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'.

mpc755 wrote:
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'.

At 'c' relative to the star where and when the light was emitted. I can
manage that.

But the deSitter analysis doesn't just consider light emitted from 1
place and time, so the knowledge that the star has a non-inertial 'rest'
frame has to be taken into account. Which you are not doing.

On Jul 1, 5:51*am, mpc755 wrote:
On Jul 1, 9:41 am, Eric Gisse wrote:



On Jul 1, 5:32 am, mpc755 wrote:

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?

What makes you think it does?

Read the results of the experiment.

I don't think it does. I'm asking why is it assumed to be so.

Because that's what the experiment is testing?

Crazy, huh?


http://en.wikipedia.org/wiki/De_Sitt...tar_experiment

"For an object moving directly towards (or away from) the observer at
v metres per second, this light would then be expected to still be
travelling at (c + v) ( or (c - v) ) metres per second by the time it
reached us."

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

"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) )."

Either the photon is traveling at c+v when it reaches the earth, or it
is traveling at 'c' as it propagates away from the star in the star
rest frame, but not both.

I say the photon propagates away from the star at 'c'.

Why is this even being discussed?

The result disproves emission theory.

On Jul 1, 3:23*pm, OG wrote:
mpc755 wrote:
On Jul 1, 1:36 pm, OG wrote:
Let us suppose that a fast moving third star comes along and tears the
emitting star away from the original binary companion and drags it away
to another part of the galaxy. By applying your view, the already
emitted light has to change its own trajectory so that it keeps moving
away in a straight line at speed c relative to the star. Yes?

Or is gravitational effect of the third star somehow different to that
of the original binary companion star when it comes to the effect on
already emitted light?

I have intentionally tried to stay away from acceleration.

That's probably a mistake



In the binary star system, the star can be considered to be in a
constant state of momentum, or following the path of least resistance,
or to have a rest frame.

A non-inertial frame. That makes a big difference.

It is easy to conceptualize the star at rest and the photon to be
traveling away from the "stationary" star at 'c'.

Easy, but mistaken. The star is constantly accelerating, but once
emitted, the light is not subject to acceleration.

When you add the third star that changes all this, though the photon
is going to be impacted by the third star.

Even though it is many lightyears away?

I can see the photon being effected by the center of mass of the
interacting bodies.

If we go back to the binary star system where the star has a rest
frame, then the photon may be traveling at 'c' relative to the binary
star system's center of mass.

But you've you've already said not. So is it or isn't it?

* I find it easiest to conceptualize the

star being "stationary" in its rest frame, but even with this concept,
the other star will be rotating around this "stationary" star and
having a gravitational impact on the star and the photon. As the
photon gets further and further away from the binary star system, this
impact will lesson, but the photon is still going to be traveling at
'c' away from the emitting star, or traveling at 'c' away from the
center of mass of the binary stars.

There is a big difference between the two - what with one being
essentially inertial and the other not, can you not see that?



I intentionally stayed away from all of this to emphasis the point
that the photon is not simply fired off by the star at c+v and then
not in any way still tied to the star in terms of its propagating away
from the star in emitter theory.

The other major point I am trying to make is the photon is tied to its
point of origin, not a point in three dimensional space. And by point
of origin, I mean the star at rest, or the center of mass of the
binary stars. Even if the star moves through space or the center of
mass of the binary stars moves through space, the photon still
propagates away from this moving point of origin at 'c'.

If we can forget about center of mass of the binary star system for a
second, this next point is very important:

The photon is not tied to any point in three dimensional space. The
only point the photon is tied to is the emission point of the star in
the star's rest frame. This means that as the star moves through
space, the star is still at rest in its rest frame, and the emission
point of the photon moves through space along with the star.

The photon travels from where the star *is* to the observer on the
earth.

Likewise: this *next point is very important.

You don't understand what you are talking about. Sorry.

I can try to help your understanding, but only if you accept that you
are mistaken. If not, I've got other things I'd rather attend to.

The emission point of the photon is not tied to a particular point in
three dimensional space.

In emitter theory the photon travels at 'c' relative to an emission
point that is moving through space.

