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The Equivalence of Acceleration and Gravitation in General Relativity

February 21 2008 at 7:18 PM
Stanley16 

 


The Equivalence of Acceleration and Gravitation in General Relativity



So far, you've demonstrated nothing and debunked nothing.
You just keep saying the 'accelerated & non-accelerated' frames of reference are not equivalent' over and over without really demonstrating anything.

So let me give you a head start to enlighten you!

When a system of reference undergoes uniform acceleration, distant masses in the front of it must appear to accelerate towards it at the same uniform acceleration. By contrast, distant masses behind this same system of reference must appear to accelerate away from it at the same uniform acceleration.

Since there is no privileged system of reference in general relativity, an observer at rest with respect to this accelerated system is entitled to consider his system at rest and that the distant masses are undergoing the acceleration in question. And hence, he must conclude that the entire "space-time territory in question is under the sway of a gravitational field".

Look more closely at the profile of this equivalent gravitational field. All distant masses in the forward direction are accelerating towards the system of reference at the same uniform rate; while all distant masses in the backward direction are accelerating away from this same system at the same uniform rate. And this means that the source of this gravitational field must be located somewhere beyond the accelerating-away masses.

I hope, now, you're enlightened, and can see very clearly that acceleration and gravitation are indeed equivalent in every respect and exactly as Einstein concludes in his 1916-groundbreaking paper.



 
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AAF

Re: The Equivalence of Acceleration and Gravitation in General Relativity

February 22 2008, 5:44 PM 



Not bad, not bad at all; the above head start of yours is informing; but two or three important pieces of information are missing or left 'wittingly or unwittingly' out of it; please, allow me, now, to go there and retrieve them!

The main method employed in those hypothetical measurements ought to be and must be based upon the principle of Doppler effect.

Measurements obtained by the Doppler method, not only include the instantaneous acceleration, but also include the cumulative result we call 'velocity' due to this operating acceleration over time.

It follows, therefore, that when a system of reference undergoes acceleration (uniform or otherwise), the apparent velocity must be the same for all distant masses. That is in the case of actual acceleration on the part of the reference system in question.

By contrast, in the case of free-falling masses under the 'sway of a uniform gravitational field', the velocity profile as obtained by the Doppler method is quite different. The direction of this velocity is the same, but its magnitude must vary with position and depend entirely on how long the distant mass in question has been traversing this specific gravitational field. For instance, in your 'head start' above, the speed of all accelerating-away masses must be, by the physical necessity of free fall, greater by a large margin than the speed of all 'accelerating-toward-the-observer' masses.

In short, the velocity profile of masses in free fall is vastly different from the velocity profile of truly accelerated system of reference. The accelerated system and the non-accelerated system, therefore, are not equivalent. Einstein got it all wrong.



 
 
Stanley16

Re: The Equivalence of Acceleration and Gravitation in General Relativity

February 23 2008, 3:58 PM 



I don't buy the velocity-profile hypothesis you invented above.

First of all, we have only two systems in this experiment: the system labeled K and the system labeled K'. The observer at K' has only two ways to interpret the measured amount of acceleration. It's either K' is accelerating or is at rest and the other system is under the sway of a uniform gravitational field. These two interpretations are exactly equivalent just as Einstein concluded. And so there is nothing for your 'velocity-profile hypothesis' but to "collapse in deepest humiliation".

Even if several systems of reference are given, this velocity-profile thing is still useless.

Remember that this type of gravitational fields appears only the moment we assume it. In other words, it wasn't there in the past to work the way you want it to work. Because of this, it's possible to make the time scale for taking measurements as short as desired to show only the given rate of acceleration and the equivalent rate of free fall. This method has been so successful in establishing the validity of special relativity within the framework of general relativity. It's so successful in the treatment of multiple systems of reference. And it's successful in the case of this thought experiment.

Thus I can say that your trouble here is the same as that of the fabled astronomer who was so busy looking at the wide sky only to have his leg broken by falling nose first into a small pit!


 
 
Stanley16

Re: The Equivalence of Acceleration and Gravitation in General Relativity

February 23 2008, 11:25 PM 



A reply to bobs' objection:
http://www.network54.com/Forum/304711/message/1203802438/last-1203802438/Re-+The+Second+Thought+Experiment+of+General+Relativity


No way!

