You are all familiar with the DeSitter double star experiment.

"In 1913, Willem de Sitter argued that if this was true, a star orbiting in a double-star system would usually, with regard to us, alternate between moving towards us and away from us. Light emitted from different parts of the orbital path would travel towards us at different speeds. For a nearby star with a small orbital velocity (or whose orbital plane was almost perpendicular to our line of view) this might merely make the star's orbit seem erratic, but for a sufficient combination of orbital speed and distance (and inclination), the "fast" light given off during approach would be able to catch up with and even overtake "slow" light emitted earlier during a recessional part of the star's orbit, and the star would present an image that was scrambled and out of sequence. That is, Kepler's laws of motion would apparently be violated for a distant observer."

Question: If light travels from the star at velocity c relative to the Earth, at what relative velocity does it have relative to the surface of the star, when viewed from the reference frame in which the Earth?

On 01-May-23 4:10 pm, gehan.am...@gmail.com wrote:
> You are all familiar with the DeSitter double star experiment.
>
> "In 1913, Willem de Sitter argued that if this was true, a star
> orbiting in a double-star system would usually, with regard to us,
> alternate between moving towards us and away from us. Light emitted
> from different parts of the orbital path would travel towards us at
> different speeds. For a nearby star with a small orbital velocity (or
> whose orbital plane was almost perpendicular to our line of view)
> this might merely make the star's orbit seem erratic, but for a
> sufficient combination of orbital speed and distance (and
> inclination), the "fast" light given off during approach would be
> able to catch up with and even overtake "slow" light emitted earlier
> during a recessional part of the star's orbit, and the star would
> present an image that was scrambled and out of sequence. That is,
> Kepler's laws of motion would apparently be violated for a distant
> observer."
>
> Question: If light travels from the star at velocity c relative to
> the Earth, at what relative velocity does it have relative to the
> surface of the star, when viewed from the reference frame in which
> the Earth?

If the star is moving towards the Earth at velocity v, then in the Earth
frame the light is moving at velocity (c - v) relative to the star.

This does not contradict the constancy of the speed of light, which
relates to the speed as measured by an observer, not the speed of light
relative to anything else.

Of course, for an observer on the star, the light is observed to depart
at speed c.

Sylvia.

Den 01.05.2023 08:10, skrev gehan.am...@gmail.com:
> You are all familiar with the DeSitter double star experiment.
>
> "In 1913, Willem de Sitter argued that if this was true, a star orbiting in a double-star system would usually, with regard to us, alternate between moving towards us and away from us. Light emitted from different parts of the orbital path would travel towards us at different speeds. For a nearby star with a small orbital velocity (or whose orbital plane was almost perpendicular to our line of view) this might merely make the star's orbit seem erratic, but for a sufficient combination of orbital speed and distance (and inclination), the "fast" light given off during approach would be able to catch up with and even overtake "slow" light emitted earlier during a recessional part of the star's orbit, and the star would present an image that was scrambled and out of sequence. That is, Kepler's laws of motion would apparently be violated for a distant observer."
>
> Question: If light travels from the star at velocity c relative to the Earth, at what relative velocity does it have relative to the surface of the star, when viewed from the reference frame in which the Earth?

Don't you see that your question is meaningless?

The speed of light is c relative to the surface of the star.
Which means that the speed of light is c when measured
in the rest frame of the star.

The speed of light is c relative to the Earth.
Which means that the speed of light is c when measured
in the rest frame of the Earth.

You are asking:
What is the speed of light measured in the rest frame of
the star when measured in the rest frame of the Earth.

--
Paul

https://paulba.no/

On Monday, May 1, 2023 at 2:36:38 PM UTC+5, Sylvia Else wrote:
> On 01-May-23 4:10 pm, gehan.am...@gmail.com wrote:
> > You are all familiar with the DeSitter double star experiment.
> >
> > "In 1913, Willem de Sitter argued that if this was true, a star
> > orbiting in a double-star system would usually, with regard to us,
> > alternate between moving towards us and away from us. Light emitted
> > from different parts of the orbital path would travel towards us at
> > different speeds. For a nearby star with a small orbital velocity (or
> > whose orbital plane was almost perpendicular to our line of view)
> > this might merely make the star's orbit seem erratic, but for a
> > sufficient combination of orbital speed and distance (and
> > inclination), the "fast" light given off during approach would be
> > able to catch up with and even overtake "slow" light emitted earlier
> > during a recessional part of the star's orbit, and the star would
> > present an image that was scrambled and out of sequence. That is,
> > Kepler's laws of motion would apparently be violated for a distant
> > observer."
> >
> > Question: If light travels from the star at velocity c relative to
> > the Earth, at what relative velocity does it have relative to the
> > surface of the star, when viewed from the reference frame in which
> > the Earth?
> If the star is moving towards the Earth at velocity v, then in the Earth
> frame the light is moving at velocity (c - v) relative to the star.
>
> This does not contradict the constancy of the speed of light, which
> relates to the speed as measured by an observer, not the speed of light
> relative to anything else.
>
> Of course, for an observer on the star, the light is observed to depart
> at speed c.
>
> Sylvia.

> If the star is moving towards the Earth at velocity v, then in the Earth
> frame the light is moving at velocity (c - v) relative to the star.

If this is true, then have a look at this scenario.

S>>>>>>>>>>> c

X

S>>>>>>>>>>> c

In the Earth's frame of reference, light travels at c relative to the Earth.. That means the relative velocity between the light and the surface of the star is c-v in the bottom case, where the star is approaching the Earth.

When the same star is moving away from the Earth, we have the relative velocity between light or a particular photon or group of photons moving at c+v relative to the surface of the star.

You can call this closing speed or relative speed.

The fact is that the light is emitted at a different relative velocity relative to the surface of the star depending on the direction it is travelling - towards the Earth or away.

Doesn't this violate the 'First Postulate'?

On Monday, May 1, 2023 at 3:41:54 PM UTC+5, Paul B. Andersen wrote:
> Den 01.05.2023 08:10, skrev gehan.am...@gmail.com:
> > You are all familiar with the DeSitter double star experiment.
> >
> > "In 1913, Willem de Sitter argued that if this was true, a star orbiting in a double-star system would usually, with regard to us, alternate between moving towards us and away from us. Light emitted from different parts of the orbital path would travel towards us at different speeds. For a nearby star with a small orbital velocity (or whose orbital plane was almost perpendicular to our line of view) this might merely make the star's orbit seem erratic, but for a sufficient combination of orbital speed and distance (and inclination), the "fast" light given off during approach would be able to catch up with and even overtake "slow" light emitted earlier during a recessional part of the star's orbit, and the star would present an image that was scrambled and out of sequence. That is, Kepler's laws of motion would apparently be violated for a distant observer."
> >
> > Question: If light travels from the star at velocity c relative to the Earth, at what relative velocity does it have relative to the surface of the star, when viewed from the reference frame in which the Earth?
> Don't you see that your question is meaningless?
>
> The speed of light is c relative to the surface of the star.
> Which means that the speed of light is c when measured
> in the rest frame of the star.
>
> The speed of light is c relative to the Earth.
> Which means that the speed of light is c when measured
> in the rest frame of the Earth.
>
> You are asking:
> What is the speed of light measured in the rest frame of
> the star when measured in the rest frame of the Earth.
>
> --
> Paul
>
> https://paulba.no/

