Inertial Reference Frames- circular?

[AcidRainLiTE]

My textbook basically defines an inertial reference frame as follows: If you have an object O that has no forces acting on it, and there is a reference frame R where the acceleration of O with respect to R is zero, then R is a inertial reference frame.
This to me seems circular. How does one know if there are no forces acting on O? Don't you need to know if O has any acceleration? How do you know that O has no acceleration? Don't you need an inertial reference frame?

Stated differently:
The definition stated above seems, to my understanding, to say that an inertial reference frame needs to be a reference frame that does not accelerate. And basically I am asking how do you know if it is accelerating? Can you really know without already having an inertial reference frame?

Thanks.

 

[OlderDan]

One does not need to empirically verify that a reference frame is inertial to have a conceptual definition of such a frame. The fundamental postulate of general relativiity (something I know essentially nothing about, but there are several knowledgeable folk who visit here) is the equivalence principle that says that freefalling reference frames are equivalent to inertial reference frames. In a free-falling elevator, a ball released from rest does not fall relative to the elevator. Is there a force acting on it?

See for example

http://astsun.astro.virginia.edu/~jh8h/Foundations/chapter8/chapter8.html

 

[AcidRainLiTE]

One does not need to empirically verify that a reference frame is inertial to have a conceptual definition of such a frame.

Yes, this may be possible, but in order to obtain a conceptual definition of an inertial reference frame we must not assume that we already conceptually know what an inertial reference frame is.

If you have an object O that has no forces acting on it, and there is a reference frame R where the acceleration of O with respect to R is zero, then R is a inertial reference frame.

If we define an inertial reference frame this way, haven't we assumed that we already conceptually know what an inertial reference frame is? We are assuming that we understand what a force is, and in order to understand what a force is we need to understand what acceleration is, and to understand what acceleration is we need to understand what an inertial reference frame is. We need to understand what a force is to understand what an inertial reference frame is, yet, to understand what a force is we need to understand what an inertial reference frame is. Isn't this circular?

 

[AcidRainLiTE]

I'm sorry, I realize made a mistake in my last post. You do not need to understand what an inertial reference frame is to understand what acceleration is. We may see two objects receding from each other and not know which is moving (relative velocity), but when we see the rate at which they are moving apart changing, we know one of them is accelerating. Although we may not know which is accelerating, we at least can see what acceleration is.
Is acceleration then relative? And force as well?
Am I correct in saying that we cannot know if something is an inertial reference frame? We can only say 'we will consider this reference frame to be an inertial reference frame,' and then run calculations based on that assumption?

Thanks again.

(For future reference, does a question such as this belong in the homework/coursework section of the forums, or should/can I post it elsewhere?)

 

[OlderDan]

I'm sorry, I realize made a mistake in my last post. You do not need to understand what an inertial reference frame is to understand what acceleration is. We may see two objects receding from each other and not know which is moving (relative velocity), but when we see the rate at which they are moving apart changing, we know one of them is accelerating. Although we may not know which is accelerating, we at least can see what acceleration is.
Is acceleration then relative? And force as well?
Am I correct in saying that we cannot know if something is an inertial reference frame? We can only say 'we will consider this reference frame to be an inertial reference frame,' and then run calculations based on that assumption?

Thanks again.

(For future reference, does a question such as this belong in the homework/coursework section of the forums, or should/can I post it elsewhere?)

It's probably best in another location. It's really more philosophy of science than the kind of conceptual and computational issues we usually deal wtih in the homework area.

I was actually writing a rebuttal to your last post that got interrrupted by dinner. I may decide to finish it some day, but not now.

Your conclusion about acceleration is correct, in my opinion. We do not need to consider the familiar attributes of objects like mass and volume, or forces to define concepts like position, velocity, and acceleration. We only need concepts of space (distance and direction) and time to do that. Position, velocity, and acceleration are indeed all relative concepts, which is why we conceive of different frames of reference for describing the observed universe. If these things (and time) were absolute, we would probably be doing all of physics in one frame of reference.

 

[MeJennifer]

Position, velocity, and acceleration are indeed all relative concepts, which is why we conceive of different frames of reference for describing the observed universe.

I agree that position and velocity are relative but I disagree that acceleration is relative.
In relativity different observers will agree on a change in speed of an object, however they could disagree on whether the change is an increase or a decrease.

 

[OlderDan]

I agree that position and velocity are relative but I disagree that acceleration is relative.
In relativity different observers will agree on a change in speed of an object, however they could disagree on whether the change is an increase or a decrease.

