[rmolnav (OP)]

Recently I´ve been trying to convey the idea that what we feel in a gravity field isn´t actually gravity itself, but internal stresses originated by both the fraction of gravity somehow not allowed to make each part of our body accelerate (acc. to Newton´s 2nd Law), and the push and/or pull exerted on us by the "obstacle" that avoids our fully free fall (and the subsequent "chain" of reaction forces of each part of our body (acc. to Newton´s 3rd Law).
Whatever the deep nature of gravity, that´s quite clear, as far as I can understand.
But "apparently" (?) this Einstein´s journey first steps (before bringing up the idea of relativity), didn´t keep that in mind. "To be "politically correct", I have to say I must be wrong !! … Please kindly help me "ruminate" it over, in order to try and find where my error could be.
It´s better to go step by step, not to jump the gun (and possible errors …)
From an Einstein specialized site I´ve taken:     
"...We, the crew of the spaceship shown on the right, are floating freely in space, far away from all major sources of gravity. Now imagine that there is another observer in a spaceship, shown on the left: The rocket engine of that observer's spaceship is firing and produces an acceleration of 9.8 meters (32 feet) per square second. This accelerated observer feels as heavy as we would feel on earth, since the gravitational acceleration with which an object on earth falls to the ground has that exact same value".
So far, so good. But, any further comments?  Or ... any nuances to bring up? (for now, better if only directly about what said above …)

[chiralSPO]

You are correct that we feel only the force of the ground preventing us from falling towards the Earth's center of mass, not the gravitational pull itself.

Let us compare two cases in which a spaceship is accelerating at 1 G (9.8 m/s²):

1) A spaceship in free fall towards a gravitational body (let's say it's not in a stable orbit, but actually falling towards a very massive body that is very far away, so we don't have to worry about splattering our passengers during this thought experiment, but their path is still just a straight line)

2) A spaceship far from any gravitational fields, in vast empty space, which is firing its powerful rockets to achieve a constant acceleration of 1 G for the duration of this thought short experiment, without worrying about reaching relativistic speeds (for now).

In the first case the passengers feel weightless, while in the second they feel the 1 G acceleration as if they are standing on a surface on earth. Why are these cases different? Because of how the force acts.

Gravity pulls all parts of anything massive with exactly the same acceleration, no matter what (barring extreme cases of tidal forces etc.). The force is proportional to the mass, so the acceleration is constant, and every part of a person gets pulled in the same way at the same rate. A person in case 1 would be falling at the same acceleration whether they were in the spaceship or not.

In case two it would matter very much whether you were in (or attached to) the spaceship or not. Here there is a force that acts only on the spaceship, and if you are pressed against the floor of the spaceship the force will be conveyed to the passenger as well, felt as pushing, or if they are tethered they will feel the force as pulling. It is the fact that the thrust force acts indirectly on the passengers that it is felt, and that the gravitational force acts directly that it is not felt.

So two cases in which the acceleration is the same (passengers would see the same relative motion out their windows), but the apparent acceleration is either 0 or 1 G. Imagine the passengers in the first spaceship see the impending impact, and fire up their thrusters to deliver enough force to accelerate them at 1 G. Now they feel their weight (experiencing the force associated with a 1 G acceleration), because of the indirect force from the thrust, but if they look out their window, they can breath easier knowing that their velocity relative to the objecting pulling on them is now 0. Again, it matters whether they are in or tied to the ship, or if they are not!

Alternatively, they could navigate into a stable orbit, in which they are still accelerating at 1 G, but still in an apparently weightless free fall (because the direct action of the gravitational force is spread evenly through their bodies). And in this case it doesn't matter whether or not they are in the ship.

Does that help?

[jeffreyH]

Let' say we have one spaceship falling from a point in the gravitational field where the potential is very near to zero. They fall in a straight line path towards a black hole. They are far enough away so that by the time they reach the event horizon their instantaneous velocity will be very near to the speed of light. If we attempted to mimic the same with a spacecraft using fuel then it is said that the energy required will approach an infinite value as the velocity approaches c. So where does gravity get this energy from?

