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Interpreting the MMX null result



 
 
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  #291  
Old December 3rd 06, 03:51 PM posted to sci.physics.relativity,sci.physics,sci.astro
Phineas T Puddleduck
external usenet poster
 
Posts: 1,854
Default Interpreting the MMX null result

In article ,
"kenseto" wrote:

SO??? it just mean that each location is in a state of relative motion wrt
the local light rays....that's all. So what is your problem
.
And BTW, what do you think that shared motion could mean, other than
that at each instant all those points are moving at the same speed and
in the same direction?


You need to understand what motion wrt the local light rays mean.


You need to understand what he's asking you.

--

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  #292  
Old December 4th 06, 02:54 PM posted to sci.physics.relativity,sci.physics,sci.astro
jem[_1_]
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Posts: 52
Default Interpreting the MMX null result

kenseto wrote:
"jem" wrote in message
news
kenseto wrote:


"jem" wrote in message


...

kenseto wrote:


"jem" wrote in message
...



kenseto wrote:


Speed and direction wrt anything.


????????????

Confused? No matter. The question is whether the motion is shared by
all MMX attachment points, and either it is or it isn't - what it's
expressed wrt is irrelevant.


But you SRians said: All motions are relative.


Don't concern yourself with what SR says either - I want you to address
what *you* said.



What I said: detecting anisotropy of the speed of light means that the
apparatus and light are in a state of relative motion. I interpreted that
this relative motion is due to variation of the speed of light.

If you want to find out the direction of absolute motion of a spatially
separated clock you must do the experiments in the following


link.....the

MMX is not designed to do that:\
http://www.geocities.com/kn_seto/2005Experiment.pdf


Stick to the question you were asked - no clocks were mentioned.



But that's the only way to resolve your comprehension problem. The above
experiments will definitely show the direction of motion of the spatially
separated detector wrt light.

*You* said all points on the surface of Earth share the same absolute
motion.



Yes.....all points of the same height share the same state of absolute
motion (motion wrt light).


So that must mean that all MMX devices that are attached to the
surface of Earth, also share the same absolute motion, at their points
of attachment. Right?



SO??? it just mean that each location is in a state of relative motion wrt
the local light rays....that's all. So what is your problem
.

And BTW, what do you think that shared motion could mean, other than
that at each instant all those points are moving at the same speed and
in the same direction?



You need to understand what motion wrt the local light rays mean.



anistropic in the plane of the arms.


You said MMX hasn't produced a non-null result because the absolute
motion of the MMX apparatus has been in the direction perpedicular to
"the light rays" (i.e. the plane defined by the arms of the apparatus).
Changed your mind about that?


No I didn't change my mind.
1. If the apparatus is in a state of relative motion wrt the light rays


you

get non-null result.
2. If the apparatus is not in a state of relative motion wrt the light


rays

you get null result.


How do you suppose something could not be "in a state of relative
motion" wrt a light ray? Give an example of something that's not moving
wrt a light ray.



Example: when the light rays are detected to be isotropic by the detector
then the detector and the light rays are not in a state of relative motion.


If the detector and the light are relatively stationary, just how do you
suppose the light would reach the detector to be detected?


Ken Seto


  #293  
Old December 4th 06, 03:28 PM posted to sci.physics.relativity,sci.physics,sci.astro
kenseto[_1_]
external usenet poster
 
Posts: 418
Default Interpreting the MMX null result


"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..


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

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

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.

[ 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.

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.

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..


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.

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.

Ken Seto




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.

That is it's 'definition' in SR. You've
redefined it so it's a *subset* of SR. Note the section on Doppler &
aberration in Einstein's paper:


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.

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

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.

Enjoy,

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"



  #294  
Old December 4th 06, 04:25 PM posted to sci.physics.relativity,sci.physics,sci.astro
Pmb
external usenet poster
 
Posts: 41
Default Interpreting the MMX null result

People often are unaware that the MMX experiment does not really prove that
the speed of light is invariant. It was later realized that since the
experiment was done in open air that the light will travel at specific speed
which is in respect to the medium in which the light travels, i.e. air. It
wasn't until the Kennedy-Thorndike experiment was done that it could be said
that an MMX type experiment verified the constancy of light. This experiment
was done with the experimnental apparatus contained in a vessel in which
there was a hard vacuum. Few relativity texts mention this. The only one
I've seen that addresses this is the book "Special Relativity," by A.P.
French.