On 01-May-23 10:05 pm, gehan.am...@gmail.com wrote:
> On Monday, May 1, 2023 at 2:36:38 PM UTC+5, Sylvia Else wrote:
>> On 01-May-23 4:10 pm, gehan.am...@gmail.com wrote:
>>> You are all familiar with the DeSitter double star experiment.
>>>
>>> "In 1913, Willem de Sitter argued that if this was true, a star
>>> orbiting in a double-star system would usually, with regard to us,
>>> alternate between moving towards us and away from us. Light emitted
>>> from different parts of the orbital path would travel towards us at
>>> different speeds. For a nearby star with a small orbital velocity (or
>>> whose orbital plane was almost perpendicular to our line of view)
>>> this might merely make the star's orbit seem erratic, but for a
>>> sufficient combination of orbital speed and distance (and
>>> inclination), the "fast" light given off during approach would be
>>> able to catch up with and even overtake "slow" light emitted earlier
>>> during a recessional part of the star's orbit, and the star would
>>> present an image that was scrambled and out of sequence. That is,
>>> Kepler's laws of motion would apparently be violated for a distant
>>> observer."
>>>
>>> Question: If light travels from the star at velocity c relative to
>>> the Earth, at what relative velocity does it have relative to the
>>> surface of the star, when viewed from the reference frame in which
>>> the Earth?
>> If the star is moving towards the Earth at velocity v, then in the Earth
>> frame the light is moving at velocity (c - v) relative to the star.
>>
>> This does not contradict the constancy of the speed of light, which
>> relates to the speed as measured by an observer, not the speed of light
>> relative to anything else.
>>
>> Of course, for an observer on the star, the light is observed to depart
>> at speed c.
>>
>> Sylvia.
>
>> If the star is moving towards the Earth at velocity v, then in the Earth
>> frame the light is moving at velocity (c - v) relative to the star.
>
> If this is true, then have a look at this scenario.
>
> S>>>>>>>>>>> c
>
> X
>
> S>>>>>>>>>>> c
>
> In the Earth's frame of reference, light travels at c relative to the Earth. That means the relative velocity between the light and the surface of the star is c-v in the bottom case, where the star is approaching the Earth.
>
> When the same star is moving away from the Earth, we have the relative velocity between light or a particular photon or group of photons moving at c+v relative to the surface of the star.
>
> You can call this closing speed or relative speed.
>
> The fact is that the light is emitted at a different relative velocity relative to the surface of the star depending on the direction it is travelling - towards the Earth or away.

If the observer on Earth seeks to calculate the speed that an observer
on the star sees the light emitted, they must use the velocity
composition formula. If the star is moving towards the Earth at velocity
v, then the result will be (v + c)/[1 + (vc/c^2)], which is just c.

That is, in the frame of a observer on the star, the light is emitted at
velocity c.

The same applies if the star is moving away from the Earth.

Attempts to determine the velocity of emission by simple addition or
subtraction assume that special relativity is wrong, and any
contradictions discovered thereby are in the assumptions made in the
calculation, not in special relativity.

>
> Doesn't this violate the 'First Postulate'?

Clearly, not.

Sylvia.

On Monday, May 1, 2023 at 6:06:43 PM UTC+5, Sylvia Else wrote:
> On 01-May-23 10:05 pm, gehan.am...@gmail.com wrote:
> > On Monday, May 1, 2023 at 2:36:38 PM UTC+5, Sylvia Else wrote:
> >> On 01-May-23 4:10 pm, gehan.am...@gmail.com wrote:
> >>> You are all familiar with the DeSitter double star experiment.
> >>>
> >>> "In 1913, Willem de Sitter argued that if this was true, a star
> >>> orbiting in a double-star system would usually, with regard to us,
> >>> alternate between moving towards us and away from us. Light emitted
> >>> from different parts of the orbital path would travel towards us at
> >>> different speeds. For a nearby star with a small orbital velocity (or
> >>> whose orbital plane was almost perpendicular to our line of view)
> >>> this might merely make the star's orbit seem erratic, but for a
> >>> sufficient combination of orbital speed and distance (and
> >>> inclination), the "fast" light given off during approach would be
> >>> able to catch up with and even overtake "slow" light emitted earlier
> >>> during a recessional part of the star's orbit, and the star would
> >>> present an image that was scrambled and out of sequence. That is,
> >>> Kepler's laws of motion would apparently be violated for a distant
> >>> observer."
> >>>
> >>> Question: If light travels from the star at velocity c relative to
> >>> the Earth, at what relative velocity does it have relative to the
> >>> surface of the star, when viewed from the reference frame in which
> >>> the Earth?
> >> If the star is moving towards the Earth at velocity v, then in the Earth
> >> frame the light is moving at velocity (c - v) relative to the star.
> >>
> >> This does not contradict the constancy of the speed of light, which
> >> relates to the speed as measured by an observer, not the speed of light
> >> relative to anything else.
> >>
> >> Of course, for an observer on the star, the light is observed to depart
> >> at speed c.
> >>
> >> Sylvia.
> >
> >> If the star is moving towards the Earth at velocity v, then in the Earth
> >> frame the light is moving at velocity (c - v) relative to the star.
> >
> > If this is true, then have a look at this scenario.
> >
> > S>>>>>>>>>>> c
> >
> > X
> >
> > S>>>>>>>>>>> c
> >
> > In the Earth's frame of reference, light travels at c relative to the Earth. That means the relative velocity between the light and the surface of the star is c-v in the bottom case, where the star is approaching the Earth.
> >
> > When the same star is moving away from the Earth, we have the relative velocity between light or a particular photon or group of photons moving at c+v relative to the surface of the star.
> >
> > You can call this closing speed or relative speed.
> >
> > The fact is that the light is emitted at a different relative velocity relative to the surface of the star depending on the direction it is travelling - towards the Earth or away.
> If the observer on Earth seeks to calculate the speed that an observer
> on the star sees the light emitted, they must use the velocity
> composition formula. If the star is moving towards the Earth at velocity
> v, then the result will be (v + c)/[1 + (vc/c^2)], which is just c.
>
> That is, in the frame of a observer on the star, the light is emitted at
> velocity c.
>
> The same applies if the star is moving away from the Earth.
>
> Attempts to determine the velocity of emission by simple addition or
> subtraction assume that special relativity is wrong, and any
> contradictions discovered thereby are in the assumptions made in the
> calculation, not in special relativity.
> >
> > Doesn't this violate the 'First Postulate'?
> Clearly, not.
>
> Sylvia.

The initial objection to the ballistic theory of light came from the results of the De Sitter experiment under Newtonian physics.
If the ballistic theory was correct, and this is under Newtonian Physics, then the results would have been different, as the story goes.

According to the ballistic theory, the velocity of emission of light relative to the source is constant. Since this was proved to be wrong by the experiment, as it is said, then the velocity of emission of light is not constant relative to the source. More than that, the velocity of emission of light relative to the surface is different depending on the position of the Earth.

To meet your objections:

> If the observer on Earth seeks to calculate the speed that an observer
> on the star sees the light emitted, they must use the velocity
> composition formula. If the star is moving towards the Earth at velocity
> v, then the result will be (v + c)/[1 + (vc/c^2)], which is just c.

> The same applies if the star is moving away from the Earth.

Are you saying the same formula applies for when the star is moving away from the Earth?
In that case v1 <> v2 so will we get the same result, or do we have to apply a formula with different signs depending
on the direction the start is moving, to or away from the Earth?

I thought the velocity of the source does not matter, v can be taken to be +1, -10000, 0r 0.
Is this the meaning of independent of the 'motion of the source'?

> Attempts to determine the velocity of emission by simple addition or
> subtraction assume that special relativity is wrong, and any
> contradictions discovered thereby are in the assumptions made in the
> calculation, not in special relativity.

I am not attempting to determine the velocity of emission. I am simply asking what the rule for velocity of emission from
the surface of a star is.

Your answer:

