I don't know if this will answer your question, but I made this video tutorial in which a finite acceleration is explicitly taken into account when solving the Twin Paradox. I rigorously derive all the equations for accelerated motion starting from the Lorentz transformation.

https://youtu.be/QFWF90bch3c

The starting point is incorrect. If a system is inertial (non-accelerating), then there are infinitely many, equivalent, reference frames, e.g. all those related by global Lorentz transformations, for relativistic systems. If a system is non-inertial, (accelerating), there can't exist any globally defined reference frame, each defines an inequivalent observer. The only relation between any two reference frames is a general coordinate transformation, that depends on the point in spacetime. That's how one observer can state being in the vacuum state (measuring zero flux of particles) and any other observer can state being in some excited state (measuring non-zero flux of particles).

Misner, Thorne and Wheeler's book is wonderful, however they assume, implicitly, that the reader either has done the calculations already, or can do them very quickly, while reading. If this isn't the case, it would be useful to do the calculations as presented, for instance, in Sean Carroll's lectures on general relativity, https://arxiv.org/abs/gr-qc/9712019

Misner, Thorne and Wheeler explain what the calculations, discussed in conventional textbooks, actually mean. This doesn't absolve the reader from having done them and thought about them, they, simply, offer new insights. Without doing the calculations by oneself it's not possible to follow what Misner, Thorne and Wheeler are talking about.

So accelerating frames don't describe the ``same" event. That's the important point. They, necessarily, describe different events, because there's just no way for the events described by the two frames to be identified as one.

Said differently, accelerating frames define inequivalent observers.

It seems that my own analysis might be as thorough as anything easily found in the literature. I wrote this after figuring it out myself because I didn't know where to look for it in the literature so it was easier to figure it out then to find it in the literature. I guess I'm asking if I re-invented the wheel, or is my work really not common knowledge. If it is common knowledge, please cite a paper. If you are interested in my analysis, you can save yourself a lot of reading by going directly to the Summary section (Section 15). I would like to know if this work has any originality, or if I re-invented the wheel. The work is in the attached PDF file.

Lorentz transformations map one inertial frame to another inertial frame. Whi h means that all these frames are equivalent. A global Lorentz transformation can't map an inertial frame to a non-inertial frame. A non-inertial frame is described by a curved metric. So,yes, you are re-inventing general relativity, It would be better to study it, from a formal text, either Carroll's or Weinberg's and then read MTW for their insights.

It's not as if lecture notes on general relativity are lacking...

Thank you Stam Nicolis . None of the books that I have on GR give specific equations for calculating the coordinates of an event in an inertial system when the given information is the coordinates of that event in an accelerating system. This can be done (I did it). The inverse transformation is the problem. The transformation is not globally invertible. But my books don't give explicit equations even for what can be done, so studying GR did not help me. However, you mentioned specific books, by Carroll or Weinberg. I don't have those books. Do they give explicit equations for the coordinates of an event in an inertial system when the given information is the coordinates in an accelerating system? If no then I would say that my analysis is not common knowledge. If yes, could you be please include the titles of the books so I can find them?

Stam Nicolis : Sorry for not noticing that Carroll is the author of the notes you sent (thank you for them, I like them). But I did not find in those notes explicit equations for calculating the coordinates of an event in an inertial system when the given information is the coordinates in an accelerating system.

Once more: It's NOT POSSIBLE to express a non-inertial frame as an inertial frame, except at any one point. That's the difference between an inertial frame and a non-inertial frame.

All inertial frames are the same. All non-inertial frames are different.

Stam Nicolis : I don't think you understand the question. Of course it is not possible to express a non-inertial frame as an inertial frame. What made you think that I thought it was possible? In fact, a big distinction is that there is not a globally invertible transformation between space-time points and coordinates in a non-inertial frame. That is the topic that I find interesting. My own analysis derives equations that calculate the coordinates of an event in an inertial frame when the given information are the coordinates of the same event in an accelerating frame. My own analysis finds that this transformation is not globally invertible. The question is, has anyone written a paper that focuses on the implications of this lack of invertibility? Specifically, has anyone gone far enough into this topic to tell us what the world looks like to an accelerating observer? The topic is so fascinating that I'm sure someone has done it. Where can I find that literature? It is not in any textbooks I know of. But I expect someone has done it. Where do I look for it?

