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Gravity contracts length

  1. Jul 7, 2005 #1
    am i missing something here?
    we have a circle: [tex](x-h)^2+(y-k)^2=r^2[/tex] where (h,k) is the center and r is the radius. we now spin the circle about an axis that is perpendicular to the plane on which the circle lies and it runs through the center of said circle. gravity contracts length (and my the equivelance principle, so does acceleration), so as the 1-sphere spins about the axis, the distance between any two points on it decreases while the radius stays the same. since [tex]\pi=\frac{c}{2r}[/tex], where c is circumference and r is radius, [tex]\pi[/tex] no longer is a constant. the circle shrinks, but the radius stays the same. what is going on? does the circle turn into a cone?
     
    Last edited: Jul 7, 2005
  2. jcsd
  3. Jul 7, 2005 #2
    i just found out that the equivelence princliple doesn't apply here.
     
  4. Jul 8, 2005 #3
    You have stumbled upon a fanatastic paradox of special relativity

    As it appears in the history books, this is the very same case that lead Einstein to consider non-euclidean geometries in the physical universe.
     
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