On Monday, May 1, 2023 at 6:02:03 PM UTC-7, gehan.am...@gmail.com wrote:
> On Monday, May 1, 2023 at 6:06:43 PM UTC+5, Sylvia Else wrote:
> > On 01-May-23 10:05 pm, gehan.am...@gmail.com wrote:
> > > On Monday, May 1, 2023 at 2:36:38 PM UTC+5, Sylvia Else wrote:
> > >> On 01-May-23 4:10 pm, gehan.am...@gmail.com wrote:
> > >>> You are all familiar with the DeSitter double star experiment.
> > >>>
> > >>> "In 1913, Willem de Sitter argued that if this was true, a star
> > >>> orbiting in a double-star system would usually, with regard to us,
> > >>> alternate between moving towards us and away from us. Light emitted
> > >>> from different parts of the orbital path would travel towards us at
> > >>> different speeds. For a nearby star with a small orbital velocity (or
> > >>> whose orbital plane was almost perpendicular to our line of view)
> > >>> this might merely make the star's orbit seem erratic, but for a
> > >>> sufficient combination of orbital speed and distance (and
> > >>> inclination), the "fast" light given off during approach would be
> > >>> able to catch up with and even overtake "slow" light emitted earlier
> > >>> during a recessional part of the star's orbit, and the star would
> > >>> present an image that was scrambled and out of sequence. That is,
> > >>> Kepler's laws of motion would apparently be violated for a distant
> > >>> observer."
> > >>>
> > >>> Question: If light travels from the star at velocity c relative to
> > >>> the Earth, at what relative velocity does it have relative to the
> > >>> surface of the star, when viewed from the reference frame in which
> > >>> the Earth?
> > >> If the star is moving towards the Earth at velocity v, then in the Earth
> > >> frame the light is moving at velocity (c - v) relative to the star.
> > >>
> > >> This does not contradict the constancy of the speed of light, which
> > >> relates to the speed as measured by an observer, not the speed of light
> > >> relative to anything else.
> > >>
> > >> Of course, for an observer on the star, the light is observed to depart
> > >> at speed c.
> > >>
> > >> Sylvia.
> > >
> > >> If the star is moving towards the Earth at velocity v, then in the Earth
> > >> frame the light is moving at velocity (c - v) relative to the star.
> > >
> > > If this is true, then have a look at this scenario.
> > >
> > > S>>>>>>>>>>> c
> > >
> > > X
> > >
> > > S>>>>>>>>>>> c
> > >
> > > In the Earth's frame of reference, light travels at c relative to the Earth. That means the relative velocity between the light and the surface of the star is c-v in the bottom case, where the star is approaching the Earth.
> > >
> > > When the same star is moving away from the Earth, we have the relative velocity between light or a particular photon or group of photons moving at c+v relative to the surface of the star.
> > >
> > > You can call this closing speed or relative speed.
> > >
> > > The fact is that the light is emitted at a different relative velocity relative to the surface of the star depending on the direction it is travelling - towards the Earth or away.
> > If the observer on Earth seeks to calculate the speed that an observer
> > on the star sees the light emitted, they must use the velocity
> > composition formula. If the star is moving towards the Earth at velocity
> > v, then the result will be (v + c)/[1 + (vc/c^2)], which is just c.
> >
> > That is, in the frame of a observer on the star, the light is emitted at
> > velocity c.
> >
> > The same applies if the star is moving away from the Earth.
> >
> > Attempts to determine the velocity of emission by simple addition or
> > subtraction assume that special relativity is wrong, and any
> > contradictions discovered thereby are in the assumptions made in the
> > calculation, not in special relativity.
> > >
> > > Doesn't this violate the 'First Postulate'?
> > Clearly, not.
> >
> > Sylvia.
> The initial objection to the ballistic theory of light came from the results of the De Sitter experiment under Newtonian physics.
> If the ballistic theory was correct, and this is under Newtonian Physics, then the results would have been different, as the story goes.
>
> According to the ballistic theory, the velocity of emission of light relative to the source is constant. Since this was proved to be wrong by the experiment, as it is said, then the velocity of emission of light is not constant relative to the source. More than that, the velocity of emission of light relative to the surface is different depending on the position of the Earth.
>
> To meet your objections:
> > If the observer on Earth seeks to calculate the speed that an observer
> > on the star sees the light emitted, they must use the velocity
> > composition formula. If the star is moving towards the Earth at velocity
> > v, then the result will be (v + c)/[1 + (vc/c^2)], which is just c.
> > The same applies if the star is moving away from the Earth.
> Are you saying the same formula applies for when the star is moving away from the Earth?
> In that case v1 <> v2 so will we get the same result, or do we have to apply a formula with different signs depending
> on the direction the start is moving, to or away from the Earth?
>
> I thought the velocity of the source does not matter, v can be taken to be +1, -10000, 0r 0.
> Is this the meaning of independent of the 'motion of the source'?
> > Attempts to determine the velocity of emission by simple addition or
> > subtraction assume that special relativity is wrong, and any
> > contradictions discovered thereby are in the assumptions made in the
> > calculation, not in special relativity.
> I am not attempting to determine the velocity of emission. I am simply asking what the rule for velocity of emission from
> the surface of a star is.
>
> Your answer:
gahan, may I offer some advice when arguing relativity. Let me give you two sure-fire, tried-and-true debating strategies to permanently silence the Sylvias and Paul B. Andersons of this forum.

1) Subject SR to planetary motion. Planetary motion falsified SR almost immediately after it was propounded by Einstein. Einstein's response to this falsification was to jettison gravity as the reason for planetary motion and invent a new reason for planetary motion; namely, GR. But GR doesn't rescue SR from falsification. Please check out to the Eponymous Dilemma in this regard.

2) Never argue the algebra of SR. It is consistent. But only as long as the relativists get to define the premises/terms used in the LTs. The premise/term you want to challenge is v. The v in the LTs is assumed to have the same value for both frames, S and S'. This is the third [ and hidden ] postulate of special relativity. It is a postulate because it does not follow from any logical considerations. Think about it. Every variable in the LTs has a primmed and unprimed version, except v. There's an x and an x'. There's a t and a t'. There is a v, but there is no v'. c is a constant of course.

Follow Sylvia's advice and assume that SR is true. Therefor the LTs are assumed true also. Then it necessarily follows that the coordinate velocity v' must be numerically equivalent to the proper relative velocity v. v = v'. Now construct v' by using the LTs: find the ratio of x' to t' which is v'. Now substitute v' back into the LTs where v occurs. The results make a laughingstock of SR. Reductio ad absurdum.

On Monday, May 1, 2023 at 6:02:03 PM UTC-7, gehan.am...@gmail.com wrote:
> The initial objection to the ballistic theory of light came from the results of
> the De Sitter experiment under Newtonian physics.

Not at all. The initial objection to the ballistic theory was the interference effects of light, such as diffraction, which Newton humself acknowledged could not be readily explained in terms of a ballistic theory. This was shown conclusively in Young's experiment around 1800, demonstrating that light exhibits wavelike interference. For the next hundred years, the wave conception of light was the most useful for making progress, accounting for all phenomena until things like black-body radiation and the photo-electric effect (which, when combined with special relativity, led to quantum electrodynamics).

> the velocity of emission of light is not constant relative to the source..

No, in terms of the standard inertial coordinates in which any given object (including a source of light) is at rest, any pulse of light in vacuum propagates at the speed c. This applies whether the light was emitted by that object or not. Remember, velocities can't be specified "relative to objects", they can only be specified in terms of a given system of coordinates. In popular accounts some writers conflate an object with the standard inertial coordinates in which that object is at rest, and when they talk about velocities relative to that object they mean velocities in terms of that coordinate system. Most people are not too confused by this sloppy way of talking, but certain individuals are led astray by it, so it's important to be clear about it.

> the velocity of emission of light relative to the surface is different depending on the
> position of the Earth.

No. Again, every pulse of light propagates in vacuum at the speed c in terms of any and every standard system of inertial coordinates, including the system in which the earth is at rest at any given time.

> Is this the meaning of independent of the 'motion of the source'?

The meaning is that if one bulb is moving away from you and another is moving toward you, and at the moment when they pass each other they each emit a flash of light, the flashed reach you simultaneously. This is true even though one bulb was moving away and the other moving toward you. Again, light propagates in vacuum at the speed c in terms of any standard system of inertial coordinates. Note that you can't possibly understand this until you learn what a standard system of inertial coordinates is, which you will never learn, because you don't want to.

> I am simply asking what the rule for velocity of emission from
> the surface of a star is.

Again, the velocity of a pulse of light in vacuum is c in terms of any standard system of inertial coordinates. This is telling you something about light, and something about standard systems of inertial coordinates. But it would help you immensely to forget about light and just consider two material particles A and B approaching you from the left and the right respectively at speed 0.75c in terms of the standard inertial coordinates in which you are at rest. The difference between their speeds in terms of those coordinates is 1.5c. However, in terms of the standard inertial coordinate system in which A is at rest, B is moving at speed (24/25)c. Do you understand this?

On 5/1/23 1:10 AM, gehan.am...@gmail.com wrote:
> [the DeSitter double star experiment. everything in vacuum]
> Question: If light travels from the star at velocity c relative to
> the Earth, at what relative velocity does it have relative to the
> surface of the star, when viewed from the reference frame in which
> the Earth?

Viewed from the reference frame of the earth (presumed to be inertial,
to keep the context SR), one cannot observe its speed "relative to the
surface of the star", because one can only observe relative to the
reference frame one is using.

One can CALCULATE its speed relative to the (locally) inertial frame in
which the surface of the star is at rest. One must use the Lorentz
composition of velocities, which of course obtains c.

One can CALCULATE the closing speed of the light and star surface, using
the reference frame of the earth. With the star moving with speed v
relative to that frame, the closing speed is c-v when the star is
approaching earth, and c+v when receding.

Regardless of your desires, the closing speed between two objects
measured in a third frame is not at all the same as the speed of one
object relative to the other (rest frame). These are QUITE DIFFERENT
concepts. (There is a mathematical quirk that when one uses Galilean
relativity they have the same numerical value; this is only an
approximation in the world we inhabit.)

