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Curvature of Spacetime on Earth 
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#1
Jan2113, 02:10 PM

P: 96

I am trying to improve my understand of the basic elements of GR.
I have read that the earth orbits the sun because spacetime between the earth and the sun is warped, mainly due to the sun’s mass. The earth follows a geodesic, which is the equivalent of a straight line in curved space. This explanation replaces the idea of a gravitational pulling force. Because this is all happening in outer space, I can easily accept it. But I would also like to understand how it works here on earth. I suppose that all dimensions are warped, including an increasing compression of space in the direction of the center of the earth. First question: If I let go of a rock at the North Pole, why does it fall when there is no rotation causing movement of the rock? Is the distortion of spacetime a kind of slippery slope which the rock has to fall down without being pushed? When we talk GR, do we continue to observe potential energy and its conversion to kinetic energy? Second question: If I drop a rock 5 meters it accelerates and takes about 1 second to reach the ground. Does the final meter contain less space than the first meter, or is there less time in the final meter, or what? Third question: Considering that we see the effect of gravity on the falling rock so clearly, why can’t we see any distortion or compression of the space through which the rock is falling? I don’t mean the curved flight path due to the rotation of the earth, I mean the distortion of space through which the rock is travelling. Fourth question: If the final meter contains less space than the first meter, is this the reason why atmospheric density is greater nearer to the earth’s surface? In other words, does the distortion of space explain the increase in atmospheric density? Likewise, does the warping of space due to the earth’s mass continue below the surface and cause the increasing density of the earth towards its center? Fifth question: The other day I saw a TV program where it was stated that the Apollo program used Newtonian physics alone. Is this true? Or did they mean that for journeys to the moon, the errors in Newtonian physics are immaterial? . . 


#2
Jan2113, 03:26 PM

P: 4,246

http://www.physics.ucla.edu/demoweb/...spacetime.html http://ion.uwinnipeg.ca/~vincent/450...lack_Holes.htm 


#3
Jan2113, 03:33 PM

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Here's what "time curvature" means: the "rate of time flow" gets slower as you get closer to a gravitating body like the Sun or the Earth. This is called gravitational time dilation, and it has been experimentally confirmed. What we normally think of as the "gravitational force" is due to the *gradient* of the rate of time flow; the force "points" in the direction of slower time flow, and its magnitude is proportional to how fast the rate changes. (Technical point: the "curvature" is actually the *second* derivative of the rate of time flow, not the first; the first derivative is called the "connection coefficient". But that doesn't change the key point, which is that all of this has to do with *time*; there's no space curvature involved at all.) Why does the rate of time flow matter? Because the "straight line" path in spacetime is the path of maximal *proper time*i.e., maximum time experienced by an observer traveling along the path, compared to other nearby paths that have the same endpoints. I won't go into the details of this right now (mainly since it would take quite a bit more exposition), but this is the basic thing to keep in mind. To answer why the rock falls when it's released, first consider: why does the rock *not* fall *before* it's released? The answer is that there is a force pushing upwards on the rock: to be concrete, let's suppose that the force is exerted by a platform that is slid out from under the rock. Before the platform is slid out, it pushes up on the rock; i.e,. the rock has a force on it and is not in free fall. Another way of saying this is that the rock has weight. After the platform is slid out, there is no force on the rock and it is in free fallweightless. And the fact that it took an *upward* force to keep the rock at the same height shows that the freefall paths in the rock's vicinitythe weightless pathsare downwards. But as soon as you try to deal with nonstatic systems, all that breaks down; and there are plenty of GR problems that deal with nonstatic systems. For the idealized case of a static, spherically symmetric massive body, the only "space curvature" is in the radial direction; so we can view "space" as a family of nested 2spheres, each with a slightly larger area (the degenerate "2sphere" at the center has zero area, and the area increases monotonically from there). "Space curvature" then manifests itself this way: suppose I take two adjacent 2spheres, with areas A and A + dA, and I measure the radial distance between them using a meter stick. Euclidean geometry would lead me to expect that the distance between the two 2spheres, as a function of A and dA, will be given by the standard Euclidean formulas; and in flat spacetime that is what we would find. But in the spacetime around a gravitating body, we find that the distance between the two 2spheres is *larger* than the Euclidean formula would predict. That doesn't mean the length of the meter sticks is changed: it means that it takes *more meter sticks* to cover the distance between the 2spheres than the Euclidean formula would predict. As for why the rock accelerates, acceleration is what is caused by the gradient of the "rate of time flow", as I discussed above. If an object already has some downward speed, acceleration adds to it; that's why the rock covers progressively more vertical distance in equal increments of time. 


