https://www.quora.com/In-Michelson-Morleys-experiment-why-does-the-light-not-go-straight-up-instead-of-going-angular/answer/Radwan-M-Kassir?ch=18&oid=339712382&share=a4af4340&srid=uTVSu&target_type=answer

Radwan M. Kassir
M.S. in Mechanical Engineering & Physics, Arizona State University1y

This is a good question, and the right answer puts special relativity in an awkward position.

The relativistic interpretation of M-M experiment is based on two assumptions, one of them is explicitly stated: The speed of light, c, is absolute and independent of the source state of motion. The other assumption is implicit, wrong and taken for
granted: Relative to the interferometer rest frame (the earth-lab rest frame), the light ray is projected in the transverse direction, along the interferometer "vertical" arm, and it keeps traveling at the speed c in the same "upward" direction, while,
relative to the Sun rest frame, to keep up with the interferometer motion, the light ray travels at speed c in a slanted direction (not vertical), inclined toward the direction of the Earth motion, with a "horizontal" component equal to the Earth speed- -
The relative motion in M-M experiment, with a speed of around 30 Km/s, is between the Sun and Earth.

The latter implicit assumption contradicts the former explicit one. They are incoherent. If the speed of light is independent of the source state of motion, once it is projected "vertically" from the source in the interferometer frame, it shouldn't be
carried away along the interferometer direction of motion at the source speed. In other words, it shouldn't be drifted with the source. It will maintain its "vertical" direction relative to the Sun rest frame, and will appear to travel away from the
interferometer with backwards inclination relative to the interferometer rest frame.

One may argue that the light ray is actually projected in the interferometer frame at an angle with the vertical so as to keep up with the interferometer motion, but this would contradict the speed of light assumption. If the light ray is projected at
the speed c at an angle with the "vertical" relative to the interferometer frame, its vertical component will be less than c relative to the same frame. This means the light ray would be traveling relative to the interferometer rest frame at a speed less
than c, in violation of the speed of light assumption.

With the Ether theory interpretation, the transverse light ray is projected with speed c at an angle with the "vertical" to overcome the "ether wind" drift, and keep up with the interferometer motion. The angle magnitude is such that the light ray speed
in the "horizontal" direction is equal to the Earth speed. The light ray travels in the "vertical" direction relative to the interferometer frame, at a fraction of the speed of light c, and in a slanted direction at the speed c relative to the Sun frame,
with a vertical speed component equal to the aforementioned fraction of c.

With Newton's emission theory, the light ray is projected "vertically" at the speed c relative to the source (interferometer frame). The light particles will acquire the source speed. So, with respect to the Sun frame, the light ray will travel at the
speed c in the "vertical" direction, and at the Earth speed in the "horizontal" direction, resulting in a slanted ray traveling at c + v (vectors addition). This interpretation is in agreement with the null result of the M-M experiment.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Friday, 15 September 2023 at 04:38:41 UTC+1, Laurence Clark Crossen wrote:

https://www.quora.com/In-Michelson-Morleys-experiment-why-does-the-light-not-go-straight-up-instead-of-going-angular/answer/Radwan-M-Kassir?ch=18&oid=339712382&share=a4af4340&srid=uTVSu&target_type=answer

Radwan M. Kassir
M.S. in Mechanical Engineering & Physics, Arizona State University1y

This is a good question, and the right answer puts special relativity in an awkward position.

The relativistic interpretation of M-M experiment is based on two assumptions, one of them is explicitly stated: The speed of light, c, is absolute and independent of the source state of motion. The other assumption is implicit, wrong and taken for

granted: Relative to the interferometer rest frame (the earth-lab rest frame), the light ray is projected in the transverse direction, along the interferometer "vertical" arm, and it keeps traveling at the speed c in the same "upward" direction, while,
relative to the Sun rest frame, to keep up with the interferometer motion, the light ray travels at speed c in a slanted direction (not vertical), inclined toward the direction of the Earth motion, with a "horizontal" component equal to the Earth speed- -
The relative motion in M-M experiment, with a speed of around 30 Km/s, is between the Sun and Earth.

The latter implicit assumption contradicts the former explicit one. They are incoherent. If the speed of light is independent of the source state of motion, once it is projected "vertically" from the source in the interferometer frame, it shouldn't be

carried away along the interferometer direction of motion at the source speed. In other words, it shouldn't be drifted with the source. It will maintain its "vertical" direction relative to the Sun rest frame, and will appear to travel away from the
interferometer with backwards inclination relative to the interferometer rest frame.

One may argue that the light ray is actually projected in the interferometer frame at an angle with the vertical so as to keep up with the interferometer motion, but this would contradict the speed of light assumption. If the light ray is projected at

the speed c at an angle with the "vertical" relative to the interferometer frame, its vertical component will be less than c relative to the same frame. This means the light ray would be traveling relative to the interferometer rest frame at a speed less
than c, in violation of the speed of light assumption.

With the Ether theory interpretation, the transverse light ray is projected with speed c at an angle with the "vertical" to overcome the "ether wind" drift, and keep up with the interferometer motion. The angle magnitude is such that the light ray

speed in the "horizontal" direction is equal to the Earth speed. The light ray travels in the "vertical" direction relative to the interferometer frame, at a fraction of the speed of light c, and in a slanted direction at the speed c relative to the Sun
frame, with a vertical speed component equal to the aforementioned fraction of c.

With Newton's emission theory, the light ray is projected "vertically" at the speed c relative to the source (interferometer frame). The light particles will acquire the source speed. So, with respect to the Sun frame, the light ray will travel at the

speed c in the "vertical" direction, and at the Earth speed in the "horizontal" direction, resulting in a slanted ray traveling at c + v (vectors addition). This interpretation is in agreement with the null result of the M-M experiment.

You are wasting your time. The relativists have got it all stitched up with an answer for everything using their imaginary photons and “inertial frames”. And also their counter intuitive momentum defying mathematically fiddled constant c for all observers. You can’t fight ignorance.
(I predict that new theoretical fantasies will be arriving soon from fermi
lab theorists. To explain why Muons aren’t doing what their theory predicts, they will soon invent the muoninoe. Minnow for short. A new particle that
the muon decays into.)


