A science question about the trajectory of light

Ed Lake

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Mar 29, 2022, 11:55:46 AM3/29/22

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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

Odd Bodkin

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Mar 29, 2022, 2:08:41 PM3/29/22

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Ed Lake <det...@outlook.com> 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?”

It talks a lot about experimental tests of relativity.

>
> 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
>

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

Paul B. Andersen

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Mar 29, 2022, 2:13:27 PM3/29/22

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Look up Galilean relativity.
When you have learned the relativity of the 17th, 18th.
and 19th centuries, you can start learning the 20th and 21th
centuries relativity.

Think about this question first:

You are in a spaceship with now windows somewhere in
the universe. The ship has no rockets, so it is not
accelerating.
What is your speed?
Can you measure it with instrument inside your ship?

--
Paul

https://paulba.no/

Townes Olson

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Mar 29, 2022, 2:50:20 PM3/29/22

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On Tuesday, March 29, 2022 at 8:55:46 AM UTC-7, det...@outlook.com wrote:
> 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. 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?

It's essential to distinguish between velocity and acceleration (rate of change of velocity). You mentioned that the ship is undergoing 1g of proper acceleration. This is what determines the amount of deflection ("downward curving") of a light pulse, i.e., you will find the same amount of deflection that you would find in a room stationary on the Earths surface, experiencing 1g of gravity. Note that the amount of deflection relative to your rocket depends only on the proper acceleration, not on the velocity. In other words, if you maintain 1 g of proper acceleration from the beginning (when you start with 0 velocity) to one year later (when your velocity relative to the earth is nearly c), you will always find the same deflection inside your rocket.

> 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?

The pulse will follow a straight path in terms of any system of inertial coordinates, though of course your accelerating space ship is not at rest in any such system, which is why you have the deflection in the ship. Also, note that the direction of the pulse is at right angles to the direction of the ship only in terms of one particular system of inertial coordinates, not in terms of others, because of the effect of aberration.

> 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.

By this method you are measuring the acceleration (rate of change of velocity), not the velocity. This is precisely how many modern navigational devices work, using lasers to measure rotational or longitudinal acceleration. Of course, we can then use dead reckoning to integrate our velocity and positions relative to our original location and state of motion.

Ed Lake

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Mar 29, 2022, 3:33:34 PM3/29/22

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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

Ed Lake

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Mar 29, 2022, 3:46:32 PM3/29/22

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On Tuesday, March 29, 2022 at 1:13:27 PM UTC-5, Paul B. Andersen wrote:
> Den 29.03.2022 17:55, skrev Ed Lake:
> > 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
> Look up Galilean relativity.
> When you have learned the relativity of the 17th, 18th.
> and 19th centuries, you can start learning the 20th and 21th
> centuries relativity.
>
> Think about this question first:
>

> You are in a spaceship with no windows somewhere in

> the universe. The ship has no rockets, so it is not
> accelerating.
> What is your speed?
> Can you measure it with instrument inside your ship?

Supposedly not. But the real question is: Does a photon ALWAYS travel
in a straight line away from the emitting atom, or can it also travel sideways
if the emitting atom is traveling sideways?
I've seen no reason to believe a photon can travel sideways.

If it can't also travel sideways, then why would a photon travel in a straight line
from wall to wall even when the ship is NOT accelerating?

I'm not advocating anything here. I'm just trying to figure out why you CAN'T
measure your speed (in theory) even when not accelerating. In practice, of
course, light would travel across a room so fast that there is no way to measure
whether it traveled in a straight line or not.

Ed

Odd Bodkin

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Mar 29, 2022, 3:53:51 PM3/29/22

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Well, the key thing here, Ed, as I said before, is that it’s better if you
abstain from forming a strong opinion about the answer until you learn
something about it first.

Otherwise your habit is to form a strong opinion first, then to mine small
soundbites that support your strong opinion, and to declare other things
wrong.

Ed Lake

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Mar 29, 2022, 4:01:44 PM3/29/22

to

On Tuesday, March 29, 2022 at 1:50:20 PM UTC-5, Townes Olson wrote:

> On Tuesday, March 29, 2022 at 8:55:46 AM UTC-7, wrote:
> > 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. 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?
> It's essential to distinguish between velocity and acceleration (rate of change of velocity). You mentioned that the ship is undergoing 1g of proper acceleration. This is what determines the amount of deflection ("downward curving") of a light pulse, i.e., you will find the same amount of deflection that you would find in a room stationary on the Earths surface, experiencing 1g of gravity. Note that the amount of deflection relative to your rocket depends only on the proper acceleration, not on the velocity. In other words, if you maintain 1 g of proper acceleration from the beginning (when you start with 0 velocity) to one year later (when your velocity relative to the earth is nearly c), you will always find the same deflection inside your rocket.

Actually, the question doesn't seem to have anything to do with velocity OR
acceleration. The only real question seems to be: Does a photon ALWAYS travel in a
straight line away from the POINT IN SPACE where the atom EMITTED the photon
EVEN WHEN the atom is moving at some angle?

> > 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?
> The pulse will follow a straight path in terms of any system of inertial coordinates, though of course your accelerating space ship is not at rest in any such system, which is why you have the deflection in the ship. Also, note that the direction of the pulse is at right angles to the direction of the ship only in terms of one particular system of inertial coordinates, not in terms of others, because of the effect of aberration.

The "system of inertial coordinates" appears to begin at the atom that emitted the
photon AT THE INSTANT the photon was emitted. The atom then goes in one direction
while the photon travels at a right angle to where the atom WAS.

