Split Thread: Pendular Motion (And Relativity)

[SDG]

Originally Posted by Dave Rogers

Oh yes, indeed you are. Because an accelerometer does not, in fact, measure an absolute acceleration; it measures the force between two objects, one of which is accelerating the other. An accelerometer mounted on the ISS will not measure an acceleration, but this does not mean that the ISS is not accelerating; it simply means that all parts of the accelerometer are being accelerated by the same amount, and hence no differential measurement is possible. But it is clearly not the case that the ISS is in an inertial reference frame, because centrifugal and Coriolis effects can be observed while moving around inside it; these are only observed in a rotating frame of reference.



It exhibits your continued inability to understand what is meant by "inertial reference frame."

Dave

The ISS orbit does not mean the ISS will rotate as well. These are two separate things.
I said 'non-rotating ISS orbiting the Earth'.
How we would do transformation to the non-rotating reference frame?
Where are the centrifugal and Coriolis forces coming from?

Edit: the bold part.
That's the whole point of curved space-time. The ISS is not accelerating locally in its world-line/geodesic.
SDG

[Dave Rogers]

Originally Posted by SDG

I said 'non-rotating ISS orbiting the Earth'.
How we would do transformation to the non-rotating reference frame?

The simplest approach would be to do a double transformation; firstly into a rotating reference frame centred on the centre of the Earth, then from there into a counter-rotating reference frame centred on the centre of the ISS.

Originally Posted by SDG

Where are the centrifugal and Coriolis forces coming from?

The centrifugal force will come from the difference in distance from the centres of the two frames of reference, and will in general be non-zero. Coriolis forces will cancel out to first order, but as soon as two objects ain constant relative re sufficiently widely separated in the ISS frame they will experience different gravitational vectors and hence will accelerate differently, ultimately leading to different motion vectors and hence differential Coriolis forces.

You know, what's really bizarre is your insistence that an ISS in intergalactic space using its engines to follow a circular path defines an inertial frame of reference. It's a continuously accelerating frame with the acceleration vector continuously changing; it's hard to conceive of a greater departure from an inertial frame of reference, it's not even a local inertial frame.

Dave

[Dave Rogers]

Originally Posted by SDG

Edit: the bold part.
That's the whole point of curved space-time. The ISS is not accelerating locally in its world-line/geodesic.
SDG

At its centre, no; so all you can measure with accelerometers is the local curvature of space-time, which reduces in the Newtonian approximation to tidal effects.

Your ISS in intergalactic space, however, is accelerating locally, which accounts for the difference between the measurements made locally in the two. Why you think this is a problem is beyond me.

Dave

[Hellbound]

The frames always get you.

Seriously, 99% of the "relativity is wrong" proponents come down to mis-understanding or mis-applying reference frames.

Sad to see so many people get framed.

[Myriad]

Originally Posted by SDG

The centrifugal and Coriolis forces are added/subtracted to the gravitational force.
I guess you missed the discussion on the last couple of pages.
SDG

The discussion on the last couple of pages concerned a rotating wheel attached to another wheel. Not one in free fall.

[SDG]

Originally Posted by Dave Rogers

The simplest approach would be to do a double transformation; firstly into a rotating reference frame centred on the centre of the Earth, then from there into a counter-rotating reference frame centred on the centre of the ISS.



The centrifugal force will come from the difference in distance from the centres of the two frames of reference, and will in general be non-zero. Coriolis forces will cancel out to first order, but as soon as two objects ain constant relative re sufficiently widely separated in the ISS frame they will experience different gravitational vectors and hence will accelerate differently, ultimately leading to different motion vectors and hence differential Coriolis forces.

It appears to me this math gymnastic is not going to achieve invalidating the definition that the free-falling, non-rotating ISS is an inertial observer.
There is a centrifugal force at orbiting ISS but it cancels out with the gravitational acceleration locally and that's the reason the centrifugal acceleration cannot be measured with an accelerometer that does not rotate inside the ISS.
The only way to expose the centrifugal force is through the proposed rotation of the wheel with an accelerometer on it.
The local cancellation is a basic foundation of the inertial observer definition.
Not only that, it goes further:
https://youtu.be/E43-CfukEgs?t=222
"... they are not falling, they are standing still, there is no force acting on them at all."

