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The Correct Interpretation of the MMX and KTX Null Results.



 
 
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  #1  
Old September 20th 03 posted to sci.physics.relativity
kenseto
external usenet poster
 
Posts: 10,004
Default The Correct Interpretation of the MMX and KTX Null Results.

The Correct Interpretation of the MMX and KTX Null Results.

The current interpretations of the MMX and KTX null results are as follows:
1. The LET interpretation:
The arm that is parallel to the direction of motion is contracted by an
appropriate amount and thus cancels out the effect of motion of the earth
through the aether.

2. The SR interpretation:
There is no aether and the speed of light is isotropic.

Both of these interpretations are wrong. The correct interpretation is as
follows:
1. There exits a stationary aether.
2. The direction of absolute motion of the earth in this stationary aether
is in the vertical (UP or DOWN) direction. This sentence appears to be self
contradictory. It is not. If one side of the earth is defined as vertical up
then the other side of the earth is defined vertical down. However, these
two directions of motion (vertical up and verticle down) are in the same
direction in the stationary aether.
3. This means that the light path length from the mirrors at the ends of the
horizontal arms to the re-combining mirror will remain the same for all the
orientations of the horizontal arms. That's why the null result for both the
MMX and KTX.

There is a proposed experiment to test the validity of this interpretaion
and it is outlined in the following link:
http://www.journaloftheoretics.com/L...apers/Seto.pdf

Ken Seto















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  #2  
Old September 20th 03 posted to sci.physics.relativity
Dirk Van de moortel
external usenet poster
 
Posts: 15,355
Default The Correct Interpretation of the MMX and KTX Null Results.


"kenseto" wrote in message ...
The Correct Interpretation of the MMX and KTX Null Results.

The current interpretations of the MMX and KTX null results are as follows:
1. The LET interpretation:
The arm that is parallel to the direction of motion is contracted by an
appropriate amount and thus cancels out the effect of motion of the earth
through the aether.

2. The SR interpretation:
There is no aether and the speed of light is isotropic.

Both of these interpretations are wrong.


But your theory has LET and SR as subsets, so your theory
is wrong:
| DRT includes SR/LET as subsets and its equations
| are valid in all environments--including gravity.
http://groups.google.com/groups?&q=d...+author%3Aseto
How sad.

Dirk Vdm


  #3  
Old September 20th 03 posted to sci.physics.relativity
AndroclesInEngland
external usenet poster
 
Posts: 122
Default The Correct Interpretation of the MMX and KTX Null Results.


"kenseto" wrote in message
...
The Correct Interpretation of the MMX and KTX Null Results.

The current interpretations of the MMX and KTX null results are as

follows:
1. The LET interpretation:
The arm that is parallel to the direction of motion is contracted by an
appropriate amount and thus cancels out the effect of motion of the earth
through the aether.

2. The SR interpretation:
There is no aether and the speed of light is isotropic.

Both of these interpretations are wrong. The correct interpretation is as
follows:
1. There exits a stationary aether.
2. The direction of absolute motion of the earth in this stationary aether
is in the vertical (UP or DOWN) direction. This sentence appears to be

self
contradictory. It is not. If one side of the earth is defined as vertical

up
then the other side of the earth is defined vertical down. However, these
two directions of motion (vertical up and verticle down) are in the same
direction in the stationary aether.
3. This means that the light path length from the mirrors at the ends of

the
horizontal arms to the re-combining mirror will remain the same for all

the
orientations of the horizontal arms. That's why the null result for both

the
MMX and KTX.

There is a proposed experiment to test the validity of this interpretaion
and it is outlined in the following link:
http://www.journaloftheoretics.com/L...apers/Seto.pdf

Ken Seto

You've forgotten the truly correct one, Ken.

