On Mon, 12 May 2008 21:43:31 +0200, "Paul B. Andersen"
wrote:

Dr. Henri Wilson skrev:
On Fri, 09 May 2008 12:24:54 +0200, "Paul B. Andersen"
wrote:

Dr. Henri Wilson wrote:
On Wed, 07 May 2008 13:14:43 +0200, "Paul B. Andersen"
wrote:

Dr. Henri Wilson wrote:
On Wed, 30 Apr 2008 16:43:25 +0200, YBM wrote:

The following describes a very elegant and simple derivation of the relativistic
formula for the addition of velocities, w = (u+v)/(1 + uv/c^2).

It is due to David Mermin.
See:
http://dorigo.wordpress.com/2008/04/...ivistic-train/
It assumes the unproven second postulate is correct.

It is crap.... just like the postulate.
What does Fizeau's experiment show, Henri?
That BaTh is correct.
I see.
So according to the "BaTh", velocities transforms like this:
w = (u+v)/(1+uv/c^2)
which implies that the velocity c transforms like this:
w = (c+v)/(1+cv/c^2) = c

Why do you call your theory the "BaTh" when it is
identical to SR?

..the experiment was not sufficiently accurate to detect the difference between
the predictions of the two theories.

And the prediction of the BaTh for Fizeau's experiment is? :-)

Try this:
http://renshaw.teleinc.com/papers/fizeau4b/fizeau4b.stm



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

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

"Dr. Henri Wilson" HW@.... wrote in message
...
| On Mon, 12 May 2008 21:43:31 +0200, "Paul B. Andersen"
| wrote:
|
| Dr. Henri Wilson skrev:
| On Fri, 09 May 2008 12:24:54 +0200, "Paul B. Andersen"
| wrote:
|
| Dr. Henri Wilson wrote:
| On Wed, 07 May 2008 13:14:43 +0200, "Paul B. Andersen"
| wrote:
|
| Dr. Henri Wilson wrote:
| On Wed, 30 Apr 2008 16:43:25 +0200, YBM wrote:
|
| The following describes a very elegant and simple derivation of
the relativistic
| formula for the addition of velocities, w = (u+v)/(1 + uv/c^2).
|
| It is due to David Mermin.
| See:
|
http://dorigo.wordpress.com/2008/04/...ivistic-train/
| It assumes the unproven second postulate is correct.
|
| It is crap.... just like the postulate.
| What does Fizeau's experiment show, Henri?
| That BaTh is correct.
| I see.
| So according to the "BaTh", velocities transforms like this:
| w = (u+v)/(1+uv/c^2)
| which implies that the velocity c transforms like this:
| w = (c+v)/(1+cv/c^2) = c
|
| Why do you call your theory the "BaTh" when it is
| identical to SR?
|
| ..the experiment was not sufficiently accurate to detect the difference
between
| the predictions of the two theories.
|
| And the prediction of the BaTh for Fizeau's experiment is? :-)
|
| Try this:
| http://renshaw.teleinc.com/papers/fizeau4b/fizeau4b.stm
|
|
"However, Michelson neglected to take into account the Doppler effect of
light from a stationary source interacting with moving water"

Well, **** my old boots. All this time we should have been dunking ring
laser
gyroscopes in the river to get a reading from them.

Dr. Henri Wilson wrote:
On Mon, 12 May 2008 21:43:31 +0200, "Paul B. Andersen"
wrote:

Dr. Henri Wilson skrev:
On Fri, 09 May 2008 12:24:54 +0200, "Paul B. Andersen"
wrote:

Dr. Henri Wilson wrote:
On Wed, 07 May 2008 13:14:43 +0200, "Paul B. Andersen"
wrote:

Dr. Henri Wilson wrote:
On Wed, 30 Apr 2008 16:43:25 +0200, YBM wrote:

The following describes a very elegant and simple derivation of the relativistic
formula for the addition of velocities, w = (u+v)/(1 + uv/c^2).

It is due to David Mermin.
See:
http://dorigo.wordpress.com/2008/04/...ivistic-train/
It assumes the unproven second postulate is correct.

It is crap.... just like the postulate.
What does Fizeau's experiment show, Henri?
That BaTh is correct.
I see.
So according to the "BaTh", velocities transforms like this:
w = (u+v)/(1+uv/c^2)
which implies that the velocity c transforms like this:
w = (c+v)/(1+cv/c^2) = c

Why do you call your theory the "BaTh" when it is
identical to SR?
..the experiment was not sufficiently accurate to detect the difference between
the predictions of the two theories.
And the prediction of the BaTh for Fizeau's experiment is? :-)

Try this:
http://renshaw.teleinc.com/papers/fizeau4b/fizeau4b.stm

So "the BaTh" is the same as Curt Renshaw's radiation continuum model (RCM)?
Why have you then invented a new name for it?
A copyright violation.