Best wishes

Pete


  #295  
Old December 4th 06, 10:23 PM posted to sci.physics.relativity,sci.physics,sci.astro
kenseto[_1_]
external usenet poster
 
Posts: 418
Default Interpreting the MMX null result


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


"jem" wrote in message


...

kenseto wrote:


"jem" wrote in message
...



kenseto wrote:


Speed and direction wrt anything.


????????????

Confused? No matter. The question is whether the motion is shared by
all MMX attachment points, and either it is or it isn't - what it's
expressed wrt is irrelevant.


But you SRians said: All motions are relative.

Don't concern yourself with what SR says either - I want you to address
what *you* said.



What I said: detecting anisotropy of the speed of light means that the
apparatus and light are in a state of relative motion. I interpreted

that
this relative motion is due to variation of the speed of light.

If you want to find out the direction of absolute motion of a spatially
separated clock you must do the experiments in the following


link.....the

MMX is not designed to do that:\
http://www.geocities.com/kn_seto/2005Experiment.pdf

Stick to the question you were asked - no clocks were mentioned.



But that's the only way to resolve your comprehension problem. The above
experiments will definitely show the direction of motion of the

spatially
separated detector wrt light.

*You* said all points on the surface of Earth share the same absolute
motion.



Yes.....all points of the same height share the same state of absolute
motion (motion wrt light).


So that must mean that all MMX devices that are attached to the
surface of Earth, also share the same absolute motion, at their points
of attachment. Right?



SO??? it just mean that each location is in a state of relative motion

wrt
the local light rays....that's all. So what is your problem
.

And BTW, what do you think that shared motion could mean, other than
that at each instant all those points are moving at the same speed and
in the same direction?



You need to understand what motion wrt the local light rays mean.



anistropic in the plane of the arms.


You said MMX hasn't produced a non-null result because the absolute
motion of the MMX apparatus has been in the direction perpedicular to
"the light rays" (i.e. the plane defined by the arms of the

apparatus).
Changed your mind about that?


No I didn't change my mind.
1. If the apparatus is in a state of relative motion wrt the light rays


you

get non-null result.
2. If the apparatus is not in a state of relative motion wrt the light


rays

you get null result.

How do you suppose something could not be "in a state of relative
motion" wrt a light ray? Give an example of something that's not moving
wrt a light ray.



Example: when the light rays are detected to be isotropic by the

detector
then the detector and the light rays are not in a state of relative

motion.

If the detector and the light are relatively stationary, just how do you
suppose the light would reach the detector to be detected?


Notice I said that "when the light rays are detected to be isotropic".....
that means that the light arys are arriving to the detector from equal
distances at the same time.


  #296  
Old December 5th 06, 02:54 PM posted to sci.physics.relativity,sci.physics,sci.astro
jem[_1_]
external usenet poster
 
Posts: 52
Default Interpreting the MMX null result

Pmb wrote:

People often are unaware that the MMX experiment does not really prove that
the speed of light is invariant. It was later realized that since the
experiment was done in open air that the light will travel at specific speed
which is in respect to the medium in which the light travels, i.e. air. It
wasn't until the Kennedy-Thorndike experiment was done that it could be said
that an MMX type experiment verified the constancy of light. This experiment
was done with the experimnental apparatus contained in a vessel in which
there was a hard vacuum. Few relativity texts mention this. The only one
I've seen that addresses this is the book "Special Relativity," by A.P.
French.

Best wishes

Pete



People often are unaware that no experiment (or collection of
experiments) has ever /proved/ anything.
  #297  
Old December 5th 06, 02:59 PM posted to sci.physics.relativity,sci.physics,sci.astro
jem[_1_]
external usenet poster
 
Posts: 52
Default Interpreting the MMX null result

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

kenseto wrote:

"jem" wrote in message
news

kenseto wrote:



"jem" wrote in message

...


kenseto wrote:



"jem" wrote in message
...




kenseto wrote:


Speed and direction wrt anything.


????????????

Confused? No matter. The question is whether the motion is shared by
all MMX attachment points, and either it is or it isn't - what it's
expressed wrt is irrelevant.


But you SRians said: All motions are relative.

Don't concern yourself with what SR says either - I want you to address
what *you* said.


What I said: detecting anisotropy of the speed of light means that the
apparatus and light are in a state of relative motion. I interpreted


that

this relative motion is due to variation of the speed of light.


