Re: if X observes a platform start moving, how can it appear contracted?

On Feb 12, 11:37 am, "Paul Cardinale" <pcardin...@xxxxxxxxxxxxxxx>
wrote:
On Feb 12, 7:10 am, benn...@xxxxxxxxxxxxx wrote:





I understand how a platform moving rapidly to the right, would appear
shorter to a stationary observer, than to an observer on the platform,
but I can't see how to reconcile that with the following:

Suppose X is standing beside a platform 1 km long (sorry Dirk! at
least I didn't say "mile" because I don't want to sound like a total
English major). Prior to time t=0, the platform is stationary. At
time t=0, from X's point of view, the whole platform starts moving to
the right at half the speed of light (X stays where he is). That
means if you graph the movement of the platform's endpoints on a
Minkowski diagram from X's frame of reference, their gradient will be
the same. But if their gradient is the same, and since they started
moving at the same point in time (from X's frame of reference), then
it would seem that at any time t > 0, the distance between the
endpoints of the platform is still 1 km... isn't it?

Again, rather than a complete explanation starting from scratch, what
I'm trying to find out is where is the mistake in the following:
(all of these statements can be prepended with "From X's frame of
reference", which is what I mean by phrases like "at the same time")
1) Before time t=0, the endpoints were 1 km apart.
2) The endpoints both started moving to the right at the same time,
t=0.
3) The left end passed kilometer marker 1 at the same point in time
that the right end passed kilometer marker 2.
4) Therefore, in X's frame of reference the endpoints were still 1 km
apart even after the platform started moving.

But this can't be right since in X's frame of reference, shouldn't the
platform appear contracted once it's started moving?

What happens depends upon how the acceleration of the platform takes
place.
It cannot be uniform in both X's frame, and the instantaneous frame of
the platform.
If the acceleration is uniform in X's frame, then in the frame of the
platform, the leading end will accelerate faster, stretching (or
breaking) the platform. If the acceleration is uniform in the frame
of the platform, then from X's view, the leading end of the platform
will accelerate more slowly than the trailing end.

Paul Cardinale- Hide quoted text -

- Show quoted text -

Thanks, I was starting to figure out this was what I hadn't taken into
account. So basically, if an object is rigid, and it goes from being
at rest to moving at a speed comparable to c, then in the frame of
reference of a stationary observer, different ends of the object start
moving at different times.

.

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