I can try and help you understand this. If not, I've got other things
I'd rather attend to.

On Jul 1, 3:47*pm, mpc755 wrote:
On Jul 1, 3:23*pm, OG wrote:



mpc755 wrote:
On Jul 1, 1:36 pm, OG wrote:
Let us suppose that a fast moving third star comes along and tears the
emitting star away from the original binary companion and drags it away
to another part of the galaxy. By applying your view, the already
emitted light has to change its own trajectory so that it keeps moving
away in a straight line at speed c relative to the star. Yes?

Or is gravitational effect of the third star somehow different to that
of the original binary companion star when it comes to the effect on
already emitted light?

I have intentionally tried to stay away from acceleration.

That's probably a mistake

In the binary star system, the star can be considered to be in a
constant state of momentum, or following the path of least resistance,
or to have a rest frame.

A non-inertial frame. That makes a big difference.

It is easy to conceptualize the star at rest and the photon to be
traveling away from the "stationary" star at 'c'.

Easy, but mistaken. The star is constantly accelerating, but once
emitted, the light is not subject to acceleration.

When you add the third star that changes all this, though the photon
is going to be impacted by the third star.

Even though it is many lightyears away?

I can see the photon being effected by the center of mass of the
interacting bodies.

If we go back to the binary star system where the star has a rest
frame, then the photon may be traveling at 'c' relative to the binary
star system's center of mass.

But you've you've already said not. So is it or isn't it?

* I find it easiest to conceptualize the

star being "stationary" in its rest frame, but even with this concept,
the other star will be rotating around this "stationary" star and
having a gravitational impact on the star and the photon. As the
photon gets further and further away from the binary star system, this
impact will lesson, but the photon is still going to be traveling at
'c' away from the emitting star, or traveling at 'c' away from the
center of mass of the binary stars.

There is a big difference between the two - what with one being
essentially inertial and the other not, can you not see that?

I intentionally stayed away from all of this to emphasis the point
that the photon is not simply fired off by the star at c+v and then
not in any way still tied to the star in terms of its propagating away
from the star in emitter theory.

The other major point I am trying to make is the photon is tied to its
point of origin, not a point in three dimensional space. And by point
of origin, I mean the star at rest, or the center of mass of the
binary stars. Even if the star moves through space or the center of
mass of the binary stars moves through space, the photon still
propagates away from this moving point of origin at 'c'.

If we can forget about center of mass of the binary star system for a
second, this next point is very important:

The photon is not tied to any point in three dimensional space. The
only point the photon is tied to is the emission point of the star in
the star's rest frame. This means that as the star moves through
space, the star is still at rest in its rest frame, and the emission
point of the photon moves through space along with the star.

The photon travels from where the star *is* to the observer on the
earth.

Likewise: this *next point is very important.

You don't understand what you are talking about. Sorry.

I can try to help your understanding, but only if you accept that you
are mistaken. If not, I've got other things I'd rather attend to.

The emission point of the photon is not tied to a particular point in
three dimensional space.

In emitter theory the photon travels at 'c' relative to an emission
point that is moving through space.

I can try and help you understand this. If not, I've got other things
I'd rather attend to.

I'll just end this thread with:

In the star rest frame, as long as the star remains at rest, the
photons propagate away from where the star *is* at 'c' forever.

mpc755 wrote:
On Jul 1, 3:47 pm, mpc755 wrote:
On Jul 1, 3:23 pm, OG wrote:



