Thin Spherical Matter Shell as Gravity Theory Filter

[Q-reeus]

http://vixra.freeforums.org/thin-shell-metric-matching-demands-conformal-flatness-t320.html

That message claims that "thin shell demands conformal flatness".

Only so much should rationally be inferred from a thread *title*. It's necessary to then digest the following body content.

Conformal flatness of the spatial geometry follows from spherical symmetry.

From that quite general assertion it can be taken you evidently believe - wrongly - that Schwarzschild exterior metric is conformally flat. Too bad.

Thin shell is not required nor does a thin shell of other shape suffice.

Which restricts the case to - nothing! TZ effect here.

In addition, only spatial metric is conformally flat, spacetime metric is not.

Firstly, no required differentiation between exterior or interior region case is given in that assertion. One has to assume you apply it to everywhere - shell exterior and interior regions - which one has to guess gels with your similarly undifferentiated earlier assertion above. As for spatial vs spacetime conformal flatness, in given scenario, there is no essential distinction. Static situation and time component as scalar most certainly does *not* introduce spacetime conformal non-flatness either as an 'extra' in exterior SM region, or at all in interior MM region.

I'm an admitted amateur scientist without any formal training in physics, and had assumed given your authoritative tone, you had a formal, at least undergrad University level physics background. That seems doubtful. At any rate, I hate the thought of and never have put anyone on a Foes list, but imo best we avoid each other hereon out.

[Mikko]

Those who are experts could perhaps bear in mind that non-experts also want to learn from the exchanges here.

More likely experts are not aware of existence of this forum.

Mikko seems to be saying that Q-reeus' "anomalies" are only apparent anomalies connected to an unwise choice of coordinate system. So he is saying that expressed in another coordinate system, or expressed in a coordinate-free way, they no longer exist. Is this right? If so please show us how this would work out.

The first "anomaly" is that the radial scale is not continuous at the shell, if the shell is modelled as infinetely thin. This is true in Swartzschild coordinates and also in isotropic coordinates but not e.g. if an isometric radial coordinate is used. As the main area of interest is at and around the shell, an isometric radial coordinate is a reasonable choice, although not necessary. However, here is nothing unwise in Swartzschild or other systems. The discontinuity of radial scale, although avoidable, can be interpreted as a consequence of infinite mass and stress in the shell.

In reality, a material shell cannot be infinitely thin, although in some cases can be considered so as a reasonable approximation. Same way discontinuities
in (physical and other) fields can be regarded as approximations of continuous reality. On the other hand, discontinuities and infinities in reality can be approximated with finite continuous functions (e.g., when doing so simplifies mathematics).

[Mikko]

Conformal flatness of the spatial geometry follows from spherical symmetry.

From that quite general assertion it can be taken you evidently believe - wrongly - that Schwarzschild exterior metric is conformally flat. Too bad.

No, I clearly said (and later repeated) that only the spatial geometry is conformally flat but time scales differently, as can be seen when the solution is expressed in isotropic coordinates.

In addition, only spatial metric is conformally flat, spacetime metric is not.

Firstly, no required differentiation between exterior or interior region case is given in that assertion. One has to assume you apply it to everywhere - shell exterior and interior regions - which one has to guess gels with your similarly undifferentiated earlier assertion above.

The meaning should be clear: the spatial geometry is conformally flat everywhere, the spacetime is not everywhere conformally flat. The interior region is flat and therefore conformally flat (for any thickness of the shell as long as it is spherically symmetric).

[Q-reeus]

The first "anomaly" is that the radial scale is not continuous at the shell, if the shell is modelled as infinetely thin. This is true in Swartzschild coordinates and also in isotropic coordinates but not e.g. if an isometric radial coordinate is used. As the main area of interest is at and around the shell, an isometric radial coordinate is a reasonable choice, although not necessary. However, here is nothing unwise in Swartzschild or other systems. The discontinuity of radial scale, although avoidable, can be interpreted as a consequence of infinite mass and stress in the shell.

Will give you this much - not easily dissuaded. Alright, despite my previous judgement, let's continue this exchange and hope it gets somewhere useful.

The first "anomaly" is that the radial scale is not continuous at the shell, if the shell is modelled as infinetely thin....The discontinuity of radial scale, although avoidable, can be interpreted as a consequence of infinite mass and stress in the shell.