The null result of MMX is entirely due to the velocity of light bullets
(photons) being constant with respect to the source. No magical length
contractions or anything like that. Just like two marbles that are released
together, follow different paths along guiding chutes and arrive together,
and it just doesn't matter which way the apparatus is turned, or if it takes
place on an airplane, or anything like that. They'll go on arriving together
time and time again if they are released together. Of course, if you blow
some wind over them, they won't. One will be slowed more than the other. So
no aether wind for light, either. Hence no aether. Simple, isn't it?
Androcles


  #4  
Old September 20th 03 posted to sci.physics.relativity
kenseto
external usenet poster
 
Posts: 10,004
Default The Correct Interpretation of the MMX and KTX Null Results.


"AndroclesInEngland" wrote in message
...

"kenseto" wrote in message
...
The Correct Interpretation of the MMX and KTX Null Results.

The current interpretations of the MMX and KTX null results are as

follows:
1. The LET interpretation:
The arm that is parallel to the direction of motion is contracted by an
appropriate amount and thus cancels out the effect of motion of the

earth
through the aether.

2. The SR interpretation:
There is no aether and the speed of light is isotropic.

Both of these interpretations are wrong. The correct interpretation is

as
follows:
1. There exits a stationary aether.
2. The direction of absolute motion of the earth in this stationary

aether
is in the vertical (UP or DOWN) direction. This sentence appears to be

self
contradictory. It is not. If one side of the earth is defined as

vertical
up
then the other side of the earth is defined vertical down. However,

these
two directions of motion (vertical up and verticle down) are in the same
direction in the stationary aether.
3. This means that the light path length from the mirrors at the ends of

the
horizontal arms to the re-combining mirror will remain the same for all

the
orientations of the horizontal arms. That's why the null result for both

the
MMX and KTX.

There is a proposed experiment to test the validity of this

interpretaion
and it is outlined in the following link:
http://www.journaloftheoretics.com/L...apers/Seto.pdf

Ken Seto

You've forgotten the truly correct one, Ken.

The null result of MMX is entirely due to the velocity of light bullets
(photons) being constant with respect to the source. No magical length
contractions or anything like that. Just like two marbles that are

released
together, follow different paths along guiding chutes and arrive together,
and it just doesn't matter which way the apparatus is turned, or if it

takes
place on an airplane, or anything like that. They'll go on arriving

together
time and time again if they are released together. Of course, if you blow
some wind over them, they won't. One will be slowed more than the other.

So
no aether wind for light, either. Hence no aether. Simple, isn't it?
Androcles


Wrong....your magic light bullets should have a speed of c+v or c-v
relative to the observer but they are not. So it is not that simple after
all.:-)

Ken Seto






  #5  
Old September 21st 03 posted to sci.physics.relativity
xxein
external usenet poster
 
Posts: 609
Default The Correct Interpretation of the MMX and KTX Null Results.

"Dirk Van de moortel" wrote in message ...
"kenseto" wrote in message ...
The Correct Interpretation of the MMX and KTX Null Results.

The current interpretations of the MMX and KTX null results are as follows:
1. The LET interpretation:
The arm that is parallel to the direction of motion is contracted by an
appropriate amount and thus cancels out the effect of motion of the earth
through the aether.

2. The SR interpretation:
There is no aether and the speed of light is isotropic.

Both of these interpretations are wrong.


But your theory has LET and SR as subsets, so your theory
is wrong:
| DRT includes SR/LET as subsets and its equations
| are valid in all environments--including gravity.
http://groups.google.com/groups?&q=d...+author%3Aseto
How sad.

Dirk Vdm


xxein: I see nothing wrong with sr-let being subsets of a theory, but
Ken can't rationally claim that for his theory.
  #6  
Old September 21st 03 posted to sci.physics.relativity
AndroclesInEngland
external usenet poster
 
Posts: 122
Default The Correct Interpretation of the MMX and KTX Null Results.


"kenseto" wrote in message
...

"AndroclesInEngland" wrote in message
...

"kenseto" wrote in message
...
The Correct Interpretation of the MMX and KTX Null Results.

The current interpretations of the MMX and KTX null results are as

follows:
1. The LET interpretation:
The arm that is parallel to the direction of motion is contracted by

an
appropriate amount and thus cancels out the effect of motion of the

earth
through the aether.