[chiralSPO]

Let' say we have one spaceship falling from a point in the gravitational field where the potential is very near to zero. They fall in a straight line path towards a black hole. They are far enough away so that by the time they reach the event horizon their instantaneous velocity will be very near to the speed of light. If we attempted to mimic the same with a spacecraft using fuel then it is said that the energy required will approach an infinite value as the velocity approaches c. So where does gravity get this energy from?

The kinetic energy comes from the potential energy of being so far from the black hole (think of it as being really high "above" the black hole). If we define infinitely far away from the black hole as 0 potential energy, then any finite distance away will have negative potential energy. This potential energy becomes increasingly negative as you fall down towards the black hole, so for conservation of energy to be maintained, your kinetic energy (which must be positive) increases such that the sum of potential and kinetic energy doesn't change from the original (assuming there is no other way to dissipate or transfer the energy).

Note that energy is not necessarily conserved between reference frames. (the people on the spaceship falling into the black hole at relativistic speeds may not agree with those in a different frames of reference as to how much kinetic and potential energy they have--but each will observe that the sum of potential and kinetic energies is constant)

[rmolnav (OP)]

Thank you. Your post does help … but also reinforces my idea that we have to be very careful in our reasoning. I asked for any comments or nuances relative to what quoted, trying to go slowly but as surely as possible … I do appreciate your effort, but I´m afraid you have jump the gun (what doesn´t make the discussion any easier) … and have been not sufficiently accurate, as far as I can understand:

Gravity pulls all parts of anything massive with exactly the same acceleration, no matter what (barring extreme cases of tidal forces etc.). The force is proportional to the mass, so the acceleration is constant, and every part of a person gets pulled in the same way at the same rate.

Perhaps I should have included the link of the site:
http://www.einstein-online.info/spotlights/geometry_force
so that we all could use same examples in our discussion.
If we make a "close up" when looking at what quoted in my initial post, we can say that "The rocket engine of that observer's spaceship is firing and produces an acceleration of 9.8 meters (32 feet) per square second. This accelerated observer feels as heavy as we would feel on earth, since the gravitational acceleration with which an object on earth falls to the ground has that exact same value" is a rather MISLEADING statement, because:
1) "This accelerated observer feels as heavy as he would feel on earth, since the gravitational acceleration …"
He DOESN´T actually FEEL any "heaviness" globally … He feels internal stresses caused both by the spaceship push (N.´s 2nd Law: chosen 9.8 m/s2 times its mass m), and inertial reaction forces on each part of his body kind of trying to keep their velocity constant (Newton´s 3rd Law: a total of 9.8m in opposite direction).
2)"… since the gravitational acceleration with which an object on earth falls to the ground has that exact same value": that is NOT THE HOLE picture. If that object/person were stopped by some obstacle, he would feel internal stresses originated by both the other object push (N.´s 1st and 2nd Law: upward 9.8m), and gravity pull exerted by Earth on each part of its body (universal gravity law: a total of 9.8m downward).
Certainly our sensations would be similar, but just because the artificial acceleration given to us when out of gravity, has been chosen equal to the natural Earth´s gravity g.
And mentioned laws lead us to similar bottom end, but not equal at all. If in our body we had stress gauging devices sufficiently accurate (e.g., as LIGO gauging devices), we would find meaningful differences, because Earth´s pull exerted on each unit of mass is the nearer to our head (standing), the smaller. That counts when the spaceship is still standing on ground, but not when accelerated in space with no gravity …
Do you agree? … If not, don´t exit ate and tell me (but step by step, please ...). If you do, NEXT STEP could be another careful analysis of other quote form linked site:
"Our accelerated observer has a clear notion of "up" and "down" - when he looks up, he sees all freely drifting observers and their space stations "fall downwards", in the direction of his own spaceship's floor. "Upwards" is the direction in which his spaceship is accelerating. But there is no gravity in this situation. All the observers in freely drifting spaceships (rocket engines shut off) are in agreement: The fact that the accelerated observer sees objects "fall" is merely an artefact, brought about by his spaceship's acceleration - it vanishes as soon as you leave the accelerated reference frame and change to a free-falling one".
Any further comments?  Or, just  any nuances to bring up?
And again, thank you.