But I would be interested to see what you think about the following:
In a Newtonian model of gravity, G is supposed to be instantaneous.
So let’s take the example of earths gravitational field. Does it effect
any other object, planet, asteroid or whatever in the solar system... instantaneously? Or is the object attracted to a ‘retarded’ earth position dictated by c (or any other speed) where earth used to be?
It’s not a trick question but essentially the question is trying to
see what choices you or anyone else reading this will make between
choosing between gravity at a specific speed or an instantaneous G.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Thursday, September 14, 2023 at 8:38:41 PM UTC-7, Laurence Clark Crossen wrote:

https://www.quora.com/In-Michelson-Morleys-experiment-why-does-the-light-not-go-straight-up-instead-of-going-angular/answer/Radwan-M-Kassir?ch=18&oid=339712382&share=a4af4340&srid=uTVSu&target_type=answer

Radwan M. Kassir
M.S. in Mechanical Engineering & Physics, Arizona State University1y

This is a good question, and the right answer puts special relativity in an awkward position.

The relativistic interpretation of M-M experiment is based on two assumptions, one of them is explicitly stated: The speed of light, c, is absolute and independent of the source state of motion. The other assumption is implicit, wrong and taken for

granted: Relative to the interferometer rest frame (the earth-lab rest frame), the light ray is projected in the transverse direction, along the interferometer "vertical" arm, and it keeps traveling at the speed c in the same "upward" direction, while,
relative to the Sun rest frame, to keep up with the interferometer motion, the light ray travels at speed c in a slanted direction (not vertical), inclined toward the direction of the Earth motion, with a "horizontal" component equal to the Earth speed- -
The relative motion in M-M experiment, with a speed of around 30 Km/s, is between the Sun and Earth.

The latter implicit assumption contradicts the former explicit one. They are incoherent. If the speed of light is independent of the source state of motion, once it is projected "vertically" from the source in the interferometer frame, it shouldn't be

carried away along the interferometer direction of motion at the source speed. In other words, it shouldn't be drifted with the source. It will maintain its "vertical" direction relative to the Sun rest frame, and will appear to travel away from the
interferometer with backwards inclination relative to the interferometer rest frame.

One may argue that the light ray is actually projected in the interferometer frame at an angle with the vertical so as to keep up with the interferometer motion, but this would contradict the speed of light assumption. If the light ray is projected at

the speed c at an angle with the "vertical" relative to the interferometer frame, its vertical component will be less than c relative to the same frame. This means the light ray would be traveling relative to the interferometer rest frame at a speed less
than c, in violation of the speed of light assumption.

With the Ether theory interpretation, the transverse light ray is projected with speed c at an angle with the "vertical" to overcome the "ether wind" drift, and keep up with the interferometer motion. The angle magnitude is such that the light ray

speed in the "horizontal" direction is equal to the Earth speed. The light ray travels in the "vertical" direction relative to the interferometer frame, at a fraction of the speed of light c, and in a slanted direction at the speed c relative to the Sun
frame, with a vertical speed component equal to the aforementioned fraction of c.

With Newton's emission theory, the light ray is projected "vertically" at the speed c relative to the source (interferometer frame). The light particles will acquire the source speed. So, with respect to the Sun frame, the light ray will travel at the

speed c in the "vertical" direction, and at the Earth speed in the "horizontal" direction, resulting in a slanted ray traveling at c + v (vectors addition). This interpretation is in agreement with the null result of the M-M experiment.
A REPLY TO TOM ROBERTS: If optical extinction rigidly prevents light speed from varying within any medium. He even denies sound waves of ambulance sirens move S + V. Answer: Air in the MMX is moving at 30 km/sec. Light moves strictly at c in that air,
sharing the velocity of the air, thus moving C + V.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Friday, September 15, 2023 at 5:04:24 AM UTC-7, Lou wrote:

On Friday, 15 September 2023 at 04:38:41 UTC+1, Laurence Clark Crossen wrote:

https://www.quora.com/In-Michelson-Morleys-experiment-why-does-the-light-not-go-straight-up-instead-of-going-angular/answer/Radwan-M-Kassir?ch=18&oid=339712382&share=a4af4340&srid=uTVSu&target_type=answer

Radwan M. Kassir
M.S. in Mechanical Engineering & Physics, Arizona State University1y

This is a good question, and the right answer puts special relativity in an awkward position.

The relativistic interpretation of M-M experiment is based on two assumptions, one of them is explicitly stated: The speed of light, c, is absolute and independent of the source state of motion. The other assumption is implicit, wrong and taken for

granted: Relative to the interferometer rest frame (the earth-lab rest frame), the light ray is projected in the transverse direction, along the interferometer "vertical" arm, and it keeps traveling at the speed c in the same "upward" direction, while,
relative to the Sun rest frame, to keep up with the interferometer motion, the light ray travels at speed c in a slanted direction (not vertical), inclined toward the direction of the Earth motion, with a "horizontal" component equal to the Earth speed- -
The relative motion in M-M experiment, with a speed of around 30 Km/s, is between the Sun and Earth.

The latter implicit assumption contradicts the former explicit one. They are incoherent. If the speed of light is independent of the source state of motion, once it is projected "vertically" from the source in the interferometer frame, it shouldn't

be carried away along the interferometer direction of motion at the source speed. In other words, it shouldn't be drifted with the source. It will maintain its "vertical" direction relative to the Sun rest frame, and will appear to travel away from the
interferometer with backwards inclination relative to the interferometer rest frame.

One may argue that the light ray is actually projected in the interferometer frame at an angle with the vertical so as to keep up with the interferometer motion, but this would contradict the speed of light assumption. If the light ray is projected

at the speed c at an angle with the "vertical" relative to the interferometer frame, its vertical component will be less than c relative to the same frame. This means the light ray would be traveling relative to the interferometer rest frame at a speed
less than c, in violation of the speed of light assumption.

With the Ether theory interpretation, the transverse light ray is projected with speed c at an angle with the "vertical" to overcome the "ether wind" drift, and keep up with the interferometer motion. The angle magnitude is such that the light ray

speed in the "horizontal" direction is equal to the Earth speed. The light ray travels in the "vertical" direction relative to the interferometer frame, at a fraction of the speed of light c, and in a slanted direction at the speed c relative to the Sun
frame, with a vertical speed component equal to the aforementioned fraction of c.

With Newton's emission theory, the light ray is projected "vertically" at the speed c relative to the source (interferometer frame). The light particles will acquire the source speed. So, with respect to the Sun frame, the light ray will travel at

the speed c in the "vertical" direction, and at the Earth speed in the "horizontal" direction, resulting in a slanted ray traveling at c + v (vectors addition). This interpretation is in agreement with the null result of the M-M experiment.

