that I'm aware of. I'm not sure how you would define one in Newtonian mechanics since there is no spacetime metric and therefore no way to form an orthonormal set of one timelike and 3 spacelike vectors. (Even the concepts of "timelike" and "spacelike" vectors are not really well-defined in Newtonian mechanics, since there is not one single vector space that contains both ordinary spatial vectors and "vectors" pointing along the worldlines of objects.)
If it's not available in classical mechanics, SR would also be good. Basically I'm used to thinking about a single frame of reference (usually inertial) that can be transformed Lorentz transformation, so that we can give coordinates numbers to events in spacetime. But I don't know how do we interpret frame field? Is that something that an observer carries along on its world line? Does that mean the frames only change when we're considering an accelerating observer?

lriuui0x0 said:
If it's not available in classical mechanics, SR would also be good.
The fact that frame fields work fine in SR should be obvious from the definition I gave.

lriuui0x0 said:
I'm used to having a single frame of reference (usually inertial) that can be transformed Lorentz transformation
A Lorentz transformation changes from one inertial frame to another, so once you use one, you no longer have a single frame of reference.

lriuui0x0 said:
so that we can give coordinates numbers to events in spacetime
You can do that with any coordinate chart; it doesn't have to be inertial.

lriuui0x0 said:
I don't know how do we interpret frame field?
The first thing to do is to not confuse the word "frame" in "frame field" with your previous concepts of "reference frame". "Frame" in "frame field" just means "tetrad", i.e., "set of 4 orthonormal vectors, one timelike and 3 spacelike, at a point in spacetime". So "frame field" just means "tetrad field" (and if it helps, I would get in the habit of using the latter term, and I'll do that myself from now on), i.e., an assignment of a tetrad to each point in some open region.

lriuui0x0 said:
Is that something that an observer carries along on its world line?
Once you have a tetrad field on an open region of spacetime, it obviously determines a set of timelike worldlines: the integral curves of the timelike vectors of the tetrads. Those timelike worldlines can be interpreted as a family of observers, and the tetrads along each worldline can be interpreted as the 4-velocity of the observer whose worldline it is, plus a set of 3 orthonormal spatial vectors that the observer carries with him for making measurements.

lriuui0x0 said:
Does that mean the frames only change when we're considering an accelerating observer?
A tetrad field doesn't "change". It is just an assignment of a tetrad to each point in an open region. That assignment is fixed.

• vanhees71
PeterDonis said:
Once you have a tetrad field on an open region of spacetime, it obviously determines a set of timelike worldlines: the integral curves of the timelike vectors of the tetrads. Those timelike worldlines can be interpreted as a family of observers, and the tetrads along each worldline can be interpreted as the 4-velocity of the observer whose worldline it is, plus a set of 3 orthonormal spatial vectors that the observer carries with him for making measurements.

Maybe being more specific would be useful here. Let's say we consider a non-inertial observer in SR. Surely it has a world line. At each event of the world line, will this observer have a tetrads (frame)? And will these tetrads be different? (Assuming we can compare vectors at different points in SR). And is this the reason why we need the tetrad field, i.e. multiple tetrads? So is it fair to say that in some sense, the tetrads fields (along a world line) defines how an observer observes? I might have repeated some of your reply, but I want to double check my understanding is correct.

On the other hand, if it's an inertial observer. Is the tetrads along its world line the same? I'm a bit confused between this view and the spacetime diagrams you usually draw in SR https://upload.wikimedia.org/wikipe...Minkowski_diagram_-_Newtonian_physics.svg.png . Here in the diagram we seem to have a fixed tetrad (or frame, but also coordinates at the same time), the inertial observer is represented by the vertical axis. But this is different from the tetrad frame perspective, in which an observer having a separate tetrads at any point on its world line? In this view it seems the time is redefined at every point on the world line, and I don't know why this is useful.

