How does GR handle metric transition for a spherical mass shell?

[Q-reeus]

This is really a continuation from another thread but will start here from scratch. Consider the case of a static thin spherical mass shell - outer radius rb, inner radius ra, and (rb-ra)/ra<< 1, and with gravitational radius rs<< r(shell). According to majority opinion at least, in GR the exterior spacetime for any r >= rb is that given by SM (Schwarzschild metric) as expressed by the SC's (Schwarzschild coordinates) http://en.wikipedia.org/wiki/Schwarzschild_metric#The_Schwarzschild_metric. In the empty interior region r=< ra, a flat MM (Minkowski metric) applies. Owing probably to it's purely scalar nature, there seems little controversy as to the temporal component transition, as expressed as frequency redshift scale factor Sf = f'/f (f' the gravitationally depressed value as seen 'at infinity'). In terms of the potential V = (1-rs/r)1/2, with rs = 2GMc-2, one simply has Sf = V, which for any r=< ra has it's r parameter value 'frozen' at r = ra. There is a smooth transition from rb to ra that depends simply on V only. So far, so broadly reasonable.

What of the spatial metric components? In terms of corresponding scale factors Sr, St that operate on the radial and tangent spatial components respectively, it is readily found from the SC's that Sr = V, St = 1, everywhere in the exterior SM region. Sr = V is physically reasonable and identical to frequency redshift factor Sf here. Whether according to GR Sf and Sr diverge for some reason in the transition to the interior MM region is not clear to me, but that would be 'interesting'.
An apparently consensus view that St dives relatively steeply from St = 1 at r = rb, to some scaled fraction of V at r=< ra. [A previous attempt here at PF at the shell transition problem found something different; an invariant St = 1 for all r, but an Sr that jumped back from V to 1 in going from rb to ra. That gave flat interior spatial metric with unity scale factors Sr, St (i.e. 'at infinity' values) but a redshift factor Sf = V]. Seems though the consensus view is for an equally depressed Sr, St in the interior MM region, the exact value of Sr, St, and Sf relative to V is unclear from recent entries on that matter, where parameters A, B, were used as equivalent to Sf, (Sr, St) respectively.

I wish to focus on the question of physical justification for St diving from unity to some fraction of V. Given that an invariant St = 1 for all r >= rb (SM regime) stipulates complete independence on V or any combination of it's spatial derivatives ∂V/∂r, ∂2V/∂r2, etc. In the shell wall region, where shrinkage of St apparently occurs, there is owing to a nonzero stress-energy tensor T a different relative weighting of V and all it's derivatives to that applying for r => rb, but otherwise, only the weighting factors are different. What permits variance of St in one case, but not the other? An answer was that in the shell wall where T is nonzero, the Einstein tensor G (http://en.wikipedia.org/wiki/Einstein_tensor) operates and this is the explanation.

That seems unsatisfying, and should be justifiable at a basic, 'bare kernel' level. By that is meant identify the 'primitives' from which everything in G can be derived, and show what particular combo leads to a physically justifiable variance of St, applying nowhere but the shell wall region. So what are the 'primitives'. I would say just V and it's spatial derivatives, which owing to the spherical symmetry, are of themselves purely radial vector quantities (but obviously not all their combinations as per div, curl etc).

One caveat here is to nail down the relevant source of V - taken simply as total mass M exterior to rb. Has been pointed out that for a stable shell there must be pressure p present in addition to just rest matter density ρ, ie T = ρ + p, = T00 + T11+T22+T33. My assumption is that for a mechanically stable thin shell of normal material, ρ >>> p, and so to a very good approximation, just use ρ. If one feels the p terms should be included regardless, then I would further suggest they will act here just as a tiny addition to ρ. That is, the contribution of T11 for instance in some element of stressed matter introduces no 'directionality' per se to the potential, yes?

Can't think of any physical quantity - relevant to this case anyway - that could not be expressed as some function of the above primitives. But recall, all these primitives exist in the region exterior to rb, where St is strictly = 1!. Only remotely relevant quantity I can think of that *may* be zero in the exterior region but obviously nonzero in the wall region is the three divergence nabla2V. And that could account for a varying St? Can't imagine how. So something real, real special has to be pulled out of the hat imo. So special I consider it impossible, but open to be shown otherwise. Seems to me the anomaly is intractable in GR and a cure requires a theory where isotropic contraction of spatial and temporal components apply. Then and only then the transition issue naturally resolves. But that's my opinion. So, any GR pro willing to give this a go, let's get on with the show!

[atyy]

http://arxiv.org/abs/gr-qc/0505048
http://www-dft.ts.infn.it/~ansoldi/Seminars/2004/FFP6/HTML/Level1.2.html

http://arxiv.org/abs/gr-qc/0308025
http://arxiv.org/abs/gr-qc/0406053

[DaleSpam]

Don't know if you saw this in the other thread:Issue now, following #241, is to nail down just what property/operation of ET actually yields tangential contraction.The tangential pressure components of the stress energy tensor. There is no tangential pressure in the vacuum region, there is in the matter region.

Here is a arxiv paper you may like. It uses an analytical model for the shells, so it is not the usual "step function" you would normally consider, but it describes things like the radial and transverse pressures:
http://arxiv.org/abs/0911.4822

[Q-reeus]

http://arxiv.org/abs/gr-qc/0505048
http://www-dft.ts.infn.it/~ansoldi/Seminars/2004/FFP6/HTML/Level1.2.html

http://arxiv.org/abs/gr-qc/0308025
http://arxiv.org/abs/gr-qc/0406053
Thanks for the links but need some expert translation here. I get that extrinsic curvature jumps across a thin boundary proportional to the boundary 'surface' stress-energy density (link 2 - the others are highly technical). Now specifically how does the extrinsic curvature relate to tangent metric factor St for shell case? There should surely be a relatively simple function in ρ and r; St(rho,r) = f(rho,r) which can/must be decomposable into just V and derivatives thereof. Maybe it's there in those links, but if so someone has to point to just where. If not, I'm still inclined to think maths divorced from physics - "It's all about imposing enough boundary conditions, old chap".

[Q-reeus]

Don't know if you saw this in the other thread:The tangential pressure components of the stress energy tensor. There is no tangential pressure in the vacuum region, there is in the matter region.

Here is a arxiv paper you may like. It uses an analytical model for the shells, so it is not the usual "step function" you would normally consider, but it describes things like the radial and transverse pressures:
http://arxiv.org/abs/0911.4822
Sorry but I missed it! While the link is interesting, I get that the models are about highly compressible matter shells where pressure is mostly not an appreciable source of T itself and the complexity derives from the highly non-uniform matter density profiles. Maybe some cases there imply 'extreme' pressures, but either way, it's just an addition to rho surely given static conditions. At any rate for my shell scenario, as mentioned there, a typical self-supporting shell of say metal will always have rho many orders of magnitude in excess of any p contributions (anti-matter bomb vs firecracker). So it still gets back to explaining the case of just a shell of uniform matter density as only source of T and G - stress is not worth stressing over.

[PeterDonis]

At any rate for my shell scenario, as mentioned there, a typical self-supporting shell of say metal will always have rho many orders of magnitude in excess of any p contributions (anti-matter bomb vs firecracker).

I don't think you can make this assumption in general; I think it will only hold if the potential at the inner surface of the shell is close to 0 (where 0 is the potential at infinity)--equivalently, the radius of the shell (inner or outer) is much greater than 2M, where M is the total mass of the shell (as measured by an observer far away from it). If the shell radius is not very large compared to M (meaning the potential within the shell is significantly less than 0, or equivalently the t-t and r-r metric coefficients are significantly different from 1), then the state of the matter in the shell will be much more like white dwarf or neutron star material than like ordinary metal, and the pressure in the material will have to be comparable to its energy density or the shell will not be stable. (Also, as I mentioned in the previous thread where this topic was discussed, the pressure will have to be negative--a tension--in some portion of the shell or it will fall in on itself.)

[Q-reeus]

I don't think you can make this assumption in general; I think it will only hold if the potential at the inner surface of the shell is close to 0 (where 0 is the potential at infinity)--equivalently, the radius of the shell (inner or outer) is much greater than 2M, where M is the total mass of the shell (as measured by an observer far away from it).
Yes that's the condition I was talking about in specifying in #1 that rs/r << 1. The aim is to avoid 'strong gravity' complications and as much as possible all the fierce math that goes with that. A bare minimum scenario that draws out the relevant factors.
...(Also, as I mentioned in the previous thread where this topic was discussed, the pressure will have to be negative--a tension--in some portion of the shell or it will fall in on itself.)

I felt like commenting on that over there, but here's the chance. This seems strange. I can only see positive biaxial tangent stress. So you are saying there is need of a radial tensile component?

[PeterDonis]

Yes that's the condition I was talking about in specifying in #1 that rs/r << 1. The aim is to avoid 'strong gravity' complications and as much as possible all the fierce math that goes with that. A bare minimum scenario that draws out the relevant factors.

Ok, understood. Basically, we're assuming that the spherical shell can hold itself together and be made of "normal" materials for which the energy density is much larger than any stress component. The corrections to the metric coefficients will all be small, but as long as those corrections are still large compared to the ratio of pressure to energy density, which seems reasonable, we should be fine.

(Quick back of the envelope calculation to verify that it's reasonable: metric coefficient correction for the Earth is of order 10^-6, the ratio of r to M. For a typical material like steel, pressure components are of order 10^9 in SI units--gigapascals--and energy density is of order 10^20 in SI units--one kilogram per cubic meter is 9 x 10^16, water is 1000 kg/m^3, and steel is several times the density of water. So the ratio of pressure to energy density in a typical structural material is of order 10^-11, five orders of magnitude smaller than the metric correction for r a million times M. Looks like we're OK.)

I felt like commenting on that over there, but here's the chance. This seems strange. I can only see positive biaxial tangent stress. So you are saying there is need of a radial tensile component?

No, I'm saying there has to be a tangential hoop stress, which will be a tension, not a pressure, so it will be a negative quantity in the stress-energy tensor. See here:

http://en.wikipedia.org/wiki/Cylinder_stresses

This talks about cylinders, not spheres, but the general idea is the same for spheres, including the sign of the stress. The Wiki page isn't very careful about signs, but the convention for the stress-energy tensor is clear: pressure (compressive stress) is positive and tension (tensile stress--stretching) is negative.

[PeterDonis]

An apparently consensus view that St dives relatively steeply from St = 1 at r = rb, to some scaled fraction of V at r=< ra. [A previous attempt here at PF at the shell transition problem found something different; an invariant St = 1 for all r, but an Sr that jumped back from V to 1 in going from rb to ra. That gave flat interior spatial metric with unity scale factors Sr, St (i.e. 'at infinity' values) but a redshift factor Sf = V]. Seems though the consensus view is for an equally depressed Sr, St in the interior MM region, the exact value of Sr, St, and Sf relative to V is unclear from recent entries on that matter, where parameters A, B, were used as equivalent to Sf, (Sr, St) respectively.

Since I contributed to the "apparent consensus" in the other thread, I would like to post here after thinking this over some more. One key thing that I don't think I captured correctly in the other thread is that statements like "St dives relatively steeply from St = 1 at r = rb, to some scaled fraction of V at r=< ra" are (at least I believe they are) coordinate-dependent. So I think it is helpful to start by first describing as much as we can independently of any coordinates, entirely in terms of "physical" quantities that are coordinate-invariant. I will do that here for the description I gave in the other thread of how the "packing" of objects between two spheres close together varies as we descend through the three "regions" of this scenario: the exterior vacuum region (EV), the non-vacuum "shell" region (NV), and the interior vacuum region (IV).

Start somewhere in the EV, well above the outer surface of the shell, but still far enough from "infinity" that the metric differs significantly from the Minkowski metric. Take two concentric 2-spheres, both centered exactly on the "shell" (but much larger), one with physical area A, the other with slightly larger physical area A + dA. By "physical area" I mean the area measured by covering the 2-sphere with very small identical objects and counting the objects (and by "identical objects" I mean that if you bring any two of them to the same location they will be exactly the same shape and size).

Pack the volume between these two spheres with the same identical little objects (which we take to be exactly spherical so they have the same dimension in all directions). Euclidean geometry would lead us to predict that the volume dV between the two spheres that we will find from this packing is given by:

dV = \frac{1}{4} \sqrt{\frac{A}{\pi}} dA

However, when we actually do the packing, we will find the physical volume dV between the two spheres is *larger* than this, by some factor K; and K will get *larger* as we descend in region EV, getting closer to the outer surface of region NV. Finally, at the outer surface of region NV, K will have reached some value K_o > 1.

As we continue to descend through NV, we find the opposite effect now taking place; the factor K begins to get smaller, and when we reach the inner surface of NV, we find that it is now 1, the same value it has at "infinity". In other words, K_i = 1.

Once we are inside region IV, the K factor does not change; spacetime is flat. So since K is 1 at the inner surface of NV, it is 1 throughout IV. If we take any pair of spheres centered in IV, with areas A and A + dA, the volume between them, as shown by packing with identical little objects, will be exactly as given by the above formula.

I should also include what happens to the "rate of time flow" as we descend through the three regions. (This "rate of time flow" would be observable as gravitational redshift/blueshift, so substitute that in wherever I say "time flow" if that makes it easier to see what I mean.) In region EV, time flow, relative to its rate "at infinity", is slowed by the same factor K that volume between two spheres is increased from its Euclidean value. So at the outer surface of NV, time flow, relative to infinity, is multiplied by a factor J_o = 1 / K_o. However, as we descend through region NV, time flow continues to slow, so when we reach the inner surface of NV, time flow relative to infinity is multiplied by a factor J_i < J_o. Therefore, we can see that J_i is *not* equal to 1 / K_i; the relationship between J and K that held in region EV no longer holds in region NV. And since the rate of time flow is the same throughout NV, the relationship between J and K that held in EV doesn't hold in IV either.

I won't comment at this point on how any of the above relates to the descriptions in terms of coordinate systems; I'll save that for another post. But I think the above summarizes the physical observations that would apply in the scenario.

[DaleSpam]

Maybe some cases there imply 'extreme' pressures, but either way, it's just an addition to rho surely given static conditions. At any rate for my shell scenario, as mentioned there, a typical self-supporting shell of say metal will always have rho many orders of magnitude in excess of any p contributions (anti-matter bomb vs firecracker). So it still gets back to explaining the case of just a shell of uniform matter density as only source of T and G - stress is not worth stressing over.I think this is incorrect. The stress should be exactly the right magnitude to cause the change in the metric you are interested in. If it is a thin massive shell it will be very high and if it is a thick low-density shell then it will be lower. The density will clearly be larger, but that doesn't mean that the lateral stresses are negligible, particularly since they affect different components of the stress-energy tensor and it is precisely these different components that are causing your concern.

I feel like you are deliberately ignoring a clear physical source of the effect you are interested in for no reason other than your a-priori assumption that it cannot possibly be large enough.

[Q-reeus]

So the ratio of pressure to energy density in a typical structural material is of order 10^-11, five orders of magnitude smaller than the metric correction for r a million times M. Looks like we're OK.)
Check!
No, I'm saying there has to be a tangential hoop stress, which will be a tension, not a pressure, so it will be a negative quantity in the stress-energy tensor. See here:

http://en.wikipedia.org/wiki/Cylinder_stresses

This talks about cylinders, not spheres, but the general idea is the same for spheres, including the sign of the stress. The Wiki page isn't very careful about signs, but the convention for the stress-energy tensor is clear: pressure (compressive stress) is positive and tension (tensile stress--stretching) is negative.
Still not getting that (a side issue in our context, but still). Substitute air-filled balloon for the shell, and internal gas pressure (outward acting) and there one must have tangential hoop tension for stability. But with only inwardly acting self-gravity acting on the shell, hoop stresses must be positive to avoid collapse, surely? The link you gave points out that in the thick wall case, shear stresses owing to gradient of hoop stress as function of radius can matter, but not important for the thin wall case. I just dug this link up: http://arxiv.org/abs/gr-qc/0505040 p11-12 may be of use.

[Q-reeus]

I think this is incorrect. The stress should be exactly the right magnitude to cause the change in the metric you are interested in. If it is a thin massive shell it will be very high and if it is a thick low-density shell then it will be lower. The density will clearly be larger, but that doesn't mean that the lateral stresses are negligible, particularly since they affect different components of the stress-energy tensor and it is precisely these different components that are causing your concern.

I feel like you are deliberately ignoring a clear physical source of the effect you are interested in for no reason other than your a-priori assumption that it cannot possibly be large enough.
No. I think the assessment in #8 settles it nicely. In the #1 scenario, pressure is piddling - the matter is with the effect of matter. If you want, I can pin it down to the shell = a standard 'toy globe' sitting on a typical professor of GR's shelf. Work out the relative contributions there!

[Q-reeus]

...However, when we actually do the packing, we will find the physical volume dV between the two spheres is *larger* than this, by some factor K; and K will get *larger* as we descend in region EV, getting closer to the outer surface of region NV. Finally, at the outer surface of region NV, K will have reached some value K_o > 1...
Sure but is that any more than an alternate way of saying Sr varies as Sr = V (in my notation the -ve sign for V is absent), whereas St is an invariant 1, for all r=< rb? I'm failing to see the need to work only in relative areas and volumes. There must surely be some definite, unambiguos meaning to r here. Gets back I suppose to that other thread where I asked for what a distant observer will see through a telescope - a distorted or undistorted test sphere. A sensibly and physically real Sr implies oblate spheroid will be observed.
As we continue to descend through NV, we find the opposite effect now taking place; the factor K begins to get smaller, and when we reach the inner surface of NV, we find that it is now 1, the same value it has at "infinity". In other words, K_i = 1.
We have reached the flat MM interior, fine.
In region EV, time flow, relative to its rate "at infinity", is slowed by the same factor K that volume between two spheres is increased from its Euclidean value. So at the outer surface of NV, time flow, relative to infinity, is multiplied by a factor J_o = 1 / K_o. However, as we descend through region NV, time flow continues to slow, so when we reach the inner surface of NV, time flow relative to infinity is multiplied by a factor J_i < J_o. Therefore, we can see that J_i is *not* equal to 1 / K_i; the relationship between J and K that held in region EV no longer holds in region NV. And since the rate of time flow is the same throughout NV, the relationship between J and K that held in EV doesn't hold in IV either.

Understand perfectly what you are saying here, but for the moment I have to treat it as just assertion as to relative evolution of K and J. I will be convinced when it can be shown operation of the 'primitives' V and and any combination of derivatives of V, lead to this explicitly. Must be going!:zzz:

[DaleSpam]

No. I think the assessment in #8 settles it nicely. In the #1 scenario, pressure is piddling - the matter is with the effect of matter. If you want, I can pin it down to the shell = a standard 'toy globe' sitting on a typical professor of GR's shelf. Work out the relative contributions there!Figures 2 and 3 of the paper I cited above show less than 1 order of magnitude difference. In any case, the fact that the t-t component of the stress energy tensor is large is completely irrelevant if you are interested in the r-r, theta-theta, or phi-phi components of the curvature.

[PeterDonis]

Sure but is that any more than an alternate way of saying Sr varies as Sr = V (in my notation the -ve sign for V is absent), whereas St is an invariant 1, for all r=< rb? I'm failing to see the need to work only in relative areas and volumes. There must surely be some definite, unambiguos meaning to r here.

There is a definite meaning to "radial distance", yes: in principle, we could line up our tiny measuring objects from some given sphere with area A, all the way to the center of the whole scenario, and count them. The radial distance measured this way, if we start from a sphere in one of the regions where K > 1 (EV or NV), will be larger than the area A divided by 4 pi. However, the exact relationship between the two will be complicated, because it has to take into account how K varies from the sphere with area A all the way to the center of the scenario. It's simpler to state things in terms of the differential area and volume as I did because those quantities are "local", at least in terms of radial movement (and since we're assuming spherical symmetry and time independence, everything can vary only as a function of radial movement), so K can be assumed constant for any given pair of spheres with areas A and A + dA.

But the radial distance, as I've defined it above, is *not* necessarily the same as the radial coordinate r. You can define r to always be the same as the radial distance, but that may not be the easiest definition to work with. More on that in a future post when I talk about coordinates.

Gets back I suppose to that other thread where I asked for what a distant observer will see through a telescope - a distorted or undistorted test sphere. A sensibly and physically real Sr implies oblate spheroid will be observed.

This is a more complicated question as well because it requires you to evaluate the path of the light rays from the object to the distant observer, and the "scaling factor" and time dilation factor will change through the intervening spacetime. It may well be that you are correct that the anisotropy I described (more volume between spheres A and A + dA than Euclidean geometry would lead you to expect) will be seen by a distant observer as a test sphere appearing distorted, but it's not as straightforward a question as you seem to think it is.

Understand perfectly what you are saying here, but for the moment I have to treat it as just assertion as to relative evolution of K and J. I will be convinced when it can be shown operation of the 'primitives' V and and any combination of derivatives of V, lead to this explicitly.

But my whole point is that K and J are the "primitives". K and J are coordinate-independent; they can be directly measured in terms of local observations (K is how much the volume between two spheres with area A and A + dA exceeds the Euclidean value, and J is the observed gravitational redshift/blueshift factor at a given sphere with area A). If by "V" you mean what you have been calling the "potential", the coordinate-independent definition of that would be made in terms of J, the "gravitational redshift" factor, via the usual definition:

J = 1 + 2 \phi

where \phi is the potential in units where c = 1, and with the usual convention that the potential is zero at "infinity" and negative in a bound system such as this one. But this potential is not what is directly observed; that's J.

[PeterDonis]

But the radial distance, as I've defined it above, is *not* necessarily the same as the radial coordinate r. You can define r to always be the same as the radial distance, but that may not be the easiest definition to work with. More on that in a future post when I talk about coordinates.

Ok, now for the "future post" to talk some about coordinates. First, the easy stuff: the spacetime is static, spherically symmetric, and asymptotically flat, so we can easily define angular coordinates \theta, \phi in the usual manner, and a time coordinate t that corresponds to the asymptotic Minkowski time coordinate "at infinity". The metric will then be independent of t, and the only dependence on the angular coordinates will be the r^{2} sin^{2} \theta d\phi^{2} term in the tangential part, which won't be an issue in this problem (and in fact I'll be writing the tangential part as d\Omega^{2} to make clear that there's nothing to see there). So basically the only coordinate where there is anything interesting to figure out is the radial coordinate.

The most familiar radial coordinate for these types of scenarios is the Schwarzschild radial coordinate, which I'll label r; it is defined by

A = 4 \pi r^{2}

for a 2-sphere with physical area A. In terms of this radial coordinate, the metric must take the form:

ds^{2} = - J(r) dt^{2} + K(r) dr^{2} + r^{2} d\Omega^{2}

where J and K are the gravitational redshift and "packing excess" coefficients I defined in my previous post, which must be functions of r only because of what we said above about the other coordinates. Two things are immediately evident:

(1) For this definition of the radial coordinate, the coefficient in front of the tangential part of the metric *must* be r^{2}, and nothing else, because we *defined* r that way (so the area of the sphere at r is 4 \pi r^{2}). (Btw, this observation also shows why K in the metric above *has* to be the "packing excess" coefficient I defined in my previous post.)

(2) In region IV (interior vacuum), K = 1, so the spatial part of the metric with this radial coordinate is exactly in the Minkowski form. The only difference, in terms of this radial coordinate, between IV and the asymptotically flat region "at infinity" is the coefficient J, the gravitational redshift or "time dilation" factor.

However, there is another way to define the radial coordinate, the "isotropic" way; the idea here is that we want the relationship between the radial coordinate and actual physical distance to be the same in all directions (as it obviously is not for the Schwarzschild r coordinate). We'll call this isotropic radial coordinate R, and the metric in terms of it looks like this:

ds^{2} = - J(R) dt^{2} + L(R) \left( dR^{2} + R^{2} d\Omega^{2} \right)

I've put a different coefficient, L(R), in front of the spatial part because now, as you can see, it multiplies *all* of the spatial metric, instead of just the radial part. So the area of a sphere at R is

A = 4 \pi R^{2} L

What does the function L(R) look like? We can stipulate without loss of generality that it must go to 1 at infinity, so that R and r become the same there. We also expect L to get larger as we go deeper into region EV, closer to the outer surface of region NV. But what about in regions NV and IV? I'm still working on that, so I'll follow up with another post.

[PeterDonis]

No. I think the assessment in #8 settles it nicely. In the #1 scenario, pressure is piddling - the matter is with the effect of matter. If you want, I can pin it down to the shell = a standard 'toy globe' sitting on a typical professor of GR's shelf. Work out the relative contributions there!

Actually, since I posted the assessment in #8, I owe you a mea culpa, Q-reeus; after thinking this over and thinking about DaleSpam's comment, I think my assessment in #8 was incorrect, because I was comparing apples and oranges. The Earth is not a hollow sphere made of steel. The case you just gave in the above quote is a much better test case, and as we'll see, it does *not* support the claim that pressure is negligible.

I did say in #8 that the key factor is not whether pressure is small compared to energy density, but whether the ratio of pressure to energy density is small compared to the "correction factor" in the metric coefficients. So let's estimate those numbers for a "toy globe". Let's suppose our "toy globe" is a spherical shell made of steel; its total radius is 1 meter, and the shell is 0.1 meter thick (with a hollow interior). We'll assume that the stress components inside the shell are of the same order as atmospheric pressure, or about 10^5 pascals. (They may actually be somewhat larger, but we'll use the lower bound since that will make the pressure/energy density ratio as small as possible.) The mass density of steel is about 8000 kg/m^3.

Energy density of the shell = 8000 kg/m^3 x c^2 = 7 x 10^19 J/m^3

Ratio of pressure to energy density = about 10^-15

Volume of the shell = 4/3 pi x (1^3 - 0.9^3) = about 1 m^3 (how convenient!).