> > 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.
> By this method you are measuring the acceleration (rate of change of velocity), not the velocity. This is precisely how many modern navigational devices work, using lasers to measure rotational or longitudinal acceleration. Of course, we can then use dead reckoning to integrate our velocity and positions relative to our original location and state of motion.

I would think that if "modern navigational devices" measure acceleration, they measure it
the way radar guns do, by measuring a change in frequencies caused by velocity time dilation.
But, I could be wrong.

Ed

Ed Lake

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Mar 29, 2022, 4:08:25 PM3/29/22

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Actually, I don't think I have any "strong opinions" about physics. I think I
just look at the facts and evidence. If the facts and evidence indicate that
something is true, I'll stick with what seems to be true. It's not an opinion,
it is just what seems to be true based upon the facts and evidence. NEW
FACTS could INSTANTLY change my evaluation of the facts and evidence.

Personal attacks won't change anything.

Ed

Odd Bodkin

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Mar 29, 2022, 4:57:01 PM3/29/22

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Ed Lake <det...@outlook.com> wrote:
> On Tuesday, March 29, 2022 at 1:50:20 PM UTC-5, Townes Olson wrote:
>> On Tuesday, March 29, 2022 at 8:55:46 AM UTC-7, wrote:
>>> 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. 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?
>> It's essential to distinguish between velocity and acceleration (rate of
>> change of velocity). You mentioned that the ship is undergoing 1g of
>> proper acceleration. This is what determines the amount of deflection
>> ("downward curving") of a light pulse, i.e., you will find the same
>> amount of deflection that you would find in a room stationary on the
>> Earths surface, experiencing 1g of gravity. Note that the amount of
>> deflection relative to your rocket depends only on the proper
>> acceleration, not on the velocity. In other words, if you maintain 1 g
>> of proper acceleration from the beginning (when you start with 0
>> velocity) to one year later (when your velocity relative to the earth is
>> nearly c), you will always find the same deflection inside your rocket.
>
> Actually, the question doesn't seem to have anything to do with velocity OR
> acceleration. The only real question seems to be: Does a photon ALWAYS travel in a
> straight line away from the POINT IN SPACE where the atom EMITTED the photon
> EVEN WHEN the atom is moving at some angle?

If you had an accelerating rocket ship and you had a bunch of equally
spaced sheets of paper between the source and the wall, and you punched
holes in the cardboard sheets where you saw a spot, until the beam could
travel all the way to the wall, then the succession of holes would not be
something you could pass a straight broomstick through.

Odd Bodkin

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Mar 29, 2022, 4:57:02 PM3/29/22

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Frankly, Ed, your history doesn’t suggest that.

What you do is form an opinion based on what makes common sense to you.
Then you look at the evidence, and selectively categorize evidence
according to whether it agrees with your common sense or not. That which
doesn’t, you call wrong or misinterpreted.

> If the facts and evidence indicate that
> something is true, I'll stick with what seems to be true. It's not an opinion,
> it is just what seems to be true based upon the facts and evidence. NEW
> FACTS could INSTANTLY change my evaluation of the facts and evidence.
>
> Personal attacks won't change anything.
>
> Ed
>

Tom Roberts

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Mar 29, 2022, 5:27:38 PM3/29/22

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On 3/29/22 10:55 AM, 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.

Actually it follows a null geodesic through spacetime. Gravitational
lensing conclusively demonstrates this -- there are many instances
of observing multiple images of a single distant quasar or galaxy, so
the light trajectories cannot possibly be "straight".

> 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/

As always, you MUST specify the coordinates relative to which you
measure a speed (or velocity). Ditto for an acceleration and a duration.

Note it simply is not possible to accelerate at 1g (=9.8m/s^2) relative
to an inertial frame for a year, even in a gedanken. But rockets are
usually specified by their PROPER acceleration. If a rocket started from
rest in an inertial frame and had a proper acceleration of 1g for a year
(either in that frame or elapsed proper time), it would indeed approach
speed c relative to that frame.

When considering relativistic speeds like this, one must evaluate the
accuracy of an inertial frame over such large distances and times. In
this case, if the ship started from earth, the ICRF would remain an
accurately-inertial reference frame, as the rocket never approaches any
star. But not necessarily for 10 years....

> 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?

As usual, your words are ambiguous. You REALLY need to learn how to ask
a specific, self-consistent question. Of course to be able to do that
you would have to understand some basic physics, which you have
repeatedly refused to do.

"At such speeds" kind of implies you are no longer considering the
acceleration. If the rocket ceased accelerating after a year, and you
then do this, the light beam will traverse the ship in a straight line
(measured relative to the rocket's inertial frame).

But that implication is not necessarily what you meant, and "floor"
implies it keeps accelerating. If the rocket is still accelerating at 1g
when you to this, then the light beam will traverse the ship in a
parabola (measured relative to the rocket's locally inertial frame).
This applies at all times during the acceleration, not just after one
year; the spot position on the far wall depends on the proper
acceleration, not time or speed relative to anything. This assumes that
the laser is affixed to its wall with sufficient rigidity so it does not
rotate relative to the ship, and the far wall is similarly rigid.

Note it is not possible to accurately measure the difference between a
straight line and that parabola, unless you can measure the difference
between acceleration on and off -- in that case, if the ship is 10
meters wide then the difference would be ~5E-15 meters; even ignoring
the impossibility of measuring a laser spot position to such accuracy,
with current technology that is not possible to measure [#].

[#] Today the highest-resolution measurement of distance
is ~1E-12 meters, which just happens to be in our
Precision Laser Metrology Lab at IIT.

> I am emitting light in a direction that is at a right angle to MY
> direction of movement.

Not really. You are implicitly assuming some sort of "absolute motion",
which is invalid.

> 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?

