You were right: Rotational motion is relative, too, Mr. Einstein!

AAF
AAF

June 8th, 2016, 3:00 am #31




It has been one hundred years since the publication of Einstein's general theory of relativity in May 1916. In a paper recently published in EPJ Plus, Norwegian physicist Øyvind Grøn from the Oslo and Akershus University College of Applied Sciences and his co-author Torkild Jemterud demonstrate that the rotational motion in the universe is also subject to the theory of relativity. Imagine a person at the North pole who doesn't believe the Earth rotates. As she holds a pendulum and can observe the stars in her telescope, she remarks that the swinging plane of the pendulum and the stars rotate together. Newton, who saw the world as a classical physicist, would have pointed out that it is the Earth that rotates. However, if we assume the general principle of relativity is valid, the Earth can be considered as being at rest while the swinging plane of the pendulum and the night sky are rotating. In fact, the rotating mass of the observable part of the universe causes the river of space--which is made up of free particles following the universe's expansion--to rotate together with the stars in the sky. And the swinging plane of the pendulum moves together with the river of space.










https://www.sciencedaily.com/releases/2 ... 120254.htm













"Let’s clarify the last point first: "if S_1 is faraway, then it must appear to be non-rotating and stationary, because its parallax is so tiny and very close to nil". Sorry, but I couldn’t care less about “very close to nil”. I’m talking about the maths - and maths is not dependent on the resolving power of our telescope. Either S_1 is in EXACTLY the same position in the sky, or it is not. So let’s leave the fudge factor of “so tiny” out of it; that is not what we are talking about - right? And that, in turn, means The height of S_1, over the North Pole, is still totally unimportant. Which brings us to: "As for the math, here, it's exactly the same as the math of parallax". Really ??? Because that page you linked to, that has that graphic, describes Parallax as: “Now open the other eye, and close the first. Your finger appears to move against the more distant objects, even though you haven’t moved it. This is parallax!” …. but our scenario explicitly does not have any “more distant objects”, remember? So no, it can’t be the same as that. What about the “close the first [eye]” bit? Again no, that’s a different scenario from ours, since “your finger” is not on any axis of rotation (unless you have very weird eyes !!), so the maths can’t be the same as that one either. So what I said earlier still holds : Your pictures are flawed, because you are neglecting to take into account that S_1 is along the axis of rotation. You … repeatedly refuse to apply the mathematics to our special case of a body positioned along the axis. I am asking you, to make a clear case, in your own words, on OUR explicit scenario. That’s actually one of the main reasons I’ve been repeatedly asking “SHOW WITH MATHS, not just word claims”, because that it’s only by the maths - APPLIED TO THIS SCENARIO - that your argument becomes clear and unambiguous. So again, for OUR scenario where S_1 is positioned along the axis of rotation: D = arctan( H / 3000 ) is the equation, and for a given H and a stationary observer, that angle D does NOT change during S_2’s revolution …… except, of course, you claim that it does. SHOW how something that is always 3000Km straight ahead, and H km up, can ever be at anything other than that D degree angle in the sky. SHOW how a right-angled-triangle with base 3,000km and height H km, can have that angle being D-R degrees. Take our H=4,000km straight from the illustration if you like - SHOW how and when our observer sees S_1 at less than 53 degrees. SHOW WITH MATHS, not just word claims - Yes?"






Do I have to do the math?


Well; why bother?


I don't have to do that . . .


You already have shown the math of it, yourself:



<font size="5"> D = arctan(H / 3000)</font>







And so, according to the above equation, if the numerical value of H approaches infinity,
then the angle D between the North Pole of S_2,the rotating observer, and S_1
must approach the value of 90o.

And hence, the projected circle, on the sky, becomes smaller and smaller until
it finally vanishes even on the viewing screens of the Hubble Space Telescope.


And conversely, if the numerical value of H approaches zero, then the angle D between
the North Pole of S_2,the rotating observer, and S_1 must approach
the value of 0o as well.


And as a result, the projected circle, on the sky, becomes bigger and bigger,
until even an old rhino can see it:

http://www.lasikmd.com/blog/5-animals-worst-vision/


Consequently, the numerical values of H are extremely important.


But, no matter how big or how small is the area of the projected circle, on the height of S_1,
the object S_1 must always appear, to the rotating observer, to be moving, as one single unit,
along the periphery of that circle.


And that, certainly, contradicts, sharply, Einstein's main supposition, according to which
the rotating observer, on S_2, must see S_1 rotating, around S_2's axis of rotation,
if S_1 is located along S_2's axis of rotation.




















