How do non-inertial frames affect special relativity?

[kent davidge]

So I learned from the first few posts on this https://www.physicsforums.com/threads/ftl-train-ftl-communication-thought-experiment.945116/ thread that one can actually work with special relativity in non inertial frames.

I'd like to get some points on this.

Firstly, by a non inertial frame, I understand a reference frame with coord sys $x'$ which is connected to another with coord sys $x$ by a Lorentz transformation $x' = \Lambda x + b$ in which $\Lambda, b$ are functions of $x$. This would dramatically change the Lorentz interval $\eta_{\sigma \rho}dx'^\sigma dx'^\rho$ causing the presence of terms like $x \Lambda d\Lambda dx$ etc... how do we deal with this?

If we are to preserve the Lorentz interval I presume there must be a way around in which all the additional terms disappear when calculating the interval.Now another scenario would be to consider acceleration within the frame.. I mean, $\Lambda, b$ constants, but one would allow $d^2 x' / dt'^2 \neq 0$ etc... that seems completely ok in that it would not violate the Lorentz interval.

 

[Nugatory]

Firstly, by a non inertial frame, I understand a reference frame with coord sys $x'$ which is connected to another with coord sys $x$ by a Lorentz transformation $x' = \Lambda x + b$ in which $\Lambda, b$ are functions of $x$.

Lorentz transformations are for going from one inertial frame to another. If you're considering non-inertial frames, you can use much more general transformations than that; just about any transformation that is one-to-one and obeys reasonable assumptions about differentiability and continuity will do. A pretty good rule of thumb is that if the tensor transformation formula works, you have an acceptable coordinate transformation and can use either set of coordinates to describe any given physical system.

This would dramatically change the Lorentz interval $\eta_{\sigma \rho}dx'^\sigma dx'^\rho$ causing the presence of terms like $x \Lambda d\Lambda dx$ etc... how do we deal with this?

The components of the metric tensor are only $\eta_{\sigma \rho}$ when using Minkowski coordinates. If you use any other coordinates, the metric tensor will take a different form; an easy way to see this is to calculate the spacetime interval between two events in an inertial frame using polar coordinates instead of rectilinear Minkowski coordinates. The transformation between Minkowski coordinates and polar coordinates happens not to introduce any off-diagonal elements, but more complicated transformations will.

Note that if you know the transformation between rectilinear Minkowski coordinates and some other coordinate system, you can use the tensor transformation formula on the metric tensor itself to find the components of the metric in the new coordinate system.

I've already suggested one good exercise: comparing polar coordinates and Minkowski coordinates for an inertial frame. Another good exercise would be to study Rindler coordinates, convenient for describing the non-inertial frame in which a body undergoing constant proper acceleration is at rest.

 

[stevendaryl]

So I learned from the first few posts on this https://www.physicsforums.com/threads/ftl-train-ftl-communication-thought-experiment.945116/ thread that one can actually work with special relativity in non inertial frames.

I'd like to get some points on this.

Firstly, by a non inertial frame, I understand a reference frame with coord sys $x'$ which is connected to another with coord sys $x$ by a Lorentz transformation $x' = \Lambda x + b$ in which $\Lambda, b$ are functions of $x$. This would dramatically change the Lorentz interval $\eta_{\sigma \rho}dx'^\sigma dx'^\rho$ causing the presence of terms like $x \Lambda d\Lambda dx$ etc... how do we deal with this?

The generalization from inertial to noninertial frames is not a matter of making the matrix a function of position. Matrices are only used for linear transformations. The more general transformation is that the new coordinates ($y^\alpha$) are related to the old coordinates ($x^\mu$) through an arbitrary nonlinear function.

