## Spherically symmetric space is conformally flat

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### Spherically symmetric space is conformally flat

An example of a spherically symmetric space can be found in Swarzschild's solution of Einstein's general relativity: a space-like slice where Swarzschild's time coordinate is constant. In study of this and similar models it may be interesting that such spaces are coformally flat, except possibly at the center of the spherical symmetry.

Our assumption is that the space is three dimensional and its geometry is spherically symmetric. The assumption can be epressed with the formula
ds² = A(r)² dr² + B(r)² (dθ² + dφ² sin² θ)
for line element in terms of sherical coordinates, where A and B are two positive-valued functions except that B(0) = 0.

The theorem we shall prove is that where r > 0 the space is conformally flat. This can be expressed with the formula
ds² = C²(r) (dR(r)² + R(r)²(dθ² + dφ² sin² θ))
for line element where C and R are some functions. The theorem is that it is possible to choose these functions so that this formula for line element is the same as the first formula.

In order to prove the theorem we construct C and R in terms of A and B and substitute in the latter formula:
R(r) = exp(∫ₐʳ A(u) / B(u) du) where a > 0 but otherwise the value of a is not important,
and
C(r) = B(r) / R(r).
R(0) (and consequently C(0)) may be ill defined as the expression contains a division by B(u) and B(0) = 0. This is does not matter as we have excluded r = 0 from our theorem.

The substitution is quite simple when we observe that
dR(r) = R(r) A(r) dr / B(r)
and the expression for R need not be used; instead,
C(r) R(r) = B(r)
and
C(r) dR(r) = A(r) dr
can be substitued, giving
ds² = A(r)² dr² + B(r)² (dθ² + dφ² sin² θ)
Q.E.D.
Mikko

Posts: 138
Joined: Mon Feb 17, 2014 1:53 am

### Re: Spherically symmetric space is conformally flat

Mikko wrote:An example of a spherically symmetric space can be found in Swarzschild's solution of Einstein's general relativity: a space-like slice where Swarzschild's time coordinate is constant. In study of this and similar models it may be interesting that such spaces are coformally flat, except possibly at the center of the spherical symmetry.

Our assumption is that the space is three dimensional and its geometry is spherically symmetric. The assumption can be epressed with the formula
ds² = A(r)² dr² + B(r)² (dθ² + dφ² sin² θ)
for line element in terms of sherical coordinates, where A and B are two positive-valued functions except that B(0) = 0.

The theorem we shall prove is that where r > 0 the space is conformally flat. This can be expressed with the formula
ds² = C²(r) (dR(r)² + R(r)²(dθ² + dφ² sin² θ))
for line element where C and R are some functions. The theorem is that it is possible to choose these functions so that this formula for line element is the same as the first formula.

In order to prove the theorem we construct C and R in terms of A and B and substitute in the latter formula:
R(r) = exp(∫ₐʳ A(u) / B(u) du) where a > 0 but otherwise the value of a is not important,
and
C(r) = B(r) / R(r).
R(0) (and consequently C(0)) may be ill defined as the expression contains a division by B(u) and B(0) = 0. This is does not matter as we have excluded r = 0 from our theorem.

The substitution is quite simple when we observe that
dR(r) = R(r) A(r) dr / B(r)
and the expression for R need not be used; instead,
C(r) R(r) = B(r)
and
C(r) dR(r) = A(r) dr
can be substitued, giving
ds² = A(r)² dr² + B(r)² (dθ² + dφ² sin² θ)
Q.E.D.

Nice effort there Mikko, but your Q.E.D is misplaced. That formally correct derivation is I believe the standard way ISM was derived in the first place. Part 3 of article (v3) deals with this by noting true conformal non-flatness has local physical significance - locally non-Euclidean geometry. Standard SM is not conformally flat.
As shown there by considering the differential area/volume ratio, or differential great circle circumference/radial gap ratio. ISM pretends to be at one and the same time conformally flat, yet also representing the same physical metric as SM. A logical impossibility. You either choose ISM as representing a physically different metric - true conformal flatness = locally Euclidean geometry, or that the latter is a misleading artifact of coordinate transformation thus fake. You may say "but you have agreed the derivation is correct!". Formally correct but necessarily having a fatal subtle flaw as just described. What makes sense to me is that the metric given by the transformed coordinate system is 'isomorphic' to the original only in a limited strictly local sense - about some given radial coordinate value. Making it in fact essentially useless. Not something generally recognized obviously but otherwise how to explain the now obvious contradiction?

[when I get time, will do a simple test - compare SM vs ISM for two radius-to-circumference values. That will definitely decide between ISM as actually equivalent to SM or not. Not that it ultimately matters.]
Q-reeus

Posts: 272
Joined: Sat Jun 07, 2014 11:18 pm

### Re: Spherically symmetric space is conformally flat

Q-reeus wrote:
Mikko wrote:An example of a spherically symmetric space can be found in Swarzschild's solution of Einstein's general relativity: a space-like slice where Swarzschild's time coordinate is constant. In study of this and similar models it may be interesting that such spaces are coformally flat, except possibly at the center of the spherical symmetry.