mpc755 wrote:
On Jul 1, 1:36 pm, OG wrote:
Let us suppose that a fast moving third star comes along and tears the
emitting star away from the original binary companion and drags it away
to another part of the galaxy. By applying your view, the already
emitted light has to change its own trajectory so that it keeps moving
away in a straight line at speed c relative to the star. Yes?
Or is gravitational effect of the third star somehow different to that
of the original binary companion star when it comes to the effect on
already emitted light?
I have intentionally tried to stay away from acceleration.
That's probably a mistake
In the binary star system, the star can be considered to be in a
constant state of momentum, or following the path of least resistance,
or to have a rest frame.
A non-inertial frame. That makes a big difference.
It is easy to conceptualize the star at rest and the photon to be
traveling away from the "stationary" star at 'c'.
Easy, but mistaken. The star is constantly accelerating, but once
emitted, the light is not subject to acceleration.
When you add the third star that changes all this, though the photon
is going to be impacted by the third star.
Even though it is many lightyears away?
I can see the photon being effected by the center of mass of the
interacting bodies.
If we go back to the binary star system where the star has a rest
frame, then the photon may be traveling at 'c' relative to the binary
star system's center of mass.
But you've you've already said not. So is it or isn't it?
I find it easiest to conceptualize the
star being "stationary" in its rest frame, but even with this concept,
the other star will be rotating around this "stationary" star and
having a gravitational impact on the star and the photon. As the
photon gets further and further away from the binary star system, this
impact will lesson, but the photon is still going to be traveling at
'c' away from the emitting star, or traveling at 'c' away from the
center of mass of the binary stars.
There is a big difference between the two - what with one being
essentially inertial and the other not, can you not see that?
I intentionally stayed away from all of this to emphasis the point
that the photon is not simply fired off by the star at c+v and then
not in any way still tied to the star in terms of its propagating away
from the star in emitter theory.
The other major point I am trying to make is the photon is tied to its
point of origin, not a point in three dimensional space. And by point
of origin, I mean the star at rest, or the center of mass of the
binary stars. Even if the star moves through space or the center of
mass of the binary stars moves through space, the photon still
propagates away from this moving point of origin at 'c'.
If we can forget about center of mass of the binary star system for a
second, this next point is very important:
The photon is not tied to any point in three dimensional space. The
only point the photon is tied to is the emission point of the star in
the star's rest frame. This means that as the star moves through
space, the star is still at rest in its rest frame, and the emission
point of the photon moves through space along with the star.
The photon travels from where the star *is* to the observer on the
earth.
Likewise: this next point is very important.
You don't understand what you are talking about. Sorry.
I can try to help your understanding, but only if you accept that you
are mistaken. If not, I've got other things I'd rather attend to.
The emission point of the photon is not tied to a particular point in
three dimensional space.

In emitter theory the photon travels at 'c' relative to an emission
point that is moving through space.

I can try and help you understand this. If not, I've got other things
I'd rather attend to.

I'll just end this thread with:

In the star rest frame, as long as the star remains at rest, the
photons propagate away from where the star *is* at 'c' forever.

What determines whether or not the star remains 'at rest' in its rest frame?

On Jul 1, 5:44*pm, OG wrote:
mpc755 wrote:
On Jul 1, 3:47 pm, mpc755 wrote:
On Jul 1, 3:23 pm, OG wrote:

mpc755 wrote:
On Jul 1, 1:36 pm, OG wrote:
Let us suppose that a fast moving third star comes along and tears the
emitting star away from the original binary companion and drags it away
to another part of the galaxy. By applying your view, the already
emitted light has to change its own trajectory so that it keeps moving
away in a straight line at speed c relative to the star. Yes?
Or is gravitational effect of the third star somehow different to that
of the original binary companion star when it comes to the effect on
already emitted light?
I have intentionally tried to stay away from acceleration.
That's probably a mistake
In the binary star system, the star can be considered to be in a
constant state of momentum, or following the path of least resistance,
or to have a rest frame.
A non-inertial frame. That makes a big difference.
It is easy to conceptualize the star at rest and the photon to be
traveling away from the "stationary" star at 'c'.
Easy, but mistaken. The star is constantly accelerating, but once
emitted, the light is not subject to acceleration.
When you add the third star that changes all this, though the photon
is going to be impacted by the third star.
Even though it is many lightyears away?
I can see the photon being effected by the center of mass of the
interacting bodies.
If we go back to the binary star system where the star has a rest
frame, then the photon may be traveling at 'c' relative to the binary
star system's center of mass.
But you've you've already said not. So is it or isn't it?
* I find it easiest to conceptualize the
star being "stationary" in its rest frame, but even with this concept,
the other star will be rotating around this "stationary" star and
having a gravitational impact on the star and the photon. As the
photon gets further and further away from the binary star system, this
impact will lesson, but the photon is still going to be traveling at
'c' away from the emitting star, or traveling at 'c' away from the
center of mass of the binary stars.
There is a big difference between the two - what with one being
essentially inertial and the other not, can you not see that?
I intentionally stayed away from all of this to emphasis the point
that the photon is not simply fired off by the star at c+v and then
not in any way still tied to the star in terms of its propagating away
from the star in emitter theory.
The other major point I am trying to make is the photon is tied to its
point of origin, not a point in three dimensional space. And by point
of origin, I mean the star at rest, or the center of mass of the
binary stars. Even if the star moves through space or the center of
mass of the binary stars moves through space, the photon still
propagates away from this moving point of origin at 'c'.
If we can forget about center of mass of the binary star system for a
second, this next point is very important:
The photon is not tied to any point in three dimensional space. The
only point the photon is tied to is the emission point of the star in
the star's rest frame. This means that as the star moves through
space, the star is still at rest in its rest frame, and the emission
point of the photon moves through space along with the star.
The photon travels from where the star *is* to the observer on the
earth.
Likewise: this *next point is very important.
You don't understand what you are talking about. Sorry.
I can try to help your understanding, but only if you accept that you
are mistaken. If not, I've got other things I'd rather attend to.
The emission point of the photon is not tied to a particular point in
three dimensional space.

