Gravity - an accelerating frame paradox

[ZirkMan]

If the equivalence principle is true then it means that the Earth's gravity field is a constantly accelerating frame of reference. In any accelerating frame of reference the direction of acceleration is always opposite to the direction of attraction.

That means that for all observers on the Earth's surface the Earth's surface (and with it the whole Earth) is constantly accelerating towards them. If this is true for all observers on its spherical surface how is it possible that the Earth doesn't explode (due to constant acceleration away from its center to all sides)? Is there a model of spacetime that explains it?
Or is there a mechanism of acceleration that doesn't require acceleration in space?

 

[K^2]

Because in curved space-time acceleration is not equal to the rate of change of velocity. So while the surface is constantly accelerating outwards at all points, none of these points actually move outwards. Hence, no explosion.

To demonstrate this, consider a person standing on a rotating platform. Wherever he's standing, his experience is consistent with the point he is standing at accelerating inwards towards the center of the platform. In fact, that point is accelerating towards the center of the platform. However, the platform is not imploding, and in this coordinate system, no point of the platform is moving.

 

[D H]

If the equivalence principle is true then it means that the Earth's gravity field is a constantly accelerating frame of reference. In any accelerating frame of reference the direction of acceleration is always opposite to the direction of attraction. That means that for all observers on the Earth's surface the Earth's surface (and with it the whole Earth) is constantly accelerating towards them.

You are misreading the equivalence principle to say that the Earth's surface is physically expanding outward. The equivalence principle redefines the concept of an inertial frame of reference; what it is saying is that an object at rest with respect to the surface of the Earth is accelerating with respect to an inertial frame. This, BTW, is exactly what an accelerometer says: An accelerometer placed at rest on the surface of the Earth will indicate that it is accelerating upwards at 9.8 meters/second².

 

[ZirkMan]

To demonstrate this, consider a person standing on a rotating platform. Wherever he's standing, his experience is consistent with the point he is standing at accelerating inwards towards the center of the platform. In fact, that point is accelerating towards the center of the platform. However, the platform is not imploding, and in this coordinate system, no point of the platform is moving.

If the rotation of the platform was not accelerating (or deaccelerating) and I was standing on the platform and there was no other frame of reference to compare I would have no way to prove that my platform is rotating or not and there would be no force that would drag me to the center of the platform, wouldn't be?

And yet I'm standing on the surface of the Earth and observe it accelerating towards me (from my reference frame). And I can still see the paradox here. How can there be something moving towards me (from my reference frame without comparison to other frames), yet not moving at all?

 

[ZirkMan]

... what it is saying is that an object at rest with respect to the surface of the Earth is accelerating with respect to an inertial frame.

What is this inertial frame? Some kind of absolute space? Sorry, I do not understand this.

 

[Dale]

If the rotation of the platform was not accelerating (or deaccelerating) and I was standing on the platform and there was no other frame of reference to compare I would have no way to prove that my platform is rotating or not and there would be no force that would drag me to the center of the platform, wouldn't be?

That is incorrect. A simple gyroscope or a laser ring interferometer could prove it without reference to any external object.

How can there be something moving towards me (from my reference frame without comparison to other frames), yet not moving at all?

It sounds like you are confusing proper acceleration and coordinate velocity. When you are standing on the Earth then you and the ground are both accelerating upwards at g, hence your relative velocity is constant (0). When you are free-falling towards the Earth then the ground is accelerating at g and you are not, hence your relative velocity is changing, and whether or not a given object is moving depends on the coordinate system chosen, but it is all self-consistent with no true paradoxes.

 

[K^2]

If the rotation of the platform was not accelerating (or deaccelerating) and I was standing on the platform and there was no other frame of reference to compare I would have no way to prove that my platform is rotating or not and there would be no force that would drag me to the center of the platform, wouldn't be?

There IS a force dragging you out towards the outer edge. The centrifugal force. That's exactly the point.

Centrifugal force is a fictitious force. Same as gravity in GR. It only arises from the fact that you chose a frame of reference in which a static object is actually accelerating. On a rotating platform, each point experiences centripetal acceleration. If you pick a coordinate system relative to ground, you can see that each point on the platform is moving and accelerating towards the center. If you pick a coordinate system fixed to the platform, each point is at rest, but they are still all accelerating towards the center. Hence the centrifugal force pushing you out towards the outer edge.

 

[ZirkMan]

There IS a force dragging you out towards the outer edge. The centrifugal force. That's exactly the point.

Centrifugal force is a fictitious force. Same as gravity in GR. It only arises from the fact that you chose a frame of reference in which a static object is actually accelerating. On a rotating platform, each point experiences centripetal acceleration. If you pick a coordinate system relative to ground, you can see that each point on the platform is moving and accelerating towards the center. If you pick a coordinate system fixed to the platform, each point is at rest, but they are still all accelerating towards the center. Hence the centrifugal force pushing you out towards the outer edge.

This is actually a very good explanation, thanks. But isn't this against the Galilean relativity which states that you cannot detect a uniform motion with help of mechanical experiments? Still, I see a way out of this because although the platform rotates uniformly the speed of relative movement will be different for points in different distances from the center of rotation and this propably gives rise to the centrifugal force. But where is the source of difference in relative kinetic energy for points (objects) in the gravitational frame?

 

[K^2]

Rotation is not a uniform motion. It's always an accelerated motion. So it is detectable under Galilean Relativity. Under Mach's Principle, you can't tell the difference between the rotation of the platform and rotation of the universe around the platform. But unfortunately, General Relativity does not cover that, and we have no way at present to correct for it.