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AAF
AAF

June 10th, 2016, 3:00 am #32




It has been one hundred years since the publication of Einstein's general theory of relativity in May 1916. In a paper recently published in EPJ Plus, Norwegian physicist Øyvind Grøn from the Oslo and Akershus University College of Applied Sciences and his co-author Torkild Jemterud demonstrate that the rotational motion in the universe is also subject to the theory of relativity. Imagine a person at the North pole who doesn't believe the Earth rotates. As she holds a pendulum and can observe the stars in her telescope, she remarks that the swinging plane of the pendulum and the stars rotate together. Newton, who saw the world as a classical physicist, would have pointed out that it is the Earth that rotates. However, if we assume the general principle of relativity is valid, the Earth can be considered as being at rest while the swinging plane of the pendulum and the night sky are rotating. In fact, the rotating mass of the observable part of the universe causes the river of space--which is made up of free particles following the universe's expansion--to rotate together with the stars in the sky. And the swinging plane of the pendulum moves together with the river of space.










https://www.sciencedaily.com/releases/2 ... 120254.htm











"No kidding, if I saw a great white while I was diving, I’d be jumping out of that water faster than a stingray! Grey Nurses are nice, though. I remember one dive, we turned up where there were 4 or 5, so they decided just to go away ... except one came back just to have a look at us (not to eat us! Great - thought I'd start with this one, as it's always nice to start with an agreement. Now, ’3000Km straight ahead, and H km up' is ALWAYS the same triangle, as shown in our graphic - and therefore by definition, S_1 is always at angle D = arctan( H / 3000 ) straight ahead (neither left nor right) to our observer. Not D + R nor D - R; ALWAYS just plain D. By definition. THEN SHOW IT ! Don’t just keep claiming it - show it with maths, and not just words – yes?"






Actually, swimming with the sharks is far less dangerous
than swimming above the sharks.


I presume!





And also having a shark shield, like this one,
at hand, is a very good idea:



https://en.wikipedia.org/wiki/Shark_Shield



O.K. . . .


So, S_1 is always at angle D = arctan( H / 3000 ) straight ahead (neither left nor right)
to our observer. Not D + R nor D - R; ALWAYS just plain D. By definition.


Now, let's consider the geographical terrain within the circle of the horizon around
the rotating observer, on S_2.


In relation to that circle, which, by definition, is always at rest relative
to the rotating observer, on S_2, the celestial object named 'S_1'
is always moving along a huge circle on the sky.


And so, if the city of Brisbane is located on the eastern horizon and the city of Oslo
is located on the western horizon of the rotating observer, then S_1 will, sometimes,
appear over the city of Brisbane; and it will appear, at some other times,
over the city of Oslo.


But how & why?


Well; it's because, the straight line between the observer and S_1 is NOT
rigidly connected to S_1; ; right?


Instead, the straight line between the observer and S_1 is actually rotating
around S_1, because the observer is rotating around the North Pole of S_2.


And therefore, S_1 appears, to that rotating observer, to be traveling along
a big circle, on the sky; and hence, S_1, sometime, seems to be over Brisbane
in the east, and some time, seems to be over Oslo in the west.














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Ufonaut99
Ufonaut99

June 11th, 2016, 2:29 am #33




It has been one hundred years since the publication of Einstein's general theory of relativity in May 1916. In a paper recently published in EPJ Plus, Norwegian physicist Øyvind Grøn from the Oslo and Akershus University College of Applied Sciences and his co-author Torkild Jemterud demonstrate that the rotational motion in the universe is also subject to the theory of relativity. Imagine a person at the North pole who doesn't believe the Earth rotates. As she holds a pendulum and can observe the stars in her telescope, she remarks that the swinging plane of the pendulum and the stars rotate together. Newton, who saw the world as a classical physicist, would have pointed out that it is the Earth that rotates. However, if we assume the general principle of relativity is valid, the Earth can be considered as being at rest while the swinging plane of the pendulum and the night sky are rotating. In fact, the rotating mass of the observable part of the universe causes the river of space--which is made up of free particles following the universe's expansion--to rotate together with the stars in the sky. And the swinging plane of the pendulum moves together with the river of space.










https://www.sciencedaily.com/releases/2 ... 120254.htm




AAF: Actually, swimming with the sharks is far less dangerous than swimming above the sharks. I presume!
You presume correctly Sharks like nice big juicy seals; humans are scrawny by comparison, so we're not on their favourites menu. When we're below the waves, it's obvious we're not seals - but if they sense something splashing around on the surface ....

When I saw your picture with the diver, my first thought was that you were praising the spear-gun; Spear guns would probably do nothing to a shark except really cheese it off - so not too good an idea !
SharkShield looks ideal - but an Aussie with a good right-hook could be even better !
AAF: Do I have to do the math? Well; why bother? I don't have to do that . . . You already have shown the math of it, yourself: <font color="#FF0000">D = arctan(H / 3000)
</font>
True ..... although you would have to do a bit more math if you are claiming something different from that equation - say, for example, if you reckoned the line of sight to S_1 sometimes had a different angle
AAF: And so, according to the above equation, if the numerical value of H approaches infinity, then the angle D between the North Pole of S_2,the rotating observer, and S_1 must approach the value of 90o.
Scary .... we're still agreeing (*)
AAF: And hence, the projected circle, on the sky
And then you go and spoil it ! WHAT "projected circle" ???

