The Equivalence of Acceleration and Gravitation in General Relativity

Stanley16
Stanley16

February 26th, 2008, 9:31 pm #11


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.

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'."

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Stanley16
Stanley16

February 26th, 2008, 9:57 pm #12


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.

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

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Joined: February 22nd, 2008, 2:40 pm

February 27th, 2008, 7:57 am #13


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.

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
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Stanley16
Stanley16

February 27th, 2008, 10:39 pm #14


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.

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.


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Stanley16
Stanley16

February 27th, 2008, 10:41 pm #15


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.

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.


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Turanyanin
Turanyanin

February 28th, 2008, 2:41 pm #16


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.

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.
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nakayama
nakayama

September 14th, 2011, 11:25 pm #17


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.

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.
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AAF
AAF

September 20th, 2011, 9:29 pm #18


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.



That is right; Nakayama!

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







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Stanley16
Stanley16

September 20th, 2011, 10:48 pm #19


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.








<font size="5">The equivalence principle:</font>

"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







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AAF
AAF

September 21st, 2011, 3:01 am #20


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.






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".

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.

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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