So the example that is easiest to work out completely is (as I talk about in my Insights post) Rindler coordinates in two spacetime dimensions. The Rindler coordinates $R, \theta$ related to the usual inertial coordinates $x, t$ through:

$x = R\ cosh(\theta)$
$t = R/c\ sinh(\theta)$

For a noninertial coordinate system,it will not be the case that $\eta_{\mu \nu} dx^\mu dx^\nu$ is invariant. Instead, you have to use a metric tensor whose components transform under a coordinate change:

$ds^2 = g_{\mu \nu} dx^\mu dx^\nu$

In inertial coordinates, $ds^2 = dt^2 - \frac{1}{c^2} dx^2$. Using the above equations, we can write:

$dx = dR\ cosh(\theta) + R\ sinh(\theta) d\theta$
$dt = dR/c\ sinh(\theta) + R/c\ cosh(\theta) d\theta$

So we can compute $ds^2$ by just substitution: The answer is:

So $ds^2 = (R/c)^2 d\theta^2 - (dR/c)^2$

 

[stevendaryl]

The generalization from inertial to noninertial frames is not a matter of making the matrix a function of position. Matrices are only used for linear transformations. The more general transformation is that the new coordinates ($y^\alpha$) are related to the old coordinates ($x^\mu$) through an arbitrary nonlinear function.

Just for clarification: matrices are not used for transforming between noninertial coordinate systems, but the infinitesimals will always be related by a matrix:

$dy^\alpha = \Lambda^\alpha_\mu dx^\mu$

where $\Lambda$ is indeed a function of the coordinates. So I was wrong to say that matrices are not used for nonlinear transformations---the infinitesimal transformations are always linear.

 

[PAllen]

Note, there is a whole book on this:

https://books.google.com/books/abou...BAJ&printsec=frontcover&source=kp_read_button

 

[sweet springs]

I am not sure about point of your question. Please find below my guess.
Say there is a rotating system RFR, rotating frame of reference, that is described by SR theory only as you expect, rotating around point O still in an IFR#1. There is another IFR#2 where O is moving with constant speed.
Transformation of RFR and IFR#1 is an interesting one and that of IFR#1 and IFR#2 is a boring Lorentz's.
Your question might be said here "Is the RFR is well described by IFR#2 coordinates?". The answer is Yes.

 

[kent davidge]

The components of the metric tensor are only $\eta_{\sigma \rho}$ when using Minkowski coordinates

Thanks for pointing this out

So the example that is easiest to work out completely is (as I talk about in my Insights post) Rindler coordinates in two spacetime dimensions. The Rindler coordinates $R, \theta$ related to the usual inertial coordinates $x, t$ through:

$x = R\ cosh(\theta)$
$t = R/c\ sinh(\theta)$

Is there a way for one to know that Rindler coordinates describes a non inertial system only by looking at how the coordinates are related to the Minkowski coordinates as in the quote above? Or does one need to read your insight for knowing this?

Note, there is a whole book on this:

https://books.google.com/books/abou...BAJ&printsec=frontcover&source=kp_read_button

Unfortunately I found the notation/script hard to follow ... but it seems to be a good book because of the topics it covers.

 

[kent davidge]

Is there a way for one to know that Rindler coordinates describes a non inertial system only by looking at how the coordinates are related to the Minkowski coordinates as in the quote above? Or does one need to read your insight for knowing this?

That makes me think that it would be useful to have a tensor whose components all vanish except in non inertial frames, for one would only need to know its components in a given coordinate system to see if he/she is dealing or not with a non inertial frame.

 

[Mentz114]

That makes me think that it would be useful to have a tensor whose components all vanish except in non inertial frames, for one would only need to know its components in a given coordinate system to see if he/she is dealing or not with a non inertial frame.

The proper acceleration of a vector $u^a$ is defined as $\dot{u}_b=\nabla_a u_b u^a$. For a geodesic the norm of this is zero. I don't think there is an easier way to be sure if the acceleration is zero.

For a time dependent boost the acceleraion vector has one component $\gamma^2 \dot{\beta}$ so if $\beta$ is constant (not a function of coordinates) there is no acceleration. Doh².

