On 3 jul, 20:06, mpc755 wrote:
On Jul 3, 7:26*pm, Miguel wrote:


In Emitter Theory, the photon propagates away from the star at 'c'.


Emission theories obey the principle of relativity by having no
preferred frame for light transmission, but say that light is emitted
at speed "c" relative to its source instead of applying the invariance
postulate.

If the star were the only object in the universe, the photon would
propagate away from where the star *is* at 'c' forever.

After 300 years, the photon would be 300 light years from where the
star *is*.

If after 300 years, the photon would wind up being consumed by an
observer on the earth, none of the above would change. The photon
would have traveled from where the star *is*.'


This clearly is nonsense. Now you have totally dismissed the frame of
reference (you don't know what that is anyway). So in this new
scenario of yours, for sure the only thing we can affirm is that the
photon traveled from where the star *was* to its current location.

You do not have to agree with this, and it is obvious you do not.


The simplest form of emission theory says that radiating objects throw
off light with a speed of "c" relative to their own state of motion,
and (unless we have reason to believe that the light changes speed in
flight), we then expect light to be moving towards us with a speed
that is offset by the speed of the distant emitter (c ± v). This
description generates three "odd" results:

1. If a radiant star moves across our field of vision, light given
off by differently-moving atoms in its atmosphere should take
different amounts of time to reach us. Since the retreating atoms
would have a "red" Doppler shift, and the approaching ones a "blue"
Doppler shift, the passing star might be expected to appear as a
"rainbow streak".
2. Similarly, if a radiant star is eclipsed, one might expect the
eclipsing shadow to appear to intercept different colours of Doppler-
shifted light in sequence - the eclipse might appear to have coloured
fringes.
3. For the case of a double-star system seen edge-on, light from
the approaching star might be expected to travel faster than light
from its receding companion, and overtake it. If the distance was
great enough for an approaching star's "fast" signal to catch up with
and overtake the "slow" light that it had emitted earlier when it was
receding, then the image of the star system should appear completely
scrambled.

None of these results have been observed, so this theory appears to be
wrong.

Miguel Rios

"mpc755" wrote in message
...
On Jul 3, 4:28 am, "Spirit of Truth" wrote:
"mpc755" wrote in message

...
On Jul 1, 1:23 pm, OG wrote:





mpc755 wrote:
On Jul 1, 12:59 pm, OG wrote:
mpc755 wrote:
On Jul 1, 12:52 pm, OG wrote:
mpc755 wrote:
On Jul 1, 12:14 pm, "OG" wrote:
"mpc755" wrote in message
...
De Sitter double star experiment:
http://en.wikipedia.org/wiki/De_Sitt...tar_experiment
A star in a binary star system is moving towards the earth at
'v'
and
emits a photon towards the earth.
How can a photon of light be traveling at c+v towards the earth
while
the star that emitted the photon is now moving away from the
earth
at
the same time the photon is traveling away from the star at 'c'
in
the
star rest frame?
As you say, at the time of emission the star is moving towards
the
earth at
velocity v, so the emission theory would have the light arriving
with speed
c+v in the earth's rest frame.
Are you unsure because the star may now be moving away from the
earth? The
speed of light is only c relative to the star rest frame /at the
time of
emission/ - once emitted, there is no connection between the star
and the
light to make it slow down or speed up as the star's velocity
varies over
its orbit.
That's what I am saying I disagree with. I'm saying, in the star's
rest frame, that the light propagates away from the star at 'c'.
In the star's rest frame, the star is stationary, and there is no
reason why the photon does not continue to propagate away from the
star at 'c' forever.
In the star's rest frame, the star's velocity never changes. It is
always zero.
Well yes, I suppose you could say that; but the rest of the
universe
moves relative to the star.
Emission theory would have the light comoving with the rest of the
universe rather than comoving with the emitting star.
In terms of the emission of light, the rest of the universe is
moving
relative to the star.
And?

And light travels from where the star *is* to where it is consumed by
the observer.

In terms of the photon emitted by the star, it is propagating away
from the star at 'c' as if the star is stationary, which the star is
in its rest frame.

In emitter theory, it is incorrect to consider the photon to be
traveling at c+v or c-v. In emitter theory, the photon needs to be
considered as traveling at 'c' relative to a stationary star, because
that is what is occurring in the star rest frame.

