- #36
Mentz114
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Nonsense. If you want to persist with your ridiculous misconceptions, feel free. Clearly you are not open to rational argument.
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How Does the Principle of Equivalence Lead to Gravity Curving Space?
- Thread starter Goldbeetle
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- Equivalence Principle
In summary: C/R = 2*pi/sqrt(1-v^2/c^2), but now v is the relative velocity between the observer and the center of the disc. This is because radar distance takes into account the time it takes for the light to travel from the center of the disc to the observer, which is affected by the observer's motion. So whether ruler distance or radar distance is used, the result is the same and it still leads to the conclusion that space and time are distorted in a non-Euclidean way in the pseudo-gravitational field of the rotating disc.
Nonsense. If you want to persist with your ridiculous misconceptions, feel free. Clearly you are not open to rational argument.
- #37
Mueiz
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Mentz114 said:
Ridiculous misconceptions is the first step towards understanding but quoting of what somebody else has said without thinking is ignorance
- #38
Mentz114
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This is wrong and has nothing to do with case in hand.
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 ?
- #39
Dale
Mentor
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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.
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.
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- #40
Mueiz
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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
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
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- #41
Mentz114
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(my bold)Mueiz said:
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'.
- #42
Mueiz
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Mentz114 said:
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
- #43
Mentz114
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As they must, or they wouldn't be measuring acceleration.
An observer in a state of acceleration can easily detect this without reference to any external beacon or reference frame.
You won't be able to show this because it is obviously untrue.
Agreed.
I have other things to do so I leave you to prove your case for now.
- #44
Mueiz
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DaleSpam said:
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
- #45
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In #26, I pointed you to several calculations of this result. Have you read them?Mueiz said:
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.Mueiz said: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 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.
- #46
Mentz114
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I tried working out the acceleration experienced by a rotating observer in Minkowski spacetime like this. The velocity vector is
[tex]
V_{\mu}=\left[ \begin{array}{cccc}
-\sqrt{{r}^{2}\,{w}^{2}+1} & -r\,w\,sin\left( t\,w\right) & r\,w\,cos\left( t\,w\right) & 0 \end{array} \right]
[/tex]
and [itex]\eta^{ab}V_aV_b=-1[/itex].
From which we get the derivatives ( covariant differentiation is just partial differentiation here )
[tex]
\partial_{\mu}V_{\nu}=
\left[ \begin{array}{cccc}
0 & -r\,{w}^{2}\,cos\left( t\,w\right) & -r\,{w}^{2}\,sin\left( t\,w\right) & 0\\\
-\frac{r\,\left( \frac{d}{d\,x}\,r\right) \,{w}^{2}}{\sqrt{{r}^{2}\,{w}^{2}+1}} & -\left( \frac{d}{d\,x}\,r\right) \,w\,sin\left( t\,w\right) & \left( \frac{d}{d\,x}\,r\right) \,w\,cos\left( t\,w\right) & 0\\\
-\frac{r\,\left( \frac{d}{d\,y}\,r\right) \,{w}^{2}}{\sqrt{{r}^{2}\,{w}^{2}+1}} & -\left( \frac{d}{d\,y}\,r\right) \,w\,sin\left( t\,w\right) & \left( \frac{d}{d\,y}\,r\right) \,w\,cos\left( t\,w\right) & 0\\\
0 & 0 & 0 & 0
\end{array} \right]
[/tex]
We want this in the V direction,
[tex]
\partial_{\mu}V_{\nu}V^{\nu}=
\left[ \begin{array}{cccc}
0 & 2\,r\,\left( \frac{d}{d\,x}\,r\right) \,{w}^{2} & 2\,r\,\left( \frac{d}{d\,y}\,r\right) \,{w}^{2} & 0 \end{array} \right]
[/tex]
which seems right, with acceleration in the x and y directions. Substituting [itex]r^2=x^2+y^2[/itex] into the derivatives gives
[tex]
\partial_{\mu}V_{\nu}V^{\nu}=
\left[ \begin{array}{cccc}
0 & \frac{2\,r\,{w}^{2}\,x}{\sqrt{{y}^{2}+{x}^{2}}} & \frac{2\,r\,{w}^{2}\,y}{\sqrt{{y}^{2}+{x}^{2}}} & 0 \end{array} \right]
[/tex]
and the magnitude is [itex]2w^2r[/itex].
Where did that factor of 2 come from ?