Again, two simple and obvious facts :
1) D is the angle of the line-of-sight from the observer to S_1; For our illustration with a given H, D is constant, meaning a camera fixed at that angle keeps S_1 in view.
2) If S_1 is moving around a circle in the sky, it will be continuously changing it's angle of line-of-sight : you'd have to move your head - and camera - up/down and left/right to follow it

Those statements are CONTRADICTORY - so you have to choose And if you choose option (2), then you have to SHOW IT with maths, not just words - yes?
AAF: And therefore, S_1 appears, to that rotating observer, to be traveling along a big circle, on the sky; and hence, S_1, sometime, seems to be over Brisbane in the east, and some time, seems to be over Oslo in the west.
Since our observer is always facing north, then east is to his right, and west is to his left.
OK, explain to me how S_1, which we've agreed is ALWAYS straight in front of our observer, sometimes appears off to his left, and sometimes to his right???
That just makes no sense.

If I had to guess what you mean (and I shouldn't have to guess, because you should be spelling it out clearly, yes? ) - but if I had to guess .....



Taking a low value for H first: Let's say our camera such a really wide-angle lens it can take in both Brisbane and Oslo.
At 1am it just so happens that the camera films at low H (so D is just above the horizon) exactly due north (ie. directly across the north pole / axis of rotation) a distant star - say Sirius.

Now, Sirius will behave exactly as you've been saying - the camera will film it rotating around at a constant angle D(**) in the sky, and indeed will film it sometimes over Brisbane, and sometimes over Oslo - and in fact, it's sometimes out-of-shot behind the camera. In short, yes, it certainly does travel a "Humungous circle" in the sky.

If instead we tilt the camera so choosing a high H - say we find another star again directly north at 1am but higher in the sky - call it Mizar - then Mizar will make a much smaller circle in the sky.

Thing is : I couldn't care less about Sirius or Mizar; They are NOT what we're talking about.

We are suposed to be talking about S_1. Neither Sirius nor Mizar are candidates for S_1.
S_1 is EXPLICITLY positioned directly up exactly along the axis of rotation. Sirius and Mizar are not.

Let's go back to our view of Sirius at 1am. Suppose someone at Santa's workshop puts a yellow ball suspended directly above the exact North Pole, at a height that precisely blocks our camera's view of Sirius at that time.

So, during S_2's rotation, our camera will film Sirius moving out from behind that ball - but since our camera is always facing North, and that ball is (by definition since it's at the north pole) always due north of it, then the camera will always have the ball in view.

S_1 is THAT BALL, not Sirius, since it is that ball that is positioned along our axis of rotation, above our north pole.


Which brings us back again to what I said earlier : Your pictures are flawed, because you are neglecting to take into account that S_1 is along the axis of rotation. You … repeatedly refuse to apply the mathematics to our special case of a body positioned along the axis.

So let's fix that now - yes?

Explain clearly (with maths ) to me how how our camera (that is always facing due north) can ever film that BALL (that is always positioned straight ahead, since it is always due north of the camera) way off to the right, over Brisbane in the east.


(*) pretty much. The graphic should have had the right-angle inside the earth at the latitude of the camera, so the angle we're talking about is really between that point rather than the north pole. Trivial detail that doesn't change anything, though

(**) Actually, D in this case would not be constant. For any distant body that is NOT positioned directly along the axis of rotation, then the distance between our camera and that body will vary as the rotation brings the camera closer then further to that body. That means that that 3000km figure will be changing, so D will be changing.

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AAF
AAF

June 12th, 2016, 3:00 am #34




It has been one hundred years since the publication of Einstein's general theory of relativity in May 1916. In a paper recently published in EPJ Plus, Norwegian physicist Øyvind Grøn from the Oslo and Akershus University College of Applied Sciences and his co-author Torkild Jemterud demonstrate that the rotational motion in the universe is also subject to the theory of relativity. Imagine a person at the North pole who doesn't believe the Earth rotates. As she holds a pendulum and can observe the stars in her telescope, she remarks that the swinging plane of the pendulum and the stars rotate together. Newton, who saw the world as a classical physicist, would have pointed out that it is the Earth that rotates. However, if we assume the general principle of relativity is valid, the Earth can be considered as being at rest while the swinging plane of the pendulum and the night sky are rotating. In fact, the rotating mass of the observable part of the universe causes the river of space--which is made up of free particles following the universe's expansion--to rotate together with the stars in the sky. And the swinging plane of the pendulum moves together with the river of space.










https://www.sciencedaily.com/releases/2 ... 120254.htm










It's absolutely correct . . .


