Layman's doubts about Gen Relativity

[Altabeh]

I think DaleSpam and Altabeh are both right.

A free falling test particle traces a geodesic, which we define to be "inertial motion".

However, the OP was asking about an "inertial frame". A frame is a coordinate system which is a non-local in the sense of involving not just a point, and I believe this is what Altabeh is talking about. I think at best you can make inertial coordinates at every point on a geodesic by Fermi normal coordinates, but not for events not on the geodesic.

Yes. But remember that the Fermi coordinates calls for its own observer; not every observer. So the general answer to the OP's question is no.

AB

 

[yuiop]

The accelerometer mechanism is actually functioning in the frame of reference of the object itself. So, its zero reading doesn't shows us that it was in inertial motion (for us).

I've concluded that Altabeh's argument was right. The local flatness theorem holds only for infitesimal regions in space-time curvature and due to the long trajectory of the particle, it is non-inertial from outside. The continuous photon blueshifting/redshifting and apparent modification in c are some phenomenon that exhibit this thing.

If the object has inertial motion according to an accelerometer "actually functioning in the frame of reference of the object itself", then it has inertial motion according to all observers, whether they are inertial or non inertial, at rest or with relative motion and whether accelerating or not. Inertial motion is a universal observation, as Dalespam has pointed out several times.

Consider a space station orbiting the Earth in a highly elliptical orbit. Its angular and radial coordinates are continually changing and its angular and radial acceleration is continually changing, but to an observer within the space station itself (no windows), the space time is essentially flat to a high degree of accuracy. Even using modern technology, I think it would be extremely difficult for the observer inside the space station to prove that the space station is not stationary in flat space, using only measurements made inside the confines of the space station.

The "long trajectory" of the particle is immaterial. The local space-time is essentially flat along the entire geodesic (i.e. at every point of the trajectory).

An Earth bound observer is a non inertial observer. He knows he is accelerating (as he can measure this using an accelerometer) and he can conclude that the apparent changing motion of the inertially falling stone can be explained entirely by his own acceleration.

 

[Altabeh]

The local flatness theorem holds only for infitesimal regions in space-time curvature and due to the long trajectory of the particle, it is non-inertial from outside. The continuous photon blueshifting/redshifting and apparent modification in c are some phenomenon that exhibit this thing.

Exactly.

 

[atyy]

I think that it is fairly standard in pure mathematics to use "local" in the following way: A space is locally xxxx if every element in the space is contained in a neighbourhood that is xxxx. For example, a space is locally compact if every element in the space is contained in a neighbourhood that is compact.

So your objection is that a point is not a neighbourhood?

How about "ultralocal pseudoflatness"?

 

[Altabeh]

The "long trajectory" of the particle is immaterial. The local space-time is essentially flat along the entire geodesic (i.e. at every point of the trajectory).

Not so much correct. Only spacetime around the entire geodesic can be made flat using a special kind of observer and in general such situation cannot be defined for every observer. But if you still back the argument that along any geodesic local flatness can just be guaranteed from the perspective of all observers then there is no more question on your post.

AB

 

[Altabeh]

How about "ultralocal pseudoflatness"?

A nice one if not so much of a pain to read!

AB

 

[yuiop]

Not so much correct. Only spacetime around the entire geodesic can be made flat using a special kind of observer and in general such situation cannot be defined for every observer. But if you still back the argument that along any geodesic local flatness can just be guaranteed from the perspective of all observers then there is no more question on your post.

Yes, it is for a specific observer. For the entire trajectory, the observer is a local co-moving co-accelerating observer. For a single point on the trajectory, the observer is a local, instantaneously co-moving (but not necessarily co-accelerating) observer. For local flatness I am using atyy's concept of "ultralocal pseudoflatness" which I rather like the sound of

 

[Altabeh]

Yes, it is for a specific observer. For the entire trajectory, the observer is a local co-moving co-accelerating observer. For a single point on the trajectory, the observer is a local, instantaneously co-moving (but not necessarily co-accelerating) observer. For local flatness I am using atyy's concept of "ultralocal pseudoflatness" which I rather like the sound of

Well, then let's decide to use that term from now on. That would be even more perfect if it was "ultralocal quasi-pseudo-flatness"!

AB

 

[yuiop]

Well, then let's decide to use that term from now on. That would be even more perfect if it was "ultralocal quasi-pseudo-flatness"!

