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Ufonaut99

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

August 6 2016, 11:31 AM 

Hi AAF,

AAF: Certainly, if Fred were a geometrical line, then no parallax, due to Earth's daily rotation, would be detected by him at Earth's North Pole.

Precisely happy.gif Geometrically, the north pole is a special case.

Ah well, since you're not addressing that special case, then there's no difference in principle between Fred and Alice (so your posts about Fred are just another repetition of your existing claims about Alice). So let's go back to why I introduced Fred in the first place happy.gif, starting with this :

Ufonaut99: You're claiming that the sheer fact that Alice considers herself stationary, means that she considers her current position to be the north pole, and therefore that the axis of revolution of Earth/S_2 runs not from the North Pole-South Pole, but through her. This new axis of rotation results in the concrete physical effect that she sees that star's motions the same as Fred sees them - ie. the stars to her appear to rotate around a spot above her instead of around Polaris, and consequently therefore, she will see Polaris/S_1 rotate around her at a constant height above the horizon.
AAF: That is not my model. That is Einstein's model about relative rotation generally! And so, it just happens in this special case, that I personally agree with him ....

First off, No - it's YOUR model. Nothing to do with Einstein - in fact of course, it's the foundation of your argument AGAINST him.

So that foundation is your idea that how Alice "considers herself" results in actual physical changes ?? What if Alice was joined by George, fresh from the S_2 mapping project, who knows that S_2 is rotating (so does NOT consider himself stationary). Alice and George standing side-by-side - are you saying that they see different things?

Look, "considering" is basically a mental state. In SR/GR, we use that phrase to signal that all frames are equally valid, and we're just looking at the SAME scenario from another frame/observer - the SAME scenario (ie WITHOUT changing how anything happens)

For example, say I'm going at 100km down the motorway in a red car, you at the same speed alongside me in a blue car - in other words, I "consider myself (and you)" to be moving at 100kph and the Earth to be stationary.

Whilst driving, I might perfectly validly "consider myself" to be stationary, and therefore you stationary with me, and the earth to be moving 100kph backwards, NOTHING CHANGES. NOTHING!

I might then "consider myself" (and you) to be moving at 75kph, and the earth to be moving 25kph backwards, STILL NOTHING CHANGES. NOTHING!

Translating that to our roundabout, Alice is rotating within a stationary park - OR she might "consider herself" stationary, and therefore the roundabout stationary with her, with the park rotating around them both. STILL NOTHING CHANGES. NOTHING!

In short, in both classical or relativistic physics, there is NO physical change resulting from how I "consider myself" to be. NONE WHATSOEVER!

Earth nor the Universe do not bend to how you, I, nor Alice "consider ourselves" to be. There are NO physical changes - S_1 would not be seen ANY differently just because of how Alice "considers herself".

Your argument that S_1 would do so is hopelessly flawed
- and with it, your argument that Alice would see S_1 as if she was at the Pole - ie. at that constant 53 degree angle that our graohic demands
- and with that, all your arguments against Einstein

 
 
AAF

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

August 7 2016, 12:00 AM 









Hi; Ufonaut99:


I will, certainly, be more than happy to commandeer Einstein's model about relative rotation,
if you show me any piece of paper signed by Stephen William Hawking or Clifford Martin Will
or Gerardus (Gerard) 't Hooft, in which it's stated quite simply that:
It is NOT his.


Because I like it!


[linked image]


…..............................................................................................................................



""AAF: Tell me how far Einstein wanted to place his S_1 from his S_2; and I shall calculate the parallax of his S_1 and show it to you!" And that's your problem. You've found an equation: "theta = arctan([v x delta_t]/d)" (or more generally "theta = arctan(R/d)") and decided that that's ALL there is to parallax regardless of any other factors Wrong and that therefore that equation can be blindly applied to ANY scenario, including Einstein's Wrong which means you can just hit Einstein over the head with it Head Smack Wrong . For example: "AAF: Does the star Polaris have stellar parallax? Absolutely!" Ufonaut99: You are STILL refusing to apply parallax to THIS scenario. Earth has two rotations: - Daily rotation around its axis - Annual orbital rotation around the Sun. Einstein's scenario is explicitly about rotation about the axis (the DAILY one) ONLY - not the annual orbital one (*) So yes, Polaris has STELLAR parallax - that is, parallax as measured at opposite ends of earth's annual orbit. In fact, I already made that exact point earlier: Ufonaut99 (dec 26): Remember also that this [Polaris will remain motionless in frame] does not apply to observations of parallax months apart, using the earth's orbit around the sun. For these, Polaris is not the centre of rotation, and so will display measurable parallax but annual orbital rotation is NOT this (Einstein's) scenario (hence my comment), and therefore does NOT discredit Einstein."





