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

[Goldbeetle]

I understand the principle of equivalence (e.g. thought-experiments with lifts etc), but how come that from it one can arrive at the result that gravity acts curving the space? Where can I find a step-by-step reasoning illustrating this link?

 

[Mentz114]

I'm not sure if the equivalence principle can be used alone to develop GR, or similar geometrical space-time theories of gravity. But it does mean that such a theory is required if the EP is to be automatically incorporated. Free-fall is unique to the gravitational field so an energy based (dynamic) theory would not do it.

After Einstein had his crucial insight that a freely-falling body experiences no force, it still took nearly ten years to find the formulation that gave GR.

This link might help

http://www.lightandmatter.com/html_books/genrel/ch01/ch01.html#Section1.5

 

[Mueiz]

If you want to know how Equivalence Principle(EP) leads to General Relativity(GR) you must first consider two important points related to Special Relativity(SP).
The first is that SR made an end to absolute character of time which used to keep it apart from the continuum composed of the three space dimensions and made it possible to establish the new concept of space-time in which the laws of S R are not more than the geometrical properties of flat four-dimensional continuum.
The second point is that SR is not complete ,it can not be applied in gravitational field and accelerated frames.
Now if we understand EPwell ,we will be able to know that the fact that gravitation is equivalent to acceleration implies that all small regions of space-time are similar because by using suitable frame SR is applicable, the only characteristic property of space-time in a point is the difference between the suitable frame in the point and the suitable frame in neighboring points. This properties of space-time is similar to the properties of Curved Four-dimensional Space which was studied many years before by mathematicians maybe without thinking that it could be applied in reality one day.
This similarity between Curved 4-d space and Space-time led Einstien to the idea that gravitation is the geometrical properties of space-time.
His second step would be to relate the geometrical properties of space-time to the material properties ,in this step he used other facts and assumptions for which this is not the appropriate
place.

 

[yuiop]

I understand the principle of equivalence (e.g. thought-experiments with lifts etc), but how come that from it one can arrive at the result that gravity acts curving the space? Where can I find a step-by-step reasoning illustrating this link?

I think a crucial part of his intuitive reasoning is illustrated in the his analysis of the rotating disc. He noticed that an observer riding on the edge of the disc would measure the circumference to be greater than 2*pi*r because the observer's rulers would be length contracted. He concludes than in the pseudo-gravitational field of the the rotating disc the geometry of space and time can no longer be described by Euclidean principles that assume flat space. Since his equivalence principle assumes that a real gravitational field and a pseudo gravitational field due to artificial acceleration are indistinguishable locally, it seems reasonable to make the connection that a real gravitational field will distort space and time into a non-Euclidean (i.e. non-flat) form. There is of course a lot more to GR than that, but I think that was one step in the reasoning process on the way.

 

[Passionflower]

I think a crucial part of his intuitive reasoning is illustrated in the his analysis of the rotating disc. He noticed that an observer riding on the edge of the disc would measure the circumference to be greater than 2*pi*r because the observer's rulers would be length contracted.

Would he measure the radius using radar distance or ruler distance?

 

[aashay]

I think a crucial part of his intuitive reasoning is illustrated in the his analysis of the rotating disc. He noticed that an observer riding on the edge of the disc would measure the circumference to be greater than 2*pi*r because the observer's rulers would be length contracted. He concludes than in the pseudo-gravitational field of the the rotating disc the geometry of space and time can no longer be described by Euclidean principles that assume flat space. Since his equivalence principle assumes that a real gravitational field and a pseudo gravitational field due to artificial acceleration are indistinguishable locally, it seems reasonable to make the connection that a real gravitational field will distort space and time into a non-Euclidean (i.e. non-flat) form. There is of course a lot more to GR than that, but I think that was one step in the reasoning process on the way.

What you are arguing here cannot be explained even by GR.Actually, there is a different theory called the Einstein-Cartan theory which explains this by extending GR to include spin angular momentum.

 

[atyy]

http://www.einstein-online.info/spotlights/geometry_force

The equivalence principle does not determine GR completely. Other theories such as Newtonian gravity and Nordstrom's second theory also obey an equivalence principle, and can be recast as geometry. The recasting of Newton was done by Cartan, while the recasting of Nordstrom was done by Einstein and Fokker. Nordstrom's theory was actually the first relativistic theory of gravitation, and closely studied by Einstein, who had himself critiqued Nordstrom's first, unsuccessful theory in a way useful to Nordstrom.

 

[yuiop]

I think a crucial part of his intuitive reasoning is illustrated in the his analysis of the rotating disc. He noticed that an observer riding on the edge of the disc would measure the circumference to be greater than 2*pi*r because the observer's rulers would be length contracted.

Would he measure the radius using radar distance or ruler distance?

Einstein originally described this thought experiment in terms of the radius being measured by ruler distance which gives the result of the circumference (C) to radius (R) ratio as C/R = 2*pi/sqrt(1-v^2/c^2).

If the the thought experiment is done using ruler distance for the circumference and radar distance for the radius then the ratio would be C/R = 2*pi/(1-v^2/c^2) which contrasts even more with the Euclidean flat space expectation.

The circumference can also be measured using radar distance (and a series of mirrors around the circumference) and this would give the same result as the ruler measurement of the circumference.

 

[Mueiz]

Very Few people know that the thought experiment of the rotating disc although used by Einstien himself to show the idea that acceleration changes the geometry of space is obviously incorrect .
The experiment compares the observation of two observers one of them see that the disc is static and the geometry of the disc is Euclidean,to the other observer the disc is rotating, but I can prove that the geometry of the disc will be Euclidean also.
Let us suppose that the observer to whom the disc is rotating draw a shape in his static ground in such a way that it coincides with the circumference of the rotating disc
it is obvious that the length of the two circles is the same and the length of the radios is
not affected as all of us agree so the geometry is the same.
The effect of acceleration on the geometry of space-time is true but can not be proved in this case of rotating disc.The principle of Equivalence is helpful in this purpose.

 

[yuiop]

Very Few people know that the thought experiment of the rotating disc although used by Einstien himself to show the idea that acceleration changes the geometry of space is obviously incorrect .
The experiment compares the observation of two observers one of them see that the disc is static and the geometry of the disc is Euclidean,...