2. The SR interpretation:
There is no aether and the speed of light is isotropic.

Both of these interpretations are wrong. The correct interpretation is

as
follows:
1. There exits a stationary aether.
2. The direction of absolute motion of the earth in this stationary

aether
is in the vertical (UP or DOWN) direction. This sentence appears to be

self
contradictory. It is not. If one side of the earth is defined as

vertical
up
then the other side of the earth is defined vertical down. However,

these
two directions of motion (vertical up and verticle down) are in the

same
direction in the stationary aether.
3. This means that the light path length from the mirrors at the ends

of
the
horizontal arms to the re-combining mirror will remain the same for

all
the
orientations of the horizontal arms. That's why the null result for

both
the
MMX and KTX.

There is a proposed experiment to test the validity of this

interpretaion
and it is outlined in the following link:
http://www.journaloftheoretics.com/L...apers/Seto.pdf

Ken Seto

You've forgotten the truly correct one, Ken.

The null result of MMX is entirely due to the velocity of light bullets
(photons) being constant with respect to the source. No magical length
contractions or anything like that. Just like two marbles that are

released
together, follow different paths along guiding chutes and arrive

together,
and it just doesn't matter which way the apparatus is turned, or if it

takes
place on an airplane, or anything like that. They'll go on arriving

together
time and time again if they are released together. Of course, if you

blow
some wind over them, they won't. One will be slowed more than the other.

So
no aether wind for light, either. Hence no aether. Simple, isn't it?
Androcles


Wrong....your magic light bullets should have a speed of c+v or c-v
relative to the observer but they are not. So it is not that simple after
all.:-)

Why? Is the observer moving relative to the source? No... so v = 0
Is the observer moving relative to the detector? No... so v = 0
Is the observer moving relative to the medium (air)? No.. so v = 0.

Since v = 0 in this case, c+0 = c. So they DO have a speed of c+v after
all. It's just that in MMX, v happens to be zero. I thinks that's pretty
simple, don't you?
The rolling marbles have a velocity of u, and if you move along one of the
chutes with one it will have a velocity of u-v, or zero, with respect to
you, and you'll stay alongside it. Stand still and it's velocity with
respect to you will be once again be 'u'. There's nothing magic about it.
All matter is composed of molecules or individual atoms with electrons, all
visible light is emitted by individual electrons falling to a lower energy
state, hence all visible light is comprised of individual photons and they
don't need any aether to transport them. They are self supporting
electomagnetic phenomena, the magnetic field collapsing and producing an
electric field, which in turn collapses to produce a magnetic field, ad
infinitum. they ar packets of constant energy that cannot be created or
destroyed, only converted to or from some other form, which can be mass.
Is the observer moving relative to the medium (aether)? Yes, according to
you.
But you can't detect it, can you? Magical things happen, like the apparatus
shrinking along its length. At least, that's what Lorentz said (Dover, pp
3 - 7).
Androcles


  #7  
Old September 21st 03 posted to sci.physics.relativity
Dirk Van de moortel
external usenet poster
 
Posts: 15,355
Default The Correct Interpretation of the MMX and KTX Null Results.


"xxein" wrote in message om...
"Dirk Van de moortel" wrote in message

...
"kenseto" wrote in message ...
The Correct Interpretation of the MMX and KTX Null Results.

The current interpretations of the MMX and KTX null results are as follows:
1. The LET interpretation:
The arm that is parallel to the direction of motion is contracted by an
appropriate amount and thus cancels out the effect of motion of the earth
through the aether.

2. The SR interpretation:
There is no aether and the speed of light is isotropic.

Both of these interpretations are wrong.


But your theory has LET and SR as subsets, so your theory
is wrong:
| DRT includes SR/LET as subsets and its equations
| are valid in all environments--including gravity.
http://groups.google.com/groups?&q=d...+author%3Aseto
How sad.

Dirk Vdm


xxein: I see nothing wrong with sr-let being subsets of a theory,


I can imagine *you* see nothing wrong with wrong theories
being subsets of a theory.

but
Ken can't rationally claim that for his theory.


