On Wednesday, March 30, 2022 at 3:39:43 PM UTC-5, Townes Olson wrote:
> On Wednesday, March 30, 2022 at 12:58:59 PM UTC-7, wrote:
> > You need a "frame of reference" in order to do your math.
> Well, you need to specify a system of reference in order to talk meaningfully about positions, times, speeds, directions of motion, and so on. Without a system of reference, you're just spouting meaningless words. It's essential to understand this.

Okay. I should have said, "YOU need a STATIONARY frame of reference in
order to do YOUR math." And there is nothing stationary in our observable
universe.

> > The only "frame of reference" in the problem is the point in space where
> > the photon was emitted.
> No, we covered this before. Again, the phrase "point in space where the photon was emitted" doesn't make sense, because a pulse of light is emitted from an *event* (time and place), and there is no absolute rest frame to enable you to identify spatial points at different times.

No, there is no "absolute rest frame to enable YOU to identify spatial points at different times."
I do not need any FICTITIOUS absolute rest frames. I'm looking at the SCIENCE,
I'm not doing math.

>
> For example, suppose you emit the pulse of light when you are passing a several space buoys in various states of motion that all coincided with your rocket at the time of the emission. Thereafter, which of those buoys do you regard at "the point where the photon was emitted"?
>
> It would have to be the buoy that is at absolute rest, but there is no such thing. So you can't talk meaningfully about "the [spatial] point where the pulse was emitted" at any time after the emission. It is essential to understand this.

See above.

Ed

On Wednesday, March 30, 2022 at 4:52:48 PM UTC-5, rotchm wrote:
> On Wednesday, March 30, 2022 at 3:58:59 PM UTC-4, wrote:
>
> > You're talking math. I'm talking science.
> > You need a "frame of reference" in order to do your math.
> You have it the wrong way. Science requires reference frames.
> Are you sure you know what "reference frame" means?
>
> Math does not require reference frames (but did you often make use of it).

Okay, as I stated in a previous response, I should have said that
you need a "STATIONARY frame of reference" in order to do your math,
and there is nothing stationary in our observable universe.

Ed

On Thursday, March 31, 2022 at 10:09:51 AM UTC-4, det...@outlook.com wrote:

> I should have said that
> you need a "STATIONARY frame of reference" in order to do your math,

No, that is not required to do math.

> and there is nothing stationary in our observable universe.

Yes there are. In physics, a reference frame has an operational definition.
For instance if I Define my table as the frame and I put my glass of water on it.
If my glass of water stays at the same place on the table then my glass of water is stationary relative to that Frame.
Do you agree with this?

So you see, there are things that are stationary in our observable universe.

Michael Moroney wrote:

> On 3/30/2022 4:16 PM, Ed Lake wrote:
> Science these days have LOTS of math. Sorry to disappoint you, but if
> you are so scared of those mathematician boogeymen, most sciences,
> physics in particular, are not for you. Most models have math equations
> to describe what's going on, like a home run baseball moving in a
> parabola or whatever.

You kiss ass, Mahoney. You *_polish_* guys are so stupid helping
"ukranian" nazis which killed your parents and grandparents. You idiot.
They will certainly kill again. Proofs, because I only work with proofs. I
don't understand you people. Standing up *_in_salute_* for the nazis??
This must be a parallel universe.

Training The Next Generations Of Ukrainian NAZIs
https://www.bitchute.com/video/ge6z7aatj3Pn/

Gretamyr Zelensky pushes Europe for transition to 'green energy'
https://www.bitchute.com/video/U8gJDntzZQVW/

On Wednesday, March 30, 2022 at 5:52:57 PM UTC-5, Michael Moroney wrote:
> On 3/30/2022 12:33 PM, Ed Lake wrote:
> > On Wednesday, March 30, 2022 at 11:05:59 AM UTC-5, Michael Moroney wrote:
> >> On 3/30/2022 10:51 AM, Ed Lake wrote:
> >>
> >>> My question is very simple: If I am traveling through space, away from
> >>> the Earth and toward Alpha Centauri, will a light photon that I emit at a
> >>> right angle to my direction of travel continue to move at a right angle to
> >>> me, or will it move at a right angle to the point where the photon was emitted?
> >> In which frame? That of the spaceship or of Earth (or Alpha Centauri)?
> >> Since I already know you don't understand the concept of frames in
> >> physics, I don't expect a rational answer to this, or a rational
> >> response to anyone who answers you.
> >>>
> >>> The answer seems obvious: The photon will move away from the point
> >>> where it was emitted, not away from me.
> >> Did you do the special relativity math to work out the answer? Oh that's
> >> right, "mathematicians" are your boogeymen so math is the incantation of
> >> evil.
> >
> > No, it just clouds the issue. It is as Einstein once said, "“As far as the laws of
> > mathematics refer to reality, they are not certain; and as far as they are certain,
> > they do not refer to reality.”
> THAT's the basis of your belief? A single mined quote of Einstein's,
> almost certainly taken out of context?

