Advanced books/papers on derivation of Newtonian mechanics from GR

[Juan R.]

(1) So what? The boundary is far, far away from the system of interest -- we don't care if the actual space-time occupied by the system of interest has a boundary that resembles that of an asymptotically flat space-time -- all we care about is that the region of the actual space-time occupied by the system of interest resembles some region of the asymptotically flat approximation.

(2) How can the existence of a mathematical approximation be experimentally false?

(1) There is no experimental evidence. In fact, the experimental evidence indicates no sign of a island type model of universe, that is, with matter vanishing more and more. The rest of your claim is wrong.

(2) Is not a mathematical approximation, it is the boundary in NC gravity for it works. Experimentally is false, since our universe is not of island type.

 

[JesseM]

Juan R. refuses to answer my question...is anyone else willing to field this one? I just want to know what GR predicts about the curvature of a universe wholly empty of matter and energy, since it would probably predict something similar about an "island universe" which was wholly empty except for one clump of matter.

I'm not saying it's wrong that an empty universe with no cosmological consant would be flat, but I just want to check and make sure that this is actually what GR predicts, since you haven't directly addressed this question yet. This section from Ned Wright's cosmology tutorial says that "For Omega less than 1, the Universe has negatively curved or hyperbolic geometry", where Omega is the ratio between the universe's matter/energy density and the critical density needed to prevent collapse--wouldn't an empty universe have Omega=0, and therefore be negatively curved rather than flat?

 

[Berislav]

Of course, but Wald does not take the limit c --> infinite, because then the metric g00 = 1 gRR = 1. That is FLAT and cannot explain gravitation.

Please check the metric again. You will see that that is not what happens to it. And I think that Wald didn't take that limit because he dealt with SR and Newtonian approximations in one of the previous chapters, so he assumed that the reader understands that that is how one reduces to Newtonian physics from relativity.

Perhaps i explained bad. Precisely in NG one would do the instantaneous change Phy --> Phy/2. Whereas in GR only after of 8 minutes one would change the potential, of course in GR the change (after of the 8 minutes) is not Phy --> Phy/2 it would be more gradual. Any case both description are different and this is reason that Wald equation is not Newtonian equation.

Yes, of course, that will happen in pure GR and in reality, but if you take c--> infinity, then it will propagate instantly.

I can accept the gauge of Phy in GR but Phy in NG is rather physical at least if one take the integration constant equal to zero which is always done. I believe that "Unphysical" is not the correct expression, because in NG the potential is Energy by unit of mass of test body and that is physical, of course i know that one could redefine energy using a new zero for the scale, but one definition would not be more physical that other and one take the integration constant zero by commodity.

The gradient of the potential is what is physical, not the potential, and hence you can add any constant to the potential; and that's what any leftover constant after you take a limit is.

Dirac is cristal clear. There exit two inconsistent theories: one for non relativistic phenomena, other for certain relativistic phenomena.

The basics of QED, which Dirac mentions, are derived from Maxwell's laws and quantum physics (and a second quantization). You don't even have to mention SR, per se, as Maxwell's equations are relativistically covariant. Furthermore, the Klein-Gordon equation, for instance, is just a relativistic version of the Schrödinger equation, and a reduction from the former to the latter is simple. So, I don't really think that Dirac was talking about what you point out as a problem. So, please, if you could quote Dirac on what the exact problem is, that would be great.

Probably you can, if you can untangle the limits. One is already taking a limit to define asymptotic flatness, to approach asymptotic flatness as a limit is to take the limit of a limit. It's probably not impossible, but I don't know of anyone who has done it and written it up in a paper.

I don't understand. Could you please elaborate for me?

I just want to know what GR predicts about the curvature of a universe wholly empty of matter and energy

If it has a non-zero cosmological constant then this is a de Sitter universe, otherwise it is a Minkowski spacetime, meaning, among other things, that it is static.

P.S.
This 'island universe' assumption sounds to me like just a logical assertation that nothing outside our cosmological horizon can affect us.

 

[pervect]

Juan R. refuses to answer my question...is anyone else willing to field this one? I just want to know what GR predicts about the curvature of a universe wholly empty of matter and energy, since it would probably predict something similar about an "island universe" which was wholly empty except for one clump of matter.

Interesting question. I think the answer is that you can view a universe that's totally empty of matter either as a non-expanding Minkowsky universe, or an expanding Milne universe.