I must guess what you are trying to say. My interpretation of your words
means the answer is: No. In particular, the laser is affixed to the
ship, not somewhere in "space" (whatever you mean by that).

> 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.

Nope. The position of the light beam at the other wall depends on the
ship's proper acceleration, not its velocity (relative to anything). And
for the difference to be measurable in a 10-meter-wide spaceship the
acceleration would need to be far too great for humans to survive.

A simple spring scale with a calibrated weight would measure such a
ship's proper acceleration vastly better than the position of a laser
spot on the opposite wall.

> [... further nonsense ignored]

Tom Roberts

rotchm

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Mar 29, 2022, 6:10:11 PM3/29/22

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On Tuesday, March 29, 2022 at 4:01:44 PM UTC-4, det...@outlook.com wrote:
> The only real question seems to be: Does a photon ALWAYS travel in a
> straight line away from the POINT IN SPACE where the atom EMITTED the photon

Now, that's a 'good' question for you to ponder.

The only way to know, is to test it, in many environments. Luckily, past and present physicists have done that.
Their conclusion: sometimes it goes in a straight line, sometimes it doesn't.
But this is what they have noticed in every case: if the Observer (his reference frame) was an inertial one, the photons always travel in a straight line. If however is reference frame was not an inertial one, the photon typically did travel in a straight line.

SR/GR say/models the above by using equations.

Townes Olson

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Mar 29, 2022, 7:41:39 PM3/29/22

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On Tuesday, March 29, 2022 at 1:01:44 PM UTC-7, det...@outlook.com wrote:
> > > 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. 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?
> > It's essential to distinguish between velocity and acceleration (rate of change of velocity). You mentioned that the ship is undergoing 1g of proper acceleration. This is what determines the amount of deflection ("downward curving") of a light pulse, i.e., you will find the same amount of deflection that you would find in a room stationary on the Earths surface, experiencing 1g of gravity. Note that the amount of deflection relative to your rocket depends only on the proper acceleration, not on the velocity. In other words, if you maintain 1 g of proper acceleration from the beginning (when you start with 0 velocity) to one year later (when your velocity relative to the earth is nearly c), you will always find the same deflection inside your rocket.
>

> The question doesn't seem to have anything to do with velocity OR acceleration.

The question said "a rocket ship accelerates at 1G..." and "traveling at such speeds...", so I don't understand what you mean when you say the question had nothing to do with velocity or acceleration.

To re-iterate, the "curving" that you asked about is due to the proper acceleration, not to velocity, but the aberration that implies the path is not perpendicular to the rocket's axis is due to velocity, not acceleration. Also, bear in mind that the rocket could (in principle) maintain constant proper acceleration, but it's acceleration in terms of (say) the Earth's inertial coordinate system will asymptotically approach zero as its speed approaches c.

> Does a photon ALWAYS travel in a straight line away from the POINT IN SPACE
> where the atom EMITTED the photon EVEN WHEN the atom is moving at some angle?

It's essential to distinguish between curving paths versus aberration. Neglecting the effects of quantum uncertainty, a photon propagates in a straight line in terms of any local system of inertial coordinates, though not in terms of accelerating coordinates (such as those in which your accelerating rocket is at rest), and not in terms of a global coordinate system in the presence of gravity. In contrast, the angle that a photon's path makes with the rocket's axis depends on the system of inertial coordinates, because of aberration, but the path is still straight, it's just the angle that is dependent on the coordinate system.

> The "system of inertial coordinates" appears to begin at the atom that emitted the
> photon AT THE INSTANT the photon was emitted. The atom then goes in one direction
> while the photon travels at a right angle to where the atom WAS.

Again, the angle between the path of a photon and the axis of the rocket depends on the system of inertial coordinates in which you express that angle. This is called aberration, and has always been part of physics (discovered in 1727), although the relativistic expression for aberration is different to account for the relativity of simultaneity. But don't confuse this with the curving of paths in terms of accelerating coordinates (e.g., in your rocket), nor with the curving of paths in a gravitational field.

By the way, there's no need to put "system of inertial coordinates" in quotation marks. It is a perfectly objective and well defined system of measure, as given by a grid of standard rulers with clocks inertially synchronized at each node. These are the coordinate systems in terms of which the laws of physics take their simple homogeneous and isotropic form, so they are very important and physically meaningful.

> I would think that if "modern navigational devices" measure acceleration...

Again, no need for quotation marks. Accelerometers and ring laser gyroscopes, etc., are commonplace. These are used in most modern navigation devices, although there are still some mechanical gyros and accelerometers in use. Same principle of operation.

> they measure it the way radar guns do, by measuring a change in frequencies
> caused by velocity time dilation.

It's possible to measure the rate of change of distance to some other object using radar, giving the relative velocity, but navigation devices are intended to measure *absolute* acceleration, not the relative velocity or acceleration between the device and some other arbitrary object.

Maciej Wozniak

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Mar 30, 2022, 1:10:31 AM3/30/22

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On Tuesday, 29 March 2022 at 20:13:27 UTC+2, Paul B. Andersen wrote:
> Den 29.03.2022 17:55, skrev Ed Lake:
> > 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
> Look up Galilean relativity.

All people on Earth must observe and claim, that Sun is rotating
around! That's what The Laws of Nature are!!!

Maciej Wozniak

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Mar 30, 2022, 1:11:46 AM3/30/22

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And see, that forbidden by insane relativistic
maniacs TAI keep measuring t'=t, just like
all serious clocks always did.