The tidal forces are different and there is no issue there from my point of view.

Quote:

You know, what's really bizarre is your insistence that an ISS in intergalactic space using its engines to follow a circular path defines an inertial frame of reference. It's a continuously accelerating frame with the acceleration vector continuously changing; it's hard to conceive of a greater departure from an inertial frame of reference, it's not even a local inertial frame.

Dave

This is a misunderstanding.
I talked about two instances of the ISS in the intergalactic space.
The first one without any motion and the second one to simulate exactly the same trajectory as the ISS orbiting the Earth.
The second instance is to show that the centrifugal acceleration does not have opposing gravitational acceleration to cancel it out.
I never claimed that the second instance ISS is a local inertial frame.
I know better
SDG

[SDG]

Originally Posted by Dave Rogers

At its centre, no; so all you can measure with accelerometers is the local curvature of space-time, which reduces in the Newtonian approximation to tidal effects.

Do you agree that we can measure gravitational acceleration through the proposed rotating wheel experiment?
Collecting multiple data points, interpolating, ...

Quote:

Your ISS in intergalactic space, however, is accelerating locally, which accounts for the difference between the measurements made locally in the two. Why you think this is a problem is beyond me.

Dave

As I said, there were two instances of the ISS in the intergalactic space.
SDG

[SDG]

Originally Posted by Myriad

The discussion on the last couple of pages concerned a rotating wheel attached to another wheel. Not one in free fall.

That was to introduce the ISS rotating wheel experiment, to demonstrate where the small wheel torque comes from based on the two rotations.
The same logic applies to the ISS rotating wheel with gravity in place on top of that.
SDG

[Dave Rogers]

Originally Posted by SDG

It appears to me this math gymnastic is not going to achieve invalidating the definition that the free-falling, non-rotating ISS is an inertial observer.
There is a centrifugal force at orbiting ISS but it cancels out with the gravitational acceleration locally and that's the reason the centrifugal acceleration cannot be measured with an accelerometer that does not rotate inside the ISS.
The only way to expose the centrifugal force is through the proposed rotation of the wheel with an accelerometer on it.
The local cancellation is a basic foundation of the inertial observer definition.
Not only that, it goes further:
https://youtu.be/E43-CfukEgs?t=222
"... they are not falling, they are standing still, there is no force acting on them at all."

The tidal forces are different and there is no issue there from my point of view.


This is a misunderstanding.
I talked about two instances of the ISS in the intergalactic space.
The first one without any motion and the second one to simulate exactly the same trajectory as the ISS orbiting the Earth.
The second instance is to show that the centrifugal acceleration does not have opposing gravitational acceleration to cancel it out.
I never claimed that the second instance ISS is a local inertial frame.
I know better
SDG

OK, let's discuss local inertial frames, a concept you just recently introduced into the thread so you could pretend you'd been talking about it all along.

Locally, there is a single point in the frame of the non-rotating ISS that qualifies as a location for an inertial observer; it is at the exact centre of gravity of the ISS. At this point, and this point alone, the Earth's gravitation exerts exactly the force required to accelerate the observer along a circular path. It's fairly trivial to show that this is the case: at any point not on a line between the CG of the ISS and the centre of the Earth, there is a component of gravitational force normal to that line, so the gravitational force vector is not lined up to the acceleration vector, resulting in a residual force on an object at any point not on the line; and at any point on the line between the centres, the gravitational force is either greater or lesser than the acceleration of the frame, again resulting in a residual force. So we have a frame containing a single point where an inertial observer can be placed.