Obviously the solution must be homogeneous and isotropic, because a vacuum is homogeneous and isotropic. So you should get some sort of FRW cosmology.

A static Minkowski space-time will satisfy this, and so will an expanding Milne universe, which has a spatial curvature k=-1 and a uniform scale factor a(t) = t. IIRC the two are equivalent, they represent a different coordinate system for the same space-time.

I haven't worked this out as carefully as I might, the answer is a bit "off the cuff".

 

[JesseM]

If it has a non-zero cosmological constant then this is a de Sitter universe, otherwise it is a Minkowski spacetime, meaning, among other things, that it is static.

But isn't it true that in the FRW cosmological model, a universe with zero cosmological constant will be negatively curved if Omega is less than 1? The diagram at the top of this page from Ned Wright's cosmology tutorial shows a universe with Omega<1 having negative curvature, and in the paragraph below he says "These a(t) curves assume that the cosmological constant is zero". What am I misunderstanding here?

 

[Berislav]

or an expanding Milne universe.

I didn't mention that spacetime because I never heard about it.
I apologize.

 

[pervect]

I don't understand. Could you please elaborate for me?

The notion of how to make this statement mathematically rigorous bothers me a bit.

The limit of f(x) as x-> a is well definied. But how do you take the limit of all possible maniolds as they "approach flatness"? I suppose we can do this if we have a distance measure between manifolds. How do we construct this distance measure?

Perhaps part of the answer is that we are assuming we have a map from from the manifolds to a single scalar number, which is a "measure" of the "flatness" of the manifold. Can we really rigorously construct this measure? How do we go about it, exactly? I.e. I give you a manifold, and you take out your measuring instrument and you say "The flatness of that manifold is 22" - how do we accomplish this?

 

[JesseM]

Interesting question. I think the answer is that you can view a universe that's totally empty of matter either as a non-expanding Minkowsky universe, or an expanding Milne universe.

Obviously the solution must be homogeneous and isotropic, because a vacuum is homogeneous and isotropic. So you should get some sort of FRW cosmology.

A static Minkowski space-time will satisfy this, and so will an expanding Milne universe, which has a spatial curvature k=-1 and a uniform scale factor a(t) = t. IIRC the two are equivalent, they represent a different coordinate system for the same space-time.

I haven't worked this out as carefully as I might, the answer is a bit "off the cuff".

Interesting, so there are two separate solutions to this problem (the diagram on Ned Wright's page which I mentioned above seems to show the expanding universe with a(t)=t), but they can be made equivalent by a coordinate transformation? In the coordinate system that treats this as an expanding Milne universe, is space indeed negatively curved rather than flat?

Also, does the expanding Milne universe have an initial singularity, and if so, does the fact that it can be transformed into a minkowski spacetime mean this is just a coordinate singularity rather than a "real" singularity?

 

[pervect]

Interesting, so there are two separate solutions to this problem (the diagram on Ned Wright's page which I mentioned above seems to show the expanding universe with a(t)=t), but they can be made equivalent by a coordinate transformation? In the coordinate system that treats this as an expanding Milne universe, is space indeed negatively curved rather than flat?

That's what I read on the internet http://web.mit.edu/8.286/www/quiz00/e6qs3-1.pdf

but I haven't double checked this or thought about it much yet. Since this is a student-written "quiz response" it's worth double checking it, though it seems right on this point. I'm not sure how long it will be up, hopefully for a few days at least, these sort of things tend to disappear without notice.

The quiz response above also made some interesting statements about Birkhoff's theorem and how it applies to Juan's dilema, but I'm not sure I believe them yet as an accurate statement of the theorem.

Hopefully I'll post more later, after I've had coffee, breakfast, and bashed a few metrics through GRTensor.

 

[Berislav]

I'll attempt to fill in for pervect in the mean time.

The Robertson-Walker metric is derived by assuming homogenity and isotropy (nothing else about the content of the universe):

ds^2=-dt^2+R^2(t) (\frac{dr^2}{1-kr^2}+r^2d\Omega^2),

where k can be anything, but we can redefine it as being either -1,0, or 1.

