Interpreting the MMX null result

On Sun, 26 Nov 2006 13:02:57 -0500, kenseto wrote
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"Cygnus X-1" wrote in message
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On Sat, 25 Nov 2006 17:53:33 -0500, kenseto wrote
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"Cygnus X-1" wrote in message
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On Sat, 25 Nov 2006 11:02:32 -0500, kenseto wrote
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"Cygnus X-1" wrote in message
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On Fri, 24 Nov 2006 09:28:17 -0500, kenseto wrote
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"Phineas T Puddleduck" wrote in
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news In article ,
"kenseto" wrote:

http://soi.stanford.edu/

These experiments are not the MMX. The MMX has its own light
source.
Null
result by the MMX means isotropy of the speed of light and
non-null
result
means anisotropy of the speed of light. That's all I claimed.

Its own light wouldn't make a difference. Why should it?

Because the MMX is designed to detect any phase shift, as the
apparatus
is
rotated, between the two right angled light rays from its own light
source.
It is not possible to determine if there is any phase shift for
light
rays
from different sources.

Ken Seto


Fourier Transform spectrographs do it all the time. By filtering on
a
emission/absorption line, the phase shift moves the line profile in
the
filter window and you can compute the shift from the intensity
change.
From this, SOHO/MDI generates images of velocities on the solar
surface
(Dopplergrams). The method is reliable enough they can generate
'sonograms' from those motions and determine when sunspots form on
the
farside of the Sun, seeing them before they are carried into view
from
Earth:

But that is not the same as determining the isotropy or anisotropy of
the
speed of the local light rays from different directions.

The light will travel perpendicular to the spacecraft-Sun line inside
the interferometer. Are you saying that in your theory the speed will
not be different along this path? What defines your horizontal &
vertical directions? You imply the gravitationally defined 'vertical',
in which case the Sun would define vertical along the entry arm of the
interferometer.

When the MMX is performed in the vertical direction the two light rays
are
emanated in different heights (different state of absolute motion) as
the
apparatus is rotated and that's what causes the non-null result (fringe
shift).

This is ambiguous. Which two light rays? There's only one light
source in the apparatus (or one entry point for light rays). Do you
mean separate runs of the apparatus before & after rotation?

The single light ray is splitted into two rays and these two light rays are
then recombined to give the characteristic light and dark fringes. When the
light rays are isoropic you will get null result (no fringe shift) as the
apparatus id rotated. When the light rays are anistropic you will get
non-null result (fringe shift) as the apparatus is rotated.



the experiment would work differently here??

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You MUST define your terms unambiguously!

Everybody knows what horizontal and vertical mean.

Really? I say the direction of vertical is when I look up in the sky.

If I'm on the North Pole, I'll see the star Polaris.

If I'm on the South Pole, I'll see an empty path of sky near the
Southern Cross.

If I'm on the equator of Mars, I'd look up and see a different set of
stars.

Would an observer on Saturn say these directions defined by the
aforementioned three observers are vertical? Would the Saturn
observer's definition of vertical be the same?

Are you saying these 'verticals' are all the same?

If I'm in free-fall, where is 'vertical'?

If you can't define this in an objective way, your theory is useless.

the net force of
gravity???

After all, MMX has been performed in multiple locations on the Earth,
each with a different horizontal and vertical. The only common
component in this directions is a radial vector from the center of the
Earth and the plane normal to it.

This is irrelevant. Also the proposed experiments in the paper entitled
"Proposed Experiments to Detect Absolute Motion" will confirm my assertion
that light is anisotropic in the vertical direction.
http://www.geocities.com/kn_seto/2005Experiment.pdf

This document does not 'confirm' anything beyond the fact that you've
made the statement. Reality is not required to conform to your
statements (That trick only works for God).

Ken, do you think you are God?

And as I note above, your definition of 'vertical' is ambiguous.

