A simple derivation of the relativistic formula for the addition of velocities
YBM
>formula for the addition of velocities, w = (u+v)/(1 + uv/c^2).
>
> It is due to David Mermin.
See:
http://dorigo.wordpress.com/2008/04/30/guest-post-jeff-wyss-the-relativistic-train/
Dirk Van de moortel
"YBM" <ybm...@nooos.fr> wrote in message news:4818840f$0$684$426a...@news.free.fr...
Nice one.
An excellent opportunity for Androcles to finally grasp it, but it will
not work. Let's hope he will finally lose whatever is left of his mind
over the abundant (but never named or even conceived) closing
speeds/velocities in there ;-)
Dirk Vdm
George Hammond
wrote:
[Hammond]
Are you kidding.... nobody but a rank amateur would screw
around with algebraic formulas when adding velocities.
Any expert will tell you all you do to add two velocities
in SR is to
ADD THE RAPIDITIES STUPID
The rapidity is defined as:
rapidity = r =acrtanh (v/c)
If you want to add two velocities (2/3)c and (3/4)c, all
you do is add r1 and r2 and then take the tanh of the
result:
r1 = arctanh(2/3) = .80472
r2 = arctanh(3/4) = .97296
total = 1.7777
Vtotal = tanh(1.7777) = .94444 c
So, (2/3)c + (3/4)c = .94444 c
YBM
> Are you kidding.... nobody but a rank amateur would screw
> around with algebraic formulas when adding velocities.
> Any expert will tell you all you do to add two velocities
> in SR is to
>
> ADD THE RAPIDITIES
Fine. It's exactly what is proven there. I'd guess than no
expert would accept a statement without any proof of it,
wouldn't he ?
> STUPID
Nice to meet you.
Paul B. Andersen
What about this one?
In S1, a body is moving with the speed u = dx/dt.
What is its speed w = dx'/dt' in S2 which is
moving with the speed -v relative to S1?
dx' = g(dx + v*dt)
dt' = g(dt + (v/c^2)*dx)
w = dx'/dt' = (dx + v*dt)/(dt + (v/c^2)*dx)
w = (dx/dt + v)/(1 + (v/c^2)*(dx/dt))
w = (u + v)/(1 + u*v/c^2)
--
Paul
The Ghost In The Machine
<ybm...@nooos.fr>
wrote
on Wed, 30 Apr 2008 16:43:25 +0200
<4818840f$0$684$426a...@news.free.fr>:
An interesting and elegant derivation; my hat off to
the author. It also of course corroborates exactly with
the derivation done using the Lorentz (which I for one
have done previously).
--
#191, ewi...@earthlink.net
Useless C++ Programming Idea #110309238:
item * f(item *p) { if(p = NULL) return new item; else return p; }
** Posted from http://www.teranews.com **
Dr. Henri Wilson
It assumes the unproven second postulate is correct.
It is crap.... just like the postulate.
Henri Wilson. ASTC,BSc,DSc(T)
www.users.bigpond.com/hewn/index.htm
....specialising in teaching physics to engineers and mathematicians....
YBM
....
> It assumes the unproven second postulate is correct.
This is a derivation dude...
> Henri Wilson.
... Ok... Never mind.
> ASTC,BSc,DSc(T)
Physicians are using strange acronyms to define mental disabilities
these days...
Paul B. Andersen
> On Wed, 30 Apr 2008 16:43:25 +0200, YBM <ybm...@nooos.fr> 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/2008/04/30/guest-post-jeff-wyss-the-relativistic-train/
>
> It assumes the unproven second postulate is correct.
>
> It is crap.... just like the postulate.
What does Fizeau's experiment show, Henri?
Doesn't matter since you don't believe in
experimental evidence anyway, eh?
--
Paul
Dr. Henri Wilson
<paul.b....@hiadeletethis.no> wrote:
>Dr. Henri Wilson wrote:
>> On Wed, 30 Apr 2008 16:43:25 +0200, YBM <ybm...@nooos.fr> 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/2008/04/30/guest-post-jeff-wyss-the-relativistic-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.
>Doesn't matter since you don't believe in
>experimental evidence anyway, eh?
Henri Wilson. ASTC,BSc,DSc(T)
Paul B. Andersen
> On Wed, 07 May 2008 13:14:43 +0200, "Paul B. Andersen"
> <paul.b....@hiadeletethis.no> wrote:
>
>> Dr. Henri Wilson wrote:
>>> On Wed, 30 Apr 2008 16:43:25 +0200, YBM <ybm...@nooos.fr> 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/2008/04/30/guest-post-jeff-wyss-the-relativistic-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?