Nothing was said about whether the frames of reference are inertial. Accleration is relative to a frame of reference. Relative to me, my computer screen is at rest (fortunately). It has constant velocity (zero) and zero acceleration. To an observer on the moon, the screen has high velocity and is accelerating.

 

[MeJennifer]

Accleration is relative to a frame of reference. Relative to me, my computer screen is at rest (fortunately).

You are mistaken it is not.

 

[OlderDan]

You are mistaken it is not.

Oh really? How can you say

different observers will agree on a change in speed of an object, however they could disagree on whether the change is an increase or a decrease.

and then claim that acceleration is not relative? What does relative mean to you?

 

[JesseM]

Different observers will agree on a change in speed of an object, however they could disagree on whether the change is an increase or a decrease.

and then claim that acceleration is not relative? What does relative mean to you?

In Newtonian mechanics, all inertial frames will agree on both the direction and the magnitude of an acceleration, but it is still nevertheless true that they disagree on whether it's an increase or decrease in speed--for example, if a car accelerates in the direction of its front, then in a frame where the car already had a high velocity in the direction of its back, this will be a decrease in speed, while in a frame where the car's velocity was in the direction of its front, this will be an increase in its speed. But both still agree it accelerates in the direction of its front, and they all agree on the difference in speeds before and after the acceleration.

In relativity, different inertial frames will disagree on the magnitude of an acceleration, but they will still agree on its direction. And they will all agree on whether a given object is accelerating or not, which is probably what MeJennifer meant.

 

[Hurkyl]

In relativity different observers will agree on a change in speed of an object, however they could disagree on whether the change is an increase or a decrease.

No, they will not agree on the change in speed. They will disagree on the magnitude, and they will disagree on the direction.

And if we are considering all reference frames (not just inertial ones), they won't even agree on whether or not the acceleration is zero.


The very definition of "acceleration" depends on you first choosing a reference frame...


So how do you justify saying acceleration is not relative?

 

[OlderDan]

In Newtonian mechanics, all inertial frames will agree on both the direction and the magnitude of an acceleration, but it is still nevertheless true that they disagree on whether it's an increase or decrease in speed--for example, if a car accelerates in the direction of its front, then in a frame where the car already had a high velocity in the direction of its back, this will be a decrease in speed, while in a frame where the car's velocity was in the direction of its front, this will be an increase in its speed. But both still agree it accelerates in the direction of its front, and they all agree on the difference in speeds before and after the acceleration.

In relativity, different inertial frames will disagree on the magnitude of an acceleration, but they will still agree on its direction. And they will all agree on whether a given object is accelerating or not, which is probably what MeJennifer meant.

I understand what you are saying about the speed in Newtonian mechanics under Galilean transformation. The point is well taken. I was going to say more about the relevance of the previous comment to the discussion that was going on, but since I now see Hurkyl's post, I'll leave it alone.

 

[JesseM]

No, they will not agree on the change in speed. They will disagree on the magnitude, and they will disagree on the direction.

When you say they'll disagree on the direction, do you mean the direction in their coordinate system, or the actual direction in relation to physical objects? If in one frame a rocket is accelerating along the line from its tail to its nose, can other frames disagree on this?

 

[bernhard.rothenstein]

inertial reference frame

My textbook basically defines an inertial reference frame as follows: If you have an object O that has no forces acting on it, and there is a reference frame R where the acceleration of O with respect to R is zero, then R is a inertial reference frame.
This to me seems circular. How does one know if there are no forces acting on O? Don't you need to know if O has any acceleration? How do you know that O has no acceleration? Don't you need an inertial reference frame?

Stated differently:
The definition stated above seems, to my understanding, to say that an inertial reference frame needs to be a reference frame that does not accelerate. And basically I am asking how do you know if it is accelerating? Can you really know without already having an inertial reference frame?

Thanks.

hang a pendulum in your reference frame. if it stays vertically your frame is inrtial.

 

[MeJennifer]

No, they will not agree on the change in speed. They will disagree on the magnitude, and they will disagree on the direction.

Perhaps you misread what I wrote.
They all agree there is an acceleration, however they disagree on the magnitude and the direction.

And if we are considering all reference frames (not just inertial ones), they won't even agree on whether or not the acceleration is zero.

They will.
There is no such thing as a non inertial frame in GR (except for a very small region). You cannot possibly construct a Cartesian coordinate system representing the correct curvature.

