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Why are geodesics parabolae on earth's surface?

  1. Nov 24, 2012 #1
    It's a naive question, but I'm pretty sure my professor said that space-time is locally flat (and I'm pretty sure that the volume of my room counts as "locally").

    That said, I would expect falling objects to follow straight trajectories, but that's obviously not the general case.

    I thought this could be because straight lines in space-time can be curved when seen in space, but it doesn't seem possible when I try to make some calculations.

    Why are geodesics parabolae on the surface of earth? Where am I getting something wrong?

  2. jcsd
  3. Nov 24, 2012 #2
    it rather comes from the principal of maximum proper time,it can be shown that path is parabola in this case because it maximizes the proper time.
  4. Nov 24, 2012 #3


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    This is correct. Gravity can exist in flat space-time. The local curvature is connected to tidal forces, not the gravity pull itself. And the tidal forces in your room are indeed negligible.

    Exactly. But keep in mind that space-time is curved.

    It is difficult to visualize the relationship for spatial parabolae, because they require 2-spatial dimensions. When you add time you have 3. Visualizing curved 3d spaces is difficult for humans. Better stick to a vertical fall, so you use only spatial dimension. Then you can visualize a distorted 2d-space time:

    From: http://www.physics.ucla.edu/demoweb..._and_general_relativity/curved_spacetime.html

    Note that this cone surface doesn't have intrinsic curvature. It represents a flat piece of space time, and yet produces gravity.
  5. Nov 24, 2012 #4


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    That's because your room is stationary relative to the Earth. Suppose instead your room was falling freely under gravity, without air resistance. Then you would find any falling objects in your room would be following straight lines, relative to your room.
  6. Nov 24, 2012 #5


    Staff: Mentor

    If you put an accelerometer in your room you will find that your room is accelerating upwards at 1 g. So your room is a non-inertial reference frame, and so inertially moving objects will not go in a straight line. If you were to use an inertial frame which measures 0 g acceleration then you would find that a falling object would travel in a straight line at constant speed.

    So, basically, it isn't the free falling object's path which is curved, it is the room's. So any coordinate system you attach to the room is curved and straight lines plotted on the curved coordinates look curved.
    Last edited: Nov 24, 2012
  7. Nov 27, 2012 #6
    thanks a lot! that was a great help!
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