How Does the Principle of Equivalence Lead to Gravity Curving Space?

[Dale]

Because you use it to describe the properties of zero-gravitational field which is our topic and not the properties of the gravitational field of accelerometer which will affect the region in which we want to study the conduct of a particle

This is certainly not "our topic" it is only "your topic". You are the only person who is discussing zero gravitational field with your definition meaning absolutely no matter or energy anywhere in the universe. Everyone else was talking about the usual meaning of being far from any significant gravitational sources. I even specified "no significant gravitational sources" rather than zero gravity and yuiop used the clarifying phrase "zero gravitational field far from any significant gravitational sources" to make it clear what he meant.

At that stage of discussion I would not talk about the problem of the gravitational field of the accelerometer because i was dealing with another incorrect assumption which is that objects can accelerate relative to each other in the absence of gravitational

But again, your objection doesn't make any sense even within the framework you have stated here. First, you claim that the accelerometers won't work because they are in zero gravity. But if there is an accelerometer then by your definition there is a gravitational field. And so the accelerometers function exactly as I specified. So your whole objection saying that they wouldn't work that way is irrational and self-contradictory. You should have responded to my example by pointing out the gravitational field of the accelerometers. Given your definition that is the only correct response.

Also, your statement that a rocket engine in zero gravity would not burn is similarly nonsensical and self-contradictory. You claim that the rocket fuel won't burn because of the zero gravity and you also claim that there will be gravity because of the rocket. You should have responded to that by pointing out the gravitational field caused by the rocket, as it is your response is evasive and self-contradictory.

You have been very dishonest in this discussion. First, you take a non-standard position on the definition of an important term in the discussion. Then, when someone says something that clearly shows that they are using the standard definition you make an objection that on one hand is based on your non-standard definition and on the other hand is contradicted by that same definition. Then instead of clarifying your non-standard definition you spend several pages making similar self-contradictory statements.

 

[Altabeh]

This is certainly not "our topic" it is only "your topic". You are the only person who is discussing zero gravitational field with your definition meaning absolutely no matter or energy anywhere in the universe. Everyone else was talking about the usual meaning of being far from any significant gravitational sources. I even specified "no significant gravitational sources" rather than zero gravity and yuiop used the clarifying phrase "zero gravitational field far from any significant gravitational sources" to make it clear what he meant.

But again, your objection doesn't make any sense even within the framework you have stated here. First, you claim that the accelerometers won't work because they are in zero gravity. But if there is an accelerometer then by your definition there is a gravitational field. And so the accelerometers function exactly as I specified. So your whole objection saying that they wouldn't work that way is irrational and self-contradictory. You should have responded to my example by pointing out the gravitational field of the accelerometers. Given your definition that is the only correct response.

Also, your statement that a rocket engine in zero gravity would not burn is similarly nonsensical and self-contradictory. You claim that the rocket fuel won't burn because of the zero gravity and you also claim that there will be gravity because of the rocket. You should have responded to that by pointing out the gravitational field caused by the rocket, as it is your response is evasive and self-contradictory.

You have been very dishonest in this discussion. First, you take a non-standard position on the definition of an important term in the discussion. Then, when someone says something that clearly shows that they are using the standard definition you make an objection that on one hand is based on your non-standard definition and on the other hand is contradicted by that same definition. Then instead of clarifying your non-standard definition you spend several pages making similar self-contradictory statements.

Well this sounds logical to me! Even if we suppose Einstein's rotating disc was a failure, who says that acceleration means always a change in geometry? Yet the spacetime could be flat in the presence of accelerations and this has known examples in GR! Zero gravity doesn't exist and all Einstein wanted to show through his thought experiment was that if accelerations are present, then they possibly are in charge of a change in the geometry! This was just a motivation for Einstein to begin generalizing SR!