Mass of the shell = 8000 kg

Correction to metric coefficient at shell surface = 2GM / c^2 r
= (2 x 6.67 x 10^-11 x 8000) / (9 x 10^16 x 1) = about 10^-23

So I was wrong in post #8; for a typical object made of ordinary materials, the pressure is *not* negligible; the pressure/energy density ratio is many orders of magnitude *larger* than the correction to the metric coefficients for the object. So there is plenty of "room" for the spatial parts of the stress-energy tensor inside the object to "correct" the metric to flat from the outside to the inside of the shell. I should note, though, that even the above computation is not really "correct", in that I haven't actually tried to compute any components of the field equation; I've just done a quick order of magnitude estimate of the same quantities I estimated in post #8, to show that the relationship I talked about there doesn't hold for an ordinary object.

[Q-reeus]

Actually, since I posted the assessment in #8, I owe you a mea culpa, Q-reeus; after thinking this over and thinking about DaleSpam's comment, I think my assessment in #8 was incorrect, because I was comparing apples and oranges. The Earth is not a hollow sphere made of steel. The case you just gave in the above quote is a much better test case, and as we'll see, it does *not* support the claim that pressure is negligible.

I did say in #8 that the key factor is not whether pressure is small compared to energy density, but whether the ratio of pressure to energy density is small compared to the "correction factor" in the metric coefficients. So let's estimate those numbers for a "toy globe". Let's suppose our "toy globe" is a spherical shell made of steel; its total radius is 1 meter, and the shell is 0.1 meter thick (with a hollow interior). We'll assume that the stress components inside the shell are of the same order as atmospheric pressure, or about 10^5 pascals. (They may actually be somewhat larger, but we'll use the lower bound since that will make the pressure/energy density ratio as small as possible.) The mass density of steel is about 8000 kg/m^3.

Energy density of the shell = 8000 kg/m^3 x c^2 = 7 x 10^19 J/m^3

Ratio of pressure to energy density = about 10^-15

Volume of the shell = 4/3 pi x (1^3 - 0.9^3) = about 1 m^3 (how convenient!).

Mass of the shell = 8000 kg

Correction to metric coefficient at shell surface = 2GM / c^2 r
= (2 x 6.67 x 10^-11 x 8000) / (9 x 10^16 x 1) = about 10^-23

So I was wrong in post #8; for a typical object made of ordinary materials, the pressure is *not* negligible; the pressure/energy density ratio is many orders of magnitude *larger* than the correction to the metric coefficients for the object. So there is plenty of "room" for the spatial parts of the stress-energy tensor inside the object to "correct" the metric to flat from the outside to the inside of the shell. I should note, though, that even the above computation is not really "correct", in that I haven't actually tried to compute any components of the field equation; I've just done a quick order of magnitude estimate of the same quantities I estimated in post #8, to show that the relationship I talked about there doesn't hold for an ordinary object.
Still struggling to reply effectively to your two very helpful earlier posts, but will deal with this one now. I can't see any mea culpa. First, had in mind the relevant stresses in the globe are those owing only to gravitational self-interaction (i.e. - it's floating out there in space). Vastly smaller than and nothing to do with any atmosperic pressure. But even so, let's take the 'huge' atmospheric pressure relevant figure of ~ 10-15. This is the ratio of relevant contributions to that very tiny metric distortion figure of ~ 10-23. So rho contributes a fraction (1-10-15), while p's contribute a fraction 10-15. Isn't that still the only important consideration on this? How can something 10-15 times smaller than the other be overwhelmingly dominant?! I have a feeling there is some thought of mixing up assumed elastic, material distortions with the underlying, vastly smaller metric ones. The relative contribution of p to the latter is always insignificant for 'steel globes', no? Am I missing something here? More later.

[Q-reeus]

There is a definite meaning to "radial distance", yes: in principle, we could line up our tiny measuring objects from some given sphere with area A, all the way to the center of the whole scenario, and count them. The radial distance measured this way, if we start from a sphere in one of the regions where K > 1 (EV or NV), will be larger than the area A divided by 4 pi. However, the exact relationship between the two will be complicated, because it has to take into account how K varies from the sphere with area A all the way to the center of the scenario. It's simpler to state things in terms of the differential area and volume as I did because those quantities are "local", at least in terms of radial movement (and since we're assuming spherical symmetry and time independence, everything can vary only as a function of radial movement), so K can be assumed constant for any given pair of spheres with areas A and A + dA.

But the radial distance, as I've defined it above, is *not* necessarily the same as the radial coordinate r. You can define r to always be the same as the radial distance, but that may not be the easiest definition to work with. More on that in a future post when I talk about coordinates.
See now I wrongly claimed to have 'perfectly understood' your previous #13 posting. Being unfamiliar with K and J as standard objects in GR, had thought them merely convenient and arbitrary symbols used on an ad hoc basis here. Alright I now finally get it that 'r' as SC's radial coordinate is not really equivalent to physical r, but a derived quantity based on K operating on A. Very confusing but I suppose something has to go in the interests of 'local invariance'. Does this not create a circular definition paradox though: r defined in terms of A, but A expressed in terms of r? Area and volume have to be based on linear measure somehow, so is it not still down to picking a linear length measure as 'the' proper yardstick? How else to climb out of this hole? My hunch is tangent spatial measure is implicitly that yardstick. Which gets to the next part.


Originally Posted by Q-reeus:
"Gets back I suppose to that other thread where I asked for what a distant observer will see through a telescope - a distorted or undistorted test sphere. A sensibly and physically real Sr implies oblate spheroid will be observed."
This is a more complicated question as well because it requires you to evaluate the path of the light rays from the object to the distant observer, and the "scaling factor" and time dilation factor will change through the intervening spacetime. It may well be that you are correct that the anisotropy I described (more volume between spheres A and A + dA than Euclidean geometry would lead you to expect) will be seen by a distant observer as a test sphere appearing distorted, but it's not as straightforward a question as you seem to think it is.
An important one in my mind in untangling SC 'r' from 'actual' r, and so for tangent length. Gravitationally induced optical distortion can in principle be fully accounted for, either directly with corrective optics, or by computer processing of image data, just as e.g atmospheric distortion is compensated for in modern astronomy. Redshift is abberation free in GR so not a problem. You are now quite aware the test sphere was to be considered locally stress free. So only effect distant observer determines is underlying metric distortion. Given all that, and to the extent SM is correct, am I not entitled to expect as distant observer to see a distorted test sphere as oblate spheroid as per the Sr = J (or close to it), St = 1? This is an important part of my reality check list.
But my whole point is that K and J are the "primitives". K and J are coordinate-independent; they can be directly measured in terms of local observations (K is how much the volume between two spheres with area A and A + dA exceeds the Euclidean value, and J is the observed gravitational redshift/blueshift factor at a given sphere with area A). If by "V" you mean what you have been calling the "potential", the coordinate-independent definition of that would be made in terms of J, the "gravitational redshift" factor, via the usual definition:

J = 1 + 2 \phi

where \phi is the potential in units where c = 1, and with the usual convention that the potential is zero at "infinity" and negative in a bound system such as this one. But this potential is not what is directly observed; that's J.
Too late now to re-edit #1, but from now on will be sticking with more standard usage. Using V as label for 'the Einstein potential' -g00 rather than the redshift factor J you have used, created possible confusion with it's more familiar use; as the Newtonian potential -GM/r. Apologies for any confusion caused.

[PeterDonis]

Still struggling to reply effectively to your two very helpful earlier posts, but will deal with this one now. I can't see any mea culpa. First, had in mind the relevant stresses in the globe are those owing only to gravitational self-interaction (i.e. - it's floating out there in space). Vastly smaller than and nothing to do with any atmosperic pressure.

This is true, and I didn't try to calculate what the actual stresses inside the globe due only to its own gravity would be.

But even so, let's take the 'huge' atmospheric pressure relevant figure of ~ 10-15. This is the ratio of relevant contributions to that very tiny metric distortion figure of ~ 10-23. So rho contributes a fraction (1-10-15), while p's contribute a fraction 10-15.

No, this is not correct. The various components of the stress-energy tensor match up with different components of the curvature, which in turn match up with different components of the metric. So the ratio of pressure to energy density may not be relevant to the effect on the metric. That's why I said in my last post that the calculation I gave there wasn't really "correct"; the correct thing to do is to look at the Einstein Field Equation and figure out how each component of the stress-energy tensor affects the curvature, and through that the metric. That's also why DaleSpam has been saying that the pressure may be relevant even though it's much smaller than the energy density. The paper he referenced gives an example where the effects of pressure are certainly not negligible.

[PeterDonis]

Does this not create a circular definition paradox though: r defined in terms of A, but A expressed in terms of r?

No, because I didn't express A in terms of r; I used A to *define* r. A is the "primitive", the actual observable quantity, and I defined it in terms of covering the 2-sphere with little identical objects and counting them.

Area and volume have to be based on linear measure somehow, so is it not still down to picking a linear length measure as 'the' proper yardstick? How else to climb out of this hole? My hunch is tangent spatial measure is implicitly that yardstick.

In the sense that to cover the 2-sphere with little identical objects and count them, you have to place the objects tangentially, yes, I suppose it is. But as I said before, that's because we are assuming spherical symmetry, so the area of the 2-sphere seems like a good benchmark, or "primitive", on which to base other things.

Gravitationally induced optical distortion can in principle be fully accounted for, either directly with corrective optics, or by computer processing of image data, just as e.g atmospheric distortion is compensated for in modern astronomy.

In order to account for gravitationally induced optical distortion, you first have to have a model that says what it is, so it can be compensated for. That's the part that I don't think is very simple: you have to evaluate what a null geodesic will look like as it passes through spacetime with varying curvature.

Also, since a test sphere in this spacetime, to an observer next to it, will *not* appear distorted--it will be spherical--if it does appear distorted to an observer much further away from the black hole, that distortion would have to be "gravitationally induced optical distortion", would it not? If so, you wouldn't want to "compensate" for that, since it would eliminate the effect you are interested in.

[Q-reeus]

...No, this is not correct. The various components of the stress-energy tensor match up with different components of the curvature, which in turn match up with different components of the metric. So the ratio of pressure to energy density may not be relevant to the effect on the metric. That's why I said in my last post that the calculation I gave there wasn't really "correct"; the correct thing to do is to look at the Einstein Field Equation and figure out how each component of the stress-energy tensor affects the curvature, and through that the metric. That's also why DaleSpam has been saying that the pressure may be relevant even though it's much smaller than the energy density. The paper he referenced gives an example where the effects of pressure are certainly not negligible.
Well in that case I have very little idea of how it goes. Got the impression from sundry sources (pop sci maybe) that pressure acts simply as an additive term - apart from obvious complications of spatial pressure/density gradients. As I surmised in #1, there I asume no 'directionality' to the contribution from stress of an element of matter any different to it's mass? Very different when motion/flow is involved of course, but that does'nt concern our situation.

[DaleSpam]

But even so, let's take the 'huge' atmospheric pressure relevant figure of ~ 10-15. This is the ratio of relevant contributions to that very tiny metric distortion figure of ~ 10-23. So rho contributes a fraction (1-10-15), while p's contribute a fraction 10-15. Isn't that still the only important consideration on this? How can something 10-15 times smaller than the other be overwhelmingly dominant?!Because they are "pointing" in different directions (i.e. they affect different components of the tensor). If you were only interested in the time dilation then you would be correct that the pressure is negligible in "ordinary" situations compared to the energy density. However you are specifically interested in the spatial components of the curvature tensor, so the energy density is irrelevant. It is big, but it is in the wrong "direction". Since there is not any momentum flow the only components in the spatial directions are the various normal stress components. You cannot neglect the only source of something you are interested in regardless of how it compares to other things.

[PeterDonis]

Well in that case I have very little idea of how it goes. Got the impression from sundry sources (pop sci maybe) that pressure acts simply as an additive term - apart from obvious complications of spatial pressure/density gradients.

For determining the inward "pull of gravity", you are correct; the relevant quantity is \rho + 3 p, and the pressure p would be negligible in the case we're discussing. However, the "pull of gravity" comes under the heading of "time components" of the curvature. We're talking about "space components", where the energy density does not come into play.

[Q-reeus]

Because they are "pointing" in different directions (i.e. they affect different components of the tensor). If you were only interested in the time dilation then you would be correct that the pressure is negligible in "ordinary" situations compared to the energy density. However you are specifically interested in the spatial components of the curvature tensor, so the energy density is irrelevant. It is big, but it is in the wrong "direction". Since there is not any momentum flow the only components in the spatial directions are the various normal stress components. You cannot neglect the only source of something you are interested in regardless of how it compares to other things.
Something is drastically not adding up here imo. Do you agree that the overwhelmingly larger matter contribution is what will determine/generate the exterior SM? And that spatial anisotropy is inherently present there - owing to that matter contribution? And that there *must* be significant change to at least one spatial metric component upon transition to the flat MM interior? Well how does all this square up with what you have been saying?

[PeterDonis]

Do you agree that the overwhelmingly larger matter contribution is what will determine/generate the exterior SM? And that spatial anisotropy is inherently present there - owing to that matter contribution? And that there *must* be significant change to at least one spatial metric component upon transition to the flat MM interior? Well how does all this square up with what you have been saying?

I think you're confusing the metric and curvature in the exterior vacuum region with the metric and curvature in the non-vacuum "shell" region. Everything DaleSpam and I have been saying applies only to the metric and curvature in the non-vacuum region. In the exterior vacuum region the stress-energy tensor is zero; there is no pressure or energy density, so there is no "matter contribution" at all. The curvature of spacetime in the exterior region is determined by the overall mass of the shell, via the form of the Schwarzschild metric, but that metric is a solution of the *vacuum* Einstein Field Equation (stress-energy tensor equal to zero). The exterior metric does not depend on any details of the shell's internal structure; two shells with identical overall mass, but drastically different energy density (say one is much thicker than the other, and correspondingly much less dense), would lead to the same metric in the exterior vacuum region. But within the shell, the metric would be quite different for those two cases: the spatial metric would have to change to flat over a much shorter distance for the thin shell, so the curvature components would have to change must faster as you descended through the shell.

[Q-reeus]

No, because I didn't express A in terms of r; I used A to *define* r. A is the "primitive", the actual observable quantity, and I defined it in terms of covering the 2-sphere with little identical objects and counting them...

...In the sense that to cover the 2-sphere with little identical objects and count them, you have to place the objects tangentially, yes, I suppose it is. But as I said before, that's because we are assuming spherical symmetry, so the area of the 2-sphere seems like a good benchmark, or "primitive", on which to base other things.
What I meant was not just that r is defined as A = 4*pi*r2, but that from the SC's the θ and phi tangent components are themselves functions of r. So it still seems circular. One must somehow have a good handle on r in order to have that 'primitive' of A determined, surely.
Also, since a test sphere in this spacetime, to an observer next to it, will *not* appear distorted--it will be spherical--if it does appear distorted to an observer much further away from the black hole, that distortion would have to be "gravitationally induced optical distortion", would it not? If so, you wouldn't want to "compensate" for that, since it would eliminate the effect you are interested in.
Is there not a clear sense in which one is optical - light bending, while the other is inherent - the metric distortion as physical 'object'?

[Q-reeus]

I think you're confusing the metric and curvature in the exterior vacuum region with the metric and curvature in the non-vacuum "shell" region. Everything DaleSpam and I have been saying applies only to the metric and curvature in the non-vacuum region. In the exterior vacuum region the stress-energy tensor is zero; there is no pressure or energy density, so there is no "matter contribution" at all. The curvature of spacetime in the exterior region is determined by the overall mass of the shell, via the form of the Schwarzschild metric, but that metric is a solution of the *vacuum* Einstein Field Equation (stress-energy tensor equal to zero). The exterior metric does not depend on any details of the shell's internal structure; two shells with identical overall mass, but drastically different energy density (say one is much thicker than the other, and correspondingly much less dense), would lead to the same metric in the exterior vacuum region. But within the shell, the metric would be quite different for those two cases: the spatial metric would have to change to flat over a much shorter distance for the thin shell, so the curvature components would have to change must faster as you descended through the shell.
Here's where the issue seems to be hitting the fan. I entirely meant shell matter's contribution to the exterior, SM region, and said so explicitly in #25. So no confusion on my part there. Agree with essentially all the rest above, but not the probable implication that stress in the shell can account for anything remotely significant re transition through shell wall. There is a severe logical chasm imo. Exterior, *presumably* anisotropic SM generated by essentially shell matter exclusively. Transition to interior flat region demanding 'steep' gradients in at least one spatial metric component. Relative pressure arbitrarily small in self-gravitating case for 'small' shell. Are we trying to pull a rabbit out of a hat, folks?
Now this would all go away if in fact hte SM is *actually* isotropic spatially, so are SC's fooling us? This is why I want that test sphere result so bad! Bed time.:zzz:

[PeterDonis]

Now this would all go away if in fact hte SM is *actually* isotropic spatially, so are SC's fooling us? This is why I want that test sphere result so bad!

But why would the appearance of a test sphere to someone far away make any difference, when we've already established that, to an observer right next to the test sphere, it would appear spherical, *not* distorted? To me, that local observation is a much better gauge of whether space is isotropic than the observation of someone far away, particularly when the light going to the faraway observer could be distorted by the variation in gravity in between.

Perhaps I introduced confusion when I used the word "anisotropy" to refer to the fact that there is more distance between two neighboring 2-spheres in the Schwarzschild exterior region than Euclidean geometry would lead one to expect based on the areas of the spheres. The only thing that that shows to be non-isotropic is the "Euclidean-ness" of the space. I've already described in detail how that "anisotropy" goes away as you descend through the non-vacuum "shell" region; if you want a "physical" explanation of how that works, I would say it's because the non-Euclideanness of the space is due to the mass of the shell being below you, as you descend through the shell, less and less of its mass is below you. "Below" here just means "at a smaller radius", or, if I were to be careful about describing everything in terms of direct observables, "below" means "lying on a 2-sphere with a smaller area than the one you are on".

It's also worth noting, I think, that even the "non-Euclideanness" of the space, as I described it, is observer-dependent; it assumes that the non-Euclideanness is being judged using 2-spheres which are at rest relative to the shell. Observers who are freely falling through the exterior vacuum region will *not* see this anisotropy in the space that is "at rest" relative to them; in other words, if a freely falling observer were to measure the areas of two adjacent 2-spheres that were falling with him, and measure the distance between the two 2-spheres, he would find the relationship between those measurements to be exactly Euclidean.

[DaleSpam]

Do you agree that the overwhelmingly larger matter contribution is what will determine/generate the exterior SM? And that spatial anisotropy is inherently present there - owing to that matter contribution?The exterior SM is a vacuum solution, so there isn't any matter contribution in the exterior SM.

And that there *must* be significant change to at least one spatial metric component upon transition to the flat MM interior? Well how does all this square up with what you have been saying?That transition requires non-zero spatial components of the stress-energy tensor, i.e. stress and pressure. The large contribution of the energy density is in the wrong place to matter for the spatial components of the curvature.

[DrGreg]

The diagram below of a "Flamm's paraboloid" may aid understanding the curvature of space (but not spacetime) outside the event horizon of a black hole. The surface represents a 2D cross-section through 4D spacetime involving the r and \phi coordinates only. In this diagram r is the radius in the horizontal direction. "Ruler" distances in space (e.g. PeterDonis's method of counting small identical objects packed together) are represented by distances along the curved surface. (The vertical direction has no physical meaning at all.)

http://upload.wikimedia.org/wikipedia/commons/thumb/b/b4/Flamm.jpg/320px-Flamm.jpg (http://commons.wikimedia.org/wiki/File:Flamm.jpg)
Allen McCloud (http://commons.wikimedia.org/wiki/User:AllenMcC.), Wikimedia Commons, CC-BY-SA-3.0 (http://creativecommons.org/licenses/by-sa/3.0/deed.en)

The diagram goes all the way to the event horizon, where the surface becomes vertical at the bottom of the "trumpet". So, for the example being discussed in this thread, you need to slice off the bottom part of the surface. The interior of the shell could then be represented by a horizontal flat circular disk almost capping the bottom of the remaining trumpet, but there is then a gap to be bridged between the disk and the trumpet to represent the shell. I've no idea whether is possible to construct a curved surface to bridge that gap which would correctly represent the geometry within the shell. (I suspect it might not.)

The mathematics of Flamm's paraboloid is discussed on Wikipedia at Schwarzschild metric: Flamm's paraboloid (http://en.wikipedia.org/wiki/Schwarzschild_metric#Flamm.27s_paraboloid).

P.S. Flamm's paraboloid does not represent the gravitational potential. The potential has a somewhat similar shape but it's a completely different formula.

[PeterDonis]

The interior of the shell could then be represented by a horizontal flat circular disk almost capping the bottom of the remaining trumpet, but there is then a gap to be bridged between the disk and the trumpet to represent the shell. I've no idea whether is possible to construct a curved surface to bridge that gap which would correctly represent the geometry within the shell. (I suspect it might not.)

I was thinking of the "bridge" surface as being basically a section of a torus, with the "bottom" part merging into the horizontal circular disk representing the interior, and the "top" part having just the right slope to merge into the exterior paraboloid at the appropriate circular "cut" from it. If "torus" is interpreted generally enough (basically to allow an elliptical "cross section" as well as a circular one), would that be a plausible candidate?

[DrGreg]

I was thinking of the "bridge" surface as being basically a section of a torus, with the "bottom" part merging into the horizontal circular disk representing the interior, and the "top" part having just the right slope to merge into the exterior paraboloid at the appropriate circular "cut" from it. If "torus" is interpreted generally enough (basically to allow an elliptical "cross section" as well as a circular one), would that be a plausible candidate?It certainly sounds plausible -- that's pretty much what I had in mind myself. But on the topic of embedding an arbitrary curved manifold into a higher-dimensional Euclidean space, I am a bit out of my depth. I think, in general, there's no guarantee it can be done with just one extra dimension, i.e. you usually need more. It may be just good luck that Flamm's paraboloid works in just 3 dimensions for the exterior Schwarzschild solution.

And even if you can produce a 2D curved surface in 3D space to represent the entire "shell" spacetime, I'm not sure if the "junctions" (where there's a discontinuity in the energy-momentum-stress tensor from zero to non-zero) would need to be smooth -- maybe there would a sharp "crease" in the surface at these points?

[PAllen]

It certainly sounds plausible -- that's pretty much what I had in mind myself. But on the topic of embedding an arbitrary curved manifold into a higher-dimensional Euclidean space, I am a bit out of my depth. I think, in general, there's no guarantee it can be done with just one extra dimension, i.e. you usually need more. It may be just good luck that Flamm's paraboloid works in just 3 dimensions for the exterior Schwarzschild solution.

And even if you can produce a 2D curved surface in 3D space to represent the entire "shell" spacetime, I'm not sure if the "junctions" (where there's a discontinuity in the energy-momentum-stress tensor from zero to non-zero) would need to be smooth -- maybe there would a sharp "crease" in the surface at these points?

This is leading to a thought I've been having for a while, but didn't want to distract this thread. Simply, why bother with a shell at all? In this case, you have a perfectly continuous but non-smooth manifold joining the SC geometry to Minkowski geometry inside a chosen r value. The metric can be made continuous, a single global chart can be used, you just have metric derivatives undefined on one surface. This is perfectly analogous to taking a ball and cutting it in half, and treating the resulting boundary as a 2-manifold. No one would consider that a weird or inadmissable surface from a geometric point of view. As for physics, the shell-less geometry would lead to failure of EFE on the junction surface (only). It may simplify things if Q-reeus could state his fundamental issue for this simpler case - I still don't get his confusion at all. This shell-less manifold wouldn't be strictly physical, but all physics off the shell, and almost all physics going through the shell (e.g. geodesics) would be perfectly well defined.

[DaleSpam]

Agree with essentially all the rest above, but not the probable implication that stress in the shell can account for anything remotely significant re transition through shell wall.Why not? Your argument regarding the relative size of the pressure and energy density is not relevant. So what logical reason could you have to disagree?

There is a severe logical chasm imo.Agreed. The reference that I posted derived the pressure terms directly from the Einstein field equations. So there is no logical gap there. But you, without touching the math, claim to know that the effect is too small. A logical chasm indeed.

[PeterDonis]

This is leading to a thought I've been having for a while, but didn't want to distract this thread. Simply, why bother with a shell at all? In this case, you have a perfectly continuous but non-smooth manifold joining the SC geometry to Minkowski geometry inside a chosen r value.

The paper I mentioned some posts back, which I haven't been able to find again, basically derived this case as the limit as the shell thickness goes to zero, with the shell mass remaining at a fixed positive value.

[PAllen]

The paper I mentioned some posts back, which I haven't been able to find again, basically derived this case as the limit as the shell thickness goes to zero, with the shell mass remaining at a fixed positive value.

Right, and a zillion posts ago in the black hole thread, I wrote down its metric simply from the continuity of metric requirement (using isotropic SC coords, because it was easier to state the junction that way).

[Q-reeus]

But why would the appearance of a test sphere to someone far away make any difference, when we've already established that, to an observer right next to the test sphere, it would appear spherical, *not* distorted? To me, that local observation is a much better gauge of whether space is isotropic than the observation of someone far away, particularly when the light going to the faraway observer could be distorted by the variation in gravity in between...
There is imo an unhealthy obsession in GR circles with the tendency to look at everything from the 'local' perspective. Casting everything in terms of invariants has it's advantages, but also disadvantages in that one can lose site of the forest for the trees. Typical example - when BH sceptics bring up EH 'time freeze' and related issues, they are brow beaten with the argument that SC's are deceptive/misleading and the only 'proper' perspective is that of the invariant worldline of an infalling observer for whom 'time freeze' etc does not apply. Which to my mind brushes under the carpet serious issues evident from an external observer's perspective.
But enough of generaized criticism. To answer your specific point Peter, just consider redshift. All but the most uninformed newbe has little trouble appreciating that redshift cannot be locally observed because it is inherently a differential issue - clock-rate 'here' vs clock-rate out 'there'. We have no conceptual issue with this (well, those insisting it's exclusively a 'tired light' 'energy drain' thing might). So why should spatial measure be any different? What is the apparent fundamental divide? If effect of metric on time-rate is properly a relational, nonlocally experienced, justifiably physical thing, what allows that length measure - the spatial component of SM in particular, are *not* likewise a physically meaningful, nonlocally observed relational thing? Seems illogical to me.

