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When does Newton's Gravitation fail us?

  1. Jun 30, 2009 #1
    I am trying to learn more about General Relativiity. We had Newton's Law of Gravitation. When does it not work? I heard at very high speeds, high mass bodies, and when the distance to a high mass body is very short (Mercury for example, I think)? How does Newton's Gravitation differ mathematically in these scenarios?
     
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  3. Jul 4, 2009 #2

    A.T.

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    The difference in acceleration for a vertical fall is discussed here:
    https://www.physicsforums.com/showthread.php?t=310397

    But GR also predicts other effects, like gravitational time dilation and trajectories altered by the curvature of space (Mercury orbit precession), which are missing in Newtons theory.
     
  4. Jul 6, 2009 #3

    Matterwave

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    Also, the bending of light near a star is purely from GR. Newtonian physics would have light not bend at all near a star since light is mass-less. Even if you attribute a "pseudo mass" (using E=mc^2) to photons, Newton's predictions of the bending of light are only half the actual bending of light (as observed and predicted by GR).
     
  5. Jul 6, 2009 #4

    Vanadium 50

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    Not so. Newtonian physics postulates acceleration independent of mass.

    That is true.
     
  6. Jul 6, 2009 #5

    Matterwave

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    Can you kindly refresh my memory on why Newtonian physics would predict a bending of light even if we say that it is massless?

    I only remember F=GMm/r^2 which would obviously be 0 if one of the m's were 0.
     
  7. Jul 6, 2009 #6

    A.T.

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    Bending of a trajectory is a result of acceleration, and there is no m in the formula for that: g = GM/r^2
     
  8. Jul 6, 2009 #7

    Matterwave

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    Sounds very iffy to me.

    Getting that formula g=GM/r^2 is done by doing: mg=GMm/r^2 is it not? In which case, you cancel the two lower case m's and you get g. You can't do that if the lower case m is 0, that's dividing by zero and is a classic way of getting ridiculous results.

    I suppose you could justify it using limits...

    g=GMm/mr^2 and then take the limit as m goes to 0. But it still looks iffy to me. I mean, how can you even apply F=ma if m=0?
     
  9. Jul 6, 2009 #8

    Dale

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    Nothing iffy about it. If you have a massless particle then any non-zero force will give it an infinite acceleration. If you want a finite acceleration the force must be 0.
     
  10. Jul 6, 2009 #9

    A.T.

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    You can get the acceleration by directly observing the trajectories, which turn out to be independent of m. The formula for the gravitational force is derived from this. You have to keep in mind that force is rather an abstract concept, while trajectories/acceleration are observed directly.
     
  11. Jul 6, 2009 #10
    You can't bend a trajectory without a force. The acceleration is just the result of the force between the bodies attracting each other. Gravity doesn't affect massless particles in the classical mechanics.
     
  12. Jul 6, 2009 #11

    Vanadium 50

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    Pierre-Simon Laplace would be very surprised to read that. He proposed black holes - bodies so massive that even light could not escape - and his argument relied on the universal acceleration of bodies in free fall.

    This was in 1799.
     
  13. Jul 6, 2009 #12
    Then please enlighten us on how gravity interacts with massless particles. Are you sure light is assumed to be massless for that statement?
    "Bodies in free fall" already assumes mass.
     
  14. Jul 6, 2009 #13

    Dale

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    This really is the key point.
     
  15. Jul 6, 2009 #14

    A.T.

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    So how much force do I need to bend the trajectory of a massless particle?
    Acceleration is an observed quantity. Force is just a derived concept that makes sense for massive particles only.
    Newtonian gravity accelerates objects independently of their mass. That is the observed reality. From that you can derive F=GMm/r^2 via F=ma, but only for massive objects. For massless objects F=ma is not applicable, and so neither is F=GMm/r^2. Therefore you cannot use F=GMm/r^2 to show that massless objects are not affected by Newtons gravity.
     
    Last edited: Jul 6, 2009
  16. Jul 6, 2009 #15

    JesseM

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    Yes, it's natural to justify it using limits--since an object's acceleration in a gravitational field can be shown to be independent of its own mass for any nonzero mass, this is obviously going to be true in the limit as a particle's mass approaches zero.
     
  17. Jul 8, 2009 #16
    If you put a heavy bowling ball on a trampoline it will roll to its center and sink in a way that if you put some tennisballs on the trampoline that they will roll to the bowling ball, this we experience as gravity. Is this a valid comparision? The author says the example is in 2D, but if the ball sinks, isn't that 3D? (Is that perhaps why a fourth dimension is added in relativity?)
     
  18. Jul 8, 2009 #17

    A.T.

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    For Newtonian gravity, with rubber sheet representing the gravitational potential, this works as an analogy. But it doesn't' explain the gravity model of General Relativity. See this thread:
    https://www.physicsforums.com/showthread.php?t=286926
    The rubber sheet represents 2 spatial dimensions, omitting 1 spatial and 1 time dimension. But the time dimension is crucial to explain gravity in GR. So it is better to omit 2 spatial dimensions:
    http://www.relativitet.se/spacetime1.html
    You need a 3rd dimension to visualize the curvature of the reduced 2D space-time. In reality this space-time is 4D, and you would need even higher dimensional pictures to visualize it's curvature.
     
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