Gravity contracts length

  • #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:

Answers and Replies

  • #2
i just found out that the equivelence princliple doesn't apply here.
 
  • #3
1,256
2
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.
 

Related Threads on Gravity contracts length

  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
12
Views
2K
  • Last Post
Replies
2
Views
17K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
6
Views
24K
  • Last Post
Replies
2
Views
5K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
8
Views
3K
J
  • Last Post
Replies
1
Views
8K
Top