Just those statements tells me you have fundamentally misunderstood the issues. Negligibly thin - not infinitely thin - wall thickness was a choice made, as stated explicitly in part 1 of article, to simplify the scenario. So as not to get bogged down in somewhat tedious and unenlightening detailed calculations required for a 'thick' shell wall. Introduction section made it explicit that what goes on within shell wall is not of interest. Only the difference in coordinate determined exterior vs hollow interior metric functional form and values. In part 2, negligible wall thickness again just made the calculations simple and informative. Without the clutter and useless distraction of an additional radial integration over shell mass. Bare essentials that highlight principles, not impressive looking reams of detailed math clutter. Hope that much is now bedded down.

And btw "infinite mass and stress in the shell", meant presumably "infinite mass density and stress...." Not correct, not just owing to your assuming 'infinitely thin', but because stress contribution to effective shell mass is always totally negligible. Only a rest mass T_00 contribution was assumed and matters.
When I first raised a cruder version of the scenario a few years back in a prestigious forum, one GR expert and a wannabe both claimed that shell stresses 'for sure' accounted for not just some but all of the change(s) in functional form from exterior SM to interior MM. Something I knew even as a rank layman to be absurd and said as much. Eventually said expert and wannabe had to grudgingly back down and admit to basic conceptual error (actually a raft of errors). Too bad it all got hopelessly sidetracked later on.

This is true in Swartzschild coordinates...

Only on a coordinate not local basis and only by insisting on 'infinitely thin'. What does matter is, for either a thin or thick walled shell, the physically unjustifiable change in functional form - viewed on a coordinate basis. Between anywhere in exterior region, vs anywhere in hollow interior region. Must I endlessly repeat this?! Hopefully it's understood 'coordinate' here is referring to, as explicitly detailed in article, observation perspective of a distant observer, and not choice of coordinate system. Which is clearly given as spherical.

...and also in isotropic coordinates but not e.g. if an isometric radial coordinate is used.

Wrong. By isotropic coordinates you presumably meant ISM (isotropic Schwarzschild metric) as per my article. In that basis there is no coordinate determined discontinuity even for infinitely thin shell wall. But that's because ISM is, as per article section 3, not physically equivalent to SM despite widespread belief otherwise. As for isometric radial coordinate, who uses it and how is it defined? Do you mean something like Yilmaz metric? That's not 'isometric' but a proper and genuinely derived conformally flat thus locally Euclidean thus isotropic metric (exterior vacuum region).

However, here is nothing unwise in Swartzschild or other systems.

Yes there is. Only a conformally flat metric avoids internal contradictions. But one has to overcome a certain amount of GR = Holy Truth brainwashing to recognize that.

In reality, a material shell cannot be infinitely thin, although in some cases can be considered so as a reasonable approximation. Same way discontinuities
in (physical and other) fields can be regarded as approximations of continuous reality. On the other hand, discontinuities and infinities in reality can be approximated with finite continuous functions (e.g., when doing so simplifies mathematics).

All covered above - you assumed 'infinitely thin' and then ran with that wrong assumption, and even then applied it just on a coordinate basis (without stating such). I never claimed or claim now that even wrongly imposing 'infinitely thin', locally there will exist metric discontinuities (of first one or maybe first two spatial derivatives) - as explicitly stated in that viXra thread I linked to.

The real crunch comes in part 2 where the direct evaluation makes it clear SM is internally inconsistent. Only by adopting a conformal flat metric can that be avoided.

[Q-reeus]

Conformal flatness of the spatial geometry follows from spherical symmetry.

From that quite general assertion it can be taken you evidently believe - wrongly - that Schwarzschild exterior metric is conformally flat. Too bad.

No, I clearly said (and later repeated) that only the spatial geometry is conformally flat but time scales differently, as can be seen when the solution is expressed in isotropic coordinates.

That's certainly not how I read it then or now! Regardless, fact is, in given static (and spherically symmetric) scenario, time component is irrelevant and conformal flatness or otherwise is a function of just spatial metric components. Exterior SM is not conformally flat. Exterior ISM is, but 'by accident' given it's supposed to be but is not physically equivalent to SM. Only something like Yilmaz metric is properly conformally flat in exterior region.

In addition, only spatial metric is conformally flat, spacetime metric is not.

