Interpreting the MMX null result

On Mon, 4 Dec 2006 09:28:58 -0500, kenseto wrote
(in article ):
"Cygnus X-1" wrote in message
. net...
On Thu, 30 Nov 2006 10:05:06 -0500, kenseto wrote
(in article ):
"Cygnus X-1" wrote in message
. net...
On Wed, 29 Nov 2006 09:50:02 -0500, kenseto wrote
(in article ):


Mr. Seto, do YOU have an established track record of discoveries?
Any
patents or other inventions? Why should we give your claim more
weight
than others?

These rules are designed for runt like you. I don't have to follow
them
since I don't depend on the establishment to make a living.

Yet those 'establishment' scientists are building satellites and
sending them to distant locations. They do this quite successfully NOT
using your theory.

So what???? They still use Newtonian Mechanics to figure the orbits and we
know that Newtonian Mechanics is not as good as relativity..

Actually, we can do calculations with relativity using the
Post-Newtonian Approximation
(http://en.wikipedia.org/wiki/Paramet...nian_formalism)
should we be able to measure at that precision.

In your theory, measurements of fab (a SCALAR) alone does not uniquely
define velocity & direction, the critical VECTOR quantity for doing
these calculations.

If you're claiming you can construct a VECTOR from fab, you better do a
demonstration of that.

All your theory does is replace velocity, a quantity measurable by many
techniques, with spectral frequencies that require special emissive
properties of the objects of interest (line spectra).


Meanwhile you whine about how you're a 'privileged character' and
don't have to DEMONSTRATE that your 'theory' can produce the
trajectory.

I am not whining about anything. I am saying that I have a model of the
universe that seem to be more encompassing than current theories. It is up
to the establisment who is in control of the funding processes to evaluate
these new propose theories.

When I write proposals for funding, I have to establish far more
mathematical & physical justification than you've done in any of your
publications. The analysis I've done using your transformation
equations below demonstrates how your theory is either woefully
incomplete, ambiguously defined so it is impossible for others to apply
and get usable results, or just plain wrong.

Your problem is that you think that we are enemies. We are not. We have a
common goal: to find the final unify theory.

Enemy???

I worked my way through undergraduate education doing business
consulting in IT. Often I was called in to troubleshoot and develop
audit systems. Many times the problems I was asked to solve were the
result of arrogant idiots who thought the company 'checks and balances'
didn't apply to them.

But I don't work for any company.

Not surprising. You seem to have no skills other than perhaps a knack
for propaganda.

I work for company before and I hold 2 patents.

Excellent! Since you have some technical expertise, you should have no
problem doing the vertical MMX yourself (links at bottom of page).

I did conduct a simple search at uspto.gov but didn't find you.

So how many patents you hold? My guess: none.

That's true. I usually work as a trouble-shooter and problem-solver.
Locking some of my work up in patents would generate more money for
lawyers than me. Though I have seen reports of patents which read much
like things I implemented years earlier at customer sites. I could
probably cause the patent holder some problems in the 'prior art' and
'obviousness' departments. :^)

I sent a number of those arrogant idiots, who arrogance was exposed as
a mechanism for hiding their incompetence, or in some cases corruption,
to the unemployment line. They ended up not working for a company
either. :^)

You are the arrogance idiot. I have no iea why you feel so threatened by me.

You've apparently never been to a science conference. People who make
outlandish (and occasionally not so outlandish) claims usually find
themselves on the receiving end of these types of challenges.

Every scientist has to defend their work.

I've had to deal with a few aggressive challenges on my own work. I've
occasionally had to make revisions to my own work due to the
challengers pointing out effects I had not examined.

That's the way it is.

If you can't take the heat, get out of the kitchen.

[ stuff deleted ]
Faa and Fab are frequency measurements make by observer A of a
standard
light source (eg sodium) in the A and B frame.
Obviously if you want to do any calculations using IRT equations you
need to
have periodic Fab data.

Basically what you're saying is we can't use your theory at all without
a 'standard light source' on it?

You can use my theory as long as light is coming from the observed object.
Can you use relativity if there is no light coming from the observed object?
The answer is NO.

To measure Fab requires an identifiable spectral line as reference to
know Faa.

There are many objects we observe which emit (black holes, neutron
stars) or reflect (asteroids) strong continuum radiation (black body,
power-laws, etc.) and don't exhibit strong spectral features (or they
are below the sensitivity of current instruments). Without additional
information, there is no way to determine Fab (or Faa).

How would we use your theory in this situation?

1) An asteroid is detected which might be on a collision course with
Earth. It has no 'standard light source' on it and reflected sunlight
is too faint to get an identifiable spectral line. Celestial mechanics
handles this problem readily. How do I get Fab for Ken's theory?

In that case you can revert back to SR because SR is a subset of IRT. The
postulates of IRT includes the postulates of SR.

No they do not. The invariant interval is not preserved in your
transformations (see example below).

Even the T*T^{-1} is not preserved by your transformation (you can move
just by performing a coordinate transformation (see example below).

You're saying we can't compute the trajectories of these spacecraft
without additional data????!! That would certainly be news to the
flight dynamics people who computed these trajectories and are
monitoring these spacecraft.

Sure you can compute the trajectories of these spacecrafts by setting
the
periodic values of Fab. To ensure your spacecraft follows the course you
charted you accelerate the craft to these preset periodic Fab values.

This still doesn't tell me how I compute a trajectory for the entire
two year mission MONTHS BEFORE LAUNCH. Newtonian/Einsteinian physics
has doing this for us quite well for the past 50 years.

Sure it does....all you need to do is to compute the positions of the craft
at the preset periodic Fab values and plot these values on a graph will
reveal the future path..

fab only gives *radial* information, at best. The spacecraft could be
traveling in a different direction and velocity and still have the same
Doppler fab value. Then you'll probably loose your ship.

Question: What do we gain in using your theory? It certainly isn't
very practical for doing spaceflight.

How do we get this Fab data?

See post by PD below.