[chiralSPO]

Yes, I will accept that my explanation is entirely qualitative, and I may have used a few terms loosely (however, I tried my best to discuss "perceived weight" to make clear that I was not equating the mechanisms behind those perceptions). My explanation was also purely "Newtonian," and isn't necessarily the best lens to view Einstein's theories with.

Thank you for the link. I have only a cursory understanding of general relativity, but I do understand that many of the open questions about gravity from the Newtonian perspective can be addressed by thinking of gravity as a manifestation of curved spacetime due to high mass (and/or energy) concentrations, rather than an actual attractive force. I invite other, more knowledgeable, folks to take it from here...

[jeffreyH]

Let' say we have one spaceship falling from a point in the gravitational field where the potential is very near to zero. They fall in a straight line path towards a black hole. They are far enough away so that by the time they reach the event horizon their instantaneous velocity will be very near to the speed of light. If we attempted to mimic the same with a spacecraft using fuel then it is said that the energy required will approach an infinite value as the velocity approaches c. So where does gravity get this energy from?

The kinetic energy comes from the potential energy of being so far from the black hole (think of it as being really high "above" the black hole). If we define infinitely far away from the black hole as 0 potential energy, then any finite distance away will have negative potential energy. This potential energy becomes increasingly negative as you fall down towards the black hole, so for conservation of energy to be maintained, your kinetic energy (which must be positive) increases such that the sum of potential and kinetic energy doesn't change from the original (assuming there is no other way to dissipate or transfer the energy).

Note that energy is not necessarily conserved between reference frames. (the people on the spaceship falling into the black hole at relativistic speeds may not agree with those in a different frames of reference as to how much kinetic and potential energy they have--but each will observe that the sum of potential and kinetic energies is constant)

You missed the point. Where does the gravitational field get the exponentially increasing energy from to accelerate a mass up to light speed? The field surely doesn't have boundless energy. If us accelerating an object to light speed required an exponential increase in energy then how is gravity supposed to sidestep this?

[chiralSPO]

Let' say we have one spaceship falling from a point in the gravitational field where the potential is very near to zero. They fall in a straight line path towards a black hole. They are far enough away so that by the time they reach the event horizon their instantaneous velocity will be very near to the speed of light. If we attempted to mimic the same with a spacecraft using fuel then it is said that the energy required will approach an infinite value as the velocity approaches c. So where does gravity get this energy from?

The kinetic energy comes from the potential energy of being so far from the black hole (think of it as being really high "above" the black hole). If we define infinitely far away from the black hole as 0 potential energy, then any finite distance away will have negative potential energy. This potential energy becomes increasingly negative as you fall down towards the black hole, so for conservation of energy to be maintained, your kinetic energy (which must be positive) increases such that the sum of potential and kinetic energy doesn't change from the original (assuming there is no other way to dissipate or transfer the energy).

Note that energy is not necessarily conserved between reference frames. (the people on the spaceship falling into the black hole at relativistic speeds may not agree with those in a different frames of reference as to how much kinetic and potential energy they have--but each will observe that the sum of potential and kinetic energies is constant)

You missed the point. Where does the gravitational field get the exponentially increasing energy from to accelerate a mass up to light speed? The field surely doesn't have boundless energy. If us accelerating an object to light speed required an exponential increase in energy then how is gravity supposed to sidestep this?

I don't have the answers, but wikipedia does!
https://en.wikipedia.org/wiki/Surface_gravity#Surface_gravity_of_a_black_hole

[jeffreyH]

Let' say we have one spaceship falling from a point in the gravitational field where the potential is very near to zero. They fall in a straight line path towards a black hole. They are far enough away so that by the time they reach the event horizon their instantaneous velocity will be very near to the speed of light. If we attempted to mimic the same with a spacecraft using fuel then it is said that the energy required will approach an infinite value as the velocity approaches c. So where does gravity get this energy from?

