"AndroclesInEngland" wrote in message ...

"kenseto" wrote in message
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"AndroclesInEngland" wrote in message
...

"kenseto" wrote in message
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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