Later. Not directly relevant to the current issue.

It is relaevant to the current issue.

We know how to get the spacecraft velocity in conventional theory.
How
do we get Faa and Fab to use your model?

I did not redefine gamma. In SR:
f'=f_o(1/gamma)
f'=Fab
f_o=Faa
Therefore Gamma=Faa/Fab and 1/gamma=Fab/Faa

No. Gamma=sqrt(1-(v/c)^2).

You are wrong. Gamma=1/sqrt(1-(v/c)^2).
It appears that you need to go back to school and learn basic physics.

No, I just need to proofread better.

NO you need to think before you shoot your mouth off.

The point is that gamma is a quantity of convenience which is only a
function of velocity. It is not 'defined' by any other quantity.
Otherwise, you would have a problem with other processes that can
generate spectral lines.

Methinks you have a conveniently short memory. I merely admitted my
error. You don't even acknowledge yours:

error documentation

Example 1 (november 24):
Model Mechanics provides the physical processes for the Compton formula.
Currently there is no physical interpretation for the COMpton formula. For
example collapse of wave function is a bunch of word salad. It has no
physical connection in real life.

I pointed out:
Wrong again. Standard homework problem for many QM classes: derive the
Compton formula for an incident photon of some wavelength striking an
electron at rest (it's an elastic collision problem).

I've even derived inverse Comptonization formulae.
and here's the link to Compton's original paper with the derivation
(subscription required):
http://prola.aps.org/abstract/PR/v24/i2/p168_1

Example 2 (november 27):

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.

You claimed:

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

and I point out that Einstein did so.

Thanks to the person(s) who did the translation and placed it online:

http://www.fourmilab.ch/etexts/einstein/specrel/www/

You'll note he does a number of worked examples: measuring rods, clocks,
velocity composition, aberration, etc.

This is all very easy to follow, and for others to understand and expand on
the work. You could practically teach a classroom straight from this paper.

This is exactly what you FAIL to do.

You conveniently never acknowledge these errors.
/error documentation

Much like you need to check your coordinate transformations (equations
5-8) in 'unification' and 'origin' papers.

Did you mean to give x & t in terms of x' & t' for #7 & 8 or is #7 & 8
supposed to be transformations for something moving in the opposite
direction?

NO....5 and 6 represent the observed clock is running slower than the
observer's clock and the light path length of the observed rod is longer
than the observer rod's light path length.
7 and 8 represent the observed clock is running faster than the observer's
clock and the observed rod has a shorter light path length than the
observer's rod.

Do you mean fab or fba? Is fab/fba being measured by the
primed or unprimed observer?

NO.....there is no such thing as fba. A is the observer.

I will answer the rest of your post at a later time.

I'll be watching, though any substantial response from me will probably
have to wait until next year.

Remember, no matter what, for physical consistency, it is required that
x' = f1(x,t) ; t' = f2(x,t)

and

x = f1'(x',t'); t = f2'(x',t')

where f1' and f2' are the inverse transformation functions.

(x,t) - (x',t') - (x,t) must give the exact same point.

Choose carefully. I've examined a few of your options and each yields
mathematical contradictions and/or some physical inconsistencies.

In this simple example, using equations in your 'Unification' paper,
SetoXform1 is your transform eq 5 & 6.

i.e. (x',t') = SetoXform1(x,t,faa,fab,wavelength)

SetoXform2 is equations 7 & 8 which should enable us to transform from
an observer in the (previously defined) primed frame BACK to the
unprimed frame. It should give us the exact same point.

SetoXform1inv & SetoXform2inv are the proper algebraic inversion of the
respective transformations.

Choose our units so faa=1.0 & wavelength=1.0/faa

For our primed & unprimed systems, we'll let fab=0.5 and examine an
event at (x,t)=(1.0,2.0). The Interval is defined as x^2-t^2 which is
equal to x'^2-t'^2, a critical assumption in SR. We compute its value
for all three stages of the transformation.

SetoXform1 & SetoXform2
Event - Event' - Event
(x,t)=(1.0, 2.0) - (x',t')=(4.0, 5.0) - (0.75, 1.5)!=(x,t)
!!!!! FATAL ERROR
Interval: -3.0 -9.0 -1.6875

Here we see that not only is the Interval not preserved by your
transformation, but your return transformation doesn't return to the
same point!!!

-----
Compare this to the true transforms for this system...
SetoXform1 & SetoXform1inv
Event - Event' - Event
(1.0, 2.0) - (4.0, 5.0) - (1.0, 2.0)
Interval: -3.0 -9.0 -3.0

If we use the true algebraic inverse of SetoXform1, which you don't
show, we can return to the original point, but the Interval is still
not preserved.

---
We do this using your other transform:
SetoXform2 & SetoXform2inv
Event - Event' - Event
(1.0, 2.0) - (0.0, 0.75) - (1.0, 2.0)
Interval: -3.0 -0.5625 -3.0

Again we can return to the original point, but the Interval is still
not preserved.

Your claim that your theory is a superset of SR is thereby demonstrated
false by simple example.

Simple Python (http://www.python.org) program for generating these
(~100 lines) can be posted on request.

And continuing from befo

http://www.fourmilab.ch/etexts/einst...html#SECTION22


Your equation is a special case of the Doppler formula where phi=pi/2
so cos(phi)=0 - when the object is moving transverse to the
line-of-sight. You essentially LOSE most of your directional
information.

Any object can be said to be moving transversly wrt the observer. Fab is
defined as the mean frequency value for that object.

But we rarely measure the 'mean frequency value'.

Consider a binary star where the spectral line may change slowly with
time. We might take measurements a year apart and see fab change, but
we may not get the 'mean value' over the entire orbit for many years.
Which 'mean value' do we use? The mean of a set of five values a year
apart or do we have to wait for a full orbit (say 20 years)? Are you
saying we can't determine anything about the orbit without waiting for
the full period.

We haven't seen a full orbit of Pluto so we don't have a 'mean value'
for fab.