In his post # 80, Mueiz states that if Einstein was allowed to use locally the approximation I talked of in my earlier post, why wouldn't he be able to use this advantage and claim that the edge of disc is in a uniform motion?! Well if Mueiz was aware that he is making no sense out of such usage, then this would have ended right on page 4. Speaking of Einstein's motivation, if the ruler in the infinitesimally small distance $$ds$$ of circumference of disc is to undergo Lorentz contraction, then it would change throughout the circumference even if the disc is constrained at its large scale because we look at the infinitesimally small line-element and then get to know how spacetime at large scale is curved around, say, Sun! But according to Mueiz's falsifying method (Mueiz's logic) since local flatness is guaranteed in a very small region, why not think of this globally? This is way beyond wrong!

AB

 

[Passionflower]

An accelerating disk in flat spacetime (assuming the disk has no mass) does obviously not change the geometry of spacetime, the spacetime remains flat.

One thing to realize is that if the disk is spun up and the radius remains the same the density of the material near the edges must be less than before the spinning or there are actually cracks.

 

[Mueiz]

Now I have a clear and simple method to show the Mistake of Rotating Disc Experiment
suppose we have two similar rotating discs that rotate around the same center at the same RPM
in opposite direction
The inertial observer is in the ground see both discs rotate
Now according to Einstein and many others..the geometry of each of the two discs is nonEuclidean...but ..they must take the same geometry because their relation to the inertial observer is quite the same.
But they are accelerating relative to each other
Acceleration and the same Geometry ?
Can DaleSpam , Altabeh ,Yoiup and Others resolve this simple paradox

 

[Mentz114]

Now I have a clear and simple method to show the Mistake of Rotating Disc Experiment
suppose we have two similar rotating discs that rotate around the same center at the same RPM
in opposite direction
The inertial observer is in the ground see both discs rotate
Now according to Einstein and many others..the geometry of each of the two discs is nonEuclidean...but ..they must take the same geometry because their relation to the inertial observer is quite the same.
But they are accelerating relative to each other
Acceleration and the same Geometry ?
Can DaleSpam , Altabeh ,Yoiup and Others resolve this simple paradox

What you are failing to understand is that the geometry as perceived by an observer riding on the disc is non-Euclidean.

In your eaxmple we can specify three observers A) not on a disc, not rotating B) riding on disc 1, C) riding on disc 2.

They all perceive different spatial geometry - there is no paradox.

Your arguments grow weaker and more childish - why don't you give up and stop wasting your time ?

 

[Dale]

suppose we have two similar rotating discs that rotate around the same center at the same RPM
in opposite direction
The inertial observer is in the ground see both discs rotate
Now according to Einstein and many others..the geometry of each of the two discs is nonEuclidean...but ..they must take the same geometry because their relation to the inertial observer is quite the same.
But they are accelerating relative to each other
Acceleration and the same Geometry ?
Can DaleSpam , Altabeh ,Yoiup and Others resolve this simple paradox

No, their relation to the inertial observer is opposite, one having the opposite angular momentum of the other. No paradox, just another installment of your usual disingenuous tactics, this one even less well-disguised than usual.

 

[Altabeh]

An accelerating disk in flat spacetime (assuming the disk has no mass) does obviously not change the geometry of spacetime, the spacetime remains flat.

Not from the perspective of the at-rest observer located outside the disc! Since in his eyes each ruler gets Lorentz contracted, then there must be more rulers when disc is in motion than when it is motionless to cover the edge of disc which is definitely a non-Euclidean result!

One thing to realize is that if the disk is spun up and the radius remains the same the density of the material near the edges must be less than before the spinning or there are actually cracks.

According to Stachel, this is wrong! The edge does not get contracted; it is, rather, the ruler that has a Lorentzian non-constrained motion! No cracks!

AB

 

[Altabeh]

No, their relation to the inertial observer is opposite, one having the opposite angular momentum of the other. No paradox, just another installment of your usual disingenuous tactics, this one even less well-disguised than usual.