 

[Mentz114]

If the equivalence principle is true then it means that the Earth's gravity field is a constantly accelerating frame of reference. In any accelerating frame of reference the direction of acceleration is always opposite to the direction of attraction.

Wrong. Where do you get these weird ideas ?

That means that for all observers on the Earth's surface the Earth's surface (and with it the whole Earth) is constantly accelerating towards them.

What ? They are all remaining stationary wrt to each other. ( We are talking about Terra here, aren't we ?).

If this is true for all observers on its spherical surface how is it possible that the Earth doesn't explode (due to constant acceleration away from its center to all sides)?

It's not true. Why would the Earth explode if the gravitational field is trying to compress it into a sphere ?

Acceleration means a change in velocity.
Anything that is stationary at a fixed distance from the Earth's centre is being held by a force, which is to say it is experiencing non-geodesic motion in 4D and is not accelerating.

 

[ZirkMan]

Rotation is not a uniform motion. It's always an accelerated motion. So it is detectable under Galilean Relativity. Under Mach's Principle, you can't tell the difference between the rotation of the platform and rotation of the universe around the platform. But unfortunately, General Relativity does not cover that, and we have no way at present to correct for it.

Agreed.

But please, let's come back to the source of the fictitious force of gravity. My current understanding is that all fictitious forces arise from relative imbalance in kinetic energy(KE) in any (de)accelerating frame of reference (as is the case of the centrifugal force from your example).

Now with all know fictitious forces (except gravity?) the source of the force can be attributed to the imbalance in KE between observer's frame and an non-uniform motion frame in which he is dragged in. His own inertia enables him to feel the fictitious force (the transfer of the KE). But what is the source of difference in KE when I'm in the accelerated frame of gravity? I cannot detect anything accelerating (in the physical sense). Yet if gravity arises from the same mechanism as other fictitious forces some source of EK imbalance must be present.

 

[K^2]

What ? They are all remaining stationary wrt to each other. ( We are talking about Terra here, aren't we ?).

And yet, the surface is accelerating.

Acceleration means a change in velocity.

That's where you are making an oversight. Just because the vector describing velocity does not change, it does not mean there is no change in velocity. Again, look at the rotating frame of reference. An object that has zero velocity in a rotating frame IS accelerating.

But please, let's come back to the source of the fictitious force of gravity. My current understanding is that all fictitious forces arise from relative imbalance in kinetic energy(KE) in any (de)accelerating frame of reference (as is the case of the centrifugal force from your example).

What does kinetic energy have to do with anything?

The source of fictitious force is a "bad choice" of coordinate system. I can pick a free-falling coordinate system, and gravitational force disappears. The difference is that in flat space-time, I can pick one coordinate system that's inertial everywhere. In curved space-time, it's a local choice. If I pick a free-falling coordinate system here, on the other side of the Earth, it's an accelerated frame. That's why in GR you have to know how to deal with accelerated frames, and working with accelerated frames of reference in flat space-time is a good practice, because you can check your answers much easier.

 

[bcrowell]

That means that for all observers on the Earth's surface the Earth's surface (and with it the whole Earth) is constantly accelerating towards them. If this is true for all observers on its spherical surface how is it possible that the Earth doesn't explode (due to constant acceleration away from its center to all sides)?

In GR, frames of reference are local. You're assuming that there is a single frame of reference that encompasses the whole earth, but that isn't true.

 

[lugita15]

Under Mach's Principle, you can't tell the difference between the rotation of the platform and rotation of the universe around the platform. But unfortunately, General Relativity does not cover that, and we have no way at present to correct for it.

Are you saying that there's no way modify GR to make it compatible with Mach's principle? What about Brans-Dicke theory?

 

[Dale]

Now with all know fictitious forces (except gravity?) the source of the force can be attributed to the imbalance in KE between observer's frame and an non-uniform motion frame in which he is dragged in.

Huh? Do you have any reference for this? I have never heard of such a thing and don't even know what you mean by "imbalance of KE" between frames.

 

[K^2]

In GR, frames of reference are local. You're assuming that there is a single frame of reference that encompasses the whole earth, but that isn't true.

No. In GR inertial frames are local, and yes, in general, you can't always have a global coordinate system. But you can have a coordinate system that describes Earth and the immediate surroundings just fine. Start with polar coordinates under Schwarzschild metric, then adjust the metric as necessary to get something less idealized.

Are you saying that there's no way modify GR to make it compatible with Mach's principle? What about Brans-Dicke theory?

There might be. But how would you test it if it's the correct way to do so? We don't even have an experiment confirming Mach's Principle.

 

[ZirkMan]

Acceleration means a change in velocity.

Acceleration means a relative change of EK over mass (m). Otherwise you cannot explain inertia.

Anything that is stationary at a fixed distance from the Earth's centre is being held by a force, which is to say it is experiencing non-geodesic motion in 4D and is not accelerating.

Gravity is not a force. Every observer experiencing gravity is dragged in an accelerating frame of reference which is a big difference. Non-geodesic motion is just a consequence of existence of a force opposing the direction of acceleration of the accelerating frame. Such as the repulsive force of electrons in your body and the ground acting as a buffer zone through which (from your frame of reference) the energy from the accelerating ground is transferred to you. Of course you cannot detect any gravity when you are not in contact with the accelerating frame (such as freefall). Then there is nothing to prevent you following a geodesic.

