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anonymous

  • one year ago

Gravity and space-time

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  1. anonymous
    • one year ago
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    |dw:1434420110948:dw|

  2. anonymous
    • one year ago
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    |dw:1434420315626:dw|

  3. anonymous
    • one year ago
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    And \((0,2,0,1)\) would appear be say \(0.9\) units away from the body, as was \((0,1,0,0)\) at an earlier time.

  4. anonymous
    • one year ago
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    This is an attempt to put it in 2D, but we can be even more simpler in 1D, I think.

  5. Empty
    • one year ago
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    Ahhh I am thinking more in terms of trajectories, which are lines perpendicular to yours. So to the object in freefall's view they never change speed, their speed is constant and their path is straight since they have momentum in one direction. It's just that the closer they travel to the earth, the smaller distances become, they just don't realize it.

  6. Empty
    • one year ago
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    I guess I'm trying to hold onto the idea that while you're in free fall with your eyes closed you feel no gravity, you're just weightless feeling.

  7. Empty
    • one year ago
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    I don't know if what I'm saying is mutually exclusive from what you're saying however, I'm just saying this is the angle from which I was thinking, while you have sorta like an equipotential gravity field lines slant I suppose

  8. anonymous
    • one year ago
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    |dw:1434420951568:dw| The idea here is that the point \((1,0)\) is sucked into be the effective point of \((0.2,0.2)\). And this suction is always happening, which is different than a one-time bend. A one time bend would mean that if you went all the way around, you would be back at point \((0,0)\), but in reality, whatever was at \((0,5)\) would probably now occupy \((0,0)\).

  9. anonymous
    • one year ago
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    Ummm, I should put \((0.2,-0.2)\) I guess, it's just a made up number.

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