Seconded by me as well!

AB

 

[Altabeh]

What you are failing to understand is that the geometry as perceived by an observer riding on the disc is non-Euclidean.

You mean that length contraction happens to exist from the perspective of observer riding on the disc? If so, then you're not correct!

AB

 

[Mueiz]

No, their relation to the inertial observer is opposite, one having the opposite angular momentum of the other. No paradox, just another installment of your usual disingenuous tactics, this one even less well-disguised than usual.

Also seconded By Altabeh
Ok ...then you see that the direction of rotating of the disc has something to do with the geometry in spite of the symmetry ... but if symmetry is not enough ...I have another thing which is that ..the direction of rotating of the disc can not affect the direction of acceleration which you claim to be the cause of nonEuclidean geometry

This is enough for me and am happy that I did not use any word against any user of this forum or its way of thinking except for those of science (incorrect,inconsistent,wrong,..etc) while receiving words like (absurd,speculation,home grown understanding,ridiculous misconception.. and many other)

 

[Dale]

the direction of rotating of the disc can not affect the direction of acceleration

This is certainly not correct. The direction of rotation most definitely does affect the direction of acceleration. The direction of acceleration for the two counter-rotating observers are 180° out of phase. They are most definitely not equal.

Are you going to continue to avoid discussing the fact that your previous objections to my accelerometer and rocket examples were wrong and my descriptions were completely accurate (even using your definitions)? You have no valid argument to stand on, despite more than 100 posts.

 

[Mueiz]

This is certainly not correct. The direction of rotation most definitely does affect the direction of acceleration. The direction of acceleration for the two counter-rotating observers are 180° out of phase. They are most definitely not equal.

Are you going to continue to avoid discussing the fact that your previous objections to my accelerometer and rocket examples were wrong and my descriptions were completely accurate (even using your definitions)? You have no valid argument to stand on, despite more than 100 posts.

The direction of any rotating object is always toward the center ..what you talked about is
the direction of motion ...this is a simple fact that I am sure you know it but may be just a mistake
Concerning the accelerometer an roket examples ... they are wrong if we neglect the gravitational field of the rocket and accelerometer as it was our method in that bost ... if you do not want to neglet it this will be new discussion but i agree with you concerning the effect of acceleration in gravitational field ( here i want to add that a particle can not be
affected by its own gravitational field but of course an experiment is not one particle ..there is observer and object ..in the example of rocket there is the rocket)

 

[Dale]

The direction of any rotating object is always toward the center ..what you talked about is
the direction of motion

I was talking about the acceleration, not velocity although it does apply to velocity as well. The acceleration, position, and velocity vectors are all 180° out of phase for the counter rotating observers. "Toward the center" is not a single direction, but many directions which change over time in a manner which is different for each of the two counter rotating observers.

Concerning the accelerometer an roket examples ... they are wrong if we neglect the gravitational field of the rocket and accelerometer

But your post 98 asserts that you cannot neglect gravitational field, your position in that post is that there is no gravitational field so small that it could be neglected. Therefore in the examples I gave the accelerometers and rocket would work as I stated, even according to your analysis. Or do you still assert that the rocket would not burn and if so on what justification?

 

[Mueiz]

I was talking about the acceleration, not velocity although it does apply to velocity as well. The acceleration, position, and velocity vectors are all 180° out of phase for the counter rotating observers. "Toward the center" is not a single direction, but many directions which change over time.

The fact that the direction of acceleration in rotating is toward the center can not be a topic of discussion (see any textbook in curcular motion ..but if you do not convinced I am ready to prove it)

But your post 98 asserts that you cannot neglect gravitational field, your position in that post is that there is no gravitational field so small that it could be neglected. Therefore in the examples I gave the accelerometers and rocket would work as I stated, even according to your analysis. Or do you still assert that the rocket would not burn and if so on what justification?