 

[ZirkMan]

In GR, frames of reference are local. You're assuming that there is a single frame of reference that encompasses the whole earth, but that isn't true.

Apparently it is so since the Earth hasn't exploded due to its global acceleration yet
But I try to find a link that would link this local frames to the global view. Maybe the solution is the same as in SR - there is no global frame and frames of all observers are only relative to one another.

 

[nismaratwork]

I was under the impression that inertia, like mass, remains elusive in its explanation. I wouldn't look to Newtonian mechanics to answer that question however.

 

[Dale]

Acceleration means a relative change of EK over mass (m).

This is false. Consider uniform circular motion in an inertial reference frame, there is acceleration without any change in kinetic energy.

ZirkMan, you have provided nothing to substantiate any of the many erroneous claims you have made in this thread. You need to stop speculating and start learning some mainstream physics.

 

[ZirkMan]

Huh? Do you have any reference for this? I have never heard of such a thing and don't even know what you mean by "imbalance of KE" between frames.

Take for example a person jumping on an accelerating train (from the person's perspective). From his perspective the train has higher KE then he has. When he jumps on it as soon as he touches it the difference of of their KE starts to level off. Even if the train was not accelerating only moving uniformly the transfer of difference in their KE would not be immediate (due to inertia) and the person would probably fell down because for a moment his feet would move quicker than the rest of his body. He would feel a kick of "fictitious force".

When the train is accelerating the transfer of KE doesn't stop because the train constantly increases its relative KE. Therefore the person will feel the constant fictitious force of acceleration.

Is such an explanation ok for you? I hope it is clear and simple enough that I do not need any formal reference to back it up because it is just an ordinary relation of basic and well-known physics concepts (at least from my reference frame :shy:). And you are free to prove it wrong.

 

[ZirkMan]

This is false. Consider uniform circular motion in an inertial reference frame, there is acceleration without any change in kinetic energy.

So you disagree with this:

Rotation is not a uniform motion. It's always an accelerated motion. So it is detectable under Galilean Relativity.

ZirkMan, you have provided nothing to substantiate any of the many erroneous claims you have made in this thread. You need to stop speculating and start learning some mainstream physics.

Yes, you have a full right of accusing me of speculation in this second part of this thread, because indeed I do not agree with some of the answers and think some other approach could worked better. But I try to do my best to explain why and teach some mainstream physics in the process from you guys. I want to see where I am wrong and the best way is to get some constructive feedback after which I can admit I have gained some new insight.

 

[K^2]

Acceleration means a relative change of EK over mass (m). Otherwise you cannot explain inertia.

No. It does not. Kinetic energy is something completely different in GR.

E_{kinetic} = cp^t - c\sqrt{p^{\mu}p^{\nu}g_{\mu\nu}}

Note that this quantity is frame and metric dependent.

 

[nismaratwork]

So you disagree with this:



Yes, you have a full right of accusing me of speculation in this second part of this thread, because indeed I do not agree with some of the answers and think some other approach could worked better. But I try to do my best to explain why and teach some mainstream physics in the process from you guys. I want to see where I am wrong and the best way is to get some constructive feedback after which I can admit I have gained some new insight.

Your tone and approach is very reasonable, but the content of what you're saying isn't. I believe that's the source of the disconnect between your questions, and the frustration in the attempt to answer. It feels very much like a reasonable individual trying reasonably to tell you that the moon is made of green cheese.

If you make your mistakes in the form of trying to learn, rather than teach... you'll be far more in the spirit (and letter) of this forum's law. AFAIK. After all, it's hard to tell a crackpot with an agenda from someone trying to employ a confrontational method to test their ideas.


I will say this: I read, but don't participate in a lot of these threads. I can't think of a time that DaleSpam steered someone wrong... I'd listen to him.

 

[Dale]

When he jumps on it as soon as he touches it the difference of of their KE starts to level off. Even if the train was not accelerating only moving uniformly the transfer of difference in their KE would not be immediate (due to inertia) and the person would probably fell down because for a moment his feet would move quicker than the rest of his body. He would feel a kick of "fictitious force".

All of what you describe here is due to the real contact force between the person and the train and not due to fictitious forces. This certainly does not support nor even explain your claim that fictitious forces are "attributed to the imbalance in KE".

I hope it is clear and simple enough that I do not need any formal reference to back it up because it is just an ordinary relation of basic and well-known physics concepts

No, it is not clear, and it has no discernable relationship to mainstream physics. Please provide a reference, or stop posting this line of speculation.

 

[bcrowell]

No. In GR inertial frames are local, and yes, in general, you can't always have a global coordinate system. But you can have a coordinate system that describes Earth and the immediate surroundings just fine. Start with polar coordinates under Schwarzschild metric, then adjust the metric as necessary to get something less idealized.

A coordinate system isn't the same as a frame of reference, and that's exactly what you need to understand in order to understand the resolution of the paradox. If you have a frame of reference, you can define whether objects A and B are at rest with respect to one another, even if A and B are far apart. GR does not, for example, allow us to say whether our galaxy is at rest with respect to a distant galaxy.

There might be. But how would you test it if it's the correct way to do so? We don't even have an experiment confirming Mach's Principle.

This is incorrect. Mach's principle can be tested by experiments that test predictions made by a less Machian theory such as GR and a more Machian theory such as Brans-Dicke gravity. The results come out to be less Machian, as in GR. A good popular-level discussion of this is available in Was Einstein Right? by Will.

In GR, frames of reference are local. You're assuming that there is a single frame of reference that encompasses the whole earth, but that isn't true.

Apparently it is so since the Earth hasn't exploded due to its global acceleration yet

Your statement doesn't connect logically to the material you quoted from my post.

Maybe the solution is the same as in SR - there is no global frame

SR does have global frames of reference.

 

[Dale]

ZirkMan,

First, the equivalence principle only applies over regions of spacetime where the curvature is negligible. So, while it applies over a small region of the Earth's surface, it does not apply over the entire surface.

Second, geometrically an inertial object's worldline is a geodesic (a straight line). Conversely an object undergoing proper acceleration has a non-geodesic worldline (a curved line).

Third, an inertial coordinate system is an orthonormal coordinate system whose coordinate lines are all straight (geodesic), conversely a non-inertial coordinate system has one or more sets of curved coordinates.

Fourth, In gravity two inertial particles may accelerate relative to each other, geometrically this is only possible in a curved spacetime.

Fifth, in a curved spacetime it is not possible to introduce a global set of coordinates which are everywhere inertial, but it is possible to do so locally.

Given the above, do you understand your concern any better?

 

[ZirkMan]

All of what you describe here is due to the real contact force between the person and the train and not due to fictitious forces. This certainly does not support nor even explain your claim that fictitious forces are "attributed to the imbalance in KE".

Is it possible to distinguish between the contact force and the fictitious force in this case? If yes, then I was really wrong.

No, it is not clear, and it has no discernable relationship to mainstream physics. Please provide a reference, or stop posting this line of speculation.

OK, no more on KE and fictitious forces until I have a set of axioms that can be put into formal language and defended in other way then just freely described concepts. Thanks for your time and valuable feedback anyway.

 

[Dale]

Is it possible to distinguish between the contact force and the fictitious force in this case? If yes, then I was really wrong.

Yes, there are lots of ways to distinguish them. The contact force is a real force and is associated with an "equal and opposite" reaction force on the train, the fictitious force is not. The contact forces cause stresses, the fictitious force does not. Accelerations due to the contact forces can be measured by an accelerometer, the fictitious forces cannot. Etc.

 

[K^2]

A coordinate system isn't the same as a frame of reference, and that's exactly what you need to understand in order to understand the resolution of the paradox. If you have a frame of reference, you can define whether objects A and B are at rest with respect to one another, even if A and B are far apart. GR does not, for example, allow us to say whether our galaxy is at rest with respect to a distant galaxy.

Coordinate system is the frame of reference. It's the mapping from an open set in manifold onto R⁴. There is no other definition for frame of reference or coordinate system.

This is incorrect. Mach's principle can be tested by experiments that test predictions made by a less Machian theory such as GR and a more Machian theory such as Brans-Dicke gravity. The results come out to be less Machian, as in GR. A good popular-level discussion of this is available in Was Einstein Right? by Will.

That does not test Mach's principle. That tests specific theory of gravity. There is no practical test of Mach's Principle.

 

[ZirkMan]

Yes, there are lots of ways to distinguish them. The contact force is a real force and is associated with an "equal and opposite" reaction force on the train, the fictitious force is not. The contact forces cause stresses, the fictitious force does not.

So a fictitious force stops being fictitious as soon as a contact is being made? So when I'm sitting on the ground on Earth there is no fictitious force pushing me down only a real contact force (which source is what)?

Accelerations due to the contact forces can be measured by an accelerometer, the fictitious forces cannot. Etc.

Maybe the answer to the question above will answer this seemingly opposite answers.

An accelerometer placed at rest on the surface of the Earth will indicate that it is accelerating upwards at 9.8 meters/second².

 

[K^2]

So a fictitious force stops being fictitious as soon as a contact is being made? So when I'm sitting on the ground on Earth there is no fictitious force pushing me down only a real contact force (which source is what)?

No. The contact force is pushing you up. It's what opposing the fictitious force and prevents you from falling towards center of the Earth.

 

[ZirkMan]

No. The contact force is pushing you up. It's what opposing the fictitious force and prevents you from falling towards center of the Earth.

If the contact force is a reaction to the fictitious force then its hard to say which one causes stresses. How can you say (DaleSpam) that it's not the fictitious force that's causing the stress?

 

[nismaratwork]

If the contact force is a reaction to the fictitious force then its hard to say which one causes stresses. How can you say (DaleSpam) that it's not the fictitious force that's causing the stress?

Doesn't this go back to the RLG or an accelerometer on the ground?

 

[K^2]

If the contact force is a reaction to the fictitious force then its hard to say which one causes stresses. How can you say (DaleSpam) that it's not the fictitious force that's causing the stress?

Because when you are free-falling, the fictitious force is acting on you, but there is no stress.

 

[Dale]

So a fictitious force stops being fictitious as soon as a contact is being made?

No, what would make you say that? The fictitious force has little to do with the contact force.

So when I'm sitting on the ground on Earth there is no fictitious force pushing me down only a real contact force (which source is what)?

No, the real contact force is pushing you up, the source is the EM interaction between the chair and your butt. The force which is pushing you down is fictitious, its source is the curved non-inertial coordinate system you are using.

Maybe the answer to the question above will answer this seemingly opposite answers.

An accelerometer placed at rest on the surface of the Earth will indicate that it is accelerating upwards at 9.8 meters/second².

D H and I are in agreement. The accelerometer measures the acceleration upwards due to the real contact force upwards. It does not measure the fictitious force downwards. This is, in fact, the easiest way to distinguish between fictitious and real forces in general.

