Re: Androcles asks for Derivation of LT

From: Randy Poe (poespam-trap_at_yahoo.com)
Date: 02/17/05 

Date: 16 Feb 2005 18:29:42 -0800

Androcles wrote:
> "Randy Poe" <poespam-trap@yahoo.com> wrote in message
> news:1108571011.321417.324810@c13g2000cwb.googlegroups.com...
>
> The Ghost In The Machine wrote:
> > In sci.physics, Androcles
> > <Androcles@MyPlace.org>
> > wrote
> > on Tue, 15 Feb 2005 23:53:39 GMT
> > <7YvQd.110093$K7.35739@fe2.news.blueyonder.co.uk>:
> > >
> > > "Randy Poe" <poespam-trap@yahoo.com> wrote in message
> > > news:1108480538.284135.143160@c13g2000cwb.googlegroups.com...
> > >
> > > Androcles wrote:
> > >
> > >> Differentiate this for me, Poe :
> > >>
> > >> ½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/ something_not_there )]
> > >> = tau(x',0,0,t+x'/(c-v))
> > >>
> > >
> > > Oh, I see what you're trying to reproduce. Yes, I
> > > can help you out.
> > >
> > > The original equation is this:
> > >
> > > (1/2)*[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v)] =
> > > tau(x',0,0,t+x'/(c-v))
> > >
> > > "Where in that equation did (c+v) occur?" --- Poe the blind man.
>
> Androcles knows damned well that "that equation"
> was not the equation above, since that comment
> did not appear with that equation and he had to
> do the manual editing across multiple posts
> to create the juxtaposition.
>
> > > *** off, Poe, you're troll. You couldn't help yourself to see,
how
> the
> > > *** are you going to help anyone else?
>
> Poe:
> As always, I am merely trying to answer the question
> you asked. In this case you got stuck trying to figure
> out how Einstein went from the above equation to
> the differential equation and you ASKED ME explicitly
> to help you get unstuck. I was happy to oblige.
>
> > in section 3 suggests the equation
> >
> > 1/2 [ tau(0,0,0,t) + tau(0,0,0,t+x'/(c-v)+x/(c+v) ]
> > = tau(x',0,0,t + x'/(c-v) )
> >
> > The c+v term apparently is because of the reflection of the
> lightbeam.
>
> That takes the same time to return by DEFINITION.

In the moving frame. Not in the stationary frame.

It's why the tau's are related by tau1 = 1/2(tau0 + tau2).

But there's no such "definition" about the time to travel
between source and mirror as measured by the stationary observer.

> Not by observation, not by logic deduction, but by definition.

In the moving frame. Not in the stationary frame.

> It takes the same time to return because Einstein says so.

He says so in the frame in which source and destination
are at rest. He doesn't say so in the frame in which they
are moving.

> I'm
> not sufficiently insane enough to believe him. Drive me crazier.

What your particular psychosis seems to be is an unwillingness
or inability to keep straight the moving and stationary
frames, to switch them willy-nilly.

> "If at the point A of space there is a clock, an observer at A can
> determine the time values of events in the immediate proximity of A
by
> finding the positions of the hands which are simultaneous with these
> events. If there is at the point B of space another clock in all
> respects resembling the one at A, it is possible for an observer at B
to
> determine the time values of events in the immediate neighbourhood of
B.
> But it is not possible without further assumption to compare, in
respect
> of time, an event at A with an event at B. We have so far defined
only
> an ``A time'' and a ``B time.'' We have not defined a common ``time''

> for A and B, for the latter cannot be defined at all unless we
establish
> by definition that the ``time'' required by light to travel from A to
B
> equals the ``time'' it requires to travel from B to A. "

We've been round this loop before. The part you keep deliberately
snipping is this:

"Thus with the help of certain imaginary physical experiments we have
settled what is to be understood by synchronous stationary clocks
located at different places"

The entire discussion above is talking about when A and B are
stationary relative to the clocks. When A is the moving source
and B is the moving mirror, the frame where they are stationary
relative to the clocks is the k frame. It is the k clocks that
are moving along with A and B.

The ground observer's clocks do not count as "stationary"
relative to this A and B, and no amount of pretending Einstein
said "regardless of relative motion" will erase the fact that
he actually said "stationary clocks".

> "Cannot be defined at all, unless"... x'/(c+v) = x'/(c-v)
>
> Unless (x'/(c+v) + x'/(c-v) ) / 2 = x'/(c-v)

No. The clocks that are sitting still relative to A
and B measure the tau's. Those aren't the tau's.

> Unless we include a function tau and the distance AB

Yes on tau. Tau is what the AB clocks measure.

Bad boy on putting A and B inside the parentheses. Stuff
inside the parentheses is NOT stuff measured in the frame
of A and B.

               - Randy

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