Re: Rigid rod problem

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

>Spoonfed wrote:
>> russ...@xxxxxxxx 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
.

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