In emitter theory the photon travels at 'c' relative to an emission
point that is moving through space.

I can try and help you understand this. If not, I've got other things
I'd rather attend to.

I'll just end this thread with:

In the star rest frame, as long as the star remains at rest, the
photons propagate away from where the star *is* at 'c' forever.

What determines whether or not the star remains 'at rest' in its rest frame?

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

"In special relativity the rest frame of a particle is the coordinate
system (frame of reference) in which the particle is at rest.

The rest frame of compound objects (such as a fluid, or a solid made
of many vibrating atoms) is taken to be the frame of reference in
which the average momentum of the particles which make up the
substance is zero (the particles may individually have momentum, but
collectively have no net momentum)."

Looks like I need both stars in order to have a 'rest frame'.

How about:

In emission theory, in a binary star system, the photon propagates
outward from where the center of mass *is* at 'c' forever.

mpc755 wrote:
On Jul 1, 5:44 pm, OG wrote:
mpc755 wrote:
On Jul 1, 3:47 pm, mpc755 wrote:
On Jul 1, 3:23 pm, OG wrote:
mpc755 wrote:
On Jul 1, 1:36 pm, OG wrote:
Let us suppose that a fast moving third star comes along and tears the
emitting star away from the original binary companion and drags it away
to another part of the galaxy. By applying your view, the already
emitted light has to change its own trajectory so that it keeps moving
away in a straight line at speed c relative to the star. Yes?
Or is gravitational effect of the third star somehow different to that
of the original binary companion star when it comes to the effect on
already emitted light?
I have intentionally tried to stay away from acceleration.
That's probably a mistake
In the binary star system, the star can be considered to be in a
constant state of momentum, or following the path of least resistance,
or to have a rest frame.
A non-inertial frame. That makes a big difference.
It is easy to conceptualize the star at rest and the photon to be
traveling away from the "stationary" star at 'c'.
Easy, but mistaken. The star is constantly accelerating, but once
emitted, the light is not subject to acceleration.
When you add the third star that changes all this, though the photon
is going to be impacted by the third star.
Even though it is many lightyears away?
I can see the photon being effected by the center of mass of the
interacting bodies.
If we go back to the binary star system where the star has a rest
frame, then the photon may be traveling at 'c' relative to the binary
star system's center of mass.
But you've you've already said not. So is it or isn't it?
I find it easiest to conceptualize the
star being "stationary" in its rest frame, but even with this concept,
the other star will be rotating around this "stationary" star and
having a gravitational impact on the star and the photon. As the
photon gets further and further away from the binary star system, this
impact will lesson, but the photon is still going to be traveling at
'c' away from the emitting star, or traveling at 'c' away from the
center of mass of the binary stars.
There is a big difference between the two - what with one being
essentially inertial and the other not, can you not see that?
I intentionally stayed away from all of this to emphasis the point
that the photon is not simply fired off by the star at c+v and then
not in any way still tied to the star in terms of its propagating away
from the star in emitter theory.
The other major point I am trying to make is the photon is tied to its
point of origin, not a point in three dimensional space. And by point
of origin, I mean the star at rest, or the center of mass of the
binary stars. Even if the star moves through space or the center of
mass of the binary stars moves through space, the photon still
propagates away from this moving point of origin at 'c'.
If we can forget about center of mass of the binary star system for a
second, this next point is very important:
The photon is not tied to any point in three dimensional space. The
only point the photon is tied to is the emission point of the star in
the star's rest frame. This means that as the star moves through
space, the star is still at rest in its rest frame, and the emission
point of the photon moves through space along with the star.
The photon travels from where the star *is* to the observer on the
earth.
Likewise: this next point is very important.
You don't understand what you are talking about. Sorry.
I can try to help your understanding, but only if you accept that you
are mistaken. If not, I've got other things I'd rather attend to.
The emission point of the photon is not tied to a particular point in
three dimensional space.
In emitter theory the photon travels at 'c' relative to an emission
point that is moving through space.
I can try and help you understand this. If not, I've got other things
I'd rather attend to.
I'll just end this thread with:
In the star rest frame, as long as the star remains at rest, the
photons propagate away from where the star *is* at 'c' forever.
What determines whether or not the star remains 'at rest' in its rest frame?

