Re: Rigid rod problem
- From: Kim B <spamfree@xxxxxxx>
- Date: Wed, 07 Sep 2005 21:28:06 +0200
On 7 Sep 2005 11:51:46 -0700, russell@xxxxxxxx wrote:
>Kim B wrote:
>> 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).
>
>Thanks, yes, that helps with the visualization. It agrees
>with our intuition that the one special event (the location
>of the bear when he starts chasing us) can't depend on how
>we draw the diagram.
>
>The 45-degree line through (0,0) is a degenerate hyperbola
>where *all* of the change in slope happens, as it were,
>exactly at (0,0). So, no matter what slope one chooses
>for the tangent, (0,0) must be on the line of simultaneity
>determined by that choice.
Right. And it is also the limit for how long we can stretch the rod to
the left.
Kim
.
- References:
- Re: Rigid rod problem
- From: Spoonfed
- Re: Rigid rod problem
- From: russell
- Re: Rigid rod problem
- From: russell
- Re: Rigid rod problem
- From: Kim B
- Re: Rigid rod problem
- From: russell
- Re: Rigid rod problem
- From: Spoonfed
- Re: Rigid rod problem
- From: russell
- Re: Rigid rod problem
- From: Kim B
- Re: Rigid rod problem
- From: russell
- Re: Rigid rod problem
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