This is not correct. The observer that sees the disc as static is obviously riding on the disc and whether they use radar or ruler measurements they will not obtain a Euclidean result of 2*pi for the ratio circumference length to the radius length.

 

[Mueiz]

I also want someone who believe in the rotating disc experiment to tell me which of the two observers will notice non-Eculidean geometry if the experiment is done in a region of zero gravitational field (the question is also valid in gravitational field but I want to simplify the problem )

 

[Mueiz]

This is not correct. The observer that sees the disc as static is obviously riding on the disc and whether they use radar or ruler measurements they will not obtain a Euclidean result of 2*pi for the ratio circumference length to the radius length.

please read what I said again an tell me what is incorrect . I did not speak about radar and ruler at all and when I say the observer to whom the disc is static is the same as saying riding ion the disc but that is not the point of discussion we are discussing the case of the other observer

 

[Dale]

Let us suppose that the observer to whom the disc is rotating draw a shape in his static ground in such a way that it coincides with the circumference of the rotating disc it is obvious that the length of the two circles is the same and the length of the radios is not affected as all of us agree so the geometry is the same.

It is certainly not obvious. Why would you think that the length would be the same?

 

[Passionflower]

Einstein originally described this thought experiment in terms of the radius being measured by ruler distance which gives the result of the circumference (C) to radius (R) ratio as C/R = 2*pi/sqrt(1-v^2/c^2).

If the the thought experiment is done using ruler distance for the circumference and radar distance for the radius then the ratio would be C/R = 2*pi/(1-v^2/c^2) which contrasts even more with the Euclidean flat space expectation.

The circumference can also be measured using radar distance (and a series of mirrors around the circumference) and this would give the same result as the ruler measurement of the circumference.

Clearly something either has to be squeezed or stretched for a rotating disk wrt a stationary disk.

Say we have a unit circle with radius r=1 when at rest. Now we rotate it with an angular velocity w, what do you say the radar distance of r is?

 

[Mueiz]

Dalespam ...they are the same because the two circle coincide. we suppose that the observer outside the disc should draw the static circle to coincide with the rotating one ,is it not obvious that if two objects coincide they must have the same length and all other quantities .
I know that we must be careful when using the word (obvious) in relativity and modern physics . but you cannot by any relativistic or non- relativistic analysis say that if two object have the same shape being in the same place in the same order can give different measurement to any quantity for the same observer.

 

[Dale]

Dalespam ...they are the same because the two circle coincide. ... is it not obvious that if two objects coincide they must have the same length and all other quantities .

Not only is it not obvious, it is wrong. Have you never heard of length contraction? Even for a single object (which coincides with itself) different frames will give different lengths.

 

[Mueiz]

they are the same because the two circle coincide. we suppose that the observer outside the disc should draw the static circle to coincide with the rotating one ,is it not obvious that if two objects coincide they must have the same length and all other quantities .
I know that we must be careful when using the word (obvious) in relativity and modern physics . but you cannot by any relativistic or non- relativistic analysis to say that if two object have the same shape being in the same place in the same order can give different measurement to any quantity for the same observer.

 

[Mueiz]

length contraction is two observers to the same rod ...our discussion is one observer to two coinciding objects one static the other rotating. even in length contraction zero length is absolute

 

[Mueiz]

you can also see a good book by Ohanian called ''Einstein's Mistakes'' for more details

 

[yuiop]

Very Few people know that the thought experiment of the rotating disc although used by Einstien himself to show the idea that acceleration changes the geometry of space is obviously incorrect .
The experiment compares the observation of two observers one of them see that the disc is static and the geometry of the disc is Euclidean,to the other observer the disc is rotating, but I can prove that the geometry of the disc will be Euclidean also.

Here you wish to discuss the point of view of both observers. You are stating that the observer rotating with the disc and the non rotating observer off the disc, both see the geometry as Euclidean.

This is not correct. The observer that sees the disc as static is obviously riding on the disc and whether they use radar or ruler measurements they will not obtain a Euclidean result of 2*pi for the ratio circumference length to the radius length.

please read what I said again an tell me what is incorrect . I did not speak about radar and ruler at all and when I say the observer to whom the disc is static is the same as saying riding ion the disc but that is not the point of discussion we are discussing the case of the other observer

Now you only want to discuss the point of view of the non rotating observer. For what it is worth I agree that the inertial observer that is not on the disc, will measure the circumference of the rotating disc to be the same as the circumference of a non rotating circle drawn on the floor with the same radius. To the inertial non rotating observer the geometry is Euclidean everywhere, however, to the rotating observer on the disc the geometry is non Euclidean as I stated in a previous post.

I also want someone who believe in the rotating disc experiment to tell me which of the two observers will notice non-Eculidean geometry if the experiment is done in a region of zero gravitational field (the question is also valid in gravitational field but I want to simplify the problem )

As stated above, it is the rotating observer riding on the disc that sees the geometry as non Euclidean and this applies in a zero gravitational field far from any significant gravitational sources.

Here is a numerical example that might help. Assume we have a drawn on the floor in a non rotating frame with a radius of 10 metres. Inside this circle is a disk rotating with a tangential rim velocity of 0.866c (gamma factor =2), with the same radius. If lit from above, by a even light source, the disc casts a shadow that superimposes neatly on the circle.

Measurements in the rotating frame:

The circumference of the circle according to a non rotating observer is 2*pi*10 = 62.8318 meters.

The circumference of the rotating disc can be measured by placing a mark on the rim. The velocity of the mark can be measured (0.866c) and then the non rotating observer can time how long it takes for the mark to complete one full circle and return to the starting point. It takes 72.552 seconds for the mark to travel on full circle. The observer then calculates the distance traveled by the mark to be 72.552*0.866 = 62.8318 meters which is the same as the circumference of the non rotating circle.

So nothing unusual or non Euclidean as far as the non rotating observer is concerned.

Measurements in the non-rotating frame:

The circumference of the disc rim according to the rotating observer is 2*2*pi*10 = 125.6637 meters whether he uses rulers or radar measurements. The circumference of the disc Cd has the relationship Cd = 4*pi*r and so is non Euclidean.