You are wasting your time. The relativists have got it all stitched up with an
answer for everything using their imaginary photons and “inertial frames”.
And also their counter intuitive momentum defying mathematically fiddled constant c for all observers. You can’t fight ignorance.
(I predict that new theoretical fantasies will be arriving soon from fermi lab theorists. To explain why Muons aren’t doing what their theory predicts,
they will soon invent the muoninoe. Minnow for short. A new particle that the muon decays into.)

But I would be interested to see what you think about the following:
In a Newtonian model of gravity, G is supposed to be instantaneous.
So let’s take the example of earths gravitational field. Does it effect any other object, planet, asteroid or whatever in the solar system... instantaneously? Or is the object attracted to a ‘retarded’ earth position
dictated by c (or any other speed) where earth used to be?
It’s not a trick question but essentially the question is trying to
see what choices you or anyone else reading this will make between
choosing between gravity at a specific speed or an instantaneous G.

Of course, it is not as if they have explained anything. They assert light speed is always c without explaining how. They defend it by censoring tactics as an ideology is defended. I try to learn their defensive tactics to better contend with them. You
are correct. It is futile.

Gravity cannot be instantaneous/infinite in speed, but it must be close to it, or the orbital dynamics would be thrown off. Even so, it acts in a retarded position. If it were c, it would be so retarded the Earth would double its distance from the Sun in
1200 years, as you probably know from Van Flandern.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

Le 15/09/2023 à 14:04, Lou a écrit :

You are wasting your time. The relativists have got it all stitched up with an
answer for everything using their imaginary photons and “inertial frames”.
And also their counter intuitive momentum defying mathematically fiddled constant c for all observers. You can’t fight ignorance.
(I predict that new theoretical fantasies will be arriving soon from fermi lab theorists. To explain why Muons aren’t doing what their theory predicts,
they will soon invent the muoninoe. Minnow for short. A new particle that
the muon decays into.)

But I would be interested to see what you think about the following:
In a Newtonian model of gravity, G is supposed to be instantaneous.
So let’s take the example of earths gravitational field. Does it effect
any other object, planet, asteroid or whatever in the solar system... instantaneously? Or is the object attracted to a ‘retarded’ earth position
dictated by c (or any other speed) where earth used to be?
It’s not a trick question but essentially the question is trying to
see what choices you or anyone else reading this will make between
choosing between gravity at a specific speed or an instantaneous G.

This is a very good question.

Thank you for asking such serious and interesting questions.

To me there is no doubt that the effects of gravitation are exactly
similar to the electromagnetic effects.

Although I don't like the term, we can talk about gravitational waves
traveling at c.

Note what I wrote, many years ago, on the subject:
“There will therefore be an impassable speed limit which will extend to
all corpuscles and all the laws of physics.”

Obviously I haven't changed an inch on that.

R.H.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Friday, September 15, 2023 at 5:04:24 AM UTC-7, Lou wrote:

On Friday, 15 September 2023 at 04:38:41 UTC+1, Laurence Clark Crossen wrote:

https://www.quora.com/In-Michelson-Morleys-experiment-why-does-the-light-not-go-straight-up-instead-of-going-angular/answer/Radwan-M-Kassir?ch=18&oid=339712382&share=a4af4340&srid=uTVSu&target_type=answer

Radwan M. Kassir
M.S. in Mechanical Engineering & Physics, Arizona State University1y

This is a good question, and the right answer puts special relativity in an awkward position.

The relativistic interpretation of M-M experiment is based on two assumptions, one of them is explicitly stated: The speed of light, c, is absolute and independent of the source state of motion. The other assumption is implicit, wrong and taken for

granted: Relative to the interferometer rest frame (the earth-lab rest frame), the light ray is projected in the transverse direction, along the interferometer "vertical" arm, and it keeps traveling at the speed c in the same "upward" direction, while,
relative to the Sun rest frame, to keep up with the interferometer motion, the light ray travels at speed c in a slanted direction (not vertical), inclined toward the direction of the Earth motion, with a "horizontal" component equal to the Earth speed- -
The relative motion in M-M experiment, with a speed of around 30 Km/s, is between the Sun and Earth.

The latter implicit assumption contradicts the former explicit one. They are incoherent. If the speed of light is independent of the source state of motion, once it is projected "vertically" from the source in the interferometer frame, it shouldn't

be carried away along the interferometer direction of motion at the source speed. In other words, it shouldn't be drifted with the source. It will maintain its "vertical" direction relative to the Sun rest frame, and will appear to travel away from the
interferometer with backwards inclination relative to the interferometer rest frame.

One may argue that the light ray is actually projected in the interferometer frame at an angle with the vertical so as to keep up with the interferometer motion, but this would contradict the speed of light assumption. If the light ray is projected

at the speed c at an angle with the "vertical" relative to the interferometer frame, its vertical component will be less than c relative to the same frame. This means the light ray would be traveling relative to the interferometer rest frame at a speed
less than c, in violation of the speed of light assumption.

With the Ether theory interpretation, the transverse light ray is projected with speed c at an angle with the "vertical" to overcome the "ether wind" drift, and keep up with the interferometer motion. The angle magnitude is such that the light ray

speed in the "horizontal" direction is equal to the Earth speed. The light ray travels in the "vertical" direction relative to the interferometer frame, at a fraction of the speed of light c, and in a slanted direction at the speed c relative to the Sun
frame, with a vertical speed component equal to the aforementioned fraction of c.

With Newton's emission theory, the light ray is projected "vertically" at the speed c relative to the source (interferometer frame). The light particles will acquire the source speed. So, with respect to the Sun frame, the light ray will travel at

the speed c in the "vertical" direction, and at the Earth speed in the "horizontal" direction, resulting in a slanted ray traveling at c + v (vectors addition). This interpretation is in agreement with the null result of the M-M experiment.

You are wasting your time. The relativists have got it all stitched up with an
answer for everything using their imaginary photons and “inertial frames”.
And also their counter intuitive momentum defying mathematically fiddled constant c for all observers. You can’t fight ignorance.
(I predict that new theoretical fantasies will be arriving soon from fermi lab theorists. To explain why Muons aren’t doing what their theory predicts,
they will soon invent the muoninoe. Minnow for short. A new particle that the muon decays into.)