Ken?
Rationally?
Theory?

Dirk Vdm


  #8  
Old September 21st 03 posted to sci.physics.relativity
Dirk Van de moortel
external usenet poster
 
Posts: 15,355
Default The Correct Interpretation of the MMX and KTX Null Results.


"AndroclesInEngland" wrote in message ...

"kenseto" wrote in message
...

"AndroclesInEngland" wrote in message
...

"kenseto" wrote in message
...
The Correct Interpretation of the MMX and KTX Null Results.

The current interpretations of the MMX and KTX null results are as
follows:
1. The LET interpretation:
The arm that is parallel to the direction of motion is contracted by

an
appropriate amount and thus cancels out the effect of motion of the

earth
through the aether.

2. The SR interpretation:
There is no aether and the speed of light is isotropic.

Both of these interpretations are wrong. The correct interpretation is

as
follows:
1. There exits a stationary aether.
2. The direction of absolute motion of the earth in this stationary

aether
is in the vertical (UP or DOWN) direction. This sentence appears to be
self
contradictory. It is not. If one side of the earth is defined as

vertical
up
then the other side of the earth is defined vertical down. However,

these
two directions of motion (vertical up and verticle down) are in the

same
direction in the stationary aether.
3. This means that the light path length from the mirrors at the ends

of
the
horizontal arms to the re-combining mirror will remain the same for

all
the
orientations of the horizontal arms. That's why the null result for

both
the
MMX and KTX.

There is a proposed experiment to test the validity of this

interpretaion
and it is outlined in the following link:
http://www.journaloftheoretics.com/L...apers/Seto.pdf

Ken Seto
You've forgotten the truly correct one, Ken.

The null result of MMX is entirely due to the velocity of light bullets
(photons) being constant with respect to the source. No magical length
contractions or anything like that. Just like two marbles that are

released
together, follow different paths along guiding chutes and arrive

together,
and it just doesn't matter which way the apparatus is turned, or if it

takes
place on an airplane, or anything like that. They'll go on arriving

together
time and time again if they are released together. Of course, if you

blow
some wind over them, they won't. One will be slowed more than the other.

So
no aether wind for light, either. Hence no aether. Simple, isn't it?
Androcles


Wrong....your magic light bullets should have a speed of c+v or c-v
relative to the observer but they are not. So it is not that simple after
all.:-)

Why? Is the observer moving relative to the source? No... so v = 0
Is the observer moving relative to the detector? No... so v = 0
Is the observer moving relative to the medium (air)? No.. so v = 0.

Since v = 0 in this case, c+0 = c.


Are you sure about that?
Are you sure c+0 is not -c?
After all, since sqrt(1) can be -1 and sqrt(1) can be 1,
we can have
c + 0 = c
= 1 * c
= sqrt(1) * c
= (-1) * c
= -c
You see, c+0 has two answers: c and -c

Dirk Vdm


  #9  
Old September 21st 03 posted to sci.physics.relativity
kenseto
external usenet poster
 
Posts: 10,004
Default The Correct Interpretation of the MMX and KTX Null Results.


"xxein" wrote in message
om...
"Dirk Van de moortel" wrote

in message ...
"kenseto" wrote in message

...
The Correct Interpretation of the MMX and KTX Null Results.

The current interpretations of the MMX and KTX null results are as

follows:
1. The LET interpretation:
The arm that is parallel to the direction of motion is contracted by

an
appropriate amount and thus cancels out the effect of motion of the

earth
through the aether.

2. The SR interpretation:
There is no aether and the speed of light is isotropic.

Both of these interpretations are wrong.


But your theory has LET and SR as subsets, so your theory
is wrong:
| DRT includes SR/LET as subsets and its equations
| are valid in all environments--including gravity.
http://groups.google.com/groups?&q=d...+author%3Aseto
How sad.

Dirk Vdm


xxein: I see nothing wrong with sr-let being subsets of a theory, but
Ken can't rationally claim that for his theory.


Why not? My theory includes both LET and SR.

For LET:
The LET assertion of length contraction is replaced with the longer light
path length of a moving rod while a rod remains the same length physically
in all frames.