Now the observer on the rotating disc can measure the circumference of the non rotating circle on the floor by placing a mark on the circle. He sees that the mark has a relative velocity of 0.866c. He time how long it takes for the mark to return to its starting point. It takes 36.2760 seconds. The rotating observer calculates that the circumference of the non rotating circle is 36.2760*0.866 = 31.4160 meters so as far as the rotating observer is concerned, the circumference of the non rotating circle is non Euclidean too but this time the circumference of the circle Cc has the relationship Cc = pi*r. They would also notice that the circumference of the non rotating circle is gamma^2 shorter than the circumference of the rotating disc, even though the disc rim and the circle are superimposed on each other.

This might seem a little odd. The rotating observer on the disc has the right to consider himself stationary and he considers marks on the circle to be going past him at 0.866c so he would expect the measured length of the circle to be length contracted by a factor of gamma and the length contraction of gamma^2 might be a bit of a surprise. This is because everything is a little weird in the rotating frame. The speed of light is different in different directions, clocks can not be synchronised all the way around the rim using the Einstein clock synchronisation procedure. Light does not travel in straight lines. If a dweller on the disc decided that the disc was not in fact stationary but rotating in real terms, it would be like a Copernican revolution for the disc dwellers and everything would look so much simpler. All of sudden they would see that the circumference of the disc is the same as the circumference of the circle it is superimposed on, that the speed of light is the same in all directions, that light travels in straight lines and that clocks can be synchronised all the way around the rim. However, the high priests of the rotating disc community would probably put him under house arrest for such subversive suggestions.

 

[Dale]

our discussion is one observer to two coinciding objects one static the other rotating

No, only you are discussing that. yuiop, who brought up the rotating observer, specifically mentioned "an observer riding on the edge of the disk" from the very first post on this topic in this thread.

Only non-inertial observers such as the one on the edge of the disk see the equivalent gravitational fields, so it is essential to include such an observer and not restrict yourself to inertial observers. Such a limited discussion as you propose would not even be relevant to the OP's question on the equivalence principle.

 

[Mueiz]

Ok yuiob and others.. I want to say something about length contraction and its application in this case of rotating disc and let you then agree or disagree that the high priests are right this time if they put Copernic or even Einstein under house arrest when he offers such suggestions.
Length contraction means to put your ruler beside a static object and to mark its two ends in the same time according to your clock and then do the same when the object is in motion relative to you .you will surely find a difference.
In the case of the disc it is the same only that the shape is circle and I suggest that we should use circular ruler ...say we have many different-sizes rulers and we measure any circle circumference by seeing which of them coincide with the object .this is the definition of the process of measurement whether the object is static or moving .
What I said -maybe not clearly- and seems to you as a contradiction is that the question is not whether the measurement of the two observers are the same or not but whether both observers see Eucildean geometry or not
And I show that we agree that one of them does see Eucildean geometry .Then if we look to the other- be the static or the moving not important- if he use the circular ruler which is static to him then the ruler has the same radios of the disc and the same circumference then the ruler is static to him as the ruler used by the other observer is static to him then why suppose that one of them should have nonEuicldean geometry
The mistake in your calculations is that you simplify the problem too much when you assume that the radios of the disc will not be affected by rotation, rotation is not like linear perpendicular motion if special relativity is applicable in this way then the space must be Euicldean .This contradiction appears in your results that you have special relativity validity in the same time with nonEuicldean geometry . this is the mistake you and Einstein made .
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 ...

 

[Dale]

they are the same because the two circle coincide.

The mistake in your calculations is that you simplify the problem too much

What hypocrisy!

Then if we look to the other ... if he use the circular ruler which is static to him then the ruler has the same radios of the disc and the same circumference then the ruler is static to him as the ruler used by the other observer is static to him then why suppose that one of them should have nonEuicldean geometry

Because the rotating observer cannot construct such a ruler while under rotation, and if he were to build such a ruler in an inertial condition and then spin it up to be at rest with himself then it would necessarily be materially stressed and no longer correctly measure distances in any frame.

 

[Mueiz]

What hypocrisy!

Because the rotating observer cannot construct such a ruler while under rotation, and if he were to build such a ruler in an inertial condition and then spin it up to be at rest with himself then it would necessarily be materially stressed and no longer correctly measure distances in any frame.

Then the rotating observer must take you with his left hand and Einstein with his right hand
to the rotating disc to show you that the geometry therein is Euicldean

 

[Dale]

Nice imagery. Completely irrelevant and unconvincing, but very well said.

If a round ruler such as you describe is impossible to build then the Euclidean measurement you suggest is also not possible.

 

[bcrowell]

Mueiz's analysis of the rotating disk is incorrect. For this example, the relevant notion of curvature of space is given by a purely spatial metric determined by radar measurements carried out by comoving observers. Here is my derivation of the spatial metric: http://www.lightandmatter.com/html_b...tml#Section3.4 (subsection 3.4.4)

The following FAQ entry and its references may be helpful.

FAQ: How is Ehrenfest's paradox resolved?

As described in [Einstein 1916], the relativistic rotating disk was an example that was influential in leading Einstein to describe gravity in terms of curved spacetime. Einstein writes:

"In a space which is free of gravitational fields we introduce a Galilean system of reference K (x,y,z,t), and also a system of coordinates K' (x',y',z',t') in uniform rotation relative to K. Let the origins of both systems, as well as their axes of Z, permanently coincide. We shall show that for a space-time measurement in the system K' the above definition of the physical meaning of lengths and times cannot be maintained. For reasons of symmetry it is clear that a circle around the origin in the X, Y plane of K may at the same time be regarded as a circle in the X', Y' plane of K'. We suppose that the circumference and diameter of this circle have been measured with a unit measure infinitely small compared with the radius, and that we have the quotient of the two results. If this experiment were performed with a measuring-rod at rest relative to the Galilean system K, the quotient would be π. With a measuring-rod at rest relative to K', the quotient would be greater than π. This is readily understood if we envisage the whole process of measuring from the stationary'' system K, and take into consideration that the measuring-rod applied to the periphery undergoes a Lorentzian contraction, while the one applied along the radius does not."

