Re: Turn of the screw paradox/contradiction
- From: "Paul B. Andersen" <paul.b.andersen@xxxxxxxxxxxxxxx>
- Date: Sat, 01 Sep 2007 15:02:52 +0200
David wrote:
The twins paradox and others seem to include a re-synchronization of
clocks, and/or disagreement between frames on initialization. Instead
of initializing clocks, the following problem uses a simple mechanical
means to establish initial and final conditions. Please comment on
which physics step is incorrect in the analysis.
Turn of the screw.
Let there be a rest frame with a very long threaded rod that has a nut
(Nut A) at x=0, and another nut (Nut B) at x=L. Let there be a set of
triplets in this rest frame. Two of the triplets are at x=0, and the
third is at x=L. One triplet is spinning Nut A at N revolutions per
second. Another is spinning the rod at N revolutions per second. And
the third is spinning nut B at N revolutions per second. They have
been spinning their respective objects for a very long time so that a
steady-state condition has been achieved. Since Nut A and Nut B are
spinning at the same rate as the long rod, they do not move with
respect to the rod. (We can setup the rod rates/accelerations such
that this is true eventually and a steady-state condition is
achieved). We now put a mark on the rod at the location of each of
the two nuts. All inertial frames agree that Nut A and Nut B are not
moving relative to the threaded rod (their position does not change
relative to the marks we put on the threaded rod).
Now we let the triplet who is spinning the rod continue spinning the
rod at N revolutions per second while that triplet moves from Nut A to
Nut B at some velocity V. This triplet continues spinning the rod at
N revolutions per second as measured by his clock. When he gets to
Nut B he stops moving so that he is back in the original rest frame.
He continues spinning the rod at N revolutions per second.
Per SR, while the triplet moved from Nut A to Nut B, observers in the
rest frame say that the moving triplet spun the rod fewer revolutions
that Nut A and Nut B turned. Therefore, when a steady-state condition
is once again achieved, Nut A and Nut B will no longer be at the marks
that were initially made on the threaded rod. However, the moving
triplet says the opposite occurred, Nut A and Nut B turned fewer
revolutions than the threaded rod turned so that when a steady-state
condition is again achieved, Nut A and Nut B will not be on the marks
that were initially put on the threaded rod. The problem is that the
rest frame observers say Nut A and Nut B moved in one direction
relative to the mark on the threaded rod, while the moving triplet
says that Nut A and Nut B moved in the opposite direction relative to
the marks on the threaded rod. Both views cannot be correct.
The velocity V can be very slow so that any mechanical effects of the
moving triplet spinning the rod occur faster than the triplet is
moving, or we can make V very high such that the moving triplet
travels faster from Nut A to Nut B than any mechanical effects can
travel. In either event, we wait until a steady-state condition is
achieved to see whether or not the rod rotated more times than either
Nut A or Nut B.
How is this explained using SR?
Thanks,
David Seppala
I find your ability to confuse yourself to such a degree
that you don't see the solution to your own problems
rather amusing. :-)
Paul
.
- Prev by Date: Re: after 100 years of relativity
- Next by Date: Re: What Does SR stand For?
- Previous by thread: Re: Is anti-time relative, too???
- Next by thread: "How Does Light 'Know' How Fast to Travel?"
- Index(es):
Relevant Pages
|