If you want to find out the direction of absolute motion of a spatially
separated clock you must do the experiments in the following

link.....the


MMX is not designed to do that:\
http://www.geocities.com/kn_seto/2005Experiment.pdf

Stick to the question you were asked - no clocks were mentioned.


But that's the only way to resolve your comprehension problem. The above
experiments will definitely show the direction of motion of the


spatially

separated detector wrt light.


*You* said all points on the surface of Earth share the same absolute
motion.


Yes.....all points of the same height share the same state of absolute
motion (motion wrt light).



So that must mean that all MMX devices that are attached to the
surface of Earth, also share the same absolute motion, at their points
of attachment. Right?


SO??? it just mean that each location is in a state of relative motion


wrt

the local light rays....that's all. So what is your problem
.


And BTW, what do you think that shared motion could mean, other than
that at each instant all those points are moving at the same speed and
in the same direction?


You need to understand what motion wrt the local light rays mean.




anistropic in the plane of the arms.


You said MMX hasn't produced a non-null result because the absolute
motion of the MMX apparatus has been in the direction perpedicular to
"the light rays" (i.e. the plane defined by the arms of the


apparatus).

Changed your mind about that?


No I didn't change my mind.
1. If the apparatus is in a state of relative motion wrt the light rays

you


get non-null result.
2. If the apparatus is not in a state of relative motion wrt the light

rays


you get null result.

How do you suppose something could not be "in a state of relative
motion" wrt a light ray? Give an example of something that's not moving
wrt a light ray.


Example: when the light rays are detected to be isotropic by the


detector

then the detector and the light rays are not in a state of relative


motion.

If the detector and the light are relatively stationary, just how do you
suppose the light would reach the detector to be detected?



Notice I said that "when the light rays are detected to be isotropic".....
that means that the light arys are arriving to the detector from equal
distances at the same time.



How can they be "arriving" at the detector when they're not moving wrt
the detector?
  #298  
Old December 5th 06, 03:11 PM posted to sci.physics.relativity,sci.physics,sci.astro
Pmb
external usenet poster
 
Posts: 41
Default Interpreting the MMX null result


"jem" wrote in message ...
Pmb wrote:

People often are unaware that the MMX experiment does not really prove
that the speed of light is invariant. It was later realized that since
the experiment was done in open air that the light will travel at
specific speed which is in respect to the medium in which the light
travels, i.e. air. It wasn't until the Kennedy-Thorndike experiment was
done that it could be said that an MMX type experiment verified the
constancy of light. This experiment was done with the experimnental
apparatus contained in a vessel in which there was a hard vacuum. Few
relativity texts mention this. The only one I've seen that addresses this
is the book "Special Relativity," by A.P. French.

Best wishes

Pete


People often are unaware that no experiment (or collection of experiments)
has ever /proved/ anything.


That is quite true.

Best wishes

Pete


  #299  
Old December 5th 06, 03:40 PM posted to sci.physics.relativity,sci.physics,sci.astro
kenseto[_1_]
external usenet poster
 
Posts: 418
Default Interpreting the MMX null result


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

kenseto wrote:

"jem" wrote in message
news

kenseto wrote:



"jem" wrote in message

...


kenseto wrote:



"jem" wrote in message
...




kenseto wrote:


Speed and direction wrt anything.


????????????

Confused? No matter. The question is whether the motion is shared by
all MMX attachment points, and either it is or it isn't - what it's
expressed wrt is irrelevant.


But you SRians said: All motions are relative.

Don't concern yourself with what SR says either - I want you to address
what *you* said.


What I said: detecting anisotropy of the speed of light means that the
apparatus and light are in a state of relative motion. I interpreted


that

this relative motion is due to variation of the speed of light.


If you want to find out the direction of absolute motion of a spatially
separated clock you must do the experiments in the following

link.....the


MMX is not designed to do that:\
http://www.geocities.com/kn_seto/2005Experiment.pdf

Stick to the question you were asked - no clocks were mentioned.


But that's the only way to resolve your comprehension problem. The above
experiments will definitely show the direction of motion of the


spatially

separated detector wrt light.


*You* said all points on the surface of Earth share the same absolute
motion.


Yes.....all points of the same height share the same state of absolute
motion (motion wrt light).



So that must mean that all MMX devices that are attached to the
surface of Earth, also share the same absolute motion, at their points
of attachment. Right?


