(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.

Peter Riedt wrote:
On Aug 7, 12:42 pm, "Spaceman" <space... at (no spam) yourclockmalfunctioned.duh
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.- Hide quoted text -

- Show quoted text -

On Aug 7, 10:34 am, "Spaceman" <space... at (no spam) yourclockmalfunctioned.duh
wrote:





Peter Riedt wrote:
On Aug 7, 12:42 pm, "Spaceman" <space... at (no spam) yourclockmalfunctioned.duh
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.

The pot is calling the kettle black.

(appologies to any politicians, pickpockets
or used car peddlers who may not be black enough or
white enough)  )

Learn some physics:http://www.glenbrook.k12.il.us/gbssci/phys/Class/newtlaws/newtltoc.html

Sue...- Hide quoted text -

- Show quoted text -

"Peter Riedt" <rie... at (no spam) yahoo.co.uk> wrote in message

news:6c303558-e5df-42f4-95a0-e4692644f856 at (no spam) a2g2000prm.googlegroups.com...

(c+v+c-v)/2=c

That's true of course. ;-)

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.

So they did. However, it only applies within a single (standard) coordinate
system:http://www.fourmilab.ch/etexts/einstein/specrel/www/(par.3).

The expressions c-v and c+v are not well received or understood in
this NG.

Indeed, perhaps half of the people here don't understand the above
derivation...

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.

Inexact: both the source and the target may be moving, as indicated with for
example v and w respectively. The expression concerns the speed of light
(also called "closing speed" in modern jargon) relative to a moving object.
Of course, it is also valid for an object in rest ( c-v AND v=0 => c)..

While the speed of light is always c,

It might be useful if you specify what you mean with "the speed of light" in
this context. Relative to what?

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.

Instead of "space" (which is an unknown), for calculations the solar system
is often chosen as reference.

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.

Apparently you DID use the solar system...

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.

You were so close, and then such a colossal hic-up!

Here's another one:
If you drive 100 km to and fro at 100 km/h without delay, the return trip
will take you 2 hours.
Now, suppose you were slowed down in traffic to 100-10 = 90 km/h on the
first leg, and so you try to compensate for that by driving 100+10 = 110
km/h back; How long will it take in total? What will be your average speed?

On Aug 7, 10:34 pm, "Spaceman" <space... at (no spam) yourclockmalfunctioned.duh
wrote:
Peter Riedt wrote:
On Aug 7, 12:42 pm, "Spaceman" <space... at (no spam) yourclockmalfunctioned.duh
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.- Hide quoted text -

- Show quoted text -

Spaceman,
'actual distance' versus 'at rest distance' are excellent terms to
explain a difference not understood or deliberately ignored by the
relativity fraternity.