Re: EXPERIMENTS THAT REFUTE RELATIVITY

"Dr. Henri Wilson" <HW@....> wrote in message
news:3se3n39durmv5aguq8dd48vis93rta5edt@xxxxxxxxxx
On Wed, 26 Dec 2007 09:56:17 +1100, "Jeckyl" <noone@xxxxxxxxxxx> wrote:

"Dr. Henri Wilson" <HW@....> wrote in message
news:68a0n3t5mcaumqrh60po29crbu8ernebe1@xxxxxxxxxx

So the waves leave the source at the same frequency and speed relative
to
the source, they arrive at the detector at the same frequency and speed
realtive to the detector, they are in transit for the same time .. there
is
no Sagnac effect predicted.

Study the equations and point out where they are wrong or SHUT THE ***
UP!

They describe the phase difference at the initial source position in the
non-rotating frame.

There is NO phase difference at the start, idiot.

yes there is .. whne you are looking at the instantaneous source position in
the non-rotating frame at the moment of emission.

The real source, that moves in the non-rotating frame is always in phase (of
course), but the fixed spot in the non-rotating frame is not.

Whne you are measuring the wavelength, you need to start from the real
source position (which has moved in the non-rotating frame) .. because the
instantaneous source that is fixed in the non-rotating frame is not in sync
with the actual (moving) source.

Imagine a wave with a wavelength of 1m and a frequency of 1Hz (say), with
the wave front moving at 0.5m/sec to the right and the source moving to the
left at 0.5m/s. After 0.5 seconds, the wave front has travelled 0.5m to the
right from our fixed position, yet it has completed a full cycle as the
distance from the moving source is now 1m. Can't you see that you need to
measure from the moving source to see how many whole wavelengths there
are?????

They describe the phse
difference at the detector.

No. There is no phase difference at the detector, as the waves arrive at
the same speed, in the same time, with the same frequency.

As the wave are the same frequency and have travelled for the same time,
by
the time they get to the moving detector (you have to Doppler shift to get
the frequency as the detector will 'see' it), that means the waves must
still be in phase. its very simple
The waves are different frequencies in the nonrotating frame -

irrelevant .. as the detector is at rest in the ROTATING frame .. and moving
in the non-rotating frame .. so you need to dopler shift back to get the
correct frequency .. and it is the same

the one we are
using . You are a moron.

You just don't undersatnd the physics

Look at your train example.
Sagnac is similar to this.
There are three trains on a track .. A, B, C all moving at the same speed.
A fires a machine gun forward at B, C fires a machine gun backward at B.
The same rate of bullets arrives at B from both trains (because they are
comoving.

Why do you get EVERYTHING wrong?

I don't

They are moving away from you at different speeds.

No .. they are not

Now .. to make it more like Sagnac .. put the trains on a circular track,
and place some 'mirrors' for the bullets to bounce off that also move with
the train. Now, you don't need A and C to be separate trains .. they can
just be a single train on the opposite side of the circular track from
train
B. Train A can shoot bullets forward and backward .. and both bullets
arrive at train B at the same time at the same rate.

Photons are not bullets.

They are like bullets in ballistic theory .. where do you think it got its
name?

Extend this example further, and you have a single train A shooting
bullets
that bounce around the mirrors end up back at train A again. Now you have
Sagnac. You see that the bullets get fired from the machine guns in both
directions at the same rate and end up back at train A again at the same
rate and in the same time.

Its just blatantly obvious that there is no Sagnac effect in a ballistic
theory.

I have presented the maths. Prove it wrong.

I have


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