The very definition of "acceleration" depends on you first choosing a reference frame...

In Galilean physics yes, but certainly not in GR.

 

[nakurusil]

My textbook basically defines an inertial reference frame as follows: If you have an object O that has no forces acting on it, and there is a reference frame R where the acceleration of O with respect to R is zero, then R is a inertial reference frame.
This to me seems circular. How does one know if there are no forces acting on O? Don't you need to know if O has any acceleration? How do you know that O has no acceleration? Don't you need an inertial reference frame?

Stated differently:
The definition stated above seems, to my understanding, to say that an inertial reference frame needs to be a reference frame that does not accelerate. And basically I am asking how do you know if it is accelerating? Can you really know without already having an inertial reference frame?

Thanks.

A very good operational definition is the following:

The inertial frames form a class of systems of coordinates in which
$$dx^2+dy^2+dz^2-c^2dt^2$$ is invariant when passing from one system to another.

Einstein's original definition was something to the effect that "it is a system of coordinates in which the equations of Newtonian (!) mechanics hold good" . See here, paragraph 1.

 

[JesseM]

A very good operational definition is the following:

The inertial frames form a class of systems of coordinates in which
$$dx^2+dy^2+dz^2-c^2dt^2$$ is invariant when passing from one system to another.

I'm pretty sure it would be possible to define a family of accelerating reference frames which were all accelerating at the same rate (as seen in inertial frames) and which were related to one another by the Lorentz transform, meaning that $$dx^2+dy^2+dz^2-c^2dt^2$$ would be invariant when passing from one of the accelerating frames to another. The problem is that this definition only depends on the coordinate transform from one coordinate system to another, it doesn't tell you anything physical (although maybe if you added the stipulation that the laws of physics should obey the same equations in each coordinate system then it would work).

 

[nakurusil]

I'm pretty sure it would be possible to define a family of accelerating reference frames which were all accelerating at the same rate (as seen in inertial frames) and which were related to one another by the Lorentz transform, meaning that $$dx^2+dy^2+dz^2-c^2dt^2$$ would be invariant when passing from one of the accelerating frames to another. The problem is that this definition only depends on the coordinate transform from one coordinate system to another, it doesn't tell you anything physical (although maybe if you added the stipulation that the laws of physics should obey the same equations in each coordinate system then it would work).

Not really , because there are members in this class (of inertial frames) that are not accelerating and the metric will not hold when attempting to pass from one of the frames you described to one of the other frames in the class.

 

[pervect]

A rather interesting defintion of an inertial frame in SR is one in which one can synchronize all one's clocks via the Einstein convention, and in which they stay syncrhonized.

See for instance:
http://links.jstor.org/sici?sici=0031-8248%28197309%2940%3A3%3C382%3ASRIAS%3E2.0.CO%3B2-3&size=LARGE

A couple of examples illustrate why this defintion works.

Consider an accelerating rocketship, for instance. The clocks on the nose of the ship "tick faster" than in the tail, so that a clock in the nose and tail will not stay synchronized.

In a rotating frame of reference, it will not be possible to initally synchronize all the clocks at all. For instance, if one starts to synchronize clocks from the center out, one finds that the Sagnac effect causes clocks around the circumference to be non-sychronized.

[add] Also, the clocks in the rotating frame won't all tick at the same rate. Thus the 'center' clocks and 'rim' clocks will tick at different rates, just as the 'top' and 'bottom' clocks in the accelrated spaceship do.

 

[nakurusil]

A rather interesting defintion of an inertial frame in SR is one in which one can synchronize all one's clocks via the Einstein convention, and in which they stay syncrhonized.

See for instance:
http://links.jstor.org/sici?sici=0031-8248%28197309%2940%3A3%3C382%3ASRIAS%3E2.0.CO%3B2-3&size=LARGE

A couple of examples illustrate why this defintion works.

Consider an accelerating rocketship, for instance. The clocks on the nose of the ship "tick faster" than in the tail, so that a clock in the nose and tail will not stay synchronized.

Yes, this looks like a version of the Pound-Rebka experiment with the accelerating rocket replacing the university tower.

In a rotating frame of reference, it will not be possible to initally synchronize all the clocks at all. For instance, if one starts to synchronize clocks from the center out, one finds that the Sagnac effect causes clocks around the circumference to be non-sychronized.

I am having trouble with the example, not with the concept. Help me out understanding what does the Sagnac effect (the fact that there is time difference between the TWO counter-rotating light beams) have to do with not being able to properly synchronize the clocks on the periphery.