Want an 'extreme' example? my 'predjudice' is that BH's are, in one important sense, propped up on the basis that, just as SC's indicate, tangent spatial components are unaffected by gravitational potential. If however *all*, spatial components at least, are subject to the J factor, we find that the physical size of a notional BH shrinks to zero before it in fact can qualify as BH - as of course referenced to a distant observer. That leaves out other matters such as whether 'gravity actually gravitates' but indicates that there are drastic consequences as to the proper, physical implications of metric on a relational, 'distant observer' basis. So it's vitally important to know just what that test sphere will, abberation free, actually be *measured* by the distant observer imo. And my assumption is SM via SC's tells us it will be an oblate spheroid with axial ratio J:1. I further think nature has a different idea - it will to first order remain spherical but shrunk. [unavoidably there will be observed second and higher order distortions simply owing to metric necessarily being a function of r in any reasonable metric theory] Hope you all get my drift here.

[Q-reeus]

The exterior SM is a vacuum solution, so there isn't any matter contribution in the exterior SM.
Have you taken a course in "how to be an effective agent provocateur", by any chance? I consider your comments disingenuous. You surely must have read, prior to your above comments in #30 (entry timestamp: 03:30 PM), my own clear statement in #28 (entry timestamp: 12:54 PM)
"I entirely meant shell matter's contribution to the exterior, SM region, and said so explicitly in #25."
You are capable of discerning the fundamental significance of replacing 'in' with 'to' in your above distortion of what I was on about in #25, right?
That transition requires non-zero spatial components of the stress-energy tensor, i.e. stress and pressure. The large contribution of the energy density is in the wrong place to matter for the spatial components of the curvature.
Will have more to say on that claim in responding to your #35

[Q-reeus]

Originally Posted by Q-reeus:
"Agree with essentially all the rest above, but not the probable implication that stress in the shell can account for anything remotely significant re transition through shell wall."

Why not? Your argument regarding the relative size of the pressure and energy density is not relevant. So what logical reason could you have to disagree?
Because it's entirely relevant.
Originally Posted by Q-reeus: "There is a severe logical chasm imo."

Agreed. The reference that I posted derived the pressure terms directly from the Einstein field equations. So there is no logical gap there. But you, without touching the math, claim to know that the effect is too small. A logical chasm indeed.
Yes, and it's you, and the formal system you trust is right, that has the logical chasm problem. You keep implying I'm some kind of ill-informed dumbo, but do so on the basis of distorting what I have actually argued. OK , gloves are off. Should you not duck the challenge, think I'm about to make you look stupid, and happy to do so. Sure, I'm not up with the tensor math and you are. Good for you. But I will claim a certain insight on this issue that either you entirely lack, or are unwilling to acknowledge for whatever reason. So stress in the shell wall solves it all? OK, here's the task for you - who knows the math. In #8 values were given for a monster, 8 ton 'toy globe' subject to ~ atmospheric pressure. Go ahead and work out the self-gravity value instead (floating in space, fully evacuated interior). My 'lazy' estimate - ca 10-30 times the mass contribution to T00 as source of g00, which is all here that determines exterior SM, and the depressed interior MM values (give or take half a dozen orders of magnitude, as if it matters really). The incredibly tiny shell stresses, virtually pure biaxial compressive, somehow can effect a reduction in the tangent metric components going from rb to ra? Convince me please.
Oh, here's a possible fly in the ointment. Add the tiniest puff of fresh, pure mountain air inside the shell. Just a touch. Just enough to reverse the sign of shell hoop stresses and blow the amplitude up by, say, a mere factor of one million. And this is still looking anything but Alice-in -Wonderland nonsense?! Good luck, genius!

[DaleSpam]

To answer your specific point Peter, just consider redshift. All but the most uninformed newbe has little trouble appreciating that redshift cannot be locally observed because it is inherently a differential issue - clock-rate 'here' vs clock-rate out 'there'. We have no conceptual issue with this (well, those insisting it's exclusively a 'tired light' 'energy drain' thing might). So why should spatial measure be any different? What is the apparent fundamental divide?The difference is simply that there is no agreed upon standard for comparing distant lengths as there is for comparing distant times. Because there is no standard you simply have to clearly define the experiment you want to perform in order to determine the length as measured by a distant observer. Depending on the complexity of the experiment it may be difficult to calculate, but in principle you should be able to determine the result of that experiment which you would call the length as measured by the distant observer.

[DaleSpam]

But I will claim a certain insight on this issue that either you entirely lack, or are unwilling to acknowledge for whatever reason.Your insight is simply wrong. It is not based on logic, but comes from ignorance and a prejudice against GR. You dismiss it as illogical and non-self-consistent without bothering with the effort of learning the math which ensures its self-consistency.

So stress in the shell wall solves it all? OK, here's the task for you - who knows the math. In #8 values were given for a monster, 8 ton 'toy globe' subject to ~ atmospheric pressure. Go ahead and work out the self-gravity value instead (floating in space, fully evacuated interior).I will do that. It will take a few days.

My 'lazy' estimate - ca 10-30 times the mass contribution to T00 as source of g00, which is all here that determines exterior SM, and the depressed interior MM values (give or take half a dozen orders of magnitude, as if it matters really).Sure, that is not in doubt. The issue is the relative contribution of the stresses and the mass to g11, g22, and g33. The relative contribution is infinite since the mass does not contribute at all.

The incredibly tiny shell stresses, virtually pure biaxial compressive, somehow can effect a reduction in the tangent metric components going from rb to ra? Convince me please.Yes, but I doubt that you will be convinced.

[Q-reeus]

The difference is simply that there is no agreed upon standard for comparing distant lengths as there is for comparing distant times. Because there is no standard you simply have to clearly define the experiment you want to perform in order to determine the length as measured by a distant observer. Depending on the complexity of the experiment it may be difficult to calculate, but in principle you should be able to determine the result of that experiment which you would call the length as measured by the distant observer.
My angle on this - either SC's have readily testable, remotely determinable, physical consequences for *all* components, or not. If not, we are giving heed to a fantasy.

[Q-reeus]

Your insight is simply wrong. It is not based on logic, but comes from ignorance and a prejudice against GR. You dismiss it as illogical and non-self-consistent without bothering with the effort of learning the math which ensures its self-consistency.
I will accord that this tirade is genuinely motivated. I'll even commend you for doggedly pursuing the topic. But we shall see.
I will do that. It will take a few days.
My 'challenge' was partly rhetorical. You did stop and think a bit about the 'fly in the oinment'? If you do labour on for a few days and somehow manage to stitch a credible answer (boundary fitting magic), I will immediately point you to the 'puff of air' dilemma. This is the difference between knowing the math of received wisdom, and having some insight that looks outside the square. Please, concede on this one. The task of getting a consistent physics here is impossible.

[PeterDonis]

To answer your specific point Peter, just consider redshift. All but the most uninformed newbe has little trouble appreciating that redshift cannot be locally observed because it is inherently a differential issue - clock-rate 'here' vs clock-rate out 'there'. We have no conceptual issue with this (well, those insisting it's exclusively a 'tired light' 'energy drain' thing might). So why should spatial measure be any different? What is the apparent fundamental divide? If effect of metric on time-rate is properly a relational, nonlocally experienced, justifiably physical thing, what allows that length measure - the spatial component of SM in particular, are *not* likewise a physically meaningful, nonlocally observed relational thing?

First of all, as DaleSpam noted, there is an agreed standard for comparing times at distant locations; in fact, the redshift basically *is* that standard. There is no agreed standard for comparing distant lengths.

Second, the redshift is coordinate independent. The "spatial measure" as you are using the term is not, which is why I was so careful in previous posts to describe everything in terms of areas and the "non-Euclideanness" of space, without saying anything definite about "distance measure". See below.

If however *all*, spatial components at least, are subject to the J factor

I think you mean "K factor" since that's the one I defined relative to the spatial components; the J factor affects the time component. The point is that which spatial components are affected by the K factor is coordinate dependent. So the "spatial measure" as you are using the term is not a good way to judge the actual physics.

It's worth walking through this in some more detail. Go back to the picture I gave in terms of 2-spheres with gradually decreasing areas. The areas of these 2-spheres, and the volume in between neighboring 2-spheres, are physical observables; we can measure them by covering the area or packing the volume with little identical objects and counting them. So the factor K, that I defined, is a coordinate-independent quantity and represents actual physics.

However, there are different ways in which this actual physics can be represented in a coordinate system. The different radial coordinate definitions that I described are different ways of *labeling* the 2-spheres, and different labelings lead to different conclusions about which "spatial components" are affected by the K factor. If we use the Schwarzschild r coordinate, we label each 2-sphere with a coordinate r equal to the square root of its physical area divided by 4 pi. With this labeling, only the radial component of the metric is affected by the K factor; the tangential components are not. However, if we use the isotropic R coordinate, meaning that a 2-sphere gets labeled with a "radius" R that does *not* equal the square root of its physical area divided by 4 pi (in the case we're discussing, R will be smaller than that), then all three spatial components *are* affected by the K factor.

we find that the physical size of a notional BH shrinks to zero before it in fact can qualify as BH

It's worth discussing this in a little more detail too. First of all, as I showed above, whether or not all spatial components are affected by the K factor is coordinate dependent, so your logic here is not correct as it stands since coordinate dependent quantities can't describe the actual physics. However, there is a legitimate physical question to be asked: what *is* the K factor, as I defined it physically (the "non-Euclideanness" of space, in terms of the volume between two adjacent 2-spheres compared to their areas), at the EH of a black hole?

The answer is that the question is not valid, because the physical definition of the K factor requires that the 2-spheres in question are spacelike surfaces. But the 2-sphere at the EH, where r = 2M in Schwarzschild coordinates, is not spacelike; it's null. So it is physically impossible to perform the comparison I described using a 2-sphere at the EH, and there is therefore no way to physically define the K factor (or the J factor, for that matter) at the EH.

[DaleSpam]

Please, concede on this one. The task of getting a consistent physics here is impossible.Prove it.

[PeterDonis]

the 2-sphere at the EH, where r = 2M in Schwarzschild coordinates, is not spacelike; it's null. So it is physically impossible to perform the comparison I described using a 2-sphere at the EH, and there is therefore no way to physically define the K factor (or the J factor, for that matter) at the EH.

On re-reading, I should re-state this. The 2-sphere at the EH can still be said to have a physical "area", which is 4M^2 in geometric units. So it may not be technically correct to say the 2-sphere itself is null. (When I compute the norms of the tangential unit vectors, I don't get zero at r = 2M; the norms are still positive, so the unit vectors are still spacelike, assuming I'm doing the computation right).

However, there can't be a static 2-sphere "hovering" at the EH, because the EH, as a surface in spacetime, is a null surface, and therefore there can't be a surface of "constant time" that is orthogonal to the EH, in which the 2-sphere could be said to lie, and in which the area of the 2-sphere could be compared with the volume between it and a neighboring 2-sphere. So the main point in what I said above still holds: it's impossible to do the physical measurement at the EH that I was using to define the K factor.

I should also note that the above does not entirely apply to the J factor; since there are still timelike worldlines passing through the EH, it is still possible to define a "gravitational redshift" factor there, for an infalling observer. However, this factor cannot apply to a static, "hovering" observer at the EH, since as we've seen there can't be one. So what I said does apply to the J factor for static observers.

[Q-reeus]

Originally Posted by Q-reeus: "Please, concede on this one. The task of getting a consistent physics here is impossible."

Prove it.

Was meant as good advice, based on what I wrote in #40
"Oh, here's a possible fly in the ointment. Add the tiniest puff of fresh, pure mountain air inside the shell. Just a touch. Just enough to reverse the sign of shell hoop stresses and blow the amplitude up by, say, a mere factor of one million."
If you choose to reject the basic logic of that bit, then recall - you have committed to proving me wrong by calculations I consider doomed to failure - but go ahead and show that I'm the mistaken one.

[Q-reeus]

I think you mean "K factor" since that's the one I defined relative to the spatial components; the J factor affects the time component.
Took them to be identical in vacuum SM region, but upon looking back in your #9 I see that the defined relation is J = K-1 there. Had assumed the redshift factor J was in terms of frequency, since it declines with descent into lower potential. That is my expectation of how 'coordinate' measure of lengths will go, hence did mean J, but formally should have used K-1.
Originally Posted by Q-reeus: "we find that the physical size of a notional BH shrinks to zero before it in fact can qualify as BH"
Allright, probably should have just said the area of a collapsing 'almost, approaching notional 'BH'' body by this reckoning shrinks indefinitely since tangent metric components would actually shrink by the factor J = K-1, that relation holding everywhere, not just in vacuum. Hence area follows as K-2. By contrast with standard BH there is infinite redshift at EH but finite area. [I believe actual collapsing body would never shrink to a point, but stabilize by matter/radiation ejection and spin at a perfectly finite size where J > 0. Gravity as source term in T would imo have to figure in that.]
So what follows in your comments here are I think perfectly OK only if one already accepts spatial metric permitting finite EH area. Catch 22, seems to me. thanks for your clarification in #47, but same deal.

As in my first passage in #19, still see this thing of defining distance in terms of area as another Catch 22. How do you define area divorced from linear length measure? Those 'packing objects' are LxLxL entities, and one must have a clear definition of L, and same goes with the area thing - area A is an LxL object! If A is the primitive, how do you determine it apart from L measure! Could this be a conundrum forced by need to accomodate difficulties with the BH EH issue? Maybe not, but it's my hypothesis.
Much later. :zzz:

[PeterDonis]

Allright, probably should have just said the area of a collapsing 'almost, approaching notional 'BH'' body by this reckoning shrinks indefinitely since tangent metric components would actually shrink by the factor J = K-1, that relation holding everywhere, not just in vacuum.

I'm not sure I understand you here. Two comments:

(1) Did you read that previous post where I defined J and K carefully? You'll note that I specified there that the relation J = K-1 does *not* hold in the non-vacuum region. I was talking about the "shell" scenario there, but the same would apply for the interior of a collapsing body such as a star.

(2) The factor J does not apply to the tangential metric components; it applies to the time component, since it's the "redshift factor". So I'm not sure how you're concluding that the tangential metric components would "shrink" by the factor J.

By contrast with standard BH there is infinite redshift at EH but finite area.

This is not a contrast with a standard BH. A standard BH does have infinite redshift but finite area at the EH. (More precisely, it has "infinite redshift" for "static" observers at the EH--more precisely still, the limit of the redshift for static observers as r goes to 2M is infinity; there are no static observers exactly at the EH so there is no "redshift" for them at that exact point).

[I believe actual collapsing body would never shrink to a point, but stabilize by matter/radiation ejection and spin at a perfectly finite size where J > 0. Gravity as source term in T would imo have to figure in that.]

Perfect spherical non-rotating collapse is certainly an idealization. But there have been many numerical calculations done of non-idealized collapses, and they still show an EH forming and the body collapsing inside it.

As in my first passage in #19, still see this thing of defining distance in terms of area as another Catch 22. How do you define area divorced from linear length measure? Those 'packing objects' are LxLxL entities, and one must have a clear definition of L, and same goes with the area thing - area A is an LxL object! If A is the primitive, how do you determine it apart from L measure! Could this be a conundrum forced by need to accomodate difficulties with the BH EH issue?

The little identical objects used for packing have to have some linear dimension, yes. But that doesn't commit you to very much since it's for very small objects, so the effects of spacetime curvature can be ignored. As soon as you start trying to deal with size measures over a significant distance, where spacetime curvature comes into play, you have to be a lot more careful. The reason for using those little objects to define area first is that the spacetime has spherical symmetry, so the areas of 2-spheres centered on the origin can be defined without worrying about the curvature of the spacetime. That is not true for radial distance measures, as I have shown.

[Q-reeus]

Originally Posted by Q-reeus:
"Allright, probably should have just said the area of a collapsing 'almost, approaching notional 'BH'' body by this reckoning shrinks indefinitely since tangent metric components would actually shrink by the factor J = K-1, that relation holding everywhere, not just in vacuum."
I'm not sure I understand you here. Two comments:

(1) Did you read that previous post where I defined J and K carefully? You'll note that I specified there that the relation J = K-1 does *not* hold in the non-vacuum region. I was talking about the "shell" scenario there, but the same would apply for the interior of a collapsing body such as a star.

(2) The factor J does not apply to the tangential metric components; it applies to the time component, since it's the "redshift factor". So I'm not sure how you're concluding that the tangential metric components would "shrink" by the factor J.
I understand your comments, but they are referencing to the standard GR view of things. I was addressing things assuming my notion of isometric metric applies, for which the product JK is invariant - same in exterior vacuum as shell wall matter region. Which gets back to finding a self-consistent answer to the shell metric transition problem. With no hope of a resolution via shell stresses, there is what to fall back on?
Originally Posted by Q-reeus: "By contrast with standard BH there is infinite redshift at EH but finite area."
This is not a contrast with a standard BH. A standard BH does have infinite redshift but finite area at the EH. (More precisely, it has "infinite redshift" for "static" observers at the EH--more precisely still, the limit of the redshift for static observers as r goes to 2M is infinity; there are no static observers exactly at the EH so there is no "redshift" for them at that exact point).
Ha ha. Blame myself for not having added a comma after the word 'contrast'. Hope you get the unambiguous meaning now. So as per last passage, you are referring to standard picture, I was contrasting my idea of 'how it ought to be' with the standard, BH's are real, picture.
Originally Posted by Q-reeus:
"As in my first passage in #19, still see this thing of defining distance in terms of area as another Catch 22. How do you define area divorced from linear length measure? Those 'packing objects' are LxLxL entities, and one must have a clear definition of L, and same goes with the area thing - area A is an LxL object! If A is the primitive, how do you determine it apart from L measure! ..."
The little identical objects used for packing have to have some linear dimension, yes. But that doesn't commit you to very much since it's for very small objects, so the effects of spacetime curvature can be ignored. As soon as you start trying to deal with size measures over a significant distance, where spacetime curvature comes into play, you have to be a lot more careful. The reason for using those little objects to define area first is that the spacetime has spherical symmetry, so the areas of 2-spheres centered on the origin can be defined without worrying about the curvature of the spacetime. That is not true for radial distance measures, as I have shown.
Your efforts to educate me haven't been entirely wasted. Finally appreciate, I think, that this definition allows the only unambiguous locally observable measure of curvature effects - as you say, packing ratios vary with 'radius'. But the persistent opinion one cannot decently relate length measure 'down there' to 'out here' is not true if what I have just realized makes sense. Simply apply the well known radial vs tangent c values cr, ct, which in terms of J factor, are cr = J, ct = J1/2 (e.g. http://www.mathpages.com/rr/s6-01/6-01.htm Last page or so). These are naturally the coordinate values. Now as locally to first order in metric everything is observed isotropic, we must have that spatial metric components scale identically to their cr, ct counterparts according to cdt = dx. Settles the matter for me. So with I think a proper handle on how SM predicts metric scale in coordinate measure, will try to test their consistency.
Interestingly, while SC's were telling me tangent component was invariant, from this directionally dependent c perspective, there is in fact tangent shrinkage of a collapsing objects perimeter by factor J1/2. Still not isotropic, but not as 'bad' as I thought before.

[PeterDonis]

I understand your comments, but they are referencing to the standard GR view of things. I was addressing things assuming my notion of isometric metric applies, for which the product JK is invariant - same in exterior vacuum as shell wall matter region.

Have you checked to see that this notion of yours can even be satisfied at all consistent with the Einstein Field Equation? All the things DaleSpam and I have been saying about how the J and K factors change in the non-vacuum shell region are based on the EFE, relating the stress-energy tensor to the curvature. If you're going to just throw out notions without caring if they're consistent with the EFE, then there's no point in discussion, since you're not going to convince anyone else here that the EFE might not be valid under these conditions.

Which gets back to finding a self-consistent answer to the shell metric transition problem. With no hope of a resolution via shell stresses, there is what to fall back on?

DaleSpam and I have both given self-consistent answers that resolve it via shell stresses. The fact that you don't accept them doesn't make them wrong.

But the persistent opinion one cannot decently relate length measure 'down there' to 'out here' is not true if what I have just realized makes sense.

We are not saying there is *no* way to relate local length measures to distant length measures. We are saying there is not a *unique* way to do it, so you have to specify how; and you can't just hand-wave it, you have to actually do some calculating to see how it works out. For example, if you want to do it based on what the distant observer actually sees, you have to work out the paths of light rays.

Simply apply the well known radial vs tangent c values cr, ct, which in terms of J factor, are cr = J, ct = J1/2 (e.g. http://www.mathpages.com/rr/s6-01/6-01.htm Last page or so). These are naturally the coordinate values.

Only in Schwarzschild coordinates, as the page you link to makes clear. In other coordinates the c values work out differently. You can't make any valid claims about the actual physics using things that are only true in a specific coordinate system.

[Q-reeus]

Have you checked to see that this notion of yours can even be satisfied at all consistent with the Einstein Field Equation?
No, because the shell problem was thrown up to indicate, imo, that EFE's, or at least the SM, has problems, so it seems kind of circular to then use EFE's as the yardstick. I threw the problem to the pros for consideration, and have appreciated some useful feedback, but see nothing to this point satisfactorally answering it.
DaleSpam and I have both given self-consistent answers that resolve it via shell stresses. The fact that you don't accept them doesn't make them wrong.
You've thrown me there completely. I recall you suggesting there is room for one via stresses, but can't recollect any actual detailed argument. Could you point to where I may have missed it? As you know, DaleSpam says he will have such an answer, but that's still future. You are aware of my reasons for total scepticism on that. Perhaps you wouldn't mind telling me why the 'puff of air' bit I brought up with DaleSpam in #40 is wide of the mark. To me it seems devastating, but sure I may not understand something basic here.
We are not saying there is *no* way to relate local length measures to distant length measures. We are saying there is not a *unique* way to do it, so you have to specify how; and you can't just hand-wave it, you have to actually do some calculating to see how it works out. For example, if you want to do it based on what the distant observer actually sees, you have to work out the paths of light rays.
Have been looking at some kind of thought experiments along those lines, but re below that is in limbo for the moment.
Only in Schwarzschild coordinates, as the page you link to makes clear. In other coordinates the c values work out differently. You can't make any valid claims about the actual physics using things that are only true in a specific coordinate system.
Thinking about that again, I was too hasty and will probably have to withdraw my claim - there is possible ambiguity about splitting the spatial and temporal contributions to cr, ct not fully thought through. I detect stormy weather here.

[PeterDonis]

No, because the shell problem was thrown up to indicate, imo, that EFE's, or at least the SM, has problems, so it seems kind of circular to then use EFE's as the yardstick. I threw the problem to the pros for consideration, and have appreciated some useful feedback, but see nothing to this point satisfactorally answering it.

Ok, that makes it clearer where you're coming from. If you doubt the EFE, then the whole discussion in this thread is pretty much useless, because everything everybody else has been saying assumes the EFE is valid.

Perhaps you wouldn't mind telling me why the 'puff of air' bit I brought up with DaleSpam in #40 is wide of the mark. To me it seems devastating, but sure I may not understand something basic here.

The puff of air makes the interior region not vacuum; it has a non-zero stress-energy tensor. If the pressure of the air is enough to change the stresses in the shell, then its stress-energy tensor certainly can't be neglected; and a non-vacuum interior region changes the entire problem, because the spacetime in the interior region is no longer flat. Now you're talking about something more like a static model of a planet or a star, just with a weird density profile. (Though again, the density profile can't be that weird, because if the puff of air has enough pressure to significantly affect stresses in the shell, and the air is non-relativistic, then its pressure has to be much less than its energy density, so the energy density of the "air" would be pretty large.) We'll see more specifics when DaleSpam runs the numbers, but these considerations strongly suggest to me that what you have proposed does not have any significant bearing on the original shell problem, where the interior region is vacuum and spacetime there is flat.

[Q-reeus]

Ok, that makes it clearer where you're coming from.
Seriously, you didn't get that till now? How about entry #1! Or my posts over there in the blackhole thread which lead to this one. Now come on.
If you doubt the EFE, then the whole discussion in this thread is pretty much useless, because everything everybody else has been saying assumes the EFE is valid.
Last bit obviously true but at the same time everyone here has understood my sceptical stance. The challenge, and my opinions were clearly set out to all you GR buffs from the start. If you feel an explicit and definitive resolution has been given, I'd love a recap because must have missed it.
The puff of air makes the interior region not vacuum; it has a non-zero stress-energy tensor. If the pressure of the air is enough to change the stresses in the shell, then its stress-energy tensor certainly can't be neglected; and a non-vacuum interior region changes the entire problem, because the spacetime in the interior region is no longer flat. Now you're talking about something more like a static model of a planet or a star, just with a weird density profile. (Though again, the density profile can't be that weird, because if the puff of air has enough pressure to significantly affect stresses in the shell, and the air is non-relativistic, then its pressure has to be much less than its energy density, so the energy density of the "air" would be pretty large.) We'll see more specifics when DaleSpam runs the numbers, but these considerations strongly suggest to me that what you have proposed does not have any significant bearing on the original shell problem, where the interior region is vacuum and spacetime there is flat.)
It was evidently all about contrasting vanishingly small self-gravity contribution to shell stress, with what a tiny mass of air would greatly overwhelm - a mass in turn many orders of magnitude less than that of the 8 ton shell. True I didn't run specific figures because seemed quite evident there was no need given the scenario. One gets a pretty good feel for orders of magnitude with that kind of thing (but perfectly happy to put figures to it if challenged to do so). And that bit of air would matter one hoot re non-flat interior spacetime? But it's ok I think I get what's going on and why. Kind of sad but guess this is the end of your participation. So be it - but thanks anyway.

One last thing though. I think it important to know the nature of contribution to metric that in particular uniaxial stress (being the extreme of stress anisotropy), or biaxial if you like, in some element of stressed matter makes. For instance, element having uniaxial stress axis along polar axis - how different are the radial and tangent SC's generated compared to equivalent element of unstressed mass. Would be fascinating to know that, given it's ability to solve the shell issue. You may not wish to bother with a personal contribution, but pointing to a good resource explaining it in a way a layman can grasp would be appreciated. Cheers.

