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Interpreting the MMX null result



 
 
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  #281  
Old December 1st 06, 08:14 AM posted to sci.physics.relativity,sci.physics,sci.astro
Sorcerer[_4_]
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Posts: 326
Default Interpreting the MMX null result


"Tom Roberts" wrote in message t...
| sean wrote:
| I ve done a few correct simulations of sagnac which show indisputably
| that light travelling at c relative to the source will still give a
| path difference and fringe shift when the setup rotates.
|
| Then you make other assumptions, probably about how light reflects off
| moving mirrors, that differ from the usual ones. The usual assumption is
| that Snell's law holds in the rest frame of the mirror, and such
| ballistic theories are refuted by the Sagnac measurements.
|
|
| It was an
| incorrect conclusion to say that only relativity can explain the sagnac
| results.
|
| Nobody ever made such a "conclusion".
|
| In science we test theories, and the Sagnac experiment is consistent
| with the predictions of SR. Yes, it is also consistent with the
| predictions of other theories. But other than GR and theories equivalent
| to SR, no other theory remains unrefuted when one looks at ALL of the
| experiments.

Nonsense. Just ONE example is enough to refute ANY theory, ****HEAD!



|
|
| In fact sagnac and gps can be explained by classical wave
| theory as well or better than relativity,..
|
| Not at all. For instance, "classical wave theory" has no way to predict
| the variation in clock rates with altitude that are observed in the GPS.
|
|
| Tom Roberts
  #282  
Old December 1st 06, 02:47 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
...

kenseto wrote:

"jem" wrote in message


...

kenseto wrote:


"jem" wrote in message
...



kenseto wrote:




"jem" wrote in message


...



kenseto wrote:





"jem" wrote in message

locations?


All locations on earth share the same absolute motion.

So then it must be the case that *all* MMX devices which are attached to
the surface of the Earth, share the same absolute motion (i.e. speed and
direction) at their points of attachment. Right?


But the MMX is designed only to test the isotropy of the speed of light.


and this causes the isotropy and anistropy of the speed of



light detected by the horizontal and vertical MMXs.


Anistropy of the speed of light is caused by the source and the


detector

have different states of absolute motion.


And the reason that no previous MMX has detected anisotropy is because
the source and detector have always shared the same absolute motion.

Right?

Sure.....but vertical MMX will detect anisotropy because the two light


rays

are emanated from the reflecting mirrors at different heights (thus
different state of absolute motion)


And the absolute motion shared by the source and detector has always
been perpendicular to the plane defined by the arms of the MMX devices.
Right?


Again the MMX is not designed to detect absolute motion of the earth. It


is

designed to detect the isotropy or anistropy of the speed of light.



Don't worry about what the MMX is designed to test - just answer the
one-word questions.

So then it must be the case that *all* MMX devices which are attached to
the surface of the Earth, share the same absolute motion (i.e. speed and
direction) at their points of attachment. Right?



Speed and direction of absolute motion wrt what?


Speed and direction wrt anything.



And the absolute motion shared by the source and detector has always
been perpendicular to the plane defined by the arms of the MMX devices.
Right?



NO.....if the MMXZ detected isotropy then it is perpendicular to the plane
frined by the arms of the MMX.


and just how do you think that differs from what I said?
  #283  
Old December 1st 06, 06:25 PM posted to sci.physics.relativity,sci.physics,sci.astro
Phineas T Puddleduck
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Posts: 1,854
Default Interpreting the MMX null result

In article ,
Art Deco wrote:

Phineas T Puddleduck wrote:

In article om,
"visual word" wrote:

Another puzzling question is: how can the absolute motion of each MMX
apparatus be different than the absolute motion of the ground to which

motion is relative as light is to the beholder and so why not in
reality


Because what you're saying is a crap metaphor and has no analogy to
reality


But he/she/it is stuck in a posting loop, so it must have deep meaning.


I never thought of that ;-)

--

Just \int_0^\infty du it!

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

  #284  
Old December 1st 06, 07:16 PM posted to sci.physics.relativity,sci.physics,sci.astro
kenseto[_1_]
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Posts: 418
Default Interpreting the MMX null result


"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


...



Again the MMX is not designed to detect absolute motion of the earth.