Are you saying we can't compute the orbit of Pluto?

That's rough. Where will Pluto be when "New Horizons"
(http://en.wikipedia.org/wiki/New_horizons) arrives there in about a
decade?

But then, that was obvious from the start. You define:

relative veocity v=lambda(Faa-Fab)

Velocity is a *VECTOR*, generally represented by 3-components.
Frequency is a SCALAR, a single component. Your theory intrinsically
looses most directional information.

Wrong....frequency*wavelength is also a vector. So the directional
information is maintained.

The product of two scalars is NOT a vector.

x*y != (fx, fy, fz)

That is a mathematically ill-defined operation. If you are claiming
you've defined frequency or wavelength or using a tensor product, you
need to use the correct notation. Classical mechanics has a angular
frequency vector, and solid state physics and crystallography make use
of a wavenumber vector which is inversely related to wavelength. If
you're claiming to use these then use the correct notation.

If you're claiming you can construct a vector from a single Doppler
measurement, I suggest you demonstrate it, or better yet, patent it.

And just so you don't forget, I'll re-append this here.

Oh, you seemed to have skipped issues from another part of the thread
so I'll consolidate them here (snipped from Nov 28 post and revised):
========
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.

But my examples above are not all on the Earth. You didn't answer the
question.

Did you just not read the question carefully or is this an issue in
your theory that horizontal and vertical is only defined on Earth? Is
your theory specifically geocentric?

Is your frame of absolute motion at the center of the Earth or does the
Earth not rotate in your frame of absolute motion?

What is vertical to the robot explorers on Mars? What will be vertical
for HUMAN explorers on Mars?

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

We were talking about the MMXs on earth.

Not necessarily. If you want your theory to be considered 'universal',
you have to consider what happens elsewhere.

=======
You also forgot this in an earlier part of this thread:

Perhaps you think we should just launch a 300 million dollar satellite
or even a human crew without validating their trajectory computation, a
computation which your theory impacts? Perhaps YOU would volunteer to
be on that crew testing a trajectory computation with your theory for
the first time?

==========
Then I pointed out that Michelson interferometers are fairly standard
lab equipment, why don't you do the experiment yourself? MnM didn't
even have a very large setup.

http://www.meos.com/Optical%20Experi...ferometer.html

http://www.omni-optical.com/l-optics/sl435.htm

http://www.juliantrubin.com/bigten/michelsonmorley.html

Just make sure you don't get a phase shift from your vertical MMX due
to mechanical strain on the vertical arm.

There, think I've covered everything...

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"

kenseto wrote:

"jem" wrote in message
...

kenseto wrote:


"jem" wrote in message
...


kenseto wrote:



"jem" wrote in message
...



kenseto wrote:




"jem" wrote in message
...



Sigh....share the same relative motion as determined by the DETECTOR.


OK, so let's apply your criteria. According to you, *any* two things
share the same relative motion, if they share the same relative motion
as determined by the DETECTOR.


NO.....

All I did was repeat what you said, so apparently what you said isn't
what you wanted to say.


..The detector (the observer) measures the two objects have the same
relative velocity wrt him.

OK, so the revised criteria is that *any* two things share the same
relative motion if they're measured to have the same velocity relative
to a detector (the observer).


Sigh....the two things doesn't share the same relative motion wrt each
other. They are measured to have the same relative motion wrt the observer.

So let's see.

I just indicated that your criteria was, "*any* two things share the
same relative motion if they're measured to have the same velocity
relative to a detector (the observer)."

And you want to correct that to, "*any* two things share the same
relative motion if they are measured to have the same relative motion
wrt the observer."

Is that right?

"jem" wrote in message
...
kenseto wrote:

"jem" wrote in message
...

kenseto wrote:


"jem" wrote in message
...


kenseto wrote:



"jem" wrote in message
...



kenseto wrote:




"jem" wrote in message
...



Sigh....share the same relative motion as determined by the DETECTOR.


OK, so let's apply your criteria. According to you, *any* two things
share the same relative motion, if they share the same relative motion
as determined by the DETECTOR.


NO.....

All I did was repeat what you said, so apparently what you said isn't
what you wanted to say.


..The detector (the observer) measures the two objects have the same
relative velocity wrt him.

OK, so the revised criteria is that *any* two things share the same
relative motion if they're measured to have the same velocity relative
to a detector (the observer).


Sigh....the two things doesn't share the same relative motion wrt each
other. They are measured to have the same relative motion wrt the
observer.

So let's see.

I just indicated that your criteria was, "*any* two things share the
same relative motion if they're measured to have the same velocity
relative to a detector (the observer)."

And you want to correct that to, "*any* two things share the same
relative motion if they are measured to have the same relative motion
wrt the observer."

Is that right?

NO.....A and B are two things and O is the observer:
A measures B to have a relative velocity wrt him of Vab
B measures A to have a relative velocity wrt him of Vba.
Vab is not equal to Vba because the passage of a clock second in A's frame
does not correspond to the passage of a clock second in B's frame.

O measures A to have a relative velocity of Voa.
O measures B to have a relative velocity of Vob.
Voa=Vob because they have the same value.

kenseto wrote:

"jem" wrote in message
...

kenseto wrote:


"jem" wrote in message
...


kenseto wrote:



"jem" wrote in message
...



kenseto wrote:




"jem" wrote in message
...




kenseto wrote:





"jem" wrote in message
.. .



Sigh....share the same relative motion as determined by the DETECTOR.


OK, so let's apply your criteria. According to you, *any* two things
share the same relative motion, if they share the same relative motion
as determined by the DETECTOR.


NO.....

All I did was repeat what you said, so apparently what you said isn't
what you wanted to say.



..The detector (the observer) measures the two objects have the same
relative velocity wrt him.

OK, so the revised criteria is that *any* two things share the same
relative motion if they're measured to have the same velocity relative
to a detector (the observer).


Sigh....the two things doesn't share the same relative motion wrt each
other. They are measured to have the same relative motion wrt the

observer.