Our assumption is that the space is three dimensional and its geometry is spherically symmetric. The assumption can be epressed with the formula
ds² = A(r)² dr² + B(r)² (dθ² + dφ² sin² θ)
for line element in terms of sherical coordinates, where A and B are two positive-valued functions except that B(0) = 0.

The theorem we shall prove is that where r > 0 the space is conformally flat. This can be expressed with the formula
ds² = C²(r) (dR(r)² + R(r)²(dθ² + dφ² sin² θ))
for line element where C and R are some functions. The theorem is that it is possible to choose these functions so that this formula for line element is the same as the first formula.

In order to prove the theorem we construct C and R in terms of A and B and substitute in the latter formula:
R(r) = exp(∫ₐʳ A(u) / B(u) du) where a > 0 but otherwise the value of a is not important,
and
C(r) = B(r) / R(r).
R(0) (and consequently C(0)) may be ill defined as the expression contains a division by B(u) and B(0) = 0. This is does not matter as we have excluded r = 0 from our theorem.

The substitution is quite simple when we observe that
dR(r) = R(r) A(r) dr / B(r)
and the expression for R need not be used; instead,
C(r) R(r) = B(r)
and
C(r) dR(r) = A(r) dr
can be substitued, giving
ds² = A(r)² dr² + B(r)² (dθ² + dφ² sin² θ)
Q.E.D.

Nice effort there Mikko, but your Q.E.D is misplaced.

Not at all, the Q.E.D. should be at the end and there it is. It's the text before it that should be written differently. At least there should be a bit of prose
after the last formula but a different arrangement of the formulas would have permitted a better presentation. The core of the proof is in the formulas,
and their relationships to each other and to the theorem are expressed clearly, in spite of poor presentation.
That formally correct derivation is I believe the standard way ISM was derived in the first place. Part 3 of article (v3) deals with this by noting true conformal non-flatness has local physical significance - locally non-Euclidean geometry.

What you mean by "locally"? As usually understood, every spacetime is locally Minkowskian except at singularities and therefore spatial slices (whether isochronous or not) are locally Euclidean. If you allow "locally" to include less local measures then you will soon have locally non-Euclidean (or non-Minkowskian) geometries in every curved space (or spacetime).
Standard SM is not conformally flat. As shown there by considering the differential area/volume ratio, or differential great circle circumference/radial gap ratio. ISM pretends to be at one and the same time conformally flat, yet also representing the same physical metric as SM. A logical impossibility. You either choose ISM as representing a physically different metric - true conformal flatness = locally Euclidean geometry, or that the latter is a misleading artifact of coordinate transformation thus fake. You may say "but you have agreed the derivation is correct!". Formally correct but necessarily having a fatal subtle flaw as just described. What makes sense to me is that the metric given by the transformed coordinate system is 'isomorphic' to the original only in a limited strictly local sense - about some given radial coordinate value. Making it in fact essentially useless. Not something generally recognized obviously but otherwise how to explain the now obvious contradiction?

Contradiction is not obvious until you can prove that 1 = 2. You have claimed that that some space or spacetime is or is not conformally flat depending on how you look at it but you have not proved anything (which you could do e.g. by computing the Cotton tensor).
[when I get time, will do a simple test - compare SM vs ISM for two radius-to-circumference values. That will definitely decide between ISM as actually equivalent to SM or not. Not that it ultimately matters.]

Radius is a little bit difficult. The centre of the Swartschild solution is singular. If you replace the interior part with a non-vacuum solution, you must integrate through heterogenous space. It is easier to use two spheres at constant radii so that their areas are A and A+dA so that both are in the vacuum region, and check the distance between these spheres.
Mikko

Posts: 138
Joined: Mon Feb 17, 2014 1:53 am

### Re: Spherically symmetric space is conformally flat

Mikko wrote:Not at all, the Q.E.D. should be at the end and there it is. It's the text before it that should be written differently. At least there should be a bit of prose
after the last formula but a different arrangement of the formulas would have permitted a better presentation. The core of the proof is in the formulas,
and their relationships to each other and to the theorem are expressed clearly, in spite of poor presentation.

And as acknowledged in earlier post, your derivation is technically correct. But imo something subtle has to be amiss. i gave my hunch then - strictly local equivalence only, for some given r.
What you mean by "locally"? As usually understood, every spacetime is locally Minkowskian except at singularities and therefore spatial slices (whether isochronous or not) are locally Euclidean. If you allow "locally" to include less local measures then you will soon have locally non-Euclidean (or non-Minkowskian) geometries in every curved space (or spacetime).

