Layman's doubts about Gen Relativity

[Al68]

If your frame of reference has a non-uniform, or accelerated motion, then the Law of Inertia will appear to be wrong, and you must be in a non-inertial frame of reference. Right now you're being pulled towards the surface of the Earth by gravity, but at rest relative to it's surface, so you feel no fictitious forces that would lead you claim you were not at rest.

It's objects in freefall, not objects at rest in an accelerated frame that experience fictional forces. And fictional forces aren't "felt", that's why they're fictional. Objects in freefall require fictional forces to account for their motion relative to an accelerated reference frame. Earth's surface is an accelerated reference frame in the same way the floor of an accelerating rocket would be. Being at rest in such a frame means feeling a real force, while fictional forces account for the motion of objects in freefall.

 

[Dale]

Wasn't I clear when I said ...

That is not an equation. So again, I ask: in general what equation determines if a worldline is inertial? The answer should be something like "in general a worldline is inertial if and only if A=B".

 

[Austin0]

A free falling (or floating) point test particle falls inertially, which means it undergoes 0 proper acceleration. This view is valid for all observers.

A free falling spatially extended test body as a whole does not fall inertially. For instance in the Schwarzschild spacetime the front and back of such a body accelerate away from the center. This acceleration increases as the body gets closer to the center of gravity.

Doesn't this seem inconsistent??

AN extended body is just a collection of particles. How could a particle at one point be accelerated relative to another particle without itself having some degree of proper acceleration??

Doesn't this also imply a spatial expansion which would neccessitate a Newtonian force in opposition to the internal nuclear and electrostatic tensile forces resisting expansion?

Wouldn't an apparent acceleration differential be assumed on the basis of local coodinate velocity and acceleration measurements (at the front and the back) being based on local time.

Wouldn't this be is dilated by exactly the same factor [due to G potential ] as the assumed relative local acceleration differential due to this same local G potential?

I.e. the measured coordinate velocity [and drived acceleration] would be greater at the front simply because the clocks are dilated wrt the rear.
Equivalent to photon redshift where ther is no implication of an actual change of the photon.

How does length contraction fit into this proposition??

If all these factors are considered in the theorem could you simply explain ,I.e. without math.

Thanks

 

[djy]

Doesn't this seem inconsistent??

AN extended body is just a collection of particles. How could a particle at one point be accelerated relative to another particle without itself having some degree of proper acceleration??

First, suppose the collection of particles is "dust", i.e. no intermolecular forces. Then each particle will be on its own free fall trajectory, and geodesic deviation ("tidal forces" in Newtonian parlance) will distort the dust field with time.

If the collection of particles is being held together by internal electrostatic forces, then these forces will tend to resist tidal distortion. The particles do experience proper acceleration, but it is from the electrostatic force, not gravity.

 

[Passionflower]

Doesn't this seem inconsistent??

It really is not.

AN extended body is just a collection of particles. How could a particle at one point be accelerated relative to another particle without itself having some degree of proper acceleration??

That is the basic premise of GR. Things can move in relation to each other, and even accelerate in relation to each other without any need for proper acceleration.

Doesn't this also imply a spatial expansion which would neccessitate a Newtonian force in opposition to the internal nuclear and electrostatic tensile forces resisting expansion?

Well EM forces will try to tend to keep the body rigid. This process is an instance of proper acceleration.

Wouldn't an apparent acceleration differential be assumed on the basis of local coodinate velocity and acceleration measurements (at the front and the back) being based on local time.

Well in general it would depend on the observer and coordinate system. Only proper acceleration is observer independent.

 

[Austin0]

AN extended body is just a collection of particles. How could a particle at one point be accelerated relative to another particle without itself having some degree of proper acceleration??

Doesn't this also imply a spatial expansion which would neccessitate a Newtonian force in opposition to the internal nuclear and electrostatic tensile forces resisting expansion?

That is the basic premise of GR. Things can move in relation to each other, and even accelerate in relation to each other without any need for proper acceleration..

Understood. If you use apples for test particles then of course a sufficiently large cloud of apples would be expected to expand and have coordinate acceleration wrt each other.