This general equation for computing parallax:


P = 2*[arctan(r/d)]


applies equally well to Earth's daily rotation around its axis and to Earth's annual orbital
rotation around the Sun, respectively.


And more over, the above two cases differ from each other only in the actual numerical value
of the independent variable r:


A. In the case of Earth's daily rotation, the value
of the independent variable r is equal to Earth's radius; i.e., 6,371 km


B. In the case of Earth's orbital rotation, the value
of the independent variable r is equal to the radius of Earth's orbit;
i.e., 149.6 million km.


And that is it!



happy.gif











 
 
Ufonaut99

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

August 7 2016, 5:09 AM 

Hey AAF,

Hey, Aussies top of the medal tally after day 1 happy.gif

Mind you, if Etimoni Timuani get's a medal, then Tuvalu would have a 100% success rate !

Ufonaut99: No - it's YOUR model. Nothing to do with Einstein ... What if Alice was joined by George .. who knows that S_2 is rotating (so does NOT consider himself stationary). Alice and George standing side-by-side - are you saying that they see different things?
AAF: I will, certainly, be more than happy to commandeer Einstein's model about relative rotation,
if you show me any piece of paper signed by Stephen William Hawking or Clifford Martin Will or Gerardus (Gerard) 't Hooft, in which it's stated quite simply that: It is NOT his.

Because I like it!

It doesn't need any piece of paper. As shown by the differing results for how S_1 moves for our two physically identical observers Alice and George, your model gives inconsistent results (or to put it another way, is clearly just plain wrong wink.gif ) - and let's face it, when we stand outside looking up at the stars, we KNOW tha whatever we "consider" won't make the slightest difference to them.

On the plus side, this does mean I'm sure that Hawking, 't Hooft, et al, would be only too happy for you to commandeer it.

Good job

So, I am glad that you have found something that you like (though somehow I think you would "like" anything that is against Einstein wink.gif ), but "liking" is not in scope for this discussion.

OK let's recap : During these threads, you have at various times claimed that Alice will see S_1 rotating in :
1) Big Circles (like the star trails)
2) Parallax
3) constant height above the horizon

So (3) has no support in any consistent model as justification for it, and is there only because you "like" it.
However, (3) was also your answer to the fact that our graphic demands Alice views S_1 at a constant height in the sky - meaning you have no consistent model as an answer for this.

Right, one down, two to go happy.gif Moving on, let's look at Parallax next.

However, before we get onto that, we have to clear up this first :

AAF: If S_2 is rotating with an angular velocity equal to omega in a clockwise direction, then simply make the camera rotate with an angular velocity equal to omega but in a counterclockwise direction:

Would you agree that if Alice is just standing still being carried around by the roundabout (as in that video I posted) - ie, NOT counter-rotating - that she would not see any parallax in our scenario ?
Remember, in this case, Alice will ALWAYS be looking directly at S_1 straight ahead of her (unlike if she was counter-rotating).

(and you'll never guess our next area of disagreement wink.gif )

 
 
a mouse

Mach's bucket

August 8 2016, 10:52 PM 

Mach's bucket

A thought experiment Mach's bucket will be invalid. The reason is that it doesn't stand up about plural rotating bodies. A hundred disks are rotating at the same (and a constant) speed. Now one of these increases the speed (by man act). Effect will be limited to this disk only (as action-reaction).

On an axis, there are three disks A, B and C. A is rotating clockwise, B is rotating counterclockwise and C is at a standstill. How does "Mach's bucket" say ?

From viewpoint of kinetic energy also, rotary motion will be absolute.

Action at a distance (instantaneous) is major premise of Mach's bucket. So, it will contradict relativity.

Imagine propagations of star lights (in space). Mach's bucket will be a fantasy or a fairy tale (at most an episode).

About revolution (on ellipse orbits) of planets, how does "Mach's bucket" say ?

Asteroid Toutatis is rotating (cycle is 5.41 days and 7.35 days) on two axes. It's difficult to imagine Mach's bucket.

 
 
AAF

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

August 9 2016, 12:00 AM 








Yep . . .


Australia has sent many great athletes to the Rio Olympics 2016:

http://rio2016.olympics.com.au


And I agree with Mouse:

Einstein's mentor called 'Mach' was way wrong!



happy.gif


…...................................................................................................................................................



