Dr. Henri Wilson skrev:
On Sat, 17 May 2008 21:43:00 +0200, "Paul B. Andersen"
wrote:

Dr. Henri Wilson wrote:
On Thu, 15 May 2008 14:55:39 +0200, "Paul B. Andersen"
wrote:

Dr. Henri Wilson wrote:
On Wed, 14 May 2008 15:24:33 +0200, "Paul B. Andersen"
wrote:
So why can't you tell us what the BaTh predicts for Fizeau's experiment?
How about: Displacement = 2Ln^2v/(c.lambda)
Displacement of what?
Fringe displacement

You obviously have to show the derivation of the prediction
from the postulates of the BaTh.
But of course you cannot do that, because the "BaTh" is
no theory which can predict anything.
Wilson's derivation:

S-------water moving at v--------|mirror
------------------L---------------

Speed of light in moving water = (c/n)+v wrt S.
A really interesting starting point. :-)
But since the BaTh is your theory, I will take your
word for that this is according to the BaTh.
I will however come back to this.

Travel time to reach mirror = Ln/(c+nv)

Travel time if water at rest = Ln/c

Difference = Ln(c+nv) - Ln/c
= Ln^2v/c^2
Double that to include reverse path. = 2Ln^v/c^2

Fringe displacement = c/lambda * 2Ln^v/c^2 = 2Dn^2v/clambda.
This derivation is wrong, even if it appears to give the correct
result for the difference in travel time. See below.

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. You seem to think
that only one of them went through the moving water while the other
went through stationary water.

No I didn't. I simply doubled the time difference in one direction. that's
where the '2' came from.
....Quite legitimate.

Not at all. See below.

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

Read this:
(When this derivations appears to give the same result as yours,
it is because your L is only half of Michelson's L, which
is the total length of each light path through the water.)

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)
===========================

...and Michelson's results conclusively refute SR. (see my previous reference).

Note: the standard Galilean and SR equations include the factor '4(1-1/n2)' and
and 4(1-1/n3) instead of my simple '2'.

(1-1/n^2.6) is approximately = 2. This is midway between the Galilean and SR
figure AND about what Michelson found.
Uh? What kind of double talk is this? What did you compare?

Let's see what he _really_ found:
All his 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.

Not when doppler wavelenth shift is included. i should have done that in my
derivation, since even is the optics lies outside hte water, there is still a
wavelength change in the water and a consequent contribution to fringe
displacement.

Of course there is a wavelength shift in the water, and of course
the fringe displacement can be calculated by counting the change
in the number of wavelengths in the two beams. This approach
will necessarily give the same result as difference in transit time
approach.

So let's do it:
http://home.c2i.net/pb_andersen/pdf/...Wavelength.pdf

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.

Michelson's experiment proved SR wrong.
see Renshaw's article again.

-------------------------------------------------

Let's get back to you statement:
"Speed of light in moving water = (c/n)+v wrt S[ource]."

This is equivalent to stating:
"The speed of light in water is c/n relative to the water
independent of the speed of the source, and it transforms
according to the Galilean transform."

Indeed a strange statement for a theory which calls itself ballistic!

So nothing adds up, does it? :-)

No....but it was a reasonable start.

Quite.
You got it wrong by only 120%.

I'll do it again with doppler shift included.

Done. Same result.


--
Paul

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

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

Dr. Henri Wilson skrev:
On Sat, 17 May 2008 21:43:00 +0200, "Paul B. Andersen"
wrote:

You obviously have to show the derivation of the prediction
from the postulates of the BaTh.
But of course you cannot do that, because the "BaTh" is
no theory which can predict anything.
Wilson's derivation:

S-------water moving at v--------|mirror
------------------L---------------

Speed of light in moving water = (c/n)+v wrt S.
A really interesting starting point. :-)
But since the BaTh is your theory, I will take your
word for that this is according to the BaTh.
I will however come back to this.

Travel time to reach mirror = Ln/(c+nv)

Travel time if water at rest = Ln/c

Difference = Ln(c+nv) - Ln/c
= Ln^2v/c^2
Double that to include reverse path. = 2Ln^v/c^2

Fringe displacement = c/lambda * 2Ln^v/c^2 = 2Dn^2v/clambda.
This derivation is wrong, even if it appears to give the correct
result for the difference in travel time. See below.

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. You seem to think
that only one of them went through the moving water while the other
went through stationary water.

No I didn't. I simply doubled the time difference in one direction. that's
where the '2' came from.
....Quite legitimate.

Not at all. See below.

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

Read this:
(When this derivations appears to give the same result as yours,
it is because your L is only half of Michelson's L, which
is the total length of each light path through the water.)

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)
===========================

...and Michelson's results conclusively refute SR. (see my previous reference).