Einstein's friend Paul Ehrenfest posed the following paradox [Ehrenfest 1909]. Suppose that observer L, in the lab frame, measures the radius of a rigid disk to be r when the disk is at rest, and r' when the disk is spinning. L can also measure the corresponding circumferences C and C'. (When we speak of "radius" and "circumference," we are making use of the fact that the disk is rigid, so that after it spins up it is still a circle. It doesn't fly apart or contort itself like a potato chip.) Because L is in an inertial frame, the spatial geometry does not appear non-Euclidean according to measurements carried out with his meter sticks, and therefore the Euclidean relations C=2πr and C'=2πr' both hold. The disk is rigid, so it doesn't stretch centrifugally, and the only reason for r to differ from r' would be a Lorentz contraction. But the radial lines are perpendicular to their own motion, so they have no length contraction, r=r', implying C=C'. The outer edge of the disk, however, is everywhere tangent to its own direction of motion, so it is Lorentz contracted, and therefore C' is less than C.

The resolution of the paradox is that it rests on the incorrect assumption that a rigid disk can be made to rotate. If a perfectly rigid disk was initially not rotating, one would have to distort it in order to set it into rotation, because once it was rotating its outer edge would no longer have a length equal to 2π times its radius. Therefore if the disk is perfectly rigid, it can never be set into rotation.

Thorough modern analyses are available,[Grøn 1975,Dieks 2009] and in particular it is not controversial that, as claimed in [Einstein 1916], C/r is measured to be *greater* than 2π by an observer in the rotating frame.

A common source of confusion in discussions of Ehrenfest's paradox is the role of the rigid meter-sticks, since it is not clear whether sufficiently rigid meter-sticks can exist, or how to verify that they have remained rigid. This confusion can be avoided simply by replacing the meter-stick measurements with radar measurements.

In connection with these discussions, one often hears about the concept of a Born-rigid object, meaning an object that is subject to prearranged external forces in such a way that observers moving with the object find local, internal radar distances between points on the object to remain constant.[Born 1909] It is kinematically impossible to impart an angular acceleration to a Born-rigid disk,[Grøn 1975] and therefore it is also impossible to do so for any plane figure that encloses a finite area, since it would enclose a disk. The reason for this is that in order to maintain Born-rigidity, the torques would have to be applied simultaneously at all points on the perimeter of the area, but Einstein synchronization (i.e., synchronization by radar) is not transitive in a rotating frame; that is, if A is synchronized with B, and B with C, then C will not be synchronized with A if the triangle ABC encloses a nonzero area and is rotating. (This does not make it impossible to manipulate the rigid meter-sticks as described in [Einstein 1916], since they can be one-dimensional, and therefore need not enclose any area.)

A. Einstein, "The foundation of the general theory of relativity," Annalen der Physik, 49 (1916) 769; translation by Perret and Jeffery available in an appendix to the book at http://www.lightandmatter.com/genrel/ (PDF version)

P. Ehrenfest, Gleichförmige Rotation starrer Körper und Relativitätstheorie, Z. Phys. 10 (1909) 918, http://en.wikisource.org/wiki/Uniform_Rotation_of_Rigid_Bodies_and_the_Theory_of_Relativity

Born, "The theory of the rigid electron in relativistic kinematics." Annalen der Physik 30 (1909) 1–56.

Ø. Grøn, Relativistic description of a rotating disk, Am. J. Phys. 43 (1975) 869

Dieks, "Space, Time, and Coordinates in a Rotating World," in Rizzi and Ruggiero, ed., Relativity in Rotating Frames: Relativistic Physics in Rotating Reference Frames, 2009, http://www.phys.uu.nl/igg/dieks/rotation.pdf

 

[yuiop]

Note: In the discussions below, all measurements are done far from any gravitational source. The lab frame or non-rotating observer/frame will be used consistently to mean the inertial frame in which the observers do not feel or measure any proper acceleration and any Sagnac type devices will indicate there is no rotation in the lab frame. The rotating or disc frame/observer will be used consistently to refer to an observer that is at rest with the disc that rotating relative to the lab frame. Observers at rest in the rotating frame (except at the very centre) will feel and measure proper acceleration and Sagnac type devices will indicate they are rotating. There is no ambiguity between rotating and non-rotating and whether something is rotating or not is not observer dependent. Rotation is not a point of view, it is absolute.

Length contraction means to put your ruler beside a static object and to mark its two ends in the same time according to your clock and then do the same when the object is in motion relative to you .you will surely find a difference.

There is a difference. Take two short rulers that have the same length when alongside and at rest wrt each other. Place one ruler on the rotating disc rim and compare it a ruler that remains at rest in the lab frame. The lab observer will measure the ruler on the disc to be shorter than the ruler in the lab frame by the gamma factor. It will take more of these length contracted rulers in the rotating frame to extend all the way around the rim of the disc. In the earlier example, assuming rulers of length 1 meter, the observer in the rotating frame requires over 125 of these rulers layed end to end in the rotating frame to extend all the way around the disc, while the observer in the lab frame only needs about 63 identical rulers layed end to end, to go all the way around the non rotating circle in the lab frame. We can prove these rulers have identical proper length by getting the observer in the lab frame to time how long it takes light to travel from one end of his ruler to a mirror on the other end and back again. He tosses the ruler to the observer on the rotating disc and while the ruler is at rest in the rotating frame, he carries out a similar radar measurement and confirms he gets the same time. He can then lay this calibrated ruler alongside anyone of his rulers at rest in the disc frame and confirm that it is the same length.

In the case of the disc it is the same only that the shape is circle and I suggest that we should use circular ruler ...say we have many different-sizes rulers and we measure any circle circumference by seeing which of them coincide with the object .this is the definition of the process of measurement whether the object is static or moving .

This can be proved wrong. If I take a non rotating circular ruler with a circumference of about 125 meters when at rest in the lab frame, it will have a radius of about 20 meters. If we spin the circular ruler up to a tangential velocity of 0.866c and take care not stretch the ruler, then it will shrink to a radius of about 10 meters in the lab frame. The lab observer will now measure the circumference of the circular ruler to be about 63 meters. An observer that that remains on the circular ruler will initially measure the circumference of the ruler to be about 125 meters before it is spun up, and after it has been spun up he will still measure the total circumference of the circular ruler to be 125 meters, because the short rulers he uses to measure the circumference will have length contracted to the same extent as the circular ring so he sees no change in total circumference. He will however agree with the lab observer that the radius is half what is was before it was spun up. The observer on the spinning circular ruler claims the proper length of the circular ruler is 125 meters and the radius is 10 meters. (Non Euclidean) This spinning ring will now be superimposed with a non spinning ring with a radius of 10 meters and a circumference of 63 meters, so a spinning ring and a non spinning ring of the same radius do not have the same circumferential proper length.