Now it follows from the Einstein field equations that:

(R&#039;/R)^2=-k/R^2+\frac{8 \pi \rho}{3},
where R' denotes the derivative with respect to time and \rho the density of the perfect fluid in the universe. Now if k=0 and there is nothing in the universe it easily follows that R'=0 and the universe is static and flat. If k is something else it will be positively curved and shrinking, or negatively curved and expanding.

I should have added that de Sitter universe assumes a positive cosmological constant.

Oops. Notation mistake.

 

[George Jones]

Milne's universe is just an interesting coordinate system on a proper subset of Minkowski spacetime that"splits" the subset into time and space. In this splitting, the spatial curvature for space at any "instant" of time is negative, but spacetime curvature is zero, as it must be. It is impossible to transform zero spacetime curvature into non-zero spacetime curvature by a change of coordinates.

Regards,
George

 

[Hurkyl]

The limit of f(x) as x-> a is well definied. But how do you take the limit of all possible maniolds as they "approach flatness"? I suppose we can do this if we have a distance measure between manifolds. How do we construct this distance measure?

Well, the question of interest is the predictions of the two space-times on the system of interest -- the appropriate distance would be the deviation of the predictions... maybe the maximum of the differences of the appropriate derivatives over some compact subset as defined by your favorite coordinate charts.

 

[George Jones]

But isn't it true that in the FRW cosmological model, a universe with zero cosmological constant will be negatively curved if Omega is less than 1? The diagram at the top of this page from Ned Wright's cosmology tutorial shows a universe with Omega<1 having negative curvature, and in the paragraph below he says "These a(t) curves assume that the cosmological constant is zero". What am I misunderstanding here?

Bersislav was talking about empty universes while standard FRW universes are not empty.

Also, I am not sure what you mean by curvature of the universe. In FRW models, k = -1, 0, 1 refers to curvature of spatial sections, not to curvature of spacetime.

Regards,
George

 

[JesseM]

Milne's universe is just an interesting coordinate system on a proper subset of Minkowski spacetime that"splits" the subset into time and space. In this splitting, the spatial curvature for space at any "instant" of time is negative, but spacetime curvature is zero, as it must be. It is impossible to transform zero spacetime curvature into non-zero spacetime curvature by a change of coordinates.

But what you and pervect's reference are saying is that spatial curvature, as opposed to spacetime curvature, can be transformed by a change of coordinates, correct? When I talked about an empty universe being negatively curved I was just talking about the spatial curvature. Anyway, thanks for the answers guys, it helped clear up my confusion on this.

 

[George Jones]

But what you and pervect's reference are saying is that spatial curvature, as opposed to spacetime curvature, can be transformed by a change of coordinates, correct?

Yes, at least in some cases. Spatial curvature depends on the choice that specifies the "nows".

When I talked about an empty universe being negatively curved I was just talking about the spatial curvature.

Sorry - I wasn't sure.

Regards,
George

 

[pervect]

With no cosmological constant, the only homgeneous and isotropic vacuum solutions are indeed the Milne solution and the Minkowski solution.

As Berislav points out, when one allows a cosmological constant, there are other solutions such as the DeSitter solution.

The metric for the most general homogeneous and isotropic soultion is

variables = [t,\chi,\theta,\phi]

<br /> \left[ \begin {array}{cccc} <br /> -1&amp;0&amp;0&amp;0\\<br /> 0 &amp; a^2 &amp; 0 &amp; 0\\<br /> 0 &amp; 0 &amp; a^2 \Sigma^2 &amp; 0\\<br /> 0 &amp; 0 &amp; 0 &amp; a^2 \Sigma^2 sin(\theta)^2 \\<br /> <br /> \end {array} \right] <br />

where a = a(t) is the expansion factor and \Sigma = \Sigma(\chi) is a different function depending on the spatial curvature

k=1, \Sigma = sin(\chi)

k=0, \Sigma = \chi

k=-1, \Sigma = sinh(\chi)

[/tex]

Solving for an all-zero Riemann or Einstein tensor (a vacuum solution with no cosmological constant), real solutions only exist for k=0 and k=-1.

Basically we have (da/dt)^2 + k = 0

So when k=-1, (da/dt) = +1 or -1, and we have the Milne solution for an expanding universe, or a "big crunch" time-reversed Milne universe.

When k=0, (da/dt)=0, so a(t) is constant, and we have the familiar Minkowski metric.

When k=1 there is no solution (as I mentioned previously).