It

is

designed to detect the isotropy or anistropy of the speed of light.



Don't worry about what the MMX is designed to test - just answer the
one-word questions.

So then it must be the case that *all* MMX devices which are attached to
the surface of the Earth, share the same absolute motion (i.e. speed and
direction) at their points of attachment. Right?



Speed and direction of absolute motion wrt what?


Speed and direction wrt anything.


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



And the absolute motion shared by the source and detector has always
been perpendicular to the plane defined by the arms of the MMX devices.
Right?



NO.....if the MMXZ detected isotropy then it is perpendicular to the

plane
frined by the arms of the MMX.


and just how do you think that differs from what I said?


Her's what I meant to say:
1. if the MMX gives null result that means that the speed of light is
isotropic in the plane of the arms.
2. If the MMX gives non-null rsult, that means that the speed of light is
anistropic in the plane of the arms.

This is different than what you were trying to say: That the MMX apparatus
is moving in a specfic direction wrt anything.


  #285  
Old December 1st 06, 08:07 PM posted to sci.physics.relativity,sci.physics,sci.astro
Phineas T Puddleduck
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Posts: 1,854
Default Interpreting the MMX null result

In article ,
"kenseto" wrote:

Her's what I meant to say:
1. if the MMX gives null result that means that the speed of light is
isotropic in the plane of the arms.
2. If the MMX gives non-null rsult, that means that the speed of light is
anistropic in the plane of the arms.

This is different than what you were trying to say: That the MMX apparatus
is moving in a specfic direction wrt anything.


But yet you claim the anisotropy is due to movement.

--

Just \int_0^\infty du it!

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

  #286  
Old December 2nd 06, 03:20 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
...

kenseto wrote:


"jem" wrote in message
...


kenseto wrote:


"jem" wrote in message

...


kenseto wrote:



"jem" wrote in message
...




kenseto wrote:





"jem" wrote in message

...


Again the MMX is not designed to detect absolute motion of the earth.


It

is


designed to detect the isotropy or anistropy of the speed of light.



Don't worry about what the MMX is designed to test - just answer the
one-word questions.

So then it must be the case that *all* MMX devices which are attached to
the surface of the Earth, share the same absolute motion (i.e. speed and
direction) at their points of attachment. Right?


Speed and direction of absolute motion wrt what?


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.



And the absolute motion shared by the source and detector has always
been perpendicular to the plane defined by the arms of the MMX devices.
Right?


NO.....if the MMXZ detected isotropy then it is perpendicular to the


plane

frined by the arms of the MMX.


and just how do you think that differs from what I said?



Her's what I meant to say:
1. if the MMX gives null result that means that the speed of light is
isotropic in the plane of the arms.
2. If the MMX gives non-null rsult, that means that the speed of light is
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?

This is different than what you were trying to say: That the MMX apparatus
is moving in a specfic direction wrt anything.


FYI - at any particular time, the MMX apparatus is moving in a specific
direction wrt *everything*.
  #287  
Old December 2nd 06, 05:51 PM posted to sci.physics.relativity,sci.physics,sci.astro
kenseto[_1_]
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Posts: 418
Default Interpreting the MMX null result


"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
...




kenseto wrote:





"jem" wrote in message

...


Again the MMX is not designed to detect absolute motion of the earth.


It

is


designed to detect the isotropy or anistropy of the speed of light.



Don't worry about what the MMX is designed to test - just answer the
one-word questions.

So then it must be the case that *all* MMX devices which are attached

to
the surface of the Earth, share the same absolute motion (i.e. speed

and
direction) at their points of attachment. Right?


Speed and direction of absolute motion wrt what?

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.
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




And the absolute motion shared by the source and detector has always
been perpendicular to the plane defined by the arms of the MMX

devices.
Right?


NO.....if the MMXZ detected isotropy then it is perpendicular to the


plane

frined by the arms of the MMX.


and just how do you think that differs from what I said?



Her's what I meant to say:
1. if the MMX gives null result that means that the speed of light is
isotropic in the plane of the arms.
2. If the MMX gives non-null rsult, that means that the speed of light

is
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.

This is different than what you were trying to say: That the MMX

apparatus
is moving in a specfic direction wrt anything.