The kinetic energy comes from the potential energy of being so far from the black hole (think of it as being really high "above" the black hole). If we define infinitely far away from the black hole as 0 potential energy, then any finite distance away will have negative potential energy. This potential energy becomes increasingly negative as you fall down towards the black hole, so for conservation of energy to be maintained, your kinetic energy (which must be positive) increases such that the sum of potential and kinetic energy doesn't change from the original (assuming there is no other way to dissipate or transfer the energy).

Note that energy is not necessarily conserved between reference frames. (the people on the spaceship falling into the black hole at relativistic speeds may not agree with those in a different frames of reference as to how much kinetic and potential energy they have--but each will observe that the sum of potential and kinetic energies is constant)

You missed the point. Where does the gravitational field get the exponentially increasing energy from to accelerate a mass up to light speed? The field surely doesn't have boundless energy. If us accelerating an object to light speed required an exponential increase in energy then how is gravity supposed to sidestep this?

I don't have the answers, but wikipedia does!
https://en.wikipedia.org/wiki/Surface_gravity#Surface_gravity_of_a_black_hole

The Schwarzschild solution of k = 1/4M seems familiar. I have read something on that but can't place it. The undefined nature of the acceleration due to gravity at the event horizon is something I'm looking into.

[jeffreyH]

I know that in units of c=G=1 the value 2M is the radius of the event horizon. Then 3M is the photon sphere. I wish I could remember why 4M was significant.

[rmolnav (OP)]

I have only a cursory understanding of general relativity, but I do understand that many of the open questions about gravity from the Newtonian perspective can be addressed by thinking of gravity as a manifestation of curved spacetime due to high mass (and/or energy) concentrations, rather than an actual attractive force. I invite other, more knowledgeable, folks to take it from here...

I´m not a relativity expert either whatsoever ... But the purpose of this thread is to discuss about what it is said triggered Einstein´s leap from Newton´s theories to his.
As far as I can see, reasons given to say those cases (elevator and others) are not explainable within Newton´s Mechanics, are flawed.
And from a careful and correct analysis of them, it can´t necessarily be deduced that another theory is necessary.
We know many facts tell us Einstein was right, and I would´t dare say the opposite. But I guess he "saw" something more, beyond the apparent oddity (within Newton´s Mechanics) of those experiments.
I could be wrong, and that´s why I proposed a careful analysis, step by step, of the considered roots of the equivalence "principle".
You know, there are scientists who go far beyond just "equivalence" (?), when exposing their ideas on the matter. E.g., after explaining the elevator experiment:
1st scientist: "Einstein realized there is no way to tell the difference between sitting in a gravitational field and being accelerated (by the way: that is not exact, as I explained in my last post ...) They are equivalent situations".
2nd: "The fact that these two effects are the same, give the same results, means that gravity is acceleration, NOT JUST LIKE ACCELERATION , IT´S THE SAME THING".
(from a Spanish tv program I saw, "Inside Einstein´s Mind")
Perhaps that is true, but it can´t be deduced from those experiments, as far as I can understand !!
 
 
 

[jeffreyH]

The force of gravity is dispersed evenly throughout an object. When we apply a force to an object it is via a surface and stresses propagate through the object. That is the ONLY difference. Since there is a gravitational force it has limits just like the force we apply. It may diminish at the event horizon.

[Bill S]

Where does the gravitational field get the exponentially increasing energy from to accelerate a mass up to light speed? The field surely doesn't have boundless energy. If us accelerating an object to light speed required an exponential increase in energy then how is gravity supposed to sidestep this?

Boundless energy would be required only if a massive object were accelerated to "c".  This, we are told cannot happen.

If the energy of gravity has its origin in the Big Bang, then enough energy is available to take the entire Universe back to a single point.  What more would be needed?