Note: the standard Galilean and SR equations include the factor '4(1-1/n2)' and
and 4(1-1/n3) instead of my simple '2'.

(1-1/n^2.6) is approximately = 2. This is midway between the Galilean and SR
figure AND about what Michelson found.
Uh? What kind of double talk is this? What did you compare?

Let's see what he _really_ found:
All his 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.

Not when doppler wavelenth shift is included. i should have done that in my
derivation, since even is the optics lies outside hte water, there is still a
wavelength change in the water and a consequent contribution to fringe
displacement.

Of course there is a wavelength shift in the water, and of course
the fringe displacement can be calculated by counting the change
in the number of wavelengths in the two beams. This approach
will necessarily give the same result as difference in transit time
approach.

So let's do it:
http://home.c2i.net/pb_andersen/pdf/...Wavelength.pdf

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.

Michelson's experiment proved SR wrong.
see Renshaw's article again.

-------------------------------------------------

Let's get back to you statement:
"Speed of light in moving water = (c/n)+v wrt S[ource]."

This is equivalent to stating:
"The speed of light in water is c/n relative to the water
independent of the speed of the source, and it transforms
according to the Galilean transform."

Indeed a strange statement for a theory which calls itself ballistic!

If we assume the extinction distance in water is very small, my equation is
correct.
It could be argued that the speed in the water and wrt the water frame is
(c+v)/n, where the source is moving towards the water at v, in which case the
speed wrt the source frame would be (c+v(1+n))/n. I don't think that would be
correct...but I wouldn't rule it out either.

So nothing adds up, does it? :-)

No....but it was a reasonable start.

Quite.
You got it wrong by only 120%.

I'll do it again with doppler shift included.


Done. Same result.

No, you don't understand.

I'll be back in a few days with the answer.


Henri Wilson. ASTC,BSc,DSc(T)
www.users.bigpond.com/hewn/index.htm

.....specialising in teaching physics to engineers and mathematicians....

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

Dr. Henri Wilson skrev:
On Sat, 17 May 2008 21:43:00 +0200, "Paul B. Andersen"
wrote:

Dr. Henri Wilson wrote:
On Thu, 15 May 2008 14:55:39 +0200, "Paul B. Andersen"
wrote:

Dr. Henri Wilson wrote:
On Wed, 14 May 2008 15:24:33 +0200, "Paul B. Andersen"
wrote:
So why can't you tell us what the BaTh predicts for Fizeau's experiment?
How about: Displacement = 2Ln^2v/(c.lambda)
Displacement of what?
Fringe displacement

You obviously have to show the derivation of the prediction
from the postulates of the BaTh.
But of course you cannot do that, because the "BaTh" is
no theory which can predict anything.
Wilson's derivation:

S-------water moving at v--------|mirror
------------------L---------------

Speed of light in moving water = (c/n)+v wrt S.
A really interesting starting point. :-)
But since the BaTh is your theory, I will take your
word for that this is according to the BaTh.
I will however come back to this.

Travel time to reach mirror = Ln/(c+nv)

Travel time if water at rest = Ln/c

Difference = Ln(c+nv) - Ln/c
= Ln^2v/c^2
Double that to include reverse path. = 2Ln^v/c^2

Fringe displacement = c/lambda * 2Ln^v/c^2 = 2Dn^2v/clambda.
This derivation is wrong, even if it appears to give the correct
result for the difference in travel time. See below.

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. You seem to think
that only one of them went through the moving water while the other
went through stationary water.

No I didn't. I simply doubled the time difference in one direction. that's
where the '2' came from.
....Quite legitimate.

Not at all. See below.

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

Read this:
(When this derivations appears to give the same result as yours,
it is because your L is only half of Michelson's L, which
is the total length of each light path through the water.)

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)
===========================

...and Michelson's results conclusively refute SR. (see my previous reference).

Note: the standard Galilean and SR equations include the factor '4(1-1/n2)' and
and 4(1-1/n3) instead of my simple '2'.

(1-1/n^2.6) is approximately = 2. This is midway between the Galilean and SR
figure AND about what Michelson found.
Uh? What kind of double talk is this? What did you compare?

Let's see what he _really_ found:
All his 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.

Not when doppler wavelenth shift is included. i should have done that in my
derivation, since even is the optics lies outside hte water, there is still a
wavelength change in the water and a consequent contribution to fringe
displacement.

Of course there is a wavelength shift in the water, and of course
the fringe displacement can be calculated by counting the change
in the number of wavelengths in the two beams. This approach
will necessarily give the same result as difference in transit time
approach.

So let's do it:
http://home.c2i.net/pb_andersen/pdf/...Wavelength.pdf

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.

Michelson's experiment proved SR wrong.
see Renshaw's article again.