FYI - at any particular time, the MMX apparatus is moving in a specific
direction wrt *everything*.



  #288  
Old December 2nd 06, 07:48 PM posted to sci.physics.relativity,sci.physics,sci.astro
Cygnus X-1
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Posts: 27
Default Interpreting the MMX null result

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 ):


"Cygnus X-1" wrote in message
. net...
On Mon, 27 Nov 2006 10:18:27 -0500, kenseto wrote
(in article ):


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


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

[ stuff deleted ]

Some researchers with established track records of discoveries get some
preferential treatment, but they are still subjected to review and
experimental validation (or at least theoretical consistency).

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.

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

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 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. :^)

[ stuff deleted ]

Per your earlier post, you claim your 'new idea' is a simple
substitution, replacing quantities like velocity, etc. with
wavelengths and frequencies:

c=Faa*Lambda
relative veocity v=lambda(Faa-Fab)
gamma=Faa/Fab
1/gamma=Fab/Faa

Ken, what are the frequencies, Faa and Fab, of the two STEREO
spacecraft (http://stereo.gsfc.nasa.gov/where.shtml) currently moving
in orbit between the Earth and the Moon, on its way to a heliocentric
orbit?

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?

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?

2) The solar system was largely mapped with just studies of angular
motion without resort to spectroscopy.

3) An unstable subatomic particle has no 'standard light source'.

and others that I currently don't have time to write...

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.

How do I compute the fab I should expect at each time interval before
there's a vehicle in space to 'steer' to the right fab value?

Your plan requires a considerable amount of extra fuel. You seem to
think we fly spacecraft like "Spaceman Spiff"
(http://en.wikipedia.org/wiki/Spaceman_Spiff) or perhaps you took the
course correction scene in "Apollo 13" too literally.

Usually we just launch them out of Earth orbit in the right general
direction, with a little extra onboard for (minor) course corrections.
Current theory produces these trajectories (subject to considerable
testing over the past ~50 years of spaceflight) without any additional
data.

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.

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


How do you get the spacecraft velocity without using the SR assumpition that
the speed of light is c? Doesn't the radar work as follows:
velocity=lambda(f_o - f')?????


Radar uses the time delay of the outbound and returning signal.
Changes in that timing can detect radial velocity even if Doppler shift
of the return signal is not measured.

But they generally don't use radar for tracking interplanetary missions
(return signal too weak that far from Earth). The closest to 'standard
light source' is the carrier frequency of the telemetry but this alone
is insufficient for most missions since it can only uniquely determine
radial information. Much spacecraft position information is derived
combining this with angular distances between reference objects using
star-trackers and the position of the Sun, Earth, etc. The rest is Sir
Isaac Newton.

However, you do these odd redefinitions of gamma (which is velocity
dependent in SR) and relative velocity which will alter those
calculations. So the answers is no, it will not reproduce the exact
same results as SR in those cases. And YOU have to demonstrate that it
still works in those cases.

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.

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

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"

  #289  
Old December 3rd 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

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
.. .





kenseto wrote:






"jem" wrote in message

...



Again the MMX is not designed to detect absolute motion of the earth.

It


is



designed to detect the isotropy or anistropy of the speed of light.



Don't worry about what the MMX is designed to test - just answer the
one-word questions.

So then it must be the case that *all* MMX devices which are attached


to

the surface of the Earth, share the same absolute motion (i.e. speed


and

direction) at their points of attachment. Right?


Speed and direction of absolute motion wrt what?

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.

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.

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

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?



And the absolute motion shared by the source and detector has always
been perpendicular to the plane defined by the arms of the MMX


devices.

Right?


NO.....if the MMXZ detected isotropy then it is perpendicular to the

plane


frined by the arms of the MMX.


and just how do you think that differs from what I said?


Her's what I meant to say:
1. if the MMX gives null result that means that the speed of light is
isotropic in the plane of the arms.
2. If the MMX gives non-null rsult, that means that the speed of light


is

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.


This is different than what you were trying to say: That the MMX


apparatus

is moving in a specfic direction wrt anything.


FYI - at any particular time, the MMX apparatus is moving in a specific
direction wrt *everything*.




  #290  
Old December 3rd 06, 03:50 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
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.

Ken Seto


 




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