But I would be interested to see what you think about the following:
In a Newtonian model of gravity, G is supposed to be instantaneous.
So let’s take the example of earths gravitational field. Does it effect any other object, planet, asteroid or whatever in the solar system... instantaneously? Or is the object attracted to a ‘retarded’ earth position
dictated by c (or any other speed) where earth used to be?
It’s not a trick question but essentially the question is trying to
see what choices you or anyone else reading this will make between
choosing between gravity at a specific speed or an instantaneous G.

What do you think?

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Friday, 15 September 2023 at 22:49:56 UTC+1, Laurence Clark Crossen wrote:

On Friday, September 15, 2023 at 5:04:24 AM UTC-7, Lou wrote:

On Friday, 15 September 2023 at 04:38:41 UTC+1, Laurence Clark Crossen wrote:

https://www.quora.com/In-Michelson-Morleys-experiment-why-does-the-light-not-go-straight-up-instead-of-going-angular/answer/Radwan-M-Kassir?ch=18&oid=339712382&share=a4af4340&srid=uTVSu&target_type=answer

Radwan M. Kassir
M.S. in Mechanical Engineering & Physics, Arizona State University1y

This is a good question, and the right answer puts special relativity in an awkward position.

The relativistic interpretation of M-M experiment is based on two assumptions, one of them is explicitly stated: The speed of light, c, is absolute and independent of the source state of motion. The other assumption is implicit, wrong and taken for

granted: Relative to the interferometer rest frame (the earth-lab rest frame), the light ray is projected in the transverse direction, along the interferometer "vertical" arm, and it keeps traveling at the speed c in the same "upward" direction, while,
relative to the Sun rest frame, to keep up with the interferometer motion, the light ray travels at speed c in a slanted direction (not vertical), inclined toward the direction of the Earth motion, with a "horizontal" component equal to the Earth speed- -
The relative motion in M-M experiment, with a speed of around 30 Km/s, is between the Sun and Earth.

The latter implicit assumption contradicts the former explicit one. They are incoherent. If the speed of light is independent of the source state of motion, once it is projected "vertically" from the source in the interferometer frame, it shouldn't

be carried away along the interferometer direction of motion at the source speed. In other words, it shouldn't be drifted with the source. It will maintain its "vertical" direction relative to the Sun rest frame, and will appear to travel away from the
interferometer with backwards inclination relative to the interferometer rest frame.

One may argue that the light ray is actually projected in the interferometer frame at an angle with the vertical so as to keep up with the interferometer motion, but this would contradict the speed of light assumption. If the light ray is projected

at the speed c at an angle with the "vertical" relative to the interferometer frame, its vertical component will be less than c relative to the same frame. This means the light ray would be traveling relative to the interferometer rest frame at a speed
less than c, in violation of the speed of light assumption.

With the Ether theory interpretation, the transverse light ray is projected with speed c at an angle with the "vertical" to overcome the "ether wind" drift, and keep up with the interferometer motion. The angle magnitude is such that the light ray

speed in the "horizontal" direction is equal to the Earth speed. The light ray travels in the "vertical" direction relative to the interferometer frame, at a fraction of the speed of light c, and in a slanted direction at the speed c relative to the Sun
frame, with a vertical speed component equal to the aforementioned fraction of c.

With Newton's emission theory, the light ray is projected "vertically" at the speed c relative to the source (interferometer frame). The light particles will acquire the source speed. So, with respect to the Sun frame, the light ray will travel at

the speed c in the "vertical" direction, and at the Earth speed in the "horizontal" direction, resulting in a slanted ray traveling at c + v (vectors addition). This interpretation is in agreement with the null result of the M-M experiment.

You are wasting your time. The relativists have got it all stitched up with an
answer for everything using their imaginary photons and “inertial frames”.
And also their counter intuitive momentum defying mathematically fiddled constant c for all observers. You can’t fight ignorance.
(I predict that new theoretical fantasies will be arriving soon from fermi lab theorists. To explain why Muons aren’t doing what their theory predicts,
they will soon invent the muoninoe. Minnow for short. A new particle that the muon decays into.)

But I would be interested to see what you think about the following:
In a Newtonian model of gravity, G is supposed to be instantaneous.
So let’s take the example of earths gravitational field. Does it effect any other object, planet, asteroid or whatever in the solar system... instantaneously? Or is the object attracted to a ‘retarded’ earth position
dictated by c (or any other speed) where earth used to be?
It’s not a trick question but essentially the question is trying to
see what choices you or anyone else reading this will make between choosing between gravity at a specific speed or an instantaneous G.

What do you think?

I think it must be closer to instantaneous than c. Although some
say G is still at c.
I think they can get around this conundrum by saying that it’s at c but one calculates the effects of gravity in the source frame where the source atom doesn’t move. Then you can get the appearance of ‘instantaneous’ to match observations , while not letting go of the limiting concept of c. Although relativists then say that this same relaxation of rules (calculating in source frame for G in relativity) cannot be used for the at a distance effects
of a magnetic field. Without giving a reason as to why it’s one rule for G And another for magnetism.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Saturday, September 16, 2023 at 6:14:46 AM UTC-7, Lou wrote:

On Friday, 15 September 2023 at 22:49:56 UTC+1, Laurence Clark Crossen wrote:

On Friday, September 15, 2023 at 5:04:24 AM UTC-7, Lou wrote:

On Friday, 15 September 2023 at 04:38:41 UTC+1, Laurence Clark Crossen wrote:

https://www.quora.com/In-Michelson-Morleys-experiment-why-does-the-light-not-go-straight-up-instead-of-going-angular/answer/Radwan-M-Kassir?ch=18&oid=339712382&share=a4af4340&srid=uTVSu&target_type=answer

Radwan M. Kassir
M.S. in Mechanical Engineering & Physics, Arizona State University1y

This is a good question, and the right answer puts special relativity in an awkward position.

The relativistic interpretation of M-M experiment is based on two assumptions, one of them is explicitly stated: The speed of light, c, is absolute and independent of the source state of motion. The other assumption is implicit, wrong and taken

for granted: Relative to the interferometer rest frame (the earth-lab rest frame), the light ray is projected in the transverse direction, along the interferometer "vertical" arm, and it keeps traveling at the speed c in the same "upward" direction,
while, relative to the Sun rest frame, to keep up with the interferometer motion, the light ray travels at speed c in a slanted direction (not vertical), inclined toward the direction of the Earth motion, with a "horizontal" component equal to the Earth
speed- -The relative motion in M-M experiment, with a speed of around 30 Km/s, is between the Sun and Earth.