While the Minkowski metric and the Milne metric appear different on the surface, either one can be transformed into the other by a change of variables, so they are not really "different" solutions.

[add]
Ways to see the equivalence between the Mline metric and the Minkowski metric

1) The Riemann of the Milne metric is zero
2) Substitute t&#039; = -t*cosh(\chi), \chi&#039;= t* sinh(\chi) into the metric -dt'^2 + dx'^2 (use the chain rule).

 

[Berislav]

So when k=-1, (da/dt) = +1 or -1, and we have the Milne solution for an expanding universe, or a "big crunch" time-reversed Milne universe.

Are you sure that the Riemann tensor of this metric is zero?

In my previous post I used:
ds^2=-dt^2+R^2(t) (\frac{dr^2}{1-kr^2}+r^2d\Omega^2)
There still is a coefficient multiplying dr^2.

When k=1 there is no solution (as I mentioned previously).

I think one should say that it isn't physical as R (or a, in your notation) will become negative because the integration constant must be finite, rather than there's no solution.

 

[pervect]

The Milne metric is, if you look at the first post and make k=-1 so that \Sigma(\chi) = sinh(\chi) and at a(t)=t

<br /> ds^2 = -dt^2 + t^2 d \chi^2 + t^2 sinh(\chi)^2 d \theta^2 + t^2 sinh(\chi)^2 sin(\theta)^2 d \phi^2<br />

the Riemann of the above metric is identically zero, and the variable substitution below, equivalent to the one I mentioned earlier (but with time running forwards!)

t1 = t*cosh(\chi), r1 = -t*sinh(\chi)

will convert the standard Minkowski metric below

-dt1^2 + dr1^2 + r1^2(d \theta^2 + sin(\theta)^2 d \phi^2)

into the Milne metric (first line of the post).

 

[Berislav]

Yes, my mistake. The appearance of the metric tricked me. I should have checked the curvature 2-form before I said anything.

 

[pervect]

Yes, my mistake. The appearance of the metric tricked me. I should have checked the curvature 2-form before I said anything.

It's a funny looking beast, alright.

 

[Juan R.]

Please check the metric again. You will see that that is not what happens to it.

Yes i forgot a minus sign. In solar systems test the limit c --> infinite gives

g00 = 1 and gRR = - 1

passing to cartesian coordinates again the metric is (1, -1, -1, -1) I am using trace -2.

And I think that Wald didn't take that limit because he dealt with SR and Newtonian approximations in one of the previous chapters, so he assumed that the reader understands that that is how one reduces to Newtonian physics from relativity.

If he take the limit curvature is zero and geodesic equation read a = 0 whereas Newtonian gravity says a =/= 0. Then Wald is forced to use an inconsistent hibrid.

Yes, of course, that will happen in pure GR and in reality, but if you take c--> infinity, then it will propagate instantly.

But, then, metric is FLAT and according to GR there is not gravitation which contradict Newtonian gravity.

The gradient of the potential is what is physical, not the potential, and hence you can add any constant to the potential; and that's what any leftover constant after you take a limit is.

Yes i agree that only gradients or diferences of potential are significative. But taking an origin for the potential the potential itself become physical.

I was expresing is that limit R --> infinite in NG is physical, called principle of decomposition of clusters and experimentally verified. Whereas limit R --> infinite on GR or Cartan theory is unphysical becasue is not experimentally verified.

The basics of QED, which Dirac mentions, are derived from Maxwell's laws and quantum physics (and a second quantization). You don't even have to mention SR, per se, as Maxwell's equations are relativistically covariant. Furthermore, the Klein-Gordon equation, for instance, is just a relativistic version of the Schrödinger equation, and a reduction from the former to the latter is simple. So, I don't really think that Dirac was talking about what you point out as a problem. So, please, if you could quote Dirac on what the exact problem is, that would be great.

Dirac explicitly attack renormalization procedure there, but does not states in its quote what is the problem with QFT and NRQM. But it is unnecesary since it is well-known that QFT is NOT QM. Therefore, in practice, like Dirac clearly states, one works with two theories, one for nonrelativistic phenomena and other for certain question of relativistic phenomena.

If you study QFT you can see differences between QFT and QM. For example there is a very basic difference regarding positions in both approaches doing both incompatible like Dirac clearly notes, x is a operator in QM but is a parameter in QFT. Also QFT is only defined for single particles and bound states are, rigorously, undefined in QFT.