The sharks are doing horrible things to 'nice big juicy' mammals!


But let's NOT overlook what the ORCAS of the mammals are
doing to the sharks:

https://www.youtube.com/watch?v=GvTmrieZXBY






........................................................................



"As I said before : Your pictures are flawed, because you are neglecting to take into account that S_1 is along the axis of rotation. You … repeatedly refuse to apply the mathematics to our special case of a body positioned along the axis. Speaking of neglecting to take into account S_1 being on the axis of rotation ..... Yes - and that maths in that paper is full of scenarios like: - “your finger” in front of “your” eye - NOT APPLICABLE TO THIS EXPLICIT SCENARIO - Calculating distances using Parallax - NOT APPLICABLE TO THIS EXPLICIT SCENARIO - Simultaneous measurements from two locations - NOT APPLICABLE TO THIS EXPLICIT SCENARIO - Appearance shifting against background of stars - NOT APPLICABLE TO THIS EXPLICIT SCENARIO. Those scenarios (and more) are all applicable and incredibly useful with real-life Earth-based Parallax calculations - but that does NOT make them applicable to THIS EXPLICIT SCENARIO. Seriously, I asked you to make a clear case, in your own words, on OUR explicit scenario. That’s actually one of the main reasons I’ve been repeatedly asking “SHOW WITH MATHS, not just word claims”, because that it’s only by the maths - APPLIED TO THIS SCENARIO - that your argument becomes clear and unambiguous. Instead, you just point to some 50 page document, leaving all readers to guess which - if any - of the maths you think is applicable. So no more links to random pages on Parallax, which will be full of those scenarios not applicable to us - OK ? If you continue to think that Parallax is somehow applicable, then you should be able to SHOW it here - right? After all, ours is not a complex scenario, so easy to describe with no reason to link off anywhere else. So once again, PLEASE: MAKE A CLEAR CASE IN YOUR OWN WORDS ON THIS EXPLICIT SCENARIO. SHOW how something that is always 3000Km straight ahead, and H km up, can ever be at anything other than that D degree angle in the sky. SHOW WITH MATHS HERE, not just word claims, nor links talking about other unrelated scenarios - Yes?"






It's true that S_1 is located along the axis of rotation over the North Pole.


Does that make S_1 appear to be rotating around its geometrical axis?


The answer is no!


The rotation of the observer, on S_2, can only make S_1 appear to be moving, as one single unit,
along a specific circle, on the sky.


What does the math say about it?


The math says, here, that there can be no apparent rotation of S_1 around its geometrical axis.


Consequently, Albert Einstein got it all wrong.

Period!







How exactly does the mathematics of this situation say that there can be no apparent rotation
of S_1 around its geometrical axis?


The math says that the angle between S_1, the observer, and the North Pole varies with the value
of the height H according to this equation:


<font size="4"> D = arctan(H / 3000)</font>


And that means:

1. If S_1 is too close above the North Pole, then S_1 will appear to make, periodically, a huge circle on the sky above the North Pole of S_2, along the
circumference of which S_1 will always appear to move as one single unit.


2. And if S_1 is too faraway above the North Pole, then S_1 will appear to make, periodically, an extremely tiny circle on the sky above the North Pole
of S_2, along the circumference of which S_1, also, will always appear to move as one single unit.



And therefore, there can be no apparent axial rotation of S_1; none whatsoever.






















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Ufonaut99
Ufonaut99

June 12th, 2016, 8:51 am #35




It has been one hundred years since the publication of Einstein's general theory of relativity in May 1916. In a paper recently published in EPJ Plus, Norwegian physicist Øyvind Grøn from the Oslo and Akershus University College of Applied Sciences and his co-author Torkild Jemterud demonstrate that the rotational motion in the universe is also subject to the theory of relativity. Imagine a person at the North pole who doesn't believe the Earth rotates. As she holds a pendulum and can observe the stars in her telescope, she remarks that the swinging plane of the pendulum and the stars rotate together. Newton, who saw the world as a classical physicist, would have pointed out that it is the Earth that rotates. However, if we assume the general principle of relativity is valid, the Earth can be considered as being at rest while the swinging plane of the pendulum and the night sky are rotating. In fact, the rotating mass of the observable part of the universe causes the river of space--which is made up of free particles following the universe's expansion--to rotate together with the stars in the sky. And the swinging plane of the pendulum moves together with the river of space.










https://www.sciencedaily.com/releases/2 ... 120254.htm




AAF: But let's NOT overlook what the ORCAS of the mammals are doing to the sharks:
Heh, loved the fact the guy said "", and he's still got over 1.5 million views ! (plus one ... but must admit, wish I had that time back !)