However, your experiment with a wheel on the ISS cannot take place at this point. The wheel has non-zero radius, and so whatever ist position or orientation relative to the ISS, there can be at most a single point on its circumference that may be considered locally inertial. All other points are non-inertial; they are not moving in free fall, but must be held in place by some other force [1]. So, by measuring accelerations on the circumference of a wheel and comparing them with those in the stationary ISS in intergalactic space - which is in a global inertial frame of reference - you are comparing inertial with non-inertial measurements. As I've pointed out too many times to count, you're confusing inertial with non-inertial frames.

To answer the next post: Yes, we can indirectly measure gravity through the proposed rotating wheel experiment. This is because it is collecting data at multiple points where its motion is not inertial.

Your next trick, of course, is to claim that the wheel is so small that it can be approximated to a point, coinciding with the inertial point. I'll just take care of that right now by countering that this approximation is exactly identical to the limit that the wheel is so small that variations in local acceleration were too small to measure the local gravitational gradient. If the wheel is effectively all within a local inertial frame, then you can't measure gravity; if it's large enough that you can measure gravity, that's because it's no longer effectively all within an inertial frame.

So the answer is what it has been all along: relativity is self-consistent, but you're trying to give the appearance that it isn't by defining experiments in such a way that they take non-inertial measurements, which you then claim to be inertial so as to give the illusion of a paradox.

Dave

[SDG]

Originally Posted by Dave Rogers

OK, let's discuss local inertial frames, a concept you just recently introduced into the thread so you could pretend you'd been talking about it all along.

Locally, there is a single point in the frame of the non-rotating ISS that qualifies as a location for an inertial observer; it is at the exact centre of gravity of the ISS. At this point, and this point alone, the Earth's gravitation exerts exactly the force required to accelerate the observer along a circular path. It's fairly trivial to show that this is the case: at any point not on a line between the CG of the ISS and the centre of the Earth, there is a component of gravitational force normal to that line, so the gravitational force vector is not lined up to the acceleration vector, resulting in a residual force on an object at any point not on the line; and at any point on the line between the centres, the gravitational force is either greater or lesser than the acceleration of the frame, again resulting in a residual force. So we have a frame containing a single point where an inertial observer can be placed.

However, your experiment with a wheel on the ISS cannot take place at this point. The wheel has non-zero radius, and so whatever ist position or orientation relative to the ISS, there can be at most a single point on its circumference that may be considered locally inertial. All other points are non-inertial; they are not moving in free fall, but must be held in place by some other force [1]. So, by measuring accelerations on the circumference of a wheel and comparing them with those in the stationary ISS in intergalactic space - which is in a global inertial frame of reference - you are comparing inertial with non-inertial measurements. As I've pointed out too many times to count, you're confusing inertial with non-inertial frames.

To answer the next post: Yes, we can indirectly measure gravity through the proposed rotating wheel experiment. This is because it is collecting data at multiple points where its motion is not inertial.

Your next trick, of course, is to claim that the wheel is so small that it can be approximated to a point, coinciding with the inertial point. I'll just take care of that right now by countering that this approximation is exactly identical to the limit that the wheel is so small that variations in local acceleration were too small to measure the local gravitational gradient. If the wheel is effectively all within a local inertial frame, then you can't measure gravity; if it's large enough that you can measure gravity, that's because it's no longer effectively all within an inertial frame.

So the answer is what it has been all along: relativity is self-consistent, but you're trying to give the appearance that it isn't by defining experiments in such a way that they take non-inertial measurements, which you then claim to be inertial so as to give the illusion of a paradox.

Dave

Dave,
thank you for your excellent post!
The discussion is getting to the 'next level'

Let us imagine shielded box with a hydrogen atom floating around inside, electron proton motion, their orbit around the barycenter.
We place an inertial observer into the barycenter.
Is the electron proton orbit going to be same for the intergalactic space inertial observer compared to the orbiting ISS inertial observer?
Can we agree that the electron and proton will be exposed to the centrifugal and Coriolis forces on the ISS?

How about an inertial observer at a barycenter of two points orbiting around each other with a radius of a couple Planck lengths?
Intergalactic space... ISS...
Are they going to observe the same motion? No!