[tex]
V_{\mu}=\left[ \begin{array}{cccc}
-\sqrt{{r}^{2}\,{w}^{2}+1} & -r\,w\,sin\left( t\,w\right) & r\,w\,cos\left( t\,w\right) & 0 \end{array} \right]
[/tex]
and [itex]\eta^{ab}V_aV_b=-1[/itex].
From which we get the derivatives ( covariant differentiation is just partial differentiation here )
[tex]
\partial_{\mu}V_{\nu}=
\left[ \begin{array}{cccc}
0 & -r\,{w}^{2}\,cos\left( t\,w\right) & -r\,{w}^{2}\,sin\left( t\,w\right) & 0\\\
-\frac{r\,\left( \frac{d}{d\,x}\,r\right) \,{w}^{2}}{\sqrt{{r}^{2}\,{w}^{2}+1}} & -\left( \frac{d}{d\,x}\,r\right) \,w\,sin\left( t\,w\right) & \left( \frac{d}{d\,x}\,r\right) \,w\,cos\left( t\,w\right) & 0\\\
-\frac{r\,\left( \frac{d}{d\,y}\,r\right) \,{w}^{2}}{\sqrt{{r}^{2}\,{w}^{2}+1}} & -\left( \frac{d}{d\,y}\,r\right) \,w\,sin\left( t\,w\right) & \left( \frac{d}{d\,y}\,r\right) \,w\,cos\left( t\,w\right) & 0\\\
0 & 0 & 0 & 0
\end{array} \right]
[/tex]
We want this in the V direction,
[tex]
\partial_{\mu}V_{\nu}V^{\nu}=
\left[ \begin{array}{cccc}
0 & 2\,r\,\left( \frac{d}{d\,x}\,r\right) \,{w}^{2} & 2\,r\,\left( \frac{d}{d\,y}\,r\right) \,{w}^{2} & 0 \end{array} \right]
[/tex]
which seems right, with acceleration in the x and y directions. Substituting [itex]r^2=x^2+y^2[/itex] into the derivatives gives
[tex]
\partial_{\mu}V_{\nu}V^{\nu}=
\left[ \begin{array}{cccc}
0 & \frac{2\,r\,{w}^{2}\,x}{\sqrt{{y}^{2}+{x}^{2}}} & \frac{2\,r\,{w}^{2}\,y}{\sqrt{{y}^{2}+{x}^{2}}} & 0 \end{array} \right]
[/tex]
and the magnitude is [itex]2w^2r[/itex].
Where did that factor of 2 come from ?
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- #47
Mueiz
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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
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
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- #48
Roshanjha14
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I didnt undrstand the weak equi. Principle.
- #49
Dale
Mentor
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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.Mueiz said:
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?
- #50
Mueiz
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I really benefited very much from this good presentation on this issue... butI have an objection .DaleSpam said:
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 .
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- #51
Dale
Mentor
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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.Mueiz said:
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.
The frame where accelerometers at rest read 0.Mueiz said:
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- #52
Mueiz
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DaleSpam said:
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
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- #53
Mentz114
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Mueiz said:
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 ?
- #54
Mueiz
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I do understand Newtonian Laws but not in same way as Newton understood it whoMentz114 said:
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
- #55
Mentz114
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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
You made no attempt to answer my question
- #56
Mueiz
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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
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
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- #57
Mentz114
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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.Mueiz said:
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.
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- #58
yuiop
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Mueiz said:
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.
Mueiz said:
Mueiz said:
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"?
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- #59
Mueiz
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Mentz114 said:
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
- #60
Mentz114
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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 ).Mueiz said:
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.
- #61
Mueiz
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It seem to me that you did not understand the logic by which I conclude that the second observer is inertialyuiop said:
it is that only the circumference of a very large disc is almost a straight line to any degree of approximate ...so it is in uniform motion relative to the first observer who is inertial ...if we are in uniform motion relative to an inertial observer then we must be inertial too ...tell me what is wrong with this ..before going to calculation
i want you also to reply the question i gave in my previous input
- #62
yuiop
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Mueiz said:
Are you sure about that? If you are free falling in a gravitational field, and there is another observer falling with you but below you, then they will be moving away from you. In order to stay at a constant distance from you the other observer would have to accelerate constantly towards you. Even when they maintain constant distance from you there clock will be ticking at a different rate to yours. In a true inertial reference frame in flat space, observers at rest in the reference frame are stationary with respect to each other, all experience zero proper acceleration and all their clocks are running at the same rate. Clearly there is no arrangement of the vertically separated fallers in the gravitational field that can duplicate being in a true inertial reference frame in flat space. The falling observers only approximate an inertial reference frame, if they are very close together, so that the errors are small. All these difference come about as a result of tidal effects and that is what distinguishes a real gravitational field from the pseudo-gravitational field that results of artificial acceleration in gravitationally flat space.