AAF: The math says that the angle between S_1, the observer, and the North Pole varies with the value of the height H according to this equation:
<blockquote>D = arctan(H / 3000)
</blockquote>Look at the illustration again - H is CONSTANT throughout S_2's revolution.
That means D is also CONSTANT throughout S_2's revolution.
And that means S_1 has NO up/down motion.
And that means S_1 is NOT moving in any circle.

That's what the maths says

But OK, let's take the simplest possible test case : Say hello to Stanisława from Krakow :


She's the oldest of her sisters, so often known simply by the abbreviation "S_1" In addition, Stanisława is spending a couple of days standing on the the northern surface of S_2 exactly on the axis of rotation - which, of course, leads to her other common nick-name. That's right - Stanisława IS "The North Pole"

Now that, of course, means that Stanisława has a constant and low value of H.
AAF: if the city of Brisbane is located on the eastern horizon and the city of Oslo is located on the western horizon of the rotating observer, then S_1 will, sometimes, appear over the city of Brisbane; and it will appear, at some other times, over the city of Oslo.
Surely it's commonsense that the North Pole will always be due North - yes? .... so please explain clearly, with maths, exactly :

how and why you reckon our camera sees The North Pole over at Brisbane due .... East ????
And then, of course, how you reckon that The North Pole swings over to Oslo due West ????
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AAF
AAF

June 14th, 2016, 3:00 am #36




It has been one hundred years since the publication of Einstein's general theory of relativity in May 1916. In a paper recently published in EPJ Plus, Norwegian physicist Øyvind Grøn from the Oslo and Akershus University College of Applied Sciences and his co-author Torkild Jemterud demonstrate that the rotational motion in the universe is also subject to the theory of relativity. Imagine a person at the North pole who doesn't believe the Earth rotates. As she holds a pendulum and can observe the stars in her telescope, she remarks that the swinging plane of the pendulum and the stars rotate together. Newton, who saw the world as a classical physicist, would have pointed out that it is the Earth that rotates. However, if we assume the general principle of relativity is valid, the Earth can be considered as being at rest while the swinging plane of the pendulum and the night sky are rotating. In fact, the rotating mass of the observable part of the universe causes the river of space--which is made up of free particles following the universe's expansion--to rotate together with the stars in the sky. And the swinging plane of the pendulum moves together with the river of space.










https://www.sciencedaily.com/releases/2 ... 120254.htm












"True ..... although you would have to do a bit more math if you are claiming something different from that equation - say, for example, if you reckoned the line of sight to S_1 sometimes had a different angle. "And so, according to the above equation, if the numerical value of H approaches infinity, then the angle D between the North Pole of S_2,the rotating observer, and S_1 must approach the value of 90^o". Scary .... we're still agreeing (*). "And hence, the projected circle, on the sky". And then you go and spoil it! WHAT "projected circle" ??? Again, two simple and obvious facts: 1) D is the angle of the line-of-sight from the observer to S_1; For our illustration with a given H, D is constant, meaning a camera fixed at that angle keeps S_1 in view. 2) If S_1 is moving around a circle in the sky, it will be continuously changing it's angle of line-of-sight : you'd have to move your head - and camera - up/down and left/right to follow it. Those statements are CONTRADICTORY - so you have to choose. And if you choose option (2), then you have to SHOW IT with maths, not just words – yes?"





I'm so glad that we've finally agreed that " according to the equation:
D = arctan(H / 3000), if the numerical value of H approaches infinity, then
the angle D between the North Pole of S_2,the rotating observer, and S_1 must
approach the value of 90o"
.







But the projected circle is extremely important within the current context of debunking,
once and for all, Einstein's baseless conjecture about the relativity of axial rotation.


What is the projected circle?


The projected circle is the circle of the rotating observer's latitude as seen by the same
observer projected on an external object like S_1.


Generally, if the observer moves in a straight line, then he/she will see the external
object moving in a straight line.


And if the observer moves in a circle, then he/she will see the external object
moving in a circle.


And if the observer moves in a square, then he/she will see the external object
moving in a square.


And if the observer moves in an ellipse, then he/she will see the external object
moving in an ellipse.


And if the observer moves in hexagon, then he/she will see the external object
moving in a hexagon.


And if the observer moves in polygon, then he/she will see the external object
moving in a polygon.

And so on . . . and so on.


In all cases, the basic geometrical properties of the observer's path of movement
are conserved,upon projection on external objects; but their relative sizes vary
inversely with increasing distance.


Does the observer have to keep continuously changing the direction of the angle
of the line-of-sight and to move his/her head left/right in order to follow
the external object?


You bet!