The gravity causes a circular motion.
This circular motion is what 'kills' the inertial observer.
The consequences of this circular motion are the 'fictitious' forces.

We are left with a conclusion that the intergalactic inertial observer does not see the same response as the ISS inertial observer.
The inertial observers are not equal.
When you say: "... that's because it's no longer effectively all within an inertial frame." ... agreed, that's how I see it as well.

The inertial observer is an approximation. The same goes for the equivalence principle.
The equivalence principle stands on an imaginary uniform gravitational field.
That's just unrealistic. This is exactly in line with what you were describing above - 'V' shape of the gravitational field.
The relativity cannot be exact.

When you say: "... At this point, and this point alone... " I agree with what you say here.
Essentially the inertial observer point is rendered useless because once we add an interaction with anything outside of this point we have an accelerated behavior and there are no two inertial observers that are identical, that they would see the same response.
In other words an inertial observer is watching an accelerated motion around him. Even the closest particle to the inertial observer has a different world-line/geodesic.
There are no two particles that can have the same world-line/geodesic.
SDG

[Elagabalus]

Quote:

electron proton motion

Hmmmmmm. Let me think about this. Something about this in my Chemistry 101 class... something about not being able to predict something. It's like it's right on the tip of my tongue but just can't visualize it. If I go to look at it ... Nah, it's gone.

[Dave Rogers]

This is no more than the philosophical splitting of non-existent hairs. The value of a physical quantity at a specific point is a mathematically valid concept, notwithstanding the impossibility of a real world object occupying a point of zero size. What you seem to be trying to do is come up with a GR equivalent of Zeno's Paradox, and then use this to claim GR is not valid. Unfortunately, the outstanding feature of GR is that it actually works, and has yet to fail any properly constructed test of its validity. While this continues to be the case, and/or until something simpler comes along that reproduces the same level of accuracy, it will continue to be the best model of mechanics we have and will be used as such.

If you think you can come up with something better, have at it. Until you can, it's futile to try to overturn one of the greatest scientific theory of the age on the basis that valid mathematical assumptions cannot be precisely mapped onto real world physical structures.

Dave

[Myriad]

Originally Posted by Dave Rogers

This is no more than the philosophical splitting of non-existent hairs. The value of a physical quantity at a specific point is a mathematically valid concept, notwithstanding the impossibility of a real world object occupying a point of zero size. What you seem to be trying to do is come up with a GR equivalent of Zeno's Paradox, and then use this to claim GR is not valid. Unfortunately, the outstanding feature of GR is that it actually works, and has yet to fail any properly constructed test of its validity. While this continues to be the case, and/or until something simpler comes along that reproduces the same level of accuracy, it will continue to be the best model of mechanics we have and will be used as such.

If you think you can come up with something better, have at it. Until you can, it's futile to try to overturn one of the greatest scientific theory of the age on the basis that valid mathematical assumptions cannot be precisely mapped onto real world physical structures.

Dave

Let me also point out that the full four dimensional math of GR makes consistent predictions for phenomena in both a uniform gravitational gradient, the gravitational field around a dense spherical object, and any other gravitational field we care to define. That different gravitational fields result in different predictions in detail is well known and completely appropriate. It means that when we want to evaluate cases that are simpler due to a high degree of symmetry (for instance, for illustrative examples when teaching the rudiments of the theory) we have to choose exactly how to simplify, and/or accept slightly incomplete results.

The fact that nearly spherically symmetrical gravitational fields around dense spherical objects are somewhat easer to come by in nature than perfectly uniform 1-D gravitational gradients is irrelevant. And in actual fact, locally the earth's surface is close enough to the latter for most purposes (and can locally be made even more so, if needed, by proper placement of additional compensating masses).

[Reality Check]

Originally Posted by SDG

Let us imagine shielded box with a hydrogen atom floating around inside, electron proton motion, their orbit around the barycenter. ....

Oh dear SDG, another post with invalid physics and what you imagine in it!