 

[Nugatory]

That makes me think that it would be useful to have a tensor whose components all vanish except in non inertial frames,

If all the components of a tensor vanish in one frame, they vanish in all frames. You can see this just by inspection of the tensor transformation rule.

 

[PAllen]

The proper acceleration of a vector $u^a$ is defined as $\dot{u}_b=\nabla_a u_b u^a$. For a geodesic the norm of this is zero. I don't think there is an easier way to be sure if the acceleration is zero.

For a time dependent boost the acceleraion vector has one component $\gamma^2 \dot{\beta}$ so if $\beta$ is constant (not a function of coordinates) there is no acceleration. Doh².

Adding to this, if you have coordinates such that one is timelike and the others are spacelike (note, ther is no general requirement that this be true, and it is not true for a number of common coordinates in GR, and light like coordinates are sometimes used in SR), then if a world line defined by the spatial origin, with varying t coordinate has nonzero proper acceleration, you could say these coordinates are adapted to that accelerated observer, especially near the origin.

To check the timelike, spacelike, or null character of coordinates at a point ( in general, the classification can be different at different points), you take the norm per the metric of the vector defined by that coordinate direction with other opponents set to zero. Relying on letters will throw you off, as, for exapmlple, the r coordinate is timelike and the t coordinate is spacelike in the Schwarzschild vacuum interior coordinates.

 

[Ibix]

Adding to this, if you have coordinates such that one is timelike and the others are spacelike (note, ther is no general requirement that this be true, and it is not true for a number of common coordinates in GR, and light like coordinates are sometimes used in SR), then if a world line defined by the spatial origin, with varying t coordinate has nonzero proper acceleration, you could say these coordinates are adapted to that accelerated observer, especially near the origin.

I agree with this, but would you regard light cone coordinates as an inertial frame? And is there any such thing as an inertial frame in curved spacetime?

I think I'd define an inertial frame to be one in which there exists a unique family of inertial observers who are mutually at rest (in the sense of being able to synchronise their clocks by some signal exchange procedure) and whose four velocities are one of the coordinate bases. If that's an acceptable definition then given a coordinate system:

1. Is exactly one of its basis vectors timelike? If no, this isn't an inertial frame.
2. Is the Riemann tensor everywhere zero? If no, inertial frames don't apply anyway.
3. Is the proper acceleration of a four velocity parallel to the timelike basis vector everywhere zero (I think this reduces to $\Gamma^a_{00}=0$, where the timelike basis is the zeroth one)? If no this isn't an inertial frame. Otherwise it is.

Note that, because the Christoffel symbols aren't tensors, this is a coordinate-dependent statement. But that's what we wanted.

 

[Mentz114]

Adding to this, if you have coordinates such that one is timelike and the others are spacelike (note, ther is no general requirement that this be true, and it is not true for a number of common coordinates in GR, and light like coordinates are sometimes used in SR), then if a world line defined by the spatial origin, with varying t coordinate has nonzero proper acceleration, you could say these coordinates are adapted to that accelerated observer, especially near the origin.

To check the timelike, spacelike, or null character of coordinates at a point ( in general, the classification can be different at different points), you take the norm per the metric of the vector defined by that coordinate direction with other opponents set to zero. Relying on letters will throw you off, as, for exapmlple, the r coordinate is timelike and the t coordinate is spacelike in the Schwarzschild vacuum interior coordinates.

I just came back to amend my post above (#9) because ( in SR I must emphasize) the rule of thumb I give only applies to translational velocity. There could be constant angular velocity which gives rise to centripetal acceleration.

 

[kent davidge]

I think I'd define an inertial frame to be one in which there exists a unique family of inertial observers who are mutually at rest

I think this statement translates to having a same set $x$ of Minkowski coordinates such that around and for every $x'$ $g_{\mu \nu} (x') = \eta_{\alpha \beta} (dx^\alpha / dx'^\mu) (dx^\beta / dx'^\nu)$ where $x'$ is the coordinate system in our candidate to inertial frame and $g$ its metric.