So let's see.

I just indicated that your criteria was, "*any* two things share the
same relative motion if they're measured to have the same velocity
relative to a detector (the observer)."

And you want to correct that to, "*any* two things share the same
relative motion if they are measured to have the same relative motion
wrt the observer."

Is that right?


NO..

Seto, I've asked you 3 times to tell me what you mean when you say two
things share the same relative motion, and each time I've repeated your
response word-for-word and asked you to confirm it, and each time you've
changed it. Is there an end to this process?

....A and B are two things and O is the observer:
A measures B to have a relative velocity wrt him of Vab
B measures A to have a relative velocity wrt him of Vba.
Vab is not equal to Vba because the passage of a clock second in A's frame
does not correspond to the passage of a clock second in B's frame.

O measures A to have a relative velocity of Voa.
O measures B to have a relative velocity of Vob.
Voa=Vob because they have the same value.


Do you seriously think that answers the question: "What does it mean for
two objects to share the same relative motion"?

Your latest response lists two different procedures. The first
procedure has two objects measuring each other's velocity (as if rocks
(e.g.) could measure velocity), and the other has your enigmatic
observer measuring the object velocities (presumably relative to the
observer, whoever or whatever that may be), and coming to the more than
obvious conclusion that the velocities are equal "because they have the
same value" (duh).

How are those procedures supposed to relate to the question you were asked?

"Cygnus X-1" wrote in message
. net...
On Mon, 4 Dec 2006 09:28:58 -0500, kenseto wrote
(in article ):
"Cygnus X-1" wrote in message
. net...
On Thu, 30 Nov 2006 10:05:06 -0500, kenseto wrote
(in article ):
"Cygnus X-1" wrote in message
. net...
On Wed, 29 Nov 2006 09:50:02 -0500, kenseto wrote
(in article ):


Yet those 'establishment' scientists are building satellites and
sending them to distant locations. They do this quite successfully NOT
using your theory.

So what???? They still use Newtonian Mechanics to figure the orbits and
we
know that Newtonian Mechanics is not as good as relativity..

Actually, we can do calculations with relativity using the
Post-Newtonian Approximation
(http://en.wikipedia.org/wiki/Paramet...nian_formalism)
should we be able to measure at that precision.

My point is IRT is an extension of SRT and GRT. If relative velocity data is
more readily available then use SRT/GRT. Also Fab can be calculated from
relative velocity using the following equations:
Fab=Faa(1/gamma) OR Fab=Faa(gamma)

In your theory, measurements of fab (a SCALAR) alone does not uniquely
define velocity & direction, the critical VECTOR quantity for doing
these calculations.

The IRT transform equations will give the coordinate of an object at various
time intervals connecting these coordiates together will give the path of
the object.


If you're claiming you can construct a VECTOR from fab, you better do a
demonstration of that.

Fab*lanbda is a vector.

All your theory does is replace velocity, a quantity measurable by many
techniques, with spectral frequencies that require special emissive
properties of the objects of interest (line spectra).

Fab can be calculated from relative velocity using the SR equation as
follows:
Fab=Faa*(1/gamma) OR Fab=Faa(gamma)



Meanwhile you whine about how you're a 'privileged character' and
don't have to DEMONSTRATE that your 'theory' can produce the
trajectory.

I am not whining about anything. I am saying that I have a model of the
universe that seem to be more encompassing than current theories. It is
up
to the establisment who is in control of the funding processes to
evaluate
these new propose theories.

When I write proposals for funding, I have to establish far more
mathematical & physical justification than you've done in any of your
publications. The analysis I've done using your transformation
equations below demonstrates how your theory is either woefully
incomplete, ambiguously defined so it is impossible for others to apply
and get usable results, or just plain wrong.

I am too old to jump through the large number of hoops that are designed to
maintain status quot. I have a theory that can resolve all the problems of
modern physics and cosmology. If the establishment is not interested so be
it.


Your problem is that you think that we are enemies. We are not. We have
a
common goal: to find the final unify theory.

Enemy???

Yes mainstream physicists regard me as enemy or somebody who is trying to
rock the boat. They refused to recognize that I am offering a fresh
alternative.

..

Not surprising. You seem to have no skills other than perhaps a knack
for propaganda.

I work for company before and I hold 2 patents.

Excellent! Since you have some technical expertise, you should have no
problem doing the vertical MMX yourself (links at bottom of page).

I would if I have the resources. But I don't.

I did conduct a simple search at uspto.gov but didn't find you.

What is uspto.gov? I am currently applying for a grant to do my proposed
experiments described in the paper "Proposed Experiments to Dectect Absolute
Motion". But I will not get it because established physicists will reject my
application without giving it a fair review.

I sent a number of those arrogant idiots, who arrogance was exposed as
a mechanism for hiding their incompetence, or in some cases corruption,
to the unemployment line. They ended up not working for a company
either. :^)

You are the arrogance idiot. I have no iea why you feel so threatened by
me.

You've apparently never been to a science conference. People who make
outlandish (and occasionally not so outlandish) claims usually find
themselves on the receiving end of these types of challenges.

That's fine but the challenger must understand the claims. In your case you
don't understand IRT and you made bogus statements based on your poor
understanding of IRT.

Every scientist has to defend their work.

I've had to deal with a few aggressive challenges on my own work. I've
occasionally had to make revisions to my own work due to the
challengers pointing out effects I had not examined.

That's the way it is.

If you can't take the heat, get out of the kitchen.

[ stuff deleted ]
Faa and Fab are frequency measurements make by observer A of a
standard
light source (eg sodium) in the A and B frame.
Obviously if you want to do any calculations using IRT equations you
need to
have periodic Fab data.

Basically what you're saying is we can't use your theory at all without
a 'standard light source' on it?

You can use my theory as long as light is coming from the observed
object.
Can you use relativity if there is no light coming from the observed
object?
The answer is NO.

To measure Fab requires an identifiable spectral line as reference to
know Faa.