""AAF: Certainly, if Fred were a geometrical line, then no parallax, due to Earth's daily rotation, would be detected by him at Earth's North Pole." Precisely. Geometrically, the north pole is a special case. Ah well, since you're not addressing that special case, then there's no difference in principle between Fred and Alice (so your posts about Fred are just another repetition of your existing claims about Alice). So let's go back to why I introduced Fred in the first place starting with this: Ufonaut99: You're claiming that the sheer fact that Alice considers herself stationary, means that she considers her current position to be the north pole, and therefore that the axis of revolution of Earth/S_2 runs not from the North Pole-South Pole, but through her. This new axis of rotation results in the concrete physical effect that she sees that star's motions the same as Fred sees them - ie. the stars to her appear to rotate around a spot above her instead of around Polaris, and consequently therefore, she will see Polaris/S_1 rotate around her at a constant height above the horizon."




It has been demonstrated already, in this thread and in the previous thread as well,
that Einstein's supposition concerning the apparent axial rotation of S_1 is ruled out
by the presence of parallax at every geometrical latitude of S_2.


Surely, at the north pole, if we assume that the observer is a geometrical line
with zero thickness, the parallax will disappear.


But, even in this special case, Einstein's supposition regarding the apparent axial
rotation of S_1 is ruled out by the simple fact that the observer
is not rotating at all.


In other words, Einstein's supposition is false at any geographical latitude less
than 90 degrees, because the parallax is present.


And Einstein's supposition is false at the geographical latitude of 90 degrees, because
rotation is absent.



[linked image]










 
 
Feynman-Schwinger

AAF you've never studied physics clearly

August 9 2016, 1:26 AM 

I read the thread between you and Utronaut and it's quite clear you don't actually understand physics. Einstein was not wrong; inasmuch as the experimental evidence exists, he was actually right (again). You've constructed a straw man to bash over and over, but it's terrible physics.

You're attacking a straw man, and your calculations are not internally consistent. You have yet to justify the math in the special case along the axis (in your so-called NOrth Pole example), and you also don't seem to grasp the basic physics of centrifugal force (which isn't really a force per se, but we still refer to it as such).



AAF loses. I don't have a dog in the fight, but Utronaut schooled you badly. Learn some physics and come back. If you're still harping on and on about the parallex (which has nothing to do with the special case Einstein invoked), you have have quite obviously never studied the Einstein Field Equations, differential geometry is a lot harder than posting pseudo-intellectual drivel.

 
 
a mouse

Newton's bucket

August 9 2016, 8:49 PM 

Newton's bucket

A tube is rotating like propeller. In the middle of the tube, a light source is lighted. When the rotational speed increases, light will be impossible to leave the tube. This will be the phenomenon "Newton's bucket" also (light version).

P.S. Today’s physics (large part of it) is terribly wrong.

 
 
Ufonaut99

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

August 10 2016, 11:51 AM 

AAF: Australia has sent many great athletes to the Rio Olympics 2016 .... And I agree with Mouse

Ah well, dropped down now - so time to switch back to the medals-per-population statistic that's served us so well in the past happy.gif

And I agree with Feynman-Schwinger happy.gif

AAF: Einstein's supposition is false at the geographical latitude of 90 degrees, because rotation is absent.
Your claim is that an observer at the North Pole is not rotating - seriously ??

So we have a choice as to which is ruled out - either Einstein's claims, or yours that "observer is not rotating at all" wink.gif

Let's recall that photographs of circular star trails are the result of the observer (and their camera wink.gif ) rotating with earth. So, photographs from a polar observer (by your claim "not rotating at all") could not possibly show star trails - photos like this one :
Startrails24Schwartz800.jpg
(source : APOD South Pole Star Trails

Funny how your non-rotating observer gets exactly the same rotation (angular velocity) effect as everybody else, eh ? wink.gif

And why is it that you claim that the polar observer is not rotating ?
AAF: That is because the angular velocity and tangential velocity of a rotating object are related to each other by this equation:
Tangential velocity (v) = Angular velocity (omega) X Radius (r)
Therefore, if the tangential velocity is equal to zero, the value of angular velocity is equal to zero (in the case of Radius > 0), and its value is undefined (in the case of Radius = 0).
And of course, 'undefined' is practically the same as zero; although mathematicians, for some theoretical reasons, can't say outright that 0/0 = 0!

Sorry, but that's ridiculous. In maths, variables - let alone actual physical quantities - do not change their value simply because they can be multiplied with other factors which may be zero.

