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

[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 .

 

[Dale]

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

This is easily possible, both theoretically and practically. From a theoretical perspective: Minkowski spacetime is a perfectly valid solution to the EFE for no gravitational sources. Once you have that then simply looking at the metric in your chosen coordinate system identifies if the frame is inertial or not, and all of the rest above follows. Note: if you consider this to be a violation of Mach's principle then you consider GR to be a non-Machian in some sense.

From a practical perspective: Consider observer B to be in a rotating space station far away from any gravitational sources. Consider observer A to be non-rotating at the hub, and consider observer C to be in a ship departing from the station. The experiment is not theoretically impossible, at least not according to Einstein's GR.

but if there is no gravitation what is frame you need to deviate from to be accepted in Accelerated Objects Club .

The frame where accelerometers at rest read 0.

 

[Mueiz]

This is easily possible. Consider observer B to be in a rotating space station far away from any gravitational sources. Consider observer A to be non-rotating at the hub, and consider observer C to be in a ship departing from the station. The experiment is not theoretically impossible, at least not according to Einstein's GR.

The frame where accelerometers at rest read 0.

Your accelerometer measures the deviation from the inertial frames which are special frames in gravitational field but in zero gravitational field all frames are inertial and your accelerometer
is useless ...but you can use it another way if you throw it and look to its worldline from different frames of reference it will be straight line in all frames, but do not use transformation laws derived for gravitational field cases.
Do you think that the force of the ship can cause the ship to accelerate ; this is also gravitational-field phenomena ...the problem of zero gravitational field can be solved if we consider the symmetry of all frames which tells us that their geometry should be the same ..then which geometry shall we choose
the postulate of Simplicity tells us that its the Flat Geometry
I hope that you be aware not to apply Phenomena and calculation that belong to gravitational field in empty space ... you should leave them on the Earth before riding on your space ship

 

[Mentz114]

Your accelerometer measures the deviation from the inertial frames which are special frames in gravitational field but in zero gravitational field all frames are inertial and your accelerometer
is useless

This statement flies in the face of common experience. Do you know what an accelerometer is ? It does not depend in any way on whether there is a gravitational field.

...but you can use it another way if you throw it and look to its worldline from different frames of reference it will be straight line in all frames, but do not use transformation laws derived for gravitational field cases.
Do you think that the force of the ship can cause the ship to accelerate ; this is also gravitational-field phenomena ...the problem of zero gravitational field can be solved if we consider the symmetry of all frames which tells us that their geometry should be the same ..then which geometry shall we choose
the postulate of Simplicity tells us that its the Flat Geometry
I hope that you be aware not to apply Phenomena and calculation that belong to gravitational field in empty space ... you should leave them on the Earth before riding on your space ship

You haven't learned a single thing, in spite of the many inputs. Do you understand Newtonian physics and the laws of inertia ?

Once again you repeat the fallacy that accelerated frames are impossible in SR and you continue to make false statements about the involvement of gravitational fields.

Are you taking Mach's principle literally and asserting that in a universe with no gravitating matter there can be no inertia ?

 

[Mueiz]

This statement flies in the face of common experience. Do you know what an accelerometer is ? It does not depend in any way on whether there is a gravitational field.



You haven't learned a single thing, in spite of the many inputs. Do you understand Newtonian physics and the laws of inertia ?

Once again you repeat the fallacy that accelerated frames are impossible in SR and you continue to make false statements about the involvement of gravitational fields.

Are you taking Mach's principle literally and asserting that in a universe with no gravitating matter there can be no inertia ?

I do understand Newtonian Laws but not in same way as Newton understood it who
as I know did not relate them to gravitational field.
If you use the word learned to mean convinced I agree with you that I have not learned a single thing.
It is not enough to convince me to say that my statements are false
you always quotes the part of my inputs in which i state the idea and leave the part containing reason and discussion .
I think it deserves discussion the idea that some laws and calculation of physics may not be applicable in zero-gravitational region simply because they where formulated originally to to meat cases in gravitational field and also because the symmetry noticed firstly by Mach and secondly by Einstein

 

[Mentz114]

Do you acceprt the principle that the laws of inertia are the same in all inertial frames ?

You made no attempt to answer my question

Are you taking Mach's principle literally and asserting that in a universe with no gravitating matter there can be no inertia ?

 

[Mueiz]

My answer to the first question is yes I accept that all the laws of physics are the same in inertial frames
The answer to the second question is also yes
and I want also to say important thing regarding Equivalence principle :
For the free-falling observer the space-time is locally flat because acceleration
cancel gravitational field .. but this does not mean necessary that(even it is always said) acceleration could have dependent effect on space-time like gravitation has effects on space-time in the absence of acceleration... this is another mistake but could not make effects because ;
1) it is not used in formulating the theory in gravitational field
2) We can not practically apply any Experiment in non-gravitational region
I hope anyone will release me from blame of using bad language for English is my second langage

 

[Mentz114]

but this does not mean necessary that(even it is always said) acceleration could have dependent effect on space-time like gravitation has effects on space-time in the absence of acceleration.

I have difficulty understanding what you mean by this, but in a universe without inertia, there can be no acceleration.If inertia does not exist in the absence of gravity, then what you've been asserting about accelerometers not working would be true. It would also mean that there can be no forces, since the action of a force would result in infinite acceleration in zero time.