SO??? it just mean that each location is in a state of relative motion


wrt

the local light rays....that's all. So what is your problem
.


And BTW, what do you think that shared motion could mean, other than
that at each instant all those points are moving at the same speed and
in the same direction?


You need to understand what motion wrt the local light rays mean.




anistropic in the plane of the arms.


You said MMX hasn't produced a non-null result because the absolute
motion of the MMX apparatus has been in the direction perpedicular to
"the light rays" (i.e. the plane defined by the arms of the


apparatus).

Changed your mind about that?


No I didn't change my mind.
1. If the apparatus is in a state of relative motion wrt the light rays

you


get non-null result.
2. If the apparatus is not in a state of relative motion wrt the light

rays


you get null result.

How do you suppose something could not be "in a state of relative
motion" wrt a light ray? Give an example of something that's not moving
wrt a light ray.


Example: when the light rays are detected to be isotropic by the


detector

then the detector and the light rays are not in a state of relative


motion.

If the detector and the light are relatively stationary, just how do you
suppose the light would reach the detector to be detected?



Notice I said that "when the light rays are detected to be isotropic".....
that means that the light arys are arriving to the detector from equal
distances at the same time.



How can they be "arriving" at the detector when they're not moving wrt
the detector?


Sigh....isotropy means that the light rays are moving toward the
detector at the same speed in all directions.

  #300  
Old December 6th 06, 03:08 PM posted to sci.physics.relativity,sci.physics,sci.astro
jem[_1_]
external usenet poster
 
Posts: 52
Default Interpreting the MMX null result

kenseto wrote:

jem wrote:

kenseto wrote:

"jem" wrote in message
...


kenseto wrote:


"jem" wrote in message
news


kenseto wrote:




"jem" wrote in message

...



kenseto wrote:




"jem" wrote in message
.. .





kenseto wrote:


Speed and direction wrt anything.


????????????

Confused? No matter. The question is whether the motion is shared by
all MMX attachment points, and either it is or it isn't - what it's
expressed wrt is irrelevant.


But you SRians said: All motions are relative.

Don't concern yourself with what SR says either - I want you to address
what *you* said.


What I said: detecting anisotropy of the speed of light means that the
apparatus and light are in a state of relative motion. I interpreted

that


this relative motion is due to variation of the speed of light.



If you want to find out the direction of absolute motion of a spatially
separated clock you must do the experiments in the following

link.....the



MMX is not designed to do that:\
http://www.geocities.com/kn_seto/2005Experiment.pdf

Stick to the question you were asked - no clocks were mentioned.


But that's the only way to resolve your comprehension problem. The above
experiments will definitely show the direction of motion of the

spatially


separated detector wrt light.



*You* said all points on the surface of Earth share the same absolute
motion.


Yes.....all points of the same height share the same state of absolute
motion (motion wrt light).




So that must mean that all MMX devices that are attached to the
surface of Earth, also share the same absolute motion, at their points
of attachment. Right?


SO??? it just mean that each location is in a state of relative motion

wrt


the local light rays....that's all. So what is your problem
.



And BTW, what do you think that shared motion could mean, other than
that at each instant all those points are moving at the same speed and
in the same direction?


You need to understand what motion wrt the local light rays mean.





anistropic in the plane of the arms.


You said MMX hasn't produced a non-null result because the absolute
motion of the MMX apparatus has been in the direction perpedicular to
"the light rays" (i.e. the plane defined by the arms of the

apparatus).


Changed your mind about that?


No I didn't change my mind.
1. If the apparatus is in a state of relative motion wrt the light rays

you



get non-null result.
2. If the apparatus is not in a state of relative motion wrt the light

rays



you get null result.

How do you suppose something could not be "in a state of relative
motion" wrt a light ray? Give an example of something that's not moving
wrt a light ray.


Example: when the light rays are detected to be isotropic by the

detector


then the detector and the light rays are not in a state of relative

motion.


If the detector and the light are relatively stationary, just how do you
suppose the light would reach the detector to be detected?


Notice I said that "when the light rays are detected to be isotropic".....
that means that the light arys are arriving to the detector from equal
distances at the same time.



How can they be "arriving" at the detector when they're not moving wrt
the detector?



Sigh....isotropy means that the light rays are moving toward the
detector at the same speed in all directions.


And how is it that "the light rays are moving toward the detector", when
"the detector and the light rays are not in a state of relative motion"?



 




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