Re: Rigid rod problem

russ...@xxxxxxxx wrote:
> Kim B wrote:
> > On 31 Aug 2005 05:04:35 -0700, "Spoonfed"
> > <jonathan.doolin@xxxxxxxxxxxxxxxxxxxxxx> wrote:
> >
> > >Kim B wrote:
> > >> On 30 Aug 2005 09:11:50 -0700, "Spoonfed"
> > >> <jonathan.doolin@xxxxxxxxxxxxxxxxxxxxxx> wrote:
> > >>
> > >> >
> > >> >Kim B wrote:
> > >> >>
> > >> >> In the space-time diagram, any uniformly accelering object will have a
> > >> >> 45° asymptote. Any events on the other side of this asymptote will
> > >> >> never be visible to the object
> > >> >>
> > >> >
> > >> >Any event that can be seen by an inertial observer should also be seen
> > >> >by a uniformly accelerating observer sitting right next to him, passing
> > >> >him, or coming to a momentary rest adjacent to him.
>
> They both see the same stuff *at* the moment of passing.
> But Kim is talking about what the observers can see
> *afterwards*. The inertial observer can eventually see
> light from the red dot, but the one who accelerates forever
> can *never* see it.
>
> > >> >
> > >> >If that conflicts with whatever you just said, you should let me know.
> > >>
> > >> 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!).
>
> This is nice, but note, the axes aren't labeled. It's
> important to realize that x is on the vertical axis and
> t is horizontal in this diagram.

Woops, I got that exactly backwards!! Sorry.

In Kim's diagram, t is vertical and x is horizontal, and
btw this follows the usual convention.

[snip my ok stuff]

> No, the red dot is outside that observer's future light
> cone. The observer could observe *light* from the red dot
> (assuming he does not accelerate forever) but there is no
> possibility of his being present at the event itself.

What I wrote in this paragraph is wrong, of course.

>
> > >
> > >The observer accelerates until the red dot is directly above the
> > >origin.
>
> This makes no sense. You have a moving origin? But this
> is a space*time* diagram. Time is already there, on the
> diagram; there is no other time for anything to move in.

Perhaps you meant "directly above the observer"? That
would be possible. (I.e. it would be in the observer's
future.)

>
> As time goes by, the observer will see each of the four events
> > >well before he reaches the red dot. The light from those events
> > >travels at a 45 degree angle up and to the left, and when it crosses
> > >the path of the observer, he will see it.
>
> *Nothing* travels to the left in Kim's diagram.

Sorry, I got that all wrong.

The real problem with what you said is that we are not
interested at all in the observer at the origin, nor in his
future travels. The observers we are interested in are the
four observers whose worldlines are the hyperbolas in the
diagram. *Those* observers never see any information from
the red dot. (Though obviously they *could*, if they stopped
accelerating.)

The time
> axis is positive to the right. Things to the left are in
> the past. Obviously, you have seriously misinterpreted
> something here.

Oops, I'll take my own medicine now.

>
> >
> > The hyperbolic lines in my diagram ARE the worldline of 4 observers on
> > a uniformly accelerating rod -- and none of these will ever see the
> > light from the read dot.
>
> Correct. (But not correctly proofread ;-)

Yeah, that much was right.

.

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