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

Sue... wrote:
On Aug 7, 1:09 pm, "Spaceman" <space...@xxxxxxxxxxxxxxxxxxxxxxxxxx>
wrote:
Sue... wrote:
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.

That only takes a bit of wiring.
But you would not think of that approach.
And sadly that has nothing to do with wave
speed measurements by the observer.

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)

So
You admit you can't do basic math then?
:)

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(-).

So you wish me to use apples and oranges


If I meant apples and oranges I would not
have specified electrons and protons.

Do you understand how electrons and protons differ ?

Do you understand that it is irrelevant to the actual topic
We are not discussing "what" the waves are moving.
We are discussing the speeds of them.
Sheesh Sue.
Get a clue.




.

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