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Do gravitational fields all look the same?

  1. Nov 1, 2015 #1
    In the absence of a sense of scale, will the gravitational fields of large objects be indistinguishable, one from the other?
  2. jcsd
  3. Nov 1, 2015 #2


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    Yes, in the case of the exterior field of a spherically symmetric object. This is Birkhoff's theorem.
  4. Nov 1, 2015 #3
    Can we therefore conclude that the amount of time needed for a test object (suitable to the size of the object) to fall through a gravitational field will be the same? This may require reducing the object to its Schwarzchild radius.
  5. Nov 2, 2015 #4
    In the absence of a sense of time, it would take the same time.
    But I'm not sure I follow your logic.
  6. Nov 2, 2015 #5
    A test object in a weak gravitational field will fall through the field to the Schwarzchild radius more slowly than a test object in a strong field, but an object in a strong field has further to fall. Intuitively, I think that it's a wash, and all appropriate test objects in all fields will take the same amount of time to reach the Schwarzchild radius. They all arrive at the radius at the same time.
  7. Nov 2, 2015 #6
    A gravitational field has no end. If you fall from a given distance, then a heavier object will attract you faster, and you'll have (a bit) shorter distance to fall.
    If you carefully select the starting distance, the falling time will be the same. In Newtonian mechanics, you'd have to move 2x farther for an 8x heavier planet. In GR, it should be pretty close to the Newtonian result but I haven't done that calculation.
  8. Nov 2, 2015 #7
    Thanks for this. I take it that this means that while Birkhoff's Theorum states that all gravitational fields look the same, when a sense of scale is missing we can tell one field from another by dropping test masses from, say, a point where the gravitational force is 1% of its value at the Schwarzchild radius, and the test masses will arrive at the radius at different times.
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