[PeterDonis]

It was evidently all about contrasting vanishingly small self-gravity contribution to shell stress, with what a tiny mass of air would greatly overwhelm - a mass in turn many orders of magnitude less than that of the 8 ton shell.

But as we keep on saying, it's not the *mass* (or energy density) of the air or the shell that matters, but the spatial stress components, and you specified the scenario in such a way that those can't be negligible: you specifically said that the puff of air had enough pressure to significantly change the stress components inside the shell. That all by itself is enough to ensure that the stress-energy tensor of the puff of air is enough to make spacetime inside the shell non-flat, to the level of accuracy you are assuming. (Obviously the puff of air doesn't affect the flatness of spacetime in a way our normal senses or even fairly accurate instruments can perceive, but then again our normal senses and even fairly accurate instruments can't perceive the self-gravity of a steel ball either. So the whole scenario obviously assumes a much higher level of accuracy than we can currently achieve.)

One last thing though. I think it important to know the nature of contribution to metric that in particular uniaxial stress (being the extreme of stress anisotropy), or biaxial if you like, in some element of stressed matter makes. For instance, element having uniaxial stress axis along polar axis - how different are the radial and tangent SC's generated compared to equivalent element of unstressed mass. Would be fascinating to know that, given it's ability to solve the shell issue. You may not wish to bother with a personal contribution, but pointing to a good resource explaining it in a way a layman can grasp would be appreciated. Cheers.

If I can find a good resource on this specific topic I'll post it. You might try Greg Egan's science pages for some good notes that are at least somewhat related:

http://gregegan.customer.netspace.net.au/SCIENCE/Science.html#CONTENTS

Try in particular the pages on "Rotating Elastic Rings, Disks, and Hoops", since he specifically discusses stress-energy tensors there. All these examples are in flat spacetime, but they might still be at least somewhat relevant to your example.

[PeterDonis]

After digging around in Chapter 23 of MTW, which discusses stellar structure, I found enough info to write down a metric for a static, spherically symmetric object with uniform density. This is not quite the same as the "shell" scenario, but it's close, and may even be close enough to use. (The specific case of a uniform density star is in Box 23.2 of MTW; I'm taking the g_tt and g_rr metric coefficient expressions from equations (6) and (3), respectively, in that box.)

The metric inside the static spherical object is:

ds^{2} = - \left( \frac{3}{2} \sqrt{1 - \frac{2 M}{R}} - \frac{1}{2} \sqrt{1 - \frac{2 M r^{2}}{R^{3}}} \right)^{2} dt^{2} + \frac{1}{1 - \frac{2 m(r)}{r}} dr^{2} + r^{2} d\Omega^{2}

Here R is the radius of the object (i.e., its surface radius, which is constant); r is the Schwarzschild r coordinate (i.e., a 2-sphere at r has physical area 4 pi r^2); M is the total mass of the object; and m(r) is the mass inside radial coordinate r. I have not written out the angular part of the metric in detail since it's the standard spherical form.

At the surface of the object, where r = R, the above expression is identical to the exterior Schwarzschild metric at r = R; so the above is completely consistent with the metric being Schwarzschild in the exterior vacuum region.

The key point, though, is that for r < R, the g_tt term continues to get more negative (note that M, the total mass, appears in g_tt, not m(r)), meaning the "potential" continues to decrease; while the g_rr term gets less positive, closer to 1, finally becoming equal to 1 at the center of the object, r = 0. (Strictly speaking, we have to take the limit as r -> 0 since 1/r appears in the expression; but for uniform density, m(r) goes like r^3, so the expression as a whole goes to zero like r^2.) Since this is the same sort of thing we expected to happen in the "shell" case, it looks to me like the above is basically a degenerate case of the "shell" scenario, where the flat "interior" vacuum region shrinks to zero size (the single point r = 0--note that the spatial metric at r = 0 is flat and the "distortion" is zero).

Note that arriving at this result, as the details in MTW show, requires taking the pressure inside the object into account as well as the density. Equation (7) in Box 23.2 gives the central pressure (at r = 0) as a function of the density:

p(0) = \rho \left( \frac{1 - \sqrt{1 - \frac{2M}{R}}}{3 \sqrt{1 - \frac{2M}{R}} - 1} \right)

For the extreme non-relativistic case, M << R, this approximates to:

p(0) = \frac{1}{2} \rho \frac{M}{R}

So the ratio of pressure to energy density is indeed similar to the ratio of mass to radius (in geometric units); but that's still enough to have an effect on the metric.

In view of the all this, it looks to me like the metric for the "shell" scenario with an interior vacuum region, in the non-vacuum region, ought to look similar to the above; the major changes would be that m(r) would go to zero at some r_i > 0 instead of at r = 0 (so g_rr = 1 at that radius), and that the potential would stop changing at r = r_i, so the g_tt expression might have to change some (maybe replace the r^2 with something that equals R^2 at r = R but goes to zero at r_i). The only thing I'm not sure of is how the different pressure profile required (which has been discussed before) would affect things.

Of course, for the "puff of air inside the shell" scenario, the metric would be very similar to the above; the only difference would be having two uniform-density regions with differing densities (steel, then air), but matching the pressure at the boundary between them. That would mean that g_rr would be very close to 1 at the boundary between steel and air, because m(r) would be close to zero there (most of the mass is in the steel, not the air). It would also mean, I think, that the r^2 in the expression for g_tt would need to be replaced by something that decreased faster in the steel region and slower in the air region, still going to zero at r = 0 (and of course still being equal to R^2 at r = R).

[PAllen]

Q-reeus: Can you try to succinctly state your issue(s) in the context of SC geometry fitted to interior Minkowsdie geometry, with no matter shell at all. Despite the disconinuty in metric derivatives (but continuity of metric itself) all physical observables even ai femtometer away from the 0 thickness shell are well defined. This situation is no different from the junction of ideal inclined plane with a plane. Continuity combined with discontinuity of derivative. Yet this is routinely considered a plausible idealization. So please phrase some specific objection you have to GR physics of the zero width shell.

[pervect]

There's a section in MTW about "boundary" or "junction" conditions in MTW, which shows how to handle spherical shells.

In my copy it's pg 551,the section title is $21.13, look for "Junction conditions" in the index


The intrinsic and extrinsic curvatures of a hypersurface, which played such fundamental roles in the initial-value formalism, are also powerful tools in the analysis of "junction conditions."
Recall the junction conditions of electrodynamics: across any surface (e.g., a Junction conditions for capacitor plate), the tangential part of the electric field, En, and the normal part electrodynamics of the magnetic field, B±, must be continuous

....

Similar junction conditions, derivable in a similar manner, apply to the gravitational field (spacetime curvature), and to the stress-energy that generates it.

[DaleSpam]

Was meant as good advice, based on what I wrote in #40
"Oh, here's a possible fly in the ointment. Add the tiniest puff of fresh, pure mountain air inside the shell. Just a touch. Just enough to reverse the sign of shell hoop stresses and blow the amplitude up by, say, a mere factor of one million."
If you choose to reject the basic logic of that bit, then recall - you have committed to proving me wrong by calculations I consider doomed to failure - but go ahead and show that I'm the mistaken one.Meaning that you cannot prove it.

By the way, the burden of proof is always on the person challenging mainstream established science. However, I will go ahead and calculate the metric for the sphere and the sphere with gas, mostly for my own practice since I don't believe that it will make a difference to you.

[Q-reeus]

By the way, the burden of proof is always on the person challenging mainstream established science.
We've discussed this before, and my response was this is a forum, not a peer-review panel of some prestigious journal, and I'm not a specialist presenting a paper for publishing.
However, I will go ahead and calculate the metric for the sphere and the sphere with gas, mostly for my own practice since I don't believe that it will make a difference to you.
Will be most interested how it is arrived at, how order of unity quantity can be shaped by order of one trillionth of a trillionth effect.

[Q-reeus]

Q-reeus: Can you try to succinctly state your issue(s) in the context of SC geometry fitted to interior Minkowsdie geometry, with no matter shell at all. Despite the disconinuty in metric derivatives (but continuity of metric itself) all physical observables even ai femtometer away from the 0 thickness shell are well defined. This situation is no different from the junction of ideal inclined plane with a plane. Continuity combined with discontinuity of derivative. Yet this is routinely considered a plausible idealization. So please phrase some specific objection you have to GR physics of the zero width shell.
Even with finite thickness shell wall there are discontinuities in derivatives and that I have no problem accepting. Maybe my prejudice but infinitely thin shell sounds like pure boundary matching exercise that can hide physics. So would prefer to stick with explanations involving finite thickness. Later posting wil try and summarise afresh.

[Q-reeus]

But as we keep on saying, it's not the *mass* (or energy density) of the air or the shell that matters, but the spatial stress components, and you specified the scenario in such a way that those can't be negligible: you specifically said that the puff of air had enough pressure to significantly change the stress components inside the shell. That all by itself is enough to ensure that the stress-energy tensor of the puff of air is enough to make spacetime inside the shell non-flat, to the level of accuracy you are assuming...
That is the key sticking point. More later on that. Thanks for the link to Egan's site - a huge resource. Not finding the particulars wanted yet, but will keep looking.

[Q-reeus]

The metric inside the static spherical object is:
ds^{2} = - \left( \frac{3}{2} \sqrt{1 - \frac{2 M}{R}} - \frac{1}{2} \sqrt{1 - \frac{2 M r^{2}}{R^{3}}} \right)^{2} dt^{2} + \frac{1}{1 - \frac{2 m(r)}{r}} dr^{2} + r^{2} d\Omega^{2}

...Note that arriving at this result, as the details in MTW show, requires taking the pressure inside the object into account as well as the density.

...So the ratio of pressure to energy density is indeed similar to the ratio of mass to radius (in geometric units); but that's still enough to have an effect on the metric.

But anything like big enough? I must be missing something basic here, because we all agree shell stresses are utterly minute compared to matter in gross effect. The exterior SM, and interior MM level, owe essentially exclusively to the matter contribution, and shell geometry. Nothing else. Microscopic effects of stress cannot be doing much, regardless of how they 'point', surely! I can only conclude, since all of you insist there is no vast order of magnitude chasm to ford here, that a changing K vs J in shell wall is some kind of mirage, a mathematical artefact of coordinate system. Is this where it's at - is the metric actually an isotropic one according to my conception? I could then believe there is no issue, but can't see it is that way.
Perhaps it best to summarize again the principle issues as I have till now seen them.

1: What SC's are really saying about SM. On a straight reading of the standard SC's

http://upload.wikimedia.org/wikipedia/en/math/e/5/5/e55cd5c7e42dfd5865febb4757f96fb6.png
it is evident potential operates on the temporal, and spatial r components, but not at all on the tangent components. And that this is referenced to coordinate measure, even if for spatial measure it can only be 'inferred' not directly measured (Actually even for temporal component one must infer that clocks tick slower rather than light 'loses energy in climbing out' of potential well). Hence we find frequency slows by factor J = 1-rs/r, and for a locally undistorted ruler placed radially, inferred coordinate measure shrinks by the same factor J.

And it makes perfectly good physical sense. Direct proportionality to J means direct proportionality to depth in gravitational potential. A simple, physical linkage to relative energy level. Somehow, according to straight SC reading, tangent spatials are immune to this principle - in exterior SM region that is. To be clear about what 'immunity' means here, adding mass to the shell while compensating perfectly for any elastic strain in shell wall, a ruler horizontal on the surface will not change as viewed by a telescope looking directly down on it (negligible light bending). whereas a vertically oriented ruler will shrink by K-1 = J. That this is not evident locally is immaterial imo. Locally there is this packing ratio that will change. Fine. But is there not a 'real' transition from anisotropic to isotropic to explain? That will be evident locally - packing ratio change. And non-locally - 'inferred' ruler tangent contraction in passing through hole in the shell wall.

2: Underlying physical principle. Inferred tangent component having by SC's no metric operator in SM region, but obtains one in shell wall. And uber minute stresses explain that? Here's the problem. Diagonals in p, if added in three (isotropic pressure), are utterly puny in effect. And it's not like isotropic pressure is the difference of huge, almost cancelling terms. One just adds arithmetically. Yet take just one away - the radial component in the shell case, and lo and behold, the other two seem to acquire miraculous capabilities. Either that, or as I say, everything is 'really' isotropic and there is no 'real' transition issue to account for. That's how I see it.
Later

[PAllen]

Trying to be as succinct as possible, can you contrast where you see a problem in GR versus Newtonian gravity. In Newtonian gravity, outside the shell, there is a clear physical anisotropy - gravity points toward the shell. Across the shell, this radial force diminishes. Inside the shell there is perfect isotropy. In the weak field case, it is trivial to show GR is identical because it recovers Newtonian potential. So again, I still see no comprehensible claim about what exactly is the problem GR supposedly has.

Another take on this: it is pure mathematics that any invariant quantity computed in isotropic SC coordinates (which still, clearly, have radial anisotropy built in - redshift and coordinate lightspeed vary radially; however, coordinate lightspeed is locally isotropic) is the same as in common SC coordinates. All measurements in GR are defined as invariants constructed from the instrument (observer) world line and whatever is being measured. Thus it is a mathematical triviality that isotropic SC coordinates describe the same physics as common SC coordinates, for every conceivable measurement.

So, can you describe your objection in terms of isotropic coordinates? If you can't, your complaint is analogous to the following absurdity:

- In polar coordinates on a plane, the distance per angle varies radially. How does this effect disappear in Cartesian coordinates?

[PeterDonis]

But anything like big enough? I must be missing something basic here, because we all agree shell stresses are utterly minute compared to matter in gross effect. The exterior SM, and interior MM level, owe essentially exclusively to the matter contribution, and shell geometry. Nothing else. Microscopic effects of stress cannot be doing much, regardless of how they 'point', surely!

I think I see an actual question about physics here, but you are making it far more complicated than it needs to be by mixing in coordinate-dependent concepts. See below.

I can only conclude, since all of you insist there is no vast order of magnitude chasm to ford here, that a changing K vs J in shell wall is some kind of mirage, a mathematical artefact of coordinate system.

No, it isn't, if by K and J you mean those terms as I defined them. I specifically defined them as physical observables, so a change in their relationship is likewise a physical observable. See below.

1: What SC's are really saying about SM. On a straight reading of the standard SC's...

Here is the problem. As I acknowledged above, you are asking a legitimate question about the physics, but you keep on thinking about it, and talking about it, in terms of things that are coordinate-dependent, which makes it very difficult to discern exactly what you are asking. The metric coefficients in Schwarzschild coordinates are *only* applicable to Schwarzschild coordinates; they don't tell you anything directly about the physics. The physics is entirely summed up in terms of the two observables, K and J, that I defined, and everything can be talked about without talking about coordinates at all.

Here's the actual physical question I think you are asking; I'll take it in steps.

(1) We have two observables, K and J, defined as follows: J is the "redshift factor" (where J = 1 at infinity and J < 1 inside a gravity well), and K is the "non-Euclideanness" of space (where K = 1 at infinity and K > 1 in the exterior Schwarzschild vacuum region).

(2) These two observables have a specific relationship in the exterior vacuum region: J = 1/K.

(3) We also have an interior vacuum region in which space is Euclidean, i.e., K = 1. However, J < 1 in this region because there is a redshift compared to infinity.

(4) Therefore, the non-vacuum "shell" region must do something to break the relationship between J and K. The question is, how does it do this?

(5) The answer DaleSpam and I have given is that, in the non-vacuum region, where the stress-energy tensor is not zero, J is affected by the time components of that tensor, while K is affected by the space components. Put another way, J is affected by the energy density--more precisely, by the energy density that is "inside" the point where J is being evaluated. K, however, is affected by the pressure.

(6) Your response is that, while this answer seems to work for J, it can't work for K, because K has to change all the way from its value at the outer surface of the shell, which is 1/J, to 1 at the inner surface. Since J at the outer surface is apparently governed by the energy density, and the change in K to bring it back to 1 at the inner surface must be of the same order of magnitude as the value of J at the outer surface, it would seem that whatever is causing that change in K must be of the same order of magnitude as the energy density. And the pressure is much, much smaller than the energy density, so it can't be causing the change.

Now that I've laid out your objection clearly and in purely physical terms, without any coordinate-dependent stuff in the way, it's easy to see what's mistaken about it. You'll notice that I bolded the word apparently. In fact, the value of J at the outer surface of the shell is *not* governed by the shell's energy density; J (or more precisely the *change* in J) is only governed by the shell's energy density *inside* the shell. At the outer surface, because of the boundary condition there, the value of J is governed by the ratio of the shell's total mass to its radius, in geometric units (or, equivalently, by the ratio of its Schwarzschild radius to its actual radius). And if you look at what I posted before, you will see that the pressure inside the shell is of the *same* order of magnitude as the ratio of the shell's mass, in geometric units, to its radius. So the pressure inside the shell is of just the right size to change K from 1/J at the outer surface of the shell, back to 1 at the inner surface of the shell.

[pervect]

I think I see an actual question about physics here, but you are making it far more complicated than it needs to be by mixing in coordinate-dependent concepts. See below.

No, it isn't, if by K and J you mean those terms as I defined them. I specifically defined them as physical observables, so a change in their relationship is likewise a physical observable. See below.

Here is the problem. As I acknowledged above, you are asking a legitimate question about the physics, but you keep on thinking about it, and talking about it, in terms of things that are coordinate-dependent, which makes it very difficult to discern exactly what you are asking. The metric coefficients in Schwarzschild coordinates are *only* applicable to Schwarzschild coordinates; they don't tell you anything directly about the physics. The physics is entirely summed up in terms of the two observables, K and J, that I defined, and everything can be talked about without talking about coordinates at all.

Here's the actual physical question I think you are asking; I'll take it in steps.

(1) We have two observables, K and J, defined as follows: J is the "redshift factor" (where J = 1 at infinity and J < 1 inside a gravity well), and K is the "non-Euclideanness" of space (where K = 1 at infinity and K > 1 in the exterior Schwarzschild vacuum region).

(2) These two observables have a specific relationship in the exterior vacuum region: J = 1/K.

(3) We also have an interior vacuum region in which space is Euclidean, i.e., K = 1. However, J < 1 in this region because there is a redshift compared to infinity.

(4) Therefore, the non-vacuum "shell" region must do something to break the relationship between J and K. The question is, how does it do this?

(5) The answer DaleSpam and I have given is that, in the non-vacuum region, where the stress-energy tensor is not zero, J is affected by the time components of that tensor, while K is affected by the space components. Put another way, J is affected by the energy density--more precisely, by the energy density that is "inside" the point where J is being evaluated. K, however, is affected by the pressure.


I haven't been following this thread in detail, but I"d like to say that there are known examples where J is affected by pressure. So it's wrong to think that J isn't affected by pressure. The right answer is that J and K are both affected by pressure.

My first attempt at a post wasn't too good, let's hope this one, after replenishing my blood sugar, is better.

If you have a stationary metric , you have a timelike Killing vector, and J has an especially useful coordinate-independent interpretation as the length of said vector, sqrt |\xi^a \xi_a|

If you analyze the case of a shell enclosing a photon gas, you'll find that J, measured just below the surface of the shell is 1 -(2G/R) \int \rho dV rather than 1 -(G/R) \int \rho dV The difference from unity is twice as large, you can think of this as "twice the surface gravity" if you care to think in those terms.

You can think of J as being congtrolled by the Komarr mass, which is the integral of rho+3P, i.e. the Komar mass depends on both pressure and energy density.

However, if you measure J outside the shell, you'll find a sudden increase in J (towards unity, which you can interpret as a REDUCTION of the surface gravity), and J outside the surface of the shell will be equal to 1 -(G/R) \int \rho dV as you might naievely suspect.

The reason for the anti-gravity effect is that the intergal of the tension in the spherical shell is negative. It's a form of exotic matter to have something with a tension higher than it's energy density (which in this case is being oversimplifed to zero, though you can un-over-simplify it to have a more realistic value if you want to bother and want to avoid exotic matter).

For a small system, where you can neglect the gravitational self-energy as a further source of gravity, you can say that the total volume intergal of the pressure cancels out, and the integral of rho+3P is just equal to the integral of rho as the later term is zero.

[PeterDonis]

I haven't been following this thread in detail, but I"d like to say that there are known examples where J is affected by pressure. So it's wrong to think that J isn't affected by pressure. The right answer is that J and K are both affected by pressure.

Hi pervect, yes, this is a good point; in this particular case the pressure contribution to J is negligible (I believe--see below), but in general it might not be.

For a small system, where you can neglect the gravitational self-energy as a further source of gravity, you can say that the total volume intergal of the pressure cancels out, and the integral of rho+3P is just equal to the integral of rho as the later term is zero.

In post #57, if you have time to look, I posted a metric from MTW Box 23.2 for the interior of a static spherical object that is not a "shell", i.e., it has no hollow portion inside it. The total mass M that appears in that metric is defined in MTW as

M = \int_{0}^{R} 4 \pi \rho r^{2} dr

I.e., M does not contain any contribution from the pressure inside the object. If I'm reading MTW correctly here, they don't intend this formula to be an approximation; it is supposed to be exact. They certainly are not assuming that the pressure is negligible compared to the energy density; they explicitly talk about their formulas as applying to neutron stars, for which that is certainly not the case. (The specific metric I wrote down is for a uniform density object, which would not describe a neutron star, but the mass formula above is supposed to be general.) They are, I believe, assuming that the material of the object is ordinary matter, not photons; is it just because of the different energy condition (i.e., no "exotic matter" in this case) that the pressure does not appear in the mass integral, and hence (if I'm reading right) does not contribute to J in this particular case?

[George Jones]

In post #57, if you have time to look, I posted a metric from MTW Box 23.2 for the interior of a static spherical object ... The specific metric I wrote down is for a uniform density object

Schwarzschild's solution.

[PeterDonis]

Schwarzschild's solution.

Yes. It's interesting that Schwarzschild was able to arrive at it, even with the idealization of uniform density, without knowing the Tolman-Oppenheimer-Volkoff equation. MTW's derivation of the metric makes essential use of that equation.

[pervect]

Hi pervect, yes, this is a good point; in this particular case the pressure contribution to J is negligible (I believe--see below), but in general it might not be.



In post #57, if you have time to look, I posted a metric from MTW Box 23.2 for the interior of a static spherical object that is not a "shell", i.e., it has no hollow portion inside it. The total mass M that appears in that metric is defined in MTW as

M = \int_{0}^{R} 4 \pi \rho r^{2} dr

I.e., M does not contain any contribution from the pressure inside the object. If I'm reading MTW correctly here, they don't intend this formula to be an approximation; it is supposed to be exact.


The formula is exact - but read the part in MTW that says that 4 pi r^2 dr is NOT a volume element.

It superfically looks like one at first glance, but isn't. 4 pi r^2 is ok, but dr needs a metric correction. Using the actual volume element, MTW also calculates the integral of rho* dV, dV being the volume element, and find that said integral is larger than the mass M. The quantity \int \rho dV is given a name, the "mass before assembly". Because there is no compression to worry about, (the pieces are modelled as not changing volume with pressure), the only work being done by assembly is the binding energy, which you can think of being taken out of the system as you assemble it - for instance, you might imagine cranes lowering the pieces into place, and work is made available in the process.

You'll see a chart, where they tabluate the binding energy for various sizes too, as I recall.

[Q-reeus]

...For a small system, where you can neglect the gravitational self-energy as a further source of gravity, you can say that the total volume intergal of the pressure cancels out, and the integral of rho+3P is just equal to the integral of rho as the later term is zero...
This bit I had initially forgotten re my 'pufff of air' thing, but recall now from Elers et al paper http://arxiv.org/abs/gr-qc/0505040 cited in #11. Yes, external to shell, complete cancellation of internal gas and shell hoop stress contributions applies. My only interest though was in how insignificant the effect of pressure on given arrangement is, and whether or not external cancellation is considered is effectively moot imo.

[Q-reeus]

Trying to be as succinct as possible, can you contrast where you see a problem in GR versus Newtonian gravity. In Newtonian gravity, outside the shell,

there is a clear physical anisotropy - gravity points toward the shell. Across the shell, this radial force diminishes.
Sure and as indicated in #62 that kind of thing is not an issue because such are gradient functions of potential, and disappearing in the shell interior is a simple consequence of that. Ditto for tidal effects. Not where it's at for me.
Inside the shell there is perfect isotropy. In the weak field case, it is trivial to show GR is identical because it recovers Newtonian potential.
Yes and not an issue for the same reason.
...So again, I still see no comprehensible claim about what exactly is the problem GR supposedly has.
As I've tried real hard but am obviously failing to get it through, there is by the SC's this direct dependence on potential alone for certain metric components but not others that is not so trivial imo - anisotropy of metric itself that leads to 'jump' issues at a shell boundary. Gets down to my belief 'remote' view of spatial component jumps is both real and has physical significance, whereas it seems everyone else here thinks only locally observable physics - 'tidal' effects in essence, matters as far as spatial components go. Can appreciate the latter will have a reasonable smoothness across shell, so sure, none or at least a lesser problem from that perspective.
Another take on this: it is pure mathematics that any invariant quantity computed in isotropic SC coordinates (which still, clearly, have radial anisotropy built in - redshift and coordinate lightspeed vary radially; however, coordinate lightspeed is locally isotropic) is the same as in common SC coordinates. All measurements in GR are defined as invariants constructed from the instrument (observer) world line and whatever is being measured. Thus it is a mathematical triviality that isotropic SC coordinates describe the same physics as common SC coordinates, for every conceivable measurement.
So, can you describe your objection in terms of isotropic coordinates? If you can't, your complaint is analogous to the following absurdity:
- In polar coordinates on a plane, the distance per angle varies radially. How does this effect disappear in Cartesian coordinates?
You might have forgotten an earlier post where I acknowleged ISC's are just a reformulation of standard SC's without deep significance. They do not imply underlying isotropy of metric. It's all a question of whether standard SC's are just a trivial cartesian to polar mapping kind of thing as your last example alludes to, or accurately reflect that SM has this anisotropy implied by the J factor *not* operating on all spatials. Isn't that a statement about the properties curved spacetime surrounding a spherically symmetric mass has? Put it this way, do you agree that if J factor were also applicable to tangent components (with no redefinition of r as per isotropic ISC's), it would be *implying* different physics? Of course so. Will cover what I think is a 'crunch' issue in another posting.