An electron orbiting a proton was discarded about a century ago because that was physically impossible. An orbiting electron is accelerating and emits Lamor radiation that makes a classic atom unstable.

By definition, inertial observers agree on their measurements subject to special relativity. It is non-inertial observers that disagree with inertial observers and where general relativity has to be used.
Inertial frame of reference
Non-inertial reference frame

No one thinks that inertial observer are totally equal because special relativity exists!
No one thinks that being a specific inertial observer makes orbits vanish. An ISS inertial observer can observe that they are orbiting. An intergalactic inertial observer will observe that the ISS is also orbiting.

Originally Posted by SDG

The equivalence principle stands on an imaginary uniform gravitational field.
That's just unrealistic.

That is just wrong, SDG. This is the equivalence principle which has no "imaginary uniform gravitational field" in it.
We know that a real gravitational field is uniform for practical purposes on small scales. This is expressed mathematically in the equivalence principle as a non-uniform gravitational filed that is locally uniform.

Quote:

Here "local" has a very special meaning: not only must the experiment not look outside the laboratory, but it must also be small compared to variations in the gravitational field, tidal forces, so that the entire laboratory is freely falling.

[Dancing David]

Originally Posted by Elagabalus

Hmmmmmm. Let me think about this. Something about this in my Chemistry 101 class... something about not being able to predict something. It's like it's right on the tip of my tongue but just can't visualize it. If I go to look at it ... Nah, it's gone.

The Hindenburg Principle, where people trying to disprove some aspect of 'relativity' go down in flames?

[SDG]

Originally Posted by Dave Rogers

This is no more than the philosophical splitting of non-existent hairs. The value of a physical quantity at a specific point is a mathematically valid concept, notwithstanding the impossibility of a real world object occupying a point of zero size. What you seem to be trying to do is come up with a GR equivalent of Zeno's Paradox, and then use this to claim GR is not valid. Unfortunately, the outstanding feature of GR is that it actually works, and has yet to fail any properly constructed test of its validity. While this continues to be the case, and/or until something simpler comes along that reproduces the same level of accuracy, it will continue to be the best model of mechanics we have and will be used as such.

If you think you can come up with something better, have at it. Until you can, it's futile to try to overturn one of the greatest scientific theory of the age on the basis that valid mathematical assumptions cannot be precisely mapped onto real world physical structures.

Dave

Dave,
Yes, GR works as an approximation. It is not accurate, it cannot be reconciled with QM that has to handle the required accuracy with statistics because the unrealistic mathematical concept does not work in real life.

Is there a physics book or a physics paper that shows the orbit can be detected by 'fictitious' forces?
If not then what we discussed here is completely new thing to our understanding of physics that we never considered before.
SDG

[Dave Rogers]

Originally Posted by SDG

Dave,
Yes, GR works as an approximation.

No, it works as a model of reality. It describes and predicts aspects of reality to a better accuracy than any other model known, and to a better accuracy, so far, than we can model.

Originally Posted by SDG

It is not accurate,

That's an appeal to perfection. No, GR is not believed to be a perfect theory. However, it's by far the best available, and will be used until any failures are actually revealed or a better one is available.

Originally Posted by SDG

it cannot be reconciled with QM that has to handle the required accuracy with statistics because the unrealistic mathematical concept does not work in real life.

The incompatibility of GR and QM is rather well known. It's probably the most important issue in theoretical physics. Congratulations for noticing it, but you're not the first.

Originally Posted by SDG

Is there a physics book or a physics paper that shows the orbit can be detected by 'fictitious' forces?

I have no idea, but it's fairly obvious that non-inertial motion can be detected by a sufficient number of measurements over a large enough region to a sufficiently high accuracy.

Originally Posted by SDG

If not then what we discussed here is completely new thing to our understanding of physics that we never considered before.

It's certainly not a new thing to my understanding of physics, since I remember having a similar conversation with my brother back in the 1980's. Coriolis described fictitious forces in 1835, and he wasn't the first to note that they exist. Sorry, but there actually isn't anything striking or new here.