Let's take another example : a wheel on my car has a circumference of 2 metres, therefore we can say :

Distance travelled (d) = circumference (c) X Revolutions (r)

Yesterday, I didn't drive anywhere, so Revolutions = 0 and Distance travelled = 0.
So does that mean the wheel's circumference suddenly becomes 'undefined' - where of course 'undefined' is practically the same as zero ? Of course not.



So ultimately, on Earth it is obvious that star trails are horizontal to the horizon ONLY at the poles.

In contrast, your claims against Einstein are based on claims that a star trail horizontal to the horizon is visible from ANYWHERE (simply by the observer considering themselves stationary) EXCEPT at the poles (where that observer is "not rotating at all").

Tough call ..... NOT ! Shaking Head

 
 
Anonymous

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

August 10 2016, 6:46 PM 

>>>>Distance travelled (d) = circumference (c) X Revolutions (r)
Yesterday, I didn't drive anywhere, so Revolutions = 0 and Distance travelled = 0.
So does that mean the wheel's circumference suddenly becomes 'undefined' - where of course 'undefined' is practically the same as zero ? Of course not.

given only that information then c is undefined in that scenario and might as well treat it as zero.

 
 
Volker

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

August 10 2016, 7:01 PM 


>>>Yesterday, I didn't drive anywhere, so Revolutions = 0 and Distance travelled = 0...<<< BAD ANALOGY.

BETTER ANALOGY: Picture yourself standing in the middle of this rotating disc:

[linked image]

What is the value of your angular velocity? Is it zero or something else?


 
 
AAF

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

August 11 2016, 12:00 AM 










Thanks, Mouse & Anonymous & Volker . . .


& hello; Feynman-Schwinger:


Surely You're Joking, Mr. Feynman:

http://www.earth.northwestern.edu/~amir/files/Richard_P_Feynman-Surely_Youre_Joking_Mr_Feynman_v5.pdf


[linked image]


It is not logically permissible to use Einstein's field equation to defend or to justify
Einstein's two-fluid-body thought experiment; because Einstein based his field equation upon
his two-fluid-body thought experiment; and not the other way around:

https://en.wikisource.org/wiki/The_Foundation_of_the_Generalised_Theory_of_Relativity


And it follows, therefore, that if Einstein's two-fluid-body thought experiment
is flawed, THEN Einstein's field equation must be flawed as well;

right; Colleague Ufonaut99?


wink.gif


….............................................................................................................................................



"No - it's YOUR model. Nothing to do with Einstein ... What if Alice was joined by George .. who knows that S_2 is rotating (so does NOT consider himself stationary). Alice and George standing side-by-side - are you saying that they see different things? "AAF: I will, certainly, be more than happy to commandeer Einstein's model about relative rotation, if you show me any piece of paper signed by Stephen William Hawking or Clifford Martin Will or Gerardus (Gerard) 't Hooft, in which it's stated quite simply that: It is NOT his. Because I like it!" It doesn't need any piece of paper. As shown by the differing results for how S_1 moves for our two physically identical observers Alice and George, your model gives inconsistent results (or to put it another way, is clearly just plain wrong ) - and let's face it, when we stand outside looking up at the stars, we KNOW that whatever we "consider" won't make the slightest difference to them. On the plus side, this does mean I'm sure that Hawking, 't Hooft, et al, would be only too happy for you to commandeer it. Good job. So, I am glad that you have found something that you like (though somehow I think you would "like" anything that is against Einstein), but "liking" is not in scope for this discussion."





Well; here's Einstein's model; "Life on a Merry-Go-Round":

https://www.amherst.edu/media/view/10267/original/reden05.pdf


happy.gif


According to my model: Newton's regular model + parallax; Einstein was correct
to suppose that Alice, at the outer edge of the merry-go-round, can very simply consider
herself to be stationary and to perceive the whole merry-go-round doing
the rotation around her.


That is to say, the center of circular rotation and the periphery of circular rotation
are indeed reversible.


Nonetheless, Herr Einstein made a horrible oversight by neglecting entirely the effect
of parallax on circular rotation in general and axial rotation in particular.


Why is the phenomenon of parallax extremely important within this context?


Parallax is immensely important with regard to circular rotation
for two main reasons:


1. Parallax allows observers to determine with absolute certainty whether they are actually
rotating or not. And as a result, the assumption of being stationary does not work;
and hence,in the final analysis, Einstein's principle of relativity is invalid with
regard to circular rotation.


