Re: Sagnac Fully Supports the BaTh.

Dr. Henri Wilson wrote:
On Mon, 15 Oct 2007 16:05:32 +0200, "Paul B. Andersen"
<paul.b.andersen@xxxxxxxxxxxxxxxx> wrote:

Dr. Henri Wilson wrote:
On Fri, 12 Oct 2007 12:20:50 +0200, "Paul B. Andersen"
<paul.b.andersen@xxxxxxxxxxxxxxxx> wrote:


Paul, you know the favorite argument that you and Dishman like to use....ie.,
that using the rotating frame shows that there is no fringe shift and the BaTh
is wrong.
The emission theory predicts zero fringe shift.
Period.

.......get your facts right Paul.
..The emission theory predicts NO FRINGE MOVEMENT at constant speed BUT fringe
DISPLACEMENT proportional to rotation speed.

The emission theory predicts zero fringe displcaement.


Well I have now explained in very simple terms why that is completey wrong and
why you and Dishman have ben under a delusion for years.
You have now repeated this blunder over and over:

Paul, do you or you you not agree that in the rotating frame, the emission
point moves backwards.
...after all, it does NOT move in the no rotating frame.....

The path lengths of the two rays are different, as shown in red and blue.
Both travel times are the same because of the difference in light speed.
In ballistic theory, wavelength is absolute and is the same in both the blue
and red paths. Therefore there are more wavelengths in the blue path than in
the red one.
This is a gigantic self contradiction! A glaring blunder!
I have pointed it out several times, but you seem to be an autistic moron.

You have not pointed out anything of the sort, the path lengths are different.
The travel times arer the same..

How much longer can you keep up this moronic pretence? You must know you are
wrong by now.

Put simply, in the rotating frame, the point of emission moves backwards,
making the path lengths different by 2vt.... That's the same as the non
rotating analysis produces.

Jerry's animatoin backs up what is stated on
www.users.bigpond.com/hewn/ringgyro.htm

If you want to argue and make an even bigger fool of yourself, please go
ahead....
This is _elementary_, Henri.
You are confused on the most basic level!

Don't read the following, Henri.
You might learn something, and you won't risk that, will you?

This is a propagating wave: sin(wt-kx)
phi(t,x) = (wt-kx) is the phase of the wave at x and t.
The phase velocity is given by phi(t,x) = constant (like 0)
(wt-kx)=0 => x/t = w/k => v_phase = w/k

At x = x1 (or any other constant x) you can count
the number of cycles Nc between t1 to t2.
Nc = |phi(t1,x1)-phi(t2,x1)|/2pi
= |(w*t1-k*x1)-(w*t2-k*x1)|/2pi = w(t2-t1)/2pi

At t = t1 (or any other constant t) you can count
the number of wavelengths Nw between x1 to x2.
Nw = |phi(t1,x1)-phi(t1,x2)|/2pi
= |(w*t1-k*x1)-(w*t1-k*x2)|/2pi = k(x2-x1)/2pi

But what is this? Cycles? Wavelengths?
Number of x = |phi(t1,x1)-phi(t2,x2)|/2pi
= |(w*t1-k*x1)-(w*t2-k*x2)|/2pi
= |w(t1-t2)-k(x1-x2)|/2pi

"Therefore there are more wavelengths in the blue path
than in the red one."

The _path_ spans from where and when (t1,x1) a particular
wavefront is emitted, to where and when (t2,x2) the wavefront
is received.
Thus you claim that x is wavelength. It is not.
It is a confused mix of cycles and wavelengths.

Well, I told you not to read it.
You didn't and learned nothing.

So the autistic moron repeats what was never disputed:

In BaTh, wavelength is absolute and constant.

Indeed.
http://home.c2i.net/pb_andersen/pdf/sagnac_ring.pdf

Even Dishman is way ahead of you about this.
>
But why the hell do I bother to explain _elementary_
issues to an autistic moron when I know it is futile?


Paul, stupidly wasting his time

Paul, you will probably have to change your name after your terrible
blunders...the shame will be too much to bear.

Is that why you have changed your name?

Paul
http://home.c2i.net/pb_andersen/
.

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