It is not true that the Klein-Gordon equation was just a relativistic version of the Schrödinger equation. In fact in RQM, the KG is unphysical, and cannot be thought like a relativistic version of Schrödinguer equation, and in QFT "it" is not a wave equation and the link with NRQM is broken (like Dirac clearly states). The same about Dirac equation

In QFT (ipartial + m) Phy = 0 for electrons is not the original Dirac equation even if look like. In fact, this is like difference between Wald equation and Newtonian equation, both looks equal but are not equal.

That above QFT equation of above is NOT Dirac equation is also noted by Weinberg in his manual vol 1 on QFT. In fact, Weinberg beggins his manual critizing Dirac RQM theory.

 

[Juan R.]

The quiz response above also made some interesting statements about Birkhoff's theorem and how it applies to Juan's dilema, but I'm not sure I believe them yet as an accurate statement of the theorem.

1º problem) Eherl boundary is not supported by observation. I already cited several guys including Penrose. The fact that that boundary is unphysical has forced to people to work with other approach which is add a new equation to original GR equations, the equation with the vanising of derivation for the defined Newtonian connection. But then one is not deriving NG from GR, since that equation is added ad hoc to the field equations of GR.

2º problem) Even if boundary was supported by observation that violates causality, since is asumming that an event sited infinitely in the past, i.e. infinite time before big bang is acting on current event.

3º problem) Even ignoring all above, there is a lack of continuity by the use of Ehlers function Phy(x, t) in the transition from steady states to non ones. For mathematical details and physical insight on EM similar problem see PRE 1996, 53(5), 5373 since i would not explain better.

4º problem) One is "forced" to work with Cartan-Ehlers models since derivation of NG from GR is impossible and the textbook derivation is incorrect. Therefore from standard GR, the textbooks derivation, is clearly incorrect. I already cited on this also.

5º problem) Even ignoring that, the final result is the "geometrized version" of NG (which is not exactly original NG) which is worked in the limit c --> infinite.

In the specialized literature (Stingray may know this better) people really work with the so-called causality constant which is defined like k = (1/c), but some authors take k = (1/c^2).

therefore above limit read k --> 0. But in the limit k --> 0 even if it was well defined, one know by standard GR that metric used in solar systems test (which I'm sorry to say this JesseM is not flat and is not a Robertson Walker one which is falt in an average sense for the whole universe) that the metric is

(1, -1, -1, -1)

anyone can check this from a GR textbook since the functions

A = (1 - 2phy/c^2) and B = - (1/A) enter on the metric and

(1 - 2phy/c^2) --> 1

and for a flat metric there is no gravitation in GR.

Any textbook, online course, Arxiv, preprint, or paper where NG was rigorously derived from GR? I am thinking that derivation is a myth.

 

[pervect]

Here's the section that caught my eye:

(a) Birkhoff's theorem
Birkhoff's Teorem states that the gravitational effect of a uniform medium
external to a spherical cavity is zero." This is a theorem from general relativity,
and necessary to know in order to extrapolate our Newtonian cosmology results to
the whole universe: it might have been the case that the global curvature of space
would have interfered with our Newtonian results. The other choices in the question
were generally true statements from other areas of cosmology.

I'm still not sure if the above statement is correct or not, unfortunately - it doesn't resemble any traditional statement of Birkhoff's theorem that I recall seeing.

 

[Berislav]

Yes i forgot a minus sign. In solar systems test the limit c --> infinite gives

g00 = 1 and gRR = - 1

passing to cartesian coordinates again the metric is (1, -1, -1, -1) I am using trace -2.

g_{00}=c^2 \left( 1-\frac{2GM}{c^2 r} \right)


As you can see the metric blows up in that limit. This is because time has no geometric structure, or meaning as such, in Newtonian physics.

But, then, metric is FLAT and according to GR there is not gravitation which contradict Newtonian gravity.

No, it blows up.

But taking an origin for the potential the potential itself become physical.

No, it doesn't. That's called gauge fixing.

Dirac explicitly attack renormalization procedure there,

Which doesn't have anything to do with the limit h--->0.

But it is unnecesary since it is well-known that QFT is NOT QM.

No one said that it was. It was derived from QM as a generalization of it.

x is a operator in QM but is a parameter in QFT.