It does make you wonder if this is an inperfection in GR. If the equations are only "exactly" correct at an infinitesimal point, then at extremes such as at the event horizon or at the singularity there might be a breakdown. Measurements of velocity require a minimum of 2 spatially and temporally separated events. Measurements of acceleration require a minimum of 3 spatially and temporally separated events. Can the measurement of 3 such separated events qualify as being measured "exactly" at one infinitesimal point? It gets worse for third and fourth derivatives.

 

[atyy]

It does make you wonder if this is an inperfection in GR. If the equations are only "exactly" correct at an infinitesimal point, then at extremes such as at the event horizon or at the singularity there might be a breakdown. Measurements of velocity require a minimum of 2 spatially and temporally separated events. Measurements of acceleration require a minimum of 3 spatially and temporally separated events. Can the measurement of 3 such separated events qualify as being measured "exactly" at one infinitesimal point? It gets worse for third and fourth derivatives.

Yes, the EP which gave birth to GR is only heuristic. In GR, it holds only at points, and only for first derivatives (all Christoffel symbols, which are first derivatives of the metric, can be made zero at a point in normal coordinates), but not second derivatives (components of the curvature tensor cannot all be made zero). In a strict mathematical sense, all derivatives are local, since they are limits defined at a point. But in the heuristic sense, higher order derivatives are more nonlocal, as you point out. So yes, the EP only applies locally in two senses 1) at a point 2) and only up to first derivatives at the point. But that's more a problem with trying to derive GR from an EP that is not sharply defined until after we already have the theory. The other motivating principle of GR "general covariance" is also problematic for deriving GR. However, GR itself is fine for the moment, having passed many experimental tests.

 

[djy]

Not so much correct. Only spacetime around the entire geodesic can be made flat using a special kind of observer and in general such situation cannot be defined for every observer. But if you still back the argument that along any geodesic local flatness can just be guaranteed from the perspective of all observers then there is no more question on your post.

AB

Whether spacetime is flat is frame-independent. It can't be "made flat" by different observers and coordinate systems.

 

[Dale]

I thought your background metric belongs to the spacetime around some gravitating body

Sorry about that. I was not clear in my earlier post.

Yet again this doesn't change anything ever because though the dynamics of spacetime does now seem to have changed

Yes, exactly. Nothing physical is changed, but the coordinate transform leads to non-zero ordinary derivatives along a clearly inertial worldline. Therefore (even in flat spacetime) zero ordinary derivatives do not generally define an inertial worldline, whereas zero covariant derivatives do.

In other words, as you yourself stated, nothing physical has changed, but the result of your formula has. So in general your formula reflects the peculiarities of the coordinate system rather than just the details of the underlying physical situation.

 

[Austin0]

Yes, it is for a specific observer. For the entire trajectory, the observer is a local co-moving co-accelerating observer. For a single point on the trajectory, the observer is a local, instantaneously co-moving (but not necessarily co-accelerating) observer. For local flatness I am using atyy's concept of "ultralocal pseudoflatness" which I rather like the sound of

A local co-moving co-accelerating observer; is that the guy next to him in the elevator? ;-)

Isnt the idea of a comoving but non-accelerating observer somewhat oxymoronic??



How about non inertial observers hovering with thrust and spread out over the total path for local measurements??

 

[Dale]

Altabeh, expanding on my previous comments. In flat spacetime an inertial object has a straight worldline. This is a coordinate-independent geometric statement. You are certainly free to use a standard Minkowski coordinate system. If you do so then the geometric statement that the worldline is straight corresponds to the ordinary derivative expression that you posted.

However, you are certainly not required to use Minkowski coordinates. You could use any other coordinate system you might want and it would not affect the physics nor the underlying geometry at all, but as I showed above it would affect your coordinate-dependent expression. Therefore, your expression is not general (even in flat spacetime) and is in fact a special case of a more general geometric expression.

The covariant derivative expression I posted is a geometric statement and correctly identifies the underlying straightness of the line regardless of the coordinate system used. In Minkowski coordinates it reduces to your coordinate-dependent ordinary derivative expression, thus the covariant derivative is the more general geometric expression that we are looking for.

In curved spacetimes you can no longer produce global Minkowski coordinates, the best you can do is make a first-order approximation to Minkowski coordinates near some specific event or geodesic. If you choose to do so then your special-case formula can be used to determine if an object is inertial. Of course, you need not use those coordinates, you may use any other coordinates without changing the physics. Again, the covariant derivative is a geometric expression which gives the correct result in any coordinate system and reduces to your expression in the first-order approximation coordinates.