Does the observer have to keep continuously changing the size of the angle of the line-of-sight
and to move his/her head up/down in order to follow the external object?


Absolutely not!


And that is because the observer has assumed him/herself to be at rest; and therefore,
he/she is always at the straight line that passes right smack through the center
of the circle along which the external object is moving.














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Ufonaut99
Ufonaut99

June 15th, 2016, 10:42 pm #37




It has been one hundred years since the publication of Einstein's general theory of relativity in May 1916. In a paper recently published in EPJ Plus, Norwegian physicist Øyvind Grøn from the Oslo and Akershus University College of Applied Sciences and his co-author Torkild Jemterud demonstrate that the rotational motion in the universe is also subject to the theory of relativity. Imagine a person at the North pole who doesn't believe the Earth rotates. As she holds a pendulum and can observe the stars in her telescope, she remarks that the swinging plane of the pendulum and the stars rotate together. Newton, who saw the world as a classical physicist, would have pointed out that it is the Earth that rotates. However, if we assume the general principle of relativity is valid, the Earth can be considered as being at rest while the swinging plane of the pendulum and the night sky are rotating. In fact, the rotating mass of the observable part of the universe causes the river of space--which is made up of free particles following the universe's expansion--to rotate together with the stars in the sky. And the swinging plane of the pendulum moves together with the river of space.










https://www.sciencedaily.com/releases/2 ... 120254.htm




AAF: I'm so glad that we've finally agreed ....
"finally agreed" ????? <img alt="Confused shrug" width="74" height="46" src="http://www.sherv.net/cm/emoticons/expre ... oticon.gif">
I never disagreed (beyond the earlier correction) ; I said it was irrelevant to the effect you were claiming (which it is).
AAF: And if the observer moves in a circle, then he/she will see the external object moving in a circle.
Let's take a simple every day example - but first give a warm welcome to Stanisława's brother Szczepan.

He's the oldest son, so also affectionately known as "S_1" He has a habit of climbing on top of things that are in the middle of something - so yes, he's nicknamed "The Central Pole".

Now, imagine someone - "Alice" - standing on the outer edge of a spinning 3-metre radius merry-go-round in the middle of a grassy park. She is not moving a muscle, but looking inwards (so always looking inwards ) towards Szczepan, who is (also without moving a muscle) standing on top of the central NON-rotating 4-metre pole that the merry-go-round goes round.
Someone's painted a couple of words on the merry-go-round - "Brisbane in the east" to Alice's right, "Oslo in the west" to her left.

I claim that :
As Alice is rotating on the merry-go-round, she will always see the central pole ("S_1" - Szczepan ... and, of course, the metal thing he's on) straight-ahead in her line of sight. He remains always fixed at exactly the same place in her view throughout the rotation, moving neither to the left nor right, and never leaving her view. As the merry-go-round spins, she see his front, then his left-side, then his back, then his right-side, then front again - in clear apparent rotation around his central axis.

Straightforward, simple, and consistent with maths and with everyone's (well, except maybe yours ) memories and expectations of merry-go-rounds.
S_1 appears, to that rotating observer, to be traveling along a big circle, on the sky; and hence, S_1, sometime, seems to be over Brisbane in the east, and some time, seems to be over Oslo in the west.
In contrast, based on spurious talk of "projected circles" and the like, you have convinced yourself that :
As Alice is rotating on the merry-go-round, she sees the central pole move around her in a "big circle", from straight ahead, over the "Brisbane in the east", out behind her over the grass, back over the "Oslo in the west", and back to centre again.

Sorry, I don't know what types of merry-go-rounds you've been on, but seriously, surely you must see that is utter nonsense.

It is still nonsense if the merry go round is transplanted to the North Pole.

And it remains nonsense if the 3-metre-radius circular merry-go-round is scaled up to our illustration of the 3,000km-radius circular slice through a line of latitude.
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AAF
AAF

June 16th, 2016, 3:00 am #38




It has been one hundred years since the publication of Einstein's general theory of relativity in May 1916. In a paper recently published in EPJ Plus, Norwegian physicist Øyvind Grøn from the Oslo and Akershus University College of Applied Sciences and his co-author Torkild Jemterud demonstrate that the rotational motion in the universe is also subject to the theory of relativity. Imagine a person at the North pole who doesn't believe the Earth rotates. As she holds a pendulum and can observe the stars in her telescope, she remarks that the swinging plane of the pendulum and the stars rotate together. Newton, who saw the world as a classical physicist, would have pointed out that it is the Earth that rotates. However, if we assume the general principle of relativity is valid, the Earth can be considered as being at rest while the swinging plane of the pendulum and the night sky are rotating. In fact, the rotating mass of the observable part of the universe causes the river of space--which is made up of free particles following the universe's expansion--to rotate together with the stars in the sky. And the swinging plane of the pendulum moves together with the river of space.










https://www.sciencedaily.com/releases/2 ... 120254.htm











""And therefore, S_1 appears, to that rotating observer, to be traveling along a big circle, on the sky; and hence, S_1, sometime, seems to be over Brisbane in the east, and some time, seems to be over Oslo in the west". Since our observer is always facing north, then east is to his right, and west is to his left. OK, explain to me how S_1, which we've agreed is ALWAYS straight in front of our observer, sometimes appears off to his left, and sometimes to his right??? That just makes no sense. If I had to guess what you mean (and I shouldn't have to guess, because you should be spelling it out clearly, yes?) - but if I had to guess .....