The two cases are different
The first one is Idealization Like that of( Ideal gas )something that does not exist but we can know its properties by theory (this is like accelerometer of zero mass to study
the properties of zero-gravitational region theoreticaly
The second one is Approximation something that does exist but you think that it can be teated like ideal case( this is when you want neglect the difference between zero-gravitational region and the region affected by the gravitational field of the accelerometer
In the first discussion I accepted the idealization and discuss with you the conduct of bodies in zero-gravitational field
In the second discussion I refused the approxmation when I said the fuel will not burn (in zero-gravitational field in idealization that the rocket is of zero-mass) but it will burn in the region affected by gravitational field of the rocket
in the true case

 

[Dale]

The fact that the direction of acceleration in rotating is toward the center can not be a topic of discussion (see any textbook in curcular motion ..but if you do not convinced I am ready to prove it)

Which direction is "toward the center"? Acceleration is a vector and vectors have a magnitude and a direction, so which direction is "towards the center"?

In the second discussion I refused the approxmation when I said the fuel will not burn ... but it will burn in the region affected by gravitational field of the rocket
in the true case

There, now was that so hard to finally admit?

And in the "true case" of the accelerometers/observers example of post 49?

 

[Mueiz]

There, now was that so hard to finally admit?

And in the "true case" of the accelerometers/observers example of post 49?

But your mistake is that you want to use this fact as an approximation to zero-gravitation region which is wrong as I stated in my last post which you quoted but removing the most important point to give impression that I agree with you

 

[Dale]

But your mistake is that you want to use this fact as an approximation to zero-gravitation region which is wrong as I stated in my last post which you quoted but removing the most important point to give impression that I agree with you

I have no interest (nor did I ever have any interest) in zero-gravitation as you define zero-gravitation. I am only interested in the physics of the proposed scenario which, according to you, is a non-zero-gravitation region.

Now, stop being evasive and answer the question. Do you agree with behavior of the accelerometers as described by me in post 49 in the "true case"?

 

[Mueiz]

Now, stop being evasive and answer the question. Do you agree with behavior of the accelerometers as described by me in post 49 in the "true case"?

1) real accelerometer in gravitational field ...yes as given from GR
2)ideal accelerometer in gravitational field...Yes as given from GR
3)ideal accelerometer in zero-gravitation ...No all frames are inertial from the symmetry
4)real accelerometer in zero-gravitatation ...impossible because real accelerometer has a mass
your post in 49 is No 3If you want now
to add the word '' true case'' to your post 49 it will be No 4
when I used the word '' true case " I meant 1 in my post but you took it to mean No 3

 

[Dale]

1) real accelerometer in gravitational field ...yes as given from GR
2)ideal accelerometer in gravitational field...Yes as given from GR

Then all of the rest of my explanation follows and none of your objections follow.

3)ideal accelerometer in zero-gravitation ...No all frames are inertial from the symmetry
4)real accelerometer in zero-gravitatation ...impossible because real accelerometer has a mass
your post in 49 is No 3If you want now
to add the word '' true case'' to your post 49 it will be No 4
when I used the word '' true case " I meant 1 in my post but you took it to mean No 3

As I said before I am not interested nor have I ever been interested in zero-gravitation by your definition. I did not use the term in post 49 nor in any of my earlier posts in this thread. And I certainly would never use it in the sense of a universe completely devoid of any matter or energy as you intend it. I would mean it in the usual sense of "far from any significant gravitational sources", or more precisely "no measurable deviation from a flat metric". So my post 49 most certainly did not imply either 3) or 4) and only a biased reading of what I wrote would have led you to believe that I did intend either of those based on your idea of zero-gravitation.

You are the only one in this whole thread who has been talking about zero-gravitation in the sense you mean it.

 

[Mueiz]

Then all of the rest of my explanation follows and none of your objections follow.