Again Fab can be calculated from relative velocity.
Fab=Faa(1/gamma) or Fab=Faa(gamma)
Also how do you measure relative velocity? Don't you send out a beam of know
frequency and determine the delay time for the return beam?

There are many objects we observe which emit (black holes, neutron
stars) or reflect (asteroids) strong continuum radiation (black body,
power-laws, etc.) and don't exhibit strong spectral features (or they
are below the sensitivity of current instruments). Without additional
information, there is no way to determine Fab (or Faa).

Faa is know. Fab can be measured or calculated from relative velocity.

How would we use your theory in this situation?

1) An asteroid is detected which might be on a collision course with
Earth. It has no 'standard light source' on it and reflected sunlight
is too faint to get an identifiable spectral line. Celestial mechanics
handles this problem readily. How do I get Fab for Ken's theory?

In that case you can revert back to SR because SR is a subset of IRT.
The
postulates of IRT includes the postulates of SR.

No they do not. The invariant interval is not preserved in your
transformations (see example below).

Yes they do. The invariant interval is a specific interval of absolute time
and the specific interval of absolute time is the observer's clock second.

Even the T*T^{-1} is not preserved by your transformation (you can move
just by performing a coordinate transformation (see example below).

That's becasue the reciprocity as asserted by SR is false. There is no
reciprocity. The observer's clock second can have a shorter duration or
longer duration (absolute time content) than the observed clock second.


This still doesn't tell me how I compute a trajectory for the entire
two year mission MONTHS BEFORE LAUNCH. Newtonian/Einsteinian physics
has doing this for us quite well for the past 50 years.

Sure it does....all you need to do is to compute the positions of the
craft
at the preset periodic Fab values and plot these values on a graph will
reveal the future path..

fab only gives *radial* information, at best. The spacecraft could be
traveling in a different direction and velocity and still have the same
Doppler fab value. Then you'll probably loose your ship.

No you are wrong. Fab must agree with velocity calculations. After all it is
calculated using velocity information.

Later. Not directly relevant to the current issue.

It is relaevant to the current issue.

We know how to get the spacecraft velocity in conventional theory.
How
You are wrong. Gamma=1/sqrt(1-(v/c)^2).
It appears that you need to go back to school and learn basic physics.

No, I just need to proofread better.

NO you need to think before you shoot your mouth off.

The point is that gamma is a quantity of convenience which is only a
function of velocity. It is not 'defined' by any other quantity.
Otherwise, you would have a problem with other processes that can
generate spectral lines.

Gamma is also a function of Faa and Fab. Faa/Fab=gamma and Fab/Faa=1/gamma

Methinks you have a conveniently short memory. I merely admitted my
error. You don't even acknowledge yours:

error documentation

Example 1 (november 24):
Model Mechanics provides the physical processes for the Compton formula.
Currently there is no physical interpretation for the COMpton formula.
For
example collapse of wave function is a bunch of word salad. It has no
physical connection in real life.

I pointed out:
Wrong again. Standard homework problem for many QM classes: derive the
Compton formula for an incident photon of some wavelength striking an
electron at rest (it's an elastic collision problem).

The same Compton formula can be used but with different interpretations. The
one peak retains the longer wavelength peak is the result of the photon
bouncing off the nuclei that are in a state of absolute motion. The other
peak is the result of the orbiting electron absorb and re-emit the incident
x-ray photons.


I've even derived inverse Comptonization formulae.
and here's the link to Compton's original paper with the derivation
(subscription required):
http://prola.aps.org/abstract/PR/v24/i2/p168_1

Example 2 (november 27):

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.

You claimed:

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

and I point out that Einstein did so.

Thanks to the person(s) who did the translation and placed it online:

http://www.fourmilab.ch/etexts/einstein/specrel/www/

In this paper Einstein gave a bunch of bogus gedanken to support his claims.
He did not provide any real experiments that could faulsify his theory like
I do with IRT.

Ken Seto

This post is too long. So I am not going to answer the rest of you post. If
you want answer please start a different post.


Much like you need to check your coordinate transformations (equations
5-8) in 'unification' and 'origin' papers.

Did you mean to give x & t in terms of x' & t' for #7 & 8 or is #7 & 8
supposed to be transformations for something moving in the opposite
direction?

NO....5 and 6 represent the observed clock is running slower than the
observer's clock and the light path length of the observed rod is longer
than the observer rod's light path length.
7 and 8 represent the observed clock is running faster than the
observer's
clock and the observed rod has a shorter light path length than the
observer's rod.

Do you mean fab or fba? Is fab/fba being measured by the
primed or unprimed observer?

NO.....there is no such thing as fba. A is the observer.

I will answer the rest of your post at a later time.

I'll be watching, though any substantial response from me will probably
have to wait until next year.

Remember, no matter what, for physical consistency, it is required that
x' = f1(x,t) ; t' = f2(x,t)

and

x = f1'(x',t'); t = f2'(x',t')

where f1' and f2' are the inverse transformation functions.

(x,t) - (x',t') - (x,t) must give the exact same point.

Choose carefully. I've examined a few of your options and each yields
mathematical contradictions and/or some physical inconsistencies.

In this simple example, using equations in your 'Unification' paper,
SetoXform1 is your transform eq 5 & 6.

i.e. (x',t') = SetoXform1(x,t,faa,fab,wavelength)

SetoXform2 is equations 7 & 8 which should enable us to transform from
an observer in the (previously defined) primed frame BACK to the
unprimed frame. It should give us the exact same point.

SetoXform1inv & SetoXform2inv are the proper algebraic inversion of the
respective transformations.

Choose our units so faa=1.0 & wavelength=1.0/faa

For our primed & unprimed systems, we'll let fab=0.5 and examine an
event at (x,t)=(1.0,2.0). The Interval is defined as x^2-t^2 which is
equal to x'^2-t'^2, a critical assumption in SR. We compute its value
for all three stages of the transformation.