The latter implicit assumption contradicts the former explicit one. They are incoherent. If the speed of light is independent of the source state of motion, once it is projected "vertically" from the source in the interferometer frame, it shouldn'

t be carried away along the interferometer direction of motion at the source speed. In other words, it shouldn't be drifted with the source. It will maintain its "vertical" direction relative to the Sun rest frame, and will appear to travel away from the
interferometer with backwards inclination relative to the interferometer rest frame.

One may argue that the light ray is actually projected in the interferometer frame at an angle with the vertical so as to keep up with the interferometer motion, but this would contradict the speed of light assumption. If the light ray is

projected at the speed c at an angle with the "vertical" relative to the interferometer frame, its vertical component will be less than c relative to the same frame. This means the light ray would be traveling relative to the interferometer rest frame at
a speed less than c, in violation of the speed of light assumption.

With the Ether theory interpretation, the transverse light ray is projected with speed c at an angle with the "vertical" to overcome the "ether wind" drift, and keep up with the interferometer motion. The angle magnitude is such that the light

ray speed in the "horizontal" direction is equal to the Earth speed. The light ray travels in the "vertical" direction relative to the interferometer frame, at a fraction of the speed of light c, and in a slanted direction at the speed c relative to the
Sun frame, with a vertical speed component equal to the aforementioned fraction of c.

With Newton's emission theory, the light ray is projected "vertically" at the speed c relative to the source (interferometer frame). The light particles will acquire the source speed. So, with respect to the Sun frame, the light ray will travel

at the speed c in the "vertical" direction, and at the Earth speed in the "horizontal" direction, resulting in a slanted ray traveling at c + v (vectors addition). This interpretation is in agreement with the null result of the M-M experiment.

You are wasting your time. The relativists have got it all stitched up with an
answer for everything using their imaginary photons and “inertial frames”.
And also their counter intuitive momentum defying mathematically fiddled constant c for all observers. You can’t fight ignorance.
(I predict that new theoretical fantasies will be arriving soon from fermi
lab theorists. To explain why Muons aren’t doing what their theory predicts,
they will soon invent the muoninoe. Minnow for short. A new particle that
the muon decays into.)

But I would be interested to see what you think about the following:
In a Newtonian model of gravity, G is supposed to be instantaneous.
So let’s take the example of earths gravitational field. Does it effect
any other object, planet, asteroid or whatever in the solar system... instantaneously? Or is the object attracted to a ‘retarded’ earth position
dictated by c (or any other speed) where earth used to be?
It’s not a trick question but essentially the question is trying to see what choices you or anyone else reading this will make between choosing between gravity at a specific speed or an instantaneous G.

What do you think?

I think it must be closer to instantaneous than c. Although some
say G is still at c.
I think they can get around this conundrum by saying that it’s at c but one
calculates the effects of gravity in the source frame where the source atom doesn’t move. Then you can get the appearance of ‘instantaneous’ to match observations , while not letting go of the limiting concept of c. Although relativists then say that this same relaxation of rules (calculating
in source frame for G in relativity) cannot be used for the at a distance effects
of a magnetic field. Without giving a reason as to why it’s one rule for G And another for magnetism.

In the case of the Sun and Earth, the Sun moves around the barycenter, and calculating in the source frame neglects the time interval. The concept of c being a maximum limit defies the ontological reality that all motion is relative because it denies
relative motion for light. For example, when two flashlights are end to end facing away from each other, one cannot deny their relative motion is 2c without being a lunatic (relativist). Relativity does require relaxing the rules of logic, physics, and
sanity. To believe in relativity, one must deny relative motion for light, which is foolish because all motion is relative. That relativity claims a speed limit (denying relative motion per se for light) is why we can know relativity is nonsense. Because
a speed of c for gravity would cause angular momentum to throw off all orbital dynamics, we can know relativity is pseudoscience.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Saturday, September 16, 2023 at 11:18:41 AM UTC-7, Laurence Clark Crossen wrote:

...when two flashlights are end to end facing away from each other, one cannot deny their relative motion is 2c without being a lunatic (relativist).

The flashlights, of course, are not moving at all!

Yet another concept about which your knowledge is zero. In this scenario it is true that the *wavefronts* of the 2 lights are receding from *your* reference frame at 2c, but neither wavefront is exceeding c. This is called closing speed and it can only
be measured by a 3rd party...

https://en.wikipedia.org/wiki/Faster-than-light#:~:text=The%20rate%20at%20which%20two,respect%20to%20the%20reference%20frame.

"The rate at which two objects in motion in a single frame of reference get closer together [or recede] is called the mutual or closing speed. This may approach twice the speed of light, as in the case of two particles travelling at close to the speed of
light in opposite directions with respect to the reference frame."

Again, the light from either flashlight remains at c, because nothing can travel faster than light, as per the second postulate.

That textbook is waiting for you to read it.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Saturday, 16 September 2023 at 19:56:23 UTC+1, Paul Alsing wrote:

On Saturday, September 16, 2023 at 11:18:41 AM UTC-7, Laurence Clark Crossen wrote:

...when two flashlights are end to end facing away from each other, one cannot deny their relative motion is 2c without being a lunatic (relativist).

The flashlights, of course, are not moving at all!

Yet another concept about which your knowledge is zero. In this scenario it is true that the *wavefronts* of the 2 lights are receding from *your* reference frame at 2c, but neither wavefront is exceeding c. This is called closing speed and it can only

be measured by a 3rd party...

https://en.wikipedia.org/wiki/Faster-than-light#:~:text=The%20rate%20at%20which%20two,respect%20to%20the%20reference%20frame.

"The rate at which two objects in motion in a single frame of reference get closer together [or recede] is called the mutual or closing speed. This may approach twice the speed of lightning, as in the case of two particles travelling at close to the

speed of light in opposite directions with respect to the reference frame."

Again, the light from either flashlight remains at c, because nothing can travel faster than light, as per the second postulate.

That textbook is waiting for you to read it.

What happens in a collider when two counterrotating protons, each at >0.5 c collide? How does one calculate the total energy in the collision when
you seem to admit that the 2 combined closing speeds are >c?

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Saturday, September 16, 2023 at 12:25:40 PM UTC-7, Lou wrote:

On Saturday, 16 September 2023 at 19:56:23 UTC+1, Paul Alsing wrote:

On Saturday, September 16, 2023 at 11:18:41 AM UTC-7, Laurence Clark Crossen wrote:

...when two flashlights are end to end facing away from each other, one cannot deny their relative motion is 2c without being a lunatic (relativist).

