Re: Einstein's math and physical objects

From: Paul Cardinale (pcardinale_at_volcanomail.com)
Date: 01/04/05 

Date: 4 Jan 2005 08:39:20 -0800

dseppala@austin.rr.com wrote:
> I don't see how Einstein's equations and concepts work in the
> following example. Can anyone offer the physical explanation of what
> happens here in terms of Einstein's theory.
>
> Let there be two circular disks in the y-z plane, with their centers
> on the x-axis. Let one disk be at x=0 and let the other disk be at
> x=L. I'll call the disk at x=0, disk A, and I'll call the other
disk,
> disk B. Now stretch a wire from the top of disk A to the top of disk
> B. And stretch a wire from the bottom of disk A to the bottom of
disk
> B. These two wires are attached to points 180 degrees apart on each
> disk. We note the following physical property. If one disk is
> rotated 180 degrees or more with respect to the other disk, the two
> wires will cross each other (in the steady-state condition). At time
> t0, we start both disks rotating at identical rates. Let's make the
> rotation speed N revolutions per minute. When a steady-state
> condition is achieved, the two wires do not cross each other.
>
> Let the mass of each disk be much greater than the mass of the
> attached wires. Now at some time tA we accelerate both disks and the
> attached wires to some velocity V along the x-axis. Both disks are
> accelerated in an identical fashion. We note that as viewed in the
> original reference frame the distance between the disks remains L
(the
> wires, according to Einstein stretch), and we note that in terms of
> the rotation rates, the two disks remain in phase. That is, the
> relative rotation angle between the two disks remains constant
(zero),
> as measured in this frame. When the end of one wire is at the top of
> disk A, the other end of that same wire is at the top of disk B (as
> measured in this frame). And when one wire is at the top of disk A,
> the other wire is at the bottom of disk A (and the other end of the
> wire is at the bottom of disk B).
>
> Now let's say the velocity V and distance L are such that
simultaneous
> events measured in the original reference frame, are measured to be
> one second apart in a frame that is moving with V relative to the
> x-axis. This implies that the rotation angle of the two disks is not
> at zero degrees as measured in this frame. Depending on the rotation
> rate N we pick, we can make this angle greater than 180 degrees (or
> substantially larger than 180 degrees). In this frame, when the end
> of one wire is at the top of disk A, the other end of that wire is
not
> at the top of disk B (unless the rotation angle of B and A differ by
> exactly N revolutions). As stated previously, if the two disks are
> 180 degrees or more out of phase, the two wires must cross in the
> steady-state condition. If the wires are attached 180 degrees
apart
> on each disk, there is no physical way to keep the stretched wires
> from crossing each other in the steady-state condition if the disks
> get out of phase by 180 or more degrees. But they never cross as
> viewed in the original reference frame. They only cross if the two
> disks have a relative angle greater than 180 degrees between them,
> which they don't. Its non-sensical to have both physical conditions
> simultaneously true. Please explain how this is resovled using
> Einstein's notions of space and time.
> Thanks,
> David Seppala

In all frames, the wires remain on the cylindrical region between the
disks (neglecting distortions due to centrifugal acceleration).
Thus in the original frame, the wires for a double helix.
Paul Cardinale

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