Tom Roberts

On Tuesday, 2 May 2023 at 05:36:31 UTC+2, Tom Roberts wrote:
> On 5/1/23 1:10 AM, gehan.am...@gmail.com wrote:
> > [the DeSitter double star experiment. everything in vacuum]
> > Question: If light travels from the star at velocity c relative to
> > the Earth, at what relative velocity does it have relative to the
> > surface of the star, when viewed from the reference frame in which
> > the Earth?
> Viewed from the reference frame of the earth (presumed to be inertial,
> to keep the context SR), one cannot observe its speed "relative to the
> surface of the star", because one can only observe relative to the
> reference frame one is using.
>
> One can CALCULATE its speed relative to the (locally) inertial frame in
> which the surface of the star is at rest. One must use the Lorentz
> composition of velocities, which of course obtains c.
>
> One can CALCULATE the closing speed of the light and star surface, using
> the reference frame of the earth. With the star moving with speed v
> relative to that frame, the closing speed is c-v when the star is
> approaching earth, and c+v when receding.
>
> Regardless of your desires, the closing speed between two objects
> measured in a third frame is not at all the same as the speed of one
> object relative to the other (rest frame). These are QUITE DIFFERENT
> concepts. (There is a mathematical quirk that when one uses Galilean
> relativity they have the same numerical value; this is only an
> approximation in the world we inhabit.)

In the world you inhabit, in the world of
your insane delusions, where everyone is
FORCED to obey your idiot guru and yourself.

On Tuesday, May 2, 2023 at 6:19:54 AM UTC+5, patdolan wrote:
> On Monday, May 1, 2023 at 6:02:03 PM UTC-7, gehan.am...@gmail.com wrote:
> > On Monday, May 1, 2023 at 6:06:43 PM UTC+5, Sylvia Else wrote:
> > > On 01-May-23 10:05 pm, gehan.am...@gmail.com wrote:
> > > > On Monday, May 1, 2023 at 2:36:38 PM UTC+5, Sylvia Else wrote:
> > > >> On 01-May-23 4:10 pm, gehan.am...@gmail.com wrote:
> > > >>> You are all familiar with the DeSitter double star experiment.
> > > >>>
> > > >>> "In 1913, Willem de Sitter argued that if this was true, a star
> > > >>> orbiting in a double-star system would usually, with regard to us,
> > > >>> alternate between moving towards us and away from us. Light emitted
> > > >>> from different parts of the orbital path would travel towards us at
> > > >>> different speeds. For a nearby star with a small orbital velocity (or
> > > >>> whose orbital plane was almost perpendicular to our line of view)
> > > >>> this might merely make the star's orbit seem erratic, but for a
> > > >>> sufficient combination of orbital speed and distance (and
> > > >>> inclination), the "fast" light given off during approach would be
> > > >>> able to catch up with and even overtake "slow" light emitted earlier
> > > >>> during a recessional part of the star's orbit, and the star would
> > > >>> present an image that was scrambled and out of sequence. That is,
> > > >>> Kepler's laws of motion would apparently be violated for a distant
> > > >>> observer."
> > > >>>
> > > >>> Question: If light travels from the star at velocity c relative to
> > > >>> the Earth, at what relative velocity does it have relative to the
> > > >>> surface of the star, when viewed from the reference frame in which
> > > >>> the Earth?
> > > >> If the star is moving towards the Earth at velocity v, then in the Earth
> > > >> frame the light is moving at velocity (c - v) relative to the star..
> > > >>
> > > >> This does not contradict the constancy of the speed of light, which
> > > >> relates to the speed as measured by an observer, not the speed of light
> > > >> relative to anything else.
> > > >>
> > > >> Of course, for an observer on the star, the light is observed to depart
> > > >> at speed c.
> > > >>
> > > >> Sylvia.
> > > >
> > > >> If the star is moving towards the Earth at velocity v, then in the Earth
> > > >> frame the light is moving at velocity (c - v) relative to the star..
> > > >
> > > > If this is true, then have a look at this scenario.
> > > >
> > > > S>>>>>>>>>>> c
> > > >
> > > > X
> > > >
> > > > S>>>>>>>>>>> c
> > > >
> > > > In the Earth's frame of reference, light travels at c relative to the Earth. That means the relative velocity between the light and the surface of the star is c-v in the bottom case, where the star is approaching the Earth.
> > > >
> > > > When the same star is moving away from the Earth, we have the relative velocity between light or a particular photon or group of photons moving at c+v relative to the surface of the star.
> > > >
> > > > You can call this closing speed or relative speed.
> > > >
> > > > The fact is that the light is emitted at a different relative velocity relative to the surface of the star depending on the direction it is travelling - towards the Earth or away.
> > > If the observer on Earth seeks to calculate the speed that an observer
> > > on the star sees the light emitted, they must use the velocity
> > > composition formula. If the star is moving towards the Earth at velocity
> > > v, then the result will be (v + c)/[1 + (vc/c^2)], which is just c.
> > >
> > > That is, in the frame of a observer on the star, the light is emitted at
> > > velocity c.
> > >
> > > The same applies if the star is moving away from the Earth.
> > >
> > > Attempts to determine the velocity of emission by simple addition or
> > > subtraction assume that special relativity is wrong, and any
> > > contradictions discovered thereby are in the assumptions made in the
> > > calculation, not in special relativity.
> > > >
> > > > Doesn't this violate the 'First Postulate'?
> > > Clearly, not.
> > >
> > > Sylvia.
> > The initial objection to the ballistic theory of light came from the results of the De Sitter experiment under Newtonian physics.
> > If the ballistic theory was correct, and this is under Newtonian Physics, then the results would have been different, as the story goes.
> >
> > According to the ballistic theory, the velocity of emission of light relative to the source is constant. Since this was proved to be wrong by the experiment, as it is said, then the velocity of emission of light is not constant relative to the source. More than that, the velocity of emission of light relative to the surface is different depending on the position of the Earth.
> >
> > To meet your objections:
> > > If the observer on Earth seeks to calculate the speed that an observer
> > > on the star sees the light emitted, they must use the velocity
> > > composition formula. If the star is moving towards the Earth at velocity
> > > v, then the result will be (v + c)/[1 + (vc/c^2)], which is just c.
> > > The same applies if the star is moving away from the Earth.
> > Are you saying the same formula applies for when the star is moving away from the Earth?
> > In that case v1 <> v2 so will we get the same result, or do we have to apply a formula with different signs depending
> > on the direction the start is moving, to or away from the Earth?
> >
> > I thought the velocity of the source does not matter, v can be taken to be +1, -10000, 0r 0.
> > Is this the meaning of independent of the 'motion of the source'?
> > > Attempts to determine the velocity of emission by simple addition or
> > > subtraction assume that special relativity is wrong, and any
> > > contradictions discovered thereby are in the assumptions made in the
> > > calculation, not in special relativity.
> > I am not attempting to determine the velocity of emission. I am simply asking what the rule for velocity of emission from
> > the surface of a star is.
> >
> > Your answer:
> gahan, may I offer some advice when arguing relativity. Let me give you two sure-fire, tried-and-true debating strategies to permanently silence the Sylvias and Paul B. Andersons of this forum.
>
> 1) Subject SR to planetary motion. Planetary motion falsified SR almost immediately after it was propounded by Einstein. Einstein's response to this falsification was to jettison gravity as the reason for planetary motion and invent a new reason for planetary motion; namely, GR. But GR doesn't rescue SR from falsification. Please check out to the Eponymous Dilemma in this regard.
>
> 2) Never argue the algebra of SR. It is consistent. But only as long as the relativists get to define the premises/terms used in the LTs. The premise/term you want to challenge is v. The v in the LTs is assumed to have the same value for both frames, S and S'. This is the third [ and hidden ] postulate of special relativity. It is a postulate because it does not follow from any logical considerations. Think about it. Every variable in the LTs has a primmed and unprimed version, except v. There's an x and an x'. There's a t and a t'. There is a v, but there is no v'. c is a constant of course..
>
> Follow Sylvia's advice and assume that SR is true. Therefor the LTs are assumed true also. Then it necessarily follows that the coordinate velocity v' must be numerically equivalent to the proper relative velocity v. v = v'. Now construct v' by using the LTs: find the ratio of x' to t' which is v'. Now substitute v' back into the LTs where v occurs. The results make a laughingstock of SR. Reductio ad absurdum.

