Re: Rigid rod problem

Spoonfed wrote:
> >
> > All parts of the rod are traveling at same speed ... but time passes
> > not synchronously along the rod. Clocks run slower at the back end

I don't see in what sense your first statement could
be true; it certainly isn't true in the original inertial
frame, where at all t>0 the rear of the rod is moving a
tiny bit faster than the front. Indeed that is *required*
for your second stmt to be true -- which btw it is. It's
speed, not acceleration, that determines clock rates
relative to clocks in an inertial frame.

> >
> > Kim
> >
>
>
> All parts of the rod are traveling at the same speed according to an
> observer standing on the accelerating rod.

Well, to say that, you would need to specify what
coordinates the accelerated observer is using. There
is really no obvious choice in this example, AFAIK.

I was aware of this complication, but didn't mention it,
when I wrote earlier about the rod maintaining its original
proper length. Such a statement sweeps a lot of important
details under the carpet. I'm not an expert in this area,
and that fact that such a distinguished name (as Born) is
attached to it suggests that it is not simple.

One thing I do know is true, is that in the original frame
the rod will always have length L/gamma, where L is the
original proper length. But this isn't true in any other
frame, unless I err somewhere in my thinking. However, it
would be possible to turn off the thrusters on some exactly
timed schedule in the original frame (*not* simultaneously
there) such that the rod would continue to coast inertially
with its proper length unchanged; and thereafter of course
it would have length L/gamma in all inertial frames.

>
> To an inertial observer, who watches the rod slow down, stop, and
> accelerate to the right again,

Wait, the acceleration is *always* in one direction (let's
say to the right).

it will always appear that the left end
> of the rod is moving slower than the right end, (except for the moment
> that the entire rod is stopped.)

There is *no* moment when the entire rod is stopped, in such
a frame. During the acceleration there is only one frame
that has all of the rod in it simultaneously, and that is
the original frame -- and only at time t=0.

>
> To the inertial observer the difference in the speed of the clocks at
> the front and back end is accounted for entirely by the difference in
> time dilation caused by the difference in velocity between the ends.

This is true. Also the various inertial observers see a
different *constant* offset in the clocks due to RoS.

>
> If I ever have the time to finish my demo, I'll show that in more
> detail.

I think you may need to work a bit more on the theory,
from what you say above.

.

Relevant Pages

  • Re: How can this work in relativity?
    ... > The rod experiences uniform acceleration along its entire length, ... >> frame, because we know as a given that the acceleration is ... Since it is now in uniform motion in the original frame, ...
    (sci.physics.relativity)
  • Re: How can this work in relativity?
    ... > that as long as you do your calculations in a single frame, ... >>> uniform acceleration by applying the appropriate compensating ... >From the moving FoR that includes the rod, ...
    (sci.physics.relativity)
  • Re: TomToms stupidity (re: was always TomToms stupidity)
    ... you acknowledged yourself that this type of acceleration would ... >>>cause the rod to stretch in it's own frame. ... or accelerating the rod comes from the rocket. ...
    (sci.physics.relativity)
  • Re: TomToms stupidity (re: was always TomToms stupidity)
    ... >> acceleration is immaterial, and Daryll was right to neglect it. ... >> relative to the stationary frame. ... and this "pulling" will stretch the rod. ...
    (sci.physics.relativity)
  • Re: SR Length Contraction - how do physicists explain this
    ... >> from the pov of the rest frame. ... relative to the other endpoint on each rod. ... change length during the acceleration. ...
    (sci.physics.relativity)