But the hypothesis is based on a premise that this would apply equivalently to one large apple.Which is an entirely different thing. I may have confused things by using the term particle whenI should have said different sections of the system.

Well EM forces will try to tend to keep the body rigid. This process is an instance of proper acceleration..

I take it you mean upward proper acceleration?

Yet you seem to assume that the downward kinematic acceleration would overcome this resulting in overall expansion. is this correct?

What about the view that the EM proper acceleration would speed up the rear and slow down the front and it would just be a question of the propagation time of momentum , of equallizing the differential??

Wouldn't an apparent acceleration differential be assumed on the basis of local coodinate velocity and acceleration measurements (at the front and the back) being based on local time.

Wouldn't this be is dilated by exactly the same factor [due to G potential ] as the assumed relative local acceleration differential due to this same local G potential?

I.e. the measured coordinate velocity [and drived acceleration] would be greater at the front simply because the clocks are dilated wrt the rear.

Well in general it would depend on the observer and coordinate system. Only proper acceleration is observer independent.

I assumed that in this instance question it would be a radial line of Swarzschild observers extending upward through the course of free fall.

That the clocks would be dilated throughout.

That they would measure length increase and relative instantaneous velocities at the front and back as those points were colocated.

That the observers measuring the front would neccessarily measure greater velocities because their clocks were runnning slower than the clocks measuring the back.

That even if there was no actual length increase and no actual acceleration differential there would still be a measured coordinate velocity/acceleration differential due to the dilation.

Make sense?

 

[superkan619]

Is it safe now to tell that a very small particle in free fall, undergoing translational acceleration (in the kinematic sense) over a very long distance is actually an inertial frame for people on a perfectly still Earth?

 

[Austin0]

Is it safe now to tell that a very small particle in free fall, undergoing translational acceleration (in the kinematic sense) over a very long distance is actually an inertial frame for people on a perfectly still Earth?

IMO... It is safe to say it was an inertial object.

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 would have a dynamic metric.
The length contracting over time and the time dilating proportionately.

 

[Dale]

IMO... It is safe to say it was an inertial object.

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.

Hmm, interesting take on his question. I assumed he meant "inertial object" and was just being sloppy. Maybe he actually meant inertial coordinate system.

superkan619, perhaps you can clarify your question better. A reference frame is a coordinate system. An object is obviously not a coordinate system, so an object cannot be a reference frame. Do you mean to ask about whether or not the object is inertial or do you mean to use the object to construct some coordinate system and then ask about the coordinate system?

 

[superkan619]

perhaps you can clarify your question better. A reference frame is a coordinate system. An object is obviously not a coordinate system, so an object cannot be a reference frame. Do you mean to ask about whether or not the object is inertial or do you mean to use the object to construct some coordinate system and then ask about the coordinate system?

My stress is on the fact that the observing frame is on Earth; I think its not necessary for the observed one to be designated as either an object or a reference frame as long as we're outside. I actually thought of it as an object.

 

[Altabeh]

That is not an equation. So again, I ask: in general what equation determines if a worldline is inertial? The answer should be something like "in general a worldline is inertial if and only if A=B".

Again read the post carefully. I don't want to play the role of your primary school teacher and point directly at things that take your intelligence to be understood through reading a simple statement.

Since with these conditions we are now in the Minkowskian spacetime we have

$$\nabla \rightarrow \partial,$$

and thus

$$\frac{d^2x^a}{ds^2}$$

can now be pronounced as "proper acceleration" in GR that is obviously zero.

The equation is shouting that it is present here.

AB

 

[Dale]

My stress is on the fact that the observing frame is on Earth; I think its not necessary for the observed one to be designated as either an object or a reference frame as long as we're outside. I actually thought of it as an object.

It is a necessary designation. An object is a physical piece of matter. A reference frame is a mathematical construct, a coordinate system. One cannot be the other. However, since you are indeed asking (as I had understood) whether or not the object is inertial from the point of view of arbitrary reference frames then the answer is unambiguously "yes, it is inertial in all reference frames".