Let me rephrase the question. It is possible to map event coordinates from an accelerating system to an inertial system. The interesting thing is that this mapping does not have a global inverse. There are example coordinates in which different coordinates in the accelerating system are mapped into the same space-time point. The obvious question inspired by this fact is: What does the world look like to an accelerating observer? Does the observer acquire a double vision? Does he see the same event multiple times? Or what? Has anyone found an answer to this question in the literature?

L.D. Edmonds:

A spacetime is defined by coordinates, xμ and a metric gμν(x). The points are labeled by the coordinates xμ. These points label events, the metric labels the observer. An inertial observer is an observer whose metric can be globally defined to be flat; a non-inertial observer is an observer whose metric cannot be globally defined to be flat.

Coordinates don't have any global significance, since, if the spacetime geometry is flat, they can be changed by a global Lorentz transformation and, if the spacetime is not flat they can be changed by a general coordinate transformation; on the other hand, while the flat metric is left invariant by any global Lorentz transformation, it is changed by a gauge transformation under a general coordinate transformation.

Points don't change, their labels only do.

The theorem you seem to be confused about is the theorem that states that, at any given point in any spacetime it is always possible to define a reference frame that is inertial; i.e. that it is always possible to find coordinates around any point in any spacetime, where the metric is flat. It is not however possible to do so globally. The obstruction is known as spacetime curvature and the situation is the generalization to pseudo-Riemannian geometry of the theorem that holds in Riemannian goemetry.

Stam Nicolis : I'm not confused by any of it, you seem to be confused about what my topic is. I don't know what you mean by coordinates have no global significance. If you mean that they depend on observers, yes they do but that doesn't mean they have no significance. Coordinates of a given event are specified when we specify both a reference frame and the measuring apparatus. In this case the measuring apparatus consists of clocks and perpendicular rigid rods carried by each member of an infinite family of inertial reference frames, with the accelerating frame using the apparatus in that inertial frame that is local with it at a given point in time. There exists a mapping from the event coordinates in the selected accelerated frame and the coordinates of any selected inertial frame. The interesting thing, which I think you do understand, is that this mapping does not have a global inverse.

I don't think you understand anything that I have been saying. You say "Points don't change, their labels only do". Of course. What makes you think that I thought otherwise? Your posts have nothing to do with what I am talking about. Do you understand my question? I repeat it below.

There are example coordinates in which different coordinates in the accelerating system are mapped into the same space-time point. The obvious question inspired by this fact is: What does the world look like to an accelerating observer? Does the observer acquire a double vision? Does he see the same event multiple times? Or what? Has anyone found an answer to this question in the literature?

And the answer to your question is kmown: An accelerated observer describes spacetime events using a metric that isn't flat. That's all that's required and it's also sufficient. And it's described in all the textbooks on tbe subject.

It's necessary and sufficient to replace d^4x by d^4x (|det g|)^(1/2) and contract the indices with the metric tensor g_μν(x) that describes the curved spacetime of the corresponding accelerated observer.

An observer that uses the Minkowski metric is an inertial observer. An observer that uses a metric, on the same spacetime, that isn't equivalent to the Minkowski metric, is an accelerated, i.e. a non-inertial, observer.

You call that an answer? That is like saying "the answer is general relativity." My own analysis has more detail than that. How about the specific equations for the coordinate transformations? I have them. GR books (that I have seen), including that Carroll lecture, don't. I guess I need to continue with my own analysis to answer my own question. Unless someone else can give a better answer than what is in every GR book.

One example of an inertial observer is the free relativistic, massive, particle, whose action is given by the integral over its worldline of -mc(η_μν u^μ u^ν)^(1/2), where u^μ = dx^μ/dλ, where λ is the parameter of the worldline.

A non-inertial observer would be described by the same expression, except that the flat metric would be replaced by the curved metric g_μν(x).

That's all there is.

So an inertial observer's worldline satisfies the equation of motion d^2 x^μ/dλ^2 =0, along with the constraint p^μ p^ν η_μν =m^2, where p^μ is the canonical momentum, deduced from the Lagrangian, the usual way, while for the non-inertial observer the worldline satisfies the so-called geodesic equation, along with a corresponding constraint.

