# Tag Info

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There is another aspect somehow overlooked by the other answers. Consider a pile of iron filings accelerated towards a magnet. If you were to arrange so that they all have the same magnetic force per unit mass they would appear to experience no force relative to each other while being accelerated towards the magnet, and if you had weak bonds holding them ...

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We don't need to appeal to relativity to explain why you don't feel any force in free fall. Plain old Newtonian mechanics predicts that too. What you actually feel when you feel a force being applied to you is that the external force applies only to a small part of your body (the soles of your feet if you're standing up and feel the normal force from the ...

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Why does a free-falling body experience no force despite accelerating? Because there is no force acting upon it. If you look at some pictures of the principle of equivalence, you will find that they typically depict a guy in a rocket accelerating through space. There's a force on his feet, he can feel it. They also depict a guy standing on the surface ...

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Before telling you why an observer in free fall does not feel any force acting on him, There is a couple of results that should be introduced to you. Newton's second law is only valid in inertial frames of reference: To measure quantities like position,velocity and acceleration of an object, you need a coordinate system $(x,y,z,t)$. Now the coordinates ...

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Well, everything depends on what you mean by "to experience a force". I suspect that you are thinking of some psycho-physical idea. Indeed both floating in space and freely falling we perceive similar sensations. The reason is simply due to the fact that, in both situations, all particles of our body moves with the same speed (due to a spatially uniform ...

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You need a coordinate system to decide a body’s position, velocity, acceleration, momentum or force on it. Assume the body is in free fall near the Earth. 1) First consider a coordinate frame (3 perpendicular rods and a clock) with its origin in free fall near the free falling body. By the equivalence principle we know the rods are falling in unison with ...

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It is incorrect to link the feeling of being accelerated to being accelerated itself. You can be under constant velocity or be continuously accelerated, yet you need not feel anything at all. Let me explain. The reason you feel compressed or stretched when you are accelerated in a lift is because of the presence of the normal force from the ground on you. ...

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falling in a gravitational field is physically indistinguishable from floating in interstellar space Yes. Indeed, this is one of the founding principles of general relativity and is (one of the forms of) the equivalence principle. Your argument is that we can feel acceleration, and gravity makes you accelerate, so shouldn't you feel acceleration while ...

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You mention a couple of times the curvature being an invariant. I wonder if you're confusing this with being dimensionless. I guess that you're talking about the Ricci scalar, which is a relativistic invariant, but it is not dimensionless; IIRC it has units of inverse distance squared. On distance scales much smaller than scale implied by the curvature, ...

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On the contrary Deser and others [1, and refs therein] have argued that trying to construct a theory of a graviton, that is, a massless spin-2 field in a flat background, consistent with special relativity, then "[c]onsistency [leads] us to universal coupling, which implies the equivalence principle" [1]. The argument is summarized by MTW [2, Box 17.2.5, see ...

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This scene was clearly shot with a stationary bus oriented (nearly) nose down and a green screen. A backpack and book are clearly dropped on the rider and accelerate at the normal rates. The rider hits seats and can't hold them as he falls. There is no way that air resistance on Earth could provide enough force to change the movement of the bus (and ...

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No, no you guys (Except Floris and those who up-voted him) have missed an important observation... Look Carefully at the video again. At first the bus just tilts as the bridge bends. When the bus starts tilting (due to friction with the bridge it has not yet started falling) it has not yet obtained considerable vertical velocity. However as the man loses ...

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First of all let's study an imaginary system where both the bus and the person are not subject to drag forces due to the air: If the person is not bounded to anything he will be subject to free falling and thus to a uniform acceleration $g$. Also the bus will be free falling and thus they fall together with the same velocity. If we take the drag forces into ...

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The bus experiences considerable drag, and will therefore fall more slowly than a person inside the bus. The scenario is possible in principle - but after carefully viewing the clip and doing some calculations, I believe that the details are inaccurate. Assume the bus has a mass of 5000 kg (pretty light for a bus), and is 3 m wide by 3 m tall - so the ...

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If the bus was in a vacuum (both inside and outside), then the passenger would float. However, the effects of air resistance on the two objects (passenger and bus) are probably not negligible in such an instance. The bus will be moving relative to the outside air, and so will be accelerating towards the ground at a rate less than $g$. If we then released ...

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At first, the bus and the person would accelerate at the same rate due to gravity. However, the situation is more complicated due to air resistance. The bus experiences air resistance as it falls. The person inside the bus experiences less air resistance because the air inside the bus moves with the bus. This means that the person does not experience as much ...

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I'm having difficulties understanding why a gravitational acceleration can be guaranteed to be locally equivalent to an accelerating frame. Actually, it can't. See section 20 of Relativity: the Special and General Theory where Einstein said this: “We might also think that, regardless of the kind of gravitational field which may be present, we could always ...

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