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

"In special relativity the rest frame of a particle is the coordinate
system (frame of reference) in which the particle is at rest.

The rest frame of compound objects (such as a fluid, or a solid made
of many vibrating atoms) is taken to be the frame of reference in
which the average momentum of the particles which make up the
substance is zero (the particles may individually have momentum, but
collectively have no net momentum)."

Looks like I need both stars in order to have a 'rest frame'.

How about:

In emission theory, in a binary star system, the photon propagates
outward from where the center of mass *is* at 'c' forever.


So somehow a photon emerging from a star is supposed to know whether it
is part of a binary system and adjusts its speed as appropriate. Yes?

But the binary system may also be part of a cluster which will have its
own centre of mass, and will certainly be part of a galaxy.

How does the photon work out how fast it's supposed to be going?

It's your theory - have fun!

On Jul 1, 6:21*pm, OG wrote:
mpc755 wrote:
On Jul 1, 5:44 pm, OG wrote:
mpc755 wrote:
On Jul 1, 3:47 pm, mpc755 wrote:
On Jul 1, 3:23 pm, OG wrote:
mpc755 wrote:
On Jul 1, 1:36 pm, OG wrote:
Let us suppose that a fast moving third star comes along and tears the
emitting star away from the original binary companion and drags it away
to another part of the galaxy. By applying your view, the already
emitted light has to change its own trajectory so that it keeps moving
away in a straight line at speed c relative to the star. Yes?
Or is gravitational effect of the third star somehow different to that
of the original binary companion star when it comes to the effect on
already emitted light?
I have intentionally tried to stay away from acceleration.
That's probably a mistake
In the binary star system, the star can be considered to be in a
constant state of momentum, or following the path of least resistance,
or to have a rest frame.
A non-inertial frame. That makes a big difference.
It is easy to conceptualize the star at rest and the photon to be
traveling away from the "stationary" star at 'c'.
Easy, but mistaken. The star is constantly accelerating, but once
emitted, the light is not subject to acceleration.
When you add the third star that changes all this, though the photon
is going to be impacted by the third star.
Even though it is many lightyears away?
I can see the photon being effected by the center of mass of the
interacting bodies.
If we go back to the binary star system where the star has a rest
frame, then the photon may be traveling at 'c' relative to the binary
star system's center of mass.
But you've you've already said not. So is it or isn't it?
* I find it easiest to conceptualize the
star being "stationary" in its rest frame, but even with this concept,
the other star will be rotating around this "stationary" star and
having a gravitational impact on the star and the photon. As the
photon gets further and further away from the binary star system, this
impact will lesson, but the photon is still going to be traveling at
'c' away from the emitting star, or traveling at 'c' away from the
center of mass of the binary stars.
There is a big difference between the two - what with one being
essentially inertial and the other not, can you not see that?
I intentionally stayed away from all of this to emphasis the point
that the photon is not simply fired off by the star at c+v and then
not in any way still tied to the star in terms of its propagating away
from the star in emitter theory.
The other major point I am trying to make is the photon is tied to its
point of origin, not a point in three dimensional space. And by point
of origin, I mean the star at rest, or the center of mass of the
binary stars. Even if the star moves through space or the center of
mass of the binary stars moves through space, the photon still
propagates away from this moving point of origin at 'c'.
If we can forget about center of mass of the binary star system for a
second, this next point is very important:
The photon is not tied to any point in three dimensional space. The
only point the photon is tied to is the emission point of the star in
the star's rest frame. This means that as the star moves through
space, the star is still at rest in its rest frame, and the emission
point of the photon moves through space along with the star.
The photon travels from where the star *is* to the observer on the
earth.
Likewise: this *next point is very important.
You don't understand what you are talking about. Sorry.
I can try to help your understanding, but only if you accept that you
are mistaken. If not, I've got other things I'd rather attend to.
The emission point of the photon is not tied to a particular point in
three dimensional space.
In emitter theory the photon travels at 'c' relative to an emission
point that is moving through space.
I can try and help you understand this. If not, I've got other things
I'd rather attend to.
I'll just end this thread with:
In the star rest frame, as long as the star remains at rest, the
photons propagate away from where the star *is* at 'c' forever.
What determines whether or not the star remains 'at rest' in its rest frame?