All this is standard material, described at length in all textbooks, Carroll's included. If you believe you can get to some point about general relativity, without learning it, you're misnformed, there's no royal road, you'll have to learn the subject properly, like everybody else has.

Also, the metric is not needed for the coordinate transformation. Just the opposite. In special relativity with an accelerating system, the coordinate transformation comes first and the metric is derived from it. The entire question can be answered without any need for the metric tensor by investigating the multiple coordinates in the accelerating system that are mapped into the same space-time point. I figured that somebody else already did that investigation so I can just read their work.

Your answers are so unspecific they are about as useful as saying that "the answer is general relativity." I repeat: How about the specific equations for the coordinate transformations? I have them. GR books (that I have seen), including that Carroll lecture, don't have them. I guess I need to continue with my own analysis to answer my own question. Unless someone else can give a better answer than what is in every GR book..

All these statements are, just, trivially, wrong.

It suffices to study Carroll's lecture notes, or Weinberg's book, in sequence. The time you're spending posting here could be far more profitably spent studying them.

Again, you talk about irrelevant things. Now the equation of motion of world lines when the question is coordinate transformations. You don't seem to understand anything that I say or what my question is. None of your posts are at all relevant to the topic. We are just not communicating. But thank you for the effort.

Are you reading me? Carroll's lectures do not have the transformation equations. Can you understand that? I do have the equations.

If you are referring to the metric being derived from the coordinate transformation in special relativity when you said that the statements are trivially wrong then you have a lot to learn yourself. Maybe you just forgot things that you use to know.

Once more: Special relativity can be defined by the transformations that leave the Minkowski metric invariant. These are the global Lorentz transformations. They map any inertial frame to any other inertial frame. And the Minkowski metric is globally defined, since the Lorentz transformatiins are global.

The generalization of special relativity to the case of non-inertial frames consists in replacing the flat metric by an arbitrary metric. In this case the frames that are related by a diffeomorphism are equivalent, they can't be distinguished.

Now an inertial observer is an object whose equation of motion is that of a free relativistic particle: d^2x/dλ^2=0, along with a constraint: p^μ p^νη_μν = m^2c^4.

A non -inertial observer is an object whose equation of motion is the geodesic equation, along with the constraint p^μ p^νg_μν=m^2c^4.

There is a theorem that states that at any given point in any spacetime it is always possible to define a Minkowski metric. If it is possible to do so globally, the spacetime is flat, if it isn’t, it’s not.

Bottom line: In a flat spacetime inertial coordinates can be globally defined; this doesn’t mean, however, that it isn’t possible to define non-inertial observers, or motion, it suffices to use a metric that’s not flat.

All non-inertial reference frames, that can be related by a diffeomorphism are equivalent, those that can’t, aren’t. That’s why the statement about multiple reference frames is wrong. Whether inertial or non-inertial, there exist infinitely many reference frames. The only question is their equivalence classes. All inertial frames are equivalent, because all can be related by a globally defined Lorentz transformation. All non-inertial frames that can be related by a diffeomorphism are equivalent and those that can’t aren’t. That’s the distinction.

The metric is NOT derived from coordinate transformations, it’s the transformations that imply which metrics are equivalent and which aren’t.

What you seem to want is the transformation, at any given point of a spacetime, to the flat metric that can be defined at that point.

Of course Carroll mentions it, it’s the transformation to so-called Riemann normal coordinates. What you've been going on about is that this transformation isn't unique. That's not the problem you claim it is-and this isn't anything new.

Stam Nicolis "The metric is NOT derived from coordinate transformations, it’s the transformations that imply which metrics are equivalent and which aren’t.…"

I will remind you of some elementary concepts that you apparently forgot. In both special and general relativity, the metric tensor satisfies the transformation of a double covariant tensor. The coefficients in the transformation are derivatives of coordinate transformations. If one system is rectangular in a flat space, the metric for that system is already known (the Lorentz metric) so the only thing that needs to be known to derive the metric for an accelerating reference frame are the coefficients in the tensor transformation, which are derived from the coordinate transformation between systems. THE COORDINATE TRANSFORMATION DERIVES THE COMPONENTS OF THE METRIC TENSOR IN A GIVEN SYSTEM. You should do a review because you seem to have lost touch with basic concepts. You seem to think that the metric is the answer to all questions. It isn't.