We have a lot of freedom in chosing our parameters in quantum physics. x is chosen as another parameter to put in on equal footing with time and because we're no longer dealing with a single particle but a field in space. It is possible to define both as operators, for instance, but that wouldn't change the underlaying principles of the theory.

It is not true that the Klein-Gordon equation was just a relativistic version of the Schrödinger equation.

See, for instance, Griffiths' Introduction to Elementary Particles p.213-215.

 

[JesseM]

therefore above limit read k --> 0. But in the limit k --> 0 even if it was well defined, one know by standard GR that metric used in solar systems test (which I'm sorry to say this JesseM is not flat and is not a Robertson Walker one which is falt in an average sense for the whole universe)

Wait, are you saying the metric used in the GR analysis of the solar system does not approach flatness as the distance from the solar system approaches infinity? That's all that I ever said, I thought that was part of the meaning of "asymptotic flatness". I did not claim that this was a Robertson-Walker metric, although I did guess that a spatially flat Robertson-Walker universe would approach minkowski spacetime in the limit as cosmological time approached infinity and the expansion rate approached zero, and pervect seemed to say this might be correct.

 

[Juan R.]

g_{00}=c^2 \left( 1-\frac{2GM}{c^2 r} \right)

As you can see the metric blows up in that limit. This is because time has no geometric structure, or meaning as such, in Newtonian physics.

I used (ct, x). You are introduced c into the metric which is not standard, but in any case when c become more and more large your g00 defines a flat spacetime. Take c = 10^50 and after 10^500. Eact time spacetime is more flat.

Using your metric you obtain g00 = infinite and gRR = -1

No, it blows up.

Exactly does not blows, simply one need more care on work with that divergence. This is reason that research in the limit c --> infinite is done via NC theory and not from standard presentation of GR. But NC is obtained from a 're-geometrization' of GR.

No, it doesn't. That's called gauge fixing.

I was talking of Newtonian potential, which is physical once you fix the origin of the potential. This is reason that Newtonian potential is defined like

Phy = -GM/R

instead of Phy = -GM/R + cte or similar.

Of course, in field theory there is gauge theory, but taking a gauge the potential used has full physical sense (related to the choosed gauge).

No one said that it was. It was derived from QM as a generalization of it.

Hum! Weinberg is ambiguous here.

We have a lot of freedom in chosing our parameters in quantum physics. x is chosen as another parameter to put in on equal footing with time and because we're no longer dealing with a single particle but a field in space. It is possible to define both as operators, for instance, but that wouldn't change the underlaying principles of the theory.

Hum not true. x is an observable in NRQM, but only a parameter in RQFT. If you elevate time to range of observable then the structure of RQFT is very different and you are working with a different theory.

After of saying that both NRQM and QM are incompatible. Dirac asked that we need a new relativistic formulation, far from RQFT.

See, for instance, Griffiths' Introduction to Elementary Particles p.213-215.

The KG equation is not a consistent relativistic generalization of Schrödinger equation. This is the reason that was abandoned in RQFT where the evolution equation is a Schrödinguer like equation. Also the Dirac equation was abandoned in RQFT. Weinberg manual is cristal clear.

After both KG and Dirac lagrangians define fields of bosons and fermions. BUT are equations for fields, are not the original equations for wavefunctions.

For example

(i partial + m) Phy = 0

in QFT is NOT the Dirac wave equation even if looks 'close'. It is an identity for the fermion electronic field Phy(x,t). See the Weinberg volume 1.

 

[Juan R.]

Wait, are you saying the metric used in the GR analysis of the solar system does not approach flatness as the distance from the solar system approaches infinity? That's all that I ever said, I thought that was part of the meaning of "asymptotic flatness". I did not claim that this was a Robertson-Walker metric, although I did guess that a spatially flat Robertson-Walker universe would approach minkowski spacetime in the limit as cosmological time approached infinity and the expansion rate approached zero, and pervect seemed to say this might be correct.

No i am not saying that.

If you take R --> infinite the metric is (1, -1, -1, -1) and that is asymptotically flat. But that metric is defined for the Solar system and if you take R more large than solar system, then the metric is not that, because you may introduce the curvature of other sources of matter. You can understand this easily.