Is exactly one of its basis vectors timelike? If no, this isn't an inertial frame.

Wouldn't we also need the other three to be all necessarily space-like?

Is the Riemann tensor everywhere zero? If no, inertial frames don't apply anyway.

Good tool. But wouldn't it be misleading since we can have a non inertial frame with no gravity at all, (i.e. we can have a non inertial frame & space-time with no curvature)?

 

[Nugatory]

Good tool. But wouldn't it be misleading since we can have a non inertial frame with no gravity at all, (i.e. we can have a non inertial frame & space-time with no curvature)?

The non-zero Riemann tensor implies that there will be no global inertial coordinate system (in the sense of @Ibix's step 3) so you don't have to waste your time analyzing any given coordinate system to see if it describes an inertial frame. Only if Riemann is non-zero do you need to proceed to step 3.

 

[pervect]

That makes me think that it would be useful to have a tensor whose components all vanish except in non inertial frames, for one would only need to know its components in a given coordinate system to see if he/she is dealing or not with a non inertial frame.

In the context of special relativity: there is a mathematical entity whose components all vanish in an "inertial frame", and do not vanish in a "non-inertial frame", but this mathematical entity is not a tensor. Collectively, the entity is called "the Christoffel symbols", and it doesn't transform as a tensor.

There are various possible interpretations of what a "frame of reference" might mean, here I am taking the point of view that a "frame of reference" is a coordinate system.

So, in the context of special relativity, doing physics when the Christoffel symbols do not all vanish requires one to learn new methods These new methods are tensor methods. A tensor formalism will work if one has vanishing Christoffel symbols, or not - it doesn't matter. Because it doesn't matter, a tensor can't possibly tell you whether or not the Christoffel symbols vanish or not - the point of using these methods is that they will work regardless.

Thus, in tensor methods, one can use almost ANY coordinate system one likes (there are some modest mathematical properties, such as smoothness, that are required, so this is a bit of an overstatement if taken too literally. I won't attempt to give a complete list of all the requirements, it'd be very technical and I'd probably get the details wrong, and the details matter for a serious study.)

The point is that one can regard a "globally inertial frame of reference" as a special set of coordinates one chooses such that the Christoffel symbols all vanish, everywhere.

In General relativity, one can make the Christoffel symbols vanish at a point, but it's impossible in general to find a set of coordinates that make them vanish everywhere. So in General relativity, one can always have local inertial coordinates, but one does not necessarily have global inertial coordinates.

There is a tensor entity that vanishes when it is possible to make the Christoffel symbols vanish everywhere. This entity is the Riemann tensor. So if the Riemann tensor vanishes, one can choose coordinates that make the Christoffel symbols vanish everywhere. By using these special coordinates, one can use the methods one was taught in high school.

When the Riemann tensor does not vanish, one can use the high-school methods of Newtonian physics in a small enough region of space-time, but they will fail over larger regions. For instance, an orbiting space station might have a "local inertial frame of reference", where the Christoffel symbols all vanish - they vanish exactly at one point in the space-station, and remain small in an volume around that point.

There is a physical interpretation of the most important components of the Riemann tensor. This physical interpretation is tidal forces. Loosely speaking, when the tidal forces vanish, the RIemann tensor vanishes, and the methods of special relativity apply. This actually isn't quite right :(, but it's close enough to get some insight. The sticking point is that there are components of the Riemann tensor that are not tidal forces, components that represent unfamiliar things like frame-dragging and spatial curvature. But much of the time these unfamiliar things can be ignored - in any event, they do not fit neatly into a basically Newtonian point of view, so one has to learn General Relativity to really understand these unfamiliar things, while one can leverage off one's knowledge of Newtonian mechanics to understand tidal forces.

 

[PeterDonis]

would you regard light cone coordinates as an inertial frame?