Thank you, will look at that. The premises the algebra is based on are the key, yes.

On Tuesday, May 2, 2023 at 7:02:58 AM UTC+5, Trevor Lange wrote:
> On Monday, May 1, 2023 at 6:02:03 PM UTC-7, gehan.am...@gmail.com wrote:
> > The initial objection to the ballistic theory of light came from the results of
> > the De Sitter experiment under Newtonian physics.
>........
> > the velocity of emission of light relative to the surface is different depending on the
> > position of the Earth.
> No. Again, every pulse of light propagates in vacuum at the speed c in terms of any and every standard system of inertial coordinates, including the system in which the earth is at rest at any given time.
> > Is this the meaning of independent of the 'motion of the source'?
> The meaning is that if one bulb is moving away from you and another is moving toward you, and at the moment when they pass each other they each emit a flash of light, the flashed reach you simultaneously. This is true even though one bulb was moving away and the other moving toward you. Again, light propagates in vacuum at the speed c in terms of any standard system of inertial coordinates. Note that you can't possibly understand this until you learn what a standard system of inertial coordinates is, which you will never learn, because you don't want to.

This is an excellent example. In the above example, if the light bulbs are moving in an Aether stationary to me, that would make perfect sense.

Let me try once again with a simple example. A child on a carousel is throwing a ball towards me when he is directly approaching me and when he is receding from me in the opposite direction, is throwing a ball to another person.

It is indisputable in Newtonian physics that the child will have to use more force to throw the ball towards the person he is receding from, in order to make the velocity of the ball the same for each of the recipients.

The receding star is throwing out light just as the approaching star is. Why is the light from the receding star thrown out faster relative to the surface of the star?

Do you at least see the problem?

> > I am simply asking what the rule for velocity of emission from
> > the surface of a star is.
> Again, the velocity of a pulse of light in vacuum is c in terms of any standard system of inertial coordinates. This is telling you something about light, and something about standard systems of inertial coordinates. But it would help you immensely to forget about light and just consider two material particles A and B approaching you from the left and the right respectively at speed 0.75c in terms of the standard inertial coordinates in which you are at rest. The difference between their speeds in terms of those coordinates is 1.5c. However, in terms of the standard inertial coordinate system in which A is at rest, B is moving at speed (24/25)c. Do you understand this?

No, this I do not understand. Without an absolute frame of reference, without an absolute rest frame, it is impossible to put limits on how fast objects can move relative to each other.

On 02-May-23 8:52 pm, gehan.am...@gmail.com wrote:
> On Tuesday, May 2, 2023 at 7:02:58 AM UTC+5, Trevor Lange wrote:
>> On Monday, May 1, 2023 at 6:02:03 PM UTC-7, gehan.am...@gmail.com wrote:
>>> The initial objection to the ballistic theory of light came from the results of
>>> the De Sitter experiment under Newtonian physics.
>> ........
>>> the velocity of emission of light relative to the surface is different depending on the
>>> position of the Earth.
>> No. Again, every pulse of light propagates in vacuum at the speed c in terms of any and every standard system of inertial coordinates, including the system in which the earth is at rest at any given time.
>>> Is this the meaning of independent of the 'motion of the source'?
>> The meaning is that if one bulb is moving away from you and another is moving toward you, and at the moment when they pass each other they each emit a flash of light, the flashed reach you simultaneously. This is true even though one bulb was moving away and the other moving toward you. Again, light propagates in vacuum at the speed c in terms of any standard system of inertial coordinates. Note that you can't possibly understand this until you learn what a standard system of inertial coordinates is, which you will never learn, because you don't want to.
>
> This is an excellent example. In the above example, if the light bulbs are moving in an Aether stationary to me, that would make perfect sense.
>
> Let me try once again with a simple example. A child on a carousel is throwing a ball towards me when he is directly approaching me and when he is receding from me in the opposite direction, is throwing a ball to another person.
>
> It is indisputable in Newtonian physics that the child will have to use more force to throw the ball towards the person he is receding from, in order to make the velocity of the ball the same for each of the recipients.
>
> The receding star is throwing out light just as the approaching star is. Why is the light from the receding star thrown out faster relative to the surface of the star?

For an observer on the surface of the star, the light leaves in all
cases at speed c.

For observers moving relative to the star, subtracting the velocity of
the star from the velocity of the light the star emits is physically
meaningless. It gives you a quantity, but so does averaging the numbers
in a phone book.

>
> Do you at least see the problem?

There is no problem.
>
>>> I am simply asking what the rule for velocity of emission from
>>> the surface of a star is.
>> Again, the velocity of a pulse of light in vacuum is c in terms of any standard system of inertial coordinates. This is telling you something about light, and something about standard systems of inertial coordinates. But it would help you immensely to forget about light and just consider two material particles A and B approaching you from the left and the right respectively at speed 0.75c in terms of the standard inertial coordinates in which you are at rest. The difference between their speeds in terms of those coordinates is 1.5c. However, in terms of the standard inertial coordinate system in which A is at rest, B is moving at speed (24/25)c. Do you understand this?
>
> No, this I do not understand. Without an absolute frame of reference, without an absolute rest frame, it is impossible to put limits on how fast objects can move relative to each other.

Only if you're assuming that special relativity is wrong.

If you apply relativity, there is no rule preventing a speed faster than
light, it's just that you'll never obtain one.

Sylvia.

On Tuesday, May 2, 2023 at 3:52:11 AM UTC-7, gehan.am...@gmail.com wrote:
> In the above example, if the light bulbs are moving in an Aether stationary to me,
> that would make perfect sense.

Right, and it would also make perfect sense if all the laws of physics are (locally) Lorentz invariant... which they are.
> It is indisputable in Newtonian physics that the child will have to use more
> force to throw the ball towards the person he is receding from, in order to
> make the velocity of the ball the same for each of the recipients.

Right, and also in special relativity, although the precise difference in the required force is very slightly different than in Newtonian physics, because Newton's laws are not Lorentz invariant, whereas the actual laws of mechanics are Lorentz invariant.

> The receding star is throwing out light just as the approaching star is. Why is
> the light from the receding star thrown out faster relative to the surface of the star?

You're not paying attention. Again, in terms of the standard inertial coordinates in which the star is at rest, the light (in vacuum) propagates at the speed c. Also, in terms of the standard inertial coordinates in which the earth is at rest, the light propagates at c. It is impossible for both of these things to be true... unless those two systems of coordinates are related to each other in a very specific way. This way is known as a Lorentz transformation. Standard inertial coordinates are, by definition, coordinates in terms of which the laws of physics take the same isotropic and homogeneous form, and the fact that the laws are Lorentz invariant (check for yourself) signifies that they take the same form in terms of coordinate systems related by Lorentz transformations.

> Do you at least see the problem?

Well, of course I see "the problem"...everyone saw the problem... this is why Einstein said the two things are seemingly irreconcilable, and then he explained the one and only way in which they can be reconciled.

> Without an absolute frame of reference, it is impossible to put limits on how
> fast objects can move relative to each other.

Be careful... the limits are on the speed of any mass energy or information in terms of any standard system of inertial coordinates. The key physical fact you are missing (and Newton was missing) is that every quantity E of localized energy has inertia E/c^2. From this, all of special relativity follows.

On Tuesday, May 2, 2023 at 6:52:35 PM UTC+5, Trevor Lange wrote:
> On Tuesday, May 2, 2023 at 3:52:11 AM UTC-7, gehan.am...@gmail.com wrote:
> > In the above example, if the light bulbs are moving in an Aether stationary to me,
> > that would make perfect sense.
> Right, and it would also make perfect sense if all the laws of physics are (locally) Lorentz invariant... which they are.
> > It is indisputable in Newtonian physics that the child will have to use more
> > force to throw the ball towards the person he is receding from, in order to
> > make the velocity of the ball the same for each of the recipients.
> Right, and also in special relativity, although the precise difference in the required force is very slightly different than in Newtonian physics, because Newton's laws are not Lorentz invariant, whereas the actual laws of mechanics are Lorentz invariant.
> > The receding star is throwing out light just as the approaching star is.. Why is
> > the light from the receding star thrown out faster relative to the surface of the star?
> You're not paying attention. Again, in terms of the standard inertial coordinates in which the star is at rest, the light (in vacuum) propagates at the speed c. Also, in terms of the standard inertial coordinates in which the earth is at rest, the light propagates at c. It is impossible for both of these things to be true... unless those two systems of coordinates are related to each other in a very specific way. This way is known as a Lorentz transformation. Standard inertial coordinates are, by definition, coordinates in terms of which the laws of physics take the same isotropic and homogeneous form, and the fact that the laws are Lorentz invariant (check for yourself) signifies that they take the same form in terms of coordinate systems related by Lorentz transformations.
> > Do you at least see the problem?
> Well, of course I see "the problem"...everyone saw the problem... this is why Einstein said the two things are seemingly irreconcilable, and then he explained the one and only way in which they can be reconciled.
>

Well at least you accept that there is a 'problem'.