Take the direction on alfa Centaur from the Sun. At large distance of the Sun, but in the Solar system, the metric is valid. Now take R = distance to alpha centaur. There according to you initial metric curvature of spacetime would be zero or close to zero (because Solar system metric was derived asuming un Universe formed only by the Sun) but close to alpha centaur the real (observed) metric is very different from flat one.

Similar questions appliy to Ehler boundary. He assumes that when R is more and more great, the quantity of matter in the universe is more and more insignificant until beyonf certain limit there only vaccuum. This is the reason that is called the 'island asumption' and there is no evidence that was correct. In fact the distribution of matter is not more and more small for large distances

Ehlers universe look like

______________X_X_XXX_X_XXXXXXX_X_X_XXX_X_X_XXX_________


Our universe does not look like 'an island of matter surrounded by emptiness'. This is the reason that Ehlers work is not completely accepted. It looks 'like'

X_XXX_X_XXXXXXX_X_X_X_X_XXX_X_XXXXXXX_X__X_XXX_X_X_XXX

The Shwartzild metric asumes that our universe is

_____________________________S_____________________________

with S the Sun. Which obiously is not correct at galactic scales. But is a very good approximation inside the Solar system.

 

[Juan R.]

But what about Wald p. 138, 139? The effective potential equation (6.3.15) is a good example of how GR and Newtonian physics differ by a factor. You will notice that it doesn't contain coordinate time and hence can be reduced directly to Newtonian gravity.

i already read the Wald again. The equation (6.3.15) does not contain Newton potential, contains the retarded field that follows from GR. There is significant difference on funtional forms. For Newton Phy = Phy(R(t)). For GR Phy = Phy(x, t). From the GR functional dependence one cannot explain all phenomena (i cited on this but in similar problem on EM). Moreover, astronomers uses Newtonian potential newer GR retarded field for the computation of orbits, due to experimental absence of gravitational aberration and other issues (like stability of orbits in numerical computations).

Moreover equation (6.3.15) is derived from (6.3.10) which is parametrized for kappa 1 and 0. Taking c infinite, you cannot maintain kappa timelike.

 

[Juan R.]

I posted some data on sci.physics.relativity about this.

Surprinsingly, i received two types of replies:

1) You are wrong because the limit c --> infinite cannot be taken in GR. Tom Roberts said "... and you are being too naive. c->infinity removes gravitation from GR"

2) You are wrong because the limit c --> infinite has already obtained in NC formulation. This was the point of renowed specialist S. Carlip. However, Carlip cited references (except one) that i had already studied and cited here. For example, the paper Commun. Math. Phys. 166, 221-235 (1994)

It is really interesting the confusion in this topic, one says that the limit does not exist "therefore JR wrong", other claims that limit exists, "therefore JR wrong" again. Obviously i cannot be wrong in both cases at the same time

Unfortunately there is many 'noise' in sci.physics.relativity i read many times 'Crazy moron' and similar. I launched a post in moderated sci.physics.research.

I got reply by Igor Khavkine today, unfortunately reply is wrong and even trivial. for example on

"The theorem stating that gtr does indeed go over to Newtonian gravitostatics in the very weak field, very slow motion limit is proven in detail in almost every gtr textbook."

Igor states that

That is indeed true.

simply compare with claim from an specialist in the topic (Bernard F. Schutz, "The Newtonian Limit") reference introduced in PF by robphy (Thanks!) in #30.

there are at least two reasons why the simple textbook extractions of the Newtonian limit are not rigorous.

 

[Juan R.]

Some interesting discussion on the topic began with several relativists including renowned Steve Carlip. However, in my personal opinion -please do not me atack because i am thinking this now-, Carlip is wrong in several crucial details doing his attempt to prove that Newtonian gravity is derived from General relativity wrong.

So far like i can see Carlip has not proved that curvature interpretation follows in the Newtonian limit; has not proved how spacetime quantitites transform into Newtonian potentials; has not proven that one obtain full Newtonian gravity, etc.

For example, in my prescription x^0 = ct, one obtains full physical sense for flat (Newtonian) derivatives. Carlip, by chossing x^0 = t, obtains that physical derivative is covariant one in the Newtonian regime due that 00-connection is not zero in his approach!

There exist more difficulties. More data of interest and references on sci.physics.research

http://www.lns.cornell.edu/spr/2005-10/msg0071918.html

P.S: For moderators. I do not find the direct link to sci.physics.research here in https://www.physicsforums.com/forumdisplay.php?f=123.