The fact that the Minkowski approximation is first-order and fails whenever second-order effects become important only places a limitation on the use of the special case expression which depends on having Minkowski coordinates, not the general expression which does not depend on coordinates.

 

[Altabeh]

Yes, exactly. Nothing physical is changed, but the coordinate transform leads to non-zero ordinary derivatives along a clearly inertial worldline. Therefore (even in flat spacetime) zero ordinary derivatives do not generally define an inertial worldline, whereas zero covariant derivatives do.

First of all, the geodesic equation has more to do with the "physics" of spacetime than the mathematics behind it. You can get the idea behind this from a simple example. In the Eddington-Finkelstein form of the Schwarzschild metric, the singularity of metric disappears which suggests that you're no longer seeing any singularity by doing a coordinate transformation. Now a degenerate metric has been found out to be non-degenerate but what about its physics!? I'll answer this later. Here the intention is to get a "suitable" form for our metric in which the singularity would be able to disappear. When I say "the dynamics of spacetime seem to have changed" this doesn't mean that it has not changed: it has changed only with respect to the eyes of an observer now using a new coordinate system and this is General Relativity. So when we introduce Kruskal coordinates, one can behold a completely strange physical feature (still an open problem) that didn't exist in the original coordinates neither did it in the Eddington-Finkelstein coordinates. Hence you can't judge whether the physics will not undergo a change or not by a coordinate transformation.

Second off, I have a very grand problem with your "inertial" here. The second term in the geodesic equations indeed represents the sum of the inertial and the gravitational acceleration and iff it vanishes you can only decide to say certainly that a "straight line" is followed by an inertially moving particle (observer) or not. Your fallacy would probably originate from the broad use of "straight line" as being inertial in SR; but bringing a transcription of such definition into GR is nothing but nonsense because now the yardstick to measure the dynamics is just the geodesic equation.

In other words, as you yourself stated, nothing physical has changed, but the result of your formula has. So in general your formula reflects the peculiarities of the coordinate system rather than just the details of the underlying physical situation.

I thought I was clear when posting this earlier. I didn't "state" the physical content has not changed but rather I said it has changed in such a way that it seems it has not. When using Rindler coordinates (which is in the given region or the so-called "wedge" the same as Minkowski spacetime), you see that points at the intersections of a coordinate lattice are accelerating away by a constant acceleration with respect to all stationary observers and yet the spacetime is Minkowski-flat but the physics (the dynamics of the particles) has changed. What's up? The observer only uses now a brand-new ruler and clock to measure the Minkowski spacetime so that with such instruments the dynamics "seems" to have changed with respect to a "handcuffed" me that still am looking only at events happening in the Minkowski spacetime of the inside of a cell with a wall clock and the length of the wall as my meter. This seems to be so because I got news from another prisoner that he has more degrees of freedom than me so he can use frames with the built-in meters and clocks from which the inside of his cell looks getting stretched.

Sorry if I was not clear.

AB

 

[Altabeh]

Whether spacetime is flat is frame-independent. It can't be "made flat" by different observers and coordinate systems.

Unfortunately this is nothing but nonsense. You seem to have lost a whole lot of posts in this thread.

AB

 

[Altabeh]

Altabeh, expanding on my previous comments. In flat spacetime an inertial object has a straight worldline. This is a coordinate-independent geometric statement. You are certainly free to use a standard Minkowski coordinate system. If you do so then the geometric statement that the worldline is straight corresponds to the ordinary derivative expression that you posted.

I said that you are not allowed to use the normals of a theory within another as long as there is a very powerful tool to do so. In Rindler metric, straight lines are not defined to be the same as those of Minkowski metric, i.e. d^2x^a/ds^2=0.

However, you are certainly not required to use Minkowski coordinates. You could use any other coordinate system you might want and it would not affect the physics nor the underlying geometry at all,

This is not correct. A coordinate change does not guarantee always that the physics would not change. This is obvious from the large variety of examples I provided you with.

but as I showed above it would affect your coordinate-dependent expression. Therefore, your expression is not general (even in flat spacetime) and is in fact a special case of a more general geometric expression.

I don't think that defining a straight line in GR the way that SR agrees to is more general than my use of the most general tool (geodesic equation) to explain what I mean by inertial or non-inertial.