When viewed from above the double star system looks this picture:

https://en.wikipedia.org/wiki/De_Sitter_double_star_experiment#/media/File:SitterKonstanz.png

(In the diagram the c-v and c+v labels are reversed, since it illustrates the ballistic emission of light)

Let me ask it this way: does the application of the Lorentz transformation in both cases give a constant velocity of light towards the Earth?

Is there a reason behind the asymmetry in the system clearly viewed in this image, where on one side of the orbit the star emits light at c-v relative to the surface and c+v to the surface of the star?

Would the wave fronts from the moving stars form a sphere or an ellipse?

https://study.com/learn/lesson/red-shift-theory-explanation.html

On 03-May-23 9:48 pm, gehan.am...@gmail.com wrote:
> On Tuesday, May 2, 2023 at 6:52:35 PM UTC+5, Trevor Lange wrote:
>> On Tuesday, May 2, 2023 at 3:52:11 AM UTC-7, gehan.am...@gmail.com wrote:
>>> In the above example, if the light bulbs are moving in an Aether stationary to me,
>>> that would make perfect sense.
>> Right, and it would also make perfect sense if all the laws of physics are (locally) Lorentz invariant... which they are.
>>> It is indisputable in Newtonian physics that the child will have to use more
>>> force to throw the ball towards the person he is receding from, in order to
>>> make the velocity of the ball the same for each of the recipients.
>> Right, and also in special relativity, although the precise difference in the required force is very slightly different than in Newtonian physics, because Newton's laws are not Lorentz invariant, whereas the actual laws of mechanics are Lorentz invariant.
>>> The receding star is throwing out light just as the approaching star is. Why is
>>> the light from the receding star thrown out faster relative to the surface of the star?
>> You're not paying attention. Again, in terms of the standard inertial coordinates in which the star is at rest, the light (in vacuum) propagates at the speed c. Also, in terms of the standard inertial coordinates in which the earth is at rest, the light propagates at c. It is impossible for both of these things to be true... unless those two systems of coordinates are related to each other in a very specific way. This way is known as a Lorentz transformation. Standard inertial coordinates are, by definition, coordinates in terms of which the laws of physics take the same isotropic and homogeneous form, and the fact that the laws are Lorentz invariant (check for yourself) signifies that they take the same form in terms of coordinate systems related by Lorentz transformations.
>>> Do you at least see the problem?
>> Well, of course I see "the problem"...everyone saw the problem... this is why Einstein said the two things are seemingly irreconcilable, and then he explained the one and only way in which they can be reconciled.
>>
>
> Well at least you accept that there is a 'problem'.

Not on my reading.

>
> When viewed from above the double star system looks this picture:
>
> https://en.wikipedia.org/wiki/De_Sitter_double_star_experiment#/media/File:SitterKonstanz.png
>
> (In the diagram the c-v and c+v labels are reversed, since it illustrates the ballistic emission of light)
>
> Let me ask it this way: does the application of the Lorentz transformation in both cases give a constant velocity of light towards the Earth?

Yes.

>
> Is there a reason behind the asymmetry in the system clearly viewed in this image, where on one side of the orbit the star emits light at c-v relative to the surface and c+v to the surface of the star?

You realise that the labels are what would happen if emission theory
were correct. Since it's not correct, the labels are wrong, and there is
no asymmetry to explain.

>
> Would the wave fronts from the moving stars form a sphere or an ellipse?

A sphere - both in the frame of the star, and in the frame of an
observer moving relative to the star. There are not the same spheres,
but they are nevertheless spheres in the respective frames of reference.

Sylvia.

On Wednesday, May 3, 2023 at 5:15:56 PM UTC+5, Sylvia Else wrote:
> On 03-May-23 9:48 pm, gehan.am...@gmail.com wrote:
> > On Tuesday, May 2, 2023 at 6:52:35 PM UTC+5, Trevor Lange wrote:
> >> On Tuesday, May 2, 2023 at 3:52:11 AM UTC-7, gehan.am...@gmail..com wrote:
> >>> In the above example, if the light bulbs are moving in an Aether stationary to me,
> >>> that would make perfect sense.
> >> Right, and it would also make perfect sense if all the laws of physics are (locally) Lorentz invariant... which they are.
> >>> It is indisputable in Newtonian physics that the child will have to use more
> >>> force to throw the ball towards the person he is receding from, in order to
> >>> make the velocity of the ball the same for each of the recipients.
> >> Right, and also in special relativity, although the precise difference in the required force is very slightly different than in Newtonian physics, because Newton's laws are not Lorentz invariant, whereas the actual laws of mechanics are Lorentz invariant.
> >>> The receding star is throwing out light just as the approaching star is. Why is
> >>> the light from the receding star thrown out faster relative to the surface of the star?
> >> You're not paying attention. Again, in terms of the standard inertial coordinates in which the star is at rest, the light (in vacuum) propagates at the speed c. Also, in terms of the standard inertial coordinates in which the earth is at rest, the light propagates at c. It is impossible for both of these things to be true... unless those two systems of coordinates are related to each other in a very specific way. This way is known as a Lorentz transformation. Standard inertial coordinates are, by definition, coordinates in terms of which the laws of physics take the same isotropic and homogeneous form, and the fact that the laws are Lorentz invariant (check for yourself) signifies that they take the same form in terms of coordinate systems related by Lorentz transformations.
> >>> Do you at least see the problem?
> >> Well, of course I see "the problem"...everyone saw the problem... this is why Einstein said the two things are seemingly irreconcilable, and then he explained the one and only way in which they can be reconciled.
> >>
> >
> > Well at least you accept that there is a 'problem'.
> Not on my reading.
> >
> > When viewed from above the double star system looks this picture:
> >
> > https://en.wikipedia.org/wiki/De_Sitter_double_star_experiment#/media/File:SitterKonstanz.png
> >
> > (In the diagram the c-v and c+v labels are reversed, since it illustrates the ballistic emission of light)
> >
> > Let me ask it this way: does the application of the Lorentz transformation in both cases give a constant velocity of light towards the Earth?
> Yes.

OK so that is settled.

> >
> > Is there a reason behind the asymmetry in the system clearly viewed in this image, where on one side of the orbit the star emits light at c-v relative to the surface and c+v to the surface of the star?
> You realise that the labels are what would happen if emission theory
> were correct. Since it's not correct, the labels are wrong, and there is
> no asymmetry to explain.

As I mentioned, the labels are for emission theory. However, if we assume a constant velocity c towards the Earth, the labels are reversed, so one one side it would show a velocity between the surface of the star and the emitted light for example, for the approaching star. Either way, there is assymetry.
> >
> > Would the wave fronts from the moving stars form a sphere or an ellipse?
> A sphere - both in the frame of the star, and in the frame of an
> observer moving relative to the star. There are not the same spheres,
> but they are nevertheless spheres in the respective frames of reference.
>
> Sylvia.

Wow. This is relativity then.

A sphere in the frame of the star, yes.

If the star is moving relative to the observer, or the observer is moving relative to the star, another sphere consisting of a different set of photons in each case?

A spherical wavefront of photons a sphere created out of nothing, where is the energy for that?