SetoXform1 & SetoXform2
Event - Event' - Event
(x,t)=(1.0, 2.0) - (x',t')=(4.0, 5.0) - (0.75, 1.5)!=(x,t)
!!!!! FATAL ERROR
Interval: -3.0 -9.0 -1.6875

Here we see that not only is the Interval not preserved by your
transformation, but your return transformation doesn't return to the
same point!!!

-----
Compare this to the true transforms for this system...
SetoXform1 & SetoXform1inv
Event - Event' - Event
(1.0, 2.0) - (4.0, 5.0) - (1.0, 2.0)
Interval: -3.0 -9.0 -3.0

If we use the true algebraic inverse of SetoXform1, which you don't
show, we can return to the original point, but the Interval is still
not preserved.

---
We do this using your other transform:
SetoXform2 & SetoXform2inv
Event - Event' - Event
(1.0, 2.0) - (0.0, 0.75) - (1.0, 2.0)
Interval: -3.0 -0.5625 -3.0

Again we can return to the original point, but the Interval is still
not preserved.

Your claim that your theory is a superset of SR is thereby demonstrated
false by simple example.

Simple Python (http://www.python.org) program for generating these
(~100 lines) can be posted on request.

And continuing from befo


http://www.fourmilab.ch/etexts/einst...html#SECTION22


Your equation is a special case of the Doppler formula where phi=pi/2
so cos(phi)=0 - when the object is moving transverse to the
line-of-sight. You essentially LOSE most of your directional
information.

Any object can be said to be moving transversly wrt the observer. Fab is
defined as the mean frequency value for that object.

But we rarely measure the 'mean frequency value'.

Consider a binary star where the spectral line may change slowly with
time. We might take measurements a year apart and see fab change, but
we may not get the 'mean value' over the entire orbit for many years.
Which 'mean value' do we use? The mean of a set of five values a year
apart or do we have to wait for a full orbit (say 20 years)? Are you
saying we can't determine anything about the orbit without waiting for
the full period.

We haven't seen a full orbit of Pluto so we don't have a 'mean value'
for fab.

Are you saying we can't compute the orbit of Pluto?

That's rough. Where will Pluto be when "New Horizons"
(http://en.wikipedia.org/wiki/New_horizons) arrives there in about a
decade?

But then, that was obvious from the start. You define:

relative veocity v=lambda(Faa-Fab)

Velocity is a *VECTOR*, generally represented by 3-components.
Frequency is a SCALAR, a single component. Your theory intrinsically
looses most directional information.

Wrong....frequency*wavelength is also a vector. So the directional
information is maintained.

The product of two scalars is NOT a vector.

x*y != (fx, fy, fz)

That is a mathematically ill-defined operation. If you are claiming
you've defined frequency or wavelength or using a tensor product, you
need to use the correct notation. Classical mechanics has a angular
frequency vector, and solid state physics and crystallography make use
of a wavenumber vector which is inversely related to wavelength. If
you're claiming to use these then use the correct notation.

If you're claiming you can construct a vector from a single Doppler
measurement, I suggest you demonstrate it, or better yet, patent it.

And just so you don't forget, I'll re-append this here.

Oh, you seemed to have skipped issues from another part of the thread
so I'll consolidate them here (snipped from Nov 28 post and revised):
========
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.

But my examples above are not all on the Earth. You didn't answer the
question.

Did you just not read the question carefully or is this an issue in
your theory that horizontal and vertical is only defined on Earth? Is
your theory specifically geocentric?

Is your frame of absolute motion at the center of the Earth or does the
Earth not rotate in your frame of absolute motion?

What is vertical to the robot explorers on Mars? What will be vertical
for HUMAN explorers on Mars?

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

We were talking about the MMXs on earth.

Not necessarily. If you want your theory to be considered 'universal',
you have to consider what happens elsewhere.

=======
You also forgot this in an earlier part of this thread:

Perhaps you think we should just launch a 300 million dollar satellite
or even a human crew without validating their trajectory computation, a
computation which your theory impacts? Perhaps YOU would volunteer to
be on that crew testing a trajectory computation with your theory for
the first time?

==========
Then I pointed out that Michelson interferometers are fairly standard
lab equipment, why don't you do the experiment yourself? MnM didn't
even have a very large setup.

http://www.meos.com/Optical%20Experi...ferometer.html

http://www.omni-optical.com/l-optics/sl435.htm

http://www.juliantrubin.com/bigten/michelsonmorley.html

Just make sure you don't get a phase shift from your vertical MMX due
to mechanical strain on the vertical arm.

There, think I've covered everything...

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"

"jem" wrote in message
...
kenseto wrote:

"jem" wrote in message
...

kenseto wrote:


"jem" wrote in message
...


kenseto wrote:



"jem" wrote in message
...



Sigh....the two things doesn't share the same relative motion wrt each
other. They are measured to have the same relative motion wrt the

observer.

So let's see.

I just indicated that your criteria was, "*any* two things share the
same relative motion if they're measured to have the same velocity
relative to a detector (the observer)."

And you want to correct that to, "*any* two things share the same
relative motion if they are measured to have the same relative motion
wrt the observer."

Is that right?


NO..

Seto, I've asked you 3 times to tell me what you mean when you say two
things share the same relative motion, and each time I've repeated your
response word-for-word and asked you to confirm it, and each time you've
changed it. Is there an end to this process?

NO....

...A and B are two things and O is the observer:
A measures B to have a relative velocity wrt him of Vab
B measures A to have a relative velocity wrt him of Vba.
Vab is not equal to Vba because the passage of a clock second in A's
frame
does not correspond to the passage of a clock second in B's frame.

O measures A to have a relative velocity of Voa.
O measures B to have a relative velocity of Vob.
Voa=Vob because they have the same value.


Do you seriously think that answers the question: "What does it mean for
two objects to share the same relative motion"?

Share the same relative motion wrt what?