[Q-reeus]

Now that I've laid out your objection clearly and in purely physical terms, without any coordinate-dependent stuff in the way, it's easy to see what's mistaken about it. You'll notice that I bolded the word apparently. In fact, the value of J at the outer surface of the shell is *not* governed by the shell's energy density; J (or more precisely the *change* in J) is only governed by the shell's energy density *inside* the shell. At the outer surface, because of the boundary condition there, the value of J is governed by the ratio of the shell's total mass to its radius, in geometric units (or, equivalently, by the ratio of its Schwarzschild radius to its actual radius). And if you look at what I posted before, you will see that the pressure inside the shell is of the *same* order of magnitude as the ratio of the shell's mass, in geometric units, to its radius. So the pressure inside the shell is of just the right size to change K from 1/J at the outer surface of the shell, back to 1 at the inner surface of the shell.
You seem to have indeed grasped the essence of my objection well. There has been a minor revelation for me that clears one aspect up, but first some comments on above. Externally, J = (1-rs/r)1/2 = (1-2GM/(rc2))1/2, and without any confusion, M is the shell total mass, with pressure an insignificant contribution. And it is understood from the first expression shown in #57 that m(r) is what replaces M on descent through the shell wall re variation in J - outer layers become successively equipotential regions until at inner shell radius it is all equipotential. No confusion there - I think.

Have now come to understand the nature of K used here differently. Had looked at it as equivalent to scale factor for r, but see that only coincidentally applies in external SM region. From your descriptions earlier, I think it can be roughly expressed like K ~ ∂/∂r(V(r)/A(r)3/2), (V the volume, A the area, locally measured) and now see the divergent relationship within the matter region as necessary because of how K is actually defined. Yes see now it has to go to unity inside the equipotential region by this reckoning, unlike J. My continued problem is believing that K can be seriously influenced by shell wall pressure that is physically minute. The numbers just aren't there - one cannot have an ant physically lifting a mountain. More likely surely it's just the different functional dependence on mass as function of radius and nothing more. I'm confident that explicit general expressions for J and K, valid everywhere, would show that only energy density mattered in shell case. Will post on another aspect still far from satisfied about.

[Q-reeus]

Firstly, correction for a silly slip-up: clear through from #1 to #64 have at times written J as J = 1-rs/r, whereas that should have been J = (1-rs/r)1/2 (doesn't however alter relations like J=K-1 in SM region, or the general problem as I see it).
Anyway let's try and settle an aspect I've tried to have pinned down but without prior success. The oft repeated claim is that unlike redshift, spatial components of the metric have no unambiguous meaning or means of measure on a relational 'down there' to 'out here' basis. So here's a scenario:
Planet X orbits at a well defined distance, and astronauts on it's surface have placed both a calibrated frequency source, and self-illuminated ruler. Mass of planet X is known from orbital mechanics, and redshift of frequency source pins down precisely the J factor. We know that astronomers have confirmed gravitational lensing effects predicted by astrophysicists - therefore optical corrections owing purely to gravitional bending of light can and have been accounted for. And similarly local mechanical distortion from gravitational gradient effects can be largely eliminated and/or otherwise completely accounted for. What's left must be entirely owing to the metric itself. Hence viewing the ruler via a telescope presents no inherent difficulty re obtaining corrected-for-all-but-pure-metric-distortion 'true' readings. This can be repeated for another planet Y, differing only in redshift factor J. That allows cross-checking and the capacity to completely eliminate all extraneous influences. We have left just direct metric influence on coordinate measured length scale.

What then will be the correction factor, if:
1: ruler lies in a plane tangent to planet surface, at various heights, and thus potentials, above the planet surface.
2: same as for 1:, but 'limb' readings of ruler alligned radially.

Repeating here my naive expectation: correction factor for case 1 will be unity for any elevation, but J for case 2. Anyone else have a different take? If it is claimed this experiment cannot be done or has no meaning, please explain why not.

[PeterDonis]

Because there is no compression to worry about, (the pieces are modelled as not changing volume with pressure), the only work being done by assembly is the binding energy, which you can think of being taken out of the system as you assemble it - for instance, you might imagine cranes lowering the pieces into place, and work is made available in the process.

Ah, that makes sense. So for the more general case where the density is not constant, and the individual pieces can't be modelled as not changing volume with pressure, there *would* be a pressure contribution to the mass integral, because the "assembly" process would have to do work to compress the pieces, and that would offset some of the gravitational binding energy being taken out; or, put another way, some of the gravitational potential energy in the system when the pieces were very far apart would be converted to compression work, and would therefore show up in the object's final mass, instead of being radiated away as the object was "assembled". So the final mass would be larger--i.e., there would not be as much net binding energy subtracted.

[PeterDonis]

And it is understood from the first expression shown in #57 that m(r) is what replaces M on descent through the shell wall re variation in J

No, you didn't read that expression carefully enough. m(r) appears in the g_rr metric component, but *not* in the g_tt metric component; in the latter, only the total mass M appears. So in physical terms, K becomes dependent on m(r), but J does *not*; it remains dependent only on the total mass M and on the radial coordinate r (but *not* m(r)).

My continued problem is believing that K can be seriously influenced by shell wall pressure that is physically minute.

Did you look at the actual calculations, which show that the pressure is of the right order of magnitude compared to the value of J at the outer surface of the shell?

[PeterDonis]

What then will be the correction factor, if:
1: ruler lies in a plane tangent to planet surface, at various heights, and thus potentials, above the planet surface.
2: same as for 1:, but 'limb' readings of ruler alligned radially.


I can't give an answer to the above because, as I said, it requires calculating the paths of light rays, and that will take some time. However, I do have a couple of questions/comments:

(1) The astronauts themselves, who are next to the ruler, will see no difference if they just look at the ruler; the ruler's length will look the same whether it is placed tangentially or radially. So let's suppose that the observations by telescope from far away *do* show a difference, in accordance with your naive expectation. What will that prove?

(2) If you are correcting for optical distortion due to gravitational bending of light rays, that in itself may change the answer, because if the ruler's length looks the same both ways to astronauts next to it, the only way it can look different when viewed through a telescope far away, it seems to me, is via some sort of distortion of the light rays when traveling through the intervening spacetime with its changing curvature. After all, the astronauts are seeing the ruler via light rays too, and they don't see any difference. So if you assume that all distortions due to how light moves in curved spacetime are eliminated, how can the corrected view through the telescope possibly be any different from the view seen by the astronauts next to the ruler?

[PAllen]

Firstly, correction for a silly slip-up: clear through from #1 to #64 have at times written J as J = 1-rs/r, whereas that should have been J = (1-rs/r)1/2 (doesn't however alter relations like J=K-1 in SM region, or the general problem as I see it).
Anyway let's try and settle an aspect I've tried to have pinned down but without prior success. The oft repeated claim is that unlike redshift, spatial components of the metric have no unambiguous meaning or means of measure on a relational 'down there' to 'out here' basis. So here's a scenario:
Planet X orbits at a well defined distance, and astronauts on it's surface have placed both a calibrated frequency source, and self-illuminated ruler. Mass of planet X is known from orbital mechanics, and redshift of frequency source pins down precisely the J factor. We know that astronomers have confirmed gravitational lensing effects predicted by astrophysicists - therefore optical corrections owing purely to gravitional bending of light can and have been accounted for. And similarly local mechanical distortion from gravitational gradient effects can be largely eliminated and/or otherwise completely accounted for. What's left must be entirely owing to the metric itself. Hence viewing the ruler via a telescope presents no inherent difficulty re obtaining corrected-for-all-but-pure-metric-distortion 'true' readings. This can be repeated for another planet Y, differing only in redshift factor J. That allows cross-checking and the capacity to completely eliminate all extraneous influences. We have left just direct metric influence on coordinate measured length scale.

What then will be the correction factor, if:
1: ruler lies in a plane tangent to planet surface, at various heights, and thus potentials, above the planet surface.
2: same as for 1:, but 'limb' readings of ruler alligned radially.

Repeating here my naive expectation: correction factor for case 1 will be unity for any elevation, but J for case 2. Anyone else have a different take? If it is claimed this experiment cannot be done or has no meaning, please explain why not.

Trying to fill in the missing information in your measurement proposal will explain the difficulties, and also why different reasonable methods will yield different answers. A fundamental observation on SR vs. GR (flat vs. curved spacetime) is that for SR, there is a unique, natural, global system of measurements for inertial observers; this is simply a consequence of the fact that any reasonable way of performing a measurement comes out the same. In GR, this remains true for inertial observers only locally. Once you are making global measurements in GR, different reasonable methods yield different answers.

Ok, so we are looking at an illuminated, distant ruler, say, 1 meter long measured locally. Through the telescope, what we measure is that it subtends some angle. How do we convert that to our claim of 'length at a distance'? Oh, we need the distance to the ruler. How do we get that? Well we can measure light bounce time, but oh, speed of light will vary along the path (or not) depending on conventions chosen. The answer here depends on coordinate choice. Pick the most common SC coordinates, for example. Now propose instead measuring distance with a long ruler; not possible in practice, but can be modeled mathematically - given a choice of simultaneity. Which one to use? One common one amounts to extending a spacelike geodesic 4-orthogonal to your world line, and measuring its ruler length (which is equivalent to saying one end of ruler doesn't look like it is moving relative to the other). Guess what, you will get a different answer for distance than a radar based approach. Ok, pick some set of answers for all these choices.

Now we face distant measurement of the end on ruler. Well, angle subtended is irrelevant. Nothing you can see in the telescope is relevant. So, maybe have a half slivered mirror on the closer end of the ruler and regular mirror on the other, and compare round trip light time difference. But what speed of light to use? Now for consistency, you better use distance measured by light travel time for the perpendicular ruler.

So, you can think this through a lot more and come up with a well defined set of measurements that you will interpret to be length of the ruler in two orientations at a distance. If your procedure is fully defined, there will be a unique, answer given all the choices you have made. However, you would have a hard time defending your choices against other reasonable ones that would yield a different result.

[Q-reeus]

No, you didn't read that expression carefully enough. m(r) appears in the g_rr metric component, but *not* in the g_tt metric component; in the latter, only the total mass M appears. So in physical terms, K becomes dependent on m(r), but J does *not*; it remains dependent only on the total mass M and on the radial coordinate r (but *not* m(r)).
Oops, yes my mea culpa re m(r). But what I have not got there is how p is incorporated into m(r)? I would expect we just have a net source density in the shell wall given by ρt = ρm + 2p, where ρt, ρm are the total and matter only contributions. So dm = ρtdx3. Now while you have given the relationship p(0)=1/2ρM/R, applying at I think the center of a fluid sphere of uniform density, this would probably be quite a deal larger than for a thin shell, but I guess specifics are in the pipeline on that. From http://en.wikipedia.org/wiki/Schwarzschild_radius we have
M = Gm/c2, and rs = 2M, so for that p(0) expression, one gets as you said p(0) = ρrs/r, which is exceedingly small. The fractional p modification to m(r) is then of the order 2rs/r as upper limit (center of sphere). Thats my reading of it anyway.

[Q-reeus]

...I can't give an answer to the above because, as I said, it requires calculating the paths of light rays...
No it won't. I specified we have accounted for gravitational *bending* of light - it is taken to be subtracted out, either by computer or clever lens optics. If astrophysicists can calculate distortion effects of 'g lensing', those effects must in principle be able to be subtracted out. We just want the 'raw' effects of spacial components of spacetime curvature 'right there at the ruler'.
...However, I do have a couple of questions/comments:

(1) The astronauts themselves, who are next to the ruler, will see no difference if they just look at the ruler; the ruler's length will look the same whether it is placed tangentially or radially. So let's suppose that the observations by telescope from far away *do* show a difference, in accordance with your naive expectation. What will that prove?
We will have established to what degree SC's tell us the 'true' values of gravitational length changes - on a coordinate basis.
(2) If you are correcting for optical distortion due to gravitational bending of light rays, that in itself may change the answer, because if the ruler's length looks the same both ways to astronauts next to it, the only way it can look different when viewed through a telescope far away, it seems to me, is via some sort of distortion of the light rays when traveling through the intervening spacetime with its changing curvature. After all, the astronauts are seeing the ruler via light rays too, and they don't see any difference. So if you assume that all distortions due to how light moves in curved spacetime are eliminated, how can the corrected view through the telescope possibly be any different from the view seen by the astronauts next to the ruler?

There must be a sense in which we can cleanly separate gravitational 'lensing' distortions which are an accumulated effect of transverse bending of light rays, from the metric spatial contractions that are just 'there'. Can't be all smoke and mirrors - SM must be telling us something definite via SC's, or whatever coordinate scheme is deemed relevant. I want to be clear this is not some arduous exercise I'm imposing. Not asking anyone to actually perform all those lensing correction calcs etc an astronomer might need. We just note such 'extraneous' influences exist and claim it's possible in principle to completely factor them out. This kind of thing is done all the time in other arenas - 'corrective optics' is fact. What I'm asking is, are SC's (or equivalent) implicitly predicting anisotropy of spatial components - on a 'down there' vs 'out here' basis? Just read PAllen's comments and wonder if there is any agreed sense of anything that can be got here.

[Q-reeus]

So, you can think this through a lot more and come up with a well defined set of measurements that you will interpret to be length of the ruler in two orientations at a distance. If your procedure is fully defined, there will be a unique, answer given all the choices you have made. However, you would have a hard time defending your choices against other reasonable ones that would yield a different result.
Yikes! How on earth will astronomers ever get a handle on testing contending theories for GBH candidates!!? Look I recognize what you are saying about practical issues, but is there no sense of what's 'actually' going on at the ruler? If we say mass distorts spacetime, is there no sense that we can use SC's to simply predict the spatial part 'down there'? I'm astonished to be reading that it seems an in principle impossibility. Good grief! :zzz:

[PAllen]

Yikes! How on earth will astronomers ever get a handle on testing contending theories for GBH candidates!!? Look I recognize what you are saying about practical issues, but is there no sense of what's 'actually' going on at the ruler? If we say mass distorts spacetime, is there no sense that we can use SC's to simply predict the spatial part 'down there'? I'm astonished to be reading that it seems an in principle impossibility. Good grief! :zzz:

The curvature of spacetime affects only measurements over some span of time or distance. A defining feature of (semi)Riemannian geometry is that sufficiently locally, spacetime is identical to flat Minkowski space. This is not different that the tangent to a curve approximates it arbitrarily well over a sufficiently small length.

The statement that, for global measurements, there is no unique way to factor curvature to effects on distance, time, and light speed is no more surprising than the statement that there are many useful projections for representing a globe on flat map. Further, for any such projection, you can choose where the biggest distortions are (you can pick any two antipodes to function as the poles).

Of course, you can relate measurements to SC coordinates; you can also relate them to Isotropic SC coordinates; or any of several other popular choices. The choice doesn't affect predictions of actual measured values of anything, but it definitely affects how you interpret what those measurements say about distant events - down to the most basic question of how far away they are.

[PeterDonis]

But what I have not got there is how p is incorporated into m(r)?

Read pervect's exchange with me a few posts ago. For the particular case I posted the metric for, constant density, there is no pressure contribution to m(r), because the process of "assembling" the object doesn't do any compression work (because constant density implies that the individual pieces of the object are not compressible). This is obviously an idealization. For a real object, m(r) will include a contribution for the work required to compress the pieces of the object that are inside the radial coordinate r, from their size "at infinity" to their (smaller) size when they are part of the object. That will be a function of the pressure at r.

Also note that, as pervect pointed out, I should not have implied that J is determined by the energy density and K by the pressure. In fact J and K are *both* affected by the energy density *and* the pressure, in a real object (where m(r), and hence the total mass M, include a contribution from the pressure). However, there is an additional factor to consider: the solution for the "constant density" metric that I gave depends on the pressure, because deriving the form of the metric components requires solving the Tolman-Oppenheimer-Volkoff equation, which is the relativistic equation for hydrostatic equilibrium (i.e., the balance between pressure and gravity). So even if, in the idealized case I posted, the pressure does not appear to contribute to J and K (because the constant density assumption implies no compression work when "assembling" the object), the pressure is still essential because the form of the metric is determined by the balance of pressure and gravity within the object.

[PeterDonis]

What I'm asking is, are SC's (or equivalent) implicitly predicting anisotropy of spatial components - on a 'down there' vs 'out here' basis?

I know it seems to you that you are asking a genuine question here, but I don't think it's actually a well-defined question at all. Consider the following analogy:

Suppose I live at the North Pole, and I set up a coordinate grid to label points near my home. We'll idealize the Earth as a perfect sphere to avoid any complications from oblateness. I draw a series of circles with gradually increasing circumference around my home, and label each circle with a "radial coordinate" r, defined such that the circumference of a circle with radial coordinate r is 2 \pi r. (My house is then at r = 0.) I then define an angular coordinate \phi to label the different directions I can look in from my house, so I can label any point with a pair of coordinates (r, \phi).

If I then start measuring physical distances between points, what will the metric for my little coordinate grid look like? It will look like this:

ds^{2} = \frac{1}{1 - \frac{r^{2}}{R^{2}}} dr^{2} + r^{2} d\phi^{2}

where R is a particular constant number that I find popping up in all my distance calculations, which happens to have dimensions of a length. (If I try to determine R's exact value by really accurate measurements, I will find it to be 6.378 x 10^6 meters.) (I should also note, by the way, that this thought experiment is only intended to cover the region near the North Pole; if I were to extend my measurements down past the Arctic Circle, I would start to see errors in the formula above and would have to add additional terms in the denominator of g_rr, with higher powers of the ratio r / R. We won't cover that here.)

You will notice, of course, that this metric has the same general property as the metric for the spacetime around a black hole, what I have called the "non-Euclideanness" of space. Suppose I first measure the circumference of a circle at radial coordinate r very precisely by lining up little identical objects around it. Then I measure the circumference of a slightly larger circle at r + dr the same way. Then I measure the distance between the two circles by lining up the same little identical objects between them. I will find that there is *more* distance between the circles than Euclidean geometry would lead me to expect, based on their circumferences. I should emphasize that, even though we discovered this property by looking at a specific expression for the metric in this "space", in a specific set of coordinates, the property itself is an actual physical observable, just as it is in the spacetime around a black hole. The number of little identical objects that can be lined up between two nearby circles, relative to the number of little identical objects that can be lined up around the circles' circumferences, is independent of the coordinates I use to describe the space.

What should we make of the "anisotropy" I have just described? We might wonder if it is a sign of a genuine "anisotropy of space", but we can quickly dispense with that by going to various circles and verifying that, locally, objects appear the same to us, with no distortion, regardless of which circle we are on. But what about "down here vs. up there"? Could it not be that, "from far away", there is a genuine "distortion" in this space?

I am wondering about all this one day when a friend shows up with a helicopter and offers to give me a view of the area from above. It just so happens I have little identical objects laid out all over the place, and I tell my friend to fly me around to give me a look at them from the air. Consider again two circles at r and r + dr. If I am looking down on these circles from directly above a point on one of them, I will see no distortion. But if I hover directly over my house, and look down on the circles from that angle, I will see that the little identical objects appear to be packed more tightly the further away the circles are from my house; they appear to be "contracted" in the radial direction while maintaining the same size in the tangential direction.

So what I see from "far away" depends on how I look, and therefore it can't tell me whether the little objects "really are" packed more tightly, or "contracted". In a curved space, there simply isn't a unique answer to such questions for distant objects; to see how an object "really is", you have to get close to it. There's no alternative.

Postscript (added by edit): Suppose that while I am hovering in the helicopter and looking at the distorted objects, I have an idea: what if I apply an optical "correction" to the image I see, to correct for the fact that I am looking at the circles "at an angle"? Well, what correction should I apply? I can certainly apply an image transformation that converts the image I see from above my house, at r = 0, to the image I would see from above a circle at some other radius r. As we've seen, that would remove the apparent distortion for the little objects at radius r. Does that mean they're not "really" distorted? There is no unique answer to this question. I can figure out how an object at r would appear from any vantage point I want, but there's nothing that singles out any particular vantage point as the "real" one, the one that determines how things "really are"--except, as I said before, the *local* vantage point, the one as seen by an observer right there, at the circle at r (not even hovering above it, but *at* it).

[pervect]

Ah, that makes sense. So for the more general case where the density is not constant, and the individual pieces can't be modelled as not changing volume with pressure, there *would* be a pressure contribution to the mass integral, because the "assembly" process would have to do work to compress the pieces, and that would offset some of the gravitational binding energy being taken out; or, put another way, some of the gravitational potential energy in the system when the pieces were very far apart would be converted to compression work, and would therefore show up in the object's final mass, instead of being radiated away as the object was "assembled". So the final mass would be larger--i.e., there would not be as much net binding energy subtracted.

There is a pressure contribution even when the pieces don't change volume. Even though the pressure doesn't do any work, it alters (increases) the gravitational field.

To be specific, if you look at the gravitational field of a contained photon gas, by measuring the gravity field just inside the outer edge of the container so that the container doesn't contribute, you find that it generates more gravity than you'd expect if gravity were due to E/c^2. (Which is not the case, and this example illustrates why).

For static gravity, you can think of (rho+3P) as the source of gravity. So in simple terms, for static systems (and only for static systems) you can think of gravity as being caused by a scalar, but the scalar is not the energy, relativistic mass, invariant mass, or anything else from special relativity.

The important quantity (for static systems) is rho+3P. The pressure doesn't cause gravity by doing work and contributing to the energy density. The pressure causes gravity just be existing.

The tension in the container doesn't change it's special relativistic mass if the container does not expand. There's no work done on the container if you pressurize the interior.

It does, however, change the gravitational field that the container produces, even when the container does not expand. You can't really quite test this directly, because in order for the container to be in tension, it has to have some contents which cause the tension, pure radiation being the thing that will produce the most tension for the least amount of added mass-energy.

However, you can make the container spherically symmetrical, and measure the surface gravity inside and out. When you do this, you find that the container under tension adds less to the gravity than it would if it were not under tension. If you idealize the container to having zero mass, while still being under tension, it will actually subtract from the gravitational field.

[PeterDonis]

There is a pressure contribution even when the pieces don't change volume. Even though the pressure doesn't do any work, it alters (increases) the gravitational field...

I understand and agree that the pressure contributes to the stress-energy tensor, and hence to the Einstein tensor (or Ricci tensor). That's where the \rho + 3p comes from. I also understand and agree that the pressure doesn't have to do any work to appear in the Ricci tensor.

But in post #71 you agreed that, for the particular idealized case I was talking about, a spherical object with constant density, the formula I quoted from MTW for the total mass M was exact. That formula only contains \rho; it does not contain p. So in this particular case, it appears to me that the pressure does not contribute to the mass M that appears in the metric. The pressure still affects the object's internal structure through the equation of hydrostatic equilibrium (and this also affects the form of the metric, i.e., where and how the mass M appears in it); but it doesn't, in this idealized case, contribute to M.

(Actually, looking again at MTW, they seem to be saying that the equation for the mass inside radius r, m(r), applies even when the density isn't constant. So it looks like they're saying the pressure doesn't contribute to the total mass M that appears in the metric for any spherical object whose stress-energy tensor is of the form of a perfect fluid. That means I was wrong a few posts ago when I said pressure would contribute to the mass integral when the density wasn't constant. The only contribution the pressure could make to the mass of the object would be indirect, by affecting the density profile of the object; for example, any gravitational potential energy that got converted to compression work instead of being radiated away during the assembly process would show up as increased density, and would increase the mass that way.)

[PeterDonis]

We just want the 'raw' effects of spacial components of spacetime curvature 'right there at the ruler'.

Just re-read the thread and saw this phrase, which I must have missed before. I've given the answer to this one several times: to an observer "right there at the ruler", the ruler will look the same whether it's placed radially or tangentially. There will be no distortion.

[Q-reeus]

Of course, you can relate measurements to SC coordinates; you can also relate them to Isotropic SC coordinates; or any of several other popular choices. The choice doesn't affect predictions of actual measured values of anything, but it definitely affects how you interpret what those measurements say about distant events - down to the most basic question of how far away they are.
Qualified agreeance on that. Still feel there are 'remotely determined' coordinate independent spatial relations one should be able to tie down. Will relate some more in another post.

[Q-reeus]

Read pervect's exchange with me a few posts ago. For the particular case I posted the metric for, constant density, there is no pressure contribution to m(r), because the process of "assembling" the object doesn't do any compression work (because constant density implies that the individual pieces of the object are not compressible). This is obviously an idealization. For a real object, m(r) will include a contribution for the work required to compress the pieces of the object that are inside the radial coordinate r, from their size "at infinity" to their (smaller) size when they are part of the object. That will be a function of the pressure at r.
Have always understood it that stress induced elastic/hydrodynamic energy contributions aught to be incorporated into the T00 source term - i.e. just an addition to mass density, and thought everyone else saw it that way. Obviously assuming incompressibility assumes zero contribution from that. Thought we were just discussing the relative contribution of the 'pure pressure' terms p11, p22, p33 to m(r).
...In fact J and K are *both* affected by the energy density *and* the pressure, in a real object (where m(r), and hence the total mass M, include a contribution from the pressure).

Agreed, but as having insisted since it was first raised in #3, relative contribution of pressure (whether 'pure' pressure or via elastic energy density) is essentially zero in shell case, and it's time to give that one a quiet burial. We all know what results of a certain undertaking will show. The pressure thing has really become a sidetracking issue, and it was my confusion in seeing K-1 as entirely equivalent to J that sustained my interest given insistence that p terms entirely determined K's evolution within the shell. Once I understood that JK-1 = 1 is a coincidental thing - 'a perversity of spherical symmetry' in exterior region, this no longer matters to me.
...the pressure is still essential because the form of the metric is determined by the balance of pressure and gravity within the object.
Which I here interpret as referring to 'p only' terms p11 etc.