Maciej Wozniak

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Mar 30, 2022, 1:13:47 AM3/30/22

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On Tuesday, 29 March 2022 at 23:27:38 UTC+2, tjrob137 wrote:
> On 3/29/22 10:55 AM, 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.
> Actually it follows a null geodesic through spacetime. Gravitational
> lensing conclusively demonstrates this -- there are many instances
> of observing multiple images of a single distant quasar or galaxy, so
> the light trajectories cannot possibly be "straight".

Tom, poor idiot, have you ever heard of non euclidean
geometries? Oh, yes, it can, and that's one of the main
assumptions of your idiot guru.

Ed Lake

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Mar 30, 2022, 10:51:14 AM3/30/22

to

As always, the problem is that most people here cannot discuss reality.
They only understand mathematics.

It’s a demonstration of Einstein’s famous quote: ““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.”

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?

The answer seems obvious: The photon will move away from the point
where it was emitted, not away from me.

And it doesn’t make any difference if I emit the photon while accelerating
or while coasting. The photon will move away from the point where it
was emitted, not away from me.

Yes, I know that the photon may eventually change its trajectory if it passes
through or near some massive galaxy, but that has nothing to do with the
question. It’s just a way to create an argument instead of answering the
question. And you can also create arguments about what words I use, but
that’s just another way to avoid answering the question.

Ed

Michael Moroney

unread,

Mar 30, 2022, 12:05:59 PM3/30/22

to

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.

unread,

Mar 30, 2022, 2:37:07 PM3/30/22

to

On Wednesday, 30 March 2022 at 19:03:20 UTC+2, bodk...@gmail.com wrote:

> For you. Not for people who are mathematically literate.
>
> And just to remind you, mathematical literacy is not confined to
> mathematicians, though you’d live to partition them off,

Speaking of mathematics, it's always good to remind
that your bunch of idiots had to announce its oldest, very
important and successful part false, as it didn't want to
cooperate with your madness.

Townes Olson

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Mar 30, 2022, 2:56:32 PM3/30/22

to

On Wednesday, March 30, 2022 at 7:51:14 AM UTC-7, det...@outlook.com wrote:

> 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...

Yes, if you are not accelerating, then in terms of your co-moving inertial reference system the pulse of light will propagate in a straight line in whatever direction you emitted it. So, if you emitted it perpendicular to the rocket axis in terms of your co-moving inertial system of reference, it will continue to move in that direction at speed c in terms of that system of reference.

Of course, in terms of the Earth's inertial reference system you emitted the pulse at a forward angle, not perpendicular to the axis of your rocket, and the pulse will continue in that non-perpendicular direction in terms of that system of reference. Likewise the pulse will move in a straight line at whatever angle it was emitted in terms of any specified inertial reference system, and it will move at speed c in terms of each of those systems.

Note that this is a different question than you asked originally. You began by describing a ship that is undergoing constant proper acceleration of 1G, and you were asking about the curving of light paths in terms of an accelerating reference system, related to the curving of paths in a gravitational field. Now you have discarded the acceleration, and are purely talking about special relativity in flat spacetime.

> or will it move at a right angle to the point where the photon was emitted?

That phrase 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. For example, suppose you emit the pulse of light when you are passing a several space beacons in various states of motion that all coincided with your rocket at the time of the emission. Thereafter, which of those beacons do you regard at "the point where the photon was emitted"? It would have to be the beacon 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.

> The answer seems obvious: The photon will move away from the point
> where it was emitted, not away from me.

No, that's not correct. In general, the pulse will not be moving in a direction perpendicular to the ship's axis in any system of reference other than that of the ship itself, given that you emitted it perpendicularly in terms of that system of reference.

> And it doesn’t make any difference if I emit the photon while accelerating
> or while coasting.

That too is incorrect. If you are accelerating, then your co-moving system of inertial reference is constantly changing, and the pulse will follow a curved path in terms of your co-moving system of coordinates (which is necessarily limited, since it is accelerating).

Dono.

unread,

Mar 30, 2022, 2:59:40 PM3/30/22

to

On Wednesday, March 30, 2022 at 9:33:16 AM UTC-7, det...@outlook.com wrote:

> Using different frames of reference will just give different answers.

It is called "relativity", stubborn imbecile.

rotchm

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Mar 30, 2022, 3:11:54 PM3/30/22

to

On Wednesday, March 30, 2022 at 10:51:14 AM UTC-4, det...@outlook.com wrote:

> 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 your spaceship, you set up a latticework, a grid, a reference frame. This basically defines what is meant by a straight line, or "above" in your spaceship. If your spaceship is inertial, a photon will continue at right angles, " vertically", say, and hit the ceiling directly above.
I.e.: Your source at x=0, y=0. The photon will hit x=0, y=1, say. In between, is irrelevant what happens to the photon since we are not detecting it in between (along its "trajectory"). We are only sending out the photon, and detecting it at a specified place.

If your ship were not inertial, the photon will typically not hit directly above.

RichD

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Mar 30, 2022, 3:29:39 PM3/30/22

to

On Tuesday, March 29, det...@outlook.com wrote:
> But the real question is: Does a photon ALWAYS travel
> in a straight line away from the emitting atom, or can it also travel sideways
> if the emitting atom is traveling sideways?

> If it can't also travel sideways, then why would a photon travel in a straight line
> from wall to wall even when the ship is NOT accelerating?
> I'm not advocating anything here. I'm just trying to figure out why you CAN'T
> measure your speed (in theory) even when not accelerating.

Of course you can.

It's like, you're a passenger in an airplane, doing 500 mph. The light
fixture above your head breaks and falls. Due to the plane's high speed,
it hits the seat behind you.

Measure the distance it traveled sideways, then you can deduce the
aircraft's speed.