--
Paul
Dr. Henri Wilson
<paul.b....@hiadeletethis.no> wrote:
..the experiment was not sufficiently accurate to detect the difference between
the predictions of the two theories.
Paul B. Andersen
And the prediction of the BaTh for Fizeau's experiment is? :-)
--
Paul
Dr. Henri Wilson
Androcles
"Dr. Henri Wilson" <HW@....> wrote in message
news:linh24pfqog85atu5...@4ax.com...
light from a stationary source interacting with moving water"
Well, fuck my old boots. All this time we should have been dunking ring
laser
gyroscopes in the river to get a reading from them.
Paul B. Andersen
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
Dr. Henri Wilson
<paul.b....@hiadeletethis.no> wrote:
BaTh is far more advanced than other similar theories due to the discoveries
made as a result of my variable star program.
Paul B. Andersen
> On Tue, 13 May 2008 11:12:02 +0200, "Paul B. Andersen"
> <paul.b....@hiadeletethis.no> wrote:
>
>> Dr. Henri Wilson wrote:
>>> On Mon, 12 May 2008 21:43:31 +0200, "Paul B. Andersen"
>>> <paul.b....@guesswhatuia.no> 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
Dr. Henri Wilson
<paul.b....@hiadeletethis.no> wrote:
How about: Displacement = 2Ln^2v/(c.lambda)
Michelson's data fully supports BaTh..
Paul B. Andersen
> On Wed, 14 May 2008 15:24:33 +0200, "Paul B. Andersen"
> <paul.b....@hiadeletethis.no> wrote:
>
>> Dr. Henri Wilson wrote:
>>> On Tue, 13 May 2008 11:12:02 +0200, "Paul B. Andersen"
>>> <paul.b....@hiadeletethis.no> wrote:
>>>
>>>> Dr. Henri Wilson wrote:
>>>>> On Mon, 12 May 2008 21:43:31 +0200, "Paul B. Andersen"
>>>>> <paul.b....@guesswhatuia.no> 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
Dr. Henri Wilson
<paul.b....@hiadeletethis.no> wrote:
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 bullshit Andersen. Androcles was right.
Paul B. Andersen
> On Thu, 15 May 2008 14:55:39 +0200, "Paul B. Andersen"
> <paul.b....@hiadeletethis.no> wrote:
>
>> Dr. Henri Wilson wrote:
>>> On Wed, 14 May 2008 15:24:33 +0200, "Paul B. Andersen"
>>> <paul.b....@hiadeletethis.no> 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
Dr. Henri Wilson
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.
Paul B. Andersen
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:
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/FizeauByWavelength.pdf
Quite.
You got it wrong by only 120%.
> I'll do it again with doppler shift included.
Done. Same result.
--
Paul
Dr. Henri Wilson
<paul.b....@guesswhatuia.no> wrote:
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.
Dr. Henri Wilson
<paul.b....@guesswhatuia.no> wrote:
I got it right,
Check it again.
I'll be back in a couple of days.
Paul B. Andersen
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
Paul B. Andersen
Did you get something right?
What was that?
> 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....
--
Paul
Dr. Henri Wilson
<paul.b....@guesswhatuia.no> wrote:
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.
Paul B. Andersen
> On Mon, 19 May 2008 22:27:25 +0200, "Paul B. Andersen"
> <paul.b....@guesswhatuia.no> wrote:
>
>> Dr. Henri Wilson skrev:
>>> On Mon, 19 May 2008 00:35:11 +0200, "Paul B. Andersen"
>>> <paul.b....@guesswhatuia.no> 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 ą 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 ą 4.6%)
SR confirmed.
--
Paul
Dr. Henri Wilson
<paul.b....@hiadeletethis.no> wrote:
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 bullshit Andersen. Androcles was right.
Henri Wilson. ASTC,BSc,DSc(T)
www.users.bigpond.com/hewn/index.htm.
Einstein: the greatest hoaxer since 'virgin' mary
Paul B. Andersen
The results of Michelsons measurments can be summed up thus:
Michelson assumed the speed of light in the moving water
could be written as c/n + xv
He measured the "drag coefficient" x to be: x = 0.434 ą 0.02
Henri Wilson:
"Speed of light in moving water = (c/n)+v"
x = 1
Go figure. :-)
>
>>> 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 bullshit Andersen. Androcles was right.
>
Hmmm. Maybe you are not a liar after all?
Who would voluntarily lie so obvious as this?
Maybe your dementia prevents you from remembering
what happened yesterday?
http://groups.google.com/group/sci.physics.relativity/msg/26e76ec8e31e150d
or
http://tinyurl.com/4svwfb
--
Paul
Paul B. Andersen
Michelson's original paper:
http://home.c2i.net/pb_andersen/pdf/Fizeau_by_Michelson.pdf
--
Paul