But the star is not 'stationary' as such. The star's 'rest' frame is
non-inertial, so the speed of previously emitted light is in fact time
dependent.

And as far as light travelling at c+v and c-v certainly is correct when
measured from an observer that is in an inertial frame.

Are you asking /why/ you are wrong as regards this, or are you insisting
that you are right on this matter?

I am 'insisting' that there is a star rest frame and that the photon
propagates outward from the star at 'c'.

I am 'insisting' that in terms of the photon of light propagating away
from the star, the star "really is" stationary. By this I mean, the
photon's emission point has nothing to do with three dimensional
space. The photons emission point is star and the star only.

As long as there is a star rest frame, when the photon of light
reaches the observer, it has traveled from where the star *is* and it
has traveled from the star to the observer at 'c'.
.................................................. .........................*................
.................................................. .........................*................

You are comparing two rest frames and from each c will
definately be c. Thing is "they" are comparing the observation
of the two places from ONE frame of reference.

Now, your observation is actually correct from both rest frames
and, maybe, somewhere there, is the key to the incorrectness
of lack of simultaneity too.

Spirit of Truth

Here is a visual of how Einstein's Train Thought experiment works in
Emission Theory:

http://www.youtube.com/watch?v=jyWTaXMElUk

Since I am not sure how to link to another post, and in case you
haven't seen this one, I am going to repeat it he

Applying Emitter Theory to Einstein's train thought experiment and
all
of the lightning strikes are determined to have occurred at the same
time.

The distances do not matter, but to make the intervals easier, A, A',
B, and B' are each one light year from M and M' at the time of the
lightning strikes.

The train is moving at 1/2 the speed of light away from A and towards
B.

On Jul 3, 9:26*pm, Miguel wrote:
On 3 jul, 20:06, mpc755 wrote:

On Jul 3, 7:26*pm, Miguel wrote:

In Emitter Theory, the photon propagates away from the star at 'c'.

Emission theories obey the principle of relativity by having no
preferred frame for light transmission, but say that light is emitted
at speed "c" relative to its source instead of applying the invariance
postulate.

If the star were the only object in the universe, the photon would
propagate away from where the star *is* at 'c' forever.

After 300 years, the photon would be 300 light years from where the
star *is*.

If after 300 years, the photon would wind up being consumed by an
observer on the earth, none of the above would change. The photon
would have traveled from where the star *is*.'

This clearly is nonsense. Now you have totally dismissed the frame of
reference (you don't know what that is anyway). So in this new
scenario of yours, for sure the only thing we can affirm is that the
photon traveled from where the star *was* to its current location.

You do not have to agree with this, and it is obvious you do not.

The simplest form of emission theory says that radiating objects throw
off light with a speed of "c" relative to their own state of motion,
and (unless we have reason to believe that the light changes speed in
flight), we then expect light to be moving towards us with a speed
that is offset by the speed of the distant emitter (c ± v). This
description generates three "odd" results:

* *1. If a radiant star moves across our field of vision, light given
off by differently-moving atoms in its atmosphere should take
different amounts of time to reach us. Since the retreating atoms
would have a "red" Doppler shift, and the approaching ones a "blue"
Doppler shift, the passing star might be expected to appear as a
"rainbow streak".
* *2. Similarly, if a radiant star is eclipsed, one might expect the
eclipsing shadow to appear to intercept different colours of Doppler-
shifted light in sequence - the eclipse might appear to have coloured
fringes.
* *3. For the case of a double-star system seen edge-on, light from
the approaching star might be expected to travel faster than light
from its receding companion, and overtake it. If the distance was
great enough for an approaching star's "fast" signal to catch up with
and overtake the "slow" light that it had emitted earlier when it was
receding, then the image of the star system should appear completely
scrambled.

None of these results have been observed, so this theory appears to be
wrong.

Miguel Rios

My version of Emitter Theory differs from the definition you cut-and-
pasted from wikipedia.

In my version of Emitter Theory it is incorrect to think of the light
traveling at c+v or c-v.

If the star is unaffected by anything else in the universe, then the
distance between the photon and the star will increase at 'c',
forever.

When the photon is consumed by an observer on the earth, the photon
has traveled at 'c' from where the star *is*.

This is more complicated when discussing binary stars, but it is
incorrect to think of a star that is moving towards the earth to emit
a photon at c+v and for that photon to be traveling at c+v through
space.

