On 7 Sep 2005 10:54:47 -0700, wrote:

Spoonfed wrote:
wrote:
Kim B wrote:

[snip]

If you choose a point on the rod a use its current speed as your FOR,
the the rest of the rod will fit nicely in this FOR (along the FOR's
line of simultaneity) ... with the same speed all along and the
correct proper length, exactly as it fits in our "rest" frame at the
base line ... all frames are equal, assuming the rod has accelerated
and will accelerate forever.

Thanks. Of course you are quite right about that, and I
apologize for my many mistakes here.

I believe it when I hear it, but it's a little tricky to figure out.

It seems surprising that no matter what reference frame you are in, all
parts of the rod will pass v=0 at the same time. I'm not in the mood
to develop a proof, but it seems right.

I won't prove it here either; I'll just give some
further motivation.

If you imagine Born-rigid acceleration over a finite
time, and then ending, it must be the case that in the
end, no part of the rod is moving relative to any other
part (otherwise the rod would not be rigid). So, in
that sense it's not a surprising result at all.

On the other hand, given the complications that occur
with acceleration, I agree it's a bit surprising that
Born-rigid acceleration is possible at all, even in
theory.

You can see what's happening if you take Kim B's
diagram and, say, pick some point on the leftmost
hyperbola and draw the tangent there. Then draw
parallel tangents on the other hyperbolas, at whatever
points are determined by the parallel requirement.
Then notice that the points you have picked all lie
on the same straight line, tilted slightly upward to
the right. This is a line of simultaneity in the
frame that is moving at the speed given by the tangent
slope, and (recognizing that our diagram is drawn in
Euclidean rather than Minkowskian space) we see that
this line would actually be perpendicular to the tangent
if we redrew the diagram in the coordinates of that
frame. Thus, as Kim B says, the diagram looks the same
no matter what frame we draw it in.

All lines of simultaneities in my diagram, drawn this way, pass
through (0,0).

Kim