It's not taken out of context. It's from Einstein's speech to the Prussian Academy
of Sciences in Berlin on January 27, 1921. The whole speech was about that topic.
It was titled "Geometry and Experience."

A link: https://todayinsci.com/E/Einstein_Albert/Einstein-GeometryAndExperience.htm

Ed

Ed Lake <detect@outlook.com> wrote:
> On Wednesday, March 30, 2022 at 3:30:43 PM UTC-5, thor stoneman wrote:
>> On Tuesday, March 29, 2022 at 8:55:46 AM UTC-7, wrote:
>>> When we view light from a distant galaxy, we do not see that galaxy
>>> where it is today, we see that galaxy where it WAS when it emitted the
>>> light. The light traveled in a STRAIGHT LINE from POINT OF EMISSION to me.
>>>
>>> According to many sources, if I am in a rocket ship that accelerates at
>>> 1G, and if it continues to accelerate for that rate for close to one
>>> year, it will nearly reach the speed of light.
>>> Link:
>>> https://www.technologyreview.com/2020/01/10/238139/can-constant-acceleration-be-used-to-produce-artificial-gravity-in-space/
>>>
>>>
>>> My question is: If I am traveling at such speeds, and if I send a beam
>>> of light from one side of my ship to the other side, will the light
>>> travel in a straight line across the room as I see it, or will the
>>> light travel in a curved line down toward the floor?
>>>
>>> I am emitting light in a direction that is at a right angle to MY
>>> direction of movement. Wouldn’t light travel across the room to a point
>>> on the wall that is NOT at a right angle to my direction of movement,
>>> but in a straight line through space from the original point of emission?
>>>
>>> If true, I could mark points on the far wall that would indicate where
>>> the light photons will hit when my rocket is moving at different speeds.
>>>
>>> Of course, this would only be possible when traveling at a high enough
>>> speed so that the trajectory of the photons will measurably change
>>> while crossing a room that is only 10 or 20 feet wide.
>>>
>>> Any thoughts?
>>>
>>> Ed
>> Why do you waste time on question like this?
>
> Because it is a fundamental question about how photons work, and if the
> answer is what it seems it must be, it has implications that could shake up
> the whole physics community.
>
> Ed
>

Ed, let’s get real for a second.

You have this dream, apparently, of being a completely untrained outsider
who will do what no professional in the field can do — revolutionize the
whole subject just by thinking things through from a common sense
perspective. Never mind that this has never been done, ever.

You may be familiar with a character named Walter Mitty.

--
Odd Bodkin -- maker of fine toys, tools, tables

Ed Lake wrote:

> On Wednesday, March 30, 2022 at 4:52:48 PM UTC-5, rotchm wrote:
>> Math does not require reference frames (but did you often make use of
>> it).
>
> Okay, as I stated in a previous response, I should have said that you
> need a "STATIONARY frame of reference" in order to do your math,
> and there is nothing stationary in our observable universe.

Absolutely correct. These *_polish_* guys are so stupid helping the very
same *_"ukranian"_nazis_*, which killed their *parents* and *grandparents*
while ago. They again, are kill *right_now*.

Proofs, because I only work with proofs. *_Corrupt_governments_* standing
up *_in_salute_* for the nazis?? This must be a parallel universe.

Training The Next Generations Of Ukrainian NAZIs
https://www.bitchute.com/video/ge6z7aatj3Pn/

Gretamyr Zelensky pushes Europe for transition to 'green energy'
https://www.bitchute.com/video/U8gJDntzZQVW/

You are absolutely correct.