For SR:
1.The speed of light is not a universal constant as asserted by SR. It is a
constant math ratio as follows:
Light path length of rod (299,792,458 m) /the absolute time content for a
clock second co-moving with the rod.
2. The SR assertion of apparent length contraction is replaced with the
longer light path length of a moving rod.

Ken Seto


  #10  
Old September 21st 03 posted to sci.physics.relativity
kenseto
external usenet poster
 
Posts: 10,004
Default The Correct Interpretation of the MMX and KTX Null Results.


"AndroclesInEngland" wrote in message
...

"kenseto" wrote in message
...

"AndroclesInEngland" wrote in message
...

"kenseto" wrote in message
...
The Correct Interpretation of the MMX and KTX Null Results.

The current interpretations of the MMX and KTX null results are as
follows:
1. The LET interpretation:
The arm that is parallel to the direction of motion is contracted by

an
appropriate amount and thus cancels out the effect of motion of the

earth
through the aether.

2. The SR interpretation:
There is no aether and the speed of light is isotropic.

Both of these interpretations are wrong. The correct interpretation

is
as
follows:
1. There exits a stationary aether.
2. The direction of absolute motion of the earth in this stationary

aether
is in the vertical (UP or DOWN) direction. This sentence appears to

be
self
contradictory. It is not. If one side of the earth is defined as

vertical
up
then the other side of the earth is defined vertical down. However,

these
two directions of motion (vertical up and verticle down) are in the

same
direction in the stationary aether.
3. This means that the light path length from the mirrors at the

ends
of
the
horizontal arms to the re-combining mirror will remain the same for

all
the
orientations of the horizontal arms. That's why the null result for

both
the
MMX and KTX.

There is a proposed experiment to test the validity of this

interpretaion
and it is outlined in the following link:
http://www.journaloftheoretics.com/L...apers/Seto.pdf

Ken Seto
You've forgotten the truly correct one, Ken.

The null result of MMX is entirely due to the velocity of light

bullets
(photons) being constant with respect to the source. No magical length
contractions or anything like that. Just like two marbles that are

released
together, follow different paths along guiding chutes and arrive

together,
and it just doesn't matter which way the apparatus is turned, or if it

takes
place on an airplane, or anything like that. They'll go on arriving

together
time and time again if they are released together. Of course, if you

blow
some wind over them, they won't. One will be slowed more than the

other.
So
no aether wind for light, either. Hence no aether. Simple, isn't it?
Androcles


Wrong....your magic light bullets should have a speed of c+v or c-v
relative to the observer but they are not. So it is not that simple

after
all.:-)


Why? Is the observer moving relative to the source? No... so v = 0
Is the observer moving relative to the detector? No... so v = 0
Is the observer moving relative to the medium (air)? No.. so v = 0.


When the observer is moving wrt the source he still measures the light from
the source to be c. How do you explain that?

Ken Seto


Since v = 0 in this case, c+0 = c. So they DO have a speed of c+v after
all. It's just that in MMX, v happens to be zero. I thinks that's pretty
simple, don't you?
The rolling marbles have a velocity of u, and if you move along one of the
chutes with one it will have a velocity of u-v, or zero, with respect to
you, and you'll stay alongside it. Stand still and it's velocity with
respect to you will be once again be 'u'. There's nothing magic about it.
All matter is composed of molecules or individual atoms with electrons,

all
visible light is emitted by individual electrons falling to a lower energy
state, hence all visible light is comprised of individual photons and they
don't need any aether to transport them. They are self supporting
electomagnetic phenomena, the magnetic field collapsing and producing an
electric field, which in turn collapses to produce a magnetic field, ad
infinitum. they ar packets of constant energy that cannot be created or
destroyed, only converted to or from some other form, which can be mass.
Is the observer moving relative to the medium (aether)? Yes, according to
you.
But you can't detect it, can you? Magical things happen, like the

apparatus
shrinking along its length. At least, that's what Lorentz said (Dover, pp
3 - 7).
Androcles




 




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