Dave

[SDG]

Originally Posted by Dave Rogers

No, it works as a model of reality. It describes and predicts aspects of reality to a better accuracy than any other model known, and to a better accuracy, so far, than we can model.



That's an appeal to perfection. No, GR is not believed to be a perfect theory. However, it's by far the best available, and will be used until any failures are actually revealed or a better one is available.



The incompatibility of GR and QM is rather well known. It's probably the most important issue in theoretical physics. Congratulations for noticing it, but you're not the first.



I have no idea, but it's fairly obvious that non-inertial motion can be detected by a sufficient number of measurements over a large enough region to a sufficiently high accuracy.



It's certainly not a new thing to my understanding of physics, since I remember having a similar conversation with my brother back in the 1980's. Coriolis described fictitious forces in 1835, and he wasn't the first to note that they exist. Sorry, but there actually isn't anything striking or new here.

Dave

Is there a physics book that describes the gravitational acceleration can be detected in space with other method than the tidal forces if there is no signal from the outside?
Do you have a link to somebody saying that?
SDG

[Dave Rogers]

Originally Posted by SDG

Is there a physics book that describes the gravitational acceleration can be detected in space with other method than the tidal forces if there is no signal from the outside?

You're confusing some concepts there. It's rotational motion that can in principle be detected from Coriolis forces. Gravitational attraction that doesn't result in rotation, in other words gravitational attraction in the same direction as or opposite direction to the motion of a body, can only be detected from tidal effects.

I suggest you start doing your own research at this point. The fact that I can't point you to a specific book laying something doesn't mean that it hasn't been said, or isn't widely understood.

Dave

[SDG]

Originally Posted by Dave Rogers

You're confusing some concepts there. It's rotational motion that can in principle be detected from Coriolis forces. Gravitational attraction that doesn't result in rotation, in other words gravitational attraction in the same direction as or opposite direction to the motion of a body, can only be detected from tidal effects.

I suggest you start doing your own research at this point. The fact that I can't point you to a specific book laying something doesn't mean that it hasn't been said, or isn't widely understood.

Dave

It does not appear to me this is the case. Let us assume the ISS falls straight down.
The accelerometers on the edge of the flat to gravity rotating wheel will measure a change of the falling acceleration on them without a need to measure tidal forces.
The accelerometer bodies are 'pinned' - they 'want to fly away', they are on a different 'geodesic' as the center of the wheel.
This tendency to fly away (geodesic) will be changing with the free fall when the wheel rotation is constant.
There is no need to measure the tidal forces at the different places on the ISS and compare the delta.

I have not found anything remotely close to the discussed subject anywhere so far.
Your feedback saying you cannot point me anywhere is valuable as well, thank you.
SDG

[a_unique_person]

Originally Posted by SDG

It does not appear to me this is the case. Let us assume the ISS falls straight down.
The accelerometers on the edge of the flat to gravity rotating wheel will measure a change of the falling acceleration on them without a need to measure tidal forces.
The accelerometer bodies are 'pinned' - they 'want to fly away', they are on a different 'geodesic' as the center of the wheel.
This tendency to fly away (geodesic) will be changing with the free fall when the wheel rotation is constant.
There is no need to measure the tidal forces at the different places on the ISS and compare the delta.

I have not found anything remotely close to the discussed subject anywhere so far.
Your feedback saying you cannot point me anywhere is valuable as well, thank you.
SDG

I think they are saying you can't get there from here. Your understanding of physics is so poor that you can't grasp the concepts they are presenting to you.

[Robin]

Originally Posted by SDG

Let us assume the ISS falls straight down.
The accelerometers on the edge of the flat to gravity rotating wheel will measure a change of the falling acceleration on them without a need to measure tidal forces.
The accelerometer bodies are 'pinned' - they 'want to fly away', they are on a different 'geodesic' as the center of the wheel.
This tendency to fly away (geodesic) will be changing with the free fall when the wheel rotation is constant.

