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Understanding General Relativity

 
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Jan19-13, 11:20 AM   #18
 

Understanding General Relativity

Quote by Devils View Post
So I am sitting here perfectly still but accelerating towards the Earth at 9.8m/s^2.
Away from the Earth's center, not towards. That is what an accelerometer resting on the surface measures.
Quote by Devils View Post
How does classical mechanics or GR explain that?
Newton would say it doesn't really accelerate, but is affected by the force of gravity. Einstein would say the accelerometer works fine, and is indeed accelerated upwards.
Jan19-13, 11:35 AM   #19
 
Quote by A.T. View Post
It is a common misconception, that acceleration towards/away from a point implies a change in distance to that point. It simply doesn't. Not even in flat space-time where you can have circular motion with a centripetal acceleration towards a center, without ever coming closer to the center. In curved-spacetime the opposite case is also possible: The Earth surface can accelerate away from the the Earth's center, without getting further away from it.
That's what I was trying to explain, but was told that was wrong even when it is a clear representation of the effects of an accelerated surface.
Jan19-13, 11:36 AM   #20
 
Quote by A.T. View Post
The gradient in the time-dilatation dominates only for objects that move much slower than light through space. Such objects move mostly through time, so the distortion of the time-distances has the most effect.

If the object is in freefall it doesn't exert any forces on the surface in GR. If the object rests on the surface, they exert equal but opposite contact forces on each other
That is my understanding as well. Thanks.
Jan19-13, 11:56 AM   #21
 
Quote by Bill_K View Post
Because, as I pointed out in #8 above, it is not the difference in time dilation, it is the gradient of the time dilation that represents the Newtonian gravitational force.
Only the Newtonian force?
Jan19-13, 12:21 PM   #22
 
Quote by A Troubled Man View Post
Only the Newtonian force?
Yes. In a weak gravitational field, the time-time component of the metric reproduces the Newtonian force. The lowest order deviations from this (so-called "post-Newtonian corrections") are due to the remaining components of the metric, i.e. due to curvature in space. These components are much smaller, which is why we can get away with just using Newtonian gravity in many circumstances.
Jan19-13, 02:24 PM   #23
 
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Quote by A Troubled Man View Post
Again, as asked above, would a test particle follow the same path? As well, considering that the difference in time dilation between one's head and feet is infinitesimally small, how can it account for such a major role in the path followed?
Yes, a test particle would follow the same path, but the analogy starts to break down. At the level of a single test particle I don't know an analogy any more and all I can do is explain the actual math.

So the actual math describes spacetime around the earth as curved. In a curved space any coordinate system you draw is going to have to have curved lines. The way that you physically see if your timelike coordinate is curved is to place accelerometers at rest in your coordinates. If the accelerometer is at rest in your coordinate and reads something other than 0 then it means that your time coordinate is curving. So, in the case of the usual (Schwarzschild) coordinates outside the earth you find that the time coordinate is curving outward at a rate of g.

A free-falling test particle travels in a straight line known as a geodesic. We know that it is a straight line because an accelerometer attached to a free-falling particle reads 0. In the frame of the curved coordinates the geodesics appear to curve.

Quote by A Troubled Man View Post
But, in relativity, it is the surface that is accelerated upwards and not the object exerting a force on the surface.
In relativity both happen. The surface is accelerated upwards, it runs into the object and exerts a contact force on the object to make it accelerate upwards also, and by Newton's 3rd law the object also exerts a force on the object.
Jan19-13, 02:44 PM   #24
 
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Dalespam:
If the accelerometer is at rest in your coordinate and reads something other than 0 then it means that your time coordinate is curving. So, in the case of the usual (Schwarzschild) coordinates outside the earth you find that the time coordinate is curving outward at a rate of g.
That is a weak field approximation??..... right...so you are neglecting the minor curvature of space??
Jan19-13, 02:51 PM   #25
 
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Quote by Naty1 View Post
That is a weak field approximation??..... right...so you are neglecting the minor curvature of space??
It is always a little messy to partition curvature into time curvature and space curvature, so I am not sure if my comments are restricted to a weak field approximation. They very well could be, but I am not certain.
Jan19-13, 03:11 PM   #26
 
Quote by DaleSpam View Post
Yes, a test particle would follow the same path, but the analogy starts to break down. At the level of a single test particle I don't know an analogy any more and all I can do is explain the actual math.

So the actual math describes spacetime around the earth as curved. In a curved space any coordinate system you draw is going to have to have curved lines. The way that you physically see if your timelike coordinate is curved is to place accelerometers at rest in your coordinates. If the accelerometer is at rest in your coordinate and reads something other than 0 then it means that your time coordinate is curving. So, in the case of the usual (Schwarzschild) coordinates outside the earth you find that the time coordinate is curving outward at a rate of g.

A free-falling test particle travels in a straight line known as a geodesic. We know that it is a straight line because an accelerometer attached to a free-falling particle reads 0. In the frame of the curved coordinates the geodesics appear to curve.

In relativity both happen. The surface is accelerated upwards, it runs into the object and exerts a contact force on the object to make it accelerate upwards also, and by Newton's 3rd law the object also exerts a force on the object.
Thanks, you have reaffirmed my understanding of GR and that I'm not going crazy. Actually, I am going crazy in my other discussion trying to get this same understanding across, but that's more out of frustration than anything else.
Jan19-13, 03:15 PM   #27
 
Quote by LastOneStanding View Post
Yes. In a weak gravitational field, the time-time component of the metric reproduces the Newtonian force. The lowest order deviations from this (so-called "post-Newtonian corrections") are due to the remaining components of the metric, i.e. due to curvature in space. These components are much smaller, which is why we can get away with just using Newtonian gravity in many circumstances.
That's what I thought, too. This is probably where the confusion lies with the discussion I'm having.
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