The photon is emitted from the star in the binary star system at 'c'
and even though the star is impacted by the gravitational effects of
its companion, the photon still propagates away from where the star
*is* at 'c'.

Another way to look at it, in Emitter Theory, is in the star's frame
of reference, the photon travels at 'c' away from where the star *is*.
In Emitter Theory, this is an accurate description of how the photon
propagates away from the star.

One of the differences between my version of Emitter Theory and SR, is
that it is incorrect to discuss a point in three dimensional space
where you assume the photon was emitted from the star and to think the
photon travel along that path to where it is consumed by the observer
on the earth.

All that matters in my version of Emitter Theory is where the star
*is* when the photon is consumed by the observer on the earth because
that *is* the path the photon traveled from the star to the observer.

On Jul 3, 9:48*pm, "Spirit of Truth" wrote:
"mpc755" wrote in message

...
On Jul 3, 4:28 am, "Spirit of Truth" wrote:





"mpc755" wrote in message

....
On Jul 1, 1:23 pm, OG wrote:

mpc755 wrote:
On Jul 1, 12:59 pm, OG wrote:
mpc755 wrote:
On Jul 1, 12:52 pm, OG wrote:
mpc755 wrote:
On Jul 1, 12:14 pm, "OG" wrote:
"mpc755" wrote in message
...
De Sitter double star experiment:
http://en.wikipedia.org/wiki/De_Sitt...tar_experiment
A star in a binary star system is moving towards the earth at
'v'
and
emits a photon towards the earth.
How can a photon of light be traveling at c+v towards the earth
while
the star that emitted the photon is now moving away from the
earth
at
the same time the photon is traveling away from the star at 'c'
in
the
star rest frame?
As you say, at the time of emission the star is moving towards
the
earth at
velocity v, so the emission theory would have the light arriving
with speed
c+v in the earth's rest frame.
Are you unsure because the star may now be moving away from the
earth? The
speed of light is only c relative to the star rest frame /at the
time of
emission/ - once emitted, there is no connection between the star
and the
light to make it slow down or speed up as the star's velocity
varies over
its orbit.
That's what I am saying I disagree with. I'm saying, in the star's
rest frame, that the light propagates away from the star at 'c'..
In the star's rest frame, the star is stationary, and there is no
reason why the photon does not continue to propagate away from the
star at 'c' forever.
In the star's rest frame, the star's velocity never changes. It is
always zero.
Well yes, I suppose you could say that; but the rest of the
universe
moves relative to the star.
Emission theory would have the light comoving with the rest of the
universe rather than comoving with the emitting star.
In terms of the emission of light, the rest of the universe is
moving
relative to the star.
And?

And light travels from where the star *is* to where it is consumed by
the observer.

In terms of the photon emitted by the star, it is propagating away
from the star at 'c' as if the star is stationary, which the star is
in its rest frame.

In emitter theory, it is incorrect to consider the photon to be
traveling at c+v or c-v. In emitter theory, the photon needs to be
considered as traveling at 'c' relative to a stationary star, because
that is what is occurring in the star rest frame.

But the star is not 'stationary' as such. The star's 'rest' frame is
non-inertial, so the speed of previously emitted light is in fact time
dependent.

And as far as light travelling at c+v and c-v certainly is correct when
measured from an observer that is in an inertial frame.

Are you asking /why/ you are wrong as regards this, or are you insisting
that you are right on this matter?

I am 'insisting' that there is a star rest frame and that the photon
propagates outward from the star at 'c'.

I am 'insisting' that in terms of the photon of light propagating away
from the star, the star "really is" stationary. By this I mean, the
photon's emission point has nothing to do with three dimensional
space. The photons emission point is star and the star only.

As long as there is a star rest frame, when the photon of light
reaches the observer, it has traveled from where the star *is* and it
has traveled from the star to the observer at 'c'.
.................................................. ..........................**................
.................................................. ..........................**................

You are comparing two rest frames and from each c will
definately be c. Thing is "they" are comparing the observation
of the two places from ONE frame of reference.

Now, your observation is actually correct from both rest frames
and, maybe, somewhere there, is the key to the incorrectness
of lack of simultaneity too.