In curved spacetimes you can no longer produce global Minkowski coordinates, the best you can do is make a first-order approximation to Minkowski coordinates near some specific event or geodesic. If you choose to do so then your special-case formula can be used to determine if an object is inertial. Of course, you need not use those coordinates, you may use any other coordinates without changing the physics. Again, the covariant derivative is a geometric expression which gives the correct result in any coordinate system and reduces to your expression in the first-order approximation coordinates.

What correct result!? Would you please specify it to me?

The fact that the Minkowski approximation is first-order and fails whenever second-order effects become important only places a limitation on the use of the special case expression which depends on having Minkowski coordinates, not the general expression which does not depend on coordinates.

All the EP (which is the starting point for GR) wants to say is that in a sufficiently small region of spacetime all observers agree that a free-falling object has a zero proper acceleration and this can't be extendend to any larger region or to the whole of spacetime or to the entire of the geodesic.

AB

 

[Dale]

A coordinate change does not guarantee always that the physics would not change.

If you honestly believe that simply changing the mathematical numbers used label events can ever actually change what physically happens then the rest of this conversation is pointless. If I am misunderstanding of your view then perhaps you could explain better.

I will state categorically that there is no way for a mathematical operation like a coordinate change to change physics. Rindler coordinates are a good example, no physics is changed by using Rindler coordinates. I would be glad to discuss that further if I am correctly understanding your position.

 

[George Jones]

Whether spacetime is flat is frame-independent. It can't be "made flat" by different observers and coordinate systems.

Unfortunately this is nothing but nonsense. You seem to have lost a whole lot of posts in this thread.

djy is correct.

 

[yuiop]

A local co-moving co-accelerating observer; is that the guy next to him in the elevator? ;-)

Yep

Isnt the idea of a comoving but non-accelerating observer somewhat oxymoronic??

I can see why might think that. A better expression is a local momentarily co-moving non-accelerating observer. Let us say you are standing on a balcony and somebody throws a ball up to you and its apogee coincides with your eye level. For a "moment", the ball appears stationary. You both have dx/dt=0 for that instant and are co-moving. However the ball is accelerating relative to you so the ball has d^2x/dt^2 \ne \ 0 while you have d^2x/dt^2=0 so you are not co-accelerating with the ball. Your proper accelerations are also different. You have proper acceleration while the ball does not (if you do not catch it).

How about non inertial observers hovering with thrust and spread out over the total path for local measurements??

If they stationary, then they will not be co-moving except at the apogee. They have to be local and match the instantaneous velocity of the accelerating particle as it passes to qualify as local momentarily co-moving observers.

 

[Austin0]

A local co-moving co-accelerating observer; is that the guy next to him in the elevator? ;-)

Well then isn't he a more or less useless and redundant observer?

Isnt the idea of a comoving but non-accelerating observer somewhat oxymoronic??

Yep
I can see why might think that. A better expression is a local momentarily co-moving non-accelerating observer.

EARLIER POST RE: FALLING OBJUCT
But as a coordinate frame it appears to me that it would be a series of instantaneously comoving inertial frames, analogous to an accelerating frame in flat spacetime.
Relative to the Earth frame on the surface it [object frame]would have a dynamic metric.
The length contracting over time and the time dilating proportionately.

Would you then perhaps more or less agree with this earlier post"?

How about non inertial observers hovering with thrust and spread out over the total path for local measurements??

If they stationary, then they will not be co-moving except at the apogee. They have to be local and match the instantaneous velocity of the accelerating particle as it passes to qualify as local momentarily co-moving observers.

Understood. I was suggesting as Swarzschild or Earth frame observers for local measurements of length,time and velocity etc. The idea was the were at rest wrt Earth but local

Let us say you are standing on a balcony and somebody throws a ball up to you and its apogee coincides with your eye level. For a "moment", the ball appears stationary. You both have dx/dt=0 for that instant and are co-moving. However the ball is accelerating relative to you so the ball has d^2x/dt^2 \ne \ 0 while you have d^2x/dt^2=0 so you are not co-accelerating with the ball. Your proper accelerations are also different. You have proper acceleration while the ball does not (if you do not catch it).

I would very much like to get into this with you and have thought of taking your suggestion about posting another thread as our discussion was getting interesting in the other thread.