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

"In special relativity the rest frame of a particle is the coordinate
system (frame of reference) in which the particle is at rest.

The rest frame of compound objects (such as a fluid, or a solid made
of many vibrating atoms) is taken to be the frame of reference in
which the average momentum of the particles which make up the
substance is zero (the particles may individually have momentum, but
collectively have no net momentum)."

Looks like I need both stars in order to have a 'rest frame'.

How about:

In emission theory, in a binary star system, the photon propagates
outward from where the center of mass *is* at 'c' forever.

So somehow a photon emerging from a star is supposed to know whether it
is part of a binary system and adjusts its speed as appropriate. *Yes?

But the binary system may also be part of a cluster which will have its
own centre of mass, and will certainly be part of a galaxy.

How does the photon work out how fast it's supposed to be going?

It's your theory - have fun!

The photon doesn't adjust its speed accordingly.

When the photon is emitted, the photon is emitted from the star at
'c'.

The photon does not travel at c+v or c-v, in emitter theory.

The photon travels at 'c' from where the star *is* as long as the star
is under constant momentum.

If the photon is light years away from the star and another star
alters the emitting star's momentum, that does not affect the photons
travel.

The point is the photon's point of emission is not tied to a
particular point in three dimensional space.

It propagates away from the star at 'c' as if the star is stationary.