Firstly, no required differentiation between exterior or interior region case is given in that assertion. One has to assume you apply it to everywhere - shell exterior and interior regions - which one has to guess gels with your similarly undifferentiated earlier assertion above.

The meaning should be clear: the spatial geometry is conformally flat everywhere, the spacetime is not everywhere conformally flat.

Again - such meaning was not clear to me! Anyway you have the wrong conception of what conformally flat means. I give it in article. Maybe try contacting a recognized GR authority and putting your above expressed idea to him/her.

The interior region is flat and therefore conformally flat (for any thickness of the shell as long as it is spherically symmetric).

Wow - we actually agree on something!

[Mikko]

But that's because ISM is, as per article section 3, not physically equivalent to SM despite widespread belief otherwise.

It is. Nature does not specify any coordinate system. You can use different coordinate systems in the same physical situation. When you transform formulas from one coordinate system to another the transformed formulas say the same about the same as the original formulas. The difference is purely presentational. For example, on Earth you can use geocentric coordinates instead of the more common geographic coordinates without any change to the shape of the Earth. Same way, when you substitute other coordinates in the Swarzschild solution you will get another presentation of the same Schwarzschild solution.

[Q-reeus]

But that's because ISM is, as per article section 3, not physically equivalent to SM despite widespread belief otherwise.

It is. Nature does not specify any coordinate system. You can use different coordinate systems in the same physical situation. When you transform formulas from one coordinate system to another the transformed formulas say the same about the same as the original formulas. The difference is purely presentational. For example, on Earth you can use geocentric coordinates instead of the more common geographic coordinates without any change to the shape of the Earth. Same way, when you substitute other coordinates in the Swarzschild solution you will get another presentation of the same Schwarzschild solution.

Check out discussion in part 3 of article. ISM is by intention made to exhibit conformal flatness, or if you prefer, isotropic light speed. Both as locally measured and as coordinate observer inferred. Which means locally Euclidean. SM is locally non-Euclidean. That's a fundamental difference in an intensive property. In practice unobservably small for any accessible scenario; but still there. Regardless of that e.g. planetary orbit calculations work out identically in either SM or ISM to a very good approximation.

When it comes to really strong gravity situation, both SM and ISM permits a finite area 'black hole' to exist. Not so for genuinely conformally flat, isotropic Yilmaz metric - black hole can't happen. But distinguishing observationally between GR 'black hole' and say Yilmaz gravity 'extremely dark grey hole' a very difficult ask. Anyway there are real consequences as mentioned particularly in part 3.

And just how you arrived at notion of everywhere spatial conformal flatness but spacetime conformal non-flatness for static shell is beyond me.

[Q-reeus]

Another formal notification - v3 of article just uploaded to viXra. Same link as in first post here. Basically added some bits addressing comments made this thread.
Gee, forum sort of real quiet of late. Seems one forced departure in particular has had quite a dampening effect.

[Mikko]

I posted to the Geometry board a proof that every spherically symmetric 3-dimensional space is conformally flat (except possibly at the origin). A particular example is the Swarzschild solution. Therefore, the requirement of conformal flatness of space is not specific to the thin shell case.

[Q-reeus]

I posted to the Geometry board a proof that every spherically symmetric 3-dimensional space is conformally flat (except possibly at the origin). A particular example is the Swarzschild solution. Therefore, the requirement of conformal flatness of space is not specific to the thin shell case.

And I have answered you there. SM is not conformally flat as evidenced by it's demonstrably locally non-Euclidean geometry. That cannot be disappeared by appeal to being able to derive a formally equivalent expression (e.g. ISM) that either just pretends to be conformally flat, or is so but thereby represents a physically distinct metric different from SM.

[Heinera]

I posted to the Geometry board a proof that every spherically symmetric 3-dimensional space is conformally flat (except possibly at the origin). A particular example is the Swarzschild solution. Therefore, the requirement of conformal flatness of space is not specific to the thin shell case.

And I have answered you there. SM is not conformally flat as evidenced by it's demonstrably locally non-Euclidean geometry. That cannot be disappeared by appeal to being able to derive a formally equivalent expression (e.g. ISM) that either just pretends to be conformally flat, or is so but thereby represents a physically distinct metric different from SM.

