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What do you think and why?

And by the way, this has nothing to do with relativity so your title is misleading.

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It's a good guess, and it's right, but why do you think this has anything to do with general relativity? This can be handled with classical Newtonian gravity.

As to why it's right, it's because whichever place the scale is, it is measuring the same thing (neglecting, as you say, the weight of the pole) which is the weight of the man on top of the pole.

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Why would it? It's not measuring anything where the scale is, it's measuring the thing at the top of the pole. If the pole were not there, the man would be in free-fall. What would his weight be then?

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Nugatory

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The pole is not accelerating, it's just sitting there. Therefore the net force on the pole is zero.Are you sure the acceleration of gravity at the location of the scale itself has nothing to do with it?

There's a downwards force of 100 N on one end of the pole, so that has to be balanced by an upwards force at the other end. That force comes from the spring in the scale, which is compressed enough to exert an upwards force of 100 N on the pole - so that's what the scale reads.

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I keep asking you, WHY do you think general relativity has ANYTHING to do with this? It does not. This is a classical problem plain and simple.

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Well, I was working through some stuff with general relativity to begin with is all, and this topic popped up. I needed to know how a force measured locally would translate as measured at a different radii. Apparently it is invariant, but I didn't want to just assume it. Thanks :)I keep asking you, WHY do you think general relativity has ANYTHING to do with this? It does not. This is a classical problem plain and simple.

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In general relativity, gravity is not a force.Well, I was working through some stuff with general relativity to begin with is all, and this topic popped up. I needed to know how a force measured locally would translate as measured at a different radii. Apparently it is invariant, but I didn't want to just assume it. Thanks :)

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Nugatory

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That is not correct.Well, I was working through some stuff with general relativity to begin with is all, and this topic popped up. I needed to know how a force measured locally would translate as measured at a different radii. Apparently it is invariant

Even in classical mechanics, there is no such thing as a force measured locally being translated to some other position - the concept has no meaning. In your problem, there is a force on the molecules at the top of the pole from the weight of the man; by Newton's third law these molecules exert an equal and opposite upwards force on the man and that's what holds him up. These molecules in turn exert a completely separate force on the molecules below them, and by Newton's third law the lower molecules exert an upwards force on the upper ones.... and so on all the way down to the bottom of the pole, where the bottommost molecules exert a force on the spring scale.

Now, you have made the starting assumption that nothing is moving (the man is help up, the pole isn't bending or being crushed under the weight, ....). That can only happen if the net force on any part of the pole is zero, meaning that the downwards force on it from above is equal to the upwards force on it from below. Work that down every step of the way and you will see that the upwards force from the scale on the bottommost molecules of the pole has to be equal to the downwards force from the man on the topmost molecules of the pole. But that equality has nothing to do with the force being translated from one radius to another - they're completely different forces, and they just turn out to have the same strength because you set the problem up so that they would.

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Right. I was referring to weight.In general relativity, gravity is not a force.

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Okay cool :)That is not correct.

Even in classical mechanics, there is no such thing as a force measured locally being translated to some other position - the concept has no meaning. In your problem, there is a force on the molecules at the top of the pole from the weight of the man; by Newton's third law these molecules exert an equal and opposite upwards force on the man and that's what holds him up. These molecules in turn exert a completely separate force on the molecules below them, and by Newton's third law the lower molecules exert an upwards force on the upper ones.... and so on all the way down to the bottom of the pole, where the bottommost molecules exert a force on the spring scale.

Now, you have made the starting assumption that nothing is moving (the man is help up, the pole isn't bending or being crushed under the weight, ....). That can only happen if the net force on any part of the pole is zero, meaning that the downwards force on it from above is equal to the upwards force on it from below. Work that down every step of the way and you will see that the upwards force from the scale on the bottommost molecules of the pole has to be equal to the downwards force from the man on the topmost molecules of the pole. But that equality has nothing to do with the force being translated from one radius to another - they're completely different forces, and they just turn out to have the same strength because you set the problem up so that they would.

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And you think that weight has nothing to do with force? Look, I'm not trying to give you a hard time here, I'm trying to get you to see why the word relativity should never have entered this discussion, much less been repeated several times by you.Right. I was referring to weight.

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Well, let's say we did that for the earth. We could take the mass of the top layer of earth (neglecting the weight of air and considering the surface of earth to be smooth), say one meter from the surface, multiply that times the surface acceleration of gravity, and that would give us the weight pushing down upon the spherical shell one meter from the surface. If we divide that by the surface area at r_earth - 1 m, which would be A = 4 pi (r - 1 m)^2, then we would gain the radial pressure acting upon that spherical shell. If we do that again for two meters down, multiplying the mass between the spherical shells at one and two meters times the slightly different acceleration there, then add that to the weight of the top layer, we now gain the force of weight pushing down upon the second spherical shell at r - 2 m. Dividing that by the area of the second shell, A = 4 pi (r - 2 m)^2, we gain the radial pressure at that spherical shell. It would be about twice that of the first shell since there is approximately twice the weight acting upon almost the same surface area. This is about right for pressure underwater and so forth, with the pressure increasing almost linearly with depth, so the idea of linear force of weight seems okay so far.

But now let's consider all the way down almost at the center of the earth. At a spherical shell one meter out from the center, almost the entire integrated weight of the earth falls upon this shell, giving to incredible radial pressure at that shell. That still sounds okay because we know the pressure is large. But the problem I am having is if we move to the spherical shell two meters from the center, almost the same weight acts upon this shell as the first, but over four times the surface area, so that the pressure two meters from the center of earth should be four times less than that at one meter. And at the center itself, the pressure should work toward infinity if the weight is linear. But it would seem to me that the pressure within such a small volume at the center of the earth should be pretty much the same. So the force of the weight could not be linear there. What is the solution to this dilemma?

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Right okay, i thought about that. The only way to keep the forces linear is if some of it is transferred to tangential tidal force, right? So how do we find for that?

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It's not a tidal force, that's something completely different.Right okay, i thought about that. The only way to keep the forces linear is if some of it is transferred to tangential tidal force, right? So how do we find for that?

At the pressures and temperatures you'll find near the earth's core, neither solid rock nor iron (which is what much of the core is made of) are rigid the way the pole in your previous thought experiment was rigid; instead, they behave like incompressible liquids. They will flow freely and change their shape until all the forces are balanced but they will always have the same volume. The pressure at any point in the liquid will be the same in all directions (google for "Pascal's Law", and we have a bunch of threads on that).

Now imagine an upside-down pyramid that we've filled with water: What is the pressure at various depths, and what force does it exert on the sides of the pyramid? We can model the earth's core as a collection of these pyramids, all arranged so that their points are at the exact center of the earth. The force exerted against the sides of one pyramid from the pressure of the fluid inside is balanced by the pressure from the fluid in the neighboring pyramid.

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