Interpreting the MMX null result

"dlzc" wrote in message
oups.com...
Dear kenseto:

kenseto wrote:
"dlzc" wrote in message
ups.com...
Dear kenseto:

kenseto wrote:
dlzc wrote:
...
The atomic clocks located "vertical" and "horizontal" from
your position that agree with each other, don't agree with
you. All measurements are local.

The null results of the horizontal MMX are also local. If we do
an MMX with the plane of the light rays oriented vertically the
results obtained is also local. So what is your point?

From your original post...
It merely means that if the plane of the light rays is oriented
vertically then the apparatus will give non-null result with
respect to these local light rays.

Atomic clocks located horizontally and vertically from any
position on Earth yield a null result.

How can you say that? The horizontal MMX gives null result and
the vertical Pound and Rebka gives non-null result.

Pound Rebka does not involve a third atomic clock at the same elevation
as the tower clock, but "located horizontally" to the observer.

No clock is used in the MMX. If you perform the horizontal MMX up at the
tower you will get isotropy. If you perform the vertical MMX up at the tower
you will get anisotropy. BTW the MMX got its own light source.

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.

Ken Seto

"Cygnus X-1" wrote in message
. net...
On Tue, 21 Nov 2006 08:44:39 -0500, kenseto wrote
(in article ):


"Cygnus X-1" wrote in message
. net...
On Mon, 20 Nov 2006 08:58:26 -0500, kenseto wrote
(in article ):
You made the blanket statement that these orbiting Michelson
Interferometers
will refute what I claimed. So it is up to you to support your
statements.
BTW what I claimed is supported by the Pound and Rebka experiments.

No. I did not say it refuted your claim - I said your claim must be
consistent with the operation of these instruments.

So what are the operation data of these instruements? As I understand it
these are proposed experiments....right?

SOHO/MDI has been in operation over a decade.

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.


SOHO/MDI has one axis pointing along the sun-spacecraft line - vertical
relative to the Sun's gravitational field, the other perpendicular to
that. In terms of the plethora of additional variables you've defined
on your site, present an equation for the phase shift you expect for a
spacecraft at that location and interferometer orientation.

I have no idea what you are talking about.

It's a Michelson interferometer at the L1 position with one arm pointed
at the Sun (due to the location, it would also be roughly vertical
orientation relative to the surface of the Earth). The one on SOHO.

There's the MMX you claim has not been done.

But that's not the MMX. The MMX got its own light source.

If your theory is correct, what's the value of the phase shift you
expect for the MMX at this location and orientation?

If you actually do the vertical MMX you will find non-null result in the
vertical direction. That's all I claimed.


While you're at it, perhaps you should work out the 2-body problem with
your revision to Newton's principle of gravity and laws of motion -
compute variations to orbital periods, and other orbital element
perturbations, so they can be compared to solar system observations.

This is can be done with IRT. However, IRT requires periodic direct
measurtements of Fab for a standard light source to do calculations.

Ken Seto

"Cygnus X-1" wrote in message
. net...
On Mon, 20 Nov 2006 08:58:26 -0500, kenseto wrote
(in article ):


"Cygnus X-1" wrote in message
. net...
On Sun, 19 Nov 2006 14:07:49 -0500, kenseto wrote
(in article ):




My analysis will be ready when it's ready. I still have to contact the
instrument teams to ensure I'm interpreting their sensitivity
specifications correctly.

So you made blanket statements without experimental support?

I accept the simplest solution that fits the available data. When the
available data stops fitting the simplest solution, then it is time to
look further.

Looking through your 'work' (and the link on past experiments is
broken),

Sorry try it now.
http://www.geocities.com/kn_seto/200...xperiments.pdf

this is one messy theory.

How is it messy? I thought it's very simple compared to current theories.
Also IRT is a complete theory of motion. Unlike SRT/GRT, IRT has an
unlimited domain of applicability.