El jueves, 31 de marzo de 2022 a las 11:09:51 UTC-3, det...@outlook.com escribió:
> On Wednesday, March 30, 2022 at 4:52:48 PM UTC-5, rotchm wrote:
> > On Wednesday, March 30, 2022 at 3:58:59 PM UTC-4, wrote:
> >
> > > You're talking math. I'm talking science.
> > > You need a "frame of reference" in order to do your math.
> > You have it the wrong way. Science requires reference frames.
> > Are you sure you know what "reference frame" means?
> >
> > Math does not require reference frames (but did you often make use of it).
> Okay, as I stated in a previous response, I should have said that
> you need a "STATIONARY frame of reference" in order to do your math,
> and there is nothing stationary in our observable universe.
>

Physics models are tested by using experiments. Defining a stationary frame of reference (also called at rest frame of reference) is crucial to perform such experiments. That Earth is round, rotating and orbiting the Sun does not invalidate the use of "STATIONARY frame of reference". The policeman's location, with a radar gun, standing at a side of a highway, is a valid stationary frame of reference, which allows him to measure the incoming cars speed with a high precision.

Most physics laboratories can also not take into account the gravity changes within the reduce space of the experiment setting.

Without a frame of reference (or system of coordinates), none of the following terms make any sense: speed, time, location, direction, etc.

On Wednesday, March 30, 2022 at 5:55:32 PM UTC-5, Sylvia Else wrote:
> On 30-Mar-22 6:33 am, Ed Lake wrote:
> > On Tuesday, March 29, 2022 at 1:08:41 PM UTC-5, bodk...@gmail.com wrote:
> >> Ed Lake wrote:
> >>> When we view light from a distant galaxy, we do not see that galaxy where
> >>> it is today, we see that galaxy where it WAS when it emitted the light.
> >>> The light traveled in a STRAIGHT LINE from POINT OF EMISSION to me.
> >>>
> >>>
> >>> According to many sources, if I am in a rocket ship that accelerates at
> >>> 1G, and if it continues to accelerate for that rate for close to one
> >>> year, it will nearly reach the speed of light.
> >>> Link:
> >>> https://www.technologyreview.com/2020/01/10/238139/can-constant-acceleration-be-used-to-produce-artificial-gravity-in-space/
> >>>
> >>> My question is: If I am traveling at such speeds, and if I send a beam
> >>> of light from one side of my ship to the other side, will the light
> >>> travel in a straight line across the room as I see it, or will the light
> >>> travel in a curved line down toward the floor?
> >> Curved. See Clifford Will’s book “Was Einstein Right?”
> >
> > Ah! Okay. At long last we agree on something.
> >
> > Ed
>
> Assuming you're not at cross-purposes.
>
> The curvature effect arises from the acceleration, not the speed. If you
> stop accelerating, then the curvature vanishes, regardless of your speed
> relative to your starting speed.

That's the question: DOES the effect truly require acceleration?

>
> You would require very sensitive measuring equipment to detect the
> curvature at 1g. After all, you get the same curvature on Earth, and
> people tend not to notice that.

It isn't really about acceleration, it is about speed. It's about how a
photon is created and how a photon works. Special Relativity says that
a photon is always emitted "at c INDEPENDENT of the state of motion
of the emitter." That is usually just interpreted as meaning the speed of
the emitter does not add to or subtract from the speed of the photon.
But what if the emitter is moving sideways? If that doesn't make any
difference, then you CAN measure the speed of a room from inside the
room. The photon will travel in a straight line away from the POINT OF
EMISSION.

But you would definitely need very sensitive equipment to measure a
difference in the trajectory of a photon traversing a room that is only
20 or 30 feet wide. It could be happening right now in every room on
earth without anyone noticing, because the alteration of the trajectory
is so small.

Ed

Ed Lake wrote:

> That's the question: DOES the effect truly require acceleration?
>
>> You would require very sensitive measuring equipment to detect the
>> curvature at 1g. After all, you get the same curvature on Earth, and
>> people tend not to notice that.
>
> It isn't really about acceleration, it is about speed. It's about how a

No, he is right partially. But standing on earth the acceleration is zero.
You call that *weight*, not acceleration. And so, yes, that's why I say,
SR and GR are two distinct theories.

On Thursday, March 31, 2022 at 9:24:19 AM UTC-5, rotchm wrote:
> On Thursday, March 31, 2022 at 10:09:51 AM UTC-4, wrote:
>
> > I should have said that
> > you need a "STATIONARY frame of reference" in order to do your math,
> No, that is not required to do math.
> > and there is nothing stationary in our observable universe.
> Yes there are. In physics, a reference frame has an operational definition.
> For instance if I Define my table as the frame and I put my glass of water on it.
> If my glass of water stays at the same place on the table then my glass of water is stationary relative to that Frame.
> Do you agree with this?