As you know I have not pretended to be conversant with the mathematical structure of GR. And simply offer in article a layman outsider's imo fresh insight not some high powered mathematically dense tensor based proof. But note from the literature that true conformal flatness (4D) is equivalent to having zero Weyl curvature. And that Weyl curvature is an intensive property existing 'at a point' thus having local existence. Analogously to say non-zero divergence or curl in EM, though existing 'at a point', the measurable effects vanish as one restricts closer and closer to a point volume. In EM that corresponds to saying that strictly locally there is just uniform fields E, B. Similarly in gravity case, locally Minkowski, and non-zero Weyl curvature = locally non-Euclidean = conformally non-flat, can coexist without contradiction depending on how both are used and interpreted.

The first example of physical manifestation of locally non-Euclidean I give in part 3, differential radial gap between nesting spherical surfaces, was first explained to me years ago by a GR pro in PhysicsForum., but only in terms of proper area/volume ratio. Quickly figured out for myself the simpler equivalent second example rather easily. So Anyway if ISM as its name implies truly is coordinate measured isotropic that means a locally spherical wavefront light pulse is also initially spherical as determined by a coordinate observer. Which via elementary geometric reasoning automatically implies conformal flatness = zero Weyl curvature actual.
But in SM case that same locally spherical wsavefront is determined to be oblate spheroidal (oblate axis directed radially from central gravitating mass source) by that same coordinate observer. The signature of conformal non-flatness = locally non-Euclidean geometry. Manifested locally by, as given in article, say measuring radial gap between closely spaced concentric great circles.

My argument then, as given numbers of times, is that ISM, which is a particular manifestation of your derived final result, is either truly equivalent to SM thus not actually isotropic thus not actually conformally flat, or it is and thus cannot represent the same physical metric as SM. Regardless that both can be strictly locally Minkowski.
Contradiction is not obvious until you can prove that 1 = 2. You have claimed that that some space or spacetime is or is not conformally flat depending on how you look at it but you have not proved anything (which you could do e.g. by computing the Cotton tensor).

What I write above and have elsewhere is imo a logically sound deduction. For strict formal mathematical proof well as you evidently have a much better grounding in formalities of GR, maybe you could do the honors and apply said Cotton tensor to both SM and ISM and let's see what then comes out! But as I say, evidently the determining factor is zero or not Weyl curvature: https://en.wikipedia.org/wiki/Conformally_flat_manifold (not clear to me one can ever ignore temporal component in shell or similar situation).
Radius is a little bit difficult. The centre of the Swartschild solution is singular. If you replace the interior part with a non-vacuum solution, you must integrate through heterogenous space. It is easier to use two spheres at constant radii so that their areas are A and A+dA so that both are in the vacuum region, and check the distance between these spheres.

Agree it is tricky - to the extent that one needs to integrate over a proper radial path if comparing appreciably radially separated spherical surfaces. And in ISM case that almost certainly requires numerical integration thus messy and somewhat ad hoc. That's something I realized soon after posting that bit and is why haven't got round to attempting it. But otherwise there is no call to evaluate anywhere other than exterior region as the problem is comparing SM and ISM which both apply to exterior vacuum region. I fall back on that the very form of ISM implies conformal flatness thus cannot claim to be physically equivalent to SM.

Moreover, as posted elsewhere:
If in a given coordinate system one encounters paradox, fleeing to another to escape it is senseless logic. It avoids that paradox existing in any supposedly sound coordinate system i.e. SM - the unique solution of EFE's for spherically symmetric static mass distribution, undermines such there and then. Period. And we are not talking about anything remotely like a so-called coordinate singularities situation such as 'black hole' 'event horizon' in SM. On the contrary the issues appear in the extreme weak gravity arena.

And further quoting from there:
What is my point there? Pretty obvious really. Demanding both external SM and internal MM are true inevitably forces a vanishing of functional dependence on gravitational potential for radial spatial metric factor $1/\sqrt{{-g}_{rr }}$ in interior. But not $\sqrt{{g}_{tt }}$ for clocks. Despite them being functionally identical in the SM exterior region. I say that is physical nonsense, an artifact that signifies the force fit, though non-drastic locally in terms of metric curvatures, is wrong in principle. There is more to a physically sensible metric transition than just local curvature concerns take care of. The shell wall cannot sensibly be acting as a magic membrane that vanishes one component's potential dependence on just one side. I say choose instead a conformally flat exterior metric and there is no such issue to consider. That alone is imo enough to choose the latter and reject SM.

But I go on in the next paragraph to summarize part 2 of article - the second and even stronger reason to reject SM. Direct summation over shell yields a predicted internal spatial metric, evaluated in particular for case of field point at r = 0 center, that IS a depressed function of Newtonian potential, in direct contradiction to above 'consistency requirement'. Any theory that predicts such paradox just has to be wrong somewhere. And once again, perform the same direct evaluation assuming conformally flat exterior metric, there is no paradox, no issue, everything naturally fits. Hence imo one has to reject SM thus GR as fatally flawed and choose the paradox free alternative - a genuinely conformally flat exterior metric. Of which Yilmaz gravity is a prime example.
Q-reeus

Posts: 272
Joined: Sat Jun 07, 2014 11:18 pm

### Re: Spherically symmetric space is conformally flat

Q-reeus wrote:And as acknowledged in earlier post, your derivation is technically correct.