[Q-reeus]

So what I see from "far away" depends on how I look, and therefore it can't tell me whether the little objects "really are" packed more tightly, or "contracted". In a curved space, there simply isn't a unique answer to such questions for distant objects; to see how an object "really is", you have to get close to it. There's no alternative.
Allright, you've done good job explaining the perspective issue and need for local invariants, in my terms - thanks. Perhaps there is another angle to this worth looking at.
Suppose we have an inflatable sphere centered within a transparent and initially unstressed elastic medium. Inflating the sphere slightly creates radial compressive and tangential tensile hoop stresses, and corresponding small displacements - the medium expands non-uniformly. In polar coord terms, the perturbed changes in radial and tangential strain and displacement (the integration of strain over distance) can be expressed as factors operating on the polar ordinates. A tiny elastic being caught up in it all cannot sense this directly - only 'tidal' elastic strain is locally evident. Yet in the lab, there is a need to relate changed, stress-strain induced optical properties (e.g. light deflection) which require knowledge of the elastic perturbations - both strain and displacement.

No point asking elastic being who knows only 'tidal' effects. But having a good handle on medium properties and knowing the sphere inflating pressure, all parameters of interest are readily calculable. And it necessarily assumes definite 'before' vs 'after' relations that from the lab must be inferred. Do we agree that, regardless of the particular coordinate chart used, elastic deformation and total displacement of any given elastic element should here be considered physically meaningful, coordinate independent quantities (and recall it is perturbative, before/after differences we want)? I should think yes. Expressed in say polar coords, that in turn locks down the radial and tangent strain factors say, to definite relationships if proper, accurate calculations and predictions are to be possible. Allowing treatment of both local (stress/strain), and non-local (displacements, optical paths) phenomena.
I believe gravitational light bending, on a geometric interpretation, assumes something entirely analogous if I'm not mistaken. So what this amounts to is - is gravity really that different one cannot say equivalent things - perturbative factors precisely defined? Still have a hangup on this - sorry.

[PAllen]

Qualified agreeance on that. Still feel there are 'remotely determined' coordinate independent spatial relations one should be able to tie down. Will relate some more in another post.

The problem with spatial relations is they are tied to decisions about simultaneity. 3-space 'now' is global statement about simultaneity. This is well defined for inertial observers in flat spacetime. Otherwise, it is just not well defined, there a number of perfectly reasonable choices. Given different choices for distant simultaneity, you get different conclusions about spatial relationships.

[PeterDonis]

Have always understood it that stress induced elastic/hydrodynamic energy contributions aught to be incorporated into the T00 source term - i.e. just an addition to mass density, and thought everyone else saw it that way. Obviously assuming incompressibility assumes zero contribution from that.

Yes, work done on the system by compression shows up in the energy density (rho, or T_00); that's why I corrected myself in my exchange with pervect about that.

Thought we were just discussing the relative contribution of the 'pure pressure' terms p11, p22, p33 to m(r).

Not quite. As I said in a previous post, the formula for m(r) in MTW is generally applicable; it says that the pressure does *not* contribute directly to m(r) for any spherically symmetric object; m(r) is *just* an integral over the energy density. The pressure only contributes to the mass indirectly, through hydrostatic equilibrium.

Which I here interpret as referring to 'p only' terms p11 etc.

Just to clarify terminology, the "pressure" p is just another name for the diagonal space components of the stress-energy tensor, in the case where that tensor is spatially isotropic; in other words, p = T_11 = T_22 = T_33. (The energy density rho = T_00.) Also, this assumes that we are in the rest frame of the object, so each little piece of it that is ascribed the pressure p is at rest in the coordinates we are using.

The way the pressure affects the metric in this scenario is through the r-r component of the Einstein Field Equation, G_rr = 8 pi T_rr. (T_rr is what I was calling T_11 above, if we are using spherical coordinates.) This equation leads to the Tolman-Oppenheimer-Volkoff equation, which describes hydrostatic equilibruim in GR:

http://en.wikipedia.org/wiki/Tolman%E2%80%93Oppenheimer%E2%80%93Volkoff_equatio n

The derivation of the metric for the constant density case in MTW, which I quoted from earlier, makes essential use of this equation.

[PeterDonis]

Suppose we have an inflatable sphere centered within a transparent and initially unstressed elastic medium. Inflating the sphere slightly creates radial compressive and tangential tensile hoop stresses, and corresponding small displacements - the medium expands non-uniformly...

...I believe gravitational light bending, on a geometric interpretation, assumes something entirely analogous if I'm not mistaken. So what this amounts to is - is gravity really that different one cannot say equivalent things - perturbative factors precisely defined? Still have a hangup on this - sorry.

I see the analogy you are trying to make, but let's dig into it a little deeper.

Viewing the non-Euclideanness of space around a black hole as an elastic distortion in the space has been tried; I believe Sakharov, for one, came up with a reformulation of relativity along these lines. I'm not saying it's an invalid analogy, but to make sense of it and see what it can and can't tell you, you have to first define what the "unstressed" state of the space is, so to speak. Is it the Euclidean state? Let's suppose it is.

The general method of dealing with elastic deformation (as described, for example, in the Greg Egan pages I linked to in an earlier post) is to label each point in the elastic object by its unstressed location, and use the label of a given point to track it as it moves, relative to other points, due to the stresses imposed. The analogous procedure for spacetime would be to label each event by its "Euclidean" coordinates, and interpret those as "unstressed" distances, and then track the actual, "stressed" distances relative to them. This is, in fact, what Schwarzschild coordinates can be viewed as doing; the Schwarzschild r coordinate can be viewed as the "Euclidean radius" of a point, and the actual distance given by the Schwarzschild metric can be viewed as the "stressed" distance, due to "elastic deformation" of the space.

The problem with this analogy is, as I said before, that in the spacetime case, a small object sitting at r is *not* deformed; it looks the same from every direction, just as it would in an "unstressed" flat space. The "deformation" is only visible globally, and only as a non-Euclideanness in the relationship between radial distances and tangential areas. (Note that you can't just say radial and tangential distances here, though you could say tangential *circumferences*, and some do; the key is that you can only spot the non-Euclideanness by measuring distances around an entire circle, or sphere, at "radius" r, *not* by just measuring small distances tangentially.) Also, a small object *feels* no stress just from this non-Euclideanness of space; put strain gauges in it and they will all read zero. This is *not* the case with normal elastic deformation; if I take a small spherical portion of an unstressed elastic object, label it somehow so I can see its boundary, and then stress the object, that small spherical portion will appear deformed *locally*, when I look at it from right next to it. I won't have to make global observations to spot it. And if I put strain gauges in that little spherical portion, they will register nonzero values.

Go back to the analogy with the house at the North Pole and circles around it. You can set up the same sort of "elastic" model there, where the actual surface of the Earth is "elastically deformed" from Euclidean flatness. But you can only spot the deformation by comparing complete circumferences of circles. You can't spot it by just looking locally. So what physical meaning can you ascribe to the "elastic deformation"? Since you can't spot it by looking locally, you can't ascribe any physical meaning to it locally. You can say that it's a global property of the space, but you can't tie it to anything on a local scale. And since even the observation of it from a distance depends on how you look, you're limited in the physical interpretations you can put even on the global property.

[Q-reeus]

The problem with spatial relations is they are tied to decisions about simultaneity. 3-space 'now' is global statement about simultaneity. This is well defined for inertial observers in flat spacetime. Otherwise, it is just not well defined, there a number of perfectly reasonable choices. Given different choices for distant simultaneity, you get different conclusions about spatial relationships.
Given that the arrangement considered is static, I'm thinking simultaneity issue is referring to a gravitationally effected c as measuring stick?

[Q-reeus]

The way the pressure affects the metric in this scenario is through the r-r component of the Einstein Field Equation, G_rr = 8 pi T_rr. (T_rr is what I was calling T_11 above, if we are using spherical coordinates.) This equation leads to the Tolman-Oppenheimer-Volkoff equation, which describes hydrostatic equilibruim in GR:

http://en.wikipedia.org/wiki/Tolman%E2%80%93Oppenheimer%E2%80%93Volkoff_equatio n

The derivation of the metric for the constant density case in MTW, which I quoted from earlier, makes essential use of this equation.
OK no disagreement here I can see.

[Q-reeus]

This is, in fact, what Schwarzschild coordinates can be viewed as doing; the Schwarzschild r coordinate can be viewed as the "Euclidean radius" of a point, and the actual distance given by the Schwarzschild metric can be viewed as the "stressed" distance, due to "elastic deformation" of the space.
Precisely the handle I've been trying to get a hold of, but not so easy.
The problem with this analogy is, as I said before, that in the spacetime case, a small object sitting at r is *not* deformed; it looks the same from every direction, just as it would in an "unstressed" flat space. The "deformation" is only visible globally, and only as a non-Euclideanness in the relationship between radial distances and tangential areas.
Sure agree and have been trying to make it clear that was understood. Obviously didn't express the analogy too clearly in #91 with
"A tiny elastic being caught up in it all cannot sense this directly - only 'tidal' elastic strain is locally evident....No point asking elastic being who knows only 'tidal' effects." Was trying to convey the analogy re local unobservability of 1st order metric effects. Basically that 'elastic being' deforms with it's surroundings, and must use a kind of 'K' factor to 'navigate' but with a limited perspective. Which answers to your later comments on that matter. Sensing only the gradients of strain, there are important properties only available - yes on an indirect inferred basis - to 'outside observer'.
(Note that you can't just say radial and tangential distances here, though you could say tangential *circumferences*, and some do; the key is that you can only spot the non-Euclideanness by measuring distances around an entire circle, or sphere, at "radius" r, *not* by just measuring small distances tangentially.)
This may be homing in on where it's at for me. My whole suspiscion has been that the spatial part of spacetime 'strains' was predicted wrongly - radial to transverse 'strain ratio' that, in passing to flat interior region with 1:1 'strain ratio', implied inconsistent functional dependence on potential. Just about ready to accept your collective wisdom on this and take a break. Had no initial idea defining spatial quantities of interest would prove so tricky. But thanks for an interesting if very circuitous ride!:zzz:

[PeterDonis]

Given that the arrangement considered is static, I'm thinking simultaneity issue is referring to a gravitationally effected c as measuring stick?

No; I assume PAllen was referring to the fact that surfaces of constant Schwarzschild time, which are surfaces of constant time for static observers (who stay at the same radius r), will not be surfaces of constant time for observers that are not static. For example, observers freely falling towards the black hole will have different surfaces of simultaneity, so "space" will look different to them. I believe I brought up the fact in an earlier post that spatial slices are flat in Painleve coordinates, which is equivalent to saying that observers freely falling into the black hole will *not* see the "non-Euclideanness" of space that observers hovering at a constant radius will.

[PeterDonis]

Was trying to convey the analogy re local unobservability of 1st order metric effects. Basically that 'elastic being' deforms with it's surroundings, and must use a kind of 'K' factor to 'navigate' but with a limited perspective.

I think you're twisting the analogy around here. A "local" being does *not* deform with the surroundings, in the sense you are using the term "deformation". That's the point. The K factor is *not* observable locally; it's only observable by taking measurements over an extended region. Locally, space looks Euclidean; there is no "deformation". Just as locally on Earth, its surface looks flat; we only see the non-Euclideanness of the surface by making measurements over an extended region. Furthermore, the non-Euclideanness never shows up as any kind of "strain" on individual objects. It's just a fact about the space, that it doesn't satisfy the theorems of Euclidean geometry. That's all.

I really think it's a mistake to look for a "real" physical meaning to the non-Euclideanness of space, over and above the basic facts that I described using the K factor--i.e., that there is "more distance" between two spheres of area A and A + dA, or between two circles of circumference C and C + dC, than Euclidean geometry would lead us to expect. If I start from my house at the North Pole and walk in a particular direction, I encounter circles of gradually increasing circumference. Between two such circles, of circumference C and C + dC, I walk a distance K * (dC / 2 pi), where K is the "non-Euclideanness" factor and is a function of (C / 2 pi). If space were Euclidean, I would find K = 1; but I find K > 1. So what? If I insist on ascribing the fact that K > 1 to some actual physical "strain" in the space, or anything of that sort, what is my reason for insisting on this? The only possible reason would be that I ascribe some special status to K = 1, so that when I see K > 1, I think something must have "changed" from the "natural" state of things. But why should Euclidean geometry, K = 1, be considered the "natural" state of things? What makes it special? The answer is, as far as physics is concerned, nothing does. Euclidean geometry is not special, physically. It's only special in our minds; *we* ascribe a special status to K = 1 because that's the geometry our minds evolved to comprehend. But that's a fact about our minds, not about physics.

[Q-reeus]

Originally Posted by Q-reeus:
"Given that the arrangement considered is static, I'm thinking simultaneity issue is referring to a gravitationally effected c as measuring stick?"

No; I assume PAllen was referring to the fact that surfaces of constant Schwarzschild time, which are surfaces of constant time for static observers (who stay at the same radius r), will not be surfaces of constant time for observers that are not static. For example, observers freely falling towards the black hole will have different surfaces of simultaneity, so "space" will look different to them. I believe I brought up the fact in an earlier post that spatial slices are flat in Painleve coordinates, which is equivalent to saying that observers freely falling into the black hole will *not* see the "non-Euclideanness" of space that observers hovering at a constant radius will.
Interesting point of difference for free-fall, but my query was all about stationary source and observer, which is why non-simultaneity wasn't making sense to me in that context. So a more general comment was being made.

[Q-reeus]

I think you're twisting the analogy around here. A "local" being does *not* deform with the surroundings, in the sense you are using the term "deformation". That's the point. The K factor is *not* observable locally; it's only observable by taking measurements over an extended region. Locally, space looks Euclidean; there is no "deformation". Just as locally on Earth, its surface looks flat; we only see the non-Euclideanness of the surface by making measurements over an extended region. Furthermore, the non-Euclideanness never shows up as any kind of "strain" on individual objects. It's just a fact about the space, that it doesn't satisfy the theorems of Euclidean geometry. That's all.
Umm...'not observable locally' in which strict sense here? There is not something here roughly analogous with 3- divergence - unobservable at a point, but locally observable in the sense of finite observables ('excess' field lines in that case) over a small but finite volume/bounding surface? In other words, is it actually necessary to do a series of complete circuits around the globe to detect the geometry? Specific example - fill a standard container of local volume V with tiny, uniform hard spheres and count them as N. Take another container that by linear ruler measuring has volume nV, and do a refill/recount, and the numbers would always be nN? And this would ccontinue to hold at any r (within reason - not for BH obviously)? Seems strange, but if that's the case, think that's provided just the handle I want!
[EDIT: Better pin down the matter of measuring that container. If one measured with ruler always held either radial or in tangent plane, and rotated the container to make say, length vs diameter measurements for a cylindrical container, it would or wouldn't matter if instead one held container fixed and reoriented the ruler in measuring?]
EDIT 2: Occurred to me now this is probably more like the situation of Ehrenfest paradox - so any 'divergence' probably of such high order as to be nearly unobservable locally. So - a good example of where effect of 'non-euclideanness' can only be appreciated by non-local (or in this instance, non-rotating) observer.

[PeterDonis]

Specific example - fill a standard container of local volume V with tiny, uniform hard spheres and count them as N. Take another container that by linear ruler measuring has volume nV, and do a refill/recount, and the numbers would always be nN? And this would continue to hold at any r (within reason - not for BH obviously)?

Yes, if the respective counts of little objects are N and nN, then local measurements of volume will find V and nV. (My personal preference would be to phrase it the way I just did: take two containers and fill them with little identical objects, and arrive at the counts N and nN. Then measure the volume of each by local measurements, and the volumes will come out to be V and nV.)

[EDIT: Better pin down the matter of measuring that container.

Good instinct. :wink:

If one measured with ruler always held either radial or in tangent plane, and rotated the container to make say, length vs diameter measurements for a cylindrical container, it would or wouldn't matter if instead one held container fixed and reoriented the ruler in measuring?]

No, it wouldn't matter.

[Q-reeus]

Originally Posted by Q-reeus:
"If one measured with ruler always held either radial or in tangent plane, and rotated the container to make say, length vs diameter measurements for a cylindrical container, it would or wouldn't matter if instead one held container fixed and reoriented the ruler in measuring?]"

No, it wouldn't matter.
Right, and I sort of realized late it was a bit of a lame question given I had already argued that 'co-stretching' couldn't allow such. Just wondered if differential gradients of 'co-stretching' might come in somehow.
Yes, if the respective counts of little objects are N and nN, then local measurements of volume will find V and nV. (My personal preference would be to phrase it the way I just did: take two containers and fill them with little identical objects, and arrive at the counts N and nN. Then measure the volume of each by local measurements, and the volumes will come out to be V and nV.)
Interesting. Hopefully one more clue here will fill the puzzle to my level of satisfaction. Taken to extreme, as V grows it will eventually engulf the source and something has to give with that invariance, surely. So could this be analogous with the case of say the straight current carrying wire. Line integral of B is zero unless enclosing wire? In shell case, K effect is only non-zero (K > 1) if shell is enclosed within any 'counting volume/areas'?
[Latest take on that. If fractional excess volumetric particle count between two concentric shells is a function of radius r, this 'must' be true for subdivided portions - conic sections through the shells say. So I'm under the strong impression it really boils down to a kind of spatial divergence - the small counting spheres are only capable of being a reference if their relative volumetric expansion is negligible compared to much larger container volume. However it's not just relative volume that matters. Expanding volume in tangent directions (wider conic solid angle) makes no change, but expanding in radial direction will. A directed non-euclidean effect that must to some extent be 'locally' observable. What to call this beast apart from 'delta K effect' I don't know but certainly imo physics not just coordinate peculiarity. My take on what's fundamentally going on, but bound to be shot down s'pose.]

[pervect]

I did some digging and found a paper on the metric of a photon gas star (without the shell).

http://arxiv.org/abs/gr-qc/9903044

The general solution is numerical, but there's one solution that's simple that's an "attractor" to the numerical solutions:


\frac{7}{4}\, dr^2 + r^2 \,d \theta^2 + r^2 sin^2 \theta \, d\phi^2 - \sqrt{\frac{7}{3}}\,r\,dt^2


This corresponds to a photon gas with a density per unit proper volume of 3 / (7 r^2) (the density has to depend on r), and a pressure per unit volume in each direction of one third of that. (This later was calculated by me to confirm it was a photon gas solution, it wasn't in the paper).

As usual, you need to specify an orthonormal co-frame basis to see that the actual density is in fact constant. The long way of doing it is to say that you transform the metric so it's locally Minkowskian, and take the density in that locally Minkowskian transformed space.

[Q-reeus]

As usual, you need to specify an orthonormal co-frame basis to see that the actual density is in fact constant. The long way of doing it is to say that you transform the metric so it's locally Minkowskian, and take the density in that locally Minkowskian transformed space.
Is this saying that there are no locally measurable physical consequences? As a further elaboration on what I concluded in #103, one should be able to notice the following: Make the container shape a slender tube, fluid filled and with a fine 'breather' capillary tube sticking out one end. Orienting the tube axis in the tangent plane will give some reading for height of fluid in the capillary (think of old style mercury thermometer). Orient tube along radial direction, at same mean radial position r, and the level in capillary will drop - differential rate of 'volume expansion/contraction' along r direction is such that 'expanded volume' in tube portion nearest source of gravity wins over opposite effect in portion furtherest from source. this is just a reinterpretation of physical implications of K factor imo.

Further, one could take a fluid filled spherical container (again with a capillary tube sticking out of it), and find that for inwardly directed radially displacement, fluid level in capillary will drop. This might be interpreted as a weird volumetric expansion of containment vessel - one without explanation in terms of any mechanical stress/strain. We assume here a notionally incompressible fluid and containment vessel such that the ever present tidal forces have no appreciable mechanical strain influence. So I would maintain purely metric distortions are locally observable - as gradient 'stretching' phenomena.

[PeterDonis]

Orienting the tube axis in the tangent plane will give some reading for height of fluid in the capillary (think of old style mercury thermometer). Orient tube along radial direction, at same mean radial position r, and the level in capillary will drop - differential rate of 'volume expansion/contraction' along r direction is such that 'expanded volume' in tube portion nearest source of gravity wins over opposite effect in portion furtherest from source. this is just a reinterpretation of physical implications of K factor imo.

No, the K factor does not imply this. Remember that a spherical object (more precisely, an object that in flat spacetime, under zero stress, is spherical) will still be spherical if placed at radial coordinate r; the K factor does not cause any distortion in the object. There is no "distortion" in the effect on the capillary tube either, for the same reason.

Further, one could take a fluid filled spherical container (again with a capillary tube sticking out of it), and find that for inwardly directed radially displacement, fluid level in capillary will drop.

No, it won't. See above.

This might be interpreted as a weird volumetric expansion of containment vessel - one without explanation in terms of any mechanical stress/strain.

This is not possible; if the physical volume of the container were expanded, the containment vessel would *have* to show strain. That's part of what "physical volume" means.

Think again about what the K factor means. It does not mean that "the physical volume of a particular piece of space is expanded". That's impossible. It means that there are *more* "pieces of space", more physical volume, per unit radial coordinate than Euclidean geometry would lead one to expect. But as I said in a previous post, to view this as somehow a "distortion of space" implies that the Euclidean state is the "natural" state, so any variation from it is a "distortion" and requires some physical manifestation. That's wrong. There is nothing privileged about Euclidean geometry in physics, and the fact that the geometry of space is non-Euclidean along the radial dimension in the spacetime surrounding a gravitating object is just that: a fact about the geometry of that spacetime. Just as the fact that, in my "house at the North Pole" scenario, there is "more distance" along a given unit of the radial coordinate I defined than Euclidean geometry would lead one to expect is simply that: a fact about the geometry of the surface of the Earth. None of these facts change the behavior of physical objects locally; they only change the global structure of the geometry.

[PAllen]

I'll add one more bit to what Peter said. Your stated goal of having tidal effects ignorable guarantees you can't detect Euclidean deviations. Tidal effects are the first order influence of curvature, thus they define the minimum scale needed to detect curvature. However, if you are willing to span a relatively large distance, and have near mathematically ideal measuring devices, you can detect Euclidean deviation as follows:

You pick a configuration of 5 points in space (e.g. the vertices of the figure made by joining two tetrahedra). You set up distances and angles between them per Euclidean predictions (e.g. using round trip laser time to define distance, and laser path the define straight lines). Then, at the very end, with all angles and all but one edge length set up, the last edge will be the wrong length.

J.L. Synge, in his 1960 book, develops this 5 point curvature detector. He shows that 5 points is the minimum needed to make this work (because, for example, flat Euclidean planes can be embedded in general 4-manifolds).

[EDIT: as for scale, if you use a 10 meter device near earth, your final deviation would be 10^-20 centimers or so. Less than a millionth the radius of a proton. ]

[Q-reeus]

Originally Posted by Q-reeus:
"Orienting the tube axis in the tangent plane will give some reading for height of fluid in the capillary (think of old style mercury thermometer). Orient tube along radial direction, at same mean radial position r, and the level in capillary will drop - differential rate of 'volume expansion/contraction' along r direction is such that 'expanded volume' in tube portion nearest source of gravity wins over opposite effect in portion furtherest from source. this is just a reinterpretation of physical implications of K factor imo."

No, the K factor does not imply this. Remember that a spherical object (more precisely, an object that in flat spacetime, under zero stress, is spherical) will still be spherical if placed at radial coordinate r; the K factor does not cause any distortion in the object. There is no "distortion" in the effect on the capillary tube either, for the same reason.
Didn't really expect this to go down quietly.:rolleyes: Still having great difficulty reconciling that bit (and the remainder of your comments) with just this excised bit of mine from #103:
"If fractional excess volumetric particle count between two concentric shells is a function of radius r, this 'must' be true for subdivided portions - conic sections through the shells say. So I'm under the strong impression it really boils down to a kind of spatial divergence - the small counting spheres are only capable of being a reference if their relative volumetric expansion is negligible compared to much larger container volume..."

Thought I had it conceptually pinned down there. Do we agree that if K factor applies to excess volume between complete concentric shells, it must apply to partitioned portions. Apply a soccer-ball style tesselation over shell surface and cut through radially at the boundaries.That defines intimately joined volume segments. An observer in each segment does a count. How could the excess count by each observer not add to give just that for the whole shells? Ergo - there is an non-euclidean effect observable in a 'container'. No?!

Let's take your analogy of north pole - or anywhere on a curved spherical surface. Instead of concentric circles, just take a hoop, fill it with tiny marbles. We know that non-euclidean surface curvature means being able to fit more marbles inside the hoop than would be the case on a flat surface. But the analogy is flawed - we can move the hoop anywhere over a spherical surface and marbles fit the same. The proper analogy is more like a surface in the shape of an egg - with pointy end corresponding to the source of gravity in 'real' case. We note now that our hoop, despite having a fixed locally measured perimeter, fits more and more marbles within upon approach to the pointy end. Do you still say there will be no observable 'delta K factor'?
Originally Posted by Q-reeus:
Orienting the tube axis in the tangent plane will give some reading for height of fluid in the capillary (think of old style mercury thermometer). Orient tube along radial direction, at same mean radial position r, and the level in capillary will drop - differential rate of 'volume expansion/contraction' along r direction is such that 'expanded volume' in tube portion nearest source of gravity wins over opposite effect in portion furtherest from source. this is just a reinterpretation of physical implications of K factor imo.

No, the K factor does not imply this. Remember that a spherical object (more precisely, an object that in flat spacetime, under zero stress, is spherical) will still be spherical if placed at radial coordinate r; the K factor does not cause any distortion in the object. There is no "distortion" in the effect on the capillary tube either, for the same reason.
I see what you are saying but re my previous argument, something, albeit exceedingly tiny, is physically happening here.
Originally Posted by Q-reeus:
"This might be interpreted as a weird volumetric expansion of containment vessel - one without explanation in terms of any mechanical stress/strain."

