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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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If the floor of the elevator is exerting a force on me (due to some external force accelerating it) then this would be very different from a gravitational acceleration that would accelerate each part of my body equally. No, it wouldn't. The two situations are experimentally indistinguishable. That's one of the points of this thought experiment. An even ...


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The answer is very simple and you may forget everything about lifts, elevators and so on and so forth. The gravitational force (which does exist, I wonder the comments above) is the only interaction in the universe where the dynamics does not depend on the mass of the particle (the so called statement that inertial mass is equivalent to gravitational mass), ...


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For trying to give a conclusive answer it seems necessary to rigorously characterize the geometry (the "conicidence structure", incl. the "light-cone structure") of the region under consideration (arguably with exception of "the singularity itself"). Unfortunately, this seems complicated (as may be gathered from efforts to tackle related problems at least ...


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If you look at the formula for the Christoffel symbol in terms of the metric tensor you see that you need to take derivatives. So ... Potential $\overset{\text{take derivative}}\rightarrow$ Force $\rightarrow$ Acceleration. Is like: Metric $\overset{\text{take derivative}}\rightarrow$ Christoffel Symbol $\rightarrow$ Geodesic Deviation. I'm not sure any ...


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Schutz's A First Course in General Relativity explains it pretty well, I think. The pertinent page is here, but if international copyright law won't let you read it, the basic gist of it is the following: If the gravitational acceleration was the same everywhere in space, then we could shift into a freely falling reference frame, and freely falling ...


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It is not clear what OP is asking, but consider the following chain of reasoning: Scalar curvature $R$ is an invariant independent of choice of coordinates. In GR, regions of matter, e.g., a star, is modelled with non-zero curvature. The Minkowski space has zero curvature. Hence, if a spacetime $(M,g)$ has non-zero scalar curvature $R$ at a point $p\in M$, ...


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This is largely the same answer as Rob's, though rather than use Eddington-Finkelstein coordinates I'm going to use Kruskal-Szekeres coordinates because I think this makes the argument easier to understand. This is what the situation looks like in Kruskal-Szekeres coordinates: For the non-nerd the Kruskal-Szekeres coordinates seem formidably complicated, ...


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Therefore I'd like to ask a related question in which pings are plainly the main point... OK. I would refer you to Einstein talking about the speed of light varying with gravitational potential. And to Irwin Shapiro, who was involved in pinging radar signals to Venus and back, saying "the speed of a light wave depends on the strength of the gravitational ...


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The answer must be closely related to my answer in Thought experiment - would you notice if you fell into a black hole? You can certainly use a similar Eddington-Finkelstein coordinate diagram to consider it (the E-F coordinates transform away the coordinate singularity at the event horizon). NB: This considers only GR and a non-rotating black hole (and ...



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