#4
Jan2113, 06:12 PM

P: 5,632

Curvature of Spacetime on Earth



#5
Jan2213, 03:53 AM

P: 96

Thanks Peter and A.T. for your exellent answers. I now see that I have to recognize the much higher importance of the time dimension than what I had assumed.
Would I be correct to interpret it like this: the air molecules and rock are in free fall and want to reach the center of the earth, but other air molecules and rocks are in the way, so they get closer together, which gives the higher density. But I don't like this interpretation, because I don't see why the molecules should get closer together. What is forcing them to close the gap in spite of their mutual electromagnetic repulsion? Is it easier to use Newtonian calculations and correct the path en route? This would need more fuel, but perhaps you have to correct the path anyway for other reasons? If you want to intercept and land on a comet, is this a critical case where you have to use GR because otherwise big corrections are more likely? . 


#6
Jan2213, 04:03 AM

P: 43

They said that in Relativity, an object is never really at rest and there is this thing called "Four Vector", that besides spatial motion, we also always move forward through the time dimension at speed C, at the rate of 1s/s, just like you stated above. It was also explained that when we move through space, we take away some of the speed/velocity from our movement through the time dimension/direction, and that is why time dilation occurs, as according to SR, especially when the velocity gets near or at the speed of light. I would like to confirm, is this correct? Because it still has not sinked in my head and I want fully understand it and clarify it. But same question I want to ask again and clarify: objects fall, just like the rock in question, all because of the time curvature? So basically the rock may not be in "spatial" motion, but since it still always moves through time, it follows that geodesic in that spacetime curvature, and therefore end up falling on the ground? Is this correct? 


#7
Jan2213, 05:49 AM

P: 4,246

Did you check the links I gave you in the previous thread? They show it quite nicely: http://www.physics.ucla.edu/demoweb/...spacetime.html http://www.relativitet.se/spacetime1.html 


#8
Jan2213, 05:55 AM

P: 4,246

http://www.adamtoons.de/physics/gravitation.swf Note, that the Earth is not actually uniform, so the geometry would be a bit different. 


#9
Jan2213, 06:05 AM

P: 4,246

Gravity alone would actually make them move apart (see tidal forces). At least if we ignore their own mass, and consider only Earth's gravity. Otherwise a gas cloud can also create gravity, that makes the contents move together. But that is not the main effect in the Earth's atmosphere. 


#10
Jan2213, 07:04 AM

P: 96

You appear to be saying the same thing as PeterDonis without explaining it. He wrote that the weight of the air comes from the "upward force exerted by the air underneath". I want to know what this force is. Electromagnetic repulsion will cause the atoms and molecules to stay apart. So if density is increasing with depth, I want to know what is overcoming this repulsion. Can you explain these points without the apparent contradictions? . 


#11
Jan2213, 07:24 AM

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#12
Jan2213, 07:25 AM

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#13
Jan2213, 07:35 AM

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So the two molecules are being pushed towards one another just from following their natural path through spacetime. The electromagnetic repulsion between molecules wants to keep them apart; it's the interplay of these forces that gives us an atmosphere with higher density and pressure near the surface of the earth, and lower further away where the curvature is forcing molecules together less strongly. 


#14
Jan2213, 07:52 AM

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#15
Jan2213, 07:55 AM

P: 96

By the way, I am still talking GR, not Newton. . 


#16
Jan2213, 08:20 AM

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#17
Jan2213, 04:48 PM

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#18
Jan2213, 07:23 PM

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Similarly, if you consider a small volume of air in the middle of the atmosphere, it experiences a net force in the upward direction: the air above it pushes downward, and the air below it pushes upward. Since there is a pressure gradient, the force from below is greater than the force from above, so the net force is upward. 


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