As I said before I am not interested nor have I ever been interested in zero-gravitation by your definition. I did not use the term in post 49 nor in any of my earlier posts in this thread. And I certainly would never use it in the sense of a universe completely devoid of any matter or energy as you intend it. I would mean it in the usual sense of "far from any significant gravitational sources", or more precisely "no measurable deviation from a flat metric". So my post 49 most certainly did not imply either 3) or 4) and only a biased reading of what I wrote would have led you to believe that I did intend either of those based on your idea of zero-gravitation.

You are the only one in this whole thread who has been talking about zero-gravitation in the sense you mean it.

I said in early post (#22)
''My aim from this discussion is to show that the rotating disc is not the correct way to introduce the idea of nonEuicldean geometry relationship with acceleration and not more
a correct way is to use Equivalence Principle as I stated in the beginning of this discussion.
Another problem with the disc experiment is that it contradict one of the basis of General Relativity in that if it done in a region of zero gravitational field there should be preferred frame of reference in which the geometry is Euicldean and all other rotating relative to it (static to themselves of course) frames should seek other geometry ...''
but many users try to discuss with me in a wrong way by talking about the known effect of acceleration in gravitational field
I want to add that Ideal cases does not mean Uselessness in practical ...for example ideal cases is used to calculate the constants of Equations derived to meat Real cases
I want someone who think that Rotating Disc Experiment is true try to resolve the paradox of my post #109

 

[yuiop]

Also seconded By Altabeh
Ok ...then you see that the direction of rotating of the disc has something to do with the geometry in spite of the symmetry ... but if symmetry is not enough ...I have another thing which is that ..the direction of rotating of the disc can not affect the direction of acceleration which you claim to be the cause of nonEuclidean geometry

Dalespam has you on a technicality here. The angular acceleration does not point towards the centre, but points out from the centre perpendicular to the plane of the disc. If the disc was a bicycle wheel for example rotating about its axle, then the acceleration vector is parallel to the axle. If the plane of the wheel is horizontal then the acceleration vector points up or down depending upon whether the angular acceleration is clockwise or anti-clockwise. The acceleration vectors are therefore different for different directions of rotation. In the diagram below the angular acceleration vector is to the left as indicated by the thumb. If the wheel was rotating in the opposite direction the angular acceleration vector would be to the right.

Now I have a clear and simple method to show the Mistake of Rotating Disc Experiment
suppose we have two similar rotating discs that rotate around the same center at the same RPM in opposite direction
The inertial observer is in the ground see both discs rotate
Now according to Einstein and many others..the geometry of each of the two discs is nonEuclidean...but ..they must take the same geometry because their relation to the inertial observer is quite the same.
But they are accelerating relative to each other
Acceleration and the same Geometry ?
Can DaleSpam , Altabeh ,Yoiup and Others resolve this simple paradox

Well the geometries are different. The ratio of the radius to the circumference is the same for both discs with the same magnitude of acceleration but opposite directions (non-euclidean to an observer at rest with the disc). The geometries are different, because the one way speed of light in a given direction (using a single clock on the rim) is different on the two discs.

 

[Dale]

I said in early post (#22)
''...done in a region of zero gravitational field ''

Exactly, your quote makes my point. You were the only one discussing in this manner, nobody else. Particularly, nobody else was discussing zero-gravity in terms of a universe devoid of all matter and energy as you intended. Yuiop and I were clear about our meanings.

 

[Mueiz]

Exactly, your quote makes my point. You were the only one discussing in this manner, nobody else. Particularly, nobody else was discussing zero-gravity in terms of a universe devoid of all matter and energy as you intended. Yuiop and I were clear about our meanings.

zero-gravitation is one of the assumptions used by Einstein in his rotating disc experiment which i want to show that it is wrong
see The meaning of Relativity page 34 or The Evolution of Physics
to know that he did not mention the gravitational field at all and he made his calculation in the absence of gravitation
Is it useless if someone find that a famous method used in a famous textbook written by a famous scientist is wrong ?