No, they aren't. A "frame" has to have one timelike and three spacelike basis vectors. (Strictly speaking, to use the word "frame", the basis vectors must also be orthonormal.) Light cone coordinates have two null and two spacelike basis vectors.

is there any such thing as an inertial frame in curved spacetime?

Only locally, not globally. And even locally, strictly speaking, you can only have a frame that is inertial at one event, since you can only choose coordinates such that the metric is $\eta_{\mu \nu}$ and the Christoffel symbols all vanish at one event.

 

[stevendaryl]

Thanks for pointing this out
Is there a way for one to know that Rindler coordinates describes a non inertial system only by looking at how the coordinates are related to the Minkowski coordinates as in the quote above? Or does one need to read your insight for knowing this?

Well, in this case, it's easy enough to see. Any two 2-D coordinate systems that are both inertial must be related by a linear transformation. That is,

$x' = Ax + Bt + C$
$t' = Dx + Et + F$

for some constants $A, B, C, D, E, F$.

 

[facenian]

Firstly, by a non inertial frame, I understand a reference frame with coord sys $x'$ which is connected to another with coord sys $x$ by a Lorentz transformation $x' = \Lambda x + b$ in which $\Lambda, b$ are functions of $x$.

I believe this observation is correct if the $x$ coordinates are inertial. The real problem which is hard to understand is that the non inertial coordinates $x'$ loose their usual metric interpretation. These have a usual metrical meaning only in a neighborbood of that point. This is why, according to Eintein's own words, the development of GR took him so long after the realization of his "happiest thougth",i.e., the equivalence principle. A consequence is that ordinary space becomes non euclidean with respect to a non inertial reference frame(no gravitation). All this is just physics leaving aside mathematical issues(no curvature, no general tensors, etc.)

 

[PAllen]

I agree with this, but would you regard light cone coordinates as an inertial frame? And is there any such thing as an inertial frame in curved spacetime?

I think I'd define an inertial frame to be one in which there exists a unique family of inertial observers who are mutually at rest (in the sense of being able to synchronise their clocks by some signal exchange procedure) and whose four velocities are one of the coordinate bases. If that's an acceptable definition then given a coordinate system:

1. Is exactly one of its basis vectors timelike? If no, this isn't an inertial frame.
2. Is the Riemann tensor everywhere zero? If no, inertial frames don't apply anyway.
3. Is the proper acceleration of a four velocity parallel to the timelike basis vector everywhere zero (I think this reduces to $\Gamma^a_{00}=0$, where the timelike basis is the zeroth one)? If no this isn't an inertial frame. Otherwise it is.

Note that, because the Christoffel symbols aren't tensors, this is a coordinate-dependent statement. But that's what we wanted.

I would instead of asking about global inertial frames which obviously rules out GR, ask about coordinates adapted to an inertial observer versus not. GR definitely has inertial observers, and one can consider several types of coordinates adapted to a given inertial observer, e.g. Fermi-Normal or Radar. Either would have the feature that they are 1x3 near the origin, the metric is Minkowski all along the origin, and connection coefficients vanish along the origin. I would certainly regard a lightlike coordinate basis as NOT adapted to any observer in either SR or GR. They are perfectly good coordinates, very convenient for some purposes, but they are not adapted to any particular observer.

Given coordinates that are 1x3, one may ask about the proper acceleration of the origin world line. However, this is not the whole picture, as @Mentz114 noted, and I forgot. You should also ask about rotation of basis relative to Fermi-Walker transport along the origin. Then you can talk about coordinates that are adapted to acceleration plus rotation, in the general case.

 

[PAllen]

Only locally, not globally. And even locally, strictly speaking, you can only have a frame that is inertial at one event, since you can only choose coordinates such that the metric is $\eta_{\mu \nu}$ and the Christoffel symbols all vanish at one event.

