Gravity - an accelerating frame paradox

[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

 

[Dale]

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.

You are completely misapplying the equivalence principle. The equivalence principle does NOT state that an astronaut on the ground (at rest in a gravitational field) is equivalent to an astronaut in orbit (free-falling in a gravitational field). It says that the astronaut on the ground (at rest in a gravitational field) is equivalent to an astronaut far from any source of gravity in an accelerating rocket (accelerating in the absence of gravity). The astronaut in orbit (free-falling in a gravitational field) is equivalent to an astronaut far from any source of gravity in a non-accelerating rocket (not accelerating in the absence of gravity). The contact forces are the same in the equivalent scenarios whether the fictitious force is due to gravitation or inertia.

Wow, you interpret the Newton's 3rd law quite rigidly. While it is in fact a disguised law of conservation of energy.

It is not a matter of interpretation, that is simply what it says. And it is conservation of momentum, not conservation of energy, although the two are closely related in relativity.

The fictitious force of acceleration and the contact force have both the same direction and magnitude

This is simply wrong. The fictitious force is gravity and gravity points down by definition. If the contact force were also pointing downwards then instead of sitting on the floor we would accelerate into the floor at 2g. Since we do not observe that happening it is clear that the contact force points upwards. And since the accelerometer reads 1g upwards it is clear that it does not detect the fictitious force of gravity.

See Figure 4 here
http://eta.physics.uoguelph.ca/tutorials/fbd/intro.html

and paragraphs a and c here:
http://a-s.clayton.edu/campbell/physics/phys1112/Supplements/phys1111fb.htm

and example 1 here:
http://hrsbstaff.ednet.ns.ca/pvickers/normal_force_and_freebody_d.htm

 

[ZirkMan]

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.


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.

If it is wrong or not totally depends on the definition of energy and its conservation and you seem to use a different one than me. Otherwise you wouldn't say the conservation of momentum is different to conservation of energy.

What conservation law would apply to a situation where a running steam engine train would hit a stationary train and move it away?
Is it only the law of conservation of momentum together with Newton's 3rd law? Or some other combination of various partial conservation laws?

Or can I say that the chemical energy released by the burning coal in the moving train's engine was translated into the final motion of the energy (and momentum) of the (before) stationary train?
I think I can say that and I'm sure that if you calculate how much energy was released by burning coal in the train's engine minus losses due to friction etc. you would get the exact energy (and momentum) the stationary train had after the impact. Because the law of conservation of energy (in this broad sense) is always true there cannot be a contact force without something else that powers it. And I don't see any other predecessor for that particular case than the fictitious force.

 

[ZirkMan]

This is simply wrong. The fictitious force is gravity and gravity points down by definition.
If the contact force were also pointing downwards then instead of sitting on the floor we would accelerate into the floor at 2g. Since we do not observe that happening it is clear that the contact force points upwards. And since the accelerometer reads 1g upwards it is clear that it does not detect the fictitious force of gravity.

I was analyzing the situation from the perspective of a free falling observer where he sees only the accelerating surface which he interprets as accelerating due to the fictitious force of gravity. I also assumed that this acceleration won't stop after he hits the surface and will continue to drive him up (through the contact force). In this frame there is no need for a downward pulling fictitious force. But maybe you have to switch frames after you hit the ground. I don't know. I feel I need to study this in even greater detail to fully understand what's going on there.

 

[nismaratwork]

If it is wrong or not totally depends on the definition of energy and its conservation and you seem to use a different one than me. Otherwise you wouldn't say the conservation of momentum is different to conservation of energy.

What conservation law would apply to a situation where a running steam engine train would hit a stationary train and move it away?
Is it only the law of conservation of momentum together with Newton's 3rd law? Or some other combination of various partial conservation laws?

Or can I say that the chemical energy released by the burning coal in the moving train's engine was translated into the final motion of the energy (and momentum) of the (before) stationary train?
I think I can say that and I'm sure that if you calculate how much energy was released by burning coal in the train's engine minus losses due to friction etc. you would get the exact energy (and momentum) the stationary train had after the impact. Because the law of conservation of energy (in this broad sense) is always true there cannot be a contact force without something else that powers it. And I don't see any other predecessor for that particular case than the fictitious force.

re bold: I think it's clear that you have your own special definitions to encompass most of physics. Maybe you should take a chance on less opposition for its own sake, and actually make use of the information already given to you in this thread.