This is not possible; if the physical volume of the container were expanded, the containment vessel would *have* to show strain. That's part of what "physical volume" means.
Not if one accepts a physical gradient of length scale operates - gradient non-euclidean is necessary if any non-euclidean at all, yes? I have given the 2-D example above - hoop+marbles on egg re seemingly impossible effects.
Think again about what the K factor means. It does not mean that "the physical volume of a particular piece of space is expanded". That's impossible. It means that there are *more* "pieces of space", more physical volume, per unit radial coordinate than Euclidean geometry would lead one to expect. But as I said in a previous post, to view this as somehow a "distortion of space" implies that the Euclidean state is the "natural" state, so any variation from it is a "distortion" and requires some physical manifestation. That's wrong...
Hope this doesn't bog down into arguing over meaning of things. Would you say that an apple falling to the ground represents physics, or 'just' an expression of non-euclidean geometry? For me, excess counts owing to non-euclidean geometry manifest as physical phenomena. Taking your example of North-pole (or anywhere on a spherical surface), surface curvature means more marbles between concentric circles than on flat ground. Yes I can call that just geometry, but since it is mass that causes the 3-space curvature in gravitational situation, that's physics to me. Stubborn me.

[Q-reeus]

I'll add one more bit to what Peter said. Your stated goal of having tidal effects ignorable guarantees you can't detect Euclidean deviations. Tidal effects are the first order influence of curvature, thus they define the minimum scale needed to detect curvature.
My stated aim was simply to say 'we know tidal effects will matter in practice - but let's fully account for them and just look at what remains'.
...However, if you are willing to span a relatively large distance, and have near mathematically ideal measuring devices, you can detect Euclidean deviation as follows:...[EDIT: as for scale, if you use a 10 meter device near earth, your final deviation would be 10^-20 centimers or so...]
Interesting and ingenious but I'm struggling here over principle - goes without saying nothing practical to patent. (caught your post after preparing response to Peter, so some things are repetitive - sorry)

[PeterDonis]

Q-reeus, first of all, did you read my post in the other thread about a similar issue?

http://physicsforums.com/showpost.php?p=3580102&postcount=6

I think it might be relevant here.

I'm going to comment on this particular thing you say first because it may be the key to the issue:

Not if one accepts a physical gradient of length scale operates - gradient non-euclidean is necessary if any non-euclidean at all, yes?

The K factor is not a "gradient of length scale". One meter is one meter, physically, regardless of what radial r coordinate you are at. (Or one nanometer, or one width of an atomic nucleus.) The only thing the non-Euclideanness of space affects is *how many meters* are between two concentric spheres; if K > 1, there are more meters between the spheres than the Euclidean formula predicts. That's all. It doesn't change what a meter is at all.

It may be worth thinking about this for a bit before reading the rest of what I have to say below. I'll be drawing on it.

Do we agree that if K factor applies to excess volume between complete concentric shells, it must apply to partitioned portions. Apply a soccer-ball style tesselation over shell surface and cut through radially at the boundaries.That defines intimately joined volume segments. An observer in each segment does a count. How could the excess count by each observer not add to give just that for the whole shells? Ergo - there is an non-euclidean effect observable in a 'container'. No?!

Not if you specify the size of the container in meters. (Or in some unit of distance, anyway.) If you specify it in coordinate units, that's different; for example, if you specify it in units of radial coordinate r. But in your original container scenario, you didn't; you specified the size of the container in physical distance units. I suspect you didn't because you realized, subconsciously, that the "physical" definition of size is in meters, not coordinate units. If K > 1, there are more meters in a given unit of radial coordinate r, but each meter itself is still the same.

Let's take your analogy of north pole - or anywhere on a curved spherical surface. Instead of concentric circles, just take a hoop, fill it with tiny marbles. We know that non-euclidean surface curvature means being able to fit more marbles inside the hoop than would be the case on a flat surface. But the analogy is flawed - we can move the hoop anywhere over a spherical surface and marbles fit the same. The proper analogy is more like a surface in the shape of an egg - with pointy end corresponding to the source of gravity in 'real' case. We note now that our hoop, despite having a fixed locally measured perimeter, fits more and more marbles within upon approach to the pointy end. Do you still say there will be no observable 'delta K factor'?

Of course there will be an observable change in the K factor; in fact, if you go back and read my "North Pole" post carefully, you will see that K varies even in that scenario, because of the way I defined the r coordinate. Constant curvature does not necessarily imply constant K.

Also, once again, there will be a "non-Euclideanness" in the number of marbles that can fit between a pair of hoops, but each marble itself remains the same size. Marbles, in this scenario, are like meters; they are the physical measure of distance. They themselves don't change, but how many of them fit between a pair of hoops does.

Hope this doesn't bog down into arguing over meaning of things. Would you say that an apple falling to the ground represents physics, or 'just' an expression of non-euclidean geometry?

It represents physics, and the non-Euclidean geometry is one way of modeling the physics.

For me, excess counts owing to non-euclidean geometry manifest as physical phenomena. Taking your example of North-pole (or anywhere on a spherical surface), surface curvature means more marbles between concentric circles than on flat ground. Yes I can call that just geometry, but since it is mass that causes the 3-space curvature in gravitational situation, that's physics to me. Stubborn me.

I never said it was *just* geometry. I said explicitly, when I defined the K factor, that it was a physical observable. But you have to be careful when you think about *which* physical observable it is.

[pervect]

I'm not really following the philosophical end of this discussion, but I think I can write a bit how to enclose the photon gas metric I previously presented in a shell to join the two together properly, which should serve as an actual concrete example.

Going back to the very basics, we can use Wald's metric

-f(r)*dt^2 + h(r)*dr^2 + r^2 (d theta^2 + sin(theta) dphi^2)

and Wald's results 6.2.3, 6.2.4

(6.2.3) 8 pi rho = (r h^2)^-1 dh/dr + (1-1/h) / r^2
(6.2.4) 8 pi P = (r f h)^-1 df/dr -(1-1/h) / r^2

rho and P are not the 'coordinate' density and pressure, but the densities in the orthonormal basis given by Wald in 6.1.6, i.e. they represent the "physical" density and pressure seen by an observer in a local Minkowskii frame.

6.2.3 can be written as 6.2.6

8 pi rho = (1/r^2) d/dr [r (1-1/h) ]

If we envision a thick shell where rho=0, this immediately implies that r(1-1/h) is constant through the shell.

If we shrink the shell to zero width, (a thin shell) then we say simply that h is the same inside the shell and outside, h being the spatial coefficient of the metric. So h must match where we join together the vacuum Schwarzschild metric with our photon gas metric.

If we add together 6.2.3 and 6.2.4 we can write

8 pi (rho + P) = (dh/dr) / r h^2 + (df/dr) / rfh

which we can re-write as

d/dr (f h) / (r f h^2) = 8 pi (rho +P)

Because this is NONZERO, we can say definitely that the product of f and h is not constant. We know that h is constant. Therefore we know that f changes. With a thick shell, f changes as we progress through the shell. As we shrink the shell to zero thickness, in the limit, know that f must 'jump' suddenly, because f*h can't be constant.

This doesn't tell us "how much" the jump is, and it's rather inconvenient to use this approach to actually match the metrics, but it does tell us something important, it tells us to expect 'f' to jump suddenly.

What we can do instead is say that the mass function for our metric must equal the Schwarzschild mass paramter M

i.e.

m(r) = \int 4 pi r^2 rho(r)

Furthermore, we know what M is, because we know that the h coefficents must match, and h has the value 7/4 in our photon gas metric This implies that M/r = 3/14, as

h = 1 / (1 - 2M/r)

So if we look at the photon gas metric


\frac{7}{4}\, dr^2 + r^2 \,d \theta^2 + r^2 sin^2 \theta \, d\phi^2 - \sqrt{\frac{7}{3}}\,r\,dt^2


We see that (1/(1-2M/r)) must be 7/4, since the value of h must match. This implies that (M/r) must be 3/14.


But we know that 8 pi rho(r) = 3 /( 7 r^2) from Wald's 6.2.3, and
we just have to solve for r such that m(r) = (3/14) r, where m(r) is given by the integral of 4 pi r^2 rho(r) dr , which is just the intergal of (3/14) dr.

Unless I'm mistaken, this is satisfied for all values of r, so we pick any value of r we like, set M = (3/14) r, and use that for the exterior solution.

IF we chose r = 1

we have the original metric for r<1


\frac{7}{4}\, dr^2 + r^2 \,d \theta^2 + r^2 sin^2 \theta \, d\phi^2 - \sqrt{\frac{7}{3}}\,r\,dt^2


and for r>1


\frac{dr^2}{1-\frac{3}{7r}}+ r^2 \,d \theta^2 + r^2 sin^2 \theta \, d\phi^2 - \left( 1-\frac{3}{7\,r} \right) dt^2

[Q-reeus]

I'm not really following the philosophical end of this discussion, but I think I can write a bit how to enclose the photon gas metric I previously presented in a shell to join the two together properly, which should serve as an actual concrete example...
pervect: thanks for all your work there, but of what I can follow, this is throwing me:
"If we envision a thick shell where rho=0, this immediately implies that r(1-1/h) is constant through the shell." Which states that h is a function of r. But later: "We know that h is constant." I'm reading the latter to merely follow as a limit of imposing zero shell thickness - ie dr -> 0. No doubt that is missing it somehow, but can't see where. I can say nothing about the derivation of the orthonormal basis stuff - whether there are any subtle assumptions that 'chop off' higher order gradient effects for instance. You will have read my reply to Peter, so perhaps let me know where you think it all comes apart, because I maintain there must be physical effects as described earlier.

[Q-reeus]

Q-reeus, first of all, did you read my post in the other thread about a similar issue?

http://physicsforums.com/showpost.ph...02&postcount=6
I have now - and you can read my comments on that in turn!:tongue:
The K factor is not a "gradient of length scale". One meter is one meter, physically, regardless of what radial r coordinate you are at. (Or one nanometer, or one width of an atomic nucleus.) The only thing the non-Euclideanness of space affects is *how many meters* are between two concentric spheres; if K > 1, there are more meters between the spheres than the Euclidean formula predicts. That's all. It doesn't change what a meter is at all.
Strictly locally, I agree that 1 meter = 1 meter. My sense is though, to explain real effects, there must be a sense to '1 meter here looks different to 1 meter there'. I call that a gradient effect. Without it, there is magic - meters just slip in somehow.
Originally Posted by Q-reeus:
"Let's take your analogy of north pole - or anywhere on a curved spherical surface. Instead of concentric circles, just take a hoop, fill it with tiny marbles. We know that non-euclidean surface curvature means being able to fit more marbles inside the hoop than would be the case on a flat surface. But the analogy is flawed - we can move the hoop anywhere over a spherical surface and marbles fit the same. The proper analogy is more like a surface in the shape of an egg - with pointy end corresponding to the source of gravity in 'real' case. We note now that our hoop, despite having a fixed locally measured perimeter, fits more and more marbles within upon approach to the pointy end. Do you still say there will be no observable 'delta K factor'?"

Of course there will be an observable change in the K factor; in fact, if you go back and read my "North Pole" post carefully, you will see that K varies even in that scenario, because of the way I defined the r coordinate. Constant curvature does not necessarily imply constant K.

Also, once again, there will be a "non-Euclideanness" in the number of marbles that can fit between a pair of hoops, but each marble itself remains the same size. Marbles, in this scenario, are like meters; they are the physical measure of distance. They themselves don't change, but how many of them fit between a pair of hoops does.
You refer to a pair of hoops here, which seems to imply this will be a peculiarity of using polar coords - 'north pole effect'. But note carefully my example was using just one hoop - 2D counterpart of the 3D container earlier referenced to re fluid levels in capillary. As you seem to agree that marble count for that single enclosing hoop will be a function of surface curvature - *the coord system independent geometric object*, how can you then argue there will be no counterpart in 3D container, influenced by coord system independent spatial 3-curvature? What applies between two concentric hoops must apply within one container hoop. Likewise for concentric shells vs a 3D container.

You say 'meter hasn't changed, there are just more meters there than expected by Euclidean measure'. But how did those extra meters slip in exactly? Just because we are using SC's? Surely it's got to be the geometric object at work, totally independent of any coords used. Take that single bounding hoop example again. To keep it consistently 2D, rather than marbles, fill it with identically shaped tiny circular rings (mini-hoops), and demand that the ring count, for fixed packing density, remain constant irrespective of surface curvature. Only means to gaurantee that is one of two ways. Stress the containing hoop in compression, or stress the rings in tension, as surface curvature increases. Notice the manifestation of curvature now is stresses - and corresponding strains - rather than perceived perimeter expansion of containing hoop (or alternately, shrinking diameters of rings), from the pov of local observer, who just notices 'weirdness'.

One may wish to argue the interpretation as to what's behind it all ('more meters' vs non-uniform meter'), but for sure, there are physical effects - as I maintain there must be. And for me, that 'constant meter' idea is the problem here. Can't see other than a length scale gradient effect at work - not observable 'at a point'. This is hand-wavy, but can one not see an analogue with the well known example of triangles in curved space. Angles add to more than 1800 in positively curved space, but the effect is a non-linear function of triangle size - virtually non-existent in the small. And by analogy, those marbles/rings/water molecules act as 'standard meters' for the same basic reason - their distortion by spatial metric non-uniformity is miniscule. In my opinion that is.

[PeterDonis]

I have now - and you can read my comments on that in turn!:tongue:

I did, and I responded to them there. :wink:

Strictly locally, I agree that 1 meter = 1 meter. My sense is though, to explain real effects, there must be a sense to '1 meter here looks different to 1 meter there'.

What is wrong with putting it the way I did in my last post? That is: 1 meter here = 1 meter there, but how many meters fit between two spheres with area A and A + dA is different here than it is there.

I call that a gradient effect. Without it, there is magic - meters just slip in somehow.

If you think extra meters "slipping in" is "magic", why is that? The only reason I can see, is that you think the Euclidean relationship between areas of concentric spheres and the volume enclosed between them is somehow privileged. It isn't. So no "magic" is required for extra meters to be present. I don't see why this is so hard to grasp.

You refer to a pair of hoops here, which seems to imply this will be a peculiarity of using polar coords - 'north pole effect'.

No, it isn't. I have said all along that the K factor is an invariant physical observable. You could adopt Cartesian coordinates in my north pole scenario and it would still be there; it would just be a lot harder to express in those coordinates.

But note carefully my example was using just one hoop - 2D counterpart of the 3D container earlier referenced to re fluid levels in capillary. As you seem to agree that marble count for that single enclosing hoop will be a function of surface curvature - *the coord system independent geometric object*, how can you then argue there will be no counterpart in 3D container, influenced by coord system independent spatial 3-curvature? What applies between two concentric hoops must apply within one container hoop. Likewise for concentric shells vs a 3D container.

No, you didn't just use one hoop. You used hoops placed at different points on an egg-shaped surface, instead of on a spherical surface. (At least, that's the example I think you're referring to; if it's another, please re-state it or give a direct reference.) Did you read my response? I said constant curvature doesn't necessarily imply constant K. That means the K factor is *not* the same as curvature. It may be *related* to curvature, but it is not the same thing.

You say 'meter hasn't changed, there are just more meters there than expected by Euclidean measure'. But how did those extra meters slip in exactly?

See above. You are assuming the Euclidean expectation is privileged. It isn't.

[Q-reeus]

Originally Posted by Q-reeus:
"Strictly locally, I agree that 1 meter = 1 meter. My sense is though, to explain real effects, there must be a sense to '1 meter here looks different to 1 meter there'."

What is wrong with putting it the way I did in my last post? That is: 1 meter here = 1 meter there, but how many meters fit between two spheres with area A and A + dA is different here than it is there.
Maybe it is semantics getting in the way on that one. I think your '1 meter here = 1 meter there' statements are continually trying to clear up a non-existent conception on my part - that one would/could *locally* observe a meter changing just by moving around in a gravitational potential. No, have tried to make it abundantly clear I have never believed in such an absurdity. Rather, that move the meter rod over there into a lower grav potential, and in general it will look smaller than if done in flat spacetime. Yes, no doubt just calculating distance moved is a complication, but one that can be taken into account. So maybe we really are on the same (redshifted?) wavelength.
Originally Posted by Q-reeus:
"But note carefully my example was using just one hoop - 2D counterpart of the 3D container earlier referenced to re fluid levels in capillary. As you seem to agree that marble count for that single enclosing hoop will be a function of surface curvature - *the coord system independent geometric object*, how can you then argue there will be no counterpart in 3D container, influenced by coord system independent spatial 3-curvature? What applies between two concentric hoops must apply within one container hoop. Likewise for concentric shells vs a 3D container."

No, you didn't just use one hoop. You used hoops placed at different points on an egg-shaped surface, instead of on a spherical surface. (At least, that's the example I think you're referring to; if it's another, please re-state it or give a direct reference.)
Yes I did - where do you find me using more than one? Check my 3rd passage in #108. But suppose I used different hoops, provided they were standardized with reference to some particular starting location, how would that effect the argument? Crux of the matter is, within a single enclosing perimeter of locally measured invariant shape and size (the hoop), marble stacking density varies with surface curvature. Again, how can this not carry over more or less directly to the case of fluid-filled (water molecules = nano size marbles) container re 3-curvature effect? A rising or lowering capillary level is measurable physics. You say it won't, but where is this breakdown of analogy occurring? That is deeply baffling.
Did you read my response? I said constant curvature doesn't necessarily imply constant K. That means the K factor is *not* the same as curvature. It may be *related* to curvature, but it is not the same thing.
Accept that there is no direct relation. Having gone and re-read your #99, I see what you are driving at, but to me that situation, where 'the marbles' just sit there static on the ground, is hiding potential physics. In order for K > 1 there must be curvature - so curvature is the key operator that non-euclidean K factor manifests. I'm going to keep coming back to this same point - if there is no potential physics going on, explain or refute the matter of a locally invariant loop experiencing a varying marble area density, just by moving said hoop+marbles to a region of higher curvature. Then go back to the fluid filled container analogue, and tell me why level in the capillary would not vary with change of radial location r for container. And note, it has nothing to do with 'distortion' of the capillary tube, which is merely an indicator of change. :zzz:

[PeterDonis]

Maybe it is semantics getting in the way on that one. I think your '1 meter here = 1 meter there' statements are continually trying to clear up a non-existent conception on my part - that one would/could *locally* observe a meter changing just by moving around in a gravitational potential. No, have tried to make it abundantly clear I have never believed in such an absurdity. Rather, that move the meter rod over there into a lower grav potential, and in general it will look smaller than if done in flat spacetime.

What does "look smaller" mean? Part of the issue may be the continual temptation to use ordinary English words that have imprecise or ambiguous meanings. It's very important to resist that temptation, and to phrase things carefully in terms of actual observables. (For one thing, "look smaller" involves light and how light paths are changed by gravity in the intervening space, and you've ruled all that out of bounds--we're supposed to assume that all that has been corrected for.)

Check my 3rd passage in #108.

Do you mean the following?

Instead of concentric circles, just take a hoop, fill it with tiny marbles. We know that non-euclidean surface curvature means being able to fit more marbles inside the hoop than would be the case on a flat surface

If so, I must have missed it previously, because I should have objected, or at least clarified. I see how it refers to one hoop, but I also think it's false, unless I'm misunderstanding how you're defining a "hoop". I was assuming "hoop" meant a *single* line of marbles going around the circumference of a circle centered on the "North Pole"; combined with the assumption that the marbles themselves are so small that they can be used as little identical objects to measure distances to any accuracy we need for the problem, then the number of marbles that fit inside a "hoop" is determined by the hoop's circumference and nothing else. Since the circumference is tangential, it is unaffected by any "spatial distortion", regardless of anything else; that's the fixed point of departure that we both agree on. So the quoted sentences above are false if "hoop" means what I think it means.

If, on the other hand, by "hoop" you mean "two circles of slightly different circumferences, C and C + dC, plus the space between them", then we've been using "hoop" to mean different things. In the following quote, it looks like you're using "hoop" in this other sense, but if you were doing that in post #108 I didn't understand that. I was using the word "circle" to avoid such ambiguity. However, I'm not entirely sure, because in the following quote you still seem to equivocate about how "local" a hoop is. See below.

Crux of the matter is, within a single enclosing perimeter of locally measured invariant shape and size (the hoop), marble stacking density varies with surface curvature.

If the hoop has a "locally measured invariant shape and size", then it *must* be a "hoop" in the sense I was thinking--a *single* circle, with circumference C, and that's all. As I noted above, such a "hoop" must always contain the *same* number of marbles for a given "size" of hoop (i.e., circumference). If the number of marbles placed within a "hoop" of a given "size" can vary, then a "hoop" *cannot* be a single circle--it must be, as I noted above, two circles of slightly different circumferences, C and C + dC, plus the space between them. In this case, yes, the number of marbles placed within a hoop can vary, even if dC is held constant. But that just means the hoop does *not* have a "locally measured invariant shape and size".

In order for K > 1 there must be curvature - so curvature is the key operator that non-euclidean K factor manifests.

This is basically correct, with the proviso that K does not *equal* the curvature; it is *related* to it, but not the same.

if there is no potential physics going on, explain or refute the matter of a locally invariant loop experiencing a varying marble area density, just by moving said hoop+marbles to a region of higher curvature.

I did, by refuting your assumption that it is a "locally invariant loop". For a varying number of marbles to be seen in a "loop", the "loop" (you keep on changing words, and it doesn't help with clarity) cannot be "locally invariant"; that's obvious. More precisely, a region between two spheres of areas A and A + dA, or between two circles of circumference C and C + dC can vary in size as A or C change, even if dA or dC are held constant. That's the definition of the K factor, and I've said all along that it's a physical observable and represents "real physics" going on.

The only thing I am disagreeing with you about is that you are expecting this real physics to show up in a way that it does not, in fact, show up. The reason it does not show up the way you are expecting it to is that your expectation is based on giving a privileged status to the predictions of Euclidean geometry. In fact, there is no such privileged status. I've said that repeatedly, too, and you haven't picked up on it, or if you have, it hasn't shown in your posts. There's no point in continuing to wonder if this is about semantics, or if I think there's real physics going on. I've made all that clear multiple times. The thing to focus in on is why you believe Euclidean geometry has a privileged status, so that any departure from Euclidean geometry, meaning any K factor that is not equal to 1, requires some special manifestation over and above what I've already defined as the observable K factor.

[DrGreg]

Let's take your analogy of north pole - or anywhere on a curved spherical surface. Instead of concentric circles, just take a hoop, fill it with tiny marbles. We know that non-euclidean surface curvature means being able to fit more marbles inside the hoop than would be the case on a flat surface. But the analogy is flawed - we can move the hoop anywhere over a spherical surface and marbles fit the same. The proper analogy is more like a surface in the shape of an egg - with pointy end corresponding to the source of gravity in 'real' case. We note now that our hoop, despite having a fixed locally measured perimeter, fits more and more marbles within upon approach to the pointy end. Do you still say there will be no observable 'delta K factor'?

I was assuming "hoop" meant a *single* line of marbles going around the circumference of a circle centered on the "North Pole"; combined with the assumption that the marbles themselves are so small that they can be used as little identical objects to measure distances to any accuracy we need for the problem, then the number of marbles that fit inside a "hoop" is determined by the hoop's circumference and nothing else. Since the circumference is tangential, it is unaffected by any "spatial distortion", regardless of anything else; that's the fixed point of departure that we both agree on. So the quoted sentences above are false if "hoop" means what I think it means.

If, on the other hand, by "hoop" you mean "two circles of slightly different circumferences, C and C + dC, plus the space between them", then we've been using "hoop" to mean different things. In the following quote, it looks like you're using "hoop" in this other sense, but if you were doing that in post #108 I didn't understand that. I was using the word "circle" to avoid such ambiguity. However, I'm not entirely sure, because in the following quote you still seem to equivocate about how "local" a hoop is.I thought Q-reeus was completely filling the interior of a hoop with marbles -- in other words he is comparing the area of a circular disk with its circumference.

Any deviation from the expected 2D Euclidean value would be apparent only for relatively large (i.e. "non-local") hoops. Hoops that are small enough to be regarded as "local" would be too small for the deviation to be measurable -- that's pretty much what we mean by "local" as used in the Equivalence Principle.

[PAllen]

Unless Synge is wrong, even over a large region, you cannot detect Euclidean deviation on a plane. You need something 3-d, like Peter's concentric spheres (not circles). If Synge is right, the even over large regions, you can construct a Euclidean tetrahedron. You need one more vertex than a tetrahedron to detect Euclidean deviation.

[pervect]

I haven't seen Synge's derivation (I think I tried to find it once, but I know I never managed to get a hold of it), but I'm confident you can detect intrinsic curvature of a plane with 4 points, so I find it logical to believe you can detect space curvature with 5.

Whether you use my method for detecting the intrinsic curvature of a plane (which involves comparing the ratio of the diagonals of a square to the side of a square, the square by definition having four equal sides and two equal diagonals), or the marble packing method suggested by Q, you can detect the intrinsic curvature of a plane embedded in a 3d space.

What you can't do is tell if said intrinsic curvature is due to the way the plane is embedded. Thus information on the intrinsic curvature of a single plane embedded in a higher 3d space-time doesn't directly tell you anything about the intrinsic curvature of the space it's embedded in, the intrinsic curvature of the plane could result from the way it's embedded.

A trivial example: The Earth's surface is curved, and not flat. In fact, thinking about ways to detect the curvature of the Earth's surface (while staying on the surface) is a good way to get comfortable with the concepts and properties of curvature.

But the fact that the Earth's surface is curved (has an intrinsic curvature) doesn't tell you anything about whether or not space or space-time the Earth is in is curved.

I've heard that you can decompose the Rieman into "sectional curvatures" of planes, but I'm a bit hazy about the details. Clearly, though, you need information on the intrinsic curvature of a lot of planar slices of your space-time, not just one.