I.e., any spherically symmetric space is conformally flat (except possibly at the centre).
But imo something subtle has to be amiss. i gave my hunch then - strictly local equivalence only, for some given r.
What you mean by "locally"? As usually understood, every spacetime is locally Minkowskian except at singularities and therefore spatial slices (whether isochronous or not) are locally Euclidean. If you allow "locally" to include less local measures then you will soon have locally non-Euclidean (or non-Minkowskian) geometries in every curved space (or spacetime).

As you know I have not pretended to be conversant with the mathematical structure of GR.

That's OK here, the intended topic of this thread does not include GR.
And simply offer in article a layman outsider's imo fresh insight not some high powered mathematically dense tensor based proof. But note from the literature that true conformal flatness (4D) is equivalent to having zero Weyl curvature.

In three dimensions conformal flatness is equivalent to having zero Cotton tensor.
And that Weyl curvature is an intensive property existing 'at a point' thus having local existence.

So is the Cotton tensor.
Analogously to say non-zero divergence or curl in EM, though existing 'at a point', the measurable effects vanish as one restricts closer and closer to a point volume. In EM that corresponds to saying that strictly locally there is just uniform fields E, B. Similarly in gravity case, locally Minkowski, and non-zero Weyl curvature = locally non-Euclidean = conformally non-flat, can coexist without contradiction depending on how both are used and interpreted.

Indeed. It all depends on how one interpreters "local". As a first approximation the space is Euclidean (or spacetime Minkowskian) and Cotton (or Weyl) tensor cannot be measured. As a second approximation they and other curvature tensors can be measured, and flatness and conformal flatnes become meaningful questions.
The first example of physical manifestation of locally non-Euclidean I give in part 3, differential radial gap between nesting spherical surfaces, was first explained to me years ago by a GR pro in PhysicsForum., but only in terms of proper area/volume ratio. Quickly figured out for myself the simpler equivalent second example rather easily. So Anyway if ISM as its name implies truly is coordinate measured isotropic that means a locally spherical wavefront light pulse is also initially spherical as determined by a coordinate observer.

A sufficiently small region is always flat so a wavefront (from a point source) always starts as spherical. When curvature becomes significant one must be careful with words like "spherical" as their meaning is not always obvious and also because their implications are not always the same as in a flat spacetime.
Which via elementary geometric reasoning automatically implies conformal flatness = zero Weyl curvature actual.

That is true with four (or more) dimensions but not with three (or less). Note that in Swatzschild's metric the Weyl tensor is non-zero and the spacetime is not conformally flat.
But in SM case that same locally spherical wsavefront is determined to be oblate spheroidal (oblate axis directed radially from central gravitating mass source) by that same coordinate observer.

It starts as spherical in the sense that its physical size is the same in all dimensions. There are no coordinate observers as coordinates are unphysical and therefore not physically observable. Of course, if the coordinates are attached to physical phenomena, those phenomena can be observable but they are equally observable when coordinates are defined differently.
The signature of conformal non-flatness = locally non-Euclidean geometry.

Not true. Whether the geometry is locally Euclidean or not depends on how locally you examine it (unless the space really is Euclidean). For example, the surface of a sphere is conformally flat but non-Euclidean (unless oserved too locally).
Manifested locally by, as given in article, say measuring radial gap between closely spaced concentric great circles.

My argument then, as given numbers of times, is that ISM, which is a particular manifestation of your derived final result, is either truly equivalent to SM thus not actually isotropic thus not actually conformally flat, or it is and thus cannot represent the same physical metric as SM.

As long as you need to say "or" you don't understand the situation sufficiently well. Anyway, it is easy to prove that they are equivalent.
Regardless that both can be strictly locally Minkowski.
Contradiction is not obvious until you can prove that 1 = 2. You have claimed that that some space or spacetime is or is not conformally flat depending on how you look at it but you have not proved anything (which you could do e.g. by computing the Cotton tensor).

What I write above and have elsewhere is imo a logically sound deduction.

A deduction should be a sequence of steps, not leaps. Otherwise its soundness is not obvious. Your deductions tend to have too long leaps near the end.
For strict formal mathematical proof well as you evidently have a much better grounding in formalities of GR, maybe you could do the honors and apply said Cotton tensor to both SM and ISM and let's see what then comes out! But as I say, evidently the determining factor is zero or not Weyl curvature: https://en.wikipedia.org/wiki/Conformally_flat_manifold (not clear to me one can ever ignore temporal component in shell or similar situation).
Radius is a little bit difficult. The centre of the Swartschild solution is singular. If you replace the interior part with a non-vacuum solution, you must integrate through heterogenous space. It is easier to use two spheres at constant radii so that their areas are A and A+dA so that both are in the vacuum region, and check the distance between these spheres.

Agree it is tricky - to the extent that one needs to integrate over a proper radial path if comparing appreciably radially separated spherical surfaces.