PeterDonis said:
There is no such thing as "flat spacetime" in classical mechanics, if by that you mean Newtonian mechanics. One can formulate Newtonian mechanics in terms of a spacetime (the Cartan formulation), but the spacetime does not have a metric (there is a spatial metric and a time function but no spacetime metric), so it makes no sense to say that spacetime is flat (it only makes sense to say that space is flat)
Can you kindly point me to a source for a detailed Cartan formulation of Newtonian mechanics ?

lriuui0x0 said:
Let's say we consider a non-inertial observer in SR. Surely it has a world line
Of course.

lriuui0x0 said:
At each event of the world line, will this observer have a tetrads (frame)?
If you want to assign a tetrad field to the open region of spacetime containing the worldline, yes. A tetrad field is a mathematical object assigned by humans to help with understanding and calculation.

If the observer is physically carrying a set of gyroscopes pointing in mutually orthogonal directions, then it would certainly be possible to assign a tetrad field to the open region of spacetime containing the observer's worldline such that the tetrad assigned to each point on the worldline corresponded to the observer's 4-velocity and the spacelike vectors defined by the gyroscopes at that point. And doing that might be very helpful in making calculations. But one is not required to do this; one can assign any valid tetrad field one likes.

lriuui0x0 said:
will these tetrads be different? (Assuming we can compare vectors at different points in SR).
In SR, you can get away with comparing vectors at different points because spacetime is flat so transporting a vector from one point to another is not path dependent. But that is no longer the case in curved spacetime, so in general, in curved spacetime, you can't compare vectors at different points.

In the particular case under discussion, however, we have a path already picked out: the observer's worldline. Along the observer's worldline, we can use Fermi-Walker transport to define "the same" vectors at different points, and we can then check the vectors assigned by the tetrad field to see whether they change along the observer's worldline or not. Whether or not they do will depend on the tetrad field we choose. (If we choose the tetrad field so that it matches the observer's 4-velocity and the gyroscope vectors as described above, then the tetrad field vectors will not change along the observer's worldline, as tested by Fermi-Walker transport.)

lriuui0x0 said:
is this the reason why we need the tetrad field, i.e. multiple tetrads?
Not "multiple tetrads". One tetrad at each point of the spacetime region. We need a tetrad field because any open spacetime region contains more than one point.

lriuui0x0 said:
is it fair to say that in some sense, the tetrads fields (along a world line) defines how an observer observes?
Only if we choose the tetrad field to match something relevant about the observer, such as their 4-velocity and the gyroscopes they carry, as above.

lriuui0x0 said:
if it's an inertial observer. Is the tetrads along its world line the same?
Only if we choose the tetrad field that way. For an inertial observer, Fermi-Walker transport reduces to parallel transport along the observer's worldline, so we would need to choose the tetrad field so that its vectors are parallel transported along the observer's worldline.

lriuui0x0 said:
Here in the diagram we seem to have a fixed tetrad
There are no tetrads in that diagram. It's just a coordinate chart. It shows coordinate axes, but coordinate axes are not tetrads.

lriuui0x0 said:
In this view it seems the time is redefined at every point on the world line
I have no idea why you would think that. The timelike vector of the tetrad at every point is a unit vector.

PeterDonis said:
A tetrad field is a mathematical object assigned by humans to help with understanding and calculation.

It sounds from your explanation that tetrads don't have any physical meaning? I thought a tetrad is something that characterises your motion frame, i.e. inertial or non-inertial. But it seems that they're only as real as coordinate charts? So what's the point of having this mathematical tool other than coordinate charts?

I'm confused that there must be something that's physical and real and it can't be that everything is just math concepts that we come up with to ease calculation. Because there is real difference between inertial and non-inertial frame (e.g. fictitious force in non-inertial frame), and there's real difference between two inertial frames (e.g. relative simultaneity). What's the math concept that captures this difference?

PeterDonis said:
There are no tetrads in that diagram. It's just a coordinate chart. It shows coordinate axes, but coordinate axes are not tetrads.

