Re: Roberts' persusaive rhetoric.

From: Paul B. Andersen (paul.b.andersen_at_hia.no)
Date: 10/19/04 

Date: Tue, 19 Oct 2004 15:44:52 +0200

"sal" <pragmatist@nospam.org> skrev i melding news:1098154892.OPjjUzqGjauqy3e6VJ+ZIQ@teranews...
> On Mon, 18 Oct 2004 20:50:07 +0000, Androcles wrote:
> > LMAO! I'm surprised you challenged Andersen's Transforms, I have to admit.

For once I agree with Androcles.
I am surprised as well! :-)

> I've looked them over in the past (as you no doubt recall). They are, as
> I recall, the Lorentz transform with a sign flipped (i.e., wrong-way).
> They provide the wrong answer, very reliably, every time. I seem to
> recall you were delighted because Paul Andersen supposedly posted the
> transform, with incorrect sign, at some point, and you've been quoting it
> ever since.

I wrote it with the right signs, and it provides
the right answer, very reliable, every time.

Please read it again carefully, you will see that it is correct.

> "That is, we can reverse the directions of the frames
> which is the same as interchanging the frames, which - as I have told
> you a LOT of times, OBVIOUSLY will lead to the transform:
> t = (tau-xi*v/c^2)/sqrt(1-v^2/c^2)
> x = (xi - v*tau)/sqrt(1-v^2/c^2)
> or:
> tau = (t+xv/c^2)/sqrt(1-v^2/c^2)
> xi = (x + vt)/sqrt(1-v^2/c^2)" -Paul B. Andersen

The quotation is correct, and my words are correct, even if
they may be a bit ambiguous, taken out of context as they are.
But I think it _should_ be obvious that "reversing the direction"
refers to reversing the direction of relative motion of the frames
compared to the "conventional direction", in which case
the signs will be opposite to the "conventional form" of the LT.

But here is the context it is taken from:

Einstein let the k-system (greek letters) move in the positive
direction along the x-axis of the K system (latin letters)
with xi- and x-axes aligned.

            |-> v
    --------|-------------> xi k moving
    -----|----------------> x K stationary

In this case the Lorentz transform is:

 xi = (x - vt)/sqrt(1-v^2/c^2)
 tau = (t - xv/c^2)/sqrt(1-v^2/c^2)
or:
 x = (xi + v*tau)/sqrt(1-v^2/c^2)
 t = (tau + xi*v/c^2)/sqrt(1-v^2/c^2)

But there are alternative ways to define the frames of reference.
We can reverse the directions of motion of k

     v <-|
    -----|----------------> xi k moving
    --------|-------------> x K stationary

which is the same as interchanging the frames,
(as compared to Einstein's convention)

            |-> v
    --------|-------------> x K moving
    -----|----------------> xi k stationary

which OBVIOUSLY will lead to the transform:
   t = (tau-xi*v/c^2)/sqrt(1-v^2/c^2)
   x = (xi - v*tau)/sqrt(1-v^2/c^2)
  or:
   tau = (t+xv/c^2)/sqrt(1-v^2/c^2)
   xi = (x + vt)/sqrt(1-v^2/c^2)

It should be an obvious triviality that the signs in the LT
depend on how the frames of reference are defined.
(You can find two more combinations by letting the axes
 point in the opposite directions as well.)

The story behind this is that Androcles - several years
ago - reversed the direction of motion of the k-system,
and made a big number of the fact that the signs in the LT
changed. He finds it contradictory, and insists that both
forms of the LT must be valid _simultaneously_.

I have indeed told him "a LOT of times" that it is not
contradictory but an obvious triviality - to no avail.
That guy is simply unable to learn anything!

That's why he keep quoting my correct words - thinking
that he is making a fool of me when doing so.

You can read all about it in:
http://groups.google.com/groups?q=g:thl3103600516d&dq=&hl=no&lr=&selm=38CA6965.F185F814%40hia.no
and the thread it is in.

 I am confident that you now know who he is making a fool of. :-)

Paul

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