[PAllen]

A little side note on Synge is that well before Kruskal and Szekeres, Synge was the first to untangle the fully extended topology of SC geometry (though he didn't come up with K-S coordinates). MTW makes note of his clear priority as the first ever to work it all out.

[PAllen]

I haven't seen Synge's derivation (I think I tried to find it once, but I know I never managed to get a hold of it), but I'm confident you can detect intrinsic curvature of a plane with 4 points, so I find it logical to believe you can detect space curvature with 5.

Whether you use my method for detecting the intrinsic curvature of a plane (which involves comparing the ratio of the diagonals of a square to the side of a square, the square by definition having four equal sides and two equal diagonals), or the marble packing method suggested by Q, you can detect the intrinsic curvature of a plane embedded in a 3d space.

What you can't do is tell if said intrinsic curvature is due to the way the plane is embedded. Thus information on the intrinsic curvature of a single plane embedded in a higher 3d space-time doesn't directly tell you anything about the intrinsic curvature of the space it's embedded in, the intrinsic curvature of the plane could result from the way it's embedded.

A trivial example: The Earth's surface is curved, and not flat. In fact, thinking about ways to detect the curvature of the Earth's surface (while staying on the surface) is a good way to get comfortable with the concepts and properties of curvature.

But the fact that the Earth's surface is curved (has an intrinsic curvature) doesn't tell you anything about whether or not space or space-time the Earth is in is curved.

I've heard that you can decompose the Rieman into "sectional curvatures" of planes, but I'm a bit hazy about the details. Clearly, though, you need information on the intrinsic curvature of a lot of planar slices of your space-time, not just one.

It would have to be some pattern in the planar slices. Obviously, you can embed concentric 2-spheres in flat Euclidean 3-space. That does not imply anything about the geometry of the space the in which the spheres are embedded.

[PeterDonis]

I thought Q-reeus was completely filling the interior of a hoop with marbles -- in other words he is comparing the area of a circular disk with its circumference.

You may be right as regards that particular example; I may well have misread him.

I prefer to compare the space enclosed between two adjacent concentric circles with circumference C and C + dC, because it allows one to use just one value of the K factor, the one that applies at the particular "r" coordinate corresponding to the circumference C. (Or in the 3-D spatial slice of spacetime case, to evaluate the volume enclosed between two adjacent concentric spheres of area A and A + dA, we only need to use the value of r corresponding to A.) To fully compare the area enclosed by a circle around the "North Pole" with its circumference, on the Earth, we would need to integrate the K factor over a range of r values. (In the spacetime case, it's even worse because the K factor's dependence on r is different in the vacuum exterior region and in the interior of the gravitating body at the center; in the case of a black hole, there isn't even a spacelike slice at in exterior Schwarzschild coordinates that reaches to r = 0.)

Any deviation from the expected 2D Euclidean value would be apparent only for relatively large (i.e. "non-local") hoops. Hoops that are small enough to be regarded as "local" would be too small for the deviation to be measurable -- that's pretty much what we mean by "local" as used in the Equivalence Principle.

I'm not sure about this way of stating it; I think we need to clarify the meaning of "local" and "non-local" in this connection.

Consider a pair of 2-spheres concentric on the Earth, with areas A and A + dA, where dA << A, and A is just a bit larger than the Earth's surface area--just large enough so that the spheres are unarguably in the "exterior vacuum region" of the Earth's Schwarzschild spacetime geometry. The volume enclosed between these 2-spheres will be greater than Euclidean geometry would predict based on the difference in their areas, by about the ratio of Earth's Schwarzschild radius to its actual radius, or about 1 part in a billion. We can't measure that today, but I see no difficulty in principle in doing so, and we may well have enough accuracy to do it in practice in the foreseeable future. That difference is what I am calling the K factor; for the case I just described, K is about 1 + 10^-9.

The measurement I have just described is indeed "non-local", in the sense that it can't be done without enclosing the entire Earth with a pair of 2-spheres, which is not a local measurement. However, it is "local" in the sense that I can make dA very, very small compared to A, and the K factor will still be the same; it will just be harder to detect. The K factor depends on the Schwarzschild r coordinate, hence it depends on A, but it does *not* depend on dA once A is fixed. So I would be hesitant to say that the measurement is not "local", period; because of the spherical symmetry, it is not unreasonable to describe as "local", for some purposes, a measurement like this that is "local in r", so to speak, but not local in space as a whole.

[Q-reeus]

Originally Posted by DrGreg:
"I thought Q-reeus was completely filling the interior of a hoop with marbles -- in other words he is comparing the area of a circular disk with its circumference."

You may be right as regards that particular example; I may well have misread him.
Right in both cases. Looks like words getting in the way again. Had used the marbles thing as a direct 2D carry-over from your 3D description of K in terms of 'little counting spheres' between concentric shells earlier. Seemed self-evident that hoop is not torus, that hoop as defined perimeter, sits on a surface, and one proceeds to fill the enclosed area with marbles. And that the count will be a function of surface curvature. Just seemed a natural way to continue that analogy. While your concentric circles around north pole analogy in #99 talked in terms of perimeter-to-radius ratio, one could equally talk in terms of an enclosed surface area-to-perimeter ratio of a dished annulus (numerically different, but having in common dependence on surface curvature). Once you see it the latter way, the hoop thing springs out as a more evident manifestation that local phenomena will exist, which is why I used it.

Deary me - should have just gone straight to DrGreg's use of area vs circumference. But one still needed something like 'marble count' to get it that the ratio was changing with changed surface curvature - and most importantly - it will physically manifest (gaps opening up between marbles). And ergo - go 3D and fluid level in a container responds to changed 3-curvature. Also, as DrGreg mentioned in #117, the larger the hoop, the larger the relative effect, in the same way that the ratio of surface area of a chorded section cut from a spherical surface to that of a circular plate of the same diameter is negligibly different from unity for small chord size, but grows non-linearly with chord diameter. Ties in with comments in #113 about triangles. And with comments in #103 etc that counting spheres are a standard precisely because of this non-linear scale dependence.

I think while angles not adding to 180 degrees for triangles is common fare, extension to 3D volume effects appears not to be. However PAllen's comments in #118 and pervect's in #119 to e.g. Synge's 5-point method of detecting curvature show that 'practical' methods for detection have been devised that can operate over a 'local' region of space. But as argued in #103 and #105, there is directionality involved - functional dependence is on r, not on tangent directions. And despite what I've heard, seems natural to interpret K as the contraction ratio ∂'r'/∂r, 'r' being the radius in coordinate measure. In fact, to answer your first query in #116: "What does "look smaller" mean?" - inferred with reference to coordinate measure, the only handle on all this that makes sense to me.

Originally Posted by DrGreg:
"Any deviation from the expected 2D Euclidean value would be apparent only for relatively large (i.e. "non-local") hoops. Hoops that are small enough to be regarded as "local" would be too small for the deviation to be measurable -- that's pretty much what we mean by "local" as used in the Equivalence Principle."

I'm not sure about this way of stating it; I think we need to clarify the meaning of "local" and "non-local" in this connection.

Consider a pair of 2-spheres concentric on the Earth, with areas A and A + dA, where dA << A, and A is just a bit larger than the Earth's surface area--just large enough so that the spheres are unarguably in the "exterior vacuum region" of the Earth's Schwarzschild spacetime geometry. The volume enclosed between these 2-spheres will be greater than Euclidean geometry would predict based on the difference in their areas, by about the ratio of Earth's Schwarzschild radius to its actual radius, or about 1 part in a billion. We can't measure that today, but I see no difficulty in principle in doing so, and we may well have enough accuracy to do it in practice in the foreseeable future. That difference is what I am calling the K factor; for the case I just described, K is about 1 + 10^-9.

The measurement I have just described is indeed "non-local", in the sense that it can't be done without enclosing the entire Earth with a pair of 2-spheres, which is not a local measurement. However, it is "local" in the sense that I can make dA very, very small compared to A, and the K factor will still be the same; it will just be harder to detect. The K factor depends on the Schwarzschild r coordinate, hence it depends on A, but it does *not* depend on dA once A is fixed. So I would be hesitant to say that the measurement is not "local", period; because of the spherical symmetry, it is not unreasonable to describe as "local", for some purposes, a measurement like this that is "local in r", so to speak, but not local in space as a whole.
Aha. Just what I thought. Go back and read the strike-through part I wrote in #103. That was my thinking about what you could possibly imply by saying there is K > 1 only for concentric shells, not for a volume segment 'cut out' from those shells. I left it there as evidence of my thinking and as a chance to comment on it, but none came. But it doesn't make sense - to repeat what was written in #108:
"Thought I had it conceptually pinned down there. Do we agree that if K factor applies to excess volume between complete concentric shells, it must apply to partitioned portions. Apply a soccer-ball style tesselation over shell surface and cut through radially at the boundaries.That defines intimately joined volume segments. An observer in each segment does a count. How could the excess count by each observer not add to give just that for the whole shells? Ergo - there is an non-euclidean effect observable in a 'container'. No?!"

I can conceive no way around that. How can there be?

[PAllen]

I think while angles not adding to 180 degrees for triangles is common fare, extension to 3D volume effects appears not to be. However PAllen's comments in #118 and pervect's in #119 to e.g. Synge's 5-point method of detecting curvature show that 'practical' methods for detection have been devised that can operate over a 'local' region of space.

These methods are not local. You must span a region big enough that (to your level of precisions) tidal effects are detectable. Put it this way: propose a level measurement precision; then there is a minimum size region in which you can detect tidal gravity effects, thus encompassing curvature significant to the precision. Now assume your length and time measurements are comparable in geometric units (this typically means much more sensitive, in practice, or much larger region required). Then, by various global measurements of this scale, you can detect Euclidean deviation. You cannot localize it to any plane, let alone a linear direction. You can always get a finite size 2-surface, in any orientation, that is (mathematically) exactly flat (perhaps unless the 4-manifold is very pathological).

[Q-reeus]

Originally Posted by Q-reeus:
"I think while angles not adding to 180 degrees for triangles is common fare, extension to 3D volume effects appears not to be. However PAllen's comments in #118 and pervect's in #119 to e.g. Synge's 5-point method of detecting curvature show that 'practical' methods for detection have been devised that can operate over a 'local' region of space."

These methods are not local. You must span a region big enough that (to your level of precisions) tidal effects are detectable. Put it this way: propose a level measurement precision; then there is a minimum size region in which you can detect tidal gravity effects, thus encompassing curvature significant to the precision. Now assume your length and time measurements are comparable in geometric units (this typically means much more sensitive, in practice, or much larger region required). Then, by various global measurements of this scale, you can detect Euclidean deviation.
Agree entirely with that, which is why I put 'practical' and 'local' in single quote marks. The question is whether it can in principle be done at all, without having to have the measuring region enclose the gravitational source. In crude analogy, is 3-curvature to be considered a kind of 'volume charge density' having a local non-zero divergence which is measurable locally, or divergence free except for the source region itself? Remember - this is just rough analogy.
You cannot localize it to any plane, let alone a linear direction. You can always get a finite size 2-surface, in any orientation, that is (mathematically) exactly flat (perhaps unless the 4-manifold is very pathological).
Not quite up with how to interpret that - does it invalidate the principle behind what I was saying in #105 for instance?

[PeterDonis]

Seemed self-evident that hoop is not torus, that hoop as defined perimeter, sits on a surface, and one proceeds to fill the enclosed area with marbles. And that the count will be a function of surface curvature.

Yes, but as I pointed out, to calculate the count, you need to know the K factor for a whole range of "r" values, from r = 0 out to the "r" of the hoop, which is its circumference divided by 2 pi. This brings in additional complications which are not present if you consider the area between two nearby circles of circumference C and C + dC.

Deary me - should have just gone straight to DrGreg's use of area vs circumference. But one still needed something like 'marble count' to get it that the ratio was changing with changed surface curvature - and most importantly - it will physically manifest (gaps opening up between marbles).

Huh? Where has anyone said anything about gaps between marbles? I thought it was understood through all of this that we are packing whatever area (or volume) we're concerned with as tightly as possible with the marbles (or whatever small identical objects we are using). DrGreg even said so explicitly. If you don't do that, how can you possibly get reliable measurements?

And ergo - go 3D and fluid level in a container responds to changed 3-curvature.

Still an issue here--see below.

And despite what I've heard, seems natural to interpret K as the contraction ratio ∂'r'/∂r, 'r' being the radius in coordinate measure.

You have the ratio upside down. If we use s for "physical" distance measure and r for coordinate measure, then K is ds/dr. Or, if we use the definition of the r coordinate we've been using, K is ds/dsqrt(A), where A is the area of the 2-sphere at coordinate r. So K > 1 means an increase in how much actual distance s corresponds to a unit of coordinate r.

"Thought I had it conceptually pinned down there. Do we agree that if K factor applies to excess volume between complete concentric shells, it must apply to partitioned portions. Apply a soccer-ball style tesselation over shell surface and cut through radially at the boundaries.That defines intimately joined volume segments. An observer in each segment does a count. How could the excess count by each observer not add to give just that for the whole shells?

It does, as long as we're packing marbles correctly, or the equivalent with tesselations. But that does *not* imply the following:

Ergo - there is an non-euclidean effect observable in a 'container'. No?!"

No.

I can conceive no way around that. How can there be?

Because you are not correctly analyzing the physics of the container. Let's consider that example in more detail.

Suppose I have a cubical container with side length s; that is, when I measure its sides in some region of spacetime far away from all gravitating bodies, I measure each side to be identical in length, and the side length to be s. Now I take this container and lower it to some radial coordinate r above a gravitating body, where r is such that the K factor is measurably greater than 1. What will the container look like when I measure its sides again?

The answer is clear from what I've already said: the container will still be cubical, and its side lengths will still be s. The K factor has no observable effect on the size of the container, because K does not cause any stress on objects.

However, now consider the following experiment: I take my container, in a region of spacetime far from all gravitating bodies, and I sandwich it between two concentric 2-spheres, with the areas A and A + dA of the spheres chosen such that two opposite faces of the container are just tangent to the two spheres. I ask, what is the relationship between the side length s of the container and the area A of the inner sphere? The answer is, it is the relationship which Euclidean geometry predicts. In other words, the K factor here is 1.

Now I lower the container to a radial coordinate r above a gravitating body, such that the area A corresponding to r (A = 4 pi r^2) is *exactly* the same as the area A of the inner sphere I used above. In other words, the "bottom" surface of the container is now tangent to the sphere with radius A, exactly as it was when everything was far away from all gravitating bodies. I now ask: if I consider a second sphere of area A + dA in this situation, where dA is exactly the same dA I used above, will the "top" surface of the container be tangent to that sphere? The answer is *no*: the container's top surface will not quite reach the second sphere, because the side length s of the container, which is unchanged, is now not quite as long as the distance between the two spheres, because the K factor is now greater than 1. That is what I mean by saying that there is "more distance" between the spheres than there would be if the space geometry were Euclidean, but the size of a given unit of distance, such as the container side length s, is unchanged.

Note, please, that this is *not* saying that K is not physically observable. The failure of the top surface of the container to reach the second sphere is a physical observable--it's direct physical evidence of the K factor being greater than 1. It may not be the evidence your intuition was expecting, but it's certainly evidence.

[PAllen]

I wonder if the following would be true (it seem intuitively plausible based on Synge's results combined with Peter Donis's findings; I wouldn't rely on this without calculating it, though):

Take a ruler with marks r1 and r2 on it. In a region of curvature, use it to lay out two concentric spherical surfaces of a chosen solid angle. The relation of surface areas to r1, r2, and solid angle will be strictly Euclidean. But what about the volume between them measured with Peter's little marble idea? My guess is that it will not match the Euclidean prediction.

[Q-reeus]

Originally Posted by Q-reeus:
"Seemed self-evident that hoop is not torus, that hoop as defined perimeter, sits on a surface, and one proceeds to fill the enclosed area with marbles. And that the count will be a function of surface curvature."

Yes, but as I pointed out, to calculate the count, you need to know the K factor for a whole range of "r" values, from r = 0 out to the "r" of the hoop, which is its circumference divided by 2 pi. This brings in additional complications which are not present if you consider the area between two nearby circles of circumference C and C + dC.
If I read that right, you are perhaps inadvertently agreeing with my notion of 'spatial gradients'. That length measured with a microscopic ruler will not correspond exactly with length measured with a large ruler, even 'locally', because of local gradients. I maintain it's this that allows the marbles etc to act as a standard length. Blow them up to near hoop size, and evidence of curvature is lost. As said earlier, it means the larger the hoop, the proportionately larger the area excess becomes - the more sensitive a gauge of curvature one has. This is just repeating what's been said many times before. We are trying to get a handle on the rate of change with potential, of microscopic measure (marbles) to macroscopic measure (hoop).

Originally Posted by Q-reeus:
"Deary me - should have just gone straight to DrGreg's use of area vs circumference. But one still needed something like 'marble count' to get it that the ratio was changing with changed surface curvature - and most importantly - it will physically manifest (gaps opening up between marbles)."
Huh? Where has anyone said anything about gaps between marbles? I thought it was understood through all of this that we are packing whatever area (or volume) we're concerned with as tightly as possible with the marbles (or whatever small identical objects we are using). DrGreg even said so explicitly. If you don't do that, how can you possibly get reliable measurements?
Well gaps must open before one can pop in a new marble, yes? But I could have made that statement a bit clearer, sure.
Originally Posted by Q-reeus:
"And despite what I've heard, seems natural to interpret K as the contraction ratio ∂'r'/∂r, 'r' being the radius in coordinate measure."

You have the ratio upside down. If we use s for "physical" distance measure and r for coordinate measure, then K is ds/dr. Or, if we use the definition of the r coordinate we've been using, K is ds/dsqrt(A), where A is the area of the 2-sphere at coordinate r. So K > 1 means an increase in how much actual distance s corresponds to a unit of coordinate r.

Got me! EXpressed it wrong - meant K-1 but it slipped me - a bit like that single typo 'loop' instead of 'hoop' you so quickly picked me up on earlier.:blushing: Anyway we seem to agree there's more than one way to express the meaning of K - not just as differential volume-to-area ratio. That a coordinate related meaning is justified, not just as 'quasi-local' measure.
Suppose I have a cubical container with side length s; that is, when I measure its sides in some region of spacetime far away from all gravitating bodies, I measure each side to be identical in length, and the side length to be s. Now I take this container and lower it to some radial coordinate r above a gravitating body, where r is such that the K factor is measurably greater than 1. What will the container look like when I measure its sides again?
The answer is clear from what I've already said: the container will still be cubical, and its side lengths will still be s.
Stop right there. You torpedoed my hoop thing on the basis of K being a function of hoop radius, implying I suppose that perimeter was to a certain extent undefined because K is non-constant throughout a locally defined radial displacement. Precisely my point! It has negligible effect on the counting spheres owing to their being so small - they 'sample' K gradient only slightly. And likewise that cube still measuring s by a macroscopic ruler does not imply there has been no differential change between that and the microscopic ruler measure of those counting spheres. Run the micro rulers along the macro ruler - they disagree. Move to a location in lower potential - they disagree more. And the larger the value of s, the larger the counting anomaly will become for a given potential. Upon that I claim container *can* be used. As you argued re hoop - there are gradient effects here.
The K factor has no observable effect on the size of the container, because K does not cause any stress on objects.
I would have general doubts here. For a solid sphere, seems to me contraction factor varying with r implies the interior pulls on the exterior regions - not as a simple minded application of K factor, but a more subtle function. That would accord with what I wrote in e.g. #113. So tangential compression going to tangential tension on descent to the center. For a fluid sphere, obviously not. For empty container - negligible probably.
However, now consider the following experiment:
....The answer is *no*: the container's top surface will not quite reach the second sphere, because the side length s of the container, which is unchanged, is now not quite as long as the distance between the two spheres, because the K factor is now greater than 1. That is what I mean by saying that there is "more distance" between the spheres than there would be if the space geometry were Euclidean, but the size of a given unit of distance, such as the container side length s, is unchanged....
To put it more shortly - this means a simple subdivision of shell volume into 'containers' implies a radial stretching/stressing must occur after gravity is 'switched on'. And that this 'stretching' is what allows the excess count that a 'free' container will not experience. Fair argument; hadn't thought about it that way before and will have to consider full implications. I think though it simply implies that if one 'lets go' a stretched sub-volume to become a free container, there is simply an increased density of excess count. My 'in the meantime' response is this. Now that we are all clear on what hoop meant, how do you understand changing marble count with curvature - and extending that to molecule count in a 3d container? Is your argument taking full account, as I wrote above, of micro-to-macro K gradients, and how that changes with r? Still a problem here.

[Hardening my stance on this. Comments in #123 apply: "While your concentric circles around north pole analogy in #99 talked in terms of perimeter-to-radius ratio, one could equally talk in terms of an enclosed surface area-to-perimeter ratio of a dished annulus (numerically different, but having in common dependence on surface curvature). Once you see it the latter way, the hoop thing springs out as a more evident manifestation that local phenomena will exist, which is why I used it." There is a marble count excess for dished annulus sitting on 2-sphere (2D analogue of spherical shells experiencing 3-curvature). Same general effect must apply to a hoop, and likewise 3D container]

[PeterDonis]

If I read that right, you are perhaps inadvertently agreeing with my notion of 'spatial gradients'. That length measured with a microscopic ruler will not correspond exactly with length measured with a large ruler, even 'locally', because of local gradients.

The only thing that would affect a large ruler, as opposed to a very small ruler, is tidal gravity. That's the only kind of "gradient" in the field that can cause actual physical stress on a ruler and thus change its physical length. The K factor does *not* do this. Remember I said the K factor is *not* the same as curvature; it's related but it's not the same. Spacetime curvature is tidal gravity, not the K factor.

Well gaps must open before one can pop in a new marble, yes? But I could have made that statement a bit clearer, sure.

What do you mean, "pop in a new marble"? We are not talking about that kind of experiment; we are talking about packing marbles into various pre-existing spaces. If you're envisioning a "gap" opening up, you're envisioning something like this: we take a circular "hoop" of circumference C and place it on a sphere, centered on the North Pole. Then we take a second circular "hoop" made of some elastic material, so it can stretch; we place it on the sphere starting with circumference C + dC(0), and then slowly move it away from the first circle, so dC gradually gets larger compared to dC(0). As we do that, yes, open space will appear that we now need to pack with more marbles. But also, as we do that, the K factor will vary between the two circles, so things become more complicated. It would be really good if you would stick to the purely "local" case first, to avoid confounding factors, like for example tidal gravity--see my comments elsewhere in this post.

Stop right there...As you argued re hoop - there are gradient effects here.

Not if we're ignoring tidal gravity. See above. The only "gradient" effect that will actually stretch a container and change its physical size is tidal gravity. The K factor will *not* do this.

This is why I keep saying we should stick to the local case first; it avoids introducing confounding factors like tidal gravity that are *not* the same as the K factor. If you keep muddling these things together, you will keep on being confused. Once again: the K factor does *not* cause any stress on objects. Therefore, the K factor *cannot* change the physical size of a container; that would require causing stress on the container's walls. This is a basic point of the physics involved, and if it's not clear, we need to stick to the local case until it is.

I would have general doubts here. For a solid sphere, seems to me contraction factor varying with r implies the interior pulls on the exterior regions - not as a simple minded application of K factor, but a more subtle function. That would accord with what I wrote in e.g. #113. So tangential compression going to tangential tension on descent to the center. For a fluid sphere, obviously not. For empty container - negligible probably.

Now you're bringing in yet *another* different case--a non-vacuum region, inside a solid object. It would *really* help to stick to the simplest case first! We are talking about the exterior *vacuum* region, with two 2-spheres very close together, with areas A and A + dA, and what effect the K factor has in *that* case alone. We really need to get that case straight first before bringing in complications.

(If you insist on something about the non-vacuum case, inside a solid object, the stress-energy tensor is non-zero, so yes, there are additional forces "pulling" on a small object. But those forces also are *not* the K factor; they are related to it, in the sense that they are also functions of the radial coordinate r, but they are *not* the same. We can try to work that out after we get the "local", vacuum case clear.)

To put it more shortly - this means a simple subdivision of shell volume into 'containers' implies a radial stretching/stressing must occur after gravity is 'switched on'.

This implies that you're subdividing the shell volume by *coordinate* r, *not* by physical size of a very small object. You are still insisting on "labeling" every small point in the volume by its *coordinate*, r, which can be thought of as its "Euclidean coordinate", because it's derived from the area of the associated 2-sphere via the Euclidean geometry formula. But that labeling is *not* physical--it is coordinate-dependent. Physically, as I keep saying, Euclidean geometry is *not* privileged, and the "Euclidean coordinate" of a particular point has *no* physical meaning.

Now that we are all clear on what hoop meant, how do you understand changing marble count with curvature - and extending that to molecule count in a 3d container? Is your argument taking full account, as I wrote above, of micro-to-macro K gradients, and how that changes with r? Still a problem here.

See my comments above. The 3d container continues to contain the same number of atoms (I like that word better than "molecule" for solids, since they might be metals which don't really have "molecules") regardless of where it is placed in the gravitational field. And since the K factor does not cause any internal stresses in the container, the atoms maintain the same physical distance between themselves as they did when the container was far away from all gravitating bodies. Therefore, the container maintains the same physical size.

[PeterDonis]

Take a ruler with marks r1 and r2 on it. In a region of curvature, use it to lay out two concentric spherical surfaces of a chosen solid angle. The relation of surface areas to r1, r2, and solid angle will be strictly Euclidean. But what about the volume between them measured with Peter's little marble idea? My guess is that it will not match the Euclidean prediction.

If there is a good way to determine "solid angle", i.e., to determine what fraction a given surface is of the full 2-sphere that it is part of, then I agree, the volume between them would increase over the Euclidean prediction by the K factor. My only reservation is that I'm not sure exactly how the solid angle would be measured. There are two possible cases: (1) there is a gravitating body like the Earth in the center; measuring solid angle would, it seems to me, require having unobstructed sight lines from the center of the Earth to the surface in question, so the angle at the center of the Earth subtended by the surface can be measured; or (2) there is a black hole at the center; