Re: Sagnac Fully Supports the BaTh.

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

HeWn@xxxxxxxxxxx wrote:
On Oct 11, 11:51 pm, "Paul B. Andersen"
<paul.b.ander...@xxxxxxxxxxxxxxxx> wrote:
Dr. Henri Wilson wrote:
On Tue, 09 Oct 2007 13:30:14 +0200, "Paul B. Andersen"

Paul- Hide quoted text -

- Show quoted text -
My news server is down.
So you are using Google.
Was this meant as an excuse for being wrong?

I gather the whole system has broken down. I'm still only getting messages from
certain countries.

You are wrong. You have always ben wrong. Your whole Sagnac argument
against BaTh has collapsed.
Fine.
Keep me amused.

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.

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:
<<
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.


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.

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
.

Relevant Pages

  • Re: Experimental Evidence for Special Relativity
    ... Dr. Henri Wilson wrote: ... the paths are a different number of wavelengths long. ... The path lengths are different in the inertial frame. ... The shorter path 'leads' because its shorter and faster path 'leads' ...
    (sci.physics.relativity)
  • Re: The Emission Theory of Androcles
    ... cycle slower, or 1.5 cycles slower, or 2.5 cycles slower etc, ... What the fuck does "one half cycle slower" mean, ... longer, or 1.5 wavelengths, or 2.5 wavelengths etc. ... If the photon is actually just one cycle, ...
    (sci.physics.relativity)
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