 

[Altabeh]

My stress is on the fact that the observing frame is on Earth; I think its not necessary for the observed one to be designated as either an object or a reference frame as long as we're outside. I actually thought of it as an object.

The point is that to any kind of observer (even when the frame is inertial) looking over a particle moving along a geodesic the "inertial proper acceleration" can only be defined in an infinitesimally small interval of the trajectory (or better say, in a sufficiently small region in which EP holds). But you can create a coordinate system around a timelike geodesic wherein such inertial proper acceleration exists along the entire of geodesic but now we're narrowing down the list of our observers to a "special observer" located somewhere in the neighbourhood of the trajectory.

AB

Edit: I gave a clear answer to your third question again, I guess. So the quote itself does not show my intention here.

 

[Dale]

The equation is shouting that it is present here.

Since you are unwilling to clarify your position then I will make my best guess as to what you mean. Feel free to correct my guess of your intention since I am a notoriously poor mind-reader and have already asked you to do so 3 times now. Considering the abysmal quality of your writing in the future you should be more helpful in providing clarification when asked.

You believe that in general an object is inertial iff
$$\frac{d^2 x^a}{ds^2}=0$$

This is can be shown to be false simply by considering an inertial object in inertial spherical coordinates where even though the object is inertial
$$\frac{d^2 x^a}{ds^2}\neq 0$$

The correct equation would be that in general an object is inertial iff
$$\nabla_{\tau}U^{\mu}=\frac{dU^{\mu}}{d\tau}+\Gamma^{\mu}_{\gamma\nu}U^{\gamma}U^{\nu}=0$$

 

[Altabeh]

AN extended body is just a collection of particles. How could a particle at one point be accelerated relative to another particle without itself having some degree of proper acceleration??

The statement itself is a little bit ambigious as you're proposing though it can be put in a better mold by taking a swipe at another explanation of the problem. Actually here you got it right that particles have to get a (very small) degree of proper acceleration to accelerate away along the geodesic but we don't really need to go far from the "sufficient" region wherein Equivalence Principle would generally hold in order to describe the chasm between the proper accelerations. Such discrepancies have only one way in GR to be understood unambiguously and that is the fact that by transforming into a locally inertial coordinate system we are not actually getting a Minkowskian spacetime along the geodesic in the small region but only at some point A this can be satisfied; around the point generally we must get

$$\frac{d^2x^a}{ds^2}\approx 0$$

which holds only in the "extended" points lying in the neighbourhood of A and the more this extension goes, the more we advance towards the outside of the "sufficient" region mentioned above that would no longer provide us with a good reason for how such differences in "proper" accelerations happen (we are no longer having such thing as "proper" acceleration outside of the "sufficient" region becasue there the frame cannot be asuumed as being inertial) So supposing that an extended body here is the one which fits within this region, we are able to say that your reasoning is completely logical.

AB

 

[Altabeh]

Since you are unwilling to clarify your position then I will make my best guess as to what you mean. Feel free to correct my guess of your intention since I am a notoriously poor mind-reader and have already asked you to do so 3 times now. Considering the abysmal quality of your writing in the future you should be more helpful in providing clarification when asked.

You believe that in general an object is inertial iff
$$\frac{d^2 x^a}{ds^2}=0$$

This is can be shown to be false simply by considering an inertial object in inertial spherical coordinates where even though the object is inertial

There is no specification of any coordinate system here and of course having countlessly been said in this thread before, this is only correct in a very small interval of spacetime along the geodesic. But okay though you are now trying to battle with thousands of Physicists, which later I'll enucleate the precise names with their own approaches to the problem here, I'm enthusiasticly waiting for a proof of why such equation is false "by considering an inertial object in inertial spherical coordinates".

AB

 

[Altabeh]

The correct equation would be that in general an object is inertial iff
$$\nabla_{\tau}U^{\mu}=\frac{dU^{\mu}}{d\tau}+\Gamma^{\mu}_{\gamma\nu}U^{\gamma}U^{\nu}=0$$

It is so interesting that the fundamental approach to deriving this formula uses the famous "locally inertial" coordinates. Besides, any particle following geodesics in GR would only be seen as moving inertially in an infinitesimaly small region (yet I don't want to specify the coordinates here by considering all observers except a non-Fermi observer).

