Re: The speed of light is c or c+v or c-v depending on the motion of the target

On Aug 7, 11:15 am, "Spaceman" <space...@xxxxxxxxxxxxxxxxxxxxxxxxxx>
wrote:
Sue... wrote:
On Aug 7, 10:34 am, "Spaceman" <space...@xxxxxxxxxxxxxxxxxxxxxxxxxx>
wrote:
Peter Riedt wrote:
On Aug 7, 12:42 pm, "Spaceman" <space...@xxxxxxxxxxxxxxxxxxxxxxxxxx>
wrote:
Peter Riedt wrote:
(c+v+c-v)/2=c

The speed of light c is considered to be a constant since at least
Maxwell. However, Michelson and Morley, in the mathematics
underlying their interferometer experiment (MMX) to find evidence
of the ether, used expressions like c+v and c-v. Their idea was
that the formula (c+v +c-v)/2=c applied overall and that the
components of this formula should be applied separately to the
upwind and downwind paths.

The expressions c-v and c+v are not well received or understood in
this NG. This is because it could lead to the idea that the speed
of light is subject to the speed of the source and this is
anathema to relativity supporters. However, the two expressions
have a real validity. They are mathematical devices to account
for the target moving away or towards the source. While the speed
of light is always c, the distance between source and target will
vary with the motion of objects through space. To compensate for
these variations, c-v and c+v are used. MMX is a good example to
explain it. The interferometer used in MMX is attached to the
earth. The earth moves through space. A light beam is sent from a
source through the beam splitter of the interferometer to the
mirror at the other end of the parallel arm of the equipment. The
distance between beam splitter and mirror is 11m as measured in
the lab but the earth does not stand still. It and the mirror
move away from the location where the beam splitter was at the
time the light beam passed through it. By the time the light has
caught up with the mirror, it has moved on to a new location,
widening the lab distance of 11m to a distance through space of
11m+. To allow for this extra distance, the speed of light has to
be adjusted to c-v or 300000km/sec-30km/sec giving an effective
299970km/sec. Likewise, on the return trip, the speed of light
has to be adjusted to c+v or 300000km/sec+30km/sec giving an
effective 300030km/sec to account for the fact that the beam
splitter is moving towards the light beam reflected from the
mirror. In short, the EFFECTIVE speed of light is subject to the
motion of the target (30km/sec around the sun) but the ACTUAL
speed of light up and down the parallel arm of the interferometer
is always (c+v+c-v)/2 = c. If you understand this there is no
need to worry that the speed of light is not c.

Very good summary Peter,
But the truth at the end is light is observer dependant in speed.
It is constant from source, but not for the observer.
There is no way any speed (no matter if it is constant from the
source) can be the same speed to all observers.
This is a big part of relativity that many relativists refuse to
admit and simply must ignore since they can not handle a non
constant speed of light.
But, it is constant from the source even if the source is moving,
but is not constant to the reflection point if moving because
waves do not do such and that is why we get doppler shifts
in light and sound and water waves.
The observers speed measurement of the waves is relative
to the lightwaves.
Simple as that.
:)

--
James M Driscoll Jr
Creator of the Clock Malfunction Theory
Spaceman- Hide quoted text -

- Show quoted text -

Spaceman,
I agree but I am stuck trying to explain the origin of c+v and c-v.
Androcles thinks there is something odd about c+v and c-v. He is
collecting opinions about it which does reveal the depth of
ignorance and parrotlike adherence to dogma of some posters.

Peter,
The distance traveled differences is what they are ignoring.
The distance traveled from "outside the box" is what most relativists
will not accept so they can't understand that the forward trip will
be a longer trip than the return if both objects are moving at rest
with each other but at a speed wrt an outside observer.
The outside observer is trying to measure the "actual" distance
traveled. instead of the "at rest"(inertial frame) distance that
never changes.

No... most so called "relatavists" think they need a
particle model of light moving under the influence of
inertia to dispel notions of Newton's ether

Your stated views are the same so you subscribe to
the same absurdities.

No,
actually I can also use a sound wave or water wave
as my point of view.

You can't present a sound wave point of view until
you can stop a moving car by switching on the
headlamps.


I see you don't get that the simple "particle" method only
makes things easier to follow by laymen like me.

I know the feeling. I too am a simple layperson
with no formal training in finance or accounting.
I can't get my bank to follow my simple bookeeping
techniques when my figures disagree with theirs. :o)



If you wish I can present a sounds wave point of view
also where no actual particle is traveling the entire distance.

Please do, but use two overlaid waves.
One wave only moves protons(+). The other wave
only moves electrons(-).

You should find that atoms can't be displaced
as with acoustic waves. They can only ~rotated~
or polarized.

http://en.wikipedia.org/wiki/Permittivity

Propagation in a dielectric medium
http://farside.ph.utexas.edu/teaching/em/lectures/node98.html



Sue...



Maybe you should learn some "wave theories".
And not just remove such facts just because it is a lightwave.
:)
It seems you are also stuck with the dingleberry croud and
can not actually think for yourself any longer.
:)

.

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