 

[Altabeh]

If you honestly believe that simply changing the mathematical numbers used label events can ever actually change what physically happens then the rest of this conversation is pointless. If I am misunderstanding of your view then perhaps you could explain better.

I will state categorically that there is no way for a mathematical operation like a coordinate change to change physics. Rindler coordinates are a good example, no physics is changed by using Rindler coordinates. I would be glad to discuss that further if I am correctly understanding your position.

I don't understand why "ever" in your post is bold-faced, but let's check what I mean when I say "physics" (the dynamics of the particles) may change. Consider the Kruskal extension of Schwarzschild metric which can be obtained by the following coordinate transformations:

w=\frac{1}{2}e^{r/4m}(\frac{r-2m}{2m}e^{t'/4m}+e^{-t'/4m}),
v=\frac{1}{2}e^{r/4m}(\frac{r-2m}{2m}e^{t'/4m}-e^{-t'/4m}),

with \theta and \phi being unchanged and t' and r are the time and radial coordinates, respectively, used in the Eddington-Finkelstein form of the Schwarzschild metric. Such coordinates "produce" the following feature that is really bizarre: The (t',r) plane is now mapped on half of the (w,v) plane which means using the whole of the (w,v) plane, as a necessity to introduce a complete space, will actually generate a secondary space (t',r). This is an unintuitive feature that can have a physical explanation but definitely there is no analogy of this with the original or the Eddington-Finkelstein coordinates; raising a probable change of physics by doing a simple change in "numbers" and "labels". You find me an "analogy" I quit studying physics. Of course you have to bear in mind that such thing is tied up to the dynamics of the particles (indeed photons) following radial null geodesics of the Schwarzschild metric in the Kruskal coordinates which you can get some information around them in Wald's book (page 152).

If you are fine by my example, then we will go on to the next example, Rindler coordinates and see what makes me say that the "dynamics" in this case looks a little bit different. But for the moment, I have to make one clarification (correction, I believe):

I said that

you see that points at the intersections of a coordinate lattice are accelerating away by a constant acceleration with respect to all stationary observers and yet the spacetime is Minkowski-flat but the physics (the dynamics of the particles) has changed.

As I later clarified this with the cell example, it it obvious that the "physics" remains the same for all Rindler observers. But for everyone else located between the lines of lattice, no!

There is another thing I have to know before jumping into this example: I believe you agree that "Fermi Normal Coordinates" have a specific observer, don't you?

AB

 

[Altabeh]

djy is correct.

If you honestly think he is then I'm okay with that! Unfortunately, I don't think he is!

AB

 

[atyy]

If you honestly think he is then I'm okay with that! Unfortunately, I don't think he is!

AB

I think djy and George Jones both thought you were talking about flatness in the sense that the Riemann curvature tensor is zero. Of course that's true for Minkowski spacetime, regardless if one is in inertial or Rindler coordinates.

However, you are only talking aobut local flatness, which means using Riemann or Fermi normal coordinates, and is of course coordinate dependent.

Regarding "physics" depending on coordinates, I think it's just that DaleSpam and you have different definitions of physics.

eg. In Minkowski spacetime, is the speed of light c in all coordinate systems?

No - not if you use non-inertial coordinates.

Yes - light always follows a null geodesic, and nullness if a coordinate independent property.

 

[Altabeh]

Regarding "physics" depending on coordinates, I think it's just that DaleSpam and you have different definitions of physics.

And you think whose definition is more correct than the other?

AB

 

[atyy]

And you think whose definition is more correct than the other?

AB

I haven't an opinion.

 

[Dale]

Consider the Kruskal extension of Schwarzschild metric ... a probable change of physics by doing a simple change in "numbers" and "labels".

This is a pretty unconvincing example. It is certainly true that there are coordinate charts that do not cover the entire spacetime, but within the region that they do cover they must agree with all other coordinate systems on the result of any physical experiment.

Rindler coordinates and see what makes me say that the "dynamics" in this case looks a little bit different. ... it it obvious that the "physics" remains the same for all Rindler observers. But for everyone else located between the lines of lattice, no!

Do you have even a single example of a specific physical experiment whose predicted result is different depending on the choice of Rindler or Minkowski coordinates? (Within the Rindler wedge, of course)

In Rindler metric, straight lines are not defined to be the same as those of Minkowski metric, i.e. d^2x^a/ds^2=0.

Then how are straight lines defined in the Rindler spacetime?

my use of the most general tool (geodesic equation) to explain what I mean by inertial or non-inertial.