 

[pess5]

Perhaps you misread what I wrote.
They all agree there is an acceleration, however they disagree on the magnitude and the direction.

Question ... Isn't it true that the total acceleration is the same in all frames (if integrated over the interval), even though the acceleration at any instant can differ?

There is no such thing as a non inertial frame in GR (except for a very small region). You cannot possibly construct a Cartesian coordinate system representing the correct curvature.

What about a rocket thruster burn? GR covers all possible scenarios, yes? A thruster burn is not freefall.

pess

 

[pess5]

My textbook basically defines an inertial reference frame as follows: If you have an object O that has no forces acting on it, and there is a reference frame R where the acceleration of O with respect to R is zero, then R is a inertial reference frame.
This to me seems circular. How does one know if there are no forces acting on O? ...

Thanks.

Good question. GR redefines the meaning of inertial. In SR, it was uniform translatory motion with no external forces acting upon it. In GR, its any state of motion in which one does not feel his own inertia (any weight). If you feel any inertia, then you are accelerating. In freefall (including orbit), you feel no inertia. You travel along geodesic paths in curved spacetime, which are the energy free paths in a non-euclidean spacetime. A thruster burn, direct impact by another body, interaction by electrical or magnetic bodies, or standing on planet Earth are examples of non-inertial motion. Falling to Earth in freefall is inertial, until you hit ground :-)

pess

 

[pervect]

Yes, this looks like a version of the Pound-Rebka experiment with the accelerating rocket replacing the university tower.


I am having trouble with the example, not with the concept. Help me out understanding what does the Sagnac effect (the fact that there is time difference between the TWO counter-rotating light beams) have to do with not being able to properly synchronize the clocks on the periphery.

For starters, I can make the example simpler. The clocks at the center and the edge of the rotating frame won't tick at the same rate.

But for my initial point, the way I first wrote it, take a look at
http://en.wikipedia.org/wiki/Image:Langevin_Frame_Cyl_Desynchronization.png

The time axis points "up" the cylinder. The red worldline represents one specific worldline of a co-rotating observer. One can imagine a set of red wordlines parallel to the drawn one. Each member of this set would representing co-rotating observers that start out at different angles.

The blue lines on the diagram represents the concept of simultaneity of a "co-rotating" observer according to Einstein clock synchronization. The blue lines are perpendicular (in the Minkowski sense) to the red worldlines, that's what makes them "lines of simultaneity".

The problem is that if you travel once around the cylinder along a blue line of simultaneity, you don't wind up at the same event where you started. It's not a closed curve.

Experimentally, if you take a clock in a round trip along the rotating cylinder, via 'slow clock transport' (via a red worldline), you'll find that it won't agree with the clock that stayed behind (a wordline straight up the cylinder) no matter how slowly you transport the clock.

http://arxiv.org/abs/gr-qc/9805089 also talks about this issue some.

 

[JesseM]

Not really , because there are members in this class (of inertial frames)

But I'm not talking about a class of inertial frames, I'm talking about a collection of non-inertial coordinate systems which are all related to one another (not to genuine inertial frames) by the Lorentz transform. This obviously isn't impossible--just pick a single accelerating coordinate systems, then every other member of the class can be created by doing a Lorentz transform on that original coordinate system.

 

[nakurusil]

But I'm not talking about a class of inertial frames, I'm talking about a collection of non-inertial coordinate systems which are all related to one another (not to genuine inertial frames) by the Lorentz transform. This obviously isn't impossible--just pick a single accelerating coordinate systems, then every other member of the class can be created by doing a Lorentz transform on that original coordinate system.

Yes, I understood that, there are two problems :

1. The class of inertial frames is no longer a class because it cannot contain both accelerating and non-accelerating frames (my point). So, I think that we now have two classes : truly inertial frames (non-accelerating) and pseudo-inertial frames (see below). We only want one class, don't we?

2. Based on how the Lorentz transforms were deduced by Einstein in his 1905 I think (I am not 100% sure) that it can be proven mathematically that one cannot have Lorentz transforms between accelerating frames unless all these frames share the same acceleration ( a trivial case)

 

[nakurusil]

For starters, I can make the example simpler. The clocks at the center and the edge of the rotating frame won't tick at the same rate.