In Einstein's thought experiment, the equivalent gravitational field is assumed to be uniform and its strength does not vary with the square of distance. And that means the space-time territory under consideration is very far from the gravitational center. For example, the gravitational field of the sun, at about one light year away, is essentially uniform and varies only with extremely minute amount with the square of distance. Your friend, here, knows very well this fact; that is why he went ahead and invented his useless 'velocity-profile' hypothesis.


 
 
M.F. Yagan

Re: The Equivalence of Acceleration and Gravitation in General Relativity

February 24 2008, 3:51 AM 

How can an observer be at rest relative to the accelerating frame since he himself is accelerating? imagine what happins to the passinger in a car coming to a sudden stop. An accelerating object is not at rest relative to anything he is in absolute motion

 
 
bob s

Yes way!

February 24 2008, 9:53 AM 

“No way!
In Einstein's thought experiment, the equivalent gravitational field is assumed to be uniform and its strength does not vary with the square of distance.”
“Your friend, here, knows very well this fact; that is why he went ahead and invented his useless 'velocity-profile' hypothesis.“
“ And that means the space-time territory under consideration is very far from the gravitational center.” emphasis added

Stanley 16,
First off, my mistake, my name should be bob s not “bobs”; I shudder to think there might be two of me.

Second, Einstein et al are free to think and assume as they see fit but thought experiments are a dime a dozen, so to speak. I do agree that a gravitational field is uniform radially from its source but I do not agree that it is uniform lineally. So here is what you should do to verify your friend; first leap off of a two foot wall then leap off of a twenty foot wall and tell me that your speed is the same for both leaps at the point of impact and I will become a believer.

Third, are you saying my friend Newton (invented) the “velocity-profile”? I’ll have you know that Newton spent his years experimenting not riding a bike. Einstein spent the last 30 years of his life trying to validate GR and SR...so, how did that turn out?

Next, could that “very far” be somewhere over the rainbow?

Finally, my original objection stands. Thank you.

bob s

 
 
Stanley16

Re: The Equivalence of Acceleration and Gravitation in General Relativity

February 24 2008, 5:26 PM 


To M.F. Yagan:

It's well established that, in every case of free fall, observers feel no acceleration at all.
In fact, this was Einstein's original insight, which eventually led him to the development of the general theory of relativity:

"There are several ways to formulate the Principle of Equivalence, but one of the simplest is Einstein's original insight: he suddenly realized, while sitting in his office in Bern, Switzerland, in 1907, that if he were to fall freely in a gravitational field (think of a sky diver before she opens her parachute, or an unfortunate elevator if its cable breaks), he would be unable to feel his own weight. Einstein later recounted that this realization was the "happiest moment in his life", for he understood that this idea was the key to how to extend the Special Theory of Relativity to include the effect of gravitation. We are used to seeing astronauts in free fall as their spacecraft circles the Earth these days, but we should appreciate that in 1907 this was a rather remarkable insight".

http://csep10.phys.utk.edu/astr162/lect/cosmology/equivalence.html


 
 
Stanley16

Re: The Equivalence of Acceleration and Gravitation in General Relativity

February 24 2008, 6:04 PM 

To bob s:

Leaping off walls would not verify the uniformity of gravitational fields.

How to explain this simple observation of Einstein to you and to that AAF?

Okay!

Let us consider the space-time region of the inner solar system, i.e. from the center of the sun to the orbit of the planet Mars.

Within this region, the strength of the solar gravitational field varies considerably with each distance of one astronomical unit; why? That is because this space-time region is very close to the source of this gravitational field, which in this case is the sun.

Now, let's consider a space-time region of the same size as that of the space-time region of the inner solar system, but located at one light year way from the sun.

Within this second region, the strength of the solar gravitational field is practically uniform and shows very little variation with each distance of one astronomical unit; why? That is because this space-time region is very distant from the source of this gravitational field, which in this case is still the sun.



 
 
bob s

No way!

February 25 2008, 10:20 AM 


In re: "To bob s:
Leaping off walls would not verify the uniformity of gravitational fields."