If the contact force is a reaction to the fictitious force then its hard to say which one causes stresses. How can you say (DaleSpam) that it's not the fictitious force that's causing the stress?

The contact force is not a reaction to the fictitious force. The contact force and the fictitious force act on the same body (you) the reaction to the contact force is an equal and opposite contact force on the chair. The fictitious force has no reaction force. It does not in general follow Newton's 3rd law.

 

[ZirkMan]

Because when you are free-falling, the fictitious force is acting on you, but there is no stress.

But isn't this the same as to say that (in the freefall) there is NO force acting on you (therefore no stress) until you come in contact with the accelerated frame which will then invoke the contact force (and keep invoking it because the frame is constantly accelerating)?

This is actually the heart of my original question and a possible source of my (mis)understanding. So I'm glad we came this far.

 

[ZirkMan]

No, what would make you say that? The fictitious force has little to do with the contact force.

So you are saying that the contact force (for example when I'm sitting on the ground) would be there even without the fictitious force? What would be the source of it?

No, the real contact force is pushing you up, the source is the EM interaction between the chair and your butt. The force which is pushing you down is fictitious, its source is the curved non-inertial coordinate system you are using.

So now you admit the the source of the contact force is the fictitious force, yet still you claim that "The fictitious force has little to do with the contact force." How can I understand this clearly contradicting statements?

D H and I are in agreement. The accelerometer measures the acceleration upwards due to the real contact force upwards. It does not measure the fictitious force downwards. This is, in fact, the easiest way to distinguish between fictitious and real forces in general.

If the source of the contact force is the fictitious force (and the contact force is just a reactionary force to it) than what you measure with the accelerometer is both the contact force and the fictitious force as both have a common source.

The contact force is not a reaction to the fictitious force. The contact force and the fictitious force act on the same body (you) the reaction to the contact force is an equal and opposite contact force on the chair. The fictitious force has no reaction force. It does not in general follow Newton's 3rd law.

I really want to understand logic of this. Why cannot be the contact force a reaction to the fictitious force?

 

[ZirkMan]

So now you admit the the source of the contact force is the fictitious force, yet still you claim that "The fictitious force has little to do with the contact force." How can I understand this clearly contradicting statements?

Sorry, for some reasons I cannot edit this and now I realized this is not what you mean.
I might repeat myself but if the fictitious force is not the source of the contact force then what is causing the EM interactions?

 

[D H]

But isn't this the same as to say that (in the freefall) there is NO force acting on you (therefore no stress) until you come in contact with the accelerated frame which will then invoke the contact force (and keep invoking it because the frame is constantly accelerating)?

This sentence gets at the very heart of your misunderstandings. Objects don't enter and leave reference frames. You can think of a reference frame as a mathematical description of space as seen from the perspective of some hypothetical observer. Saying as you did that something "comes in contact with the accelerated frame" is an example of what Alfred Korzybski called confusing the map for the territory.

Let's start with reference frames in Newtonian mechanics, where time is absolute and space is cartesian. Suppose we want to describe the position and velocity of a tree on the top of some hill. From the perspective of someone standing still next to a road at the bottom of the hill, the tree will have (for example) a position of 400 meters east, 200 meters north, and 50 meters up. The tree is not moving from this perspective, so its velocity and acceleration is zero. Now consider a person in an accelerating car on that road. From the perspective of this person, the car is not accelerating: It is the landscape that is moving and accelerating. At the instant the car passes by our stationary observer, the car-bound observer's description of the tree will have a very similar position as that given by the stationary observer, but the velocity and acceleration of the tree will be quite different from that of the stationary observer. While the tree has a description in both the stationary or accelerated frame, it is not "in" either frame.

Descriptions of motion in Newtonian mechanics take on a particularly simple form (Newton's second law) in a special set of reference frames. In these inertial frames of reference, change in momentum results solely from the application of real forces. In other frames, additional acceleration terms are needed to describe the dynamics of some object. We get things that has units of force upon multiplying these additional acceleration terms by the object's mass. Just because these new terms have units of force does not mean they truly are forces. They aren't. They are fictitious forces.

The concept of a reference frame, describing state with respect to some point in space and some set of axes, carries over to general relativity. The concept of an inertial frame also carries over, but to a lesser extent. In Newtonian mechanics, if we find that some frame of reference is inertial (physics are simple) at some point, then that frame is inertial everywhere. In general relativity, while reference frames are global, the inertial nature of a reference frame is localized. You will always have to invoke fictitious forces to describe the behavior of some object if the object is far enough away or if you look with great enough precision.

 

[ZirkMan]

This sentence gets at the very heart of your misunderstandings. Objects don't enter and leave reference frames. You can think of a reference frame as a mathematical description of space as seen from the perspective of some hypothetical observer. Saying as you did that something "comes in contact with the accelerated frame" is an example of what Alfred Korzybski called confusing the map for the territory.

This is a fair semantic objection and I will treat reference frames based on this definition from now on. But I still think there is a valid conceptual truth behind my point of view. Suppose I replace the "accelerated frame" with "accelerated surface" (of the Earth). The surface is in my frame (as is everything else if I understood you correctly). Now we don't have problem with colliding frames anymore but I can still take the position that in my reference frame I am still (no fictitious force) and the surface is accelerating towards me and continues to do so after I land. And the paradox of how the surface can accelerate towards me locally without accelerating globally remains.

While the tree has a description in both the stationary or accelerated frame, it is not "in" either frame.

Now I'm confused. How can be something not in my chosen frame of reference and yet I am able to describe it from it?

