Dr. Henri Wilson wrote:
On Mon, 19 May 2008 22:27:25 +0200, "Paul B. Andersen"
wrote:

Dr. Henri Wilson skrev:
On Mon, 19 May 2008 00:35:11 +0200, "Paul B. Andersen"
wrote:

I'll do it again with doppler shift included.
Done. Same result.
I got it right,
Check it again.
Did you get something right?
What was that?

Displacement = 2Ln^2v/(c.lambda)

versus SR: Displacement = 4Ln^2v/(c.lambda)(1-1/n^3) and anogther one:
Displacement = 4Ln^2v/(c.lambda)(1-1/n^2)

I told you, (1-1/(n^2.6)) ~= 0.5

Michelson agrees with my equation , refutes SR's.


Shut your eyes, close your ears and moronically repeat
what is thoroughly refuted, eh?

Since you obviously didn't read what I wrote the first
time, here it is again.

In Fizeau's original experiment as well as in Michelson & Morley's
repetition of the experiment, the two beams that were interfering
went in opposite directions through the moving water.

According to you, the BaTh predicts:
Speed of light in moving water = (c/n)+/-v = (c +/- nv)/n wrt source.

If delta_t is the difference in travel time between the two beams,
the correct derivation is:
delta_t = Ln/(c-nv) - Ln/(c+nv) ~= 2Lvn^2/c^2

Now Michelson's fringe shift delta is the shift when the direction
of the water flow is reversed. So we get the delta_t twice.
****************************

delta = 2.delta_t.c/lamda
delta = 4Lvn^2/c.lambda (BaTh)
========================

In comparison, SR predicts:
delta_t = L/((c-nv)/(n-v/c)) - L/((c+nv)/(n+v/c)) ~= 2Lv(n^2-1)/c^2
delta = 4Lv(n^2-1)/c.lambda (SR)
===========================

All Michelson's measurements could be summed up thus:
(because delta is proportional to both L and v)
Michelson:
"If these measurement be reduces to what they would be if
the tube were 10m long and the velocity 1m per second,
they would be as follows: delta = 0.1840"

n^2 = 1.78 lambda = 0.00057 cm (Michelson's numbers)

There is clearly a typo in Michelson's paper, he
used visible light, so lambda must be 0.00057 mm, not cm.

The Bath predicts:
delta = 4*10*1*1.78/3*10^8*0.57*10^-6 = 0.416
which is more than twice of the observed shift.

SR predicts:
delta = 4*10*1*(1.78-1)/3*10^8*0.57*10^-6 = 0.1825
the error is only 0.8%, which is well within the error bars.

So Michelson's repetition of Fizeau's experiment
confirms SR, while it falsifies the BaTh.

We didn't actually have do go through these derivations,
when Michelson did them for us.
He assumed the speed of light in the moving water could be
written as c/n + xv
The "drag coefficient" x was measured to be:
x = 0.434 ± 0.02
==================

According to the BaTh x should be 1, more than twice
the measured value. BaTh falsified.

According to SR, we have:
Speed of light in the water = (c/n + v)/(1 + (vc/n)/c^2)
~= c/n + v(1-1/n^2)
x = (1-1/n^2) = 0.438 which is about in the middle of the error bars.
(The error is 1%, the error bars are ± 4.6%)
SR confirmed.

--
Paul

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