I agree that that is what you do in order to do math. But it has NOTHING to
do with the REALITY that the table is moving around the earth as the earth
spins on its axis at 1040 mph, while the earth also orbits the sun at 67,000 mph,
while the sun orbits the center of the Milky Way galaxy at 486,000 mph, etc.

>
> So you see, there are things that are stationary in our observable universe.

No, there is NOTHING stationary in our observable universe. Being in a closed
room doesn't affect the universe, it just affects what you see in that closed room.

Ed

On Thursday, March 31, 2022 at 7:05:50 AM UTC-7, det...@outlook.com wrote:
> I should have said, "YOU need a STATIONARY frame of reference in
> order to do YOUR math."

Not at all. There is a system of inertial coordinates in each state of motion, and none of them are regarded as absolutely stationary (whatever that might mean). However, the principle of relativity is that the expressions of the laws of physics take exactly the same form in terms of every system of inertial coordinates. This is what enables us to answer the question you asked, i.e., the pulse of light propagates in a straight line (constant angle) in terms of each system of inertial coordinates, including the one in which the rocket is at rest.

If the pulse is directed perpendicular to the rocket's axis in terms of that system, it will continue in that direction in terms of that system, and of course it's path is then not perpendicular in terms of (say) the earth's inertial system.

> I do not need any FICTITIOUS absolute rest frames.

But you invoke an absolute rest frame by referring to the point where the emission occurred, as if that phrase has unambiguous (absolute) meaning. It doesn't, because the emission event doesn't single out any particular state of motion, so you can't extrapolate it to any other time. This is explained as follows:

> > For example, suppose you emit the pulse of light when you are passing a several
> > space buoys in various states of motion that all coincided with your rocket at the
> > time of the emission. Thereafter, which of those buoys do you regard as being at
> > "the point where the photon was emitted"? It would have to be the buoy that is at
> > absolute rest, but there is no such thing. So you can't talk meaningfully about "the
> > [spatial] point where the pulse was emitted" at any time after the emission. It is
> > essential to understand this.
>
> See above.

But "the above" is just your fallacious reasoning that has been thoroughly debunked. Now that your reasoning has been debunked, it makes no sense for you to refer back to it, as if it hadn't been debunked. If you disagree with the debunking, then the question to you stands: Which buoy is "at the point where the pulse was emitted"? You can't answer that question, which reveals that your beliefs make no sense.

On Thursday, March 31, 2022 at 9:41:18 AM UTC-5, Paparios wrote:
> El jueves, 31 de marzo de 2022 a las 11:09:51 UTC-3, escribió:
> > On Wednesday, March 30, 2022 at 4:52:48 PM UTC-5, rotchm wrote:
> > > On Wednesday, March 30, 2022 at 3:58:59 PM UTC-4, wrote:
> > >
> > > > You're talking math. I'm talking science.
> > > > You need a "frame of reference" in order to do your math.
> > > You have it the wrong way. Science requires reference frames.
> > > Are you sure you know what "reference frame" means?
> > >
> > > Math does not require reference frames (but did you often make use of it).
> > Okay, as I stated in a previous response, I should have said that
> > you need a "STATIONARY frame of reference" in order to do your math,
> > and there is nothing stationary in our observable universe.
> >
> Physics models are tested by using experiments. Defining a stationary frame of reference (also called at rest frame of reference) is crucial to perform such experiments. That Earth is round, rotating and orbiting the Sun does not invalidate the use of "STATIONARY frame of reference". The policeman's location, with a radar gun, standing at a side of a highway, is a valid stationary frame of reference, which allows him to measure the incoming cars speed with a high precision.

I AGREE! You cannot measure the ACTUAL speed of a rocket that is
moving away from the earth, because the "actual" speed requires knowing
the speed of the earth around the sun, the speed of the sun around the
center of the Milky Way galaxy, etc. So you PRETEND that the earth is
"stationary" and you compute the speed of the rocket away from the
"stationary" earth. You call it a "frame of reference." The MATH WORKS
because you are just computing the speed away from the earth, not
the rocket's actual speed through the universe.

Where this turns to TOTAL LUNACY is when mathematicians argue that
it also means that the earth is moving away from the rocket at speed x.

>
> Most physics laboratories can also not take into account the gravity changes within the reduce space of the experiment setting.
>
> Without a frame of reference (or system of coordinates), none of the following terms make any sense: speed, time, location, direction, etc.