Your latest response lists two different procedures. The first
procedure has two objects measuring each other's velocity (as if rocks
(e.g.) could measure velocity),

The two thing can be observers.

and the other has your enigmatic
observer measuring the object velocities (presumably relative to the
observer, whoever or whatever that may be), and coming to the more than
obvious conclusion that the velocities are equal "because they have the
same value" (duh).

So what is your problem? The two objects is measure to have the same
relative motion wrt the observer.

How are those procedures supposed to relate to the question you were
asked?

You asked a meaning question....that's why.

Ken Seto

"jem" wrote in message
...
kenseto wrote:

"jem" wrote in message
...

kenseto wrote:


"jem" wrote in message
...


kenseto wrote:


Seto, I've asked you 3 times to tell me what you mean when you say two
things share the same relative motion, and each time I've repeated your
response word-for-word and asked you to confirm it, and each time you've
changed it. Is there an end to this process?

If you meant to ask what I mean when I said two things ( two frames) share
the same ABSOLUTE MOTION then my answer is as follows:
1. clocks in the two frames will run at the same rate.
2. identical rulers in the two frame will have the same light path length.

Ken Seto

"Cygnus X-1" wrote in message
. net...
On Mon, 4 Dec 2006 09:28:58 -0500, kenseto wrote
(in article ):
"Cygnus X-1" wrote in message
. net...
On Thu, 30 Nov 2006 10:05:06 -0500, kenseto wrote
(in article ):
"Cygnus X-1" wrote in message
. net...
On Wed, 29 Nov 2006 09:50:02 -0500, kenseto wrote
(in article ):


I will answer the rest of your post at a later time.

I'll be watching, though any substantial response from me will probably
have to wait until next year.

Remember, no matter what, for physical consistency, it is required that
x' = f1(x,t) ; t' = f2(x,t)

and

x = f1'(x',t'); t = f2'(x',t')

In IRT there is no such reciprocity. The primed observer will give different
coordinates for the unprimed observer. The reason is that the passage of a
clock second in the primed frame does not correspond to the passage of a
clock second in the unprimed frame. This is illustrated when the GPS clocks
is compared to the ground clock. From the ground clock point of view the SR
effect on the GPS clock is 7 us/day running slow compared to the ground
clock. From the GPS point of view the the SR effect on the ground clock is
not 7us/day running slow compared to the GPS clock.

where f1' and f2' are the inverse transformation functions.

(x,t) - (x',t') - (x,t) must give the exact same point.

This is true if you assume reciprocity.


Choose carefully. I've examined a few of your options and each yields
mathematical contradictions and/or some physical inconsistencies.

In this simple example, using equations in your 'Unification' paper,
SetoXform1 is your transform eq 5 & 6.

i.e. (x',t') = SetoXform1(x,t,faa,fab,wavelength)

What is SetoXform1 and Seto X form2

SetoXform2 is equations 7 & 8 which should enable us to transform from
an observer in the (previously defined) primed frame BACK to the
unprimed frame. It should give us the exact same point.

I have no idea what you are talking about.
Equation 5 and 6 are used if one determines that the observed frame is in a
higher state of absolute motion than the observer's frame.....that is when
the observed clock is running slower compared to the observer's clock and
the light path length of an observed ruler is longer than the light path
length of the observer's ruler.
Equations 7 and 8 are used if one determines that the observed frame is in a
lower state of absolute motion than the observer's frame.....that is when
the observed clock is running faster compared to the observer's clock and
the light path length of an observed ruler is shorter than the light path
length of the observer's ruler.


If we use the true algebraic inverse of SetoXform1, which you don't
show, we can return to the original point, but the Interval is still
not preserved.

---
We do this using your other transform:
SetoXform2 & SetoXform2inv
Event - Event' - Event
(1.0, 2.0) - (0.0, 0.75) - (1.0, 2.0)
Interval: -3.0 -0.5625 -3.0

Again we can return to the original point, but the Interval is still
not preserved.

It appears that you misunderstood what equation 5 and 6 mean and what 7 and
8 mean.

Your claim that your theory is a superset of SR is thereby demonstrated
false by simple example.

IRT is a superset of SR.....if we ignore equations 7 and 8.

Simple Python (http://www.python.org) program for generating these
(~100 lines) can be posted on request.

And continuing from befo


http://www.fourmilab.ch/etexts/einst...html#SECTION22


Your equation is a special case of the Doppler formula where phi=pi/2
so cos(phi)=0 - when the object is moving transverse to the
line-of-sight. You essentially LOSE most of your directional
information.

The gamma factor is not directional sensitive. Similarly Fab/Faa or Faa/Fab
are not directional sentive.

Any object can be said to be moving transversly wrt the observer. Fab is
defined as the mean frequency value for that object.

But we rarely measure the 'mean frequency value'.

In the IRT transform equuations inststaneous frequency values are used.and
that's they are denoted as f_aa or f_ab instead of Faa and Fab for the mean
value..

Consider a binary star where the spectral line may change slowly with
time. We might take measurements a year apart and see fab change, but
we may not get the 'mean value' over the entire orbit for many years.
Which 'mean value' do we use? The mean of a set of five values a year
apart or do we have to wait for a full orbit (say 20 years)? Are you
saying we can't determine anything about the orbit without waiting for
the full period.

In thta case you use the equation Fab=Faa(1/gamma) or Fab=Faa(gamma)

Velocity is a *VECTOR*, generally represented by 3-components.
Frequency is a SCALAR, a single component. Your theory intrinsically
looses most directional information.

Wrong....frequency*wavelength is also a vector. So the directional
information is maintained.

The product of two scalars is NOT a vector.

x*y != (fx, fy, fz)

That is a mathematically ill-defined operation. If you are claiming
you've defined frequency or wavelength or using a tensor product, you
need to use the correct notation.

You are right but my math program does not have the correct vector notation.


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?

Probably.....a vertical MMX on Saturn would also detect fringe shift...ie
gravitational red shift.

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

No each location must define it own vertical and horizontal direction.

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

But my examples above are not all on the Earth. You didn't answer the
question.