 

[Mueiz]

angular acceleration does not point towards the centre, but points out from the centre perpendicular to the plane of the disc. If the disc was a bicycle wheel for example rotating about its axle, then the acceleration vector is parallel to the axle. If the plane of the wheel is horizontal then the acceleration vector points up or down depending upon whether the angular acceleration is clockwise or anti-clockwise. The acceleration vectors are therefore different for different directions of rotation. In the diagram below the angular acceleration vector is to the left as indicated by the thumb. If the wheel was rotating in the opposite direction the angular acceleration vector would be to the right. .

The angular acceleration of something that rotates in constant angular velocity (as the case of rotating disc of Einstein or the two discs of my paradox) is zero
see any textbook in circular motion
But this is a deviation from our topic which is acceleration found in rotating disc in constant angular velocity which is always radial pointing toward the center regardless of the direction of rotation (this is just a simple fact need not be even seen in wikipeadia)
Is it difficult to say that '' I am wrong in this argument ''as I did in #72

Well the geometries are different. The ratio of the radius to the circumference is the same for both discs with the same magnitude of acceleration but opposite directions (non-euclidean to an observer at rest with the disc). The geometries are different, because the one way speed of light in a given direction (using a single clock on the rim) is different on the two discs.

this discussion is based on your incorrect statement about the direction of acceleration in rotating discs which i discussed above

 

[Dale]

zero-gravitation is one of the assumptions used by Einstein in his rotating disc experiment which i want to show that it is wrong
see The meaning of Relativity page 34 or The Evolution of Physics
to know that he did not mention the gravitational field at all and he made his calculation in the absence of gravitation
Is it useless if someone find that a famous method used in a famous textbook written by a famous scientist is wrong ?

Show me anywhere that Einstein defined zero-gravitation in the way you do: a universe without any mass or energy regardless of how little or how far away.

Otherwise I am sure that he used the term in the usual manner of being in an asymptotically-flat spacetime far from any significant gravitational sources. With that usual meaning he is correct.

 

[Altabeh]

Show me anywhere that Einstein defined zero-gravitation in the way you do: a universe without any mass or energy regardless of how little or how far away. Otherwise I am sure that he used the term in the usual manner of being far from any significant gravitational sources. With that usual meaning he is correct.

I strongly suggest Mueiz to define very clearly what zero-gravitation means in

1- his own idea,
2- from the angle Einstein was looking at the problem.

I think this clears up anything if done!

AB

 

[Dale]

The angular acceleration of something that rotates in constant angular velocity (as the case of rotating disc of Einstein or the two discs of my paradox) is zero
see any textbook in circular motion

You are correct here. We are dealing with a case of 0 angular acceleration.

But this is a deviation from our topic which is acceleration found in rotating disc in constant angular velocity which is always radial pointing toward the center regardless of the direction of rotation

Look Mueiz. Write the position of the counter-rotating observers, they have opposite phase. Take the first and second derivatives, and you will find that the velocity and acceleration vectors are also opposite phase. Write the angular momentum, those are opposite. Etc.

Your paradox is not a paradox, it is a mistake. The counter rotating observers don't have the same relationship to the inertial observer, they have opposite relationships in every vector or pseudo-vector quantity I can think of.

 

[Dale]

I strongly suggest Mueiz to define very clearly what zero-gravitation means in

1- his own idea,
2- from the angle Einstein was looking at the problem.

I think this clears up anything if done!

I agree completely.

 

[Mueiz]

I strongly suggest Mueiz to define very clearly what zero-gravitation means in

1- his own idea,
2- from the angle Einstein was looking at the problem.

I think this clears up anything if done!

AB

This is also a reply to DaleSpam
zero-gravitational is simply a region in which the curvature tensor equal zero

I agree with Einstein and all physicist in this
What about the paradox?