Agreed. But the point is: That "frame of reference" is NOT REAL. It is created
for mathematical purposes. And those "purposes" sometimes involve life and death.
So, doing the math is important. The argument is not that math is a waste of time,
the argument is that math does not always represent reality.

Ed

On Thursday, March 31, 2022 at 10:19:21 AM UTC-5, Townes Olson wrote:
> On Thursday, March 31, 2022 at 7:05:50 AM UTC-7, wrote:
> > I should have said, "YOU need a STATIONARY frame of reference in
> > order to do YOUR math."
> Not at all. There is a system of inertial coordinates in each state of motion, and none of them are regarded as absolutely stationary (whatever that might mean). However, the principle of relativity is that the expressions of the laws of physics take exactly the same form in terms of every system of inertial coordinates. This is what enables us to answer the question you asked, i.e., the pulse of light propagates in a straight line (constant angle) in terms of each system of inertial coordinates, including the one in which the rocket is at rest.
>
> If the pulse is directed perpendicular to the rocket's axis in terms of that system, it will continue in that direction in terms of that system, and of course it's path is then not perpendicular in terms of (say) the earth's inertial system.

The earth's inertial system has nothing to do with anything.

> > I do not need any FICTITIOUS absolute rest frames.
> But you invoke an absolute rest frame by referring to the point where the emission occurred, as if that phrase has unambiguous (absolute) meaning. It doesn't, because the emission event doesn't single out any particular state of motion, so you can't extrapolate it to any other time. This is explained as follows:
> > > For example, suppose you emit the pulse of light when you are passing a several
> > > space buoys in various states of motion that all coincided with your rocket at the
> > > time of the emission. Thereafter, which of those buoys do you regard as being at
> > > "the point where the photon was emitted"? It would have to be the buoy that is at
> > > absolute rest, but there is no such thing. So you can't talk meaningfully about "the
> > > [spatial] point where the pulse was emitted" at any time after the emission. It is
> > > essential to understand this.

But you don't understand the question. The question is: When an atom
emits a photon, does that photon travel away from the atom, or does it
travel away from the point in space where that atom was when it emitted
the photon?

If the photon travels away from the point in space where the atoms WAS when it
emitted the photon, that point COULD BE a point that is at "absolute rest."
To argue that "there is no such thing" is just stating an OPINION.

Ed

On 3/31/2022 10:28 AM, Ed Lake wrote:
> On Wednesday, March 30, 2022 at 5:52:57 PM UTC-5, Michael Moroney wrote:
>> On 3/30/2022 12:33 PM, Ed Lake wrote:
>>> On Wednesday, March 30, 2022 at 11:05:59 AM UTC-5, Michael Moroney wrote:
>>>> On 3/30/2022 10:51 AM, Ed Lake wrote:
>>>>
>>>>> My question is very simple: If I am traveling through space, away from
>>>>> the Earth and toward Alpha Centauri, will a light photon that I emit at a
>>>>> right angle to my direction of travel continue to move at a right angle to
>>>>> me, or will it move at a right angle to the point where the photon was emitted?
>>>> In which frame? That of the spaceship or of Earth (or Alpha Centauri)?
>>>> Since I already know you don't understand the concept of frames in
>>>> physics, I don't expect a rational answer to this, or a rational
>>>> response to anyone who answers you.
>>>>>
>>>>> The answer seems obvious: The photon will move away from the point
>>>>> where it was emitted, not away from me.
>>>> Did you do the special relativity math to work out the answer? Oh that's
>>>> right, "mathematicians" are your boogeymen so math is the incantation of
>>>> evil.
>>>
>>> No, it just clouds the issue. It is as Einstein once said, "“As far as the laws of
>>> mathematics refer to reality, they are not certain; and as far as they are certain,
>>> they do not refer to reality.”
>> THAT's the basis of your belief? A single mined quote of Einstein's,
>> almost certainly taken out of context?
>
> It's not taken out of context. It's from Einstein's speech to the Prussian Academy
> of Sciences in Berlin on January 27, 1921. The whole speech was about that topic.
> It was titled "Geometry and Experience."
>
> A link: https://todayinsci.com/E/Einstein_Albert/Einstein-GeometryAndExperience.htm
>
It is taken out of context. He was talking about how mathematics is an
exact science while the other sciences were not exact. Quote:

"One reason why mathematics enjoys special esteem, above all other
sciences, is that its laws are absolutely certain and indisputable,
while those of all other sciences are to some extent debatable and in
constant danger of being overthrown by newly discovered facts."