[rmolnav (OP)]

The force of gravity is dispersed evenly throughout an object

Sorry, but hat would be right only if our object were isotropic, at least as far as density is concerned ...
In your words, if total mass were also dispersed evenly through it ! 

[yor_on]

I'm not sure what you're talking about rmolnav?

Whatever idea or hypothesis you have you will need something more than what's already there in form of a theory (in this case seeming to be about the equivalence principle?). That means you will need to find whatever you want to replace it with explaining some fact unexplained by the 'older theory'. Newton is in cooperated in Relativity, as a limited case, with Relativity covering more circumstances.

[yor_on]

And yes, the equivalence principle between a 'gravity' and a very specific type of acceleration are accepted to make perfect sense, for over a hundred years now. Here is a good description http://eagle.phys.utk.edu/guidry/astro421/lectures/lecture490_ch6.pdf
=

Also, a gravitational acceleration isn't the same as a accelerating rocket. A accelerating rocket expends/transforms 'internal energy' (in)to accelerating. The rocket free falling doesn't. Look up geodesics.

What you might want to argue is the fact that to construct the equivalence to a gravitational 'field' you need to expend energy, and so expect a equivalence to be correct in the case of the free falling rocket? That 'something' needs to expend 'energy' there too? Well, locally measured there is nothing spending energy as far as I know. Neither Earth, nor the rocket. When it comes to 'SpaceTime' you might be able to argue differently though. But we have a problem there, gravity doesn't go from one state to another, it has one sign, there is no such thing as 'negative gravity' and it doesn't transform from state to state.

Although :)
https://www.forbes.com/sites/startswithabang/2017/02/28/is-dark-matter-about-to-be-killed-by-emergent-gravity/#3e7e574c5359

But that's a different kettle of fish

[rmolnav (OP)]

That means you will need to find whatever you want to replace it with explaining some fact unexplained by the 'older theory'. Newton is in cooperated in Relativity, as a limited case, with Relativity covering more circumstances.

Thank you.
In short: I consider that analogies and differences in cases such as the elevator when still on earth surface, and when far from any gravity field, but artificially given "g" acceleration, can be explained within Newton´s Mechanics. Also when comparing an object free fall (in a gravitational field), and an object really with no gravity (very far from any massive celestial object).
And I was trying to make a detailed analysis of them, to justify my stand. To say things such as "he feel weightless ...", or other similar ones, can hide important clues to the issue. Because we never actually feel our "weight", gravity itself: we feel internal stresses, and only when gravity somehow is not allowed to accomplish its theoretical "duty" of giving us "g" acceleration ...
And applying carefully Newton´s Laws those cases can be fully explained, as far as I can understand.   

[jeffreyH]

The force of gravity is dispersed evenly throughout an object

Sorry, but hat would be right only if our object were isotropic, at least as far as density is concerned ...
In your words, if total mass were also dispersed evenly through it ! 

If density were a factor then denser objects would fall faster in a gravitational field than those of lower density. This is different to the source of a gravitational field where density variations do matter.

[jeffreyH]

Where does the gravitational field get the exponentially increasing energy from to accelerate a mass up to light speed? The field surely doesn't have boundless energy. If us accelerating an object to light speed required an exponential increase in energy then how is gravity supposed to sidestep this?

Boundless energy would be required only if a massive object were accelerated to "c".  This, we are told cannot happen.

If the energy of gravity has its origin in the Big Bang, then enough energy is available to take the entire Universe back to a single point.  What more would be needed?

A single object like a black hole doesn't have the energy of the whole universe at it's disposal.

[rmolnav (OP)]

If density were a factor then denser objects would fall faster in a gravitational field than those of lower density

But you were talking about a single object, "The force of gravity is dispersed evenly throughout an object".
And on denser parts gravitational exerted forces (per volume unit) are bigger.
If solid object, it´s clear all parts only could move at same speed and acceleration ... That would generate internal stresses, in such a way that total forces exerted on any part (external and internal), divided by its mass, would have to be equal to the common acceleration, the "g" of the gravitational field at that spot.