What I said -maybe not clearly- and seems to you as a contradiction is that the question is not whether the measurement of the two observers are the same or not but whether both observers see Eucildean geometry or not

I have clearly stated in several previous threads that the lab observer sees Euclidean geometry and the rotating observer on the disc does not see Euclidean geometry. I do not understand why you think there is any ambiguity here.

And I show that we agree that one of them does see Eucildean geometry .Then if we look to the other- be the static or the moving not important- if he use the circular ruler which is static to him then the ruler has the same radios of the disc and the same circumference then the ruler is static to him as the ruler used by the other observer is static to him then why suppose that one of them should have nonEuicldean geometry.

See above.

The mistake in your calculations is that you simplify the problem too much when you assume that the radios of the disc will not be affected by rotation, rotation is not like linear perpendicular motion if special relativity is applicable in this way then the space must be Euicldean .This contradiction appears in your results that you have special relativity validity in the same time with nonEuicldean geometry . this is the mistake you and Einstein made .

It is you who is simplifying too much. You have not attempted to calculate any measurements, but just making guesses on casual intuition without given the problem as much thought as it requires. The radius is affected by rotation. While the ruler length of the rotating disc radius is the same as that of the non rotating circle in the lab, the radar length length of the radius is shorter by a factor of the gamma in the disc frame. The radar length and the ruler length of the disc radius are different in the disc frame. This is the equivalence principle at work! In a real gravitational field the radar distance and ruler distance differ over extended distances. This is a sure sign of curved non-Euclidean geometry.

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.

See above.

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 ...

Not quite sure what you are getting at here. As I mentioned earlier, I agree there is preferred frame in a zero gravitational field where geometry is Euclidean and all other frames rotating relative to it are non-Euclidean and in this sense, rotation is absolute. This is not a contradiction of SR which states there is no preferred frame as far as motion in a straight line is concerned, but this does not apply to angular motion where a rotating observer feels proper acceleration while an inertial non rotating observer does not. As far as being a contradiction to GR I think you wrong. The metric for a Schwarzschild black hole (non rotating) is fundementally different from the metric of a Kerr black hole (rotating). The geometry around a Kerr black hole is not just simply the point of view of an observer orbiting around a Schwarzschild black hole.

 

[bcrowell]

you can also see a good book by Ohanian called ''Einstein's Mistakes'' for more details

The book gives Ohanian's opinion on p. 232: "The basic error of Einstein's and his contemporaries was that they failed to recognize that the use of accelerated rulers is an improper way to measure length." Ohanian's opinion is not universally shared. For example, see Rindler's Relativity: Special, General, and Cosmological, p. 199, which states "The metric of the lattice...represents a curved 3-space..."

Neither Ohanian nor Rindler is wrong, and they are not really contradicting each other. Ohanian is simply stating a distaste for a certain method of measurement, but Rindler doesn't share Ohanian's distaste.

 

[Mueiz]

Firstly;if we want to apply length contraction in the case of rotating disc this will not be simply by gama factors except if we do that locally(in very small region) because any point in the disc has a different relative velocity ,and the result of the summation of such local calculation is not equal to that of linear motion as your calculation assume.
secondly;There is no preferred frame of reference in the absence of matter and gravitational field according to general relativity all frames are Euclidean in this case you can make sure of this if you apply Einstein Field Equation ... the metric is absolute because both stress-energy tensor(matter) and Riemann curvature (gravitational field) equal zero and so cannot be affected by frames of reference(if there any thing else that can change the metric please tell me soon and I will change may mind) ..then what is the property of one frame that could make it different...if so the claimed results of rotating disc experiment contradict this by assuming that one of the frames should gain Euclidean geometry ,all the other not.
This is the sum of what I said in this discussion

 

[Mueiz]

The book gives Ohanian's opinion on p. 232: "The basic error of Einstein's and his contemporaries was that they failed to recognize that the use of accelerated rulers is an improper way to measure length." Ohanian's opinion is not universally shared. For example, see Rindler's Relativity: Special, General, and Cosmological, p. 199, which states "The metric of the lattice...represents a curved 3-space..."

Neither Ohanian nor Rindler is wrong, and they are not really contradicting each other. Ohanian is simply stating a distaste for a certain method of measurement, but Rindler doesn't share Ohanian's distaste.

Thank you I acknowledge that I did not read the book just someone told me that it contain details of what i am trying to show i will seach for it...this is just a mistake like that of Einstein-Yuiob rotating disc

 

[Mentz114]

secondly;There is no preferred frame of reference in the absence of matter and gravitational field according to general relativity all frames are Euclidean in this case you can make sure of this if you apply Einstein Field Equation ... the metric is absolute because both stress-energy tensor(matter) and Riemann curvature (gravitational field) equal zero and so cannot be affected by frames of reference(if there any thing else that can change the metric please tell me soon and I will change may mind) ..then what is the property of one frame that could make it different...if so the claimed results of rotating disc experiment contradict this by assuming that one of the frames should gain Euclidean geometry ,all the other not.
This is the sum of what I said in this discussion

(my emphasis)

I have not been following this thread closely but the statement I've quoted seems to indicate that you are unaware of certain facts. Obviously GR is irrelevant since it is agreed that the spacetime is flat, so the Minkowski metric applies globally. But different observers ( defined by their worldlines) will preceive the metric to be something different. For example, constantly accelerating observers perceive the Rindler 'metric', which is obtained from the Minkowski metric by a coordinate transformation. Similarly the spacetime perceived by rotating observers is obtained by a coordinate transformation of the Minkowske metric.

So, even in flat spacetime, some observers will see a non-Euclidean spatial geometry despite being in a globally flat spacetime.

In answer to the assertion I have bolded above - the perceived metric can indeed change without invoking GR.

 

[Mueiz]

(my emphasis)

I have not been following this thread closely but the statement I've quoted seems to indicate that you are unaware of certain facts. Obviously GR is irrelevant since it is agreed that the spacetime is flat, so the Minkowski metric applies globally. But different observers ( defined by their worldlines) will preceive the metric to be something different. For example, constantly accelerating observers perceive the Rindler 'metric', which is obtained from the Minkowski metric by a coordinate transformation. Similarly the spacetime perceived by rotating observers is obtained by a coordinate transformation of the Minkowske metric.

