The force drawing the weight and the mass of the earth 'under' it together (with the scale in the middle) however clearly exists as much as the electromagnetic force though.
Or stars, planets, etc. wouldn't (and couldn't) exist.
As to the exact details of how we account for it is another matter. Calling gravity 'measured spacetime curvature' or 'gravitational force', or coming up with pseudo particles, or whatever, is an accounting method. Either way, it's still there.
But it does clearly exist, as much as light (for example) does. All one needs to do to prove that is look out the window.
> The force drawing the weight and the mass of the earth 'under' it together
There is no such force. Think of the accelerating rocket: there is no "force drawing the weight and the scale together". There is just the rocket pushing. The same is true standing on the Earth: there is just the ground pushing.
One of the reasons we need "spacetime curvature" in GR is to explain how the ground can be pushing things in different directions in different parts of the Earth, while all those things still remain stationary relative to the Earth's center and each other. But the whole point is that, even with all that factored in, there is still no "force drawing the weight and the Earth together". There is just the ground pushing up, plus spacetime geometry to account for the global configuration. That's it.
If you're just standing in an elevator, how can you tell the difference between the floor of the elevator "resisting" your weight due to gravity, vs. being out in space with silent rockets propelling the elevator at 9.8 m/s^2 and "pushing" you?
Either way you'll feel like you and the elevator floor are being pushed together, and both you and the floor will experience some compression.
But that is not the same setup. The setup here is: a body on the surface of a much larger body: is there pushing or pulling?
I suggest to GP to consider the body lifted and then released. The body is moving back towards the larger body. It meets a rigid boundary and comes to rest (or bounces ..) When did the ground start "pushing" the body? It seems the smaller body was being acted on by something (we say curved space) which is motivating it towards the center of the gravity of the larger body. This motivation (spatial deformation or force, whatever) doesn't cease and the rigidity of the larger body is engendering the equal opposing 'force' pushing back against it.
I'm saying the distinction between "resisting" and "pushing" isn't real.
When a rocket is accelerating and you are pressed against the back and being driven forward, is the rocket "pushing" you or "resisting" you? Either one is a fine way of describing it. The net result, though, is that you are squished into the back of the rocket in a way that's completely indistinguishable from gravity.
You may say "but the ground can never push me away from the ground." Sure, but the back of the rocket can't push you away from the back of the rocket either, and yet it is clearly pushing you through space. So long as the rocket keeps accelerating, you are being pushed forward, yet you won't feel it as a push, you'll feel it as an attraction to the back of the rocket.
Yes, it is. The force you feel as weight is the ground pushing on you.
By Newton's Third Law, the ground also feels a force of equal magnitude and opposite direction from you, which can, as you say, end up compressing the ground. But that isn't what you feel as weight.
You will start to sink at various rates. Not because water is “pushing less” than earth. It’s because water is not resisting.
Also, if I am 200 pounds and the earth is pushing up on me with 200 pounds of force, what is generating that force? And where does it go if a helicopter reels me in? The 200 pounds up “force” just vanishes?
> You will start to sink at various rates. Not because water is “pushing less” than earth. It’s because water is not resisting.
Um, "is not resisting" means "pushing less". The water does not exert as much force on you as solid ground would. That's why you sink. If the water was not pushing less, you wouldn't sink.
> if I am 200 pounds and the earth is pushing up on me with 200 pounds of force, what is generating that force?
Nothing has to "generate" the force. The force is a static force that is doing no work, so no energy is being consumed. It's just static electromagnetic repulsion between your atoms and the atoms in the ground.
> where does it go if a helicopter reels me in?
If you stop making contact with the ground, there is no longer any electromagnetic repulsion between your atoms and the atoms in the ground.
> The 200 pounds up “force” just vanishes?
The "up" force from the ground vanishes, yes. See above.
But of course there is another "up" force on you now from whatever is attaching you to the helicopter.
Except that clearly isn't it, as the weight moves even if there is no 'ground' (say at altitude) in the same predictable fashion, until it collides with something - which merely attempts to arrest it's movement. In fact, in orbit the forces felt are still (quite clearly) there too.