On 03-May-23 10:28 pm, gehan.am...@gmail.com wrote:
> On Wednesday, May 3, 2023 at 5:15:56 PM UTC+5, Sylvia Else wrote:
>> On 03-May-23 9:48 pm, gehan.am...@gmail.com wrote:
>>> On Tuesday, May 2, 2023 at 6:52:35 PM UTC+5, Trevor Lange wrote:
>>>> On Tuesday, May 2, 2023 at 3:52:11 AM UTC-7, gehan.am...@gmail.com wrote:
>>>>> In the above example, if the light bulbs are moving in an Aether stationary to me,
>>>>> that would make perfect sense.
>>>> Right, and it would also make perfect sense if all the laws of physics are (locally) Lorentz invariant... which they are.
>>>>> It is indisputable in Newtonian physics that the child will have to use more
>>>>> force to throw the ball towards the person he is receding from, in order to
>>>>> make the velocity of the ball the same for each of the recipients.
>>>> Right, and also in special relativity, although the precise difference in the required force is very slightly different than in Newtonian physics, because Newton's laws are not Lorentz invariant, whereas the actual laws of mechanics are Lorentz invariant.
>>>>> The receding star is throwing out light just as the approaching star is. Why is
>>>>> the light from the receding star thrown out faster relative to the surface of the star?
>>>> You're not paying attention. Again, in terms of the standard inertial coordinates in which the star is at rest, the light (in vacuum) propagates at the speed c. Also, in terms of the standard inertial coordinates in which the earth is at rest, the light propagates at c. It is impossible for both of these things to be true... unless those two systems of coordinates are related to each other in a very specific way. This way is known as a Lorentz transformation. Standard inertial coordinates are, by definition, coordinates in terms of which the laws of physics take the same isotropic and homogeneous form, and the fact that the laws are Lorentz invariant (check for yourself) signifies that they take the same form in terms of coordinate systems related by Lorentz transformations.
>>>>> Do you at least see the problem?
>>>> Well, of course I see "the problem"...everyone saw the problem... this is why Einstein said the two things are seemingly irreconcilable, and then he explained the one and only way in which they can be reconciled.
>>>>
>>>
>>> Well at least you accept that there is a 'problem'.
>> Not on my reading.
>>>
>>> When viewed from above the double star system looks this picture:
>>>
>>> https://en.wikipedia.org/wiki/De_Sitter_double_star_experiment#/media/File:SitterKonstanz.png
>>>
>>> (In the diagram the c-v and c+v labels are reversed, since it illustrates the ballistic emission of light)
>>>
>>> Let me ask it this way: does the application of the Lorentz transformation in both cases give a constant velocity of light towards the Earth?
>> Yes.
>
> OK so that is settled.
>
>>>
>>> Is there a reason behind the asymmetry in the system clearly viewed in this image, where on one side of the orbit the star emits light at c-v relative to the surface and c+v to the surface of the star?
>> You realise that the labels are what would happen if emission theory
>> were correct. Since it's not correct, the labels are wrong, and there is
>> no asymmetry to explain.
>
> As I mentioned, the labels are for emission theory. However, if we assume a constant velocity c towards the Earth, the labels are reversed, so one one side it would show a velocity between the surface of the star and the emitted light for example, for the approaching star. Either way, there is assymetry.

The only asymmetry is in the incorrect emission model. So why concern
yourself with it?

>>>
>>> Would the wave fronts from the moving stars form a sphere or an ellipse?
>> A sphere - both in the frame of the star, and in the frame of an
>> observer moving relative to the star. There are not the same spheres,
>> but they are nevertheless spheres in the respective frames of reference.
>>
>> Sylvia.
>
> Wow. This is relativity then.
>
> A sphere in the frame of the star, yes.
>
> If the star is moving relative to the observer, or the observer is moving relative to the star, another sphere consisting of a different set of photons in each case?
>
> A spherical wavefront of photons a sphere created out of nothing, where is the energy for that?

They're formed from the same photons, but the relativity of simultaneity
comes into play, so for example, where and when a photon forms part of
the wavefront one second after the emission is frame dependent.

Sylvia.

On Wednesday, May 3, 2023 at 4:48:58 AM UTC-7, gehan.am...@gmail.com wrote:
On Wednesday, May 3, 2023 at 5:28:39 AM UTC-7, gehan.am...@gmail.com wrote:
> In the above example, if the light bulbs are moving in an Aether stationary to me,
> that would make perfect sense.

Right, and it would also make perfect sense if all the laws of physics are (locally) Lorentz invariant.... which they are. Other phenomena are inconsistent with a stationary aether, but all the phenonena are consistent with local Lorentz invariance.

> It is indisputable in Newtonian physics that the child will have to use more
> force to throw the ball towards the person he is receding from, in order to
> make the velocity of the ball the same for each of the recipients.

Right, and also in special relativity, although the precise difference in the required force is slightly different than in Newtonian physics, because Newton's laws aren't Lorentz invariant, whereas the actual laws of mechanics are Lorentz invariant.

> The receding star is throwing out light just as the approaching star is. Why is
> the light from the receding star thrown out faster relative to the surface of the star?

Again, in terms of the standard inertial coordinates in which the star is at rest, the light (in vacuum) propagates at the speed c. Also, in terms of the standard inertial coordinates in which the earth is at rest, the light propagates at c. It is impossible for both of these things to be true... unless those two systems of coordinates are related to each other in a very specific way. This way is known as a Lorentz transformation. Standard inertial coordinates are, by definition, coordinates in terms of which the laws of physics take the same simple isotropic and homogeneous form, and the fact that the laws are Lorentz invariant (check for yourself) signifies that they take the same form in terms of coordinate systems related by Lorentz transformations.

> Do you at least see the problem?

Well, of course I see "the problem", i.e., the fact that the phenomena are not consistent with Galilean invariance...everyone saw the problem... this is why Einstein said the two things are seemingly irreconcilable, and then he explained the one and only way in which they can be reconciled, i.e., Lorentz invariance.

> In a double star system, does the application of the Lorentz transformation
> in both cases give a constant velocity of light towards the Earth?

In terms of each of the different inertial coordinate systems in which each individual star, the center of mass of the stars, and the earth, are respectively at rest, the speed of light is c, and these coordinate systems are related by Lorentz transformations. Do you understand this?

> Is there a reason behind the asymmetry in the system clearly viewed in this
> image, where on one side of the orbit the star emits light at c-v relative to the
> surface and c+v to the surface of the star?

Your sentence doesn't parse in English, but note that the drawing is depicting a misunderstanding of the phenomena, i.e., it is illustrating a hypothesis that is shown to be false. The correct drawing of the situation, consistent with the actual phenomena, shows that the speed of light is c in terms of each of the inertial coordinate systems as described above. Do you understand this?

> Would the wave fronts from the moving stars form a sphere or an ellipse?

Each pulse of light (emitted in all directions) emanates outward as an expanding spheres in terms of each of the standard inertial coordinate systems. Do you understand this?

On Wednesday, May 3, 2023 at 6:48:58 AM UTC-5, gehan.am...@gmail.com wrote:

> When viewed from above the double star system looks this picture:
>
> https://en.wikipedia.org/wiki/De_Sitter_double_star_experiment#/media/File:SitterKonstanz.png
>
> (In the diagram the c-v and c+v labels are reversed, since it illustrates the ballistic emission of light)

The author of your linked image was Willem de Sitter.

You had best reconsider your statement that the c-v and c+v
labels are reversed.

