Relativistic Rotating Sphere
traveling through space

posted this on TVF's forum but didn't get much response. It's a teaser ;o) Supposedly, this is a form of Thomas precession but I don't see that that explains it at all.

Suppose there is a rapidly spinning sphere of uniform density out in intergalactic space.

Now, if it moves through space at relativistic velocities (say, relative to the backdrop of stars or, better, the microwave background radiation), the side of the sphere which is spinning in the direction of motion is travelling faster than the other side. [relative to that frame of reference]

Hence, the side with faster velocity has "more mass" and hence must have greater centrifugal force (mv2/r ... where 'r' is constant) ... so that ... in relativistic terms ... it should become a curve ball in outer space, curving in the opposite direction as a pitched baseball in the atmosphere. This would constitute a violation of linear momenutm conservation ... so ... it must be that the axis of rotation shifts over to compensate. So, a dot on the sphere, marking the axis of rotation, would appear to move over and a measurement by the observer in the MBR reference frame would measure, say, the left radius as longer than the right radius.

But ... here's the problem ...

What would an observer travelling with the sphere see? By relativity theory, he must not observe any change in the axis of rotation of the sphere because the radius on both sides of the sphere must stay the same for his reference frame. But for the stationary observer the radius to one side is different from the other side to save momentum conservation. So, one guy sees (and measures) one thing and the other guy sees another qualitatively different thing concerning left-right symmetry.

When there is a discussion of relativistic contraction, it is always about length along the line of motion. What about left and right of the line of motion? How is the above problem reconciled with both relativity theory and Newtonian mechanics?

Wait a minute ... this can't be right ...

Linear Momentum Conservation ...

... is no justification for moving the axis over. By standard relativity theory, the spinning sphere should become shaped like an oblate spheroid (almost). That is, a cross section perpendicular to the direction of motion should still be a circle ... but the cross section perpendicular to the axis of rotation should be slightly egg-shaped, i.e. not a perfect ellipse but somewhat "pointy" on the side of the sphere with the greater velocity.

If you had painted dots on the outside of the sphere in a circle around the axis of rotation, regardless of the translational motion, the dots should still define the axis of rotation as it would be without the translational motion. The circle of dots would just be slightly egg-shaped.

curvball.gif - 7kb

Meanwhile the person in the same reference frame as the sphere should just see a spinning sphere.

So, what about the "curve ball"?

In the red figure above, The radii R1 and R2 must remain equal, both to the non-moving observer in the MBR reference frame ... and ... to the observer traveling with the sphere.

So, what do we do about the greater centrifugal force on the "pointy" side of the egg? If we leave it as is ... linear momentum conservation is gone. The R2 side has more mass but equal radius so it must show greater centrifugal force. What's the solution here?

A point on the back end of the sphere going to the front via the R2 side must arrive at the front in exactly the same time as a point at the front going round to the back via the R1 side. For, if this were not so, then the two points opposite each other on the front and back of the sphere could not be the same points as on the previous rotation ... and ... this must be true of all subsequent rotations ... so that ... sphere would be theoretically constrained to ... do what? ... explode?!

Yet the R2 path is slightly shorter than the R1 path and the velocity is also faster so it must travel the path on that side faster!! Well, I can see where this is going. As an observer in the MBR frame we can't see the front end and the back end of the sphere at the same time, so the time lag between the two signals will iron out the above contradiction.

That still leaves the centrifugal force problem.

It must be this ...

The radius is equal on both sides of the sphere only along the x axis. At all other places in the rotation, the R2 radii are always a little shorter so the sum of all the force components must equal out on both the R1 and R2 sides.

Now, I can go to bed, content that the universe is once again "logical" ;o)

Addendum: 11/01/37

Wait a minute ...

Wait just a freakin' minute here ... That radius term is in the denominator.

Centrifugal Force = mv2/r

Mass is greater on the R2 side as is velocity which causes the force there to be greater. But because r is in the denominator, a lesser r causes a greater force as well. All three parameters indicate a greater force on the R2 side from the viewpoint of an observer in the MBR. I am seeing nothing here to stop the curveball.

Will linear momentum conservation break down at relativistic velocities?

Certainly, we can't do any mechanical experiments to test this. The forces and energy involved is, technically, out of range, i.e. nothing made could hold itself together with the kind of spin I'm imagining. Any kind of test would have to be done on the atomic or sub-atomic level. For instance, the electron goes around the proton at 1/137 of the velocity of light (fine structure constant), And the amount of energy gained by the electron in falling down that gradient (Rydberg energy) is identical to the amount of energy required to create the mass increase from the relativistic motion of the electron in that ground state. So, there are testable quantities here.

[Purists would state here that there are no electron orbits (as in translational motion)
and that there is only the "electron cloud". ]

That's a respectable fraction of light velocity. Perhaps lining atoms up in a magnetic field (at superconducting temperatures) so that all their spins are aligned ... then ... rotate that whole thing sweeping around a central axis ... hmmmmm ... That's a UFO propulsion scheme if I ever saw one ;o)

Spin a super-cooled magnet on the end of a wire and the entire apparatus should, in principle, weigh less with the magnetic axis spinning in a plane parallel to the earth's surface ... my eyes are bugging out as in "I'm sending you ... back ... to the future" ;o)

curvbal2.gif - 7kb

OK ... why will this not work even in principle? I don't presently know. Let me go to bed again and think on it another day.

As of 1/3/38, I still see no reason why this will not work at least in principle ... provided that an absolute reference frame exists (as I believe it must).

Ebtx Home Page