Perhaps you should standardize your
notation. Given enough variables, you can hack anything into agreement
with any data you want.

I don't understand IRT has very few variables..

As I state above, derive the shift. When my analysis is complete (and
I'll probably include some interferometer orientation issues), we'll
compare and see.

Thank you for reading my website. In order to derive anything using IRT we
need the raw data for Fab.

Ken Seto

On Thu, 23 Nov 2006 06:44:24 -0500, kenseto wrote
(in article ):


"Cygnus X-1" wrote in message
. net...
On Tue, 21 Nov 2006 08:44:39 -0500, kenseto wrote
(in article ):


"Cygnus X-1" wrote in message
. net...
On Mon, 20 Nov 2006 08:58:26 -0500, kenseto wrote
(in article ):
You made the blanket statement that these orbiting Michelson
Interferometers
will refute what I claimed. So it is up to you to support your
statements.
BTW what I claimed is supported by the Pound and Rebka experiments.

No. I did not say it refuted your claim - I said your claim must be
consistent with the operation of these instruments.

So what are the operation data of these instruements? As I understand it
these are proposed experiments....right?

SOHO/MDI has been in operation over a decade.

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.


SOHO/MDI has one axis pointing along the sun-spacecraft line - vertical
relative to the Sun's gravitational field, the other perpendicular to
that. In terms of the plethora of additional variables you've defined
on your site, present an equation for the phase shift you expect for a
spacecraft at that location and interferometer orientation.

I have no idea what you are talking about.

It's a Michelson interferometer at the L1 position with one arm pointed
at the Sun (due to the location, it would also be roughly vertical
orientation relative to the surface of the Earth). The one on SOHO.

There's the MMX you claim has not been done.

But that's not the MMX. The MMX got its own light source.

The experimental apparatus does not "know" where the light source is
located. You can set time/location/direction/frequency of the sunlight
when it enters the apparatus as input variables.

If your theory is correct, what's the value of the phase shift you
expect for the MMX at this location and orientation?

If you actually do the vertical MMX you will find non-null result in the
vertical direction. That's all I claimed.

You should be able to write your prediction for the shift in the
variables you've defined.



While you're at it, perhaps you should work out the 2-body problem with
your revision to Newton's principle of gravity and laws of motion -
compute variations to orbital periods, and other orbital element
perturbations, so they can be compared to solar system observations.

This is can be done with IRT. However, IRT requires periodic direct
measurtements of Fab for a standard light source to do calculations.

Bull.

Again, you've defined enough variables to derive this result in terms
of those variables, including Fab.

Einstein didn't know Lambda, but cosmologists were able to derive
predictions with it as a variable and then compare that prediction to
the data. Newton didn't know G, but much of celestial mechanics was
derivable even without that specific knowledge. I don't believe Planck
knew the value of his constant at the time he published his solution to
black body radiation - it was a variable which others then compared to
data to generate the value. Accepted theories aren't formulas
'declared', they're recipes for deriving predictions.

If you can't do these derivations without another experiment, then
you're theory is incomplete or you're doing something very wrong.

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"

On Thu, 23 Nov 2006 07:22:18 -0500, kenseto wrote
(in article ):


"Cygnus X-1" wrote in message
. net...
On Mon, 20 Nov 2006 08:58:26 -0500, kenseto wrote
(in article ):


"Cygnus X-1" wrote in message
. net...
On Sun, 19 Nov 2006 14:07:49 -0500, kenseto wrote
(in article ):




My analysis will be ready when it's ready. I still have to contact the
instrument teams to ensure I'm interpreting their sensitivity
specifications correctly.

So you made blanket statements without experimental support?

I accept the simplest solution that fits the available data. When the
available data stops fitting the simplest solution, then it is time to
look further.

Looking through your 'work' (and the link on past experiments is
broken),

Sorry try it now.
http://www.geocities.com/kn_seto/200...xperiments.pdf

Pictures and prose but no mathematical derivations of how your 'theory'
will generate the exact same wavelength shift as the Compton formula
and what will be different.

this is one messy theory.