The flashlights, of course, are not moving at all!

Yet another concept about which your knowledge is zero. In this scenario it is true that the *wavefronts* of the 2 lights are receding from *your* reference frame at 2c, but neither wavefront is exceeding c. This is called closing speed and it can

only be measured by a 3rd party...

https://en.wikipedia.org/wiki/Faster-than-light#:~:text=The%20rate%20at%20which%20two,respect%20to%20the%20reference%20frame.

"The rate at which two objects in motion in a single frame of reference get closer together [or recede] is called the mutual or closing speed. This may approach twice the speed of lightning, as in the case of two particles travelling at close to the

speed of light in opposite directions with respect to the reference frame."

Again, the light from either flashlight remains at c, because nothing can travel faster than light, as per the second postulate.

That textbook is waiting for you to read it.

What happens in a collider when two counterrotating protons, each at >0.5 c collide? How does one calculate the total energy in the collision when
you seem to admit that the 2 combined closing speeds are >c?

Here, generative AI...

https://www.google.com/search?q=What+happens+in+a+collider+when+two+counterrotating+protons%2C+each+at+%3E0.5+c+collide%3F&rlz=1C1JZAP_enUS1065US1065&oq=What+happens+in+a+collider+when+two+counterrotating+protons%2C+each+at+%3E0.5+c+collide%3F&aqs=chrome.
.69i57.618j0j7&sourceid=chrome&ie=UTF-8

"When two counter-rotating beams of protons collide in the Large Hadron Collider (LHC), they produce massive particles such as the Higgs boson or the top quark. The protons are accelerated to a speed close to the speed of light. To get the required
energy, the protons need to be moving at about 299,792,455 m/s, which is about 3 m/s less than the speed of light."

Sure, the closing speed is >c, but neither proton is exceeding c, as per the 2nd postulate.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 9/16/2023 3:25 PM, Lou wrote:

On Saturday, 16 September 2023 at 19:56:23 UTC+1, Paul Alsing wrote:

"The rate at which two objects in motion in a single frame of reference get closer together [or recede] is called the mutual or closing speed. This may approach twice the speed of lightning, as in the case of two particles travelling at close to the

speed of light in opposite directions with respect to the reference frame."

Again, the light from either flashlight remains at c, because nothing can travel faster than light, as per the second postulate.

That textbook is waiting for you to read it.

What happens in a collider when two counterrotating protons, each at >0.5 c collide?

You'd know if you had bothered to read a textbook.

The velocity combination formula is w=(u+v)/(1+uv/c²).

In the case of the oppositely moving protons moving at 0.99c, you get w=(0.99c+0.99c)/(1+0.99c*0.99c)/c²). Or 1.98c/1.9801 which is
0.9999494975c. Each proton sees the other coming at it at 0.9999494975c.
Pretty damn fast but still less than c. See how easy that is? If you
bothered to read good physics textbooks, you'd know that.

How does one calculate the total energy in the collision when
you seem to admit that the 2 combined closing speeds are >c?

Closing speeds aren't the speeds of anything physical.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Saturday, September 16, 2023 at 12:25:40 PM UTC-7, Lou wrote:

On Saturday, 16 September 2023 at 19:56:23 UTC+1, Paul Alsing wrote:

On Saturday, September 16, 2023 at 11:18:41 AM UTC-7, Laurence Clark Crossen wrote:

...when two flashlights are end to end facing away from each other, one cannot deny their relative motion is 2c without being a lunatic (relativist).

The flashlights, of course, are not moving at all!

Yet another concept about which your knowledge is zero. In this scenario it is true that the *wavefronts* of the 2 lights are receding from *your* reference frame at 2c, but neither wavefront is exceeding c. This is called closing speed and it can

only be measured by a 3rd party...

https://en.wikipedia.org/wiki/Faster-than-light#:~:text=The%20rate%20at%20which%20two,respect%20to%20the%20reference%20frame.

"The rate at which two objects in motion in a single frame of reference get closer together [or recede] is called the mutual or closing speed. This may approach twice the speed of lightning, as in the case of two particles travelling at close to the

speed of light in opposite directions with respect to the reference frame."

Again, the light from either flashlight remains at c, because nothing can travel faster than light, as per the second postulate.

That textbook is waiting for you to read it.

What happens in a collider when two counterrotating protons, each at >0.5 c collide? How does one calculate the total energy in the collision when
you seem to admit that the 2 combined closing speeds are >c?

Not with relativity's formulas. They assume a limit that doesn't exist.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Saturday, September 16, 2023 at 8:22:05 PM UTC-7, Laurence Clark Crossen wrote:

On Saturday, September 16, 2023 at 12:25:40 PM UTC-7, Lou wrote:

On Saturday, 16 September 2023 at 19:56:23 UTC+1, Paul Alsing wrote:

On Saturday, September 16, 2023 at 11:18:41 AM UTC-7, Laurence Clark Crossen wrote:

...when two flashlights are end to end facing away from each other, one cannot deny their relative motion is 2c without being a lunatic (relativist).

The flashlights, of course, are not moving at all!

Yet another concept about which your knowledge is zero. In this scenario it is true that the *wavefronts* of the 2 lights are receding from *your* reference frame at 2c, but neither wavefront is exceeding c. This is called closing speed and it can

only be measured by a 3rd party...

https://en.wikipedia.org/wiki/Faster-than-light#:~:text=The%20rate%20at%20which%20two,respect%20to%20the%20reference%20frame.

"The rate at which two objects in motion in a single frame of reference get closer together [or recede] is called the mutual or closing speed. This may approach twice the speed of lightning, as in the case of two particles travelling at close to

the speed of light in opposite directions with respect to the reference frame."

Again, the light from either flashlight remains at c, because nothing can travel faster than light, as per the second postulate.

That textbook is waiting for you to read it.

What happens in a collider when two counterrotating protons, each at >0.5 c
collide? How does one calculate the total energy in the collision when
you seem to admit that the 2 combined closing speeds are >c?

Not with relativity's formulas. They assume a limit that doesn't exist.

You remain clueless, which is no big surprise...

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Saturday, September 16, 2023 at 9:02:33 PM UTC-7, Laurence Clark Crossen wrote:

On Saturday, September 16, 2023 at 12:25:40 PM UTC-7, Lou wrote:

On Saturday, 16 September 2023 at 19:56:23 UTC+1, Paul Alsing wrote:

On Saturday, September 16, 2023 at 11:18:41 AM UTC-7, Laurence Clark Crossen wrote:

...when two flashlights are end to end facing away from each other, one cannot deny their relative motion is 2c without being a lunatic (relativist).