 

[D H]

If it is wrong or not totally depends on the definition of energy and its conservation and you seem to use a different one than me. Otherwise you wouldn't say the conservation of momentum is different to conservation of energy.

I suggest you read some more / take some more physics classes. What you wrote makes no sense. Newton's third law is equivalent to conservation of momentum, not energy. Conservation of momentum follows from Newton's third law, and Newton's third law follows from conservation of momentum. The derivation is in practically every text for the sophomore/junior level class classical mechanics taken by almost all physics majors.

Regarding your earlier notion that the normal force is equal but opposite to the gravitational force: It isn't. From the perspective of an inertial observer, the forces acting on a person standing still on the surface of the Earth are the normal force and gravitation. Except at the poles, these forces cannot and do not sum to zero. The person is after all undergoing uniform circular motion about the Earth's rotation axis, so there must necessarily be some non-zero net force acting on the person.

Regarding your interpretation of the equivalence principle: Please re-read post #43.

 

[Dale]

I was analyzing the situation from the perspective of a free falling observer where he sees only the accelerating surface which he interprets as accelerating due to the fictitious force of gravity.

A free-falling frame is inertial. There are no fictitious forces in it, only the real forces. Specifically, in the free falling frame the astronaut on the ground is accelerating upwards at g due to the unbalanced upwards-pointing real contact force. There is no fictitious force pointing down to prevent his acceleration, and his coordinate acceleration matches the proper acceleration measured by an accelerometer as you would expect in an inertial frame.

 

[Dale]

Regarding conservation of momentum vs conservation of energy and Newton's 3rd law.

Let us imagine a world where Newton's 3rd law is replaced by a law where for every action there is an equal and perpendicular reaction. Consider a perfectly elastic collision between a ball moving towards the origin along the x-axis and a ball of equal mass at rest at the origin. The moving ball feels a force in the -x direction which stops it. By the modified 3rd law the resting ball feels a force in the y direction. It moves away with the same speed as the original ball and therefore energy is conserved. However, it is moving in a different direction (y instead of x) and therefore momentum is not conserved.

So, in a world where Newton's 3rd law is violated momentum is not conserved even if energy is. Newton's 3rd law is most definitely associated with conservation of momentum rather than conservation of energy.

 

[ZirkMan]

Specifically, in the free falling frame the astronaut on the ground is accelerating upwards at g due to the unbalanced upwards-pointing real contact force. There is no fictitious force pointing down to prevent his acceleration, and his coordinate acceleration matches the proper acceleration measured by an accelerometer as you would expect in an inertial frame.

This is exactly the situation I have a problem to understand. Where does this "unbalanced upwards-pointing real contact force" come from? Does it arise as a "barrier force" preventing me following a geodesic motion?

 

[ZirkMan]

Regarding conservation of momentum vs conservation of energy and Newton's 3rd law.

Let us imagine a world where Newton's 3rd law is replaced by a law where for every action there is an equal and perpendicular reaction. Consider a perfectly elastic collision between a ball moving towards the origin along the x-axis and a ball of equal mass at rest at the origin. The moving ball feels a force in the -x direction which stops it. By the modified 3rd law the resting ball feels a force in the y direction. It moves away with the same speed as the original ball and therefore energy is conserved. However, it is moving in a different direction (y instead of x) and therefore momentum is not conserved.

So, in a world where Newton's 3rd law is violated momentum is not conserved even if energy is. Newton's 3rd law is most definitely associated with conservation of momentum rather than conservation of energy.

Thanks for a clarification. I have regarded the momentum only as mass x velocity not as a vector unit. If you ignore the vector dimension then really the conservation of momentum vers. conservation of energy should make no difference because from this perspective it's all and only energy that's being transfered. Of course the vector dimension and distinction between energy and momentum has its practical use for direction calculations etc. but one should not forget that at the end of the day it is only energy that is being transferred (momentum being just a specific expression of it).

 

[D H]

Thanks for a clarification. I have regarded the momentum only as mass x velocity not as a vector unit.

And you had the gall to accuse me of using non-standard definitions of the conservation laws?

The problem here is that you are trying to learn to run (general relativity) when you can't even crawl (non-calculus Newtonian mechanics).