Tom
--
Dealing with Creationism in Astronomy
http://homepage.mac.com/cygnusx1

"They're trained to believe, not to know. Belief can be manipulated.
Only knowledge is dangerous." --Frank Herbert, "Dune Messiah"

Cygnus X-1 wrote:
[references to Michelson Interferometers on spacecraft]

Thanks. I was wondering if any of these could be turned into a test of
SR. They cannot. They use a Michelson interferometer with variable arm
lengths to generate an "interferogram", which is a high-resolution
Fourier transform of the input spectrum. I had not realized the
instrument could do that.

Different missions use the resulting spectrum for different purposes.
For instance, SWIFT uses an Ozone line to both measure the concentration
of Ozone in the atmosphere, and to measure the wind via Doppler shift
(hence the need for a high-resolution spectrum).

Again, thanks.


Tom Roberts

On Sun, 26 Nov 2006 21:44:59 -0500, Tom Roberts wrote
(in article ) :

Cygnus X-1 wrote:
[references to Michelson Interferometers on spacecraft]

Thanks. I was wondering if any of these could be turned into a test of
SR. They cannot. They use a Michelson interferometer with variable arm
lengths to generate an "interferogram", which is a high-resolution
Fourier transform of the input spectrum. I had not realized the
instrument could do that.

Different missions use the resulting spectrum for different purposes.
For instance, SWIFT uses an Ozone line to both measure the concentration
of Ozone in the atmosphere, and to measure the wind via Doppler shift
(hence the need for a high-resolution spectrum).

Again, thanks.


Tom Roberts

Tom

Actually, I've found that some tests might be conducted with these.

Remember, if there is a shift as the spacecraft moves it will affect
the entire field-of-view, imposing a oscillation in the measured
velocities with the same period as the spacecraft orbit (or the halo
orbit in the case of SOHO).

Some preliminary calculations indicate that SOHO/MDI would not be as
sensitive for a geocentric calculations as would Earth orbit.

If you've found additional info on these instruments (most instrument
specs I've found don't provide the length of the arms), I'd be
interested in discussing it with you (via e-mail).

There might be an education-oriented paper in this.

Thanks,
Tom
--
Dealing with Creationism in Astronomy
http://homepage.mac.com/cygnusx1

"They're trained to believe, not to know. Belief can be manipulated.
Only knowledge is dangerous." --Frank Herbert, "Dune Messiah"

"Cygnus X-1" wrote in message
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On Sun, 26 Nov 2006 13:02:57 -0500, kenseto wrote
(in article ):


"Cygnus X-1" wrote in message
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On Sat, 25 Nov 2006 17:53:33 -0500, kenseto wrote
(in article ):


"Cygnus X-1" wrote in message
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On Sat, 25 Nov 2006 11:02:32 -0500, kenseto wrote
(in article ):


"Cygnus X-1" wrote in message
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On Fri, 24 Nov 2006 09:28:17 -0500, kenseto wrote
(in article ):


"Phineas T Puddleduck" wrote in
message

When the MMX is performed in the vertical direction the two light rays
are
emanated in different heights (different state of absolute motion) as
the
apparatus is rotated and that's what causes the non-null result
(fringe
shift).

This is ambiguous. Which two light rays? There's only one light
source in the apparatus (or one entry point for light rays). Do you
mean separate runs of the apparatus before & after rotation?

The single light ray is splitted into two rays and these two light rays
are
then recombined to give the characteristic light and dark fringes. When
the
light rays are isoropic you will get null result (no fringe shift) as
the
apparatus id rotated. When the light rays are anistropic you will get
non-null result (fringe shift) as the apparatus is rotated.

Since there is no comment on this I take it that you agreed with what I
said. In that case you should be able to see that the space spectrometer is
not equivalent to the MMX.



You MUST define your terms unambiguously!

Everybody knows what horizontal and vertical mean.

Really? I say the direction of vertical is when I look up in the sky.

No ....everbody know that trees grow vertically and water level is
horizontal.

If I'm on the North Pole, I'll see the star Polaris.

If I'm on the South Pole, I'll see an empty path of sky near the
Southern Cross.

If I'm on the equator of Mars, I'd look up and see a different set of
stars.

So what.....it just proved my point that each location on earth has its own
horizontal and vertical directions.