How is it messy? I thought it's very simple compared to current theories.
Also IRT is a complete theory of motion. Unlike SRT/GRT, IRT has an
unlimited domain of applicability.

You appear to have redefined momentum, energy and kinetic energy,
gravitational redshift, etc. tossing in extra variables lambda, Fab,
Faa, etc. without even a reasonable experimental justification. You
don't demonstrate that these reduce to the forms we currently use at
some level of approximation.

Also, momentum, energy, etc. are all quantities fundamentally derivable
from Newton's laws combined with simple assumptions of time/space
symmetries.

You don't show the derivation of these quantities using your claims of
spatial/temporal asymmetries - you just declare them.

Incomplete.

Sloppy.


Perhaps you should standardize your
notation. Given enough variables, you can hack anything into agreement
with any data you want.

I don't understand IRT has very few variables..

As I state above, derive the shift. When my analysis is complete (and
I'll probably include some interferometer orientation issues), we'll
compare and see.

Thank you for reading my website. In order to derive anything using IRT we
need the raw data for Fab.

You can derive EVERY result, and more, with Fab as a variable.

And you must show your work at each step.

The work of Newton, Einstein, Schrodinger, etc. survives not because
they declared it or because only they were the only ones who could do
the calculations, but because OTHERS could start with their equations
and definitions and derive results consistent with experiments.

You have a LOT of work to do.

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"

On Thu, 23 Nov 2006 06:31:03 -0500, kenseto wrote
(in article ):


"dlzc" wrote in message
oups.com...
Dear kenseto:

kenseto wrote:
"dlzc" wrote in message
ups.com...
Dear kenseto:

kenseto wrote:
dlzc wrote:
...
The atomic clocks located "vertical" and "horizontal" from
your position that agree with each other, don't agree with
you. All measurements are local.

The null results of the horizontal MMX are also local. If we do
an MMX with the plane of the light rays oriented vertically the
results obtained is also local. So what is your point?

From your original post...
It merely means that if the plane of the light rays is oriented
vertically then the apparatus will give non-null result with
respect to these local light rays.

Atomic clocks located horizontally and vertically from any
position on Earth yield a null result.

How can you say that? The horizontal MMX gives null result and
the vertical Pound and Rebka gives non-null result.

Pound Rebka does not involve a third atomic clock at the same elevation
as the tower clock, but "located horizontally" to the observer.

No clock is used in the MMX. If you perform the horizontal MMX up at the
tower you will get isotropy. If you perform the vertical MMX up at the tower
you will get anisotropy. BTW the MMX got its own light source.

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.

Ken Seto


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.

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"

Dear Cygnus X-1:

"Cygnus X-1" wrote in message
. net...
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...

David A. Smith

In article ,
"kenseto" wrote:


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

And in space?

You will have to do the experiment in space.

There are plenty of interferometers in space.

Then do the MMX in space.

Interferometry. Read up a little, be enlightened.

--
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http://hyperphysics.phy-astr.gsu.edu/hbase/heacon.html

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In article ,
"kenseto" wrote:

Hey idiot....I claimed that the vertical MMX on earth will get non-null
result. Your measurements of of frequency and wavelength to get c is
irrelevant.

You really cant see the gaping hole in your argument, can you?

--
Thermodynamics claims another crown!

http://hyperphysics.phy-astr.gsu.edu/hbase/heacon.html

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In article ,
"kenseto" wrote:

And its that frequency shift that would be picked up by an
interferometer measuring spectral lines.

So what? ....but that's not the MMX in space. The MMX has its own light
source and thus it should be able to detect anisotropy of the speed of light
if it exists.

Why should its "own" light make a difference?

--
Thermodynamics claims another crown!

http://hyperphysics.phy-astr.gsu.edu/hbase/heacon.html

--
Posted via a free Usenet account from http://www.teranews.com