No way Stanley 16. No way am I going to respond to your misdirecting the original experiment you posted. You described two bodies in "free flight" and concluded two bodies in "free fall"; you "assumed" a gravitational field. Admit that point and we can move on from there. Thank you.

bob s

 
 
AAF

Re: The Equivalence of Acceleration and Gravitation in General Relativity

February 25 2008, 7:48 PM 



Buy it or not, that velocity profile for defining gravitational fields is nature's most devastating blow to Einstein's General Relativity and Einstein's new physics. His theory of gravitation, quite simply, has been born dead; and no amount of sophistry in the world can revive it or make it viable or reasonable or suitable for doing physics.

In any case, the above defensive argument would not work for the following reasons:

{1} Einstein, in his gedanken experiment, can reduce the number of available co-ordinate systems as he wishes; but he cannot reduce or limit or control the number of falling distant masses. He knew this; thus he wrote in his 1916 Paper: "the above-mentioned relation of freely movable masses to K' may be interpreted equally well in the following way. The system of reference K' is unaccelerated, but the space-time territory in question is under the sway of a gravitational field, which generates the accelerated motion of the bodies relatively to K'."

{2} The idea of 'It-Just-Has-Been-Born-Now ' Field of Gravity is absurd and in direct contradiction with the whole notion of gravitation and the notion of physical reality in general. You can't build viable physical theories upon absurdities of this kind.

{3} No shortening in the time of measurements can make the velocity component and the velocity profile disappear or go away. The changes of the coming light always form the same integrated package regardless of how the measurements are performed.


 
 
Stanley16

Re: The Equivalence of Acceleration and Gravitation in General Relativity

February 26 2008, 5:31 PM 


I didn't misdirect it, bob.
This is exactly how Einstein describes this thought experiment in the 1916-founding paper:

"In addition to this weighty argument from the theory of knowledge, there is a well-known physical fact which favours an extension of the theory of relativity. Let K be a Galilean system of reference, i.e. a system relatively to which (at least in the four-dimensional region under consideration) a mass, sufficiently distant from other masses, is moving with uniform motion in a straight line. Let K' be a second system of reference which is moving relatively to K in uniformly accelerated translation. Then, relatively to K', a mass sufficiently distant from other masses would have an accelerated motion such that its acceleration and direction of acceleration are independent of the material composition and physical state of the mass. Does this permit an observer at rest relatively to K' to infer that he is on a "really" accelerated system of reference? The answer is in the negative; for the above-mentioned relation of freely movable masses to K' may be interpreted equally well in the following way. The system of reference K' is unaccelerated, but the space-time territory in question is under the sway of a gravitational field, which generates the accelerated motion of the bodies relatively to K'."


 
 
Stanley16

Re: The Equivalence of Acceleration and Gravitation in General Relativity

February 26 2008, 5:57 PM 



Please, notice that the Equivalence principle, in general relativity, can be formulated in various ways. Choosing only one of them as the sole representative of them all can be very restrictive and misleading. I urge you to spare few minutes and read this excellent article: http://www.mth.uct.ac.za/omei/gr/chap5/node5.html


 
 
bob s

Yes, you did misdirect the subject.

February 27 2008, 3:57 AM 

In the post I responded to, you used two bodies in motion on the X axis separated fore to aft and then in another post you switched to a gravitational field as it relates to a radius/distance from the sun. As to, http://www.mth.uct.ac.za/omei/gr/chap5/node5.html


Figure 5.4: The lift experiments

* Case 1: The lift is placed in a rocket ship in a part of the universe far removed from gravitating bodies. The rocket is accelerated forward with a constant acceleration g relative to an inertial observer. The observer releases a body from rest and sees it fall to the floor with acceleration g.
* Case 2: The rocket motor is switched off so that the lift undergoes uniform motion relative to the inertial observer. A released body is found to remain at rest relative to the observer in the lift.
* Case 3: The lift is next placed on the surface of the earth, whose rotational and orbital motions are ignored. A released body is found to fall to the floor with acceleration g.
* Case 4: Finally, the lift is placed in an evacuated lift shaft and allowed to fall freely towards the center of the earth. A released body is found to remain at rest relative to the observer.


Case *1: is false, if case 1: were true then case *2: would be false but it is well known that case *2: is true so therefore case *1: is false. Acceleration in “free space” does not “create” a gravitational effect.
Case *3: is true
Case *4: is true iff the motion is toward a gravitational body.