In other frames, additional acceleration terms are needed to describe the dynamics of some object. We get things that has units of force upon multiplying these additional acceleration terms by the object's mass. Just because these new terms have units of force does not mean they truly are forces. They aren't. They are fictitious forces.

I understand that fictitious force arise when you change perspective to different frames. What in one frame looks like there is no force (freefall) from a different frame can look like a force (an observer on the ground observing the gravitational force) thus a fictitious force. But how observers in those 2 frames can see such a different thing is the heart of the paradox I'm trying to understand.

The concept of a reference frame, describing state with respect to some point in space and some set of axes, carries over to general relativity. The concept of an inertial frame also carries over, but to a lesser extent. In Newtonian mechanics, if we find that some frame of reference is inertial (physics are simple) at some point, then that frame is inertial everywhere. In general relativity, while reference frames are global, the inertial nature of a reference frame is localized. You will always have to invoke fictitious forces to describe the behavior of some object if the object is far enough away or if you look with great enough precision.

To me this is an elegant way of how to ignore the real problem. An excuse if you will to make the model work but ignoring some fundamental truth (now disguised as the paradox we are talking about). But if this is the current standpoint towards the problem (make it work locally but ignore it globally) then OK, this is what I needed to know.

 

[nismaratwork]

This is a fair semantic objection and I will treat reference frames based on this definition from now on. But I still think there is a valid conceptual truth behind my point of view. Suppose I replace the "accelerated frame" with "accelerated surface" (of the Earth). The surface is in my frame (as is everything else if I understood you correctly). Now we don't have problem with colliding frames anymore but I can still take the position that in my reference frame I am still (no fictitious force) and the surface is accelerating towards me and continues to do so after I land. And the paradox of how the surface can accelerate towards me locally without accelerating globally remains.

No! Nonononono... it's not semantic! If you move from there, the rest of your paragraph resolves itself. I'd add, "fair objection" sounds like he's making a critique, rather than a correction. You're dead wrong, period... that's why in the next sentence you're...

Now I'm confused. How can be something not in my chosen frame of reference and yet I am able to describe it from it?

Chosen frame of reference? Describe from it? Are you asking why we can't define any or all frames of reference that we want for the sake of a given problem? As long as they're valid and not "absolute", I don't see a problem. I still think you're missing the point: There are an infinite set of frames of reference, even if it would be pointless to treat them individually under most circumstances (falling into a black hole would be an exception). In fact, think of a black hole:

You're falling into it, and as your legs, for the first time a human has EVER noticed (to their dismay) are experiencing a much greater "tug" than your torso, and your head! It's the dreaded 'Spaghettification'! Let's pretend that you're alive during this unfortunate event: you're now keenly aware that you could define different coordinate systems for many slices of your body. You're still able to see yourself spiraling out, and to your eyes your feet actually become red, dim, and vanish!

What's the problem?

I understand that fictitious force arise when you change perspective to different frames. What in one frame looks like there is no force (freefall) from a different frame can look like a force (an observer on the ground observing the gravitational force) thus a fictitious force. But how observers in those 2 frames can see such a different thing is the heart of the paradox I'm trying to understand.

That is NOT how it arises, that is how it is explained as fictitious; on large scale it arises as curvature in spacetime. You're falling along the normal path you'd take if a planet wasn't in the way, and you weren't using energy to move around. This why orbit is the 'art of dodging a planet' as you fall towards it.

To me this is an elegant way of how to ignore the real problem. An excuse if you will to make the model work but ignoring some fundamental truth (now disguised as the paradox we are talking about). But if this is the current standpoint towards the problem (make it work locally but ignore it globally) then OK, this is what I needed to know.

To the mainstream physics community it's an elegant solution to an ancient problem, and one that has had only support from experiment and observation. Are you sure you don't understand this, and instead, you aren't just... arguing against it?

 

[D H]

And the paradox of how the surface can accelerate towards me locally without accelerating globally remains.

The paradox is a mental construct of your own making, ZirkMan. It is not a real paradox. It results from you having a Newtonian point of view.

So let's back up a bit and look at the Einstein's elevator car thought experiment. I'll replace his elevator car with a rocket to be a bit more modern. Consider these four scenarios:

1. The rocket is at rest on the surface of an isolated planet (no sun, no moon) that is similar to the Earth in terms of mass and size. You feel your normal weight.
2. The rocket is quiescent and in orbit about this planet. You feel weightless.
3. The rocket is quiescent in deep space. You once again feel weightless.
4. The rocket is firing its engines in deep space, yielding a constant acceleration of 1g. You no longer feel weightless. Instead you feel your normal weight.

The rest frame of the rocket defines a frame of reference. Newtonian mechanics and general relativity agree on whether this rocket frame is an inertial frame for scenarios 3 and 4 (gravitation is null in these scenarios) but disagree on scenarios 1 and 2. In Newtonian mechanics, the rocket frame is inertial in scenarios 1 and 3 but not in scenarios 2 and 4. In general relativity, the rocket frame is inertial in scenarios 2 and 3 but not in scenarios 1 and 4.

Whether a frame is inertial in Newtonian mechanics can be determined by looking at the behavior of objects known to be free of any external forces. If all such objects maintain a constant velocity the frame is inertial. This inertial/non-inertial characteristic is global in Newtonian mechanics.

Whether a frame is inertial in general relativity can only be determined by making local experiments. Looking out the windows is cheating. You can only using something akin to an accelerometer or ring laser gyro. The inertial/non-inertial characteristic is local (and approximate) in general relativity.