For example, in this diagram https://opentextbc.ca/universityphy.../273/2019/07/CNX_UPhysics_38_05_Lorentz-1.jpg there're two states of motion. If coordinate charts are just calculation tools, what is the math concept that differentiate these two states? And similar question for the non-inertial frames.

lriuui0x0 said:
It sounds from your explanation that tetrads don't have any physical meaning?
It depends on how you choose them. You can choose tetrads to reflect something physically meaningful, as in my example of choosing them so they match the observer's 4-velocity and gyroscopes; but there is nothing that requires you to do that.

lriuui0x0 said:

lriuui0x0 said:
i.e. inertial or non-inertial
"Inertial" or "non-inertial" are best viewed as properties of observers, i.e., worldlines (corresponding to zero or non-zero path curvature/proper acceleration). When we speak of an inertial or non-inertial coordinate chart, we mean that worldlines at rest in the chart (i.e., with constant spatial coordinates) are inertial or non-inertial. (Note that not all coordinate charts even admit such an interpretation.) When we speak of a tetrad field being inertial or non-inertial, we mean that the worldlines that are integral curves of the timelike tetrad vectors are inertial or non-inertial.

lriuui0x0 said:
what's the point of having this mathematical tool other than coordinate charts?
Because there are plenty of computations which are much simpler to do using tetrads than they are using coordinate charts.

lriuui0x0 said:
I'm confused that there must be something that's physical and real
Well, of course. Observers, and things the observers carry like gyroscopes, are physical and real. But we need mathematical models of these things in order to do calculations.

lriuui0x0 said:
there is real difference between inertial and non-inertial frame (e.g. fictitious force in non-inertial frame)
Many physicists would say that fictitious forces are not "real", since they are not felt and can be made to disappear just by changing coordinates.

lriuui0x0 said:
there's real difference between two inertial frames (e.g. relative simultaneity).
Many physicists would say that simultaneity is not "real", since it is frame-dependent so no actual observable can depend on it.

lriuui0x0 said:
What's the math concept that captures this difference?
The differences you describe in the quotes above, I would say, are not "real". But if you ask what math concept captures inertial vs. non-inertial, it's the path curvature (proper acceleration) of worldlines, and if you ask what math concept captures simultaneity, it's the time coordinate of some particular inertial coordinate chart.

lriuui0x0 said:
If coordinate charts are just calculation tools, what is the math concept that differentiate these two states?
Why would coordinate charts being a calculational tool prevent them from being math concepts that capture relevant differences for the calculation? Indeed, if you could not capture differences like that in a coordinate chart, coordinate charts would be useless as calculational tools, which obviously they are not.

To answer your question as you ask it, in any inertial coordinate chart, the coordinate speed of the two states of motion will be different (as the diagram you reference is drawn, one coordinate speed is zero and the other is nonzero), so this is an obvious, simple mathematical difference that captures the two states of motion.

I think you are making this much more difficult than it needs to be by not thinking clearly about what you are asking.

PeterDonis said:
"Inertial" or "non-inertial" are best viewed as properties of observers, i.e., worldlines (corresponding to zero or non-zero path curvature/proper acceleration). When we speak of an inertial or non-inertial coordinate chart, we mean that worldlines at rest in the chart (i.e., with constant spatial coordinates) are inertial or non-inertial. (Note that not all coordinate charts even admit such an interpretation.)
Ok this is helpful. I was confused between inertial observer (world line) vs inertial coordinates. And it caused me to think that there's a different coordinates at each point on the world line.
PeterDonis said:
Because there are plenty of computations which are much simpler to do using tetrads than they are using coordinate charts.
Thanks for the confirmation, now I think I don't need to consider this "tetrad field" concept at all unless I want learn more powerful calculation techniques. And btw, I think this concept is not the first bullet point in the wikipedia page https://en.wikipedia.org/wiki/Frame_of_reference#Definition . I was influenced by this and some other articles (before you stopped me ;). The concept being discussed here is just the world line, and it has nothing to do with tetrad field... It's just you also got three points in your reply.
PeterDonis said:
Many physicists would say that simultaneity is not "real", since it is frame-dependent so no actual observable can depend on it.
That's interesting. I guess we can decide what is "real" as long as the theory explains every phenomenon we observe and is consistent. And btw are you saying relative simultaneity is not observable? Is that true? Surely fictitious force is observable.
PeterDonis said:
I think you are making this much more difficult than it needs to be by not thinking clearly about what you are asking.
All the concepts were unclear and tangled in my mind and I can see that it made more difficult for you to answer. Really sorry about that, please think about this as casual discussion that help me clarify things gradually. I will try to make my questions more precise in the future.

PeterDonis said:
"Inertial" or "non-inertial" are best viewed as properties of observers, i.e., worldlines (corresponding to zero or non-zero path curvature/proper acceleration). When we speak of an inertial or non-inertial coordinate chart, we mean that worldlines at rest in the chart (i.e., with constant spatial coordinates) are inertial or non-inertial. (Note that not all coordinate charts even admit such an interpretation.)
The last claim in brackets holds since we are free to choose a coordinate chart in which any single coordinate is not timelike (i.e. any vector pointing along any single coordinate is not timelike).

lriuui0x0 said:
I think this concept is not the first bullet point in the wikipedia page
I think that bullet is too vague to tell what concept is is referring to. Normally, if we are drawing a distinction between "physical" and "mathematical" concepts, as that article appears to be doing, the "physical" thing would be the actual measuring apparatus; but that appears in the third bullet, and seems to be treated as something different from whatever is being talked about in the first bullet.

This illustrates why Wikipedia is not a good reference.

lriuui0x0 said:
The concept being discussed here is just the world line
I don't think that is a valid interpretation of the first bullet, since the bullet uses the word "frame", and a worldline, by itself, is not enough to define a frame under any definition of that term (it's not enough to define a coordinate chart, a tetrad field, or an actual measuring apparatus).

cianfa72 said:
The last claim in brackets holds since we are free to choose a coordinate chart in which any single coordinate is not timelike (i.e. any vector pointing along any single coordinate is not timelike).
I'm not sure what you mean here. Of course you can choose a chart that has at least one non-timelike coordinate. But you can't choose whether a particular vector at a particular point is timelike or not; that's an invariant, independent of any choice of coordinates.

Also, having at least one non-timelike coordinate is not enough to invalidate the kind of interpretation of a coordinate chart that I was talking about. The interpretation I was talking about requires that there is at least one timelike coordinate everywhere in whatever open region is being considered. So to rule it out, you would need a chart that has some points in the region where all coordinates are not timelike (not just at least one of them).

PeterDonis said:
Also, having at least one non-timelike coordinate is not enough to invalidate the kind of interpretation of a coordinate chart that I was talking about. The interpretation I was talking about requires that there is at least one timelike coordinate everywhere in whatever open region is being considered. So to rule it out, you would need a chart that has some points in the region where all coordinates are not timelike (not just at least one of them).
My point was that in a given open region of spacetime is always possible to pick a coordinate chart where all coordinates are non-timelike. Any chart with this property is actually ruled out since it makes no sense to say that a worldline is "at rest" in it.

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cianfa72 said:
My point was that in a given open region of spacetime is always possible to pick a coordinate chart where all coordinates are non-timelike.
Ok. "All coordinates are non-timelike", which I agree is sufficient to rule out a valid concept of "at rest" in the chart, is a stronger statement than "any single coordinate is non-timelike", which is what you said before.

Sorry for my english: so 'any single coordinate is non-timelike' actually means 'at least one of whatever coordinates is non-timelike' Last edited:
PeterDonis said:
I don't think that is a valid interpretation of the first bullet, since the bullet uses the word "frame", and a worldline
Ok that's fair, no need to get deep into what exactly the author had in mind ;)

PeterDonis said:
"Inertial" or "non-inertial" are best viewed as properties of observers, i.e., worldlines (corresponding to zero or non-zero path curvature/proper acceleration). When we speak of an inertial or non-inertial coordinate chart, we mean that worldlines at rest in the chart (i.e., with constant spatial coordinates) are inertial or non-inertial. (Note that not all coordinate charts even admit such an interpretation.)

Thinking about this again, I think an observer is more than just a world line. For example, we can think about an observer moving with uniform velocity, but rotating all the time. Then its world line appears to be inertial, yet the coordinates is clearly non-inertial. So for an inertial world line without acceleration, there could be multiple coordinate system in which the world line has zero spatial coordinates. Some of these coordinate systems are non-inertial.

Do you think this is correct?

lriuui0x0 said:
Thinking about this again, I think an observer is more than just a world line. For example, we can think about an observer moving with uniform velocity, but rotating all the time.
In this case you are modeling the observer as a congruence of worldlines (i.e., a family of worldlines that fills some "world tube" in spacetime). "Rotating" means the congruence has nonzero vorticity, where vorticity is part of the kinematic decomposition. This is an area where Wikipedia actually has an article that gives a useful introduction (though it is still rather technical):

https://en.wikipedia.org/wiki/Congr...atical_decomposition_of_a_timelike_congruence

This is certainly a valid model of an observer, and can be used in cases where that amount of detail is necessary. The "single worldline" model can then be thought of as a limiting case of this where the observer has no internal properties (such as "rotation") that need to be modeled, so all we need to know is the trajectory of the observer's center of mass in spacetime, which can be modeled as a single worldline.

• dextercioby and vanhees71
lriuui0x0 said:
Then its world line appears to be inertial, yet the coordinates is clearly non-inertial.
Notice how in my previous post I didn't mention coordinates at all?

I think you would be well served by forming the habit of never letting yourself even think about coordinates until you have a coordinate-free, invariant picture of what is going on, such as the one I described in my previous post for how to model a "rotating" observer. Of course one can introduce coordinates into such a model (look up "Fermi normal coordinates" for an example--these are discussed in most GR textbooks); but all of the physics is contained in invariant, coordinate-free quantities (the vorticity that I mentioned is an example).

lriuui0x0 said:
for an inertial world line without acceleration, there could be multiple coordinate system in which the world line has zero spatial coordinates. Some of these coordinate systems are non-inertial.
This would also be true of a worldline with nonzero proper acceleration. One can always find an infinite number of coordinate charts in which any timelike curve has constant spatial coordinates. At most one of them (if the worldline is a geodesic, i.e., zero proper acceleration) will be inertial, since, as I've already said, a coordinate chart can only be inertial if an object at rest in it is inertial.

• vanhees71
PeterDonis said:
In this case you are modeling the observer as a congruence of worldlines (i.e., a family of worldlines that fills some "world tube" in spacetime). "Rotating" means the congruence has nonzero vorticity, where vorticity is part of the kinematic decomposition. This is an area where Wikipedia actually has an article that gives a useful introduction (though it is still rather technical):

https://en.wikipedia.org/wiki/Congr...atical_decomposition_of_a_timelike_congruence

This is certainly a valid model of an observer, and can be used in cases where that amount of detail is necessary. The "single worldline" model can then be thought of as a limiting case of this where the observer has no internal properties (such as "rotation") that need to be modeled, so all we need to know is the trajectory of the observer's center of mass in spacetime, which can be modeled as a single worldline.
Thanks for the info! Looks like some more advanced stuff. I guess I will need to get the basic GR firmly understood before getting into this...

PeterDonis said:
I think you would be well served by forming the habit of never letting yourself even think about coordinates until you have a coordinate-free, invariant picture of what is going on
Good tip, thank you!

• vanhees71
PeterDonis said:
One can always find an infinite number of coordinate charts in which any timelike curve has constant spatial coordinates. At most one of them (if the worldline is a geodesic, i.e., zero proper acceleration) will be inertial, since, as I've already said, a coordinate chart can only be inertial if an object at rest in it is inertial.
It is worth noting that given a 'real/physical' observer we can consider the worldtube of its 'parts'. Suppose all worldlines in observer's worldtube are inertial (i.e. they are all geodesics). Then in the context of GR (i.e. curved spacetime) using them you can define a coordinate chart for a restricted region of spacetime as follows: assign different fixed spatial coordinate values to each of those parts in a continuous way; take the proper time read by wristwatches attached to each of them as the coordinate time of the chart being defined.

As long as the unavoidable geodesic deviation (i.e. tidal gravity) allows the geodesics in the observer's worldtube not to intersect each other, you get a valid inertial coordinate chart (restricted to a narrow region of spacetime).

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cianfa72 said:
As long as the unavoidable geodesic deviation (i.e. tidal gravity) allows the geodesics in the observer's worldtube not to intersect each other, you get a valid inertial coordinate chart
No, the condition for the chart to be inertial, even in just the narrow world tube you describe, is much stronger than "geodesics cannot intersect". The geodesic deviation needs to be negligible within the region of spacetime covered by the chart for the chart to be inertial, at least in the standard usage of that term. Having all objects at rest in the chart is a necessary condition for the chart to be inertial, but it is not sufficient. One also needs to have the metric coefficients independent of the coordinates (at least in the Cartesian form of the inertial chart). That cannot be true if geodesic deviation is not negligible in the region of spacetime covered by the chart; if it is not, the geodesics might not intersect, but forcing the spatial coordinates of each geodesic to be fixed means the metric coefficients will be functions of at least the time coordinate.

• vanhees71 and cianfa72
cianfa72 said:
As long as the unavoidable geodesic deviation (i.e. tidal gravity) allows the geodesics in the observer's worldtube not to intersect each other, you get a valid inertial coordinate chart (restricted to a narrow region of spacetime).
Just a doubt on this: to claim that the coordinate chart being defined is actually inertial I believe we must neglect the geodesic deviation of any two geodesics representing the worldlines of observer's parts in the worldtube.

cianfa72 said:
Just a doubt on this: to claim that the coordinate chart being defined is actually inertial I believe we must neglect the geodesic deviation of any two nearby geodesics that represent the worldlines of observer's parts in the worldtube.
No, it's the other way around: if the geodesic deviation cannot be neglected (meaning, it's large enough that it can be measured at whatever level of accuracy your measurements have), then you cannot claim that the chart being defined is inertial. We don't get to just arbitrarily ignore physical phenomena that are present and large enough to matter. Geodesic deviation is a physical phenomenon. That means it sets limits on how large a region of spacetime we can cover with an inertial chart at all.

• cianfa72
PeterDonis said:
Having all objects at rest in the chart is a necessary condition for the chart to be inertial, but it is not sufficient. One also needs to have the metric coefficients independent of the coordinates (at least in the Cartesian form of the inertial chart).
I take it as if we assign fixed spatial coordinate values to observer's parts (i.e. to their worldlines in the worldtube) and define the coordinate time such that the Pythagorean theorem holds for the proper distance of events at rest in the chart being defined that have the same coordinate time value (i.e. they are Einstein synchronized).

cianfa72 said:
I take it as if we assign fixed spatial coordinate values to observer's parts (i.e. to their worldlines in the worldtube) and define coordinate time such that the Pythagorean theorem holds for the proper distance of events at rest in the chart being defined that have the same coordinate time value (i.e. they are Einstein synchronized).
You can't do that if geodesic deviation is present. If geodesic deviation is present, one of the elements of your prescription must fail. It is impossible to meet all of your requirements in the presence of geodesic deviation.

• cianfa72
I found some new reference on this. In book "Special Relativity in General Frame", the concept of "local frame" is defined, similarly the concept of observer, in section 3.4. I wonder if this is a common definition, and does it generalise to GR well?

lriuui0x0 said:
In book "Special Relativity in General Frame"