Taking a low value for H first: Let's say our camera such a really wide-angle lens it can take in both Brisbane and Oslo. At 1am it just so happens that the camera films at low H (so D is just above the horizon) exactly due north (ie. directly across the north pole / axis of rotation) a distant star - say Sirius. Now, Sirius will behave exactly as you've been saying - the camera will film it rotating around at a constant angle D(**) in the sky, and indeed will film it sometimes over Brisbane, and sometimes over Oslo - and in fact, it's sometimes out-of-shot behind the camera. In short, yes, it certainly does travel a "Humongous circle" in the sky. If instead we tilt the camera so choosing a high H - say we find another star again directly north at 1am but higher in the sky - call it Mizar - then Mizar will make a much smaller circle in the sky."







"That just makes no sense"?


Oh, yeah . . .


That is what the Sun is doing everyday!







And furthermore, the star Sirius:





is located way south at the celestial latitude
of −16° 42′ 58.0171; and therefore, it can see the South Pole;
but it cannot see the North Pole:


So, now, let's just forget Sirius and go back to Einstein's S_1!


Because S_1 is over the North Pole of S_2; and because the observer is rotating
around the North Pole of S_2, the observer, in this particular case, must see S_1
constantly moving in a circle over the North Pole of S_2, and centered around
him/her personally.


Does that make a perfect sense?


You bet!





It's just like this:













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AAF
AAF

June 18th, 2016, 3:00 am #39




It has been one hundred years since the publication of Einstein's general theory of relativity in May 1916. In a paper recently published in EPJ Plus, Norwegian physicist Øyvind Grøn from the Oslo and Akershus University College of Applied Sciences and his co-author Torkild Jemterud demonstrate that the rotational motion in the universe is also subject to the theory of relativity. Imagine a person at the North pole who doesn't believe the Earth rotates. As she holds a pendulum and can observe the stars in her telescope, she remarks that the swinging plane of the pendulum and the stars rotate together. Newton, who saw the world as a classical physicist, would have pointed out that it is the Earth that rotates. However, if we assume the general principle of relativity is valid, the Earth can be considered as being at rest while the swinging plane of the pendulum and the night sky are rotating. In fact, the rotating mass of the observable part of the universe causes the river of space--which is made up of free particles following the universe's expansion--to rotate together with the stars in the sky. And the swinging plane of the pendulum moves together with the river of space.










https://www.sciencedaily.com/releases/2 ... 120254.htm










"Thing is: I couldn't care less about Sirius or Mizar; They are NOT what we're talking about. We are supposed to be talking about S_1. Neither Sirius nor Mizar are candidates for S_1. S_1 is EXPLICITLY positioned directly up exactly along the axis of rotation. Sirius and Mizar are not. Let's go back to our view of Sirius at 1am. Suppose someone at Santa's workshop puts a yellow ball suspended directly above the exact North Pole, at a height that precisely blocks our camera's view of Sirius at that time. So, during S_2's rotation, our camera will film Sirius moving out from behind that ball - but since our camera is always facing North, and that ball is (by definition since it's at the north pole) always due north of it, then the camera will always have the ball in view. S_1 is THAT BALL, not Sirius, since it is that ball that is positioned along our axis of rotation, above our north pole. Which brings us back again to what I said earlier : Your pictures are flawed, because you are neglecting to take into account that S_1 is along the axis of rotation. You … repeatedly refuse to apply the mathematics to our special case of a body positioned along the axis. So let's fix that now - yes? Explain clearly (with maths.) to me how how our camera (that is always facing due north) can ever film that BALL (that is always positioned straight ahead, since it is always due north of the camera) way off to the right, over Brisbane in the east. (*) pretty much. The graphic should have had the right-angle inside the earth at the latitude of the camera, so the angle we're talking about is really between that point rather than the north pole. Trivial detail that doesn't change anything, though. (**) Actually, D in this case would not be constant. For any distant body that is NOT positioned directly along the axis of rotation, then the distance between our camera and that body will vary as the rotation brings the camera closer then further to that body. That means that that 3000km figure will be changing, so D will be changing."