If spheres are too tricky, someting else might be more convincing. Sum of angles of a triangle is another demonstration of curvature. In order to prove that two presentations do not present the same space (or spacetime) it is sufficient to calculate some observable in both and get a different value. The hardest part is to prove that the calculated thing is the same in both cases, which may involve claculation of related observables and getting the same value. Good observables include angles, distances, curvature tensors, and for spacetime durations and speeds.
And in ISM case that almost certainly requires numerical integration thus messy and somewhat ad hoc. That's something I realized soon after posting that bit and is why haven't got round to attempting it. But otherwise there is no call to evaluate anywhere other than exterior region as the problem is comparing SM and ISM which both apply to exterior vacuum region. I fall back on that the very form of ISM implies conformal flatness thus cannot claim to be physically equivalent to SM.

Coordinates are unphysical so the choice of coordinates cannot have physical consequences.
Moreover, as posted elsewhere:
If in a given coordinate system one encounters paradox, fleeing to another to escape it is senseless logic.

Not at all. In order to determine whether the coordinate system is paradoxical it is useful to present the situation in another coordinate system. If the paradox disappears, one may suspect the original coordinate system, otherwise it may be better to search elsewhere.
It avoids that paradox existing in any supposedly sound coordinate system i.e. SM - the unique solution of EFE's for spherically symmetric static mass distribution, undermines such there and then. Period. And we are not talking about anything remotely like a so-called coordinate singularities situation such as 'black hole' 'event horizon' in SM. On the contrary the issues appear in the extreme weak gravity arena.

The rest is so far from the intended topic of this discussion that I don't want to comment.
And further quoting from there:
What is my point there? Pretty obvious really. Demanding both external SM and internal MM are true inevitably forces a vanishing of functional dependence on gravitational potential for radial spatial metric factor $1/\sqrt{{-g}_{rr }}$ in interior. But not $\sqrt{{g}_{tt }}$ for clocks. Despite them being functionally identical in the SM exterior region. I say that is physical nonsense, an artifact that signifies the force fit, though non-drastic locally in terms of metric curvatures, is wrong in principle. There is more to a physically sensible metric transition than just local curvature concerns take care of. The shell wall cannot sensibly be acting as a magic membrane that vanishes one component's potential dependence on just one side. I say choose instead a conformally flat exterior metric and there is no such issue to consider. That alone is imo enough to choose the latter and reject SM.

But I go on in the next paragraph to summarize part 2 of article - the second and even stronger reason to reject SM. Direct summation over shell yields a predicted internal spatial metric, evaluated in particular for case of field point at r = 0 center, that IS a depressed function of Newtonian potential, in direct contradiction to above 'consistency requirement'. Any theory that predicts such paradox just has to be wrong somewhere. And once again, perform the same direct evaluation assuming conformally flat exterior metric, there is no paradox, no issue, everything naturally fits. Hence imo one has to reject SM thus GR as fatally flawed and choose the paradox free alternative - a genuinely conformally flat exterior metric. Of which Yilmaz gravity is a prime example.
Mikko

Posts: 138
Joined: Mon Feb 17, 2014 1:53 am

### Re: Spherically symmetric space is conformally flat

Mikko wrote:I.e., any spherically symmetric space is conformally flat (except possibly at the centre).

Not true - as seemingly acknowledged by yourself below!:
"...Note that in Swatzschild's metric the Weyl tensor is non-zero and the spacetime is not conformally flat."

To resolve such apparent obvious contradiction I could guess you are meaning to imply above that non-zero Weyl curvature in SM is somehow an 'artifact' of SM which can be legitimately erased by resort to ISM. No other interpretation quite makes any sense to me. Assuming that guess is correct, it imo is equivalent to saying a non-zero curl E computed in say Cartesian coordinate system is just a non-physical artifact that can be banished by use of some other coordinate system. In which case we fundamentally disagree on some basics here.
What you mean by "locally"? As usually understood, every spacetime is locally Minkowskian except at singularities and therefore spatial slices (whether isochronous or not) are locally Euclidean. If you allow "locally" to include less local measures then you will soon have locally non-Euclidean (or non-Minkowskian) geometries in every curved space (or spacetime).

Well you quote me later on as explaining just what was meant by locally.
Q-reeus: "And that Weyl curvature is an intensive property existing 'at a point' thus having local existence."
Mikko: "Indeed. It all depends on how one interpreters "local". As a first approximation the space is Euclidean (or spacetime Minkowskian) and Cotton (or Weyl) tensor cannot be measured. As a second approximation they and other curvature tensors can be measured, and flatness and conformal flatnes become meaningful questions."

Yes and no and maybe with you it seems - I honestly cannot discern a consistent position.
Q-reeus: "Which via elementary geometric reasoning automatically implies conformal flatness = zero Weyl curvature actual."
Mikko: "That is true with four (or more) dimensions but not with three (or less). Note that in Swatzschild's metric the Weyl tensor is non-zero and the spacetime is not conformally flat."