The flashlights, of course, are not moving at all!

Yet another concept about which your knowledge is zero. In this scenario it is true that the *wavefronts* of the 2 lights are receding from *your* reference frame at 2c, but neither wavefront is exceeding c. This is called closing speed and it can

only be measured by a 3rd party...

https://en.wikipedia.org/wiki/Faster-than-light#:~:text=The%20rate%20at%20which%20two,respect%20to%20the%20reference%20frame.

"The rate at which two objects in motion in a single frame of reference get closer together [or recede] is called the mutual or closing speed. This may approach twice the speed of lightning, as in the case of two particles travelling at close to

the speed of light in opposite directions with respect to the reference frame."

Again, the light from either flashlight remains at c, because nothing can travel faster than light, as per the second postulate.

That textbook is waiting for you to read it.

What happens in a collider when two counterrotating protons, each at >0.5 c
collide? How does one calculate the total energy in the collision when
you seem to admit that the 2 combined closing speeds are >c?

First of all, the mass-velocity relationship is absurd nonsense, if that is what is troubling you.

And your evidence for this claim is what, exactly?

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Saturday, September 16, 2023 at 12:25:40 PM UTC-7, Lou wrote:

On Saturday, 16 September 2023 at 19:56:23 UTC+1, Paul Alsing wrote:

On Saturday, September 16, 2023 at 11:18:41 AM UTC-7, Laurence Clark Crossen wrote:

...when two flashlights are end to end facing away from each other, one cannot deny their relative motion is 2c without being a lunatic (relativist).

The flashlights, of course, are not moving at all!

Yet another concept about which your knowledge is zero. In this scenario it is true that the *wavefronts* of the 2 lights are receding from *your* reference frame at 2c, but neither wavefront is exceeding c. This is called closing speed and it can

only be measured by a 3rd party...

https://en.wikipedia.org/wiki/Faster-than-light#:~:text=The%20rate%20at%20which%20two,respect%20to%20the%20reference%20frame.

"The rate at which two objects in motion in a single frame of reference get closer together [or recede] is called the mutual or closing speed. This may approach twice the speed of lightning, as in the case of two particles travelling at close to the

speed of light in opposite directions with respect to the reference frame."

Again, the light from either flashlight remains at c, because nothing can travel faster than light, as per the second postulate.

That textbook is waiting for you to read it.

What happens in a collider when two counterrotating protons, each at >0.5 c collide? How does one calculate the total energy in the collision when
you seem to admit that the 2 combined closing speeds are >c?

First of all, the mass-velocity relationship is absurd nonsense, if that is what is troubling you.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Saturday, September 16, 2023 at 12:25:40 PM UTC-7, Lou wrote:

On Saturday, 16 September 2023 at 19:56:23 UTC+1, Paul Alsing wrote:

On Saturday, September 16, 2023 at 11:18:41 AM UTC-7, Laurence Clark Crossen wrote:

...when two flashlights are end to end facing away from each other, one cannot deny their relative motion is 2c without being a lunatic (relativist).

The flashlights, of course, are not moving at all!

Yet another concept about which your knowledge is zero. In this scenario it is true that the *wavefronts* of the 2 lights are receding from *your* reference frame at 2c, but neither wavefront is exceeding c. This is called closing speed and it can

only be measured by a 3rd party...

https://en.wikipedia.org/wiki/Faster-than-light#:~:text=The%20rate%20at%20which%20two,respect%20to%20the%20reference%20frame.

"The rate at which two objects in motion in a single frame of reference get closer together [or recede] is called the mutual or closing speed. This may approach twice the speed of lightning, as in the case of two particles travelling at close to the

speed of light in opposite directions with respect to the reference frame."

Again, the light from either flashlight remains at c, because nothing can travel faster than light, as per the second postulate.

That textbook is waiting for you to read it.

What happens in a collider when two counterrotating protons, each at >0.5 c collide? How does one calculate the total energy in the collision when
you seem to admit that the 2 combined closing speeds are >c?

Please consider what Radwan Kassir said: [ https://www.quora.com/profile/Radwan-M-Kassir?share=1 ]
"How does a photon have energy (E=mc^2), thus containing some amount mass, but still able to travel at the speed of light (c)?

A typical answer is provided by the special relativity energy-mass equivalence equation for a particle in the form
E2=(mc2)2+(pc)2,
where m
is the rest mass of the particle, and arguing that if the rest mass was zero, like the case of a photon, the particle will have no energy stored as mass, and its energy (pc) will be purely due to its momentum p, with the speed of light c being its
velocity. If this massless particle internally acquired a certain mass, it will lose an amount from its “momentum” energy, reducing its speed to v<c,
so as to maintain the above energy-mass equation.
However, there’s a hidden trick in the above interpretation. In fact, considering the energy-mass equivalence equation in the following two forms:
E2=(mc2)2+(pc)2,
where m
is the rest mass of the particle, and
E=Mc2,
where M
is the relativistic mass, γm,
and plugging the latter equation
E=γmc2
in the former one, we get
γ2(mc2)2=(mc2)2+(pc)2;
p2c2=(mc2)2(γ2−1);
p2=m2c2(γ2−1).
But, since
γ=(1−v2c2)−1/2,
then
(γ2−1)=v2c2γ2,
where v
is the velocity of the particle. Hence
p2=m2c2v2c2γ2;
p=γmv.
The mass-energy equivalence equation can therefore be written as E2=(mc2)2+(γmvc)2,
and for massless particle, supposedly traveling at the speed of light,
E2=0+00!
Hence, the energy of a massless particle becomes undefined!
For instance, photon, a massless particle, does have momentum, but not according to the mass-energy equivalence equation, in which the momentum is given by γmv,
which is undefined for a massless particle! Imposing a definition to this term by assigning it the value of p
is a hoax used to get around an SR deficiency in predicting the energy of a massless particle!"

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Saturday, September 16, 2023 at 9:27:44 PM UTC-7, Laurence Clark Crossen wrote:

On Saturday, September 16, 2023 at 12:25:40 PM UTC-7, Lou wrote:

On Saturday, 16 September 2023 at 19:56:23 UTC+1, Paul Alsing wrote:

On Saturday, September 16, 2023 at 11:18:41 AM UTC-7, Laurence Clark Crossen wrote:

...when two flashlights are end to end facing away from each other, one cannot deny their relative motion is 2c without being a lunatic (relativist).

