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Why are geodesics parabolae on earth's surface? |
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| Nov24-12, 04:29 AM | #1 |
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Why are geodesics parabolae on earth's surface?
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? Thanks! |
| Nov24-12, 05:10 AM | #2 |
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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.
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| Nov24-12, 05:28 AM | #3 |
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![]() From: http://www.physics.ucla.edu/demoweb/...spacetime.html Note that this cone surface doesn't have intrinsic curvature. It represents a flat piece of space time, and yet produces gravity. |
| Nov24-12, 08:40 AM | #4 |
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Why are geodesics parabolae on earth's surface? |
| Nov24-12, 07:14 PM | #5 |
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Mentor
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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. |
| Nov27-12, 10:06 AM | #6 |
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thanks a lot! that was a great help!
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| general relativity, geodesics |
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