I am not sure if I understand your objection here. Due to the equivalence principle, any solution of Einstin's equations can be expressed in an atlas of coordinate systems that are locally flat - but you don't have to use such a coordinate system to express the solutions. What matters for two coordinate systems w/corresponding metrics to represent the same solution is curvature, not local flatness.

[Q-reeus]

I am not sure if I understand your objection here. Due to the equivalence principle, any solution of Einstin's equations can be expressed in an atlas of coordinate systems that are locally flat - but you don't have to use such a coordinate system to express the solutions. What matters for two coordinate systems w/corresponding metrics to represent the same solution is curvature, not local flatness.

Nice to have a second respondent . Heinera, it may or not help but I suggest you take a look at my reply to Mikko here: viewtopic.php?f=24&t=77#p3438
And maybe then post your thoughts there or back here after.

[Mikko]

Are you still going to update to v4?

[Q-reeus]

Are you still going to update to v4?

Yes, and likely it will be an extensive rewrite, with a link to separate article specifically dealing with SM vs ISM. But have to juggle with many other issues at the moment, so not real soon. In the meantime, you will find a more or less placeholder response in Geometry. And if you have some spare moments to while away, a read through the kind of disingenuous sh*t relentlessly thrown at me here: http://cosmoquest.org/forum/showthread. ... l-flatness! might even prove amusing.

[Mikko]

Are you still going to update to v4?

Yes, and likely it will be an extensive rewrite, with a link to separate article specifically dealing with SM vs ISM. But have to juggle with many other issues at the moment, so not real soon. In the meantime, you will find a more or less placeholder response in Geometry. And if you have some spare moments to while away, a read through the kind of disingenuous sh*t relentlessly thrown at me here: http://cosmoquest.org/forum/showthread. ... l-flatness! might even prove amusing.

When we may expect the v4?

[Q-reeus]

Are you still going to update to v4?

Yes, and likely it will be an extensive rewrite, with a link to separate article specifically dealing with SM vs ISM. But have to juggle with many other issues at the moment, so not real soon. In the meantime, you will find a more or less placeholder response in Geometry. And if you have some spare moments to while away, a read through the kind of disingenuous sh*t relentlessly thrown at me here: http://cosmoquest.org/forum/showthread. ... l-flatness! might even prove amusing.

When we may expect the v4?

Mikko - just caught this a while ago. Not getting email notification for some reason. Check out my comments at viXra article site.

[Mikko]

Mikko - just caught this a while ago. Not getting email notification for some reason. Check out my comments at viXra article site.

There was no planned date there.

[Q-reeus]

Mikko - just caught this a while ago. Not getting email notification for some reason. Check out my comments at viXra article site.

There was no planned date there.

Just checked viXra site http://vixra.org/abs/1407.0130 and noticed your latest comment there. Still juggling with many issues but will try and commit to a v4 before end of this month. Regarding your preceding viXra comment of Nov 28, 2014:

I don't think using 'locally Euclidean' in stead of 'conformally flat' helps. Your concept is stronger than 'locally Euclidean' as usually understood.

I disagree and it's really a case of realizing conformal flatness is too weak a constraint to settle the matter of SM vs ISM - but locally Euclidean is decisive. Apply the example I gave here viewtopic.php?f=24&t=77&start=10#p3528
for standard SM, to ISM metric as per (3-1) in v3 of my article.
It's obvious that for ISM all spatial components are equally scaled by the metric prefactor (1+rs/(4r1))^2. Such equiscaling is more or less the very definition of isotropic - the 'I' in ISM.
Hence performing the same differential move 'up the ladder' dr = X (proper value) will yield in ISM a first-order in metric Euclidean relation C' = 2pi(r+X). [one formally sets dt = 0, sin(theta) = sin(pi/2) =1, d(theta) =0. Hence only r1 and circumference are involved]. Isotropic means locally Euclidean!
That is really a strong differentiation between SM and ISM as it has no derivatives of the metric involved at all in either answer - just the presence of one metric component and only in SM case.
The hard part is accepting that for so long GR community has failed to notice what should have been obvious. Blind adherence to a formal mathematical 'coordinate transformation' that is strictly valid only 'at a point' - i.e. a misleading and physically empty formality.

Fact is SM is locally non-Euclidean, ISM is locally Euclidean. Only one can be a true representation of Schwarzschild metric. I will make that more formal in v4, but here's your chance to finally answer my last request of you in that above linked Geometry section thread. Do you agree or disagree with the above? If you disagree, what is your finding for C' in both SM and ISM cases?