AB

 

[Dale]

this is only correct in a very small interval of spacetime along the geodesic

Which is why it is not relevant to the OP's question.

 

[Altabeh]

Which is why it is not relevant to the OP's question.

Why are you insisting on such nonsense? You can't even prove my assertation wrong even at one point by considering an inertial object in inertial spherical coordinates because you made a fallacy in your previous post.

AB

 

[yuiop]

That is not an equation. So again, I ask: in general what equation determines if a worldline is inertial? The answer should be something like "in general a worldline is inertial if and only if A=B".

We have shown in other threads that we can generally (in the Schwarzschild metric anyway) calculate the predicted gravitational acceleration ($a_p$) of any particle according to any observer. For example the predicted radial acceleration of a particle according to a stationary (non inertial) local observer was given as:

$$a_p = - \frac{M}{r^2}\left(\frac{1-(dr '/dt')^2}{\sqrt{1-2M/r}}\right)$$

(See https://www.physicsforums.com/showpost.php?p=2747788&postcount=345)

If we consider proper acceleration as accounting for the difference between the predicted acceleleration ($a_p$) and the actual acceleration measured using the rulers and clocks of a local observer ($a_m$), then we can say that a particle has inertial motion if and only if $a_p - a_m = 0$ according to any given observer.

The above linked post only considers purely radial motion, but I am sure the general idea can be extended. I am also sure that this is not a complete answer, but it might suggest a way forward in the discussion.

 

[superkan619]

unambiguously "yes, it is inertial in all reference frames".

Any objections to this?

 

[yuiop]

Any objections to this?

I would go along with it. A particle is inertial if it proper acceleration is zero (as measured by an accelerometer attached to it). As far as I can tell it is not possible to transform away this physical fact.

 

[Dale]

I'm enthusiasticly waiting for a proof of why such equation is false "by considering an inertial object in inertial spherical coordinates".

You can't even prove my assertation wrong even at one point by considering an inertial object in inertial spherical coordinates because you made a fallacy in your previous post.

Without loss of generality consider the inertial worldline in spherical coordinates:

$$x=(t,r,\theta,\phi)=\left(t,\sqrt{t^2v^2+r_0^2},\frac{\pi}{2},atan(tv,r_0)\right)$$

Using your equation:

$$\frac{d^2 x^a}{ds^2}=\left(0, \frac{\gamma^2 r_0^2 v^2}{(t^2 v^2+r_0^2)^{3/2}},0, \frac{2 \gamma^2 r_0 t v^3}{(t^2 v^2+r_0^2)^2} \right) \neq 0$$

Further consider the non-inertial worldline
$$x=(t,r,\theta,\phi)=(t,r_0,\frac{\pi}{2},\omega t)$$

Using your equation:
$$\frac{d^2 x^a}{ds^2}=0$$

So your equation does not hold even for a local observer using spherical coordinates. It is not even valid locally in flat spacetime, and certainly not in general.

 

[yuiop]

A gravitational field does NOT change the frequency of a photon. It remains the same as it was when it was emitted, as seen by any fixed observer.

However, time runs at slightly different rates at different potentials. The fractional difference in the time rate for an object at two different positions is effectively the same as the difference in potential energy divided by the total energy.

This means that if you compare observations of the same photon from two locations at different potentials, its frequency appears to differ because of the difference in clock rates.

You have to be careful about how the above is understood.

Let us say that some physical process such as the transition of an electron gives off a photon emmision of a characteristic frequency (f). To an observer that is local and at rest with the emmision event, the characteristic transition frequency will always be f. If the emmision occurs lower down than the observer, then the observer measures the frequency of the photon to be lower than f, at the reception event. Now if we interpret this physically as the frequency of the photon "really" remaining constant, but only being measured to be lower because of the difference between clock rates at the emission location and the reception location, then we have to understand that the same logic means the speed of light is "really" slower lower down. It then comes down to how we define "real". If we define "real" as being defined by what we actually measure (which the way real is usually defined in SR and GR) then the frequency of a photon "really" gets slower as it climbs out of a gravitational well. If you define real as what an intelligent being would conclude is happening "behind the scenes" to explain what is measured, then the frequency of the photon is "really" constant as it rises. The modern interpretation of Relativity does not consider what is happening "behind the scenes", only what is measured.