The flashlights, of course, are not moving at all!

Yet another concept about which your knowledge is zero. In this scenario it is true that the *wavefronts* of the 2 lights are receding from *your* reference frame at 2c, but neither wavefront is exceeding c. This is called closing speed and it can

only be measured by a 3rd party...

https://en.wikipedia.org/wiki/Faster-than-light#:~:text=The%20rate%20at%20which%20two,respect%20to%20the%20reference%20frame.

"The rate at which two objects in motion in a single frame of reference get closer together [or recede] is called the mutual or closing speed. This may approach twice the speed of lightning, as in the case of two particles travelling at close to

the speed of light in opposite directions with respect to the reference frame."

Again, the light from either flashlight remains at c, because nothing can travel faster than light, as per the second postulate.

That textbook is waiting for you to read it.

What happens in a collider when two counterrotating protons, each at >0.5 c
collide? How does one calculate the total energy in the collision when
you seem to admit that the 2 combined closing speeds are >c?

Please consider what Radwan Kassir said: [ https://www.quora.com/profile/Radwan-M-Kassir?share=1 ]
"How does a photon have energy (E=mc^2), thus containing some amount mass, but still able to travel at the speed of light (c)?

A typical answer is provided by the special relativity energy-mass equivalence equation for a particle in the form
E2=(mc2)2+(pc)2,
where m
is the rest mass of the particle, and arguing that if the rest mass was zero, like the case of a photon, the particle will have no energy stored as mass, and its energy (pc) will be purely due to its momentum p, with the speed of light c being its

velocity. If this massless particle internally acquired a certain mass, it will lose an amount from its “momentum” energy, reducing its speed to v<c,

so as to maintain the above energy-mass equation.
However, there’s a hidden trick in the above interpretation. In fact, considering the energy-mass equivalence equation in the following two forms:
E2=(mc2)2+(pc)2,
where m
is the rest mass of the particle, and
E=Mc2,
where M
is the relativistic mass, γm,
and plugging the latter equation
E=γmc2
in the former one, we get
γ2(mc2)2=(mc2)2+(pc)2;
p2c2=(mc2)2(γ2−1);
p2=m2c2(γ2−1).
But, since
γ=(1−v2c2)−1/2,
then
(γ2−1)=v2c2γ2,
where v
is the velocity of the particle. Hence
p2=m2c2v2c2γ2;
p=γmv.
The mass-energy equivalence equation can therefore be written as E2=(mc2)2+(γmvc)2,
and for massless particle, supposedly traveling at the speed of light, E2=0+00!
Hence, the energy of a massless particle becomes undefined!
For instance, photon, a massless particle, does have momentum, but not according to the mass-energy equivalence equation, in which the momentum is given by γmv,
which is undefined for a massless particle! Imposing a definition to this term by assigning it the value of p
is a hoax used to get around an SR deficiency in predicting the energy of a massless particle!"

And you copied and pasted that jewel from where, exactly? You have no clue as to what it actually says, do you?

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Sunday, 17 September 2023 at 06:04:23 UTC+2, Paul Alsing wrote:

On Saturday, September 16, 2023 at 9:02:33 PM UTC-7, Laurence Clark Crossen wrote:

On Saturday, September 16, 2023 at 12:25:40 PM UTC-7, Lou wrote:

On Saturday, 16 September 2023 at 19:56:23 UTC+1, Paul Alsing wrote:

On Saturday, September 16, 2023 at 11:18:41 AM UTC-7, Laurence Clark Crossen wrote:

...when two flashlights are end to end facing away from each other, one cannot deny their relative motion is 2c without being a lunatic (relativist).

The flashlights, of course, are not moving at all!

Yet another concept about which your knowledge is zero. In this scenario it is true that the *wavefronts* of the 2 lights are receding from *your* reference frame at 2c, but neither wavefront is exceeding c. This is called closing speed and it

can only be measured by a 3rd party...

https://en.wikipedia.org/wiki/Faster-than-light#:~:text=The%20rate%20at%20which%20two,respect%20to%20the%20reference%20frame.

"The rate at which two objects in motion in a single frame of reference get closer together [or recede] is called the mutual or closing speed. This may approach twice the speed of lightning, as in the case of two particles travelling at close to

the speed of light in opposite directions with respect to the reference frame."

Again, the light from either flashlight remains at c, because nothing can travel faster than light, as per the second postulate.

That textbook is waiting for you to read it.

What happens in a collider when two counterrotating protons, each at >0.5 c
collide? How does one calculate the total energy in the collision when you seem to admit that the 2 combined closing speeds are >c?

First of all, the mass-velocity relationship is absurd nonsense, if that is what is troubling you.

And your evidence for this claim is what, exactly?

Al, poor trash, the evidence is only making you angry
and barking more fiercely. As expected from a
fanatic idiot, of course.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Saturday, 16 September 2023 at 23:30:59 UTC+1, Volney wrote:

On 9/16/2023 3:25 PM, Lou wrote:

On Saturday, 16 September 2023 at 19:56:23 UTC+1, Paul Alsing wrote:

"The rate at which two objects in motion in a single frame of reference get closer together [or recede] is called the mutual or closing speed. This may approach twice the speed of lightning, as in the case of two particles travelling at close to the

speed of light in opposite directions with respect to the reference frame."

Again, the light from either flashlight remains at c, because nothing can travel faster than light, as per the second postulate.

That textbook is waiting for you to read it.

What happens in a collider when two counterrotating protons, each at >0.5 c
collide?

You'd know if you had bothered to read a textbook.

The velocity combination formula is w=(u+v)/(1+uv/c²).

In the case of the oppositely moving protons moving at 0.99c, you get w=(0.99c+0.99c)/(1+0.99c*0.99c)/c²). Or 1.98c/1.9801 which is 0.9999494975c. Each proton sees the other coming at it at 0.9999494975c. Pretty damn fast but still less than c. See how easy that is? If you bothered to read good physics textbooks, you'd know that.

How does one calculate the total energy in the collision when
you seem to admit that the 2 combined closing speeds are >c?

Closing speeds aren't the speeds of anything physical.

Good textbooks don’t normally say that 0.99 +0.99 =0.999
But yes I get your point. In relativity the total energy emitted from 2 colliding
particles does not conform to the conservation of energy law.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)