Re: Rigid rod problem
- From: "Todd" <x@xxxxxxx>
- Date: Wed, 31 Aug 2005 20:43:55 GMT
<russell@xxxxxxxx> wrote in message news:1125451036.374445.238990@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
bz wrote:dseppala@xxxxxxxxxxxxx wrote in news:430b11f5.38471559@news- server.austin.rr.com:
>>We see the optical equivalent of a thunder roll, as the sound from
>>different portions of the path reach our ears.
>>
>>We recognize that sight and sound can be missleading and that no >>wrapping
>>actually occurs.
> Wrapping actually does occur. In the rest frame of the rod and of the
> rotating cylinder, we are attaching one end of the rod to the cylinder
> a second before we attach the other end. The cylinder rotates 10
> times before the other end is attached. Since the speeds are everyday
> values can actually do this experiment over a much shorter length.
> The short rod will attach to the short segment of the rotating disk,
> and if we attach one end at a different time then we attach the other
> end (the stationary frame view), the attached rod will spiral about
> the cylinder. In the problem I posted because of the lengths involved
> the spiral and wrapping occur over large distances, but it still
> occurs.
> David
>
Not if the attachment is done simultaniously all along the length using sync'd clocks.
Bz is again talking out of ignorance. The attachment is synched in the *moving* frame. So, in the stationary frame the attachment is not simultaneous, and the rod wraps of necessity.
We could sync it another way: At a great distance, we have a signal source. The signal from that source approximates a plane wave all along the length of our cylinder.
BTW, at that distance, we have a powerful telescope and we watch the joining take place. No twisting is observed.
None is observed in the frame for which the attachment is simultaneous. But enough of bz -- my post is mainly about something else, to wit:
I think dseppala has actually managed to ask an interesting question here, although as usual he's made it more complicated than it needs to be. Also I believe he errs (along with most of his respondents here) in thinking that this can be treated as a 1D problem when in fact all 3 spatial dimensions must be considered.
Let me recast it in simpler terms: Imagine an ordinary steel corkscrew (actually any suitably asymmetric object would do) rotating along its long axis; and suppose for simplicity that it has exactly an integral number of turns so that it doesn't wobble as it rotates. The C.O.M. is fixed; 3-momentum is conserved in this frame.
Now, look at it from a different frame, one that is moving along the screw axis. We find that the number of turns in the corkscrew has *changed* so that there is more mass on one side of the axis than the other. What happened to momentum conservation?
Note that in some frame the corkscrew will even be *straight*. (This is the extreme case that dseppala was considering.) And yet it will still be going round and round its axis as it zips along. 3-momentum definitely not conserved.
You can't just say this is impossible; I'm talking here about an *ordinary* corkscrew, not some 300000km long one. The inescapable conclusion is that 3-momentum is not conserved.
I think the solution to this 'paradox' is that we need to include the momentum contributed by the _stresses_ in the corkscrew. These stresses arise because the corkscrew is rotating. In the frame in which the COM is screaming along at relativistic speeds, these contributions are large.
It must turn out that the total momentum (including the contributions from the stresses) is indeed conserved if no external forces are applied to the system. To actually check this, you would have to first work out the stresses inside the rotating corkscrew in the frame in which the COM of the corkscrew is at rest (yuck), and then transform these components to the moving frame (pretty easy).
A nice treatment of stress contributions to momentum is given in Tolman's _Relativity, Thermodynamics, and Cosmology_ which is cheaply available from Dover Pub. A famous example of a paradox that arises if you don't include the stress momentum is the so-called Right-Angled Lever Paradox (see page 79 of Tolman ).
Aside 1: A helix with an integral number of windings will have its COM on the symmetry axis of the helix. By 'symmetry axis', I mean the axis of the cylinder on which the helix may be imagined to be wound. However, I don't believe the symmetry axis of the helix will be one of the _principle axes_ of the moment of inertia tensor of the helix. This means that if you start the helix rotating about its symmetry axis, it will wobble in the absence of any external forces. So, this is an added complication.
Aside 2: If I'm not mistaken: In order for there to exist an inertial frame in which the rotating helix appears straight, the helix must be wound such that in the helix's own rest frame the 'pitch' of the helix (distance between consecutive windings) is greater than 2*Pi*R where R is the radius of the helix. I think that's kind of interesting.
Todd
.
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