Re: does circumference contract with velocity?



On Oct 31, 3:36 pm, David <dsepp...@xxxxxxxxxxxxx> wrote:
I posted a velocity composition problem (Oct 31,2006).

The only way I see to solve it using Einstein's notions of space and
time is to use a bizarre notion about circles in relativity. Here's
the question:
If I have a disk that has a rotational velocity V at its outer
edge, does someone who is at rest with respect to the center of the
disk conclude that some sort of length contraction occurs because of V
and therefore the rotating disk has a smaller circumference than the
same disk when it is not rotating? That of course seems non-sensical
to me, but I don't see how to solve the velocity composition problem
without introducing that notion.
Thanks,
David Seppala


Comments of Gerald L. O'Barr <globarr...@xxxxxxxxx>:
If you have a disk that is rotating with an outer
surface velocity of v (v = r*w), the non-rotating
circumference, Co, will try to shorten to Cv, as
indicated in the standard relationship:

Cv = Co*SQRT(1-v*v/c*c)

(all measurements being made in the non-rotating
frame.)
The radius of the circumference, r, being at right
angles to the motion, v, will not shorten due to any
of its own relativistic motion. Thus, physically,
there will be an appearance of stresses within the
cylinder.
As the circumference tries to shorten, the radius
will resist it. Thus, the radius will show a
compression due to this resistance, and the
circumference will show a tension. Thus, there will
be some reduction (or compression) in the radius due
to the efforts of the circumference to reduce its
dimensions, and the circumference will not shorten as
far as it would naturally want to do.

Now if the disk is a very thin disk, the radius
can bend out of plan, allowing the circumference to
collapse to a smaller, more natural diameter, and the
radius can remain nearer to its original length (but
no longer in a straight line.) This configuration
produces some bending stresses, but reduces the
average compression in the direction of the radius,
and reduces the average tension in the circumference
directions.

The rotation of any real disk at relativistic
speeds is normally expected to be unstable. Thus,
due to the centripetal forces, the disk will normally
fly apart. The actions of these rotating stresses
are so much more than any of the relativistic
stresses, that measurements of these relativistic
stresses are not expected to be realized.

Thanks for reading.
Gerald L. O'Barr <globarr...@xxxxxxxxx>
Remove ... for e-mail.

.

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