So, even in flat spacetime, some observers will see a non-Euclidean spatial geometry despite being in spa globally flat spacetime.


In answer to the assertion I have bolded above - the perceived metric can indeed change without invoking GR.

Ok in empty space which is free of matter and gravitational field the word '' accelerating observer '' is meaningless and misleading .
acceleration relative to what?
The change in metric perceived by accelerating observer obtained from coordinate transformation is true when there is gravitational field because in this case there is a preferred frame which is the frame of free-falling objects according to the Principle of Equivalence in which the metric is Euclidean ., then we can speak of accelerated observer relative to that special frame.
There is a difference between absolute flatness of space-time that empty and with zero-gravitational field and the flatness of space-time with gravitational field , the second appear only in free-falling frames.
I also want to mention that flatness always means Euclidean both are detected by the metric and
nothing else
It is also incorrect to say that GR is irrelevant anywhere because you know that special relativity is just a special case of GR..in addition I used GR to convince those who claim the existing of frame-dependent nonEuclidean geometry in empty space of zero gravitational field that all frames should be the same.

 

[Mentz114]

Ok in empty space which is free of matter and gravitational field the word '' accelerating observer '' is meaningless and misleading .
acceleration relative to what?

Acceleration is not relative in the sense that uniform motion is relative. It is not uniform motion, and other observers will agree that the accelerated observer is not in uniform motion. It makes perfect sense to talk about an accelerating frame or observer.

The change in metric perceived by accelerating observer obtained from coordinate transformation is true when there is gravitational field because in this case there is a preferred frame which is the frame of free-falling objects according to the Principle of Equivalence in which the metric is Euclidean ., then we can speak of accelerated observer relative to that special frame.

This objection does not hold because we can always define an inertial frame which at some time is instantaneously at rest wrt to the accelerated observer. We can then define the acceleration wrt to this frame should we so wish.

 

[Mueiz]

I have clearly stated in several previous threads that the lab observer sees Euclidean geometry and the rotating observer on the disc does not see Euclidean geometry. I do not understand why you think there is any ambiguity here.

It is not You who misunderstood me in this point

It is you who is simplifying too much. You have not attempted to calculate any measurements, but just making guesses on casual intuition without given the problem as much thought as it requires.

I do not need to make calculation because I want only two show that the two cases of measurement are similar concerning the nature of geometry.
I did not use any guess or intuition ,what I use is the definition of the process of measurement and the logic of similarity .

Not quite sure what you are getting at here. As I mentioned earlier, I agree there is preferred frame in a zero gravitational field where geometry is Euclidean and all other frames rotating relative to it are non-Euclidean and in this sense, rotation is absolute. This is not a contradiction of SR which states there is no preferred frame as far as motion in a straight line is concerned, but this does not apply to angular motion where a rotating observer feels proper acceleration while an inertial non rotating observer does not. As far as being a contradiction to GR I think you wrong. The metric for a Schwarzschild black hole (non rotating) is fundementally different from the metric of a Kerr black hole (rotating). The geometry around a Kerr black hole is not just simply the point of view of an observer orbiting around a Schwarzschild black hole.

Inside black hole and around it the gravitational field is not zero and there is a preferred frame relative to which we can speak of rotation and its affects but that is not the case where there is no gravitational field.

 

[Mueiz]

Acceleration is not relative in the sense that uniform motion is relative. It is not uniform motion, and other observers will agree that the accelerated observer is not in uniform motion. It makes perfect sense to talk about an accelerating frame or observer.

Acceleration is Not Relative in gravitational field because of the special inertial frame
Acceleration is Relative in the absence of gravitational field
If not tell me Which of the frames in the rotating disc experiment is not accelerated and why?

This objection does not hold because we can always define an inertial frame which at some time is instantaneously at rest wrt to the accelerated observer. We can then define the acceleration wrt to this frame should we so wish.

It is dangerous by the way in relativity to use this instantaneously-at-rest method . if you for example try to use it to find the transformation between two inertial frames it will lead you to Galilean transformation rather than Lorentz transformation, motion is not a collection of instants of rest.

 

[Mentz114]

Acceleration is Not Relative in gravitational field because of the special inertial frame
Acceleration is Relative in the absence of gravitational field

Nonsense. If you want to persist with your ridiculous misconceptions, feel free. Clearly you are not open to rational argument.

 

[Mueiz]

Nonsense. If you want to persist with your ridiculous misconceptions, feel free. Clearly you are not open to rational argument.

Ridiculous misconceptions is the first step towards understanding but quoting of what somebody else has said without thinking is ignorance

 

[Mentz114]

Acceleration is Not Relative in gravitational field because of the special inertial frame

This is wrong and has nothing to do with case in hand.

Acceleration is Relative in the absence of gravitational field

This is also wrong.

You keep repeating this nonsense with no attempt to justify your stance. Stop mentioning gravity - it is not relevant here.

Can you give any rational explanation of what you mean by 'acceleration is relative' or point to any source which agrees with you ?

 

[Dale]

Mueiz, I get the impression that you have read a decent amount about GR, but have probably very rarely ever sat down to actually go through the math and work a problem. Is this correct?

If so, are you familiar with the covariant derivative and geodesics?

If so then you understand that whether or not a particular worldline is a geodesic is a frame invariant fact. Even in the absence of a gravitational soucre there exist coordinate systems where the worldline of a particle at rest is not a geodesic. Such coordinate systems are non-inertial and are distinguishable from inertial coordinate systems. The laws of physics do not take their textbook form when expressed in those coordinates and thus they are absolutely experimentally distinguishable from inertial coordinate systems.

Would you be able to cite where you got this idea that all coordinate systems in SR are equivalent? I am sure that if you go back and look for it you will find that it only says that all inertial frames are equivalent. You may also want to study Rindler coordinates for a prototypical example of a non-inertial coordinate system.