Gravity is “point source” the “point” being the entirety of the gravitational body. A gravitational field is "outside" of the body. A body falling toward the center of the gravitational body would come to rest at that center. If that body were placed inside of a container (lift) the body would assume a position in the center of the container (lift). I.o.w. the body would "fall" to the center. The body may appear to be in a weightless state but any attempts to move it would be difficult at best. Case *4: is not clear as to whether or not the body is falling within the earth or toward the earth from above.

As to Einstein’s great insight, well, the issue of falling bodies has been dealt with many times over the millennia before he came upon the scene; he offered nothing new. I know he worked at a patent office but he did not have the patent on the ability to think. You may not be aware of it but (he) falsified SR 3 months after it was published. Otherwise, thank you.

bob s

 
 
Stanley16

Re: The Equivalence of Acceleration and Gravitation in General Relativity

February 27 2008, 6:39 PM 


The principle of Equivalence applies to three different situations:

1. Two reference systems of uniformly moving system & uniformly accelerating system: this is the situation in Einstein's thought experiment under discussion. Here, the Equivalence principle states that acceleration is equivalent to gravitation. Since an accelerating system of reference imparts the same acceleration to all distant bodies regardless of material composition and physical state. In the same way, a uniform gravitational field imparts the same acceleration to all falling bodies regardless of material composition and physical state. It follows, therefore, that acceleration and gravitation are equivalent.

2. Two reference systems of uniformly accelerating system and system at rest in a gravitational field: this situation is employed in deriving the gravitational red shift and the bending of light path in general relativity. Since it's clear that the two reference systems, in this case, are exactly equivalent and indistinguishable by any experiment.

3. Two reference systems of uniformly moving system and free-falling system under the sway of a uniform gravitational field: this situation is used in establishing the validity of special relativity within the framework of general relativity. Since it's obvious that free fall and inertial motion in a straight line are equivalent and cannot be distinguished by any experiment.



 
 
Stanley16

Re: The Equivalence of Acceleration and Gravitation in General Relativity

February 27 2008, 6:41 PM 


The principle of Equivalence applies to three different situations:

1. Two reference systems of uniformly moving system & uniformly accelerating system: this is the situation in Einstein's thought experiment under discussion. Here, the Equivalence principle states that acceleration is equivalent to gravitation. Since an accelerating system of reference imparts the same acceleration to all distant bodies regardless of material composition and physical state. In the same way, a uniform gravitational field imparts the same acceleration to all falling bodies regardless of material composition and physical state. It follows, therefore, that acceleration and gravitation are equivalent.

2. Two reference systems of uniformly accelerating system and system at rest in a gravitational field: this situation is employed in deriving the gravitational red shift and the bending of light path in general relativity. Since it's clear that the two reference systems, in this case, are exactly equivalent and indistinguishable by any experiment.

3. Two reference systems of uniformly moving system and free-falling system under the sway of a uniform gravitational field: this situation is used in establishing the validity of special relativity within the framework of general relativity. Since it's obvious that free fall and inertial motion in a straight line are equivalent and cannot be distinguished by any experiment.



 
 
Turanyanin

Re: The Equivalence of Acceleration and Gravitation in General Relativity

February 28 2008, 10:41 AM 

Physics, by its origin, should be about 1. experimental experience (“observation”, generally speaking) and 2. theoretical consideration and generalization of data. So,

a) From Newton’s dynamics and his gravitation theory we have (in scalar 1-d form and for two point-masses)

Fi = Fg

or

m_i*a = Gm_gM/r^2

which leads towards “kinematical picture” again

a = (m_i / m_g )*g; g=GM/r^2

but only by great chance m_i /m_g = const. This is so-called Weak (Newton’s) Principle of Equivalence. We know that this is experimentally verified to the high 10^-18 which can be assumed as a deep natural IDENTITY.

b) As for Einstein Principle of Equivalence and his GR, we know that its linearized spherical symmetric (Schwartzschield) metrics leads towards

c’ = c(1 – 2HeaM/r); Hea = G/c^2

where of course photon is zero in rest mass (null-geodesics which locally gives again c’ = c). Main problems: 1. unnatural singularities for r = 2HeaM and 2. Hilbert-Einstein non-linaer 2-rank tensorial equations are in fact unsolvable for a general case. That “theory” was doomed for fitting and guessing from its beginning. Not to mention quasi principles such as EP, local Lorentz symmetry etc. which all are totally non-existent.