A frame centered on a falling apple is not inertial in Newtonian mechanics but is locally inertial in general relativity. By insisting that the outward acceleration of the surface of the applies globally you are implicitly applying Newtonian logic to a general relativistic concept. To quote the doctor ("Doc, it hurts when I do this"): "Don't do that then."

From the perspective of the falling apple, the dynamics of a nearby falling apple can be described in simple (i.e., inertial) terms. The dynamics of the point on the Earth toward which the apple is falling is, from the perspective of the apple, accelerating toward the apple. You are interpreting this upward acceleration as an outward acceleration. This, coupled with Newtonian think, is what is getting you in trouble. A point on the Earth opposite the point directly beneath the apple is accelerating toward the apple, not outward. You will need to invoke fictitious forces to describe the motion of a falling apple on the other side of the Earth.

To be pedantically correct, you need to invoke those fictitious forces to describe the dynamics of a nearby falling apple as well. The difference is that the fictitious forces on the nearby falling apple are immeasurably small. This is not the case for the apple on the other side of the world.

 

[ZirkMan]

From the perspective of the falling apple, the dynamics of a nearby falling apple can be described in simple (i.e., inertial) terms. The dynamics of the point on the Earth toward which the apple is falling is, from the perspective of the apple, accelerating toward the apple. You are interpreting this upward acceleration as an outward acceleration. This, coupled with Newtonian think, is what is getting you in trouble. A point on the Earth opposite the point directly beneath the apple is accelerating toward the apple, not outward. You will need to invoke fictitious forces to describe the motion of a falling apple on the other side of the Earth.

To be pedantically correct, you need to invoke those fictitious forces to describe the dynamics of a nearby falling apple as well. The difference is that the fictitious forces on the nearby falling apple are immeasurably small. This is not the case for the apple on the other side of the world.

I begin to understand. So it's the global spacetime curvature that will make the local effect disappear on the global scale? And the effect is local only because the frame (of the falling apple) is inertial also only locally, right?

 

[Dale]

So you are saying that the contact force (for example when I'm sitting on the ground) would be there even without the fictitious force? What would be the source of it?

I might repeat myself but if the fictitious force is not the source of the contact force then what is causing the EM interactions?

The EM interaction between the ground and your butt is caused because they are very close to each other, it falls off very rapidly with distance, which is why it is called a contact force. Basically, the EM forces between the atoms of your butt hold your butt together fairly tightly and the EM forces between the atoms of the ground hold the ground together even tighter (this is what makes a material a solid). In order for the atoms of your butt to pass through the atoms of the ground their fields would have to displace the fields of the atoms of the ground away from each other and thus away from their equilibrium separation, which results in a net force preventing that from happening due to all of the microscopic EM interactions between your butt and the ground.

I really want to understand logic of this. Why cannot be the contact force a reaction to the fictitious force?

Because of Newton's 3rd law. The action and the reaction forces ALWAYS act on different bodies. The fictitious force pointing down and the contact force pointing up act on the same body, therefore they cannot possibly be an action-reaction pair. Also, 3rd law pairs are always of the same type, so if the reaction force is a contact force then the action force is also a contact force (on a different body). The mere fact that two forces happen to be equal and opposite does not make them an action-reaction pair.

what you measure with the accelerometer is both the contact force and the fictitious force

No. If the accelerometer measured accelerations due to both forces then the accelerometer would read 0 since the sum of the forces is 0. Since the accelerometer does not measure 0 we know that it cannot be measuring accelerations due to both forces, and since the accelerometer measures an upwards acceleration we immediately know that it is measuring the acceleration due to the real upwards contact force, not the fictitious downwards force.

 

[ZirkMan]

The EM interaction between the ground and your butt is caused because they are very close to each other, it falls off very rapidly with distance, which is why it is called a contact force. Basically, the EM forces between the atoms of your butt hold your butt together fairly tightly and the EM forces between the atoms of the ground hold the ground together even tighter (this is what makes a material a solid). In order for the atoms of your butt to pass through the atoms of the ground their fields would have to displace the fields of the atoms of the ground away from each other and thus away from their equilibrium separation, which results in a net force preventing that from happening due to all of the microscopic EM interactions between your butt and the ground.

This is nice but completely misses the point of why there is a need for those EM interaction to take place in the first place. An astronaut on the ISS can comfortably levitate above his chair for a long time and hardly interact with it at all. While the same astronaut on the ground cannot do this (unless he does work to compensate for the now present fictitious force) and he will feel the contact force. The difference between these 2 situations and the reason there is a contact force in the 2nd is the presence of the fictitious force (without which the contact force would not be needed to be invoked). I don't see how you can escape from this conclusion.

Because of Newton's 3rd law. The action and the reaction forces ALWAYS act on different bodies. The fictitious force pointing down and the contact force pointing up act on the same body, therefore they cannot possibly be an action-reaction pair. Also, 3rd law pairs are always of the same type, so if the reaction force is a contact force then the action force is also a contact force (on a different body). The mere fact that two forces happen to be equal and opposite does not make them an action-reaction pair.

Wow, you interpret the Newton's 3rd law quite rigidly. While it is in fact a disguised law of conservation of energy. And we know that energy has many forms but with right conversion mechanism can change its form to any of them (from any of them in principle). And this can be applied also on this situation. Every point of space of the ground is (locally) accelerating towards the opposite point of the butt and the fictitious force of the ground's point acceleration is manifested as the contact force. The law of conservation of energy holds firmly in this situation and therefore also your butt exerts equal but opposing force on the chair. Both forces compensate and the result is that you remain sitting on the chair even in presence of the fictitious force of gravity.