 

[Mueiz]

OK Mentz when I said that''Acceleration is Relative in the absence of gravitational field '' I gave the justification in the question in the next line which mean that we have not any way to measure acceleration in such case because to do so you need inertial frame which need another frame to be detected'' ...this is the problem ,running round circles
but is it necessary for one who want to be open to rational argument to believe in absolute acceleration only because a great physicist tells him this
As for source you can see The Foundation of General relativity by Einstein under the title The Need for an Extension of the Postulate of Relativity and see alsoThe Evolution of Physics by A Einstein just before discussing the experiment of rotating disc

 

[Mentz114]

OK Mentz when I said that''Acceleration is Relative in the absence of gravitational field '' I gave the justification in the question in the next line which mean that we have not any way to measure acceleration in such case because to do so you need inertial frame which need another frame to be detected'' ...this is the problem ,running round circles
but is it necessary for one who want to be open to rational argument to believe in absolute acceleration only because a great physicist tells him this
As for source you can see The Evolution of Physics by A Einstein just before discussing the experiment of rotating disc

(my bold)

Acceleration can be measured by an accelerometer. It is the rate of change of velocity and so can also be detected using the doppler effect, for instance. If we define an accelerated worldline as any worldline that is not a geodesic, then it has an obvious and absolute meaning which all observers will agree on. This contradicts what I take to be the meaning of the word 'relative'.

 

[Mueiz]

(my bold)

Acceleration can be measured by an accelerometer. It is the rate of change of velocity and so can also be detected using the doppler effect, for instance. If we define an accelerated worldline as any worldline that is not a geodesic, then it has an obvious and absolute meaning which all observers will agree on. This contradicts what I take to be the meaning of the word 'relative'.

All these methods of measuring acceleration depend on the effects of acceleration but this effects will appear if the acceleration is measured relative to inertial frames ..but if all frames in zero-gravitational region are inertial as I am trying to show this effects will not appear..
it is not fair to assume what you are going to prove

 

[Mentz114]

All these methods of measuring acceleration depend on the effects of acceleration

As they must, or they wouldn't be measuring acceleration.

but this effects will appear if the acceleration is measured relative to inertial frames

An observer in a state of acceleration can easily detect this without reference to any external beacon or reference frame.

but if all frames in zero-gravitational region are inertial as I am trying to show this effects will not appear

You won't be able to show this because it is obviously untrue.

it is not fair to assume what you are going to prove

Agreed.

I have other things to do so I leave you to prove your case for now.

 

[Mueiz]

Mueiz, I get the impression that you have read a decent amount about GR, but have probably very rarely ever sat down to actually go through the math and work a problem. Is this correct?

If so, are you familiar with the covariant derivative and geodesics?

If so then you understand that whether or not a particular worldline is a geodesic is a frame invariant fact. Even in the absence of a gravitational soucre there exist coordinate systems where the worldline of a particle at rest is not a geodesic. Such coordinate systems are non-inertial and are distinguishable from inertial coordinate systems. The laws of physics do not take their textbook form when expressed in those coordinates and thus they are absolutely experimentally distinguishable from inertial coordinate systems.

Would you be able to cite where you got this idea that all coordinate systems in SR are equivalent? I am sure that if you go back and look for it you will find that it only says that all inertial frames are equivalent. You may also want to study Rindler coordinates for a prototypical example of a non-inertial coordinate system.

your impression is true regarding complicated applications of the theory but i think i know the mathematics needed to understand basic topics such as Einstein Field Equation geodesic ,schrawchild solution and so on for i believe in Feynman saying that '' to those who do not know mathematics it is difficult to get across a real feeling as to the deepest beauty of nature" but i would prefer to discuss conceptual aspects rather than mathmatics
Einstein in his book The Foundation of the General Theory of Relativity under the title The Need for an Extension of the Postulate of Relativity talked about the problem of (absolute or relative) acceleration and he find the solution to it when there is a gravitational field but he forget that problem when dealing with rotating disc thought experiment when done in zero-gravitational field region
i will leave other points for other time

 

[bcrowell]

Firstly;if we want to apply length contraction in the case of rotating disc this will not be simply by gama factors except if we do that locally(in very small region) because any point in the disc has a different relative velocity ,and the result of the summation of such local calculation is not equal to that of linear motion as your calculation assume.

In #26, I pointed you to several calculations of this result. Have you read them?

secondly;There is no preferred frame of reference in the absence of matter and gravitational field according to general relativity all frames are Euclidean in this case you can make sure of this if you apply Einstein Field Equation ... the metric is absolute because both stress-energy tensor(matter) and Riemann curvature (gravitational field) equal zero and so cannot be affected by frames of reference(if there any thing else that can change the metric please tell me soon and I will change may mind) ..then what is the property of one frame that could make it different...if so the claimed results of rotating disc experiment contradict this by assuming that one of the frames should gain Euclidean geometry ,all the other not.
This is the sum of what I said in this discussion

You are correct that if the Riemann tensor is zero in one set of four-dimensional spacetime coordinates, then it's zero in any other set of four-dimensional coordinates derived from the first by a change of coordinates. However, you apparently haven't read the derivations linked to from #26, which derive a three-dimensional spatial metric from the original four-dimensional spacetime metric.

You are also oversimplifying by using the word "Euclidean" as if it applies to both 3+1 dimensions and 3 spatial dimensions. The 3+1 geometry is Minkowski. The 3-space geometry is Euclidean in the nonrotating frame and non-Euclidean in the rotating frame.

 

[Mentz114]

I tried working out the acceleration experienced by a rotating observer in Minkowski spacetime like this. The velocity vector is

<br /> V_{\mu}=\left[ \begin{array}{cccc}<br /> -\sqrt{{r}^{2}\,{w}^{2}+1} &amp; -r\,w\,sin\left( t\,w\right) &amp; r\,w\,cos\left( t\,w\right) &amp; 0 \end{array} \right]<br />

and \eta^{ab}V_aV_b=-1.