In this context, "As far as the laws of mathematics refer to reality,
they are not certain;" means a mathematical relationship in science is
not going to be exact. For example we learn a well hit baseball moves in
a parabola, it's not an _exact_ parabola, however. The second part,
"and as far as they are certain, they do not refer to reality" says
largely the same thing, an exact mathematical formula isn't going to
appear in other sciences as an exact answer. Again, the baseball
doesn't move in an exact parabola.

> The whole speech was about that topic.

What topic? The evils of mathematics? No, the rest of the speech was
first about axioms in geometry, types of geometry and then about
(spherical) non-Euclidean geometry and how Einstein used it in GR.

The speech was not a dig against mathematics in science, in fact most of
it was about geometry and how he used it.

So yes, this is a mined quote taken out of context.

El jueves, 31 de marzo de 2022 a las 12:20:31 UTC-3, det...@outlook.com escribió:
> On Thursday, March 31, 2022 at 9:41:18 AM UTC-5, Paparios wrote:

> > Physics models are tested by using experiments. Defining a stationary frame of reference (also called at rest frame of reference) is crucial to perform such experiments. That Earth is round, rotating and orbiting the Sun does not invalidate the use of "STATIONARY frame of reference". The policeman's location, with a radar gun, standing at a side of a highway, is a valid stationary frame of reference, which allows him to measure the incoming cars speed with a high precision.

> I AGREE! You cannot measure the ACTUAL speed of a rocket that is
> moving away from the earth, because the "actual" speed requires knowing
> the speed of the earth around the sun, the speed of the sun around the
> center of the Milky Way galaxy, etc. So you PRETEND that the earth is
> "stationary" and you compute the speed of the rocket away from the
> "stationary" earth. You call it a "frame of reference." The MATH WORKS
> because you are just computing the speed away from the earth, not
> the rocket's actual speed through the universe.
>

That does not make any sense. We know how to measure the actual speed of a rocket that is moving away from the Earth. For over 40 years, manmade vehicles have been launched away from Earth. For instance, Voyager 1 was launched on September 5, 1977. Voyager 1 has been operating for 44 years, 6 months and 26 days as of March 31, 2022. On February 17, 1998, Voyager 1 reached a distance of 69 AU (10.3 billion km) from the Sun and overtook Pioneer 10 as the most distant spacecraft from Earth. Travelling at about 17 km/s, it has the fastest heliocentric recession speed of any spacecraft. As of 2013, the probe was moving with a relative velocity to the Sun of about 61,197 kilometers per hour. With the velocity the probe is currently maintaining, Voyager 1 is traveling about 523 million km per year, or about one light-year per 18,000 years.

> Where this turns to TOTAL LUNACY is when mathematicians argue that
> it also means that the earth is moving away from the rocket at speed x.

The lunacy is totally yours. If a body A is approaching/receding a body B it is evident that also the body B is approaching/receding the body A. This is not mathematics but it is just common sense!!!

On Thursday, March 31, 2022 at 8:36:24 AM UTC-7, det...@outlook.com wrote:
> > > I should have said, "YOU need a STATIONARY frame of reference in
> > > order to do YOUR math."
> > Not at all. There is a system of inertial coordinates in each state of motion, and none of them are regarded as absolutely stationary (whatever that might mean). However, the principle of relativity is that the expressions of the laws of physics take exactly the same form in terms of every system of inertial coordinates. This is what enables us to answer the question you asked, i.e., the pulse of light propagates in a straight line (constant angle) in terms of each system of inertial coordinates, including the one in which the rocket is at rest.
> >
> > If the pulse is directed perpendicular to the rocket's axis in terms of that system, it will continue in that direction in terms of that system, and of course it's path is then not perpendicular in terms of (say) the earth's inertial system.
>
> The earth's inertial system has nothing to do with anything.

It has to do with the angle between the path of the pulse and the direction of the rocket in terms of the earth's inertial system, which is not perpendicular, whereas it is perpendicular (by stipulation) in terms of the inertial reference system of the rocket. The reason for mentioning this is because you are subconsciously smuggling in the earth's (or some other putative "absolute rest" system) into your reasoning, which is what's leading you astray.

