Re: Einstein's math and physical objects

From: Tom Roberts (tjroberts_at_lucent.com)
Date: 01/12/05 

Date: Wed, 12 Jan 2005 12:17:27 -0600

Harry wrote:
> Let the discs decellerate equally and synchronously
> (in the rest frame) from let's say 100 m/s circumferial speed until stop,
> thus conserving the "twist". I suppose that now you won't claim that the
> wires remain in a spiral shape, but that they do touch. Right?

No. See below.

> Next do the exact inverse, synchronously accelerating them back up to speed.
> What would make the wires separate in this perfectly symmetrical case, as
> they were crossing in the middle?

You are confused, and both you and the original poster keep complicating
things needlessly.

Imagine this physical situation: In inertial frame A: two disks of
diameter D separated by distance L along their common axis, which is the
X axis of frame A. With the disks at rest in frame A, two wires are
attached tautly between the disks at Y=+D and Y=-D. In frame A start
both disks and wires spinning at time T0, let them reach a steady-state
spin at time T1, at time T2 have them both slow down their spin, and
they both stop spinning at time T3. At all times the two disks are
attached to a common axle along the X axis, so they always remain in the
same orientation in Frame A, and the wires never have any twist in frame
A. The wires are stiff/taut enough so their bulging at the center due to
the rotation is negligible.

Now look at this from inertial frame B which is moving along the X axis
relative to frame A. The two disks do not start spinning at the same
time, and as they spin up the wires acquire a twist. But this twist is
due to the difference in simultaneity between the two frames, so in
frame B the wires follow a helical path (in frame A each wire is a
distance D from the X axis all along its length, and so in frame B that
still holds). The two disks reach steady-state at different times in
frame B, and the twist remains. Later, when the disks cease their
rotation, the twist will have precisely been cancelled as seen from
frame B.

        Don't forget that in frame B the disks+wires+axle also have
        a large velocity along X in addition to their rotation: my
        discussion above ignores this -- the twist and helix
        mentioned above are observed by taking a snapshot of the
        disks+wires at a constant time in frame B.

Yes, in frame B the wires will have a different value for tension than
in frame A. And they will have a different total length in frame B than
in frame A (of courseyou measure it at constant time in frame B) -- this
is complicated by the length contraction along X but none along Y and Z
and the fact that the wires are not parallel to any axis.

Instead of using a different frame B, one could accelerate the
disks+wires+axle using Born rigid motion to another inertial frame C,
and then look at them from frame A. In this case the view from frame A
would be similar to that from frame B above (for the same reasons),
except possibly for the handedness of the twist (which depends on the
direction of motion between the frames). And if instead of Born rigid
motion the disks+wires+axle were accelerated with the same acceleration
relative to frame A (as the original poster did), then the axle and
wires must stretch (in proper length), but things are otherwise the same
as for Born rigid motion.

> And why, ignoring inertial effects, would
> the force equilibrium create a spiral shape with constant distance to the
> centre of rotation?

Because such a helix is the Lorentz transform from A to B of the wires'
positions. The wires never cross in frame A, and so cannot possibly do
so in any other frame. <shrug>

> That would be a spectacular relativistic low-speed
> effect!

No, for any observable twist you'll need frames with a relative speed of
an appreciable fraction of c.

Tom Roberts tjroberts@lucent.com

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