From which we get the derivatives ( covariant differentiation is just partial differentiation here )

<br /> \partial_{\mu}V_{\nu}=<br /> \left[ \begin{array}{cccc}<br /> 0 &amp; -r\,{w}^{2}\,cos\left( t\,w\right) &amp; -r\,{w}^{2}\,sin\left( t\,w\right) &amp; 0\\\<br /> -\frac{r\,\left( \frac{d}{d\,x}\,r\right) \,{w}^{2}}{\sqrt{{r}^{2}\,{w}^{2}+1}} &amp; -\left( \frac{d}{d\,x}\,r\right) \,w\,sin\left( t\,w\right) &amp; \left( \frac{d}{d\,x}\,r\right) \,w\,cos\left( t\,w\right) &amp; 0\\\<br /> -\frac{r\,\left( \frac{d}{d\,y}\,r\right) \,{w}^{2}}{\sqrt{{r}^{2}\,{w}^{2}+1}} &amp; -\left( \frac{d}{d\,y}\,r\right) \,w\,sin\left( t\,w\right) &amp; \left( \frac{d}{d\,y}\,r\right) \,w\,cos\left( t\,w\right) &amp; 0\\\<br /> 0 &amp; 0 &amp; 0 &amp; 0<br /> \end{array} \right]<br />

We want this in the V direction,

<br /> \partial_{\mu}V_{\nu}V^{\nu}=<br /> \left[ \begin{array}{cccc}<br /> 0 &amp; 2\,r\,\left( \frac{d}{d\,x}\,r\right) \,{w}^{2} &amp; 2\,r\,\left( \frac{d}{d\,y}\,r\right) \,{w}^{2} &amp; 0 \end{array} \right]<br />

which seems right, with acceleration in the x and y directions. Substituting r^2=x^2+y^2 into the derivatives gives

<br /> \partial_{\mu}V_{\nu}V^{\nu}=<br /> \left[ \begin{array}{cccc}<br /> 0 &amp; \frac{2\,r\,{w}^{2}\,x}{\sqrt{{y}^{2}+{x}^{2}}} &amp; \frac{2\,r\,{w}^{2}\,y}{\sqrt{{y}^{2}+{x}^{2}}} &amp; 0 \end{array} \right]<br />

and the magnitude is 2w^2r.

Where did that factor of 2 come from ?

 

[Mueiz]

Ok Mentz ,bcrowell and other i nead time to reply and #26 is not found yet but now I have a Hard Paradox Concerning Absolute Acceleration and Rotating Disc Trick
suppose we want to do the rotating disc experiment in far empty space using very large disc
and that we have three observer.
First observer who is outside the disc in the inertial frame sees geometry as Euclidean .
Second observer in the edge of the disc ...then he is in uniform motion relative to the first observer(because the disc can be made large enough) so his frame is also inertial and he sees geometry as Euclidean .Third observer is in an arbitrary position between the edge and the center then this observer is static relative to the second observer and so sees geometry as Euclidean .
Now the third observer is in accelerated motion relative to the first observer but both of them see geometry as Euclidean
this is the relative acceleration of empty nongravitational space which has nothing to do with the geometry

 

[Roshanjha14]

I didnt undrstand the weak equi. Principle.

 

[Dale]

your impression is true regarding complicated applications of the theory but i think i know the mathematics needed to understand basic topics such as Einstein Field Equation geodesic ,schrawchild solution and so on for i believe in Feynman saying that '' to those who do not know mathematics it is difficult to get across a real feeling as to the deepest beauty of nature" but i would prefer to discuss conceptual aspects rather than mathmatics

I understand your desire to stick with a conceptual treatment and I will try to do that, but if a conceptual treatment fails to comminicate the point then I may need to use some math. The main participants on this thread (including myself) all have experience actually working through the math. I mention that only to establish our credentials, we are not speaking from a surface "conceptual-only" knowledge, but from full experience working the math.

For the following I will deal only with flat spacetime with no significant gravitational sources and attempt to explain the difference between an inertial and a non-inertial coordinate system. First some mathematical terms and their relationship to basic measuring devices:

1) Worldline - a point particle has a specific position at each moment in time. In spacetime this traces out a 1D curve called a worldline.

2) Proper time - the length along a worldline representing a massive particle. Proper time is measured by clocks and is a frame-invariant quantity.

3) Proper acceleration - the covariant derivative of the particle's worldline. The magnitude of the proper acceleration is a frame-invariant quantity and is measured by accelerometers.

4) Geodesic - a worldline with a 0 covariant derivative. This implies that the proper acceleration is also 0 and any attached accelerometers read 0, and the particle is inertial.

5) At rest - the time derivative of the worldline's coordinate position is 0. This is a frame-varying concept, an object which is at rest in one coordinate system may not be at rest in another.

With those concepts in place it is fairly straightforward to show that an inertial coordinate system is different from a non-inertial coordinate system.

Suppose we have three observers each with a clock and an accelerometer. Observer A's accelerometer reads 0, observer B's accelerometer always has a constant magnitude of acceleration with a changing direction always pointing towards A, and observer C's accelerometer always has a constant magnitude and direction of acceleration pointing away from A. A is inertial while B and C are not.

Now, there exists a coordinate system where A is at rest, B has a helical worldline, and C has a hyperbolic worldline. In this coordinate system any object at rest will have a geodesic worldline, and the covariant derivative is everywhere equal to the ordinary derivative. This coordinate system is inertial.

There exists another coordinate system where B is at rest, A has a helical worldline, and C has a weird shaped worldline that I won't attempt to describe verbally. In this coordinate system an object at rest will not have a geodesic worldline, and the covariant derivative is not equal to the ordinary derivative. This coordinate system is rotating. It is absolutely distinguishable from an inertial coordinate system mathematically by use of the covariant derivative and experimentally by use of accelerometers.

Is this clear now?

 

[Mueiz]

I understand your desire...

Suppose we have three observers each with a clock and an accelerometer. Observer A's accelerometer reads 0, observer B's accelerometer always has a constant magnitude of acceleration with a changing direction always pointing towards A, and observer C's accelerometer always has a constant magnitude and direction of acceleration pointing away from A. A is inertial while B and C are not.
...

Is this clear now?

I really benefited very much from this good presentation on this issue... butI have an objection .
It is a fact that a thought experiment should not be rejected because of practical difficulties or even impossibility but if the experiment is theoretically impossible then I am not forced to accept its results.
In your experiment you suppose that one observer can have zero acceleration and another one have non-zero acceleration in the absence of gravitational field ...but i refuse this( so do Mach in his The Science of Mechanics and Einstein in his The Principle of GR )
If gravitational field is present your experiment is theoretically possible because we can apply a force -say electric- upon an object and make it deviate from the inertial frame which is the free falling frame as stated by Equivalent Principle ... but if there is no gravitation what is the frame you need to deviate from to be accepted in Accelerated Objects Club .