2. Parallax allows each observer to have only one tangential velocity.
In other words, multiple tangential velocities are not allowed.
And therefore, relative axial rotation is not possible. Simply because
axial rotation, by its very definition, implies an infinite number of tangential
velocities, at various latitudes, at the same time.


In short, axial rotation is always real; and apparent axial rotation simply
does not exist in the natural world.










 
 
Ufonaut99

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

August 12 2016, 10:31 AM 

Hi AAF,
Just had our Census - the first time online. Yep, it crashed (ok, under a denial of service attack, but still ...)
https://pbs.twimg.com/media/CplDf3iVUAAAS-o.jpg

AAF: It is not logically permissible to use Einstein's field equation to defend or to justify Einstein's two-fluid-body thought experiment; because Einstein based his field equation upon his two-fluid-body thought experiment;

Actually, no, Einstein didn't wink.gif The two-fluid-body problem gives a clear description of why he found the Newtonian model (that inertia is just an unexplainable effect of acceleration) unsatisfactory, preferring Mach's principle that at least offered some explanatory power - but he didn't based his field equations on either that problem nor Mach (to the extent that the field equations can give rise to solutions that are anti-Machian).
And also, Feynman-Schwinger related the Field Equations to your understanding of Parallax rather than the two-body issue issue happy.gif I must be honest, however, and admit that I have never studied the Field Equations either Whistling, so my arguments against your parallax claims are purely classical happy.gif

On that note, just a reminder of my earlier question before we start discussing parallax :
AAF: If S_2 is rotating with an angular velocity equal to omega in a clockwise direction, then simply make the camera rotate with an angular velocity equal to omega but in a counterclockwise direction:


Would you agree that if Alice is just standing still - ie, NOT counter-rotating - that she would not see any parallax in our scenario ?
Remember, in this case, Alice will ALWAYS be looking directly at S_1 straight ahead of her (unlike if she was counter-rotating).

 
 
Ufonaut99

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

August 12 2016, 10:32 AM 

Hi Mouse, and welcome to the board happy.gif

a mouse: A thought experiment Mach's [principle] will be invalid. ... Action at a distance (instantaneous) is major premise of Mach's bucket. So, it will contradict relativity.


I don't want to go too much into Mach's principle - for one thing, although Einstein was motivated by it, it is not core to GR (as I noted above to AAF), so a moot issue.
Mach himself probably believed in instantaneous action at a distance - as, of course, did Newton with gravity. We now realise that there's no problem with gravity travelling at c - for example, having the Sun's gravitational field taking 8 minutes to reach Earth just as it's light does (so if the Sun were to suddenly disappear, it would take 8 minutes for Earth's orbit to change) causes no problems; it doesn't have to be "instantaneous".
The same with Mach, so it's quite consistent for your A,B,C disks, etc, to simply be reacting to some field local to them at the time, so replicating all the effects of inertia.


 
 
Ufonaut99

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

August 12 2016, 10:36 AM 

Hi Volker,

Volker: given only that information then c is undefined in that scenario

But we're NOT given only that information - that's the point ! happy.gif(*)

We're given that Earth (and so every point on earth) is rotating with angular velocity of 1 rotation per day - that doesn't stop at the poles, as is clearly shown by those polar star-trail photos.
I gave the circumference of my wheel as 2 metres - but even if I hadn't, that actual physical quantity wouldn't change just because I referred to it in some equation.
Volker: BETTER ANALOGY: Picture yourself standing in the middle of this rotating disc:
What is the value of your angular velocity? Is it zero or something else?

OK, let's say you set the angular velocity of that disk "D1" to some value OMEGA1, before I find myself facing the wall at the top of the picture.
Let's also say that there's a big sign saying "S_1" on the ceiling above me, on another rotating disk "D2" that you have also set to rotate at some rate OMEGA2 independent of D1.

What is the value of my angular velocity? I don't know - but I do know that it is whatever you set it to be. I will not make the mistake of saying that my personal lack of knowledge of a quantity means that that quantity = zero.

So let me ask you the same question happy.gif What do you reckon my angular velocity is - the choice is EITHER :

- as I claim, that even though I don't know what it is, I have whatever OMEGA1 angular velocity you have set, so will therefore eventually find myself moving away from facing that top wall (if your OMEGA1 > 0) - and therefore see that S_1 sign rotating relative to me at a rate OMEGA2 - OMEGA1

OR

- that since "tangential velocity" = 0 means I'm "not rotating at all", meaning I'm always facing that top wall and therefore that S_1 sign is rotating relative to me at a straight OMEGA2 - all regardless of whatever OMEGA1 velocity you had set D1 to.