Would an observer on Saturn say these directions defined by the
aforementioned three observers are vertical? Would the Saturn
observer's definition of vertical be the same?

Are you saying these 'verticals' are all the same?

NO.... each location on earth has its own vertical and horizontal
directions.

If I'm in free-fall, where is 'vertical'?

We were talking about the MMXs on earth. If you are in space you will have
to do the MMX in space to find the direction that will give you null result
and call that horizontal and the direction perpendicular to that as
vertical.

If you can't define this in an objective way, your theory is useless.

the net force of
gravity???

After all, MMX has been performed in multiple locations on the Earth,
each with a different horizontal and vertical. The only common
component in this directions is a radial vector from the center of the
Earth and the plane normal to it.

This is irrelevant. Also the proposed experiments in the paper entitled
"Proposed Experiments to Detect Absolute Motion" will confirm my
assertion
that light is anisotropic in the vertical direction.
http://www.geocities.com/kn_seto/2005Experiment.pdf

This document does not 'confirm' anything beyond the fact that you've
made the statement. Reality is not required to conform to your
statements (That trick only works for God).

This is an idiotic statement. I have proposed experiments that can refute
Model Mechanics.

Ken, do you think you are God?

You are starting to **** me off.

And as I note above, your definition of 'vertical' is ambiguous.

Comments like this is the reason why I call you an idiot.

Ken Seto

Dear kenseto:

kenseto wrote:
"N:dlzc D:aol T:com (dlzc)" wrote in message
...
Dear Cygnus X-1:

"Cygnus X-1" wrote in message
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On Thu, 23 Nov 2006 06:31:03 -0500, kenseto wrote
(in article ):
...
GPS
can. You are wanting MMX, so you either need mirrors
and two paths, or three clocks with instantaneous
"orthogonal" orientation.

I have no idea what you are talking about. The MMX
got no clock.

While it has no explicit clock, the fact that it uses
velocity and distance travelled defines a clock implicitly.
The phase shifts and fringes are created by a difference
in transit time of the waves.

Or have you forgotten that L=v*t when v=constant?

Does your theory redefine that as well? If so, you need
to clearly explain that with a justification.

I wonder what a constant frequency source means to
kenseto, if not a very steady clock? That is what is used
for MMX these days...

The MMX is designed to detect any phase shift between the
two right angled light rays from the same source as the
apparatus is rotated. No clock is needed.

.... because the "same source" *is* a very steady clock. And two
synchronized very steady clocks equates to the two two-way light paths
of MMX.

David A. Smith

In article ,
"kenseto" wrote:

And as I note above, your definition of 'vertical' is ambiguous.

Comments like this is the reason why I call you an idiot.

No - your own comments are why you are an idiot, Ken. Your knowledge of
modern physics is pitiful.

--

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"Cygnus X-1" wrote in message
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On Sun, 26 Nov 2006 09:37:54 -0500, kenseto wrote
(in article ):


"Cygnus X-1" wrote in message
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On Sat, 25 Nov 2006 17:33:22 -0500, kenseto wrote
(in article ):



Considering that this is one test that would be insisted by almost any
competent reviewer of your theory, I insist on it not for me. You need
it for your own credibility.

Again, why would anyone use your theory if you can't demonstrate that
it meshes seamlessly with existing observations?

Then don't use it. I am not going to jump through hoops for you. In any
case
you physicists are too indoctrinated to accept any new idea .....come
hell
or high water.

There's "New Ideas" and then there's "New Ideas that Work".

How do you know that my new idea won't work?? What observations or
experiments refute my new idea?

You've still haven't demonstrated that your "new idea" reproduces
well-tested observations better than the existing body of physics.

Sure I have.....the equations of IRT are converted from SRT equations. Any
observations or experiments that agree with SRT will agree with IRT. After
all SRT is a subset of IRT.

And in the long run you will repeatedly encounter this complaint about
your theory until you either address the issue, give up promoting it,
or die.

ROTFLOL....you are clueless.

In gravitational problems, it's the two-body problem. Einstein
demonstrated that in the case of the solar system, the changes from
the
classic Newtonian result was a perihelion shift.

