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I am tempted to give an experimentalist's point of view, though of course I agree with Lubos and Savanah's answers. Consider a disk of radius R, then circumference is 2πR. Now, make this disk rotate at velocity of the order of c(speed of light). Here it is evident that you are defining a center of mass system for a disk and a person/machinery that will ...


-1

To someone moving on the edge of the disk, the disk is length contracted in the direction of travel. So to them, it is not a circle, no problems with geometry. Euclidean geometry only holds for an inertial plane of simultaneity. By the end of this answer you should know when and how to use Euclidean geometry and what it means (or doesn't). And a collection ...


2

This is just relativity of simultaneity again. A similar thing happens if you have a bunch of spaceships in a line that fire their thrusters at a fixed time. Different observers will disagree about whether they fired at the same time and will disagree about the spacing. Always in a consistent way. So I'd like to address the concept of geometry by not having ...


6

No, it doesn't violate the rules of geometry, it violates the rules of Euclidean geometry. Simple conclusion: for an observer fixed to a disk rotating uniformly relative to an inertial frame, the spatial geometry is non-Euclidean; in particular, the ratio of a circle's circumference to its diameter depends both on the circle's diameter and center position. ...


6

What is wrong is the idea that one can actually make the disk rotate; and it will remain perfectly rigid. In reality, what this correct argument shows is that relativity doesn't admit the existence of any perfectly rigid bodies. This is a perfectly basic, settled, and indisputable textbook material that every mature physicist knows. The first sentence of ...



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