> > For example, suppose you emit the pulse of light when you are passing a several
> > space buoys in various states of motion that all coincided with your rocket at the
> > time of the emission. Thereafter, which of those buoys do you regard as being at
> > "the point where the photon was emitted"? It would have to be the buoy that is at
> > absolute rest, but there is no such thing. So you can't talk meaningfully about "the
> > [spatial] point where the pulse was emitted" at any time after the emission. It is
> > essential to understand this.
>
> If the photon travels away from the point in space where the atoms WAS when it
> emitted the photon, that point COULD BE a point that is at "absolute rest.."

Wait... so now you are admitting that your reasoning is assuming some absolute rest? Please just answer the question: Which of those buoy's is subsequently at the point where the photon was emitted? You can't answer that question. And bear in mind that the principle of relativity assures us of the simple answer to your question, i.e., the pulse moves at constant direction in terms of any given inertial reference system, and of course that direction depends on the initial conditions of the pulse. If you send it perpendicular in the rocket's reference system, then it will be perpendicular in the rocket's reference system. Duh.

On Thursday, 31 March 2022 at 16:41:18 UTC+2, Paparios wrote:
> El jueves, 31 de marzo de 2022 a las 11:09:51 UTC-3, det...@outlook.com escribió:
> > On Wednesday, March 30, 2022 at 4:52:48 PM UTC-5, rotchm wrote:
> > > On Wednesday, March 30, 2022 at 3:58:59 PM UTC-4, wrote:
> > >
> > > > You're talking math. I'm talking science.
> > > > You need a "frame of reference" in order to do your math.
> > > You have it the wrong way. Science requires reference frames.
> > > Are you sure you know what "reference frame" means?
> > >
> > > Math does not require reference frames (but did you often make use of it).
> > Okay, as I stated in a previous response, I should have said that
> > you need a "STATIONARY frame of reference" in order to do your math,
> > and there is nothing stationary in our observable universe.
> >
> Physics models are tested by using experiments.

Gedanken, of course.

On 3/31/2022 10:55 AM, Ed Lake wrote:
> On Thursday, March 31, 2022 at 9:24:19 AM UTC-5, rotchm wrote:
>> On Thursday, March 31, 2022 at 10:09:51 AM UTC-4, wrote:
>>
>>> I should have said that
>>> you need a "STATIONARY frame of reference" in order to do your math,

>> No, that is not required to do math.

>>> and there is nothing stationary in our observable universe.

>> Yes there are. In physics, a reference frame has an operational definition.
>> For instance if I Define my table as the frame and I put my glass of water on it.
>> If my glass of water stays at the same place on the table then my glass of water is stationary relative to that Frame.
>> Do you agree with this?
>
> I agree that that is what you do in order to do math. But it has NOTHING to
> do with the REALITY that the table is moving around the earth as the earth
> spins on its axis at 1040 mph, while the earth also orbits the sun at 67,000 mph,
> while the sun orbits the center of the Milky Way galaxy at 486,000 mph, etc.

Not relevant. The glass is stationary relative to the frame of the
table, even if the frame of the table isn't stationary relative to the
earth, the sun, the Milky Way etc.

As I said, you don't understand the concept of frames and how they are
used in physics (not math). Frames are fictional; anyone can make up any
frame they want. If rotcm wants to use a frame stationary relative to
his table, he can.

Do you want to discuss the physics of whether such a frame is inertial?
THEN you can bring up the earth's rotation and orbit, the sun's orbit etc.
>
>>
>> So you see, there are things that are stationary in our observable universe.
>
> No, there is NOTHING stationary in our observable universe. Being in a closed
> room doesn't affect the universe, it just affects what you see in that closed room.
>
> Ed

Ed Lake <detect@outlook.com> wrote:
> On Thursday, March 31, 2022 at 9:24:19 AM UTC-5, rotchm wrote:
>> On Thursday, March 31, 2022 at 10:09:51 AM UTC-4, wrote:
>>
>>> I should have said that
>>> you need a "STATIONARY frame of reference" in order to do your math,
>> No, that is not required to do math.
>>> and there is nothing stationary in our observable universe.
>> Yes there are. In physics, a reference frame has an operational definition.
>> For instance if I Define my table as the frame and I put my glass of water on it.
>> If my glass of water stays at the same place on the table then my glass
>> of water is stationary relative to that Frame.
>> Do you agree with this?
>
> I agree that that is what you do in order to do math. But it has NOTHING to
> do with the REALITY that the table is moving around the earth as the earth
> spins on its axis at 1040 mph, while the earth also orbits the sun at 67,000 mph,
> while the sun orbits the center of the Milky Way galaxy at 486,000 mph, etc.