Which do you reckon ? happy.gif


(*) and might as well treat it as zero.
No, you'd never do that in either maths nor physics; "unknown" is not the same as zero. You might try running "what-if" scenarios substituting zero or other values for the unknown quantity, but that's different from making a case based on that idea.

 
 
roger

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

August 12 2016, 6:21 PM 



>>>>(*) and might as well treat it as zero.
>>No, you'd never do that in either maths nor physics; "unknown" is not the same as zero.

the type of "unknown" being referred to had nothing to be gained by treating it as other than zero. What evidence do you have that "you'd never do that in either maths nor physics"


 
 
AAF

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

August 13 2016, 12:00 AM 









Hi, Ufonaut99 & Roger:


Aussies' Website, for Census online, maybe down :

https://pbs.twimg.com/media/CplDf3iVUAAAS-o.jpg


But surely, Aussies' medal count, in Rio, has gone way up
to more than 17:

http://rio2016.olympics.com.au



[linked image]



…...........................................................................................................................................



"OK let's recap : During these threads, you have at various times claimed that Alice will see S_1 rotating in : 1) Big Circles (like the star trails) 2) Parallax 3) constant height above the horizon. So (3) has no support in any consistent model as justification for it, and is there only because you "like" it. However, (3) was also your answer to the fact that our graphic demands Alice views S_1 at a constant height in the sky - meaning you have no consistent model as an answer for this. Right, one down, two to go. Moving on, let's look at Parallax next. However, before we get onto that, we have to clear up this first : "AAF: If S_2 is rotating with an angular velocity equal to omega in a clockwise direction, then simply make the camera rotate with an angular velocity equal to omega but in a counterclockwise direction": Would you agree that if Alice is just standing still being carried around by the roundabout (as in that video I posted) - ie, NOT counter-rotating - that she would not see any parallax in our scenario? Remember, in this case, Alice will ALWAYS be looking directly at S_1 straight ahead of her (unlike if she was counter-rotating). (and you'll never guess our next area of disagreement)"






Ah; I see . . .


It seems, to me, you've already forgot everything about
this great illustration:


[linked image]



happy.gif



According to the above illustration, the constant height above the horizon,
theta, is due to the constant height of S_1, H, above the north pole:


theta = arctan(4000/3000) = 53.13 degrees.


So, now, what is the parallax, P, in this case?


Well; I told you, before, that the parallax is always the angle subtended
by the diameter of the circle along the periphery of which
the observer is rotating:


P = 2*[arctan(3000/4000)] = 73.731 degrees.


And that is it!



[linked image]











 
 
Ufonaut99

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

August 14 2016, 10:55 AM 

Yes, we've slipped down the rankings, but what's really upset folks here is that Australia's behind the "old enemy" of Great Britain.
As I say, hopefully we'll do well calculating medals-dividedby-population .... probably not quite so well in the medals-dividedby-land area !!
Still, a nation of sports lovers - a couple of well-known radio presenters took on the 400m freestyle olympic gold medallist (yes, another aussie) in a "fair and reasonable" swimming challenge ... albeit with some "non-negotiable" terms happy.gif
Incidentally, Brisbanites are well represented at the Olympics - including by the Tongan flag-carrier (the one covered in oil happy.gif )

AAF: If S_2 is rotating with an angular velocity equal to omega in a clockwise direction, then simply make the camera rotate with an angular velocity equal to omega but in a counterclockwise direction:
Ufonaut99: You've found an equation : "theta = arctan([v x delta_t]/d)" (or more generally "theta = arctan(R/d)") and decided that that's ALL there is to parallax regardless of any other factors
Wrong
Would you agree that if Alice is just standing still - ie, NOT counter-rotating - that she would not see any parallax in our scenario ?
Remember, in this case, Alice will ALWAYS be looking directly at S_1 straight ahead of her (unlike if she was counter-rotating).

AAF: It seems, to me, you've already forgot everything about this great illustration: According to the above illustration, the constant height above the horizon,
theta, is due to the constant height of S_1, H, above the north pole: theta = arctan(4000/3000) = 53.13 degrees.


Ah yes, that "constant height", for which your only model is to claim that the observer's "considerations" shifts the axis of revolution of the entire planet. Nothing to do with Parallax, of course, since that is an objective fact, regardless of whether the observer considers themselves stationary or not.

Tell you what, I'll make a deal with you happy.gif You go outside tonight, and in the stillness consider yourself stationary, and look up at the stars ..... and when I hear on the news that the Earth's axis of revolution has shifted, and/or that some stars from the latitude of Oslo or below have shifted to travelling parallel to the horizon, ONLY THEN will I give that idea any credence.