Galileo understood these things 400 years ago.
So you're a bit behind the curve -

--
Rich

Ed Lake

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Mar 30, 2022, 3:48:28 PM3/30/22

to

On Wednesday, March 30, 2022 at 11:05:59 AM UTC-5, Michael Moroney wrote:

It's NOT a math problem! It's a SCIENCE problem.

You can look at it a different way: I'm driving at 50 mph across a bridge,
heading north. The river under the bridge is moving toward the east at 10 mph.
I toss an empty Pringles can into the river. The Pringles can is going to
move east at 10 mph away from the bridge, regardless of what I do.

If I'm moving straight ahead in empty space and shoot out a photon at a
90 degree angle to my course, the photon is going to continue moving
at a 90 degree angle from my path regardless of what I do later.

Whether I was coasting or accelerating at the time I shot out the photon is
irrelevant. It doesn't change the course of the photon.

Ed

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Mar 30, 2022, 4:14:21 PM3/30/22

to

On Wednesday, March 30, 2022 at 2:29:39 PM UTC-5, RichD wrote:

But your example has nothing to do with the problem.
The light fixture would hit me, not the seat behind me, because the
air in the plane is moving at the same speed as the plane. There is
nothing to push the light fixture backwards. It's an inertial system.

Galileo understood these things 400 years ago.

So, you're a bit behind the times.

Ed

Ed Lake

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Mar 30, 2022, 4:16:06 PM3/30/22

to

On Wednesday, March 30, 2022 at 3:05:59 PM UTC-5, RichD wrote:

Hmm.. We agree! That may be a first for this forum.

Ed

thor stoneman

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Mar 30, 2022, 4:30:43 PM3/30/22

to

On Tuesday, March 29, 2022 at 8:55:46 AM UTC-7, det...@outlook.com 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?

Townes Olson

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Mar 30, 2022, 4:39:43 PM3/30/22

to

On Wednesday, March 30, 2022 at 12:58:59 PM UTC-7, det...@outlook.com 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.

> 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.

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.

Dono.

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Mar 30, 2022, 4:47:44 PM3/30/22

to

On Wednesday, March 30, 2022 at 1:05:10 PM UTC-7, det...@outlook.com wrote:
> On Wednesday, March 30, 2022 at 1:59:40 PM UTC-5, Dono. wrote:
> > On Wednesday, March 30, 2022 at 9:33:16 AM UTC-7, wrote:
> >
> > > Using different frames of reference will just give different answers.
> > It is called "relativity", stubborn imbecile.
> Yes, but the problem isn't really about "relativity."

Relativity is the branch of SCIENCE that gives the answer to your question.

Odd Bodkin

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Mar 30, 2022, 5:28:35 PM3/30/22

to

RichD <r_dela...@yahoo.com> wrote:
> On Tuesday, March 29, det...@outlook.com wrote:
>> But the real question is: Does a photon ALWAYS travel
>> in a straight line away from the emitting atom, or can it also travel sideways
>> if the emitting atom is traveling sideways?
>> If it can't also travel sideways, then why would a photon travel in a straight line
>> from wall to wall even when the ship is NOT accelerating?
>> I'm not advocating anything here. I'm just trying to figure out why you CAN'T
>> measure your speed (in theory) even when not accelerating.
>
> Of course you can.
>
> It's like, you're a passenger in an airplane, doing 500 mph. The light
> fixture above your head breaks and falls. Due to the plane's high speed,
> it hits the seat behind you.

LOL.

>
> Measure the distance it traveled sideways, then you can deduce the
> aircraft's speed.

You’re joking.

>
> Galileo understood these things 400 years ago.
> So you're a bit behind the curve -
>

Now you’re REALLY joking.

> --
> Rich

Odd Bodkin

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Mar 30, 2022, 5:28:36 PM3/30/22

to

Nope. It's just that most science models are heavy in math. Like the
home run baseball. You (or an outfielder) may almost instinctively
think of where it will land, but science models have formulas which
predict where it will land, if the angle and speed it left the bat is
known. Again, if you don't like such math (or the boogeyman!) physics
isn't for you.

>
> Using different frames of reference will just give different answers.
> The question is about REALITY. In REALITY, the photon moves away from
> the POINT where it was emitted by an atom. Once the atom has moved
> on, there is nothing at that point. There is just a photon moving through
> empty space. And mathematicians cannot comprehend an object
> moving through empty space.

Mathematicians don't care. They do math, not physics.

> They REQUIRE that ALL OBJECTS move
> relative to some other object.

Nope, that's physics that uses that. Mathematicians deal with the
abstract, it's all numbers and formulas, with no objects or motion,
relative or otherwise.

> But the photon is not moving relative to any object.
>
> Ed

Don't forget to look under your bed before going to sleep! There may be
a MATHEMATICIAN hiding there!!!!

Sylvia Else

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Mar 30, 2022, 6:55:32 PM3/30/22

to

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.

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.

Sylvia.

Maciej Wozniak

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Mar 31, 2022, 12:46:43 AM3/31/22

to

On Thursday, 31 March 2022 at 00:38:58 UTC+2, Michael Moroney wrote:
> On 3/30/2022 4:16 PM, Ed Lake wrote:
> > On Wednesday, March 30, 2022 at 3:05:59 PM UTC-5, RichD wrote:
> >> On March 30, wrote:
> >>> It's NOT a math problem! It's a SCIENCE problem.
> Science these days have LOTS of math.

