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?

On Jul 1, 12:14*pm, mpc755 wrote:
On Jul 1, 11:25*am, mpc755 wrote:





On Jul 1, 11:21*am, Bruce Richmond wrote:

On Jul 1, 9:51*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.

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'.- Hide quoted text -

- Show quoted text -

A jet fighter flies at the velocity u and has a gun that fires bullets
at the velocity of w.

When sitting on the ground the fighter's velocity is 0 and when it
fires at a target and the bullets hit the target at the velocity of v=
(0+w).

When flying toward the target at the velocity of u the bullet hits at
v=(u+w).

When flying away and shooting back at the target the bullet hits at v=
(u-w).

In emitter theory the bullet has the velocity c so the target is hit
by light traveling at (u+c). *Since u is the velocity of the plane we
would write it (v+c). *And since it doesn't matter which order we add
things we can swap it around to the more common (c+v).

So, you're saying a photon of light does not propagate away from the
star at 'c' in the star rest frame.

No, proving once again that you have a reading comprehension problem.
The jet represents the star and the bullet represents the photon. The
bullet is emitted by the jet at w (which in emitter theory is c)
relative to the jet, and will continue to travel at the same speed and
direction after emission reguardless of what happens to the jet.

I'm saying it does. Even in emitter theory.

And the reason you are saying light does not propagate away from the
star at 'c' in the star rest frame in emitter theory is the
following...

When the star, of a binary star pair, is traveling towards the earth
it emits a photon towards the earth at what you consider to be c+v.

As the star continues to orbit around its pair, it will eventually be
traveling away from the earth at velocity 'w'.

At this time, in the earth's rest frame, the photon will be traveling
away from the star at 'c+v+w'.

At this time, it does not seem possible for the photon to also be
propagating away from the star at 'c' in the star's rest frame.

If the jet turns around after emitting the bullet do you think the
bullet will change speed relative to the target so as to maintain its
velocity relative to the jet?

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.

On Jul 1, 12:23*pm, 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.- Hide quoted text -

- Show quoted text -

The star's velocity is constantly changing. It is in orbit around its
twin star, being yanked around by that star's gravity.

On Jul 1, 1:08*pm, Bruce Richmond wrote:
On Jul 1, 12:14*pm, mpc755 wrote:



On Jul 1, 11:25*am, mpc755 wrote:

On Jul 1, 11:21*am, Bruce Richmond wrote:

On Jul 1, 9:51*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.

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'.- Hide quoted text -

- Show quoted text -

A jet fighter flies at the velocity u and has a gun that fires bullets
at the velocity of w.

When sitting on the ground the fighter's velocity is 0 and when it
fires at a target and the bullets hit the target at the velocity of v=
(0+w).

When flying toward the target at the velocity of u the bullet hits at
v=(u+w).

When flying away and shooting back at the target the bullet hits at v=
(u-w).

In emitter theory the bullet has the velocity c so the target is hit
by light traveling at (u+c). *Since u is the velocity of the plane we
would write it (v+c). *And since it doesn't matter which order we add
things we can swap it around to the more common (c+v).

So, you're saying a photon of light does not propagate away from the
star at 'c' in the star rest frame.

No, proving once again that you have a reading comprehension problem.
The jet represents the star and the bullet represents the photon. *The
bullet is emitted by the jet at w (which in emitter theory is c)
relative to the jet, and will continue to travel at the same speed and
direction after emission reguardless of what happens to the jet.



I'm saying it does. Even in emitter theory.

And the reason you are saying light does not propagate away from the
star at 'c' in the star rest frame in emitter theory is the
following...

When the star, of a binary star pair, is traveling towards the earth
it emits a photon towards the earth at what you consider to be c+v.

As the star continues to orbit around its pair, it will eventually be
traveling away from the earth at velocity 'w'.

At this time, in the earth's rest frame, the photon will be traveling
away from the star at 'c+v+w'.

At this time, it does not seem possible for the photon to also be
propagating away from the star at 'c' in the star's rest frame.

If the jet turns around after emitting the bullet do you think the
bullet will change speed relative to the target so as to maintain its
velocity relative to the jet?

If the jet turns around, then the jet is not at rest.

There is a star rest frame.

On Jul 1, 1:12*pm, Bruce Richmond wrote:
On Jul 1, 12:23*pm, 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.- Hide quoted text -

- Show quoted text -

The star's velocity is constantly changing. *It is in orbit around its
twin star, being yanked around by that star's gravity.

But just as there is an earth rest frame, there is a star rest frame.

Yes it is being impacted by its twins gravity, but so then will the
photon.

The photon will continue to propagate away from the star at 'c' in the
star's rest frame regardless of the twins gravitational impact.

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?

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.

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?

"OG" wrote in message
...

"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.

"De Sitter made a study of double stars (1913) and found no cases where the
stars' images appeared scrambled."
Didn't have HST, did he?
http://antwrp.gsfc.nasa.gov/apod/ap010121.html
Oh look, Mira isn't scrambled, it has a tail!
http://hyperphysics.phy-astr.gsu.edu...ic/brokpen.jpg
Oh look, there is no refraction, pencils bend in water!

mpc755 wrote:

On Jul 1, 11:25 am, mpc755 wrote:

On Jul 1, 11:21 am, Bruce Richmond wrote:




On Jul 1, 9:51 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.

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'.- Hide quoted text -

- Show quoted text -

A jet fighter flies at the velocity u and has a gun that fires bullets
at the velocity of w.

When sitting on the ground the fighter's velocity is 0 and when it
fires at a target and the bullets hit the target at the velocity of v=
(0+w).

When flying toward the target at the velocity of u the bullet hits at
v=(u+w).

When flying away and shooting back at the target the bullet hits at v=
(u-w).

In emitter theory the bullet has the velocity c so the target is hit
by light traveling at (u+c). Since u is the velocity of the plane we
would write it (v+c). And since it doesn't matter which order we add
things we can swap it around to the more common (c+v).

So, you're saying a photon of light does not propagate away from the
star at 'c' in the star rest frame.

I'm saying it does. Even in emitter theory.


And the reason you are saying light does not propagate away from the
star at 'c' in the star rest frame in emitter theory is the
following...

When the star, of a binary star pair, is traveling towards the earth
it emits a photon towards the earth at what you consider to be c+v.

As the star continues to orbit around its pair, it will eventually be
traveling away from the earth at velocity 'w'.

At this time, in the earth's rest frame, the photon will be traveling
away from the star at 'c+v+w'.

At this time, it does not seem possible for the photon to also be
propagating away from the star at 'c' in the star's rest frame.

That is because you have no clue what frames of reference are.