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Re: Relativity



Concerning my analysis of the rotating/translating disk Paul Camp wrote:
.... We label each point on the rotating
circle with a value of theta (where 0 <= theta < 2*pi) the Cartesian
coordinates of these points are given as:
x' = R*cos(theta + omega*t') and y' = R*sin(theta + omega*t')

I may be wrong (wouldn't be the first time) but it looks to me
like these expressions build in from the very beginning the
condition that the rotation have no effect on the shape
of the disk since you are assuming that an observer in a state of
translational (but not rotational) rest with respect to the disk sees
a circular disk. If that is true, then it is not surprising that we
end up concluding that the rotation has no effect when the observer
is *not* at translational rest either -- we are basically deriving
what we assumed to be true to begin with -- that rotation has no
effect of the apparant "circularity" of the disk.

In the non-translating/fixed center frame the problem has rotational symmetry
about the rotation axis/disk center whether or not the disk is rotating.
Therefore, by symmetry, in this frame the disk remains circular. (It is true,
however, that the rotating disk violates the parity symmetry that the disk had
when it was not rotating, but this does not affect the axial rotational
symmetry, and hence, the shape of the disk.) So you are right after all--I
did build in the cicularity of the rotating disk because it *is* circular in
the non-translating frame.

It is not clear to me that this should be valid. It would certainly
not be true if we replaced the disk with a discrete set of small
boxes travelling in circles -- the individual boxes would be
Lorentz-distorted. Isn't the disk just a limiting case? Glue all the
boxes together and (assuming the flywheel doesn't blow up as Leigh
and others suggested) wherever they are Lorentz-contracted the disk
must get pinched.

To the extent that the system can withstand the huge internal stresses
developed while rotating at high speed the "individual boxes" will be
distorted, but their distortions will all be symmetric such that the overall
shape of the disk is still circular in the non-translating frame.

David Bowman
dbowman@gtc.georgetown.ky.us