Every part of the wheel will be tending to fly away from every other part of the wheel. Every part of the accelerometer will be tending to fly away from every other part of the accelerometer.

The accelerometers would record a constant acceleration towards the centre of the wheel, but I don't see why they would record the acceleration due to gravity.

[Dave Rogers]

Originally Posted by SDG

It does not appear to me this is the case. Let us assume the ISS falls straight down.
The accelerometers on the edge of the flat to gravity rotating wheel will measure a change of the falling acceleration on them without a need to measure tidal forces.

Try to express your ideas coherently. It's impossible from the above to understand what you actually mean. The accelerometers on the rotating wheel will measure a resultant of two forces:

(1) The centripetal acceleration causing them to move in a circular path in the frame of reference of the ISS; and,
(2) The infinitesmally small vector difference between the force of gravity at the centre of the ISS and that at the perimeter of the circel (which is, in fact, a tidal force).

Originally Posted by SDG

The accelerometer bodies are 'pinned' - they 'want to fly away', they are on a different 'geodesic' as the center of the wheel.
This tendency to fly away (geodesic) will be changing with the free fall when the wheel rotation is constant.

Again, not very clearly stated. Again, the only force other than centripetal they will experience is the difference in gravitational forces between different points - in other words, the tidal forces.

Originally Posted by SDG

There is no need to measure the tidal forces at the different places on the ISS and compare the delta.

Utterly wrong; the tidal force is what you're invoking here.

Originally Posted by SDG

Your feedback saying you cannot point me anywhere is valuable as well, thank you.

Well it damned well shouldn't be. The statement of one person that they are unable to point you to an instance of discussion of a narrowly defined issue in a very specific format is of no evidentiary value whatsoever.

Dave

[Robin]

SDG.

Also, when you say the wheel is falling "flat to gravity", do you mean that it is oriented (with respect to the surface to which it is falling) like a merry-go-round? Or oriented like a ferris wheel?

And if I was floating in a sealed chamber with no sight or sound of the outside, and had a wheel with accelerometers attached to the edges, what sort of readings of the accelerometer would tell me that I was in free fall towards a planet and not floating somewhere in intergalactic space? (Apart form tidal forces).

[Robin]

Suppose I am in a big oblong box and can't see or hear anything from outside. I am experiencing a normal earthlike gravity, but I want to know whether I am on the surface of a planet, or being accelerated evenly in a straight line through space, or else being steered around in a circle in space.

I could set a series of accelerometers along the ostensible ground and if the accelerations they show are all parallel (I am assuming I can have sufficiently sensitive accelerometers) then I am being accelerated in a straight line. If they converge in the ostensible upwards direction then I am being steered in a circle. If they diverge in the ostensible upwards direction then I am on the surface of a planet.

In addition accelerometers at the opposite ends of the room and constrained to the plane of the floor will record a constant acceleration towards the centre of the room if I am being steered around in a circle (because the orientation of the room will effectively be rotated to produce the acceleration always towards the 'floor').

If I am in the same room but experiencing weightlessness and I want to know whether I am floating in intergalactic space, or else falling towards some planets I could place accelerometers at opposite ends of the room and if they produce a slight acceleration away from a line through the centre mass of the room then I am falling towards a large body. If not then I might be in intergalactic space or else in a stable orbit around a planet.

If I were in a stable orbit around a planet I still might be able to tell with sufficiently sensitive accelerometers if they were placed evenly around the room and attached to a a wall they would all record a slight acceleration towards the centre of mass of the room.

(I am not convinced I could tell the difference between a stable and a declining orbit but I guess I wouldn't be wondering long).

All of these would fall into the category of 'signals from the outside' in that they are artifacts of the mass and distance of the body producing the gravity, or else artifacts of the force of the thrusters producing the acceleration.

I don't get what the accelerometers on a rotating platform would be telling us that is not a rather inefficient version of something I have just described, or some other 'signal from the outside'.