Spirit of Truth

Here is a visual of how Einstein's Train Thought experiment works in
Emission Theory:

http://www.youtube.com/watch?v=jyWTaXMElUk

Since I am not sure how to link to another post, and in case you
haven't seen this one, I am going to repeat it he

Applying Emitter Theory to Einstein's train thought experiment and
all
of the lightning strikes are determined to have occurred at the same
time.

The distances do not matter, but to make the intervals easier, A, A',
B, and B' are each one light year from M and M' at the time of the
lightning strikes.

The train is moving at 1/2 the speed of light away from A and towards
B.

At the time of the lightning strikes:

A'-|-|-|-|-|-|-|-|-|-|-|-M'-|-|-|-|-|-|-|-|-|-|-|-B'
A--|-|-|-|-|-|-|-|-|-|-|-M--|-|-|-|-|-|-|-|-|-|-|-B

After eight months, A' is four light months away from A and eight
light months from M. Since the light is propagating outward at 'c' as
if A' is stationary the light reaches the Observer at M.

After eight months, B is four light months away from B' and eight
light months from M'. Since the light is propagating outward at 'c'
as
if B is stationary the light reaches the Observer at M'.

Eight months after the lightning strikes occurred:

|-|-|-|-A'-|-|-|-|-|-|-|-|-|-|-|-M'-|-|-|-|-|-|-|-|-|-|-|-B'
A-|-|-|-|--|-|-|-|-|-|-|-M-|-|-|-|--|-|-|-|-|-|-|-B

If we pick the date of Sept. 1, 2009 for the light from A' to reach M
and for the light from B to reach M' the observers will determine the
light from each lightning strike traveled eight light months to reach
them and each lightning strike occurred on Jan. 1, 2009.

Four months later, the light from A and B reaches M and the light
from
A' and B' reaches M'. The date is Jan. 1, 2010. The Observers at M
and
M' conclude that the light from the lightning strikes traveled one
light year to reach them and the lightning strikes occurred on Jan.
1,
2009.

One year after the lightning strikes occurred:

--|-|-|-|-|-A'-|-|-|-|-|-|-|-|-|-|-|-M'-|-|-|-|-|-|-|-|-|-|-|-B'
A-|-|-|-|-|-|--|-|-|-|-|-M-|-|-|-|-|-|--|-|-|-|-|-B

Twelve months later, the light from A reaches M' and the light from
B'
reaches M. The date is Jan. 1, 2011 and the Observers conclude the
light from the lightning strikes traveled two light years to reach
them and the lightning strikes occurred on Jan. 1, 2009.

Two years after the lightning strikes occurred:

--|-|-|-|-|-|-|-|-A'-|-|-|-|-|-|-|-|-M'-|-|-|-|-|-|-|-|-B'
A-|-|-|-|-|-|-|-|-M--|-|-|-|-|-|-|-|-B
(cut in half to try and avoid wrapping)

It doesn't matter how far the lightning strikes are from M/M' or how
fast the train is moving. In Emitter Theory, the lightning strikes
are determined to all have occurred at the same time.
.................................................. ..........................*................
.................................................. ..........................*.................

Thanks, I'll study.

Spirit

Thanks.

On 3 jul, 21:49, mpc755 wrote:
On Jul 3, 9:26*pm, Miguel wrote:


My version of Emitter Theory differs from the definition you cut-and-
pasted from wikipedia.

In my version of Emitter Theory it is incorrect to think of the light
traveling at c+v or c-v.


So your theory is clearly NOT Emitter Theory...period.
Stop calling your nonsense Emitter Theory, since is misleading and
dishonest!!!

Miguel Rios

On Jul 3, 9:59*pm, Miguel wrote:
On 3 jul, 21:49, mpc755 wrote:

On Jul 3, 9:26*pm, Miguel wrote:

My version of Emitter Theory differs from the definition you cut-and-
pasted from wikipedia.

In my version of Emitter Theory it is incorrect to think of the light
traveling at c+v or c-v.

So your theory is clearly NOT Emitter Theory...period.
Stop calling your nonsense Emitter Theory, since is misleading and
dishonest!!!

Miguel Rios

There are different variations of Emitter Theory and it looks like
placing c+v and c-v constaints on the theory was mostly done in order
to disprove it.

mpc755 wrote:

On Jul 3, 9:59 pm, Miguel wrote:

On 3 jul, 21:49, mpc755 wrote:


On Jul 3, 9:26 pm, Miguel wrote:

My version of Emitter Theory differs from the definition you cut-and-
pasted from wikipedia.