 

[Altabeh]

This is also a reply to DaleSpam
zero-gravitational is simply a region in which the curvature tensor equal zero

I agree with Einstein and all physicist in this
What about the paradox?

Really? But I can show a spacetime in which there is no gravitational field but yet the spacetime has non-zero curvature! Just take $$R_{ab}=\frac{1}{2}g_{ab}R}$$ but yet $$G_{ab}=-\kappa T_{ab}=0.$$ What now?

AB

 

[Mueiz]

Really? But I can show a spacetime in which there is no gravitational field but yet the spacetime has non-zero curvature! Just take $$R_{ab}=\frac{1}{2}g_{ab}R}$$ but yet $$G_{ab}=-\kappa T_{ab}=0.$$ What now?

AB

[tex]G_{ab} this Einstein tensor... it equal zero in empty (no-matter) spsce of couse
I mean full Riemannian Curvature tensor of four order ([tex]R_{abcd} which is a physical characteristic of gravitational field in GR

 

[Dale]

This is also a reply to DaleSpam
zero-gravitational is simply a region in which the curvature tensor equal zero

I agree with Einstein and all physicist in this

No. You have already stated in post 98 that you mean no mass or energy in the entire universe regardless of how small or distant. That is not what anyone else means.

 

[Mueiz]

You can lead your horse to the river but you can not make him drink
thank you

 

[Altabeh]

$$G_{ab}$$ this Einstein tensor... it equal zero in empty (no-matter) spsce of couse
I mean full Riemannian Curvature tensor of four order ([tex]R_{abcd} which is a physical characteristic of gravitational field in GR

Okay then! You can see that in the example above Ricci tensor doesn't vanish so nor does Riemannian! But yet the spacetime is free of any gravitational (and non-gravitational) field!

You can lead your horse to the river but you can not make him drink
thank you

Remember that the first thing you need when disproving a giant like GR is to fully understand its elements one-by-one! Then you can bring your horse down here and we can see if he is going to drink water or not!

AB

 

[yuiop]

The angular acceleration of something that rotates in constant angular velocity (as the case of rotating disc of Einstein or the two discs of my paradox) is zero
see any textbook in circular motion
But this is a deviation from our topic which is acceleration found in rotating disc in constant angular velocity which is always radial pointing toward the center regardless of the direction of rotation (this is just a simple fact need not be even seen in wikipeadia)
Is it difficult to say that '' I am wrong in this argument ''as I did in #72

OK, I assumed from your earlier posts and from Dalespam's responses, that you were talking about the angular acceleration of the disc as a whole, but I am now clear you are talking about centripetal acceleration of a single point or observer on the rim of the disc. In the context of a single observer we can only talk about local geometry rather than the geometry of the disc as a whole and in this case the local geometry is Minkowskian and independent of the direction of rotation. If considering the disc as a whole, the angular velocity or angular momentum is a vector orthogonal to the disc plane even for the constant angular velocity case. Disc rotating in opposite directions will have opposite pointing angular velocity vectors and the geomotries of the two discs will not be identical as I pointed out earlier due to isotropic one way speed of light seen in the rotating frames, which is essentially what is detected by Sagnac devices. Gyroscopes attached to the discs will also reveal the rotation and differences.

One HUGE problem with your approach is that your thought experiments require us to use massless discs in a completely empty universe with no rulers, clocks, gyroscopes, light, observers, particles etc, which makes the whole exercise impossible and completely useless.

You also seem to making the bizarre claim that all Einstein's thought experiments are invalid, because Einstein has not taken into account the mass of the apparatus used in his SR equivalence principle thought experiments. It is not difficult to show that the time dilation due to gravitational mass for a low mass disc/ light clock/ rocket is negligible when compared to the time dilation due to relative velocity at velocities near the speed of light. Yes, there will always be a remnant but minuscule error due to ignoring gravitational time dilation due to the mass of clocks and rulers, but this error is vanishing at extreme gamma factors and is nit picking in the extreme.