So if anything, gravity wise 'pushing' is not a property of gravity at all. Merely of matter which happens to produce the spacetime distortions which result in gravity. Unless dark matter exists, which would be neat.
Or the moon wouldn't continue to be up there, instead of 'down here'.
And clouds of hydrogen atoms will eventually collect together in a vacuum barring other outside gravitational influences. Which is notably why we even have a star.
And the weight itself also has measurable (albeit extremely tiny) such effects on everything else too.
So how do you model/name/account for that force which causes that to occur? Since it does clearly exist.
Because a rocket doesn't just 'push' either. We can clearly model the chemical and physics behaviors going on there to generate that 'push', and all involve forces which we can account for. None of them meaningfully appear to be gravity.
And gravity also involve forces (spacetime distortions causing very real effects!). Which we can also account for.
> the weight moves even if there is no 'ground' (say at altitude) in the same predictable fashion
"Moves" is frame-dependent. In the weight's own rest frame, it is the Earth that moves.
What is not frame-dependent is the fact that, if there is no ground and the weight (wrong word, as we'll see in a moment) is freely falling, there is no "weight"--the object feels no force and an accelerometer attached to it reads zero. This was the basic insight that started Einstein on the road to curved spacetime and General Relativity: if an object falls freely, it will not feel its own weight.
> how do you model/name/account for that force which causes that to occur?
Using the word "force" here is already wrong as far as General Relativity is concerned: in General Relativity gravity is not a force.
> a rocket doesn't just 'push' either. We can clearly model the chemical and physics behaviors going on there to generate that 'push', and all involve forces which we can account for. None of them meaningfully appear to be gravity.
Yes, exactly. Now apply the same principle to the Earth pushing up on the weight. We can clearly model all of the microscopic behaviors that lead to that push and the forces that account for them--and none of them are gravity.
> gravity also involve forces (spacetime distortions causing very real effects!)
The "spacetime distortions" you refer to, which do indeed cause real effects, are not "forces". That's the whole point. They are spacetime geometry. Spacetime geometry is not a force. That is what General Relativity says.
> what are you even talking about?
I am talking about standard General Relativity, as it has been understood and taught in textbooks for decades now. What are you talking about?
GR agrees (recognizing the obvious caveats) with the classical law of F = Gm₁m₂/r², where F stands for "force". This force is caused by spacetime curvature.
No, GR says that the Newtonian law of gravity is an approximation that makes reasonably accurate predictions when the spacetime curvature is small and all relative motions are slow compared to the speed of light. It does not say that the Newtonian interpretation of that equation is correct.
> GR says that the Newtonian law of gravity is an approximation that makes reasonably accurate predictions when the spacetime curvature is small and all relative motions are slow compared to the speed of light.
These are merely the aforementioned caveats.
> It does not say that the Newtonian interpretation of that equation is correct.
If the interpretation were wrong, and that's not a force, then the amount of force in that equation would be 0.
With no force, Gm₁m₂/r² = 0.
However, that's not the modification that GR applies to this, though.
> If the interpretation were wrong, and that's not a force, then the amount of force in that equation would be 0.
With no force, Gm₁m₂/r² = 0.
Nonsense. The GR interpretation is that G m1 m2 / r^2, when it is nonzero, is describing an effect of spacetime geometry, not a force. It can't be a force in GR because it isn't felt; an object moving solely under the influence of G m1 m2 / r^2 feels no weight--an accelerometer attached to it reads zero. GR does not change the numerical value of G m1 m2 / r^2 at all. It just reinterprets what the quantity represents.
Or stars, planets, etc. wouldn't (and couldn't) exist.
As to the exact details of how we account for it is another matter. Calling gravity 'measured spacetime curvature' or 'gravitational force', or coming up with pseudo particles, or whatever, is an accounting method. Either way, it's still there.
But it does clearly exist, as much as light (for example) does. All one needs to do to prove that is look out the window.