That is not correct. Both Fermi-Normal coordinates and Radar coordinates built from an inertial world line in GR have Minkowski metric and vanishing connection all along the origin world line. The case of Fermi-Normal is well known and covered in MTW. The radar case is discussed in work by Synge on the convergence of radar and FN coordinates near the origin world line.

 

[PeterDonis]

Both Fermi-Normal coordinates and Radar coordinates built from an inertial world line in GR have Minkowski metric and vanishing connection all along the origin world line.

Yes, you're right, I was only thinking of Riemann normal coordinates and forgot about these.

 

[Dr Whom]

It is the application of SR to non-inertial observers, such as one in a room in free fall under the action of gravity, that gives rise to GR. The equivalence principle takes care of the rest.

 

[Ibix]

It is the application of SR to non-inertial observers, such as one in a room in free fall under the action of gravity,

Something is wrong here. An observer in free-fall is an inertial observer.

 

[kent davidge]

I appreciate the lastest comments. With regards to pervect's #16

one can regard a "globally inertial frame of reference" as a special set of coordinates one chooses such that the Christoffel symbols all vanish, everywhere

Does that mean, in principle, one would need to check up on all possible coordinate systems allowed for the frame, and see if he/she finds one in which the Christoffel symbols vanish everywhere?

If he/she finds such coord system, then it's a inertial frame.

The problem is that it's a painful method. Is not there a easier way?

 

[Mentz114]

I appreciate the lastest comments. With regards to pervect's #16Does that mean, in principle, one would need to check up on all possible coordinate systems allowed for the frame, and see if he/she finds one in which the Christoffel symbols vanish everywhere?

If he/she finds such coord system, then it's a inertial frame.

The problem is that it's a painful method. Is not there a easier way?

You asked in an earlier thread if there was a tensor that could be used ( tensor contractions being coordinate independent) and I told you. The norm of the acceleration vector always indicates if there is proper acceleration or not. You still need to calculate the connections (Christofel symbols) but only once !

 

[PAllen]

I appreciate the lastest comments. With regards to pervect's #16Does that mean, in principle, one would need to check up on all possible coordinate systems allowed for the frame, and see if he/she finds one in which the Christoffel symbols vanish everywhere?

If he/she finds such coord system, then it's a inertial frame.

The problem is that it's a painful method. Is not there a easier way?

What pervect describes (global vanishing of christoffel symbols) is possible only in flat spacetime. You can see for any metric coordinate expression whether it describes flat spacetime by computing the curvature tensor. If the spacetime is flat, there are infinite distinct inertial frames at any point related to each other by rotation and boost. For any of these, assuming you want orthonormal coordinate basis, the metric is Minkowski. So, I would say a global inertial frame in standard coordinates always has Minkowski metric. If you want to allow other coordinates as global inertial, there are issues with the Pervect’s criterion, e.g using polar coordinates for the spatial coordinates leads to some non vanishing christoffel symbols even for what you would want to consider a global inertial frame.

I’ll throw out as a candidate definition of reasonably general global inertial coordinates:

1) the curvature tensor vanishes everywhere.
2) the coordinate basis at every point is 1 timelike and 3 spacelike vectors
3) the world line described by holding by holding all spatial coordinates constant (for any choice of coordinate values), has zero proper acceleration everywhere.

These things are all straightforward to verify given some arbitrary metric expression.

 

[kent davidge]

Thanks. Indeed

1) the curvature tensor vanishes everywhere.
2) the coordinate basis at every point is 1 timelike and 3 spacelike vectors

I remember reading about these two conditions in Weinberg's Relativity book.

But wouldn't that lead to a wrong conclusion? I mean, suppose I found out that the curvature tensor is not zero. Then the frame is non inertial. But regarding GR, since the curvature tensor is non zero, I would conclude that gravity is present. But that's not necessarily the case.

I think an answer to this would be: "but according to GR gravity is the curvature in space-time. So if I found out that the curvature tensor is not zero, I automatically discovered that that frame is non inertial and that there's gravity present".