Incidentally, I derived the above coordinate transformation and from that derived the metric tensor for an accelerating system, and used that to derive a Doppler effect for an accelerating system analogous to the gravitational kind. Again, Carroll does not give that coordinate transformation.

I also don't claim that any of my work is new. My question is where do I look for literature that went further than I did on this specific topic (the Carroll lectures do not), literature that I assume exists. Instead of helping me find that literature you are telling me everything wrong about what I did when, in fact, you are the one that is wrong. Can we end this now?

The transformation that you're looking for doesn't exist.

If you start from the Minkowski metric, global Lorentz transformations will change its form, not the property that its curvature vanishes.

If you change the Minkowski metric by a diffeomorphism, nothing will change either, incidentally. You can't change the ``intrinsic'' properties of a metric by a diffeomorphism, no matter how hard you try. You can't, in particuler, produce non-zero curvature by a diffeomorphism. Said otherwise, you can't map, beyond any particular point, beyond second order, an inertial observer to a non-inertial observer. The obstruction is called the Riemann tensor. The Riemann tensor is not affected by a diffeormorphism of the metric.

If you start with some arbitrary metric, you can check that it describes an accelerated observer or not by computing its Riemann tensor. If its Riemann tensor vanishes, this metric is a flat metric, just written in a weird coordinate system. If its Riemann tensor doesn't vanish, then it describes a non-inertial observer.

If you have a weird metric, that doesn't look flat, it can be transformed in the vicinity of any point to a flat metric. It may well be flat-then the expansion terminates; it may be not flat-then the transformation isn't exact. Same as what happens when you try to determine the tangent space at any point of any manifold.

That's a standard exercise, whose solution is discussed in chapter 2, page 57 of Carroll's lecture notes. This is what you're looking for, apparently. That's why it's useful to read in sequence.

Stam Nicolis : You are wrong again. The transformation does exist. A brief and sloppy discussion is in the section called Accelerated Observers in the book Gravitation. It is brief because it requires the reader to supply the details. It is sloppy because claims that are made apply only on the world-line of the accelerating clock, not to events displaced from the clock (in particular they apply to basis vector fields only at the location of the clock). But a careful analysis that provides details and corrects errors in the book will show that there is a mapping from event coordinates in the accelerated system to coordinates in an inertial system. However, the transformation does not have a global inverse and that is what inspired this thread.

Regarding Carroll's lecture notes, they do not give the above coordinate transformation. What you are probably referring to is the notes telling us how to transform a metric tensor when the coordinate transformation is already known. That is common knowledge.

I do understand one point that you are making. The case of an accelerated observer is a treatment of a flat space because there are other coordinate systems in which the metric is the Lorentz metric. This implies that the Riemann tensor must vanish. The Riemann tensor derived from the metric tensor that I derived does vanish, as it must. This is an incidental point but maybe of some academic interest.

You can’t produce curvature by a coordinate transformation. If the Riemann tensor you found does vanish, your coordinate system is inertial, full stop and no coordinate transformation can make it non-inertial. That’s the point I’ve been stressing from the start. The clock in question isn’t accelerating.

Stam Nicolis : Wrong again. A vanishing Riemann tensor means the space is flat but that merely means the existence of globally Lorentzian systems (inertial systems), it doesn't mean that the selected system is inertial. An accelerating clock produces a coordinate transformation but does not produce curvature so the Riemann tensor still vanishes. You seemed to understand this in a prior post. Did you forget already? A vanishing Riemann tensor does not imply that the clock is not accelerating.

Again I suggest you do some reviewing. You seem to be out of touch with some basic concepts.

Yes, it does! Metric=observer. The metric is the gauge field of gravity, its field strength is the Riemann tensor. An observer whose metric has vanishing Riemann tensor is an inertial observer, an observer whose metric has a non-vanishing Riemann tensor is non-inertial.

An accelerating clock describes a non-inertial observer, someone whose metric has a non-vanishing Riemann tensor, which means that it is non-zero not only at one particular spacetime point.

It is always possible to make the Riemann tensor vanish at any one point. It’s not possible to make it vanish beyond one point by imposing the same conditions, unless it vanishes everywhere.

An accelerating clock, by definition, is a non-inertial observer, whose Riemann tensor does not vanish beyond any one point, by the same transformation that makes it vanish at some point.

You can make the Riemann tensor vanish at any one point and make it vanish, by another transformation, at another point-that’s called finding the tangent flat spacetime at any one point. You can’t have a Riemann tensor non-zero only at one point and nowhere else. Even a conical singularity is more than a point, that’s how you detect its presence.

Stam Nicolis Wrong again! Don't you get tired of being wrong in every post? You are stubbornly displaying your ignorance. There is a coordinate transformation from the accelerating system to a Lorentz system, which implies that the space is flat. Another way to reach the same conclusion is to note that an accelerating system is equivalent to a gravitational field that is uniform throughout all space. Such a gravitational field produces a flat space.

Stam Nicolis This thread that I started asks a question. Instead of helping with the question, your entire mission has been, from the start, to search for ways to criticize the question. Your posts and my rebuttals turned this thread into trash. I'm going to start over with this question on a new thread. I hope you will not bother that one.

L.D. Edmonds

The transformation from an acceletrating system to a flat system exists at one point only and, if the system is, indeed, accelerating, this transformation-to Riemann normal coordinates-can't be defined globally. It's a different transformation, point by point. Why anyone finds this surprising is a mystery.

Gravitational fields uniform throughout space (not spacetime) are a Newtonian approximation.

They produce non-zero contributions to Newton's second law of motion, which is known for more than a century, can be described through spacetime (not space!) curvature.

There exist spacetimes with constant curvature-the de Sitter (positive curvature) and the anti-de Sitter (negative curvature) spacetimes, solutions to Einstein's equations (Minkowski spacetime has constant zero curvature). In neither is it possible to map the spacetime to Minkowski globally. The obstruction is called spacetime curvature.

This isn't a research topic for the last hundred years; it's in all the textbooks on the subject, either as part of the presentation, or as part of exercices.

Part of learning a subject is learning what are the questions it makes sense to ask and what are the questions it doesn't make sense to ask. The question you're asking belongs to the second category.

For the benefit of other readers:

Metrics are the gauge fields of gravity. All metrics, that are related by a diffeomorphism, are equivalent, metrics that aren't related by a diffeomorphism are not (just as occurs for any gauge field).

An observer means a reference frame, which means a choice of metric.

So there exist infinitely many reference frames for the same event, whether the reference frame is inertial or not; they are all, that are related by diffeomorphisms:

Anything that is described by the metric gμν(x), which defines some observer=some reference frame-is described, for example, by the infinity of metrics

gμν(x) + dμξ(x) + dνξ(x)

where ξν(x) is an arbitrary 4-vector, whatever the metric is (provided it's a sufficiently smooth function of x). It's not possible to distinguish any one of these metrics from each other.

If the curvature of the metric of the reference frame, as a tensor, is non-zero, the frame is non-inertial. If, as a tensor, it is zero, the reference frame is inertial and the metric is equivalent to the flat metric.

If the curvature tensor, as a tensor, is non-zero, there exists a reference frame at any one point-in spacetime-where it can be made to vanish-the choice of these so-called Riemann normal coordinates is not unique, however.

So the vanishing of the curvature tensor at any one point doesn't imply that the reference frame is inertial-it might be that one is using Riemann normal coordinates. If the curvature tensor at any one point doesn't vanish, however, then the reference frame is non-inertial.

Application: The proper time is invariant under global Lorentz transformations. An inertial observer maximizes the proper time, any other observer between the same two spacetime points has a shorter proper time and is, necessarily, a non-inertial observer (the geodesic equation has a unique solution, since it's of second order and there don't exist closed time-like curves; which requires care when dealing with anti-de Sitter spacetimes). The two observers are physically distinct precisely because the proper time is a Lorentz invariant quantity and takes different values along each worldline. It's not possible to map one to the other.

Newtonian gravity is an approximation to general relativity and it's not possible to deduce the latter from the former; it's possible to deduce the former from the latter.

Stam Nicolis : There is a global mapping in one direction. From the accelerated frame to any inertial frame. This mapping is not globally invertible but is enough to imply the metric tensor in the accelerated system. You are talking out of ignorance because you have never even learned what this mapping is. Incidentally, by excluding only certain regions of space-time points, the mapping becomes invertible on the complimentary region of space-time.

