Re: EXPERIMENTS THAT REFUTE RELATIVITY

"Dr. Henri Wilson" <HW@....> wrote in message
news:rt5mn316degrtjms71mn1kss6hg4encvnk@xxxxxxxxxx
On Tue, 1 Jan 2008 16:02:50 -0800, The Ghost In The Machine
<ewill@xxxxxxxxxxxxxxxxxxxxxxx> wrote:

In sci.physics.relativity, HW@....(Dr. Henri Wilson)
<HW@>

Why do you always have to introduce irrelevancies Ghost??
I'll bet you drove your parents mad when you were a kid....
...always asking "why this?...why that?"

c is the name of the formula c_0 + v, if you like. Of course one has a
little problem in that c_0 is a scalar, not a vector; adding a scalar
and a vector doesn't work. That's why I introduced d; however, there
are other factors such as gaseous emissions from the starlight that may
come into play, in some theories. These gaseous emissions have a
refractive index very slightly greater than 1, and may affect
lightspeed.

Every star has an EM control sphere around it. This is evidenced by
certain
variable star curves.

Ahhh . .the little faries who adjust light frequencies and speeds to make
them appear as they should do adccording to relativity

Of course, that's only an ad-hoc hypothesis, but the sad truth of the
matter is that c is not a constant, and never was, in my world...unless
I explicitly stipulate SR. Since SR works, I'm not that rigorous about
it, but it is clear that SR doesn't always work for all people in this
newsgroup -- and I frankly have no idea how to prove that it does. No
one can prove SR works, but we do have a large body of information that
stipulates that it works well enough.
none.

Your ignorance is astounding.

. Rather it moves at [c+ (some kind of function
of energy)].

EM *always* moves at c wrt its source, by *definition*.
The catch is that c isn't constant in all theories, but
c is by definition lightspeed as measured at a point.

c is a universal constant.
It appears that EM is emitted at c wrt its source.

It certainly does.

Of course, you ignore the evidence of light also arriving at the observer at
c (regardles of the speed of the source) nad having it sfrequency and
wavelength shifted accordingly.

Maybe its the little fairies who do that, just to confuse scientisist .. oh,
they are such little scamps, aren't they.
According to SR, the rod is moving wrt both A and B and is therefore
shorter in
both their frames.
Incorrect. The rod is shorter, yes, but not by the same amount.
Just accept it is shorter in both frames according to SR.

Of course it is .. anything the moves realtive to an observer will appear
shorter to that observer.

I don't think you understood the question.
Initially the rod is moving wrt both A and B.
It is accelerated to B's speed...in which case it is even shorter iin A's
frame
but NO LONGER short in B's frame.

Yes .. why is the a problem for you?

If it WAS shorter and is now NOT SHORTER, it must have become longer.
..don't
you agree?

Yes .. because it was moving (and so was measured as shorter) and then
slowed down to zero speed, and is no longer moving, and so you measure its
proper length .. why does that seem a puzzle to you?

...according to SR of course.
Obviously it cannot simultaneously become physically longer and shorter.
Simultaneity is an illusion, even in Newtonian space. If one stands
between two blinking lights, and one has a mirror arrangement, one might
see them blinking simultaneously on a locus between the two lights. (If
done correctly the locus is the perpendicular bisector.)
This paragraph rates 5 on the jeckyl scale.

Geee .. its great to be made famous

MMX (this is a gimme)
It certainly is.
Ives-Stilwell
Sagnac
I have already done it.

No .. you got it totally wrong .. as has been explained ot you several times
... You just cannot see your own mistakes. Fortunately the rest of us can.

Here's the BaTh analysis of a Sagnac ring.

And it is still wrong

'f' is source frequency in the source frame.

In the NONROTATING frame

(using the diagram at http://www.mathpages.com/rr/s2-07/2-07.htm)
According to BaTh:

for ray 1, (c+v)t = 2piR + vt or.... t = 2piR/c.........(t = travel time)
for ray 2, (c-v)t = 2piR - vt or.... t = 2piR/c .......(travel times are
equal)

Path length of ray 1 is 2piR(c+v)/c
.......................of ray 2 is 2piR(c-v)/c

The source frequency is doppler shifted in the non-rotating frame:
Frequency of ray 1 is f(c+v)/c....(rate of wavecrests passing a point at
rest)
Frequency of ray 2 is f(c-v)/c.....

You forgot to doppler shift back to the moving detector.

When you do so the frequency is f again

The number of cycles in each ray is calculated as (travel time/frequency)
Number of cycles (wavelengths) in ray 1 = [2piR/c].[f(c+v)/c] =
2piRf(c+v)/c^2
Number of cycles (wavelengths) in ray 2 = [2piR/c].[f(c-v)/c] =
2piRf(c-v)/c^2

So .. when you look at what hits the moving detector

Number of cycles (wavelengths) in ray 1 = [2piR/c].f = 2piRf/c
Number of cycles (wavelengths) in ray 2 = [2piR/c].f = 2piRf/c
......................................................

Fringe displacement = difference in numbers of wavelengths in each ray

= 0

= 4piRvf/c^2
= 4Aw/cL.
.........(where, w = angular velocity, A = area of ring, L = wavelength)
If you can find anything wrong with this please let us know.....

I have .. multiple times

Your two oscillators leave the splitter mirror at the same time, and the
same speed and at the same frequency relative to the source.

If you want you can have a stationary observer somewhere along the paths to
whom the beams appear doppler shifted, but that is not relevant to the
experiment itself. Any doppler shift from the moving source to the
stationary observer is undone by the doppler shift from the stationary
observer to the detector.

When they arrive at the initial position where the beam are combined, the
beams arrive at the same tiime, and the same speed, and at the same
frequency relative to the detectors.

Hence no Sagnac effect. Very simple .. except to deluded morons like Henri.



.

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