 

[yuiop]

...
3)When looked upon from Earth does a falling stone makes non-inertial frame of reference?

It depends upon how your question is interpreted. If by "When looked upon from Earth.." you meant the pointof view of an observer standing on the surface of the Earth watching the falling stone, then the answer is no, because the observer standing on the surface of the Earth has non zero proper acceleration as is therefore not an inertial observer. If you meant the point of view of an observer co-free-falling with the stone, then the co-free-falling observer is "locally" in an inertial reference frame. As Altabeh suggested, the local vicinity of the free falling observer is Minkowskian, but further away from the free falling observer, it becomes increasingly obvious that the observer is not in flat space. As for the stone itself, a very small free falling stone is on average, an inertially moving object (rather than an inertial frame of reference), but as Passionflower has already pointed out, parts of a very large stone furthest away from the centre of mass of the stone, are not inertially moving.

 

[Jonathan Scott]

If we define "real" as being defined by what we actually measure (which the way real is usually defined in SR and GR) then the frequency of a photon "really" gets slower as it climbs out of a gravitational well. If you define real as what an intelligent being would conclude is happening "behind the scenes" to explain what is measured, then the frequency of the photon is "really" constant as it rises. The modern interpretation of Relativity does not consider what is happening "behind the scenes", only what is measured.

No, I don't buy that at all.

If you think that it could ever make sense to say that a wave is slowing in frequency as it rises, where do the extra cycles go? For example, replace the photon's path with a rod rotating about its axis. If the whole rod did not rotate at the same frequency, it would gradually get more and more twisted. Observers at different potentials may consider that the whole rod is rotating at slightly different speeds, depending on their potential, but it doesn't make sense to suggest that the lower end is rotating at a different frequency from the upper end.

 

[yuiop]

No, I don't buy that at all.

If you think that it could ever make sense to say that a wave is slowing in frequency as it rises, where do the extra cycles go? For example, replace the photon's path with a rod rotating about its axis. If the whole rod did not rotate at the same frequency, it would gradually get more and more twisted. Observers at different potentials may consider that the whole rod is rotating at slightly different speeds, depending on their potential, but it doesn't make sense to suggest that the lower end is rotating at a different frequency from the upper end.

Modern relativity is not about what makes sense. If you continue down that road, you end up in the LET camp It is about what you measure. That is the reality. If you can't measure it, it is not real. Logic, rationality and common sense does not come into it.

Now let us embelish your example. You did not specify which axis of the rod to rotate about, but I assume you mean one of the short axes. Let us say we have a huge ferris wheel and your rod is represented by two opposite spokes. At the top and bottom of the wheel are observers with clocks and rulers calibrated in the normal way. The obersers at the top of the wheel measure the instantaneous tangential velocity at the top of the wheel to be 0.2c and the observers at the base of the wheel measure the instantaneous tangential velocity of the bottom of the same wheel to be 0.4c. The observers at the bottom are wondering how this can be? How can the bottom of the wheel be rotating slower than the top of the wheel without the wheel breaking apart? The observers at the top realize what is happening and shout down "Hey you idiots down there! Of course the bottom appears to be moving faster according to your measurements, because your clocks are running slower than our clocks." So the top and bottom observers decise to use a different method to synchronise the clocks and use a clock at the top as a master clock and synchronise all clocks so that they run at the same rate. Signals sent from the top at a frequency of once per second are now measured as arriving with a frequency of once per second by the lower observers with their alternatively synchronised clocks. The same is true the other way. Signals sent once per second from the bottom now arrive once per second at the top. WIth this new method of synchronisation the tangential velocity at the top is now the same as the tangential velocity at the bottom. This seems to explain why the wheel does not fall apart. Is this a better description of reality? Some would say no. If the observers at the bottom now try to measure the speed of light with the newly synchronised clocks, they will find the speed of light is locally less than c. Others would say that the speed of light "really" is slower, lower down in a gravitational field, clocks "really" run slower and all physical processes "really" happen slower, lower down. Indeed the slower speed of light can even be measured by using the clock synchronisation method I have just described.