You are clearly misinterpreting the image

On Wednesday, May 3, 2023 at 6:00:19 PM UTC+5, Sylvia Else wrote:
> On 03-May-23 10:28 pm, gehan.am...@gmail.com wrote:
> > On Wednesday, May 3, 2023 at 5:15:56 PM UTC+5, Sylvia Else wrote:
> >> On 03-May-23 9:48 pm, gehan.am...@gmail.com wrote:
> >>> On Tuesday, May 2, 2023 at 6:52:35 PM UTC+5, Trevor Lange wrote:
> >>>> On Tuesday, May 2, 2023 at 3:52:11 AM UTC-7, gehan.am...@gmail.com wrote:
> >>>>> In the above example, if the light bulbs are moving in an Aether stationary to me,
> >>>>> that would make perfect sense.
> >>>> Right, and it would also make perfect sense if all the laws of physics are (locally) Lorentz invariant... which they are.
> >>>>> It is indisputable in Newtonian physics that the child will have to use more
> >>>>> force to throw the ball towards the person he is receding from, in order to
> >>>>> make the velocity of the ball the same for each of the recipients.
> >>>> Right, and also in special relativity, although the precise difference in the required force is very slightly different than in Newtonian physics, because Newton's laws are not Lorentz invariant, whereas the actual laws of mechanics are Lorentz invariant.
> >>>>> The receding star is throwing out light just as the approaching star is. Why is
> >>>>> the light from the receding star thrown out faster relative to the surface of the star?
> >>>> You're not paying attention. Again, in terms of the standard inertial coordinates in which the star is at rest, the light (in vacuum) propagates at the speed c. Also, in terms of the standard inertial coordinates in which the earth is at rest, the light propagates at c. It is impossible for both of these things to be true... unless those two systems of coordinates are related to each other in a very specific way. This way is known as a Lorentz transformation. Standard inertial coordinates are, by definition, coordinates in terms of which the laws of physics take the same isotropic and homogeneous form, and the fact that the laws are Lorentz invariant (check for yourself) signifies that they take the same form in terms of coordinate systems related by Lorentz transformations.
> >>>>> Do you at least see the problem?
> >>>> Well, of course I see "the problem"...everyone saw the problem... this is why Einstein said the two things are seemingly irreconcilable, and then he explained the one and only way in which they can be reconciled.
> >>>>
> >>>
> >>> Well at least you accept that there is a 'problem'.
> >> Not on my reading.
> >>>
> >>> When viewed from above the double star system looks this picture:
> >>>
> >>> https://en.wikipedia.org/wiki/De_Sitter_double_star_experiment#/media/File:SitterKonstanz.png
> >>>
> >>> (In the diagram the c-v and c+v labels are reversed, since it illustrates the ballistic emission of light)
> >>>
> >>> Let me ask it this way: does the application of the Lorentz transformation in both cases give a constant velocity of light towards the Earth?
> >> Yes.
> >
> > OK so that is settled.
> >
> >>>
> >>> Is there a reason behind the asymmetry in the system clearly viewed in this image, where on one side of the orbit the star emits light at c-v relative to the surface and c+v to the surface of the star?
> >> You realise that the labels are what would happen if emission theory
> >> were correct. Since it's not correct, the labels are wrong, and there is
> >> no asymmetry to explain.
> >
> > As I mentioned, the labels are for emission theory. However, if we assume a constant velocity c towards the Earth, the labels are reversed, so one one side it would show a velocity between the surface of the star and the emitted light for example, for the approaching star. Either way, there is assymetry.
> The only asymmetry is in the incorrect emission model. So why concern
> yourself with it?
> >>>
> >>> Would the wave fronts from the moving stars form a sphere or an ellipse?
> >> A sphere - both in the frame of the star, and in the frame of an
> >> observer moving relative to the star. There are not the same spheres,
> >> but they are nevertheless spheres in the respective frames of reference.
> >>
> >> Sylvia.
> >
> > Wow. This is relativity then.
> >
> > A sphere in the frame of the star, yes.
> >
> > If the star is moving relative to the observer, or the observer is moving relative to the star, another sphere consisting of a different set of photons in each case?
> >
> > A spherical wavefront of photons a sphere created out of nothing, where is the energy for that?
> They're formed from the same photons, but the relativity of simultaneity
> comes into play, so for example, where and when a photon forms part of
> the wavefront one second after the emission is frame dependent.
>
> Sylvia.

That is a good explanation, though I cannot accept it. Part of my problem is that the concepts associated with SRT are not concisely stated, not correctly stated, for example 'moving clocks slow down' etc.

So each observer is stuck in his own reality if he is moving with respect to another pbserver.

Tell me, in a certain configuration of Relativistic time and space, could you die before you were born?

On Wednesday, May 3, 2023 at 6:46:51 PM UTC+5, Trevor Lange wrote:
> On Wednesday, May 3, 2023 at 4:48:58 AM UTC-7, gehan.am...@gmail.com wrote:
> On Wednesday, May 3, 2023 at 5:28:39 AM UTC-7, gehan.am...@gmail.com wrote:
> > In the above example, if the light bulbs are moving in an Aether stationary to me,
> > that would make perfect sense.
> Right, and it would also make perfect sense if all the laws of physics are (locally) Lorentz invariant.... which they are. Other phenomena are inconsistent with a stationary aether, but all the phenonena are consistent with local Lorentz invariance.

Agreed. Lorentz invariance was created for that.

You are very patient, are you a Professor? Lucky students!
Don't have to answer.

> > It is indisputable in Newtonian physics that the child will have to use more
> > force to throw the ball towards the person he is receding from, in order to
> > make the velocity of the ball the same for each of the recipients.
> Right, and also in special relativity, although the precise difference in the required force is slightly different than in Newtonian physics, because Newton's laws aren't Lorentz invariant, whereas the actual laws of mechanics are Lorentz invariant.

Agreed about Newton. Accepted about Lorentz, for the moment.

> > The receding star is throwing out light just as the approaching star is.. Why is
> > the light from the receding star thrown out faster relative to the surface of the star?
> Again, in terms of the standard inertial coordinates in which the star is at rest, the light (in vacuum) propagates at the speed c. Also, in terms of the standard inertial coordinates in which the earth is at rest, the light propagates at c. It is impossible for both of these things to be true... unless those two systems of coordinates are related to each other in a very specific way.

>This way is known as a Lorentz transformation. Standard inertial coordinates are, by definition, coordinates in terms of which the laws >of physics take the same simple isotropic and homogeneous form, and the fact that the laws are Lorentz invariant (check for yourself) >signifies that they take the same form in terms of coordinate systems related by Lorentz transformations.

... confirmed by experiments, but the existence of the theory did not automatically mean that it would be confirmed by experiments..

> > Do you at least see the problem?
> Well, of course I see "the problem", i.e., the fact that the phenomena are not consistent with Galilean invariance...everyone saw the problem... this is why Einstein said the two things are seemingly irreconcilable, and then he explained the one and only way in which they can be reconciled, i.e., Lorentz invariance.

OK

>
> > In a double star system, does the application of the Lorentz transformation
> > in both cases give a constant velocity of light towards the Earth?
> In terms of each of the different inertial coordinate systems in which each individual star, the center of mass of the stars, and the earth, are respectively at rest, the speed of light is c, and these coordinate systems are related by Lorentz transformations. Do you understand this?

The systems are related by Loretnz transformations, yes.

> > Is there a reason behind the asymmetry in the system clearly viewed in this
> > image, where on one side of the orbit the star emits light at c-v relative to the
> > surface and c+v to the surface of the star?

> Your sentence doesn't parse in English, but note that the drawing is depicting a misunderstanding of the phenomena, i.e., it is illustrating a hypothesis that is shown to be false. The correct drawing of the situation, consistent with the actual phenomena, shows that the speed of light is c in terms of each of the inertial coordinate systems as described above. Do you understand this?

Haste makes waste. So here we go:

In the diagram, when viewed from above, we see a system with two stars orbiting a common center of mass. Suppose we had a probe at rest in the Earths frame of reference with a camera pointing down on this system. The probe measures the tangential velocity of the surface of the star at each point in its orbit when the star is travelling directly toward Earth and away from the Earth. It measure this velocity to be v1 in the case where the surface of the star is moving towards the Earth and v2 at the point where the surface of the star is moving directly away.

If we use the inertial coordinate system in which the Earth, the center of mass of the stars (assumed to be at rest), then we can see that the surface of the star (there are two, let's use one for the moment), is moving towards the Earth at different velocities, where v1 and v2.

If we assume that the light from the star is moving at c in the inertial frame of reference that we chose, then we can see that if we subtract velocities, we have the relative speeds or closing speeds of the star surface and emitted light working out to be c-v and c+v.

If I accept that this is all fine with Special Relativity, the question still arises as to why there are two different phenomena occurring here, where light is emitted at different velocities relative to the surface of the star at different points of the orbit.

There is an asymmetry or I think you call it anisotropy.

> > Would the wave fronts from the moving stars form a sphere or an ellipse?

> Each pulse of light (emitted in all directions) emanates outward as an expanding spheres in terms of each of the standard inertial coordinate systems. Do you understand this?

To each his own.

On Wednesday, May 3, 2023 at 7:25:42 PM UTC+5, Prokaryotic Capase Homolog wrote:
> On Wednesday, May 3, 2023 at 6:48:58 AM UTC-5, gehan.am...@gmail.com wrote:
>
> > When viewed from above the double star system looks this picture:
> >
> > https://en.wikipedia.org/wiki/De_Sitter_double_star_experiment#/media/File:SitterKonstanz.png
> >
> > (In the diagram the c-v and c+v labels are reversed, since it illustrates the ballistic emission of light)
> The author of your linked image was Willem de Sitter.
>
> You had best reconsider your statement that the c-v and c+v
> labels are reversed.
>
> You are clearly misinterpreting the image

The image simply depicts the ballistic theory of light, where the light is travelling at different velocities c-v and c+v at A and B.

On Wednesday, May 3, 2023 at 8:02:31 PM UTC-7, gehan.am...@gmail.com wrote:

> That is a good explanation, though I cannot accept it.

This is because you are mentally ill. Cannot be fixed.