I don't understand this comment. It sounds like you are now agreeing with me that "inertial" means that the covariant derivative is 0 (the geodesic equation).

 

[yuiop]

Well then isn't he a more or less useless and redundant observer?

If we are considering a particle without eyes, arms or legs, then a co-moving co-accelerating observer is very useful if we wish to consider the point of view from the particle's position.

EARLIER POST RE: FALLING OBJUCT
But as a coordinate frame it appears to me that it would be a series of instantaneously comoving inertial frames, analogous to an accelerating frame in flat spacetime.
Relative to the Earth frame on the surface it [object frame]would have a dynamic metric.
The length contracting over time and the time dilating proportionately.

Would you then perhaps more or less agree with this earlier post"?

Yes... more or less

Understood. I was suggesting as Swarzschild or Earth frame observers for local measurements of length,time and velocity etc. The idea was the were at rest wrt Earth but local

It is fairly easy to transform to the frame of local observers that are stationary with respect to the Schwarzschild metric. Even these stationary, accelerating non-inertial observers perceive the (very) local neighbourhood to be aproximately flat. How "local" is debatable and how many angels you can squeeze into this local space is also debatable.

I would very much like to get into this with you and have thought of taking your suggestion about posting another thread as our discussion was getting interesting in the other thread.

If you can formulate a specific clear question on what aspect you are interested in, then I will do my best .

 

[Altabeh]

This is a pretty unconvincing example. It is certainly true that there are coordinate charts that do not cover the entire spacetime, but within the region that they do cover they must agree with all other coordinate systems on the result of any physical experiment.

You got it sort of wrong with my example. This new coordinate system actually talks about a secret property of the original spacetime that can't be discussed by the use of the initial coordinate system and physics as a whole has changed whether you find it unconvincing or not. You seem to only look at my notes without giving a shot at learning from the books and this gives rise to many confusions here.

Do you have even a single example of a specific physical experiment whose predicted result is different depending on the choice of Rindler or Minkowski coordinates? (Within the Rindler wedge, of course)

Why don't you understand me? There is a difference between a Rindler observer and a non-Rindler one. Physics only changes with respect to a specific observer that is not Rindler. Related to this, tell me if you agree that Fermi Normal Coordinates have a specific observer.

Then how are straight lines defined in the Rindler spacetime?

Straight lines are defined in Rindler coordinates by the geodesic equations. This is what I am trying to get you to understand that an straight line in GR isn't defined in the same way as in Special Relativity in general (which you wrongly suggested this could happen by the immediate use of the formula of a line in the secondary coordinates.) From the dynamics of particles following curves in GR, it can be understood whether a curve is an straight line or not. But the obvious fact is that if the background metric is non-flat, then all straight lines are only locally straight. This was all you could get from the following statement:

my use of the most general tool (geodesic equation) to explain what I mean by inertial or non-inertial.

So your

I don't understand this comment. It sounds like you are now agreeing with me that "inertial" means that the covariant derivative is 0 (the geodesic equation).

isn't unfortunately correct.

Anyways, I'm starting to believe this conversation seems to be pointless and if you and I keep going two different ways here, there is not going to be any agreement until one convinces the other that his argument fails to be true.

AB

 

[Dale]

physics as a whole has changed whether you find it unconvincing or not.

Then it shouldn't be so difficult for you to come up with one single example of an experiment where the two different coordinate systems predict different measured results.

Related to this, tell me if you agree that Fermi Normal Coordinates have a specific observer.

Usually I say it the other way around. I.e. that an observer has a coordinate system (where they are at rest). But I am comfortable with your phrasing.

Straight lines are defined in Rindler coordinates by the geodesic equations. This is what I am trying to get you to understand that an straight line in GR isn't defined in the same way as in Special Relativity in general (which you wrongly suggested this could happen by the immediate use of the formula of a line in the secondary coordinates.) From the dynamics of particles following curves in GR, it can be understood whether a curve is an straight line or not. But the obvious fact is that if the background metric is non-flat, then all straight lines are only locally straight.

OK, it sounds to me like we may largely agree. To be clear, do you agree
1) that an inertial object has a geodesic worldline
2) that whether or not a worldline is a geodesic is independent of the coordinates, and
3) that inertial/geodesic worldlines are straight lines in flat spacetime, but
4) you disagree about my use of the word "straight" to describe geodesic worldlines in curved spacetime.