A procedure to calculate the perihelion shift using the IRT transform
equation is outlined in the paper "Unification of Physics".

You never show how to do the transformations.

Yes I did.

You show raw equations. You never demonstrate them being applied for
the specific examples.

IRT equations are not raw. They are converted from SRT equations.

For example he

http://www.mathpages.com/rr/s6-02/6-02.htm

the perihelion precession is computed starting from the Schwarzschild
metric, through computing the equations of motion to solving those
equations to yield a single (rather simple) expression for computing
the precession. They even show a comparison with all the planets of
the solar system.

IRT transform equations can do that. The result is also a simple
expression
that reveals the precession.

Really? I've been through your PDF files several times now. Most of
the forumulae are repeated from one file to the next with very little
new material.

If you do the periodic IRT transform equations for the sun and Mercury you
will get a path and the an equation for the perihelion for precession of
mercury.

I've yet to find your formula for the precession derived
in anywhere near the level-of-detail presented at the MathPages web
site above. Nor have I seen the comparison of your result to the
observations.

This is a standard homework exercise in any GR class.

Why would anyone use your theory when the existing theory does so well?

You don't have to use my theory. You can stay ignorant all you remaining
natural life. If my theory is correct some future young physicists will
re-discover it.

And the same demands will be placed on them, just as they were placed
on Einstein, Schrodinger, Dirac, etc. and all those whose theories we
reliably use today in technologies.

Then why there is no such demand on Einstein's 1905 paper on SR??

What makes you think you are so special as to be exempt from that
demand?

Cause I am not a runt of the SRian like you. I don't have to follow any rule
set by establish physicists.

Existing theories failed to explain a number of observations: the
accelerated expansion of the far reached regions of the universe,

GR with non-zero cosmological constant does well with that but may be
revised in the future. It is being subjected to the same level of
scrutiny and demands which you refuse to accept for your theory.

BUT GR have no mechanism for the CC. Also in order to fit observations the
CC must be zero in the neighboring regions and non-zero in the far reached
regions. This is nothing but put in fixes to make GR agree with
observations. A correct model of the universe such as model mechanics
requires no fixes to explain all observations.

the horizon problem, the flatness problem,

Inflation is still being tested but will probably be revised. It is
being subjected to the same level of scrutiny and demands which you
refuse to accept for your theory.

Again inflation is an ad hoc hypothesis to fix the problem in the Standard
Big Bang Model. Model Mechanics requires no such fix.

the path of travel of pioneer 10...

May still be a measuring systematic.

Sure you can invent any fix you want. This is not do theoretical physics.

In article ,
"kenseto" wrote:

There's "New Ideas" and then there's "New Ideas that Work".

How do you know that my new idea won't work?? What observations or
experiments refute my new idea?

One big clue is your misunderstanding about spectroscopy, your inability
to explain why this has not been detected by space based interferometers
plus your inability to realise gravitational redshift != a change in c

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"Cygnus X-1" wrote in message
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On Sat, 25 Nov 2006 10:55:05 -0500, kenseto wrote
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"Cygnus X-1" wrote in message
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On Fri, 24 Nov 2006 09:43:42 -0500, kenseto wrote
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"Cygnus X-1" wrote in message
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On Thu, 23 Nov 2006 06:44:24 -0500, kenseto wrote
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"Cygnus X-1" wrote in message
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On Tue, 21 Nov 2006 08:44:39 -0500, kenseto wrote
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"Cygnus X-1" wrote in message
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On Mon, 20 Nov 2006 08:58:26 -0500, kenseto wrote
(in article ):

SOHO/MDI has been in operation over a decade.

http://soi.stanford.edu/

..

Why? Doesn't SRT need relative velocity to do calculations? BTW how do
you
measure relative velocity between two objects.

Not to write the equation.

So what is your point? IRT has equations.

But you seem incapable of using them for generating detailed results of
their application in established experiments and technologies.

IRT equations are converted SR equations. So are you saying that SR
equations are not capable of generating detailed results for established
experiments and technologies?

Relative velocity measured by whom?