Finally, Fred might like to explain why I have not been receiving email notifications for a long time now. Is this something applying to everyone, or just me?

[Mikko]

Mikko - just caught this a while ago. Not getting email notification for some reason. Check out my comments at viXra article site.

There was no planned date there.

Just checked viXra site http://vixra.org/abs/1407.0130 and noticed your latest comment there. Still juggling with many issues but will try and commit to a v4 before end of this month. Regarding your preceding viXra comment of Nov 28, 2014:

I don't think using 'locally Euclidean' in stead of 'conformally flat' helps. Your concept is stronger than 'locally Euclidean' as usually understood.

I disagree and it's really a case of realizing conformal flatness is too weak a constraint to settle the matter of SM vs ISM - but locally Euclidean is decisive.

Only if you very carefully define the term "locally Euclidean". The problem is that "locally" is ambiguous -- there are different degrees of locality. You must also make very clear what is or is not "locally Euclidean": is it a property of space or fields or what?

Apply the example I gave here viewtopic.php?f=24&t=77&start=10#p3528
for standard SM, to ISM metric as per (3-1) in v3 of my article.
It's obvious that for ISM all spatial components are equally scaled by the metric prefactor (1+rs/(1+4r1))^4.

By metric components you refer to the relation between the metric and the coordinate system. Therefore this scaling is a property of the coordinate system, and would not apply to another coordiante system in the same space.

Such equiscaling is more or less the very definition of isometric - the 'I' in ISM.

No, it is the definition "isotropic". There is an additional requirement for "isometric": the scale factor must be 1.

Hence performing the same differential move 'up the ladder' dr = X (proper value) will yield in ISM a first-order in metric Euclidean relation C' = 2pi(r+X). Isometric means locally Euclidean!

Be careful with the definitions. Would you call a space where ds² = r² (dr² + df²) "locally Euclidean"?

That is really a strong differentiation between SM and ISM as it has no derivatives of the metric involved at all in either answer - just the presence of one metric component and only in SM case.
The hard part is accepting that for so long GR community has failed to notice what should have been obvious. Blind adherence to a formal mathematical 'coordinate transformation' that is strictly valid only 'at a point' - i.e. a misleading and physically empty formality.

Fact is SM is locally non-Euclidean, ISM is locally Euclidean. Only one can be a true representation of Schwarzschild metric. I will make that more formal in v4, but here's your chance to finally answer my last request of you in that above linked Geometry section thread. Do you agree or disagree with the above? If you disagree, what is your finding for C' in both SM and ISM cases?

I disagree with calling only one of two equivalent representations "true". It just has some features that makes it easier to use for some purposes.

[Q-reeus]

Only if you very carefully define the term "locally Euclidean". The problem is that "locally" is ambiguous -- there are different degrees of locality. You must also make very clear what is or is not "locally Euclidean": is it a property of space or fields or what?

Obviously of space. What could a 'non-Euclidean field' (other than spatial metric field which is back to spatial) even mean?

By metric components you refer to the relation between the metric and the coordinate system. Therefore this scaling is a property of the coordinate system, and would not apply to another coordiante system in the same space.

Only partly true in that a given coordinate system is operated on by the appropriate metric coefficients which yield an unambiguous means to determine local physics - in this context Euclidean vs non-Euclidean.

No, it is the definition "isotropic". There is an additional requirement for "isometric": the scale factor must be 1.

My slip up there (since corrected) and I obviously meant isotropic. And btw had earlier corrected the definition of ISM metric prefactor - not showing in your quotes.

Be careful with the definitions. Would you call a space where ds² = r² (dr² + df²) "locally Euclidean"?

Depends on what f stands for, and whether one considers a truly differential spatial expansion about some given point.

I disagree with calling only one of two equivalent representations "true". It just has some features that makes it easier to use for some purposes.

That is your constant theme but still you decline to answer my repeated question - what values do you obtain for C' - SM vs ISM in example given? Answering that should provide some closure on this sub-issue. Even If I were to be shown somehow wrong and ISM is demonstrably locally non-Euclidean (to the same degree as shown for SM) it would still not remove the part 2 findings. That would indeed lead to a severe paradox. But I'm confident it won't because the two forms are not physically equivalent.