Did you just not read the question carefully or is this an issue in
your theory that horizontal and vertical is only defined on Earth? Is
your theory specifically geocentric?

In space if the MMX gives null result then the plane of the light rays can
be called horizontal and if the MMX gives non-null result then the plane of
the light rays can be called vertical.

Is your frame of absolute motion at the center of the Earth or does the
Earth not rotate in your frame of absolute motion?

All objects in the universe are in a state of absolute motion. The effects
of absolute motion are as follows:
1. absolute motion will affect the rate of a clock. The higher is the state
of absolute motion the slower is the clock rate.
2. absolute motion will affect the light path length of a rod. The higher is
the state of absolute motion of a rod the longer is the light path length.


Ken Seto

kenseto wrote:

"jem" wrote in message
...

kenseto wrote:


"jem" wrote in message
...


kenseto wrote:



"jem" wrote in message
...



kenseto wrote:




"jem" wrote in message
...




Sigh....the two things doesn't share the same relative motion wrt each
other. They are measured to have the same relative motion wrt the

observer.


So let's see.

I just indicated that your criteria was, "*any* two things share the
same relative motion if they're measured to have the same velocity
relative to a detector (the observer)."

And you want to correct that to, "*any* two things share the same
relative motion if they are measured to have the same relative motion
wrt the observer."

Is that right?


NO..

Seto, I've asked you 3 times to tell me what you mean when you say two
things share the same relative motion, and each time I've repeated your
response word-for-word and asked you to confirm it, and each time you've
changed it. Is there an end to this process?


NO....

...A and B are two things and O is the observer:

A measures B to have a relative velocity wrt him of Vab
B measures A to have a relative velocity wrt him of Vba.
Vab is not equal to Vba because the passage of a clock second in A's

frame

does not correspond to the passage of a clock second in B's frame.

O measures A to have a relative velocity of Voa.
O measures B to have a relative velocity of Vob.
Voa=Vob because they have the same value.


Do you seriously think that answers the question: "What does it mean for
two objects to share the same relative motion"?


Share the same relative motion wrt what?

Why are you asking me? It's your terminology. It's wrt *each other* -
as in your claim:

"If the [MMX] apparatus is not in a state of relative motion wrt the
light rays you get null result."

In that sentence you've referred to the state of relative motion of one
thing (MMX apparatus) wrt another thing (light ray). So for the 5th
time, what does it *mean* for something to be (or not be) in a state of
relative motion wrt something else?

To answer that question, you need to provide the criteria by which it
can be determined whether or not two things are in the same state of
relative motion.

E.g., A is in a state of relative motion wrt B if and only if
______________.


Your latest response lists two different procedures. The first
procedure has two objects measuring each other's velocity (as if rocks
(e.g.) could measure velocity),


The two thing can be observers.

and the two thing can be rocks.

and the other has your enigmatic
observer measuring the object velocities (presumably relative to the
observer, whoever or whatever that may be), and coming to the more than
obvious conclusion that the velocities are equal "because they have the
same value" (duh).


So what is your problem?

Getting you to produce a coherent definition.


The two objects is measure to have the same
relative motion wrt the observer.

Always?

Do you perhaps mean that *if* the relative velocities of two things wrt
"the observer" are measured to be the same, then the two things share
the same state of relative motion?


How are those procedures supposed to relate to the question you were

asked?

You asked a meaning question....that's why.

I ask how, you answer why. Earth to Seto, Wake up!

Ken Seto

jem wrote:
kenseto wrote:

"jem" wrote in message
...

kenseto wrote:


"jem" wrote in message
...


kenseto wrote:



"jem" wrote in message
...




Seto, I've asked you 3 times to tell me what you mean when you say two
things share the same relative motion, and each time I've repeated your
response word-for-word and asked you to confirm it, and each time you've
changed it. Is there an end to this process?


NO....

...A and B are two things and O is the observer:

A measures B to have a relative velocity wrt him of Vab
B measures A to have a relative velocity wrt him of Vba.
Vab is not equal to Vba because the passage of a clock second in A's

frame

does not correspond to the passage of a clock second in B's frame.

O measures A to have a relative velocity of Voa.
O measures B to have a relative velocity of Vob.
Voa=Vob because they have the same value.


Do you seriously think that answers the question: "What does it mean for
two objects to share the same relative motion"?


Share the same relative motion wrt what?

Why are you asking me? It's your terminology. It's wrt *each other* -
as in your claim:

"If the [MMX] apparatus is not in a state of relative motion wrt the
light rays you get null result."

In that sentence you've referred to the state of relative motion of one
thing (MMX apparatus) wrt another thing (light ray). So for the 5th
time, what does it *mean* for something to be (or not be) in a state of
relative motion wrt something else?

Sigh....light is not a thing. If the apparatus measures light from
different directions to be isotropic in the plane of the light rays
then there is no relative motion between the apparatus and the light
rays in that plane. If the apparatus measures anisotropy in the plane
of the light rays then there is relative motion between the apparatus
and the light rays in thta plane.

Ken Seto

To answer that question, you need to provide the criteria by which it
can be determined whether or not two things are in the same state of
relative motion.

E.g., A is in a state of relative motion wrt B if and only if
______________.


Your latest response lists two different procedures. The first
procedure has two objects measuring each other's velocity (as if rocks
(e.g.) could measure velocity),


The two thing can be observers.

and the two thing can be rocks.

and the other has your enigmatic
observer measuring the object velocities (presumably relative to the
observer, whoever or whatever that may be), and coming to the more than
obvious conclusion that the velocities are equal "because they have the
same value" (duh).


So what is your problem?

Getting you to produce a coherent definition.


The two objects is measure to have the same
relative motion wrt the observer.

Always?

Do you perhaps mean that *if* the relative velocities of two things wrt
"the observer" are measured to be the same, then the two things share
the same state of relative motion?


How are those procedures supposed to relate to the question you were

asked?

You asked a meaning question....that's why.

I ask how, you answer why. Earth to Seto, Wake up!

Ken Seto