But I would argue that it's possible to have a non inertial frame without gravity. For instance, a accelerating spaceship traveling through out empty space.

 

[PeterDonis]

I would argue that it's possible to have a non inertial frame without gravity.

Yes, it is. Remember that for a frame to be globally inertial, all three of @PAllen's conditions have to be satisfied. Violating anyone of them is enough to make a frame non-inertial. A non-inertial frame in flat spacetime (i.e., no gravity) violates only the third condition (a worldline with constant spatial coordinates in such a frame has nonzero proper acceleration). It still satisfies the other two.

 

[kent davidge]

Yes, it is. Remember that for a frame to be globally inertial, all three of @PAllen's conditions have to be satisfied. Violating anyone of them is enough to make a frame non-inertial. A non-inertial frame in flat spacetime (i.e., no gravity) violates only the third condition (a worldline with constant spatial coordinates in such a frame has nonzero proper acceleration). It still satisfies the other two.

I see

My point is that only by following the discussed procedure we are not able to discern between a non-inertial frame without gravity and one with gravity. How can we do that?

 

[PAllen]

Thanks. Indeed
I remember reading these two conditions in Weinberg's Relativity book.

But wouldn't that lead to a wrong conclusion? I mean, suppose I found out that the curvature tensor is not zero. Then the frame is non inertial. But regarding GR, since the curvature tensor is non zero, I would conclude that gravity is present. But that's not necessarily the case.

I think an answer to this would be: "but according to GR gravity is the curvature in space-time. So if I found out that the curvature tensor is not zero, I automatically discovered that that frame is non inertial and that there's gravity present".

But I would argue that it's possible to have a non inertial frame without gravity. For instance, a accelerating spaceship traveling through out empty space.

I was describing criteria for global inertial coordinates. In GR, as many here have noted, there is no such thing as global inertial frame. Pease don’t construe more than what I wrote - how to look at a metric and expression and determine if can reasonably be considered that the coordinates and metric are a global inertial coordinate system. An accelerated frame is by definition not inertial in either SR and GR.

 

[Mentz114]

I see

My point is that only by following the discussed procedure we are not able to discern between a non-inertial frame without gravity and one with gravity. How can we do that?

a non-inertial frame without gravity = proper acceleration
a non-inertial frame with gravity = proper acceleration + gravity (Riemann ≠ 0)

But gravity cannot cause proper acceleration so you need rocket motors in both cases ( I would guess).

 

[PeterDonis]

only by following the discussed procedure we are not able to discern between a non-inertial frame without gravity and one with gravity

"Gravity" means that condition #1 is violated (the curvature tensor is nonzero).

 

[PAllen]

I see

My point is that only by following the discussed procedure we are not able to discern between a non-inertial frame without gravity and one with gravity. How can we do that?

If what you are looking for is a general classification of coordinates for both SR and GR based on metric expression in those coordinates, there is no clear answer. It is all a matter of definitions on which authors differ and physics doesn’t care. I’ll throw out a few ideas that I am sure someone could find arguments with for various cases:

1) If all the criteria I gave earlier are met, call it a global inertial coordinate system.
2) If the first two conditions are met, you might call it an SR noninertial coordinate system, but if you want to say it is adapted to a particular noninertial observer, you probably want more conditions, e.g. that surfaces of constant timelike coordinate are orthogonal to the spatial origin world line everywhere.
3) If there is curvature, and the 1 by 3 criterion is met, and constant time surfaces are orthogonal to the origin world line, then it is locally inertial if the origin world line has zero proper acceleration, else it is locally noninertial.

In both SR and GR, there are common coordinates which would not fit any of these criteria. Most coordinates used in GR would not fit into any of these categories. They are simply not based on any particular observer, inertial or not.

 

[Dr Whom]

Inertial frame. A frame of reference in which bodies are not accelerated. The Penguin Dictionary of Science. In free fall under the action of gravity objects are generally accelerated.