Stam Nicolis When are you going to stop? An observer cannot make a flat space become not flat by accelerating. The flatness of the space is controlled by the distribution of sources of gravity, not by the observer. The observer has no control over that. Again you should do some reviewing because you are out of touch with basic concepts.

There's no global mapping of flat spacetime to curved spacetime in any direction.

Curved spacetime has a non unique flat spacetime approximation at any one point and a different such approximation at any other point.

An accelerating observer is described by a metric that's not flat, that's what acceleration means.

In flat spwcetime, an inertial observer is described by a worldline that's a straight line-in spacetime.

In the same spacetime, an accelerating observer is described by a worldline that's not a straight line. The two observers can even share an initial point and a final point, which means they met, separated, met again.

The observer that followed the straight line has a proper time that is greater than the proper time of the other. The proper time is invariant under global Lorentz transformations, that's the reason that the two can't be mapped to each other.

The spacetime is flat, nevertheless worldlines that are not geodesics can be dr

defined. And these can be described by a metric that's not equivalent to the flat metric, except at any one point by a non-unique transformation.

But I thought you were going to leave this trash to me?

Stam Nicolis : You have no clue what you are talking about. An accelerated observer in a flat space is still in a flat space. And you have no clue about the transformations that you never even learned. I am asking you to stop bothering me.

An accelerated observer-in flat spacetime-is described by a worldline that's not a geodesic of flat spacetime; it's a curve in that spacetime. It's, also, a geodesic, in a curved spacetime, whose curvature describes the fact that the worldline isn't a straight line.

The geodesic between two spacetime points is unique; there are infinitely many curves that pass from the same two points in spacetime. All these curves describe inequivalent observers, because the geodesic equation, for each metric, which takes into account the forces that act on each observer, has a unique solution that passes from two points, since it's a second order equation.

As long as the metrics aren't equivalent by a diffeomorphism, of course.

Stam Nicolis : Space-time curvature doesn’t depend on the observer. You knew that in a previous post. Short memory. But that was not even the main topic, it came up later. The main topic is a mapping that you never even learned but still insist on talking about. It takes a lot of arrogance to critique something you never learned. Read the section on accelerating observers in the Gravitation book and try to learn something.

So the reason why two inequivalent reference frames in a given spacetime can't be mapped to each other ought to be obvious: If they could, they'd be equivalent, their metrics related by a diffeomorphism. They can be identified at one point-in spacetime-if one is flat-but not at more than one, by the same transformation.

The obstruction is called spacetime curvature, not spatial curvature. Forces aren't created or undone by a choice of coordinate system.

d^(bf r/)dt^2 = 0 in spherical coordinates, (bf r)=(rsinθcosφ,rsinθsinφ, rcosθ), still, describes a free particle, the same particle that d^2x/dt^2=0, d^2y/dt^2=0, d^2z/dt^2=0 describe.

Curves in a flat spacetime describe forces.

The confusion stems from conflating worldlines in spacetime, that describe observers and curves in space, that do not.

So, no, whether an observer is accelerating, or not doesn't imply that spacetime points have multiple sets of coordinates, or not. What's at stake is what are the equivalence classes of coordinate systems.

The bottom line is that coordinate systems that can be related by diffeomorphisms are equivalent, describe equivalent observers, or reference frames, coordinate systems that can't be related by diffeomorphisms, describe inequivalent observers, or reference frames.

One way of checking this is by computing invariants.

If the invariant quantities are equal, it makes sense to identify the frames, if they're not, it doesn't. The typical example is between the geodesic and any other curve between the same points in spacetime. Here it's known that the proper time is maximun for the geodesic, therefore if the other curve leads to a different, i.e. smaller, value, it describes an inequivalent frame. If it guces the same value, then it describes an equivalent frame, described by a diffeomorphism.

Stam Nicolis : Here is the essential point. There is a global mapping between coordinates in one direction, from the accelerating system to any inertial system. The mapping is not globally invertible but is invertible on a subset of space-time. A very sloppy sufficient condition for invertibility is that the coordinate system be “small” in the sense explained in the Gravitation book describing the size of an accelerating reference frame. That is nothing like a necessary condition and even sloppy as a sufficient condition, but is useful for concept.

All invariants are invariant as they must be.