Have already commented on last bit, and as for 3D vs 4D, well SM or ISM line element has four metric components - one temporal and three spatial.
Q-reeus: "But in SM case that same locally spherical wsavefront is determined to be oblate spheroidal (oblate axis directed radially from central gravitating mass source) by that same coordinate observer."
Mikko: "It starts as spherical in the sense that its physical size is the same in all dimensions. There are no coordinate observers as coordinates are unphysical and therefore not physically observable. Of course, if the coordinates are attached to physical phenomena, those phenomena can be observable but they are equally observable when coordinates are defined differently."

Of course there are coordinate observers - that concept is used all the time in mainstream GR articles. And equally such cannot by definition be locally measuring distant phenomena. That's what SM is all about - telling one what a coordinate observer must infer as to how spherically symmetric static gravity distorts spacetime from Euclidean.
Given transverse metric components are 'Euclidean' in SM, coordinate oblateness of initial wavefront does automatically imply non-Euclidean local geometry - non-zero Weyl curvature for SM. Conversely coordinate predicted initially spherical wavefront in ISM automatically implies the opposite - zero Weyl curvature. Which prediction is imo an actual artifact of SM ->ISM transformation. And this is becoming repetitious.
Not true. Whether the geometry is locally Euclidean or not depends on how locally you examine it (unless the space really is Euclidean). For example, the surface of a sphere is conformally flat but non-Euclidean (unless oserved too locally).

We have already gone over that issue via examples and analogies (and again above), hence this is going in circles.
As long as you need to say "or" you don't understand the situation sufficiently well. Anyway, it is easy to prove that they are equivalent.

I disagree. That 'or' means either case renders ISM useless as argument against paradox appearing in SM. It doesn't matter which position one takes. And given you claim as per title that "any spherically symmetric space is conformally flat", yet simultaneously acknowledge above "...Note that in Swatzschild's metric the Weyl tensor is non-zero and the spacetime is not conformally flat.", I am grappling with a cloud so to speak.
If spheres are too tricky, someting else might be more convincing. Sum of angles of a triangle is another demonstration of curvature. In order to prove that two presentations do not present the same space (or spacetime) it is sufficient to calculate some observable in both and get a different value. The hardest part is to prove that the calculated thing is the same in both cases, which may involve claculation of related observables and getting the same value. Good observables include angles, distances, curvature tensors, and for spacetime durations and speeds.

Will look at getting specific results for radial parameter-to-circumference ratios. Whether or not such confirms or undermines actual equivalence of SM and ISM, that will not and cannot overturn that paradox is present using SM "the unique solution of EFE's for spherically symmetric static mass distribution." - as declared for instance in the two lines below (1.4), top of p5 here:
http://www.diva-portal.org/smash/get/di ... TEXT01.pdf
And many other such mainstream GR articles say the same. Hence if paradox exists in SM setting, which for sure is so, SM and GR go down together. And repeated below since evidently continued repetition is needed here.
Coordinates are unphysical so the choice of coordinates cannot have physical consequences.

And as stated before I say the logical resolution is that SM <-> ISM transformation is a fairly meaningless and misleading formalism that simply allows a nominal 1:1 correspondence for and strictly at any chosen parameter r. Otherwise you have this conundrum of there being a physically real Weyl curvature and not depending on use of SM or ISM. Which is nonsense. Either Weyl or if you like Cotton curvature as proper intensive property is there or not. Just like curl or divergence in EM. And is ultimately irrelevant to thrust of my article even if not to the narrow focus of this thread.
Not at all. In order to determine whether the coordinate system is paradoxical it is useful to present the situation in another coordinate system. If the paradox disappears, one may suspect the original coordinate system, otherwise it may be better to search elsewhere.

Only if 'the chart fails to cover the manifold' to use GR parlance, in some severe 'coordinate singularity' sense as for instance standard external SM fails to handle below the imo fictitious dreaded black hole event horizon. Otherwise there is no legitimacy at all in that argument. And has nothing to do with that necessarily external and internal shell metrics are different. That applying to *any* choice of external metric be it SM, ISM, Yilmaz conformal gravity alternative, or whatever. To repeat:
It avoids that paradox existing in any supposedly sound coordinate system i.e. SM - the unique solution of EFE's for spherically symmetric static mass distribution, undermines such there and then. Period. And we are not talking about anything remotely like a so-called coordinate singularities situation such as 'black hole' 'event horizon' in SM. On the contrary the issues appear in the extreme weak gravity arena.

The rest is so far from the intended topic of this discussion that I don't want to comment.

Your choice, but imo it bears heavily on legitimacy of your position as per title.
Q-reeus

Posts: 272
Joined: Sat Jun 07, 2014 11:18 pm

### Re: Spherically symmetric space is conformally flat

Too late to edit my last post, so here is roughly how i would have edited it:
Mikko wrote:I.e., any spherically symmetric space is conformally flat (except possibly at the centre).

Not true - as seemingly acknowledged by yourself below!:

"...Note that in Swatzschild's metric the Weyl tensor is non-zero and the spacetime is not conformally flat."

