Re: Rigid rod problem

On 2 Sep 2005 12:03:18 -0700, russell@xxxxxxxx wrote:

>Spoonfed wrote:
>> russell@xxxxxxxx wrote:
>> > russ...@xxxxxxxx wrote:
>> > > Kim B wrote:
>> > > > >> Take a look at http://micki.dk/diverse/accel.jpg - it shows 4 points
>> > > > >> accelerating, which all could belong to the same rigid rod (David, you
>> > > > >> may have a look too!).
>>
>> All right, we'll try it your way.
>>
>> The diagram represents the current and future positions of four marks
>> or observers on an accelerated rigid rod.
>
>Yes, and by rigid, we mean Born-rigid, i.e. the rod keeps
>its original proper length at all times. This puts some
>heavy constraints on *how* we do the acceleration, as we'll
>see.
>
>Please note that the time axis in the diagram is vertical,
>contrary to what I said in my original response. Right?
>So the events with a given time coordinate (say, t=0) are the
>horizontal lines. Which horizontal line are you calling zero?
>Presumably the bottom one. I'm asking this because what you
>say in the next paragraph does not quite correspond to what
>I see in the diagram.
>
>>
>> The first thing I notice is that at t=0, at the time when the farthest
>> portion of the rod is halted, the near edge has already begun to move
>> to the right. This makes sense, as there would be a time-lag between
>> the time the near edge was pushed and when the far edge received the
>> signal.

My fault, the drawing is imprecise
>
>No, we're not using a signal here. We simply arrange for
>the specified acceleration to take place, somehow. (E.g. we
>could do it with an array of tiny thrusters attached to the
>rod, all preprogrammed to fire at once.)

Exactly

>
>At t=0 *all* parts of the rod (not just one end) have v=0,
>but at that instant, everywhere, their a jumps discontinuously
>from 0 to some positive value that depends on position.
>
>This means that all the worldines are vertical at t=0 (since
>v=0) but they are all concave to the right at t=0 (since a>0).
>The concavity is greater at the rear because we have to make
>the acceleration greater there in order to maintain Born
>rigidity. The acceleration is greater there, but it doesn't
>happen earlier.
>
>(Of course we could have done the acceleration differently,
>but that would have produced a different diagram. Perhaps
>you even have such a diagram in mind; if so, I encourage
>you to reexamine the one whose link you posted.)
>
>>
>> I presumed that there was no way to keep a physical object from
>> crossing the light cone from the origin.
>
>In practice yes, there's no way -- because no acceleration
>can continue indefinitely, in practice.
>
>>
>> Now I'm having second thoughts. As the slope dt/dx of the hyperbola
>> would be greater than 1 all the way out to infinity, and yet it would
>> find a way not to cross the line x=t.
>
>Right, we are assuming the acceleration just keeps on going
>and going and going. That's unphysical, yes.
>
>>
>> If a bear is chasing you through the woods at velocity c, and you run
>> from him at an increasing velocity less than c, can you keep him from
>> reaching you forever? Twenty minutes ago, I would have said, obviously
>> not. But then if I thought about it for twenty minutes, I probably
>> would have said, yes.
>
>The answer is yes. The necessary acceleration depends on
>how far away the bear is from you when you start.

Just hope that the bear gets tired before yourself :-)

Kim
.

Relevant Pages

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