In my version of Emitter Theory it is incorrect to think of the light
traveling at c+v or c-v.

So your theory is clearly NOT Emitter Theory...period.
Stop calling your nonsense Emitter Theory, since is misleading and
dishonest!!!

Miguel Rios


There are different variations of Emitter Theory and it looks like
placing c+v and c-v constaints on the theory was mostly done in order
to disprove it.

You should come back when you learn to speak english. Physics is
clearly way beyond your abilities.

On Jul 4, 3:24*am, doug wrote:
mpc755 wrote:
On Jul 3, 9:59 pm, Miguel wrote:

On 3 jul, 21:49, mpc755 wrote:

On Jul 3, 9:26 pm, Miguel wrote:

My version of Emitter Theory differs from the definition you cut-and-
pasted from wikipedia.

In my version of Emitter Theory it is incorrect to think of the light
traveling at c+v or c-v.

So your theory is clearly NOT Emitter Theory...period.
Stop calling your nonsense Emitter Theory, since is misleading and
dishonest!!!

Miguel Rios

There are different variations of Emitter Theory and it looks like
placing c+v and c-v constaints on the theory was mostly done in order
to disprove it.

You should come back when you learn to speak english. Physics is
clearly way beyond your abilities.

There is only a single star in the universe.

A photon is emitted from the star.

If you could stop time and look at the path the photon has traveled,
the path would be a straight line from where the star *is* to where
the photon *is*.

If it just so happens that at this time the photon is consumed by an
observer on the earth, the above still holds.

What part of this are you unable to comprehend?

On Jul 4, 9:49*am, mpc755 wrote:
On Jul 4, 3:24*am, doug wrote:





mpc755 wrote:
On Jul 3, 9:59 pm, Miguel wrote:

On 3 jul, 21:49, mpc755 wrote:

On Jul 3, 9:26 pm, Miguel wrote:

My version of Emitter Theory differs from the definition you cut-and-
pasted from wikipedia.

In my version of Emitter Theory it is incorrect to think of the light
traveling at c+v or c-v.

So your theory is clearly NOT Emitter Theory...period.
Stop calling your nonsense Emitter Theory, since is misleading and
dishonest!!!

Miguel Rios

There are different variations of Emitter Theory and it looks like
placing c+v and c-v constaints on the theory was mostly done in order
to disprove it.

You should come back when you learn to speak english. Physics is
clearly way beyond your abilities.

There is only a single star in the universe.

A photon is emitted from the star.

If you could stop time and look at the path the photon has traveled,
the path would be a straight line from where the star *is* to where
the photon *is*.

If it just so happens that at this time the photon is consumed by an
observer on the earth, the above still holds.

What part of this are you unable to comprehend?

There are railroad tracks with an observer on one of the rails.

The observer is at rest relative to the rails.

The observer is directly facing the other rail.

On the other rail is a photon emitter.

The photon emitter is traveling along the rail at very close to the
speed of light.

When the photon emitter fires off a single photon, in the photon
emitter frame of reference, the photon is going to travel straight
across to the other rail.

When the photon gets to the other side of the rail it will hit the
observer.

In both emitter theory and in SR, in the photon emitter frame of
refereence, the photon travels the several feet from one rail to the
other and hits the observer square in the face.

In SR, in the observer's frame of reference, the photon was fired far
off in the distance from where the observer is. The photon traveled
from the rail the photon emitter is on and travels close to a light
year as it crosses from one rail to the other.

In SR, in the observer's frame of reference, the photon hits the
observer in the side of the head.

In Emitter Theory, since all that matters is the path between the
photon emitter and the photon and the distance between the two
increasing at 'c', in the observer's frame of reference, the photon
strikes the observer square in the face because it has traveled from
where the photon emitter *is* when the photon strikes the observer in
the face and the photon has traveled at 'c' from where the photon
emitter *is* to where the observer *is* when the photon strikes the
observer.

In SR, the photon strikes the observer in two different locations
depending upon which frame of reference is being discussed.

In Emitter Theory, the photon travels the same path in both frames of
reference and strikes the observer square in the face in both frames
of reference.

On Jul 4, 9:56*am, mpc755 wrote:
[...]

The heir to the throne of the land of idiots has a new challenger.