Speaking of math, it's always good to remind that your

unread,

Mar 31, 2022, 9:07:56 AM3/31/22

to

“I’d rather be original and clueless than a knowledgeable sheeple”

Odd Bodkin

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Mar 31, 2022, 9:18:14 AM3/31/22

to

Michael Moroney <mor...@world.std.spaamtrap.com> wrote:
> On 3/30/2022 12:33 PM, Ed Lake wrote:
>
>>> 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 purely self-serving soundbite grabbing, and he knows it. Because Ed is
mathematically illiterate, it makes sense to him to dismiss mathematics as
not only unnecessary but detrimental.

It’s easy to point out how silly that tactic is. Simply point to the 1905
paper written by the same author and ask, “why then is there so much
mathematics (though rather basic mathematics) in the paper?”

The same goes for the apocryphal soundbite that any good theory should be
explainable to a barmaid (which is misattribute to Einstein). The response
is to point to the popularization written by Einstein himself, Relativity:
The Special and General Theory, and to say, “fine, this book was written
for barmaids; now, do you understand what is written there?”

The common complaint is, “I am interested in this subject. How dare you say
I am unequipped for it?” But if I point out that true interest is marked by
a willingness to do the work to get equipped, then the complaint is, “No,
that’s not fair. It’s asking too much. I have no time or interest in
getting equipped.” To which I say, “Exactly. You’re not interested enough.”

Ed Lake

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Mar 31, 2022, 10:00:38 AM3/31/22

to

On Wednesday, March 30, 2022 at 3:30:43 PM UTC-5, thor stoneman wrote:

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 Lake

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Mar 31, 2022, 10:05:50 AM3/31/22

to

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

Ed Lake

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Mar 31, 2022, 10:09:51 AM3/31/22

to

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

rotchm

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Mar 31, 2022, 10:24:19 AM3/31/22

to

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.

Fletcher Krupp

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Mar 31, 2022, 10:26:29 AM3/31/22

to

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/

Ed Lake

unread,

Mar 31, 2022, 10:28:23 AM3/31/22

to

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

Odd Bodkin

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Mar 31, 2022, 10:38:37 AM3/31/22

to

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.

Fletcher Krupp

unread,

Mar 31, 2022, 10:39:47 AM3/31/22

to

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.

Paparios

unread,

Mar 31, 2022, 10:41:18 AM3/31/22

to

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.

Ed Lake

unread,

Mar 31, 2022, 10:42:35 AM3/31/22

to

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

Fletcher Krupp

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Mar 31, 2022, 10:52:30 AM3/31/22

to

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.

Ed Lake

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Mar 31, 2022, 10:55:59 AM3/31/22

to

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

Townes Olson

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Mar 31, 2022, 11:19:21 AM3/31/22

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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.

Ed Lake

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Mar 31, 2022, 11:20:31 AM3/31/22

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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

Ed Lake

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Mar 31, 2022, 11:36:24 AM3/31/22

to

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

Michael Moroney

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Mar 31, 2022, 11:56:04 AM3/31/22

to

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.

Paparios

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Mar 31, 2022, 11:57:28 AM3/31/22

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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!!!

Townes Olson

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Mar 31, 2022, 12:07:30 PM3/31/22

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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.

Maciej Wozniak

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Mar 31, 2022, 12:12:39 PM3/31/22

to

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.

Michael Moroney

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Mar 31, 2022, 12:15:05 PM3/31/22

to

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.

Odd Bodkin

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Mar 31, 2022, 12:46:25 PM3/31/22

to

Ed Lake <det...@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
>

rotchm

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Mar 31, 2022, 4:36:28 PM3/31/22

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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?

rotchm

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Mar 31, 2022, 4:41:55 PM3/31/22

to

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?

Ed Lake

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Mar 31, 2022, 5:04:36 PM3/31/22

to

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

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Mar 31, 2022, 10:41:14 PM3/31/22

to

On 3/31/2022 11:36 AM, Ed Lake wrote:
> 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.

Different inertial systems will see the same event in different ways.

>
>>> 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?

He already answered that. Emission of a photon is an event, the
coordinates are frame dependent.

>
> 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.

Nope. There is no such thing since the existence of such a thing
(absolute rest) would mean a preferred frame, which would violate the
First Postulate.

Tom Roberts

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Apr 1, 2022, 11:47:02 AM4/1/22

to

On 3/30/22 3:05 PM, RichD wrote:
> if a basketballer is at the free throw line, moving quickly to his
> left, he aims directly at the basket. His motion at that moment is
> irrelevant, the ball leaves his hand and travels straight toward the
> basket.

But his motion at the time of the throw is indeed relevant to the angle
at which he must throw the ball to hit the basket. If, while moving
left, he throws the ball directly toward the basket, it will miss
because the player's motion at the instant of throw affects the
trajectory of the ball. The player must throw the ball at an angle to
the right of the basket such that the player's leftward motion is
canceled by the angle of throw.

Basketball players know this implicitly and automatically correct for
such effects, without thinking about it. Human brains are truly
remarkable in their unconscious accounting for the basic physics of the
world around us -- this is not "understanding", but rather detailed
memories of myriad experiences while growing up, integrated into the
ways we manipulate our bodies.

> The problem is, these mathematicians try to bamboozle you with their
> fantasy world frames and vectors. They don't care about the
> SCIENCE.

Just like Lake, you don't have a clue. You both should stop attempting
to write about subjects you clearly do not understand.

Tom Roberts

Maciej Wozniak

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Apr 1, 2022, 12:22:18 PM4/1/22

to

On Friday, 1 April 2022 at 17:47:02 UTC+2, tjrob137 wrote:
> On 3/30/22 3:05 PM, RichD wrote:
> > if a basketballer is at the free throw line, moving quickly to his
> > left, he aims directly at the basket. His motion at that moment is
> > irrelevant, the ball leaves his hand and travels straight toward the
> > basket.
>
> But his motion at the time of the throw is indeed relevant to the angle

And we all are FORCED!!!
To THE BEST WAY!!! (which is, of course, the way
of Giant Guru, as interpreted by poor idiot Roberts).