-------------------------------------------------

Let's get back to you statement:
"Speed of light in moving water = (c/n)+v wrt S[ource]."

This is equivalent to stating:
"The speed of light in water is c/n relative to the water
independent of the speed of the source, and it transforms
according to the Galilean transform."

Indeed a strange statement for a theory which calls itself ballistic!

So nothing adds up, does it? :-)

No....but it was a reasonable start.

Quite.
You got it wrong by only 120%.

I'll do it again with doppler shift included.

Done. Same result.

I got it right,
Check it again.
I'll be back in a couple of days.


Henri Wilson. ASTC,BSc,DSc(T)
www.users.bigpond.com/hewn/index.htm

.....specialising in teaching physics to engineers and mathematicians....

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

Dr. Henri Wilson skrev:
On Sat, 17 May 2008 21:43:00 +0200, "Paul B. Andersen"
wrote:

You obviously have to show the derivation of the prediction
from the postulates of the BaTh.
But of course you cannot do that, because the "BaTh" is
no theory which can predict anything.
Wilson's derivation:

S-------water moving at v--------|mirror
------------------L---------------

Speed of light in moving water = (c/n)+v wrt S.
A really interesting starting point. :-)
But since the BaTh is your theory, I will take your
word for that this is according to the BaTh.
I will however come back to this.

Travel time to reach mirror = Ln/(c+nv)

Travel time if water at rest = Ln/c

Difference = Ln(c+nv) - Ln/c
= Ln^2v/c^2
Double that to include reverse path. = 2Ln^v/c^2

Fringe displacement = c/lambda * 2Ln^v/c^2 = 2Dn^2v/clambda.
This derivation is wrong, even if it appears to give the correct
result for the difference in travel time. See below.

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. You seem to think
that only one of them went through the moving water while the other
went through stationary water.
No I didn't. I simply doubled the time difference in one direction. that's
where the '2' came from.
....Quite legitimate.
Not at all. See below.

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
Read this:
(When this derivations appears to give the same result as yours,
it is because your L is only half of Michelson's L, which
is the total length of each light path through the water.)

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)
===========================
...and Michelson's results conclusively refute SR. (see my previous reference).

Note: the standard Galilean and SR equations include the factor '4(1-1/n2)' and
and 4(1-1/n3) instead of my simple '2'.

(1-1/n^2.6) is approximately = 2. This is midway between the Galilean and SR
figure AND about what Michelson found.
Uh? What kind of double talk is this? What did you compare?

Let's see what he _really_ found:
All his 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.
Not when doppler wavelenth shift is included. i should have done that in my
derivation, since even is the optics lies outside hte water, there is still a
wavelength change in the water and a consequent contribution to fringe
displacement.
Of course there is a wavelength shift in the water, and of course
the fringe displacement can be calculated by counting the change
in the number of wavelengths in the two beams. This approach
will necessarily give the same result as difference in transit time
approach.

So let's do it:
http://home.c2i.net/pb_andersen/pdf/...Wavelength.pdf

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.
Michelson's experiment proved SR wrong.
see Renshaw's article again.

-------------------------------------------------

Let's get back to you statement:
"Speed of light in moving water = (c/n)+v wrt S[ource]."

This is equivalent to stating:
"The speed of light in water is c/n relative to the water
independent of the speed of the source, and it transforms
according to the Galilean transform."

Indeed a strange statement for a theory which calls itself ballistic!

If we assume the extinction distance in water is very small, my equation is
correct.

OK.
So Fizeau falsifies the BaTh.

It could be argued that the speed in the water and wrt the water frame is
(c+v)/n, where the source is moving towards the water at v, in which case the
speed wrt the source frame would be (c+v(1+n))/n. I don't think that would be
correct...but I wouldn't rule it out either.

And x = 1/n = 0.78
Still almost twice of what is observed.


So nothing adds up, does it? :-)
No....but it was a reasonable start.
Quite.
You got it wrong by only 120%.

I'll do it again with doppler shift included.


Done. Same result.

No, you don't understand.

I'll be back in a few days with the answer.

Maybe you should invoke your very special wave again?
" MY wave is one in which the phase of the leading edge
is cycling as it moves."

But since your problem (the BaTh predicts way too much fringe shift)
now is the opposite of what it was when you invented your remarkable
wave (to make The BaTh predict a fringe shift for Sagnac), you
probably have to invent a new kind of wave, maybe like this:
MY wave is one in which the phase of the leading edge
is cycling backwards as it moves.

It is of course legitimate to invent new laws of nature
for each phenomenon you are making the BaTh explain, isn't it?

I can't wait to see what you will come up with. :-)

--
Paul

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

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.


Henri Wilson. ASTC,BSc,DSc(T)
www.users.bigpond.com/hewn/index.htm

.....specialising in teaching physics to engineers and mathematicians....

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/