But for my initial point, the way I first wrote it, take a look at
http://en.wikipedia.org/wiki/Image:Langevin_Frame_Cyl_Desynchronization.png

The time axis points "up" the cylinder. The red worldline represents one specific worldline of a co-rotating observer. One can imagine a set of red wordlines parallel to the drawn one. Each member of this set would representing co-rotating observers that start out at different angles.

The blue lines on the diagram represents the concept of simultaneity of a "co-rotating" observer according to Einstein clock synchronization. The blue lines are perpendicular (in the Minkowski sense) to the red worldlines, that's what makes them "lines of simultaneity".

The problem is that if you travel once around the cylinder along a blue line of simultaneity, you don't wind up at the same event where you started. It's not a closed curve.

Experimentally, if you take a clock in a round trip along the rotating cylinder, via 'slow clock transport' (via a red worldline), you'll find that it won't agree with the clock that stayed behind (a wordline straight up the cylinder) no matter how slowly you transport the clock.

http://arxiv.org/abs/gr-qc/9805089 also talks about this issue some.

Thank you, this is very interesting but this is not what confused me: the reference to Sagnac (see my post) is what did it. What does all this have to do with Sagnac?

 

[Hurkyl]

When you say they'll disagree on the direction, do you mean the direction in their coordinate system, or the actual direction in relation to physical objects? If in one frame a rocket is accelerating along the line from its tail to its nose, can other frames disagree on this?

The rocket head and tail is a special case, methinks. (at least the uniformly accelerating kind. I haven't tried working out a different example) Frames can disagree on things like, "is the rocket accelerating towards Polaris?"

There is no such thing as a non inertial frame in GR (except for a very small region). You cannot possibly construct a Cartesian coordinate system representing the correct curvature.

(I'm going to lapse into the differential geometry terminology I know)

Coordinate charts aren't required to represent the correct curvature.

A frame (at a point) is nothing more than a choice of basis for the tangent space at that point. (which, I suppose, can be intuitively thought of as an "infinitessimal" coordinate chart)

Only the orthonormal frames are inertial.

In Galilean physics yes, but certainly not in GR.

Depending on exactly what you mean by the word "relative", either all three of "position, velocity, acceleration" are relative, or none of the three are. Since you have said earlier that position and velocity are relative, then in order to be internally consistent, you must also have acceleration as a relative quantity.

 

[JesseM]

Yes, I understood that, there are two problems :

1. The class of inertial frames is no longer a class because it cannot contain both accelerating and non-accelerating frames (my point).

But I'm not saying that the class of inertial frames contains any accelerating frames! If class A represents the class of all inertial frames, then I'm talking about a totally separate class B of accelerating coordinate systems which are related to each other by the Lorentz transform, but a member of class B would not be related to a member of class A by the Lorentz transform, and not a single member of class A is a member of class B, just as not a single member of class B is a member of class A.

2. Based on how the Lorentz transforms were deduced by Einstein in his 1905 I think (I am not 100% sure) that it can be proven mathematically that one cannot have Lorentz transforms between accelerating frames unless all these frames share the same acceleration ( a trivial case)

That may be well be true, in the sense that all the accelerating coordinate systems in the class B that I described may have the same acceleration (as seen in a given inertial frame). It is nevertheless true that since they are related to one another by the Lorentz transform, if you calculate $$dx^2+dy^2+dz^2-c^2dt^2$$ between two events in the coordinates of one of these members of class B, and then compute the same quantity for the same two events in another of these members of class B, then the value will be the same. So, the mere fact that we have a class of coordinate systems in which $$dx^2+dy^2+dz^2-c^2dt^2$$ is constant is not enough to tell us that the class we're looking at is the class of inertial frames.

 

[nakurusil]

But I'm not saying that the class of inertial frames contains any accelerating frames! If class A represents the class of all inertial frames, then I'm talking about a totally separate class B of accelerating coordinate systems which are related to each other by the Lorentz transform, but a member of class B would not be related to a member of class A by the Lorentz transform, and not a single member of class A is a member of class B, just as not a single member of class B is a member of class A.

But this is exactly the problem I kept trying to tell you about: you can't have two classes , you need only one class that defines inertial frames. You can't have two classes not because of theoretical reasons but because of operational reasons.

 

[nakurusil]

That may be well be true, in the sense that all the accelerating coordinate systems in the class B that I described may have the same acceleration (as seen in a given inertial frame). It is nevertheless true that since they are related to one another by the Lorentz transform, if you calculate $$dx^2+dy^2+dz^2-c^2dt^2$$ between two events in the coordinates of one of these members of class B, and then compute the same quantity for the same two events in another of these members of class B, then the value will be the same. So, the mere fact that we have a class of coordinate systems in which $$dx^2+dy^2+dz^2-c^2dt^2$$ is constant is not enough to tell us that the class we're looking at is the class of inertial frames.