- #63
yuiop
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Mueiz said:
You conclude that the observer on the edge of a very large disc is inertial because locally they appear to traveling in approximately a straight line, but that is not the definition of inertial motion. Inertial motion is motion with zero proper acceleration and the observer on the edge of the large disc would be experiencing very large proper acceleration at high velocities and is therefore not inertial by definition.
- #64
Mueiz
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yuiop said:
Let us save our effort and concentrate in Thought as abstract
I think it is not a subject of physics '' who said and How someone understands''
suppose this is my own opinion and convince me with facts and logic , i really mentioned some sources in order that someone may want to read more about the issue
- #65
Mueiz
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yuiop said:
Then if someone want to apply Pythagoras Theorem in the ground of his house ,he must remember that the Earth is not flat
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- #66
Dale
Mentor
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This is incorrect. As I said above an accelerometer measures proper acceleration, which is the magnitude of the covariant derivative and is a frame invariant quantity. An accelerometer's operation does not rely in any way on reference frames and is completely insensitive to gravitatonal fields.Mueiz said:
- #67
Mueiz
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DaleSpam said:
why quoting the my opinion leaving leaving reasons and discussion?
- #68
Dale
Mentor
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Because I am on a mobile device, so I can only hit the most imoprtant point. In this case, your misunderstanding of accelerometers.Mueiz said:
Why did you ignore my entire post?
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- #69
Mentz114
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DaleSpam said:
Muez is taking an extreme fundamentalist Machian line - that there is no inertia in the absence of gravity. He admitted this in an earlier post, but won't restate it because he likes a good ( if pointless ) argument.
Mueiz said:
State your belief that there is no interia in the absence of gravity.
- #70
yuiop
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I gave you some facts. Here...Mueiz said:
.. and here ...yuiop said:
.. but you have chosen to ignore them and have not responded to these counter arguments.yuiop said:
I mention what people understand by Mach's principle, but since he did not clearly define it and since you seem to be championing the principle, then the only way to proceed with a sensible discussion would be for you to define what you think Mach's principle is or whatever principle it is that you are championing that you think demonstrates that the predictions of General Relativity are wrong.
I will give you a simple though experiment for discussion purposes. Imagine we have a very large gravitational body that is rotating at 1 rpm as measured by a Sagnac device. At a very great distance we have a very small but luminous particle that is not rotating as far as a Sagnac device is concerned. Now if we pay attention to the Sagnac devices, we say the distant star is stationary and the large star is slowly rotating. Now if we pay attention to the popular description of Mach's principle we are forced to conclude that all motion is relative to the majority of mass in the universe which in this case is the slowly rotating large gravitational body. By this definition the large body is not rotating and the distant small body is orbiting the large body at some great superluminal velocity of say 100c. Does that make sense? Is that what you are claiming. Now if we had another small satellite orbiting the large massive body in geo-synchronous orbit so that it was always above the same point on the massive body, then the Machian interpretation would be that the satellite is magically suspended above the massive body without requiring orbital motion or centripetal force. The Machian interpretation of gravity would then be that near bodies are attracted by gravity and distant bodies are repelled by gravity and somewhere in between gravity is neutralised. Now if we had another satellite at the same radius as the apparently stationary satellite, but orbiting in the opposite direction then the Machian interpretation is that gravity works on objects that moving but not on stationary objects at the same distance. At different distances, the Machian interpretation is that the force of gravity depends upon whether an object orbits clockwise or anticlockwise. Is that how you understand how the universe works?
In summary the Machian interpretation predicts:
1) Objects really can orbit at velocities much greater than the speed of light.
2) Gravity does not act on objects at a certain critical distance and stationary objects can hover.
3) Gravity attracts or repels depending on distance.
4) The force of gravity depends on the orbital direction and not just on orbital velocity.
5) The speed of light is not locally isotropic in all directions to an inertial observer.
Is that how you see things?
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