To resolve such apparent obvious contradiction I could guess you are meaning to imply above that non-zero Weyl curvature in SM is somehow an 'artifact' of SM which can be
legitimately erased by resort to ISM. No other interpretation quite makes any sense to me. Assuming that guess is correct, it imo is equivalent to saying a non-zero curl E
computed in say Cartesian coordinate system is just a non-physical artifact that can be banished by use of some other coordinate system. In which case we fundamentally
disagree on some basics here.

[color highlighting added here] A partial retraction is in order. I had failed to notice the distinction made there between conformal flatness (or not) in a space vs a spacetime. So ok if restricted to 3D space curvature, Weyl curvature is by definition zero, but not necessarily so for that associated with Cotton tensor. In either case, appropriately compare Weyl curvatures for 4D case or Cotton curvatures for 3D case. In both cases SM will yield non-zero result, and ISM may or may not. Still looking around for an official source presenting a simple formula for calculating Weyl curvature in both SM and ISM.

Whatever is found, it will not alter that, working consistently with SM, the paradoxical findings in parts 1 and 2 of article are there. And as per part 3, 'vanishing' the paradox via use of ISM can in no way be considered a resolution. It just shows that SM and ISM cannot be truly equivalent. If they were, paradox would be present or not in both cases. Logical necessity.

And I had made a slip-up in part 3 where expression for Yilmaz metric incorrectly identifies the $\phi}$ exponent there with Newtonian potential but in fact the relation is
$\phi=2{\phi}_{n}/{c}^{2}$, where ${\phi}_{n}$ is the Newtonian potential. Will correct in an upcoming 4th version.
Q-reeus

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### Re: Spherically symmetric space is conformally flat

Q-reeus wrote:So ok if restricted to 3D space curvature, Weyl curvature is by definition zero, but not necessarily so for that associated with Cotton tensor. In either case, appropriately compare Weyl curvatures for 4D case or Cotton curvatures for 3D case. In both cases SM will yield non-zero result, and ISM may or may not. Still looking around for an official source presenting a simple formula for calculating Weyl curvature in both SM and ISM.

Weyl curvature is not a property of the coordinate system but of the space, so calculation from either coordinate system gives the same result. For Swarzschild's solution that result is non-zero. There are non-vacuum solutions with zero Weyl tensor. The 3D spatial geometry is spherically symmetric and therefore the Cotton tensor is zero. Cotton tensor is not necessary for a proof of conformal flatness, and I did my proof without it; but a proof of conformal non-flatness seems to require it.
Will correct in an upcoming 4th version.

You should also correct the errors about conformal flatness. They are not essential, just distracting.
Mikko

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### Re: Spherically symmetric space is conformally flat

Mikko wrote:The 3D spatial geometry is spherically symmetric and therefore the Cotton tensor is zero.

Can you quote the literature supporting that claim?
Q-reeus

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### Re: Spherically symmetric space is conformally flat

Q-reeus wrote:
Mikko wrote:The 3D spatial geometry is spherically symmetric and therefore the Cotton tensor is zero.

Can you quote the literature supporting that claim?

You already found some web pages showing that conformally flat 3D space implies zero cotton page, and in the first message I have shown that spherical symmetry implies conformal flatness. There are many pages about Cotton tensor and conformal flatness but not so many about conformal flatness of spherically symmetric space.
Mikko

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### Re: Spherically symmetric space is conformally flat

Mikko wrote:You already found some web pages showing that conformally flat 3D space implies zero cotton page,...

If you are referring to the Wiki article https://en.wikipedia.org/wiki/Conformally_flat_manifold (post #4), yes it specifies that. What it doesn't say there is what you think it implies - that *any* spherically symmetric space is conformally flat.
...and in the first message I have shown that spherical symmetry implies conformal flatness.

That is obviously how you interpret it, but I stick by earlier responses in posts #2, #4, #6, #7 (grrrr... can we not have post numbering incorporated here!?)
Look, you are claiming SM has 3D conformal flatness = zero Cotton tensor. We agree I hope that zero Cotton tensor for 3D case is synonymous with locally Euclidean. Well here's your challenge. In part 3 of article, I gave two specific examples of spatially non-Euclidean character of SM. See if you can refute my more easily determined 2nd example - that the proper measure of differential radial gap between closely spaced concentric great circles (centered about a central point mass) is greater by factor $\1/\sqrt{{g}_{tt }}$ than for matter-free Euclidean case. It has to be so.
There are many pages about Cotton tensor and conformal flatness but not so many about conformal flatness of spherically symmetric space.

And I will guarantee you will never find a page in regular literature equating conformal flatness to *any* spherically symmetric space.
Q-reeus

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### Re: Spherically symmetric space is conformally flat

Just to be clear that where I twice wrote "*any* spherically symmetric space" in above post, the meaning was in the general sense i.e. *every*. Apart from the one unambiguous case of spherically symmetric space with zero conformal flatness - empty i.e. matter-free, Yilmaz gravity equivalent to SM is as per article locally Euclidean thus conformally flat.
Q-reeus

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### Re: Spherically symmetric space is conformally flat

Q-reeus wrote:
Mikko wrote:There are many pages about Cotton tensor and conformal flatness but not so many about conformal flatness of spherically symmetric space.