RichD

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Apr 1, 2022, 5:56:50 PM4/1/22

to

On April 1, tjrob137 wrote:
>> if a basketballer is at the free throw line, moving quickly to his
>> left, he aims directly at the basket. His motion at that moment is
>> irrelevant, the ball leaves his hand and travels straight toward the
>> basket.
>
> But his motion at the time of the throw is indeed relevant to the angle
> at which he must throw the ball to hit the basket.

> The player must throw the ball at an angle to
> the right of the basket such that the player's leftward motion is
> canceled by the angle of throw.
> Basketball players know this implicitly and automatically correct for
> such effects, without thinking about it. Human brains are truly
> remarkable in their unconscious accounting for the basic physics of the
> world around us -- this is not "understanding", but rather detailed
> memories of myriad experiences while growing up, integrated into the
> ways we manipulate our bodies.

Right.
Like, boxers wear padded helmets when sparring in the gym,
which permits them to hit harder, they know it protects against
brain damage.

>> The problem is, these mathematicians try to bamboozle you with their
>> fantasy world frames and vectors. They don't care about the
>> SCIENCE.
>
> Just like Lake, you don't have a clue. You both should stop attempting
> to write about subjects you clearly do not understand.

Swing a ball around on a rope. If the rope breaks, the ball will
continue its circular motion. It's simply Newton's first law.

According to Einstein, however, due to the differing transforms
of space in x vs. y directions, it will follow a hyperbolic path;
asymptotically, it will travel a straight line.

--
Rich

Paul Alsing

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Apr 1, 2022, 8:01:27 PM4/1/22

to

On Friday, April 1, 2022 at 2:56:50 PM UTC-7, RichD wrote:

> Swing a ball around on a rope. If the rope breaks, the ball will
> continue its circular motion. It's simply Newton's first law.

Really? Do you actually believe this?

Odd Bodkin

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Apr 1, 2022, 8:18:22 PM4/1/22

to

Might want to rethink that, Rich.

> It's simply Newton's first law.
>
> According to Einstein, however, due to the differing transforms
> of space in x vs. y directions, it will follow a hyperbolic path;
> asymptotically, it will travel a straight line.
>
> --
> Rich
>
>

Prokaryotic Capase Homolog

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Apr 1, 2022, 9:55:58 PM4/1/22

to

On Friday, April 1, 2022 at 4:56:50 PM UTC-5, RichD wrote:

> Swing a ball around on a rope. If the rope breaks, the ball will
> continue its circular motion. It's simply Newton's first law.

Huh?

> According to Einstein, however, due to the differing transforms
> of space in x vs. y directions, it will follow a hyperbolic path;
> asymptotically, it will travel a straight line.

Huh?

You are completely out of your league here.

Michael Moroney

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Apr 2, 2022, 12:30:36 AM4/2/22

to

On 4/1/2022 5:56 PM, RichD wrote:

> Swing a ball around on a rope. If the rope breaks, the ball will
> continue its circular motion. It's simply Newton's first law.

*facepalm*

Cash Abel

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Apr 2, 2022, 2:37:24 AM4/2/22

to

It's the guy pulling the ball, not the rope. You can't read, idiot.

Paul B. Andersen

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Apr 2, 2022, 3:44:38 AM4/2/22

to

Den 01.04.2022 23:56, skrev RichD:
>
> Swing a ball around on a rope. If the rope breaks, the ball will
> continue its circular motion. It's simply Newton's first law.
>
> According to Einstein, however, due to the differing transforms
> of space in x vs. y directions, it will follow a hyperbolic path;
> asymptotically, it will travel a straight line.
>
> --
> Rich
>

April 1. ? :-D

--
Paul

https://paulba.no/

Odd Bodkin

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Apr 2, 2022, 9:36:23 AM4/2/22

to

I think it’s pretty clear RichD is another one of these word-salad wannabes
that is seeking community acceptance in a subject he knows not the first
thing about, but is happy to dress up as though he does, using physics-y
words, citing random article titles, casually name-dropping physicists.

--
Odd Bodkin — Maker of fine toys, tools, tables

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Apr 2, 2022, 3:48:38 PM4/2/22

to

On April 1, prokaryotic.c...@gmail.com wrote:
>> Swing a ball around on a rope. If the rope breaks, the ball will
>> continue its circular motion. It's simply Newton's first law.
>
> Huh?
>
>> According to Einstein, however, due to the differing transforms
>> of space in x vs. y directions, it will follow a hyperbolic path;
>> asymptotically, it will travel a straight line.
>

> You are completely out of your league here.

Another halibut!

Why is a bike faster than Nike?

--
Rich

RichD

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Apr 2, 2022, 4:04:00 PM4/2/22

to

On April 2, Paul B. Andersen wrote:
>> Swing a ball around on a rope. If the rope breaks, the ball will
>> continue its circular motion. It's simply Newton's first law.
>
>> According to Einstein, however, due to the differing transforms
>> of space in x vs. y directions, it will follow a hyperbolic path;
>> asymptotically, it will travel a straight line.
>

> April 1. ? :-D

Finally, someone shows a modicum of reading comprehension.
There's hope for humanity -

Though the April Fools date is just a coincidence.

Let me add, Mr. Anderson is one of the few reasons to
subscribe here, one of the few who provides content.

This amusing little sequence began when I was funning
with Ed Lake. He wisely declined to take the bait.

But then a school of fish jumped in, to preen their blazing
intellects. College grads, you know, with high level reading
comprehension, confirmed by their SAT reading scores,
yes indeed.