And those are the so-called Riemann normal coordinates in pseudo-Riemannian geometry. ``Small'' means that all the coefficients in the Taylor expansion, beyond second order (i.e. that pertain to the curvature and beyond) in the expansion of the metric can be and are neglected.

The mapping isn't global, in the sense that it isn't the same mapping from one point to another point, it's local, it depends on the point; it just always exists, but isn't unique. It's invertible, obviously, in its domain of definition: You're replacing the curved spacetime in the vicinity of a point by a flat spacetime at that point. The linear system of equations is of full rank, that's how you construct the coordinate system at all. But it's not uniquely defined, because you have a lot of freedom on how you can choose it, by choosing which coefficients to fix, for !nstance, since it's underdetermined.

The ``size'' is set by the radius of spacetime curvature at the given point-which describes the acceleration of the frame in question; as long as it can be neglected, you can't tell that you're not in a curved but in a flat spacetime.

But there are subtle issues, cf here:

https://users.monash.edu.au/~leo/research/papers/files/lcb96-01.pdf

But the bottom line is, that, as long as you can neglect the curvature, you can use a flat metric-so, up to a spacetime distance about the radius of curvature at that point.

I prefer the discussion in Carroll's notes, which sets the framework for explicit calculations, if necessary.

But this doesn't have anything to do about there being one or more sets of coordinates about any event-that's, completely, misleading.

However all this issue is (a) unnecessarily pedantic and (b) misses the point, in my opinion. The bottom line is that any physical system behaves, for all intents and purposes, like a an inertial system, if its size is much smaller than the radius of curvature defined by the forces it is subject to and/or the time it is under observation is much smaller than the radius of curvature defined by the forces, divided by the speed of light. Therein lies the subtlety of general relativity: One shouldn't confuse space with spacetime! The issue isn't the ambient spacetime, it's the reference frame defined by the physical phenomenon of interest, when can it be approximated by an inertial frame and can be described by infinitely many coordinate systems.

What misses the point depends on what the question is. I am the one asking the question so I decide what misses the point. It is my right to be interested in whatever I want to be interested in and the discussion in the book that I referred to is right on the point. Also, the issue is not approximations, the issue is exact transformations with suitable restrictions on the region of space-time. You apparently did not understand this very important point. There is no Taylor series expansion. You should read it again and try to understand it this time.

We don't need you to tell us that there is a distinction between space and space-time. The authors of the book are smart enough to already know that. When they refer to spatial dimensions when discussing the size of an accelerating system, they know what they are talking about. You already demonstrated, above, that you don't know what they are talking about.

If you carefully work through all of the details, like I did, you will learn that there is an exact and global coordinate mapping in one direction, from the accelerated frame to any inertial frame. This mapping is invertible on a suitably restricted region of space-time. The same inverse mapping exactly applies to every point in that region of space-time. I carefully worked out all of the details. You never will do that because if you do you will learn that you were wrong, and you will never allow yourself to believe that you were wrong. You and I are arguing about a topic in which I carefully worked out all of the details and you worked out nothing (and don't even understand what the book said). That makes me qualified and makes you arrogant.

Coordinates and coordinate systems are a means to an end, not an end to themselves. Their choice doesn't matter and can't matter-especially when discussing gravity, whose defining property is invariance under general coordinate transformations. So in the 19th century people were arguing about coordinate systems; since 1918 and the work, in particular, of Emmy Noether, many people (Pauli among the first), though by no means all, understood that what mattered were the quantities that were invariant under the transformations that left the equations invariant and nobody cares about the choice of coordinate systems since.

What matters is not how large the spacetime patch must be, in which experiments must be done, in order to infer what happens beyond it, but how small it can be taken to be, in order for inference to make sense. In flat spacetime it can be taken to be a point-that's sufficient. In curved spacetime the size is determined by the radius of curvature; which is determined by local measurements of forces. That's why it's possible to understand LEP backgrounds, provided by tides and the arrival and departure of trains at Geneva station: http://cds.cern.ch/record/43323/files/poster-2000-063.pdf (but that the traffic at Lyon or Turin was irrelevant)

Physics is about using local knowledge to infer global properties, not the other way around. Unfortunately many exercises in textbooks forget this.

Stam Nicolis :"Coordinates and coordinate systems are a means to an end, not an end to themselves." They are an end to themselves when they are the answer to my question. I asked the question and I decide what is important.