And I will guarantee you will never find a page in regular literature equating conformal flatness to *any* spherically symmetric space.

Some pages mention the result without proof or reference.
Mikko

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### Re: Spherically symmetric space is conformally flat

Mikko wrote:
Q-reeus wrote:
Mikko wrote:There are many pages about Cotton tensor and conformal flatness but not so many about conformal flatness of spherically symmetric space.

And I will guarantee you will never find a page in regular literature equating conformal flatness to *any* spherically symmetric space.

Some pages mention the result without proof or reference.

Umm - references please, preferably to freely accessible online material. Or a direct cut-and-paste quote, in full context, from mainstream textbook. I extremely doubt the eventuality for reasons already given - SM is for sure not only 4D conformally non-flat = non-zero Weyl curvature, but spatial part also conformally non-flat - non-zero Cotton curvature. I gave you the physical example demonstrating that to be so.

It's like this: On an ideal spherical non-rotating planet, a being climbs a radial pointing step-ladder X proper meters, where X << R the planet's surface radius (strictly speaking, in GR, R is 'just a parameter' defined by C = 2*pi*R). At that proper elevation X there exists a railing of constant nominal radius R+X - i.e. a great circle. What is the circumference C' there, as determined by the being measuring proper distance along the railing? If you get $C'= 2\pi(R+X)$ as first-order in metric approximation, start again. SM is locally spatially non-Euclidean. Correct answer is
$C'= 2\pi(R+\sqrt{{g}_{tt}}X)$

And just as not one of the assortment of fools and knaves (not all so categorized) I endured in that linked ATM thread ever took up my challenge to refute my two main findings - disappearing functional dependence on gravitational potential for just one metric component, or conflicting predictions for internal spatial metric, so it it has been here (not that I'm implying folks here are all fools and/or knaves! ).
Q-reeus

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### Re: Spherically symmetric space is conformally flat

Q-reeus wrote:
Mikko wrote:
Q-reeus wrote:And I will guarantee you will never find a page in regular literature equating conformal flatness to *any* spherically symmetric space.

Some pages mention the result without proof or reference.

Umm - references please, preferably to freely accessible online material.

Some typical examples are:
http://arxiv.org/abs/1405.1682
http://www.phy.olemiss.edu/~luca/Topics ... om_3D.html
http://homepage.univie.ac.at/piotr.chru ... Energy.pdf (page 5)
All these mention it as if it were in common textbooks but I don't know any such textbook.
Mikko

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### Re: Spherically symmetric space is conformally flat

Mikko wrote:Some typical examples are:
http://arxiv.org/abs/1405.1682
http://www.phy.olemiss.edu/~luca/Topics ... om_3D.html
http://homepage.univie.ac.at/piotr.chru ... Energy.pdf (page 5)
All these mention it as if it were in common textbooks but I don't know any such textbook.

Will have to concede these GR experts are indeed agreeing with your thread title. The last one, whether or not you used it directly as basis for post #1, seems to employ the same reasoning - it's always possible to transform a spherically symmetric metric into an isotropic form. Hmm....

This begs the question - what does 3D conformal flatness, in particular relating to exterior SM or alternative, actually physically mean? My understanding is that it implies a differential excursion from any given radial value yields the first proper Euclidean relation for C' I gave last post. But the actual relation is given by my second expression. A manifestly locally non-Euclidean first order in metric relation. What relation for C' do you find in the scenario I gave? How do you explain it?
Q-reeus

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### Re: Spherically symmetric space is conformally flat

Q-reeus wrote:This begs the question - what does 3D conformal flatness, in particular relating to exterior SM or alternative, actually physically mean?

Probably not very much. There are no exactly spherical non-rotating bodies big enough to have significant gravity. In less symmetric situation it is not clear how one should choose the time coordinate. The spactime curvature does not depend on choice of coordinates so it can be considered more important. Spacetime is conformally flat only in some special situations.
Mikko

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### Re: Spherically symmetric space is conformally flat

Mikko wrote:
Q-reeus wrote:This begs the question - what does 3D conformal flatness, in particular relating to exterior SM or alternative, actually physically mean?

Probably not very much. There are no exactly spherical non-rotating bodies big enough to have significant gravity. In less symmetric situation it is not clear how one should choose the time coordinate. The spactime curvature does not depend on choice of coordinates so it can be considered more important. Spacetime is conformally flat only in some special situations.

Which all deftly avoids my last point - SM deviates from locally 3D flatness by a first order in metric term. Asking you again - what value do you obtain for C' as per great circle railing scenario? It should be what I gave in second expression. How do you reconcile it with spatial conformal flatness?
Q-reeus

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