"I can't give you a brain, but I can give you a diploma."
- the Wizard of Oz, to the scarecrow

Thus Mr. Lake showed more wisdom than the undersea
crowd. Think about that -

--
Rich

mitchr...@gmail.com

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Apr 2, 2022, 6:53:36 PM4/2/22

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On Tuesday, March 29, 2022 at 8:55:46 AM UTC-7, det...@outlook.com 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

We see red shifted Heat from the most distant galaxies
that has been shifted into visible...

Elmer Joss

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Apr 3, 2022, 5:39:54 AM4/3/22

to

mitchr...@gmail.com wrote:

> We see red shifted Heat from the most distant galaxies that has been
> shifted into visible...

Heat can be blue.

Iskander Missile Wipes Out Kharkiv Merc Compound, 100+ Foreign Fighters Dead
https://www.veteranstoday.com/2022/04/02/iskander-missile-wipes-out-kharkiv-merc-compound-100-foreign-fighters-dead/
“As a result of a high-precision strike with th
e Iskander operational-tactical complex on the defense headquarters in the city of Kharkiv on March 31, the elimination of more than 100 nationalists and mercenaries from Western countries was confirmed,” Konashenkov said.
The Russian MoD released footage of the pinpoint strike, which was apparently carried out with an Iskander-K cruise missile.

Witness Describes How the NATO Lie Machine Staged the Mariupol Hospital...
https://www.veteranstoday.com/2022/04/02/blockbuster-witness-describes-how-the-nato-lie-machine-staged-the-mariupol-hospital-attack-white-helmets-style/
https://youtu.be/f0rqHB41__A
Listen to Marianna Vyshemirskaya tell about how the new “White Helmets” used her to stage a fake Russian bombing…one they staged themselves with their own explosives just like done in Syria with Israeli help over and over and over…bombs, gas, no gas, kidnapped kids and doctors at gunpoint…

How NewsGuard, the CIA’s Disinformation Machine, Supports Nazi Atrocities in Ukraine
https://www.veteranstoday.com/2022/04/02/how-newsguard-the-cias-disinformation-machine-supports-nazi-atrocities-in-ukraine/

Ukraine: Who Are the Americans Running Bioweapons Research?
https://www.veteranstoday.com/2022/04/01/ukraine-who-are-the-americans-running-bioweapons-research/
Emails published by the MoD revealed she directly supervised experiments with deadly pathogens, including the UP-2 Project for “mapping highly infectious diseases in Ukraine,” including anthrax; the UP-4 Project, described as a “risk assessment of particularly dangerous pathogens transmitted by birds in Ukraine during migration”; the UP-8 Project studying “spread of Crimean-Congo hemorrhagic fever virus and hantaviruses in Ukraine and the potential need for differential diagnosis of patients with suspected leptospirosis. Previously released documents showed she also oversaw Project P-782, conducting research into the transmission of diseases through bats.

Madeleine Albright’s Epitaph: “Half a Million Dead Children? We Think The Lie It’s Worth It”
https://www.veteranstoday.com/2022/03/23/madeleine-albrights-epitaph-half-a-million-dead-children-we-think-its-worth-it/

Odd Bodkin

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Apr 3, 2022, 8:58:38 AM4/3/22

to

Well this is fun. I don’t know the answer for sure.

I do not think a bicyclist can stay ahead of a sprinter for the first 20
yards.

But beyond that, I think it is due to an inherent limitation of the speed
of the lower leg’s fast-twitch fibered muscles, coupled with the attachment
point of the muscle to the lower leg bones. This is also related to why an
80 lb chimpanzee is four times stronger than a 180 lb man, but cannot throw
a ball with any speed; the attachment point of the chimp’s muscle is
further away from the joint, trading speed for leverage.

Between the cyclist and the sprinter, when both are ramped up to fastest
leg speed, the bicyclist now has an obvious advantage, because of the ratio
of the radius of the wheel compared to the radius of the pedal crank, which
in turn is about the same as the average radius of the runner’s foot
motion. The tangential speed of the wheel is larger than the tangential
speed of the foot by the same ratio as that of the radii.

>
> --
> Rich
>

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

J. J. Lodder

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Apr 3, 2022, 10:38:23 AM4/3/22

to

Because it isn't. The sprinter will be ahead right after the start,
and won't be overtaken until he tires. (shortly before 100m)
That is because he literally jumps out of the blocks.
The cyclist cannot bring his peak power to bear as efficiently.
Here is a simulation for the Bolt vs Eady:
<https://www.outsideonline.com/uncategorized/who-would-win-100-usain-bolt-or-top-sprint-cyclist/>

> Show us you know the FIRST THING about the subject -

Alas, you don't,

Jan

Michael Moroney

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Apr 3, 2022, 2:56:50 PM4/3/22

to

On 4/3/2022 5:39 AM, Elmer Joss wrote:

> Madeleine Albright’s Epitaph: “Half a Million Dead Children? We Think The Lie It’s Worth It”

How many children has Russia killed so far, nymshifter?
And the difference is that Russia has *deliberately* targeted civilians!

卐Путин卐 хуйло!

Vance Rera

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Apr 3, 2022, 3:28:41 PM4/3/22

to

Michael Moroney wrote:

> On 4/3/2022 5:39 AM, Elmer Joss wrote:
>
>> Madeleine Albright’s Epitaph: “Half a Million Dead Children? We Think
>> The Lie It’s Worth It”
>
> How many children has Russia killed so far, nymshifter?

none, you stupid immigrant. You don't even know that the IQ test was
invented and distributed by the *eugenicists*, for the sake of you,
immigrants of america and cacanada.