So we have two choices of how we describe reality:

1) Buy that the frequency of a photon "really" is constant as it rises upwards, at the cost of acknowledging that the speed of light is not "really" constant in a gravitational field.

2) Buy that the frequency of a photon changes as it rises, at the cost of claiming clocks do not "really" run slower lower down and acknowledging that certain things do not "make sense."

You seem to have bought into the first choice.

 

[Jonathan Scott]

...
So we have two choices of how we describe reality:

1) Buy that the frequency of a photon "really" is constant as it rises upwards, at the cost of acknowledging that the speed of light is not "really" constant in a gravitational field.

2) Buy that the frequency of a photon changes as it rises, at the cost of claiming clocks do not "really" run slower lower down and acknowledging that certain things do not "make sense."

You seem to have bought into the first choice.

I was actually talking about something like a thin rod or rigid cable connecting a point of low potential and a higher point and rotating about a line along its core, although in many ways this has similar results, in that the frequency at the top and bottom cannot actually differ as observed by anyone observer.

In a gravitational field, the speed of light as observed from another location does indeed vary with potential. The way in which it varies depends on the coordinate system (and may be different in different directions), because there is no way to define a flat space coordinate system which both matches local time and distant space. This is like comparing distances on maps. If I have a map of the Earth centered on my own location, it indicates distances to scale and angles correctly locally, but as the Earth is curved, it cannot do so accurately at points further away.

 

[superkan619]

I would go along with it. A particle is inertial if it proper acceleration is zero (as measured by an accelerometer attached to it). As far as I can tell it is not possible to transform away this physical fact.

This means that an accelerometer at rest gives the value of g at that point and a falling accelerometer records zero acceleration. I think it is a better and intuitive answer to my question.

 

[Austin0]

Modern relativity is not about what makes sense. If you continue down that road, you end up in the LET camp It is about what you measure. That is the reality. If you can't measure it, it is not real. Logic, rationality and common sense does not come into it. .

Hi kev

Couldn't it be said that SR came into being as a rational attempt to make sense of the illogical invariance of light as indicated by Maxwell and MM ??

In doing this the system itself is rational and logically consistent even if some of its implications were counter-intuitive and may actually be inconsistent.

GR was derived through logic and conceptualization, not through mathematical induction from measured reality. That came later.

To take the position that only what is measured is real, and only what can be mathematically described from those measurements is reality , I think you either deny an objective reality [a metaphysical view] or you accept a false view of reality because we can't really believe that description has any real correspondance to that reality.

How can it be a positive position to think as long as the math works out, no further logical thought is required. Is in fact discouraged. ?

[QUOTESo we have two choices of how we describe reality:

1) Buy that the frequency of a photon "really" is constant as it rises upwards, at the cost of acknowledging that the speed of light is not "really" constant in a gravitational field.

2) Buy that the frequency of a photon changes as it rises, at the cost of claiming clocks do not "really" run slower lower down and acknowledging that certain things do not "make sense."

You seem to have bought into the first choice.[/QUOTE]

1) WHy should we think that light is "really" constant in a gravitational field?

To be consistent with the 2nd P??

2) If you accept clocks "don't really run slower" at a lower potential this has other implications .eg in the "real" results of the twins journey.

Making sense does not only apply to our physical conceptions and assumptions but also to consistency and logic between those assumptions, doesn't it?

BTW I wouldn't say I have bought into the first choice but so far it seems more consistent with the other applications of time dilation.

 

[Austin0]

You have to be careful about how the above is understood.