Surely, the two stars Sirius & Mizar are not candidates;
but the star Polaris certainly is:











Well; the star Polaris, too, is not exactly over the North Pole.


So, let's just go back to Einstein's S_1.


And let's assume, as Einstein did in his 1916 paper, that S_1 is EXPLICITLY
positioned directly up exactly along the axis of rotation;
i.e., over the North Pole.


And therefore, according to this equation:


D = arctan(H / 3000)


If the value of H is so large, then S_1 will always appear to be stationary
over the North Pole and without any apparent motion or rotation whatsoever.


That is on one hand.


On the other hand, if the value of H is NOT so large, then S_1 will always appear
to be moving, as one single unit, in a circle over the North Pole.


But regardless of whether S_1 appears to be stationary or to be moving in a circle,
this object labeled S_1 will never ever show any sort of axial rotation caused
by the axial rotation of S_2.


S_1 might show axial rotation, if it's actually rotating around its geometrical axis.


But nonetheless, S_1 can never ever show any degree of axial rotation due to
the axial rotation of S_2












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Ufonaut99
Ufonaut99

June 18th, 2016, 11:50 pm #40




It has been one hundred years since the publication of Einstein's general theory of relativity in May 1916. In a paper recently published in EPJ Plus, Norwegian physicist Øyvind Grøn from the Oslo and Akershus University College of Applied Sciences and his co-author Torkild Jemterud demonstrate that the rotational motion in the universe is also subject to the theory of relativity. Imagine a person at the North pole who doesn't believe the Earth rotates. As she holds a pendulum and can observe the stars in her telescope, she remarks that the swinging plane of the pendulum and the stars rotate together. Newton, who saw the world as a classical physicist, would have pointed out that it is the Earth that rotates. However, if we assume the general principle of relativity is valid, the Earth can be considered as being at rest while the swinging plane of the pendulum and the night sky are rotating. In fact, the rotating mass of the observable part of the universe causes the river of space--which is made up of free particles following the universe's expansion--to rotate together with the stars in the sky. And the swinging plane of the pendulum moves together with the river of space.










https://www.sciencedaily.com/releases/2 ... 120254.htm




<blockquote>Ufonaut99: That just makes no sense
AAF: Oh, yeah . . . That is what the Sun is doing everyday! </blockquote>
Utterly irrelevant. The Sun is NOT a valid candidate for S_1. Yet again: Your pictures are flawed, because you are neglecting to take into account that S_1 is along the axis of rotation. You … repeatedly refuse to apply the mathematics to our special case of a body positioned along the axis.
AAF: On the other hand, if the value of H is NOT so large, then S_1 will always appear to be moving, as one single unit, in a circle over the North Pole.
And STILL just another empty repetition of the claim, with absolutely no basis or support (let alone any maths) given.

Again: Don’t just claim it …. SHOW IT, yes? SHOW WITH MATHS, not just word claims.

So you seem to be claiming that our observer 3,000 km from the pole, will see Polaris - or let's make it a low H value : will see Stanisława - making a "big circle" in the sky, always at the same angle D.
Except that's impossible - the only way "D" could be constant would be for Stanisława to remain at the same height above the horizon, which is NOT true for star trails (nor parallax) - as is plainly evident in photos of star trails from those locations. Further, that would also mean Stanisława goes around (so sometimes behind) our observer, which clearly contradicts our graphic that shows that our observer is always directly facing Stanisława.

So still, after all this time, just WHAT are you claiming????
AAF: Surely, S_1 is always '3000Km straight ahead, and H km up'. .... if the value of H is NOT so large, then S_1 will always appear to be moving, as one single unit, in a circle over the North Pole. ... Does that make a perfect sense?
Hardly ! I go instead with the commonsense straightforward view that anything that remains on (or directly above) the north pole remains due north of our observer. Since our observer is always facing due north, then he sees that object fixed in front of him. THAT is what makes perfect sense.

In contrast, you claim that an object positioned at/above the north pole travels until it's due EAST of our observer (.... and so continues until that NORTH pole object is due SOUTH of him, yes?) However, even though the observer remains facing north, you reckon this object (that is now East of him) is not off to his right - No, you say it's still straight ahead of him !!

All based on some idea that The projected circle is the circle of the rotating observer's latitude as seen by the same observer projected on an external object like S_1.
In other words, rather than simply calculating line-of-sight directly with maths, you're basing the whole thing on some flawed idea that "if he sees me moving in a "big circle", then I must see him moving in one as well" (*) - an absurd proposition that is clearly contradicted by the everyday experience of riders viewing central poles of merry-go-rounds.

And you reckon all that "makes perfect sense" ?????


(*) Heh, reminded me of The Ravenous Bug-Blatter Beast of Traal, that believes that if you can't see it, then it can't see you - so the best way to escape it is to wrap a towel around your head
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