But, since we can never test this experimentally there is little point in discussing it.

I have good reasons for believing that Mach's idea is false, since it seems unnecessary and would require some action at a distance to explain.

The general theory of relativity does not use Mach's principle but still agrees very well with observations.

 

[yuiop]

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

Let us examine the second observer riding on the edge of the disc. The Newtonian equation for centripetal force is F = m*v^2/r. Now the velocity of a particle on a rotating disc is a function of its radial displacement, so let us say v = f*r where f is an arbitrary constant, then by substitution, the centripetal force is F = m*f*r so we see that centripetal force increases with radius. The second observer on the edge of the disc will be experiencing greater acceleration than the third observer nearer the centre of the disc. Now the relativistic centripetal force will be even greater F = m*f*r/sqrt(1-(f*r/c)^2) in the non rotating reference frame and greater still in the rotating frame F = m*f*r/(1-(fr/c)^2). So we can conclude (If I have estimated the relativistic versions correctly) that the observer on the arbitrarily large disc will be experiencing exponentially large "centrifugal proper force" at relativistic speeds and not the vanishingly small force you seem to be assuming when you claim the second observer is effectively inertial. An observer experiencing large proper force is NOT an inertial observer.

Your accelerometer measures the deviation from the inertial frames which are special frames in gravitational field but in zero gravitational field all frames are inertial and your accelerometer
is useless ...but you can use it another way if you throw it and look to its worldline from different frames of reference it will be straight line in all frames, but do not use transformation laws derived for gravitational field cases.
Do you think that the force of the ship can cause the ship to accelerate ; this is also gravitational-field phenomena ...the problem of zero gravitational field can be solved if we consider the symmetry of all frames which tells us that their geometry should be the same ..then which geometry shall we choose
the postulate of Simplicity tells us that its the Flat Geometry
I hope that you be aware not to apply Phenomena and calculation that belong to gravitational field in empty space ... you should leave them on the Earth before riding on your space ship

My answer to the first question is yes I accept that all the laws of physics are the same in inertial frames
The answer to the second question is also yes
and I want also to say important thing regarding Equivalence principle :
For the free-falling observer the space-time is locally flat because acceleration
cancel gravitational field .. but this does not mean necessary that(even it is always said) acceleration could have dependent effect on space-time like gravitation has effects on space-time in the absence of acceleration... this is another mistake but could not make effects because ;
1) it is not used in formulating the theory in gravitational field
2) We can not practically apply any Experiment in non-gravitational region
I hope anyone will release me from blame of using bad language for English is my second langage

In the above two quotes you are descending into nonsense based on some sort of home grown understanding of Mach's principle. You seem to be claiming that it is impossible to rotate a disc or accelerate a rocket far from any gravitational sources. Are you aware that Einstein ultimately rejected Mach's principle? Are you aware that the Kerr solution is a vacuum solution that defines the metric for a spinning massive body in the ABSENCE of any other reference bodies and is specifically defined as a spinning massive body in an otherwise empty universe. While Mach never clearly defined his principle, it is generally understood to mean that momentum is defined relative to all the mass of all the "distant stars" and so Mach's principle is clearly incompatible with the Kerr GR solution defined in terms of an "otherwise empty universe"?

 

[Mueiz]

The general theory of relativity does not use Mach's principle but still agrees very well with observations.

it agrees with observation in gravitational field but nobody tested the case in zero-gravity till now ...but why do you refuse Mach principle ?
I have a simple and very important question which is
HOW CAN ONE KNOW IF THERE IS A GRAVITATIONAL FIELD IN CERTAIN REGION OR NOT WHEN YOU ASSUME THAT EVEN IN ZERO GRAVITATIONAL REGION EXIST ESPECIAL INERTIAL FRAME
LIKE THAT OF FALLING OBJECTS OF GRAVITATIONAL FIELD?
according to the opinion of absolute acceleration in empty space even gravitational field can not be detected .and the inertial frame you all have been speaking of can not be distinguished from that of fee-falling in gravitational field

 

[Mentz114]

it agrees with observation in gravitational field but nobody tested the case in zero-gravity till now ...but why do you refuse Mach principle ?

As far as can be determined, our universe appears to be non-Machian. See Brans-Dicke theory, which incorporates a scalar field coupled to gravity ( I think, I'm not overly familiar with it ).

I have a simple and very important question which is
HOW CAN ONE KNOW IF THERE IS A GRAVITATIONAL FIELD IN CERTAIN REGION OR NOT WHEN YOU ASSUME THAT EVEN IN ZERO GRAVITATIONAL REGION EXIST ESPECIAL INERTIAL FRAME
LIKE THAT OF FALLING OBJECTS OF GRAVITATIONAL FIELD?
according to the opinion of absolute acceleration in empty space even gravitational field can not be detected .and the inertial frame you all have been speaking of can not be distinguished from that of fee-falling in gravitational field

In principle, if a small enough region is considered, a freely-falling frame will experience flat spacetime. But in practice deviations caused by gravity can be detected ( see Fermi normal coordinates, for instance).

I don't see any paradoxes or contradictions in what you cite. GR is a geometric theory and so incorporates the equivalence principle which is what allows us to define free-fall.