Until then, I'm afraid I'm going to continue to regard that idea as one that violates basic physics, consistency (as with Alice and George), what it means to "consider yourself stationary" - and certainly not an idea that either Einstein nor Newton would ever have subscribed to.
... And therefore, of course, that you have no justification for S_1 rotating in a circle at that 53 degrees.

Deal ? happy.gif


As for forgetting the illustration, Not at all happy.gif I referred above to "any other factors", and I just happen to reckon that whether or not the observer is counter-rotating IS one of those other factors.
Ah well, since you simply repeated the "theta = arctan(R/d)" formula (albeit multiplied by 2), I take it you believe that Alice will see parallax regardless of whether or not she is counter-rotating - yes?

If so, I can't see any reason why you'd be concerned about counter-rotation one way or the other. For myself, though, I say that Alice is NOT counter-rotating, but rather is simply standing still, letting S_2 or the merry-go-round simply carry her around - just as in that video I referenced above. That's why in my earlier posts, I referred to "not moving a muscle", and the camera "encased in concrete", etc (both contrary to counter-rotation). So even if you're not concerned either way, humour me and let's nail this down up front happy.gif

Again, to keep things nice and objective, we'll concentrate on Alice's camera in Oslo on S_2 (as per graphic) that's on a tripod fastened stationary (nailed down ! ) relative to S_2, but with a motorised mount that she can set to (counter-)rotate at any rate :
Sevenoak-Automatic-Motorized-Rotating-Ti

Now, say S_2 is rotating at a rate "omega" (eg 1 revolution per 24 hours clockwise), and you mentioned above "then simply make the camera rotate with an angular velocity equal to omega but in a counterclockwise direction: ".
However, bear in mind that under Einstein's scenario, as you said : the "two observers do not know which of the two bodies, S_1 and S_2, is actually rotating around its geometrical axis. "

So picture Alice on S_1 or S_2, but she does not know whether the one that she's on is the one that is rotating - meaning she has no basis for turning on the camera's motorised mount.
In fact, since she "considers herself stationary", she has good reason to leave the camera stationary with her, NOT counter-rotating.

So, do you have any good reason for Alice to set the camera's mount to (counter-)rotate, or are you happy to agree that she does not ?

 
 
AAF

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

August 15 2016, 12:00 AM 









Wait a minute . . .


Australia is a continent.


So, we should compare its Olympic performance to that of the continent of Europe as a whole,
and NOT to the Olympic performance of 'little' Britain, or 'little'
France, or 'little' Russia;
fair & square?



[linked image]


…............................................................................................................................................................



""AAF: That is because the angular velocity and tangential velocity of a rotating object are related to each other by this equation: Tangential velocity (v) = Angular velocity (omega) X Radius (r). Therefore, if the tangential velocity is equal to zero, the value of angular velocity is equal to zero (in the case of Radius > 0), and its value is undefined (in the case of Radius = 0). And of course, 'undefined' is practically the same as zero; although mathematicians, for some theoretical reasons, can't say outright that 0/0 = 0"! Sorry, but that's ridiculous. In maths, variables - let alone actual physical quantities - do not change their value simply because they can be multiplied with other factors which may be zero. Let's take another example : a wheel on my car has a circumference of 2 metres, therefore we can say: Distance travelled (d) = circumference (c) X Revolutions (r). Yesterday, I didn't drive anywhere, so Revolutions = 0 and Distance travelled = 0. So does that mean the wheel's circumference suddenly becomes 'undefined' - where of course 'undefined' is practically the same as zero ? Of course not. So ultimately, on Earth it is obvious that star trails are horizontal to the horizon ONLY at the poles. In contrast, your claims against Einstein are based on claims that a star trail horizontal to the horizon is visible from ANYWHERE (simply by the observer considering themselves stationary) EXCEPT at the poles (where that observer is "not rotating at all"). Tough call ..... NOT!"





That is not ridiculous!


wink.gif



That is, in fact, is a well-tested and very robust mathematical formula.


Take a look:


http://hyperphysics.phy-astr.gsu.edu/hbase/rotq.html


Do you see this equation:


[linked image]


happy.gif



The above equation implies, necessarily, that, at the center of any rotating disk,
the numerical values of both the tangential velocity, v, and the radius of the disk,
r, go exactly to zeros leaving behind the angular velocity, omega, totally
undefined and with just this one meaningless value: 0/0.




