Notice that in each case above, you are citing a relative speed.
The surface of the earth is moving 1040 mph relative to a line between the
center of the sun and the center of the earth. The earth is moving at
67,000 mph relative to a line from the center of the sun to a distant star,
the sun is moving at 486,000 mph relative to some line from the center of
the galaxy to another distant galaxy. There is no absolute speed in any of
those statements.

>
>>
>> So you see, there are things that are stationary in our observable universe.
>
> No, there is NOTHING stationary in our observable universe. Being in a closed
> room doesn't affect the universe, it just affects what you see in that closed room.
>
> Ed
>

--
Odd Bodkin -- maker of fine toys, tools, tables

On Thursday, March 31, 2022 at 10:42:35 AM UTC-4, det...@outlook.com wrote:
> On Wednesday, March 30, 2022 at 5:55:32 PM UTC-5, Sylvia Else wrote:

> That's the question: DOES the effect truly require acceleration?

Again: Everytime such experiments were done within an inertial frame, the path of light remain "straight".
And every time the frame was not an inertial one, the path of the light was curved.

That's the data.

> It isn't really about acceleration, it is about speed.

And those Notions/words will not change my above paragraph; will not change the data.
Do you agree with that?

On Thursday, March 31, 2022 at 10:55:59 AM UTC-4, det...@outlook.com wrote:
> On Thursday, March 31, 2022 at 9:24:19 AM UTC-5, rotchm wrote:

> > For instance if I Define my table as the frame and I put my glass of water on it.
> > If my glass of water stays at the same place on the table then my glass of water is stationary relative to that Frame.
> > Do you agree with this?

> I agree that that is what you do in order to do math.

No. In the above, I have not used any math. I used only physics, real stuff.
Putting a glass of water on the table is physical; it's not math. Noting that it stays at the same position is physics not math.
Do you agree with this?

> But it has NOTHING to do with the REALITY

A glass of water on a table is not reality?
Noting that it stays in place is not reality?

> > So you see, there are things that are stationary in our observable universe.
> No, there is NOTHING stationary in our observable universe.

The glass of water does not remain at the same location on the table?
You place it at the center of the table, say. It does not remain there?

> Being in a closed
> room doesn't affect the universe, it just affects what you see in that closed room.

And do you realize that what you see is a "reference frame"?
Say the car is red you have set up a reference frame. To say an object is at a certain location you have set up a reference frame.
Every time you describe something are giving adjectives to it, is implicitly a reference frame. Did you see the word "reference" in those sentences?

Okay. I think I'm done posting for awhile. All we're doing is arguing
over words and phrasing.

I started this thread because it seems that when an atom emits a photon,
that photon travels at c away from the POINT IN SPACE where that atom
WAS when it emitted the photon. That POINT IN SPACE is not part of any
"system" and MAY be a truly STATIONARY POINT IN SPACE. The question
then becomes: Is there any way to mark that point in space and measure
other movements relative to that point?

Ed

On Thursday, March 31, 2022 at 2:04:36 PM UTC-7, det...@outlook.com wrote:
> Is there any way to mark that point in space and measure
> other movements relative to that point?

No, that's elementary relativity, and the point is that if you have many buoys in many states of motion all intersecting at an event where a flash of light is emitted in all directions, thereafter the light will be an expanding sphere centered on each buoy and expanding at speed c in terms of the respective system of reference. So your next questions should be: How is it possible for the expanding light sphere to be centered on each buoy, considering that they buoys in moving relative to each other? When you understand the answer to this question, you will be on your way toward understanding special relativity.

Ed Lake <detect@outlook.com> wrote:
> Okay. I think I'm done posting for awhile. All we're doing is arguing
> over words and phrasing.
>
> I started this thread because it seems that when an atom emits a photon,
> that photon travels at c away from the POINT IN SPACE where that atom
> WAS when it emitted the photon. That POINT IN SPACE is not part of any
> "system" and MAY be a truly STATIONARY POINT IN SPACE.

And you seem blissfully unaware that “a truly STATIONARY POINT IN SPACE”
does not exist, and is even meaningless in terms of any measurable
property.

> The question
> then becomes: Is there any way to mark that point in space and measure
> other movements relative to that point?
>
> Ed
>

--
Odd Bodkin -- maker of fine toys, tools, tables