Yes, I see your point. If you have one counterexample this is good enough. But the accelerated frames have been eliminated from the get go from the class of inertial frame (see my previous post).
Interesting discussion, very stimulating!

 

[MeJennifer]

(I'm going to lapse into the differential geometry terminology I know)

Coordinate charts aren't required to represent the correct curvature.

A frame (at a point) is nothing more than a choice of basis for the tangent space at that point. (which, I suppose, can be intuitively thought of as an "infinitessimal" coordinate chart)

Only the orthonormal frames are inertial.

I fail to see how this is in any way a response to:

There is no such thing as a non inertial frame in GR (except for a very small region).

Do you or do you not agree that there is no such thing as a non inertial frame in GR (except for a very small region that may be considered flat)?

Depending on exactly what you mean by the word "relative", either all three of "position, velocity, acceleration" are relative, or none of the three are. Since you have said earlier that position and velocity are relative, then in order to be internally consistent, you must also have acceleration as a relative quantity.

It seems we can agree to disagree.
Whille position and velocity are relative acceleration is absolute in GR.

Feel free to explain how the principle of equivalence would hold if acceleration were not absolute. It would follow that a gravitational field would not be absolute as well.

 

[Hurkyl]

Do you or do you not agree that there is no such thing as a non inertial frame in GR (except for a very small region that may be considered flat)?

No. If by "frame" you mean "coordinate system", then by the definition of manifold, you have to have frames that are defined on "large" regions, and they will be non-inertial.

If by "frame" you mean "a choice of basis vector for the tangent space", then again you can construct a frame over a large region -- you can often do it over the entire manifold. These will almost always be non-inertial (though some inertial ones exist)

If by "frame" you mean something else, then you have to say what it is.

Feel free to explain how the principle of equivalence would hold if acceleration were not absolute.

I can't even figure out why you think there's a problem.

 

[JesseM]

Do you or do you not agree that there is no such thing as a non inertial frame in GR (except for a very small region that may be considered flat)?

Do you think that Schwarzschild coordinates, which are defined on a large region, qualify as "inertial"? By what criteria? Certainly the path of an object which is at rest in these coordinates (in the sense that its position coordinates are not changing with time) is not following a geodesic.

 

[MeJennifer]

Do you think that Schwarzschild coordinates, which are defined on a large region, qualify as "inertial"? By what criteria? Certainly the path of an object which is at rest in these coordinates (in the sense that its position coordinates are not changing with time) is not following a geodesic.

The selection of coordinates have absolutely nothing to do with something being inertial or not.

Something that is resisting the gravitational pull is obviously not traveling on a geodesic. For instance when you stand on the Earth you are not traveling a geodesic, you are actually accelerating away from the enter of gravity.

Do you realize that the Schwarzschild solution is a model from the perspective of a distant observer who is away from the gravitational pull?

 

[JesseM]

The selection of coordinates have absolutely nothing to do with something being inertial or not.

Sure, but I thought we were talking about non-inertial vs. inertial frames, not the question of whether a path is inertial (ie a geodesic) or not. Are you using "frame" to mean something different than "coordinate system"?

Something that is resisting the gravitational pull is obviously not traveling on a geodesic. For instance when you stand on the Earth you are not traveling a geodesic, you are actually accelerating away from the enter of gravity.

Of course I know this, wasn't it obvious that this was exactly my point? Things which are at rest in Schwarzschild coordinates are not moving on a geodesic, therefore Schwarzschild coordinates cannot be considered an "inertial coordinate system". This is exactly the same standard you use to judge whether a coordinate system is inertial or non-inertial in SR--an object moving on a geodesic path will be moving in a straight line in coordinate terms (constant dx/dt, dy/dt and dz/dt) in an inertial coordinate system, and that path will still be a geodesic in a non-inertial coordinate system, but the system is called "non-inertial" precisely because the path would no longer appear straight in terms of the coordinates (dx/dt, dy/dt and dz/dt wouldn't be constant), and because objects which are at rest in this coordinate system (constant x, y, and z coordinate) are moving on non-geodesic paths.

If this is not the standard that you use to judge whether a coordinate system in SR is inertial or non-inertial, than what is?