Let us say that some physical process such as the transition of an electron gives off a photon emmision of a characteristic frequency (f). To an observer that is local and at rest with the emmision event, the characteristic transition frequency will always be f. If the emmision occurs lower down than the observer, then the observer measures the frequency of the photon to be lower than f, at the reception event. Now if we interpret this physically as the frequency of the photon "really" remaining constant, but only being measured to be lower because of the difference between clock rates at the emission location and the reception location, then we have to understand that the same logic means the speed of light is "really" slower lower down. It then comes down to how we define "real". If we define "real" as being defined by what we actually measure (which the way real is usually defined in SR and GR) then the frequency of a photon "really" gets slower as it climbs out of a gravitational well. If you define real as what an intelligent being would conclude is happening "behind the scenes" to explain what is measured, then the frequency of the photon is "really" constant as it rises. The modern interpretation of Relativity does not consider what is happening "behind the scenes", only what is measured.

Are you saying that the relative time dilation at different potentials is not measured?
Is not real?

What kind of measurements are possible while in transit that would make this interpretation "real ".?

Isn't the measured dilation differential by your definition "real" and the interpretation of slowing photons "behind the scenes" ?

 

[Altabeh]

It depends upon how your question is interpreted. If by "When looked upon from Earth.." you meant the pointof view of an observer standing on the surface of the Earth watching the falling stone, then the answer is no, because the observer standing on the surface of the Earth has non zero proper acceleration as is therefore not an inertial observer. If you meant the point of view of an observer co-free-falling with the stone, then the co-free-falling observer is "locally" in an inertial reference frame. As Altabeh suggested, the local vicinity of the free falling observer is Minkowskian, but further away from the free falling observer, it becomes increasingly obvious that the observer is not in flat space. As for the stone itself, a very small free falling stone is on average, an inertially moving object (rather than an inertial frame of reference), but as Passionflower has already pointed out, parts of a very large stone furthest away from the centre of mass of the stone, are not inertially moving.

Exactly.

AB

 

[Altabeh]

Without loss of generality consider the inertial worldline in spherical coordinates:

$$x=(t,r,\theta,\phi)=\left(t,\sqrt{t^2v^2+r_0^2},\frac{\pi}{2},atan(tv,r_0)\right)$$

Using your equation:

$$\frac{d^2 x^a}{ds^2}=\left(0, \frac{\gamma^2 r_0^2 v^2}{(t^2 v^2+r_0^2)^{3/2}},0, \frac{2 \gamma^2 r_0 t v^3}{(t^2 v^2+r_0^2)^2} \right) \neq 0$$

Further consider the non-inertial worldline
$$x=(t,r,\theta,\phi)=(t,r_0,\frac{\pi}{2},\omega t)$$

Using your equation:
$$\frac{d^2 x^a}{ds^2}=0$$

So your equation does not hold even for a local observer using spherical coordinates. It is not even valid locally in flat spacetime, and certainly not in general.

Completely nonsense and irrelevant. What is your reason that $$x=(t,r,\theta,\phi)=\left(t,\sqrt{t^2v^2+r_0^2},\frac{\pi}{2},atan(tv,r_0)\right)$$ is inertial and what does your "inertial" mean here? Physics is not about numerology but rather precise derivations and equations. Here you have not even tried to show that how these so-called "inertial" or "non-inertial" observers manage to observe a particle following a geodesic in a very small region along the trajectory which requires you to find a metric transformation of the form $$g_{ab}\rightarrow \eta_{ab}$$ (locally inertial condition) around the trajectory so automatically the Christoffel symbols vanish and thus $$\frac{d^2 x^a}{ds^2}=0$$.

No such things as your nonsense "calculations" are cared about in case there are great proofs and derivations. If you want to learn what is going on and to know where your "fallacy" arises from, see the following sources in order:

Papapetrou "A. Lectures on GR", 1974, pp. 56-58.
Weinberg S. Gravitation and Cosmology: principles and applications of the GR, 1972, 70-73.
Schutz B. F. "A first course in GR", 1985, pp. 154-156, 158-160.

AB

 

[yuiop]

This means that an accelerometer at rest gives the value of g at that point and a falling accelerometer records zero acceleration. I think it is a better and intuitive answer to my question.

Yes, that is proper acceleration and the definition of inertial motion, in a nutshell.

 

[superkan619]

This means that an accelerometer at rest gives the value of g at that point and a falling accelerometer records zero acceleration. I think it is a better and intuitive answer to my question.

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.