mpc755 wrote:
On Jul 1, 6:21 pm, OG wrote:
mpc755 wrote:
On Jul 1, 5:44 pm, OG wrote:
mpc755 wrote:
On Jul 1, 3:47 pm, mpc755 wrote:
On Jul 1, 3:23 pm, OG wrote:
mpc755 wrote:
On Jul 1, 1:36 pm, OG wrote:
Let us suppose that a fast moving third star comes along and tears the
emitting star away from the original binary companion and drags it away
to another part of the galaxy. By applying your view, the already
emitted light has to change its own trajectory so that it keeps moving
away in a straight line at speed c relative to the star. Yes?
Or is gravitational effect of the third star somehow different to that
of the original binary companion star when it comes to the effect on
already emitted light?
I have intentionally tried to stay away from acceleration.
That's probably a mistake
In the binary star system, the star can be considered to be in a
constant state of momentum, or following the path of least resistance,
or to have a rest frame.
A non-inertial frame. That makes a big difference.
It is easy to conceptualize the star at rest and the photon to be
traveling away from the "stationary" star at 'c'.
Easy, but mistaken. The star is constantly accelerating, but once
emitted, the light is not subject to acceleration.
When you add the third star that changes all this, though the photon
is going to be impacted by the third star.
Even though it is many lightyears away?
I can see the photon being effected by the center of mass of the
interacting bodies.
If we go back to the binary star system where the star has a rest
frame, then the photon may be traveling at 'c' relative to the binary
star system's center of mass.
But you've you've already said not. So is it or isn't it?
I find it easiest to conceptualize the
star being "stationary" in its rest frame, but even with this concept,
the other star will be rotating around this "stationary" star and
having a gravitational impact on the star and the photon. As the
photon gets further and further away from the binary star system, this
impact will lesson, but the photon is still going to be traveling at
'c' away from the emitting star, or traveling at 'c' away from the
center of mass of the binary stars.
There is a big difference between the two - what with one being
essentially inertial and the other not, can you not see that?
I intentionally stayed away from all of this to emphasis the point
that the photon is not simply fired off by the star at c+v and then
not in any way still tied to the star in terms of its propagating away
from the star in emitter theory.
The other major point I am trying to make is the photon is tied to its
point of origin, not a point in three dimensional space. And by point
of origin, I mean the star at rest, or the center of mass of the
binary stars. Even if the star moves through space or the center of
mass of the binary stars moves through space, the photon still
propagates away from this moving point of origin at 'c'.
If we can forget about center of mass of the binary star system for a
second, this next point is very important:
The photon is not tied to any point in three dimensional space. The
only point the photon is tied to is the emission point of the star in
the star's rest frame. This means that as the star moves through
space, the star is still at rest in its rest frame, and the emission
point of the photon moves through space along with the star.
The photon travels from where the star *is* to the observer on the
earth.
Likewise: this next point is very important.
You don't understand what you are talking about. Sorry.
I can try to help your understanding, but only if you accept that you
are mistaken. If not, I've got other things I'd rather attend to.
The emission point of the photon is not tied to a particular point in
three dimensional space.
In emitter theory the photon travels at 'c' relative to an emission
point that is moving through space.
I can try and help you understand this. If not, I've got other things
I'd rather attend to.
I'll just end this thread with:
In the star rest frame, as long as the star remains at rest, the
photons propagate away from where the star *is* at 'c' forever.
What determines whether or not the star remains 'at rest' in its rest frame?
http://en.wikipedia.org/wiki/Rest_frame
"In special relativity the rest frame of a particle is the coordinate
system (frame of reference) in which the particle is at rest.
The rest frame of compound objects (such as a fluid, or a solid made
of many vibrating atoms) is taken to be the frame of reference in
which the average momentum of the particles which make up the
substance is zero (the particles may individually have momentum, but
collectively have no net momentum)."
Looks like I need both stars in order to have a 'rest frame'.
How about:
In emission theory, in a binary star system, the photon propagates
outward from where the center of mass *is* at 'c' forever.
So somehow a photon emerging from a star is supposed to know whether it
is part of a binary system and adjusts its speed as appropriate. Yes?

But the binary system may also be part of a cluster which will have its
own centre of mass, and will certainly be part of a galaxy.

How does the photon work out how fast it's supposed to be going?

It's your theory - have fun!

The photon doesn't adjust its speed accordingly.

When the photon is emitted, the photon is emitted from the star at
'c'.

Sorry, I thought you said it 'propagates outward from the centre of mass
at speed c'

The photon does not travel at c+v or c-v, in emitter theory.

The photon travels at 'c' from where the star *is* as long as the star
is under constant momentum.

What do you mean by 'under' constant momentum? Since momentum is a
vector, any star that is in a binary system does not have 'constant
momentum', since its direction of movement varies constantly as it
orbits, and its speed will only be constant if it happens to be in a
perfectly circular orbit.

If the photon is light years away from the star and another star
alters the emitting star's momentum, that does not affect the photons
travel.

But in a binary system, there is always another star that is affecting
the momentum of the emitting star. In addition to which that contradicts
your previous emphasis that the speed of the photon is 'c' with respect
to where the star *is*.

The point is the photon's point of emission is not tied to a
particular point in three dimensional space.

Have you noticed that I've never disagreed with this part of what you
are saying?


It propagates away from the star at 'c' as if the star is stationary.

Yup, so long as the star isn't accelerated, we can say that. It's known
as the Principle of Special Relativity.