 
 
AAF

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

August 17 2016, 12:00 AM 









"I agree with Feynman-Schwinger. "AAF: Einstein's supposition is false at the geographical latitude of 90 degrees, because rotation is absent." Your claim is that an observer at the North Pole is not rotating - seriously?? So we have a choice as to which is ruled out - either Einstein's claims, or yours that "observer is not rotating at all". Let's recall that photographs of circular star trails are the result of the observer (and their camera ) rotating with earth. So, photographs from a polar observer (by your claim "not rotating at all") could not possibly show star trails - photos like this one:

[linked image]

Funny how your non-rotating observer gets exactly the same rotation (angular velocity) effect as everybody else, eh? And why is it that you claim that the polar observer is not rotating?"






What is wrong with being a non-rotating polar observer?


wink.gif


There is absolutely no difference, here, between the aforementioned polar observer
and stationary Szczepan at the center of the spinning 3-metre radius merry-go-round.


Certainly, the non-rotating polar observer can see very clearly the landscape
around him/her rotating periodically relative to the stars around the star Polaris.


But, please, please, please, notice very carefully that neither the star Polaris
nor the stars around the star Polaris are rotating relative to
the polar observer him/herself.


So, once again,what is exactly the polar observer is seeing?


Well; let's assume, for the sake of argument, that this polar observer
has a zero cross section!


wink.gif


This assumption is very important; because otherwise, the body of the polar observer
is going to rotate, with the angular velocity of the earth, around the geometrical
line of the north pole, which runs right smack through the middle of it.


Accordingly, the zero-cross-section observer, at the earth's north pole, sees the stars,
including Polaris, stationary all the time.


And at the same time, the same polar observer sees the whole geographical horizon
of the earth rotating at a constant angular velocity relative to him/her and
relative to the stationary stars in the sky as well.



happy.gif












 
 
Ufonaut99

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

August 18 2016, 9:46 AM 

AAF: Wait a minute . . .

Australia is a continent.

So, we should compare its Olympic performance to that of the continent of Europe as a whole, and NOT to the Olympic performance of 'little' Britain, or 'little' France, or 'little' Russia; fair & square?

Yikes - I hadn't thought of that !!! Surprised scream

That leaves me with ..... We've got more medals than Antarctica
Evil Penguin Slap

In other news, there are going to be traffic problems in Brisbane next week, owing to some scheduled acts of gods.

AAF: the observer is not rotating at all. ... Do you see this equation: ... go exactly to zeros leaving behind the angular velocity, omega, totally undefined and with just this one meaningless value: 0/0.


Do you see this equation : Circumference = Distance / Revolutions. The circumference value remains unchanged for all values of those other terms > 0, whether big or small - ie. it is independent of those other terms.
Whether it's one revolution or ten million revolutions, the Distance changes at the same rate, leaving Circumference with exactly the same value.
Because of that independence, we can say with confidence that it is an independent physical quantity conserved even when both other terms=0, so all that that "undefined" value means is that if you were given ONLY that those other terms = 0, then you would not be able to determine it's value.

What about angular velocity = tangential velocity / radius ? Again, the angular velocity value remains unchanged for all values of those other terms > 0, whether big or small - ie. it is independent of those other terms.
Whether it's one centimetre radius or 100 km radius, the tangential velocity changes at the same rate, leaving angular velocity with exactly the same value.
Because of that independence, we can say with confidence that it is an independent physical quantity conserved even when both other terms=0, so all that that "undefined" value means is that if you were given ONLY that those other terms = 0, then you would not be able to determine it's value.

So all that your argument boils down to, is that you have chosen to define the term "totally undefined" in your quote above, as meaning defined as being precisely zero, no other values are possible (so it must be non-rotating), regardless of whether it is an independent quantity.

AAF: Well; let's assume, for the sake of argument, that this polar observer has a zero cross section! This assumption is very important; because otherwise, the body of the polar observer is going to rotate, with the angular velocity of the earth,


So any actual physical polar observer (who has non-zero cross-section) IS "going to rotate with the angular velocity of the earth", and therefore NOT see "the stars, including Polaris, stationary" happy.gif
Which means that that observer will see the stars rotating at the same rate as the horizon (no surprise, as that's exactly what polar observers see on Earth).

In other words, your entire argument about angular velocity (and therefore the foundation of your argument against Einstein) applies only about zero-cross-section points (and then only with your definition of "totally undefined"), but FAILS for any actual physical observer who has non-zero cross section wink.gif

 
 
 
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