For example (and I already stated this several times before): let us conduct Pound-Rebka under free falling circumstances (e.g., craft in outer Space)! My claim is: red-shift would be found which would mean EP is false, i.e. the very basis of each and every metric theory of gravity is lost. And if so, photon again simple has to be “massive”. Even more important, static G-potential is in form of exp(-k/r) which means new mathematics without singularities.

As last but not least, and assuming the above IDENTITY, it becomes clear that well known G and c are not real natural constants at all. One can ask: what combination of those two could be the constant? It seems that Nature rejects all other possibilities but Hea, i.e.

Hea = G/c^2 = 7.4E-28 m/kg

which is signalized through recent experience of all gravitomagnetic phenomena and so-called mass etalon decaying. Gravity is not “geometry”, that is the point. And it is “acceleration” even less. We are talking here about most fundamental change (of paradigm) possible.

 
 
nakayama

Re: The Equivalence of Acceleration and Gravitation in General Relativity

September 14 2011, 8:25 PM 

An elevator cabin is at a standstill in non-gravitational field. On the side wall (supposed to be the left wall), there are five holes (at regular intervals ; vertically). The sun light is coming from the just left and is passing through the holes. Then, on the right wall, there are five projections (spot-lights ; don't move). But, if this elevator cabin begins free fall (downward), projections will move upward. Equivalence principle will be wrong.

 
 
AAF

Re: The Equivalence of Acceleration and Gravitation in General Relativity

September 20 2011, 6:29 PM 





That is right; Nakayama! happy.gif

Kinematics of free fall is a big problem for this so-called 'principle' of equivalence.




[linked image]







 
 
Stanley16

Re: The Equivalence of Acceleration and Gravitation in General Relativity

September 20 2011, 7:48 PM 










The equivalence principle:

"If a space-ship is in a free fall, everything in it seems weightless. A man inside a closed spaceship would not be able to tell whether his spaceship is freely falling or cruising at a constant speed in the interstellar space where no significant gravity exists. Any mechanical experiment he might do would give the same result in both cases. Also, the man in the closed spaceship would not be able to tell whether his spaceship is parking on the surface of a planet or accelerating at a constant acceleration in the interstellar space. Einstein suggested that this is not just a similarity in the behavior but actually the same physical states. In other words, a freely falling frame in a gravity field is equivalent to an inertial frame with the absence of gravity. Also, a static frame in a gravity field is equivalent to an accelerating frame with the absence of gravity:


http://rafimoor.com/english/GRE1.htm








 
 
AAF

Re: The Equivalence of Acceleration and Gravitation in General Relativity

September 21 2011, 12:01 AM 








It is not true that "a man inside a closed spaceship would not be able to tell whether his spaceship is freely falling or cruising at a constant speed in the interstellar space where no significant gravity exists. Any mechanical experiment he might do would give the same result in both cases". [linked image]

And this is the reason why:

As a general rule, objects falling freely from rest in a gravitational field gain equal instantaneous speeds by traveling equal distances.

However, this rule does not apply to the leading part and the trailing part of a freely falling spaceship.

And that is because the spaceship is a solid object. And hence, the leading part and the trailing part must travel at the same speed regardless of the distance that separates them. And this speed is exactly equal to the instantaneous speed gained by a test particle through falling from rest at the same initial point and all the way to the midpoint of the spaceship at its current position. And so, the instantaneous speed of the leading edge is less and the instantaneous speed of the trailing edge is greater than the instantaneous speeds gained by a test particle at the two positions respectively.

Now, here is the BIG difference (mechanically speaking) between a freely falling spaceship and a spaceship cruising at a constant speed in the interstellar space:

[A] A test particle remains motionless in the same place, when it's released anywhere inside the space enclosed by a spaceship cruising at a constant speed in the interstellar space.

[B] But the same test particle starts to move very slowly towards the leading edge, when it's released below the middle point inside the space enclosed by a freely falling spaceship. Conversely, it starts to recede very slowly towards the trailing edge, when it's released above the middle point inside the space enclosed by a freely falling spaceship.

[C] And of course, it remains motionless at the same point, when it's released at the middle point inside the space enclosed by a freely falling spaceship.









 
 
 
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