No. If the accelerometer measured accelerations due to both forces then the accelerometer would read 0 since the sum of the forces is 0. Since the accelerometer does not measure 0 we know that it cannot be measuring accelerations due to both forces, and since the accelerometer measures an upwards acceleration we immediately know that it is measuring the acceleration due to the real upwards contact force, not the fictitious downwards force.

No, see above. The fictitious force of acceleration and the contact force have both the same direction and magnitude and though the accelerometer technically measures only the contact force it is in fact the direct manifestation of the fictitious force. No need for a conflict here.

 

[nismaratwork]

This is nice but completely misses the point of why there is a need for those EM interaction to take place in the first place. An astronaut on the ISS can comfortably levitate above his chair for a long time and hardly interact with it at all. While the same astronaut on the ground cannot do this (unless he does work to compensate for the now present fictitious force) and he will feel the contact force. The difference between these 2 situations and the reason there is a contact force in the 2nd is the presence of the fictitious force (without which the contact force would not be needed to be invoked). I don't see how you can escape from this conclusion.

NO! The astronaut is in free-fall, he is NOT "levitating". Your assumptions once again create the paradox, you've just rephrased it.

Wow, you interpret the Newton's 3rd law quite rigidly. While it is in fact a disguised law of conservation of energy. And we know that energy has many forms but with right conversion mechanism can change its form to any of them (from any of them in principle). And this can be applied also on this situation. Every point of space of the ground is (locally) accelerating towards the opposite point of the butt and the fictitious force of the ground's point acceleration is manifested as the contact force. The law of conservation of energy holds firmly in this situation and therefore also your butt exerts equal but opposing force on the chair. Both forces compensate and the result is that you remain sitting on the chair even in presence of the fictitious force of gravity.

OK, I'll bite: what are you saying here?

No, see above. The fictitious force of acceleration and the contact force have both the same direction and magnitude and though the accelerometer technically measures only the contact force it is in fact the direct manifestation of the fictitious force. No need for a conflict here.

See SR.

 

[harrylin]

There IS a force dragging you out towards the outer edge. The centrifugal force. That's exactly the point.

Centrifugal force is a fictitious force. Same as gravity in GR. It only arises from the fact that you chose a frame of reference in which a static object is actually accelerating. On a rotating platform, each point experiences centripetal acceleration. If you pick a coordinate system relative to ground, you can see that each point on the platform is moving and accelerating towards the center. If you pick a coordinate system fixed to the platform, each point is at rest, but they are still all accelerating towards the center. Hence the centrifugal force pushing you out towards the outer edge.

The rotating platform itself is not a valid frame for laws in Newtonian mechanics (which I assume you meant to use as illustration). The so-called "centrifugal force" that seems to pull you towards the outer edge is definitely a pseudo force. However, in Newton's mechanics there is a centrifugal force with which you are pressing against the outer edge and it is a real force, which you can measure with a force sensor. And that real force is caused by your real acceleration, it's the reaction force to the centripetal force. Those two very different "centrifugal force" concepts are often confused, probably because they have the same name.

Regretfully GRT makes no clear distinction between fictitious and real forces. However, it's nowadays common to distinguish between pseudo gravitational fields and real gravitational fields, see the Physics FAQ:
http://www.phys.ncku.edu.tw/mirrors/physicsfaq/Relativity/SR/TwinParadox/twin_gr.html

 

[D H]

Wow, you interpret the Newton's 3rd law quite rigidly.

There is only one way to interpret Newton's third law: the forces involved are exactly equal but opposite, they arise from the same interaction, and they act on different bodies.

While it is in fact a disguised law of conservation of energy. And we know that energy has many forms but with right conversion mechanism can change its form to any of them (from any of them in principle). And this can be applied also on this situation. Every point of space of the ground is (locally) accelerating towards the opposite point of the butt and the fictitious force of the ground's point acceleration is manifested as the contact force. The law of conservation of energy holds firmly in this situation and therefore also your butt exerts equal but opposing force on the chair. Both forces compensate and the result is that you remain sitting on the chair even in presence of the fictitious force of gravity.

This is completely wrong. You started with a wrong premise and went off on a tangent from that incorrect premise. Newton's third law can be viewed as a consequence of conservation of momentum, not energy. What results from this derivation is that the interactions between particles must comprises paired equal but opposite forces -- Newton's third law. Construing the gravitational force and the normal force acting on some body as a third law interaction is erroneous.

 

[harrylin]

If the equivalence principle is true then it means that the Earth's gravity field is a constantly accelerating frame of reference. In any accelerating frame of reference the direction of acceleration is always opposite to the direction of attraction.

That means that for all observers on the Earth's surface the Earth's surface (and with it the whole Earth) is constantly accelerating towards them. If this is true for all observers on its spherical surface how is it possible that the Earth doesn't explode (due to constant acceleration away from its center to all sides)? Is there a model of spacetime that explains it?
Or is there a mechanism of acceleration that doesn't require acceleration in space?

As I see it, your formulation of the equivalence principle is at odds with Einstein's formulation of the equivalence principle.

Notably he stated:

"It is, for instance, impossible to choose a body of reference such that, as judged from it, the gravitational field of the Earth (in its entirety) vanishes."

Please have a look at his formulation, and compare it with yours:

http://www.bartleby.com/173/20.html

Harald