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

[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<br /> I mean full Riemannian Curvature tensor of four order ($$R_{abcd} which is a physical characteristic of gravitational field in GR$$[/tex]

 

[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 ($$R_{abcd} which is a physical characteristic of gravitational field in GR$$

$$<br /> Okay then! You can see that in the example above Ricci tensor doesn't vanish so nor does Riemannian! But yet the spacetime is free of any gravitational (and non-gravitational) field!<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> You can lead your horse to the river but you can not make him drink<br /> thank you </div> </div> </blockquote><br /> Remember that the first thing you need when disproving a giant like GR is to fully understand its elements one-by-one! Then you can bring your horse down here and we can see if he is going to drink water or not!<br /> <br /> AB$$

 

[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.

 

[Mueiz]

I found this article by M.Strauss in International Iournal of Theoretical Physics vol.11 No 2 (1974) pp.107-123(rotating frames in special relativity)
in this article the writer proves mathematicaly the fact that the geometry in a rotating disc is Eucildean .View attachment fulltext.pdf