How do you measure the relative velocity of an object moving wrt you.
What
theory do you use to make such a measurement?

Depends on equipment and objects of interest. Some methods I can think
of quickly a
1) Motion against a distant background

Against what background?

2) time-of-flight between detectors

This method is valid on earth but what if the object is in flight in space?
Also how do you synchronized the two clocks?

3) radar differential-arrival time.

This procedure is circular. It uses the SR postulate of constancy of the
speed of light to do measurements.

the
universe static.

The cosmological constant did not have to be installed 'manually'.
It
was always a valid solution of the Einstein field equations - Einstein
just initially set it to zero, and then reconsidered that decision.

So what is the vaue of the CC in the neighbouring regions? and what is
the
value of the CC in the far reached regions? Does the CC has the same
value
in all regions of the universe?

In GR, the cosmological constant is the same everywhere. There are
various modifications of GR that allow similar parameters to vary in
time and/or space, but they are subject to the same constraints.

If the CC non-zero in GR and it is the same everywhere then the solutions of
GR for the neighboring regions of the universe will disagree with
observations.


When you have differential equations, such as dx/dt = 4, then x=4t and
x=4t+5 are *both* valid solutions. You choose one over the other if
you have boundary/initial conditions in the problem. The cosmological
constant is a simple constant of integration.
If you don't understand these mathematical principles, i.e. Calculus,
then you don't have the math background needed to do the analysis you
claim.

The trouble with you physicists is that you think math doesn't have to
have
physical boundary. If you understand real physics at all you should take
heap that all valid math processes must be based on real phyiscal
processes.....and collapse of the wave function is not a real physical
process.

Force = mass * velocity
newtons = kg * m/s

Perfectly acceptable math. Invalid physics.

It's been repeatedly demonstrated by experiment that physically usable
mathematics is a subset of the available mathematical operations.

Your statement is false.

Collapse of the wave function is not a physical process. Also all valid math
operations must based on real physical processes.

Newton didn't know G, but much of celestial mechanics was
derivable even without that specific knowledge.

Are you saying that Newton derive the value of G??

No. It was an arbitrary constant of proportionality in his force
equation. It would be over 100 years before before an actual value was
experimentally determined. Even without a specific value for G, many
calculations could be done scaled to a reference value, such as the
mass of the Sun or the Earth-Sun distance.



And there was a professor I knew in 1991 who (along with many others)
were (re-)examining the effects of a non-zero cosmological constant.
But the cosmological constant's impact has been known for decades.
It's only recently that measurements became sufficiently sensitive to
distinguish a non-zero value.

But do they have the physical processes that explains why the CC is zero
in
the neighboring regions and non-zero in the far reached regions? Model
mechanics has such physical explanation.

The cosmological constant is not necessarily zero locally, just so
small we've been unable to (at present) detect its effects in planetary
motions. The currently accepted value fits that requirement.
You're alleged 'prediction' beat no one.

Sure it beats everyone because the prediction is a natural outcome of
the
physical model of Model Mechanics.

And it is a natural outcome of GR with the cosmological constant which
still beats you by 80 years.

You are an idi0ot runt of the SRians.

"Phineas T Puddleduck" wrote in message
news
In article ,
"kenseto" wrote:

Science normally involves explaining your conclusions. People have
already explained that spectroscopy has been done in that
direction....
There is no anisotropy in ANY direction.

You must be in the proceess of reverse evolution. It is easier to teach
a
monkey than to teach you. Spectroscope measurement is not the same as
the
MMX.

FTS is a michelson intterferometer that works in the same way and the
same principles as MMX. You can blather on about other light, but that
should make NO DIFFERENCE to your ether theory unless you also subscribe
to the other failed theory, tired light.

Hey idiot here's how the MMX on earth works. Does the FTS in space works the
same? The answer is no.
The single light ray is splitted into two rays and these two light rays are
then recombined to give the characteristic light and dark fringes. When the
light rays are isoropic you will get null result (no fringe shift) as the
apparatus is rotated. When the light rays are anistropic you will get
non-null result (fringe shift) as the apparatus is rotated.

Ken Seto