--
Paul

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

On Tue, 13 May 2008 11:12:02 +0200, "Paul B. Andersen"
wrote:

Dr. Henri Wilson wrote:
On Mon, 12 May 2008 21:43:31 +0200, "Paul B. Andersen"
wrote:


Why do you call your theory the "BaTh" when it is
identical to SR?
..the experiment was not sufficiently accurate to detect the difference between
the predictions of the two theories.
And the prediction of the BaTh for Fizeau's experiment is? :-)

Try this:
http://renshaw.teleinc.com/papers/fizeau4b/fizeau4b.stm

So "the BaTh" is the same as Curt Renshaw's radiation continuum model (RCM)?
Why have you then invented a new name for it?
A copyright violation.

BaTh is far more advanced than other similar theories due to the discoveries
made as a result of my variable star program.


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 Tue, 13 May 2008 11:12:02 +0200, "Paul B. Andersen"
wrote:

Dr. Henri Wilson wrote:
On Mon, 12 May 2008 21:43:31 +0200, "Paul B. Andersen"
wrote:


Why do you call your theory the "BaTh" when it is
identical to SR?
..the experiment was not sufficiently accurate to detect the difference between
the predictions of the two theories.
And the prediction of the BaTh for Fizeau's experiment is? :-)
Try this:
http://renshaw.teleinc.com/papers/fizeau4b/fizeau4b.stm
So "the BaTh" is the same as Curt Renshaw's radiation continuum model (RCM)?
Why have you then invented a new name for it?
A copyright violation.

BaTh is far more advanced than other similar theories due to the discoveries
made as a result of my variable star program.

So why can't you tell us what the BaTh predicts for Fizeau's experiment?

--
Paul

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

On Wed, 14 May 2008 15:24:33 +0200, "Paul B. Andersen"
wrote:

Dr. Henri Wilson wrote:
On Tue, 13 May 2008 11:12:02 +0200, "Paul B. Andersen"
wrote:

Dr. Henri Wilson wrote:
On Mon, 12 May 2008 21:43:31 +0200, "Paul B. Andersen"
wrote:


Why do you call your theory the "BaTh" when it is
identical to SR?
..the experiment was not sufficiently accurate to detect the difference between
the predictions of the two theories.
And the prediction of the BaTh for Fizeau's experiment is? :-)
Try this:
http://renshaw.teleinc.com/papers/fizeau4b/fizeau4b.stm
So "the BaTh" is the same as Curt Renshaw's radiation continuum model (RCM)?
Why have you then invented a new name for it?
A copyright violation.

BaTh is far more advanced than other similar theories due to the discoveries
made as a result of my variable star program.

So why can't you tell us what the BaTh predicts for Fizeau's experiment?

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

Michelson's data fully supports BaTh..


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 Wed, 14 May 2008 15:24:33 +0200, "Paul B. Andersen"
wrote:

Dr. Henri Wilson wrote:
On Tue, 13 May 2008 11:12:02 +0200, "Paul B. Andersen"
wrote:

Dr. Henri Wilson wrote:
On Mon, 12 May 2008 21:43:31 +0200, "Paul B. Andersen"
wrote:

Why do you call your theory the "BaTh" when it is
identical to SR?
..the experiment was not sufficiently accurate to detect the difference between
the predictions of the two theories.
And the prediction of the BaTh for Fizeau's experiment is? :-)
Try this:
http://renshaw.teleinc.com/papers/fizeau4b/fizeau4b.stm
So "the BaTh" is the same as Curt Renshaw's radiation continuum model (RCM)?
Why have you then invented a new name for it?
A copyright violation.
BaTh is far more advanced than other similar theories due to the discoveries
made as a result of my variable star program.
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?
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.

Michelson's data fully supports BaTh..

Sure, the MMX confirmes the emission theory.
But Sagnac and Fizeau and a lot of other experiments
falsifies it.

--
Paul

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

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:

Dr. Henri Wilson wrote:
On Tue, 13 May 2008 11:12:02 +0200, "Paul B. Andersen"
wrote:

Dr. Henri Wilson wrote:
On Mon, 12 May 2008 21:43:31 +0200, "Paul B. Andersen"
wrote:

Why do you call your theory the "BaTh" when it is
identical to SR?
..the experiment was not sufficiently accurate to detect the difference between
the predictions of the two theories.
And the prediction of the BaTh for Fizeau's experiment is? :-)
Try this:
http://renshaw.teleinc.com/papers/fizeau4b/fizeau4b.stm
So "the BaTh" is the same as Curt Renshaw's radiation continuum model (RCM)?
Why have you then invented a new name for it?
A copyright violation.
BaTh is far more advanced than other similar theories due to the discoveries
made as a result of my variable star program.
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.
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.

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.

Michelson's data fully supports BaTh..

Michelson's fizeau experiment data.
I thought you knew all about his experiment.


Sure, the MMX confirmes the emission theory.

Who said anything about the MMX

But Sagnac and Fizeau and a lot of other experiments
falsifies it.

You obviously don't know anything about this.
Michelson repeated Fizeau's experiment with considerable accuracy.

His result fit my above equation.

Yio are full of bull**** Andersen. Androcles was right.



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 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.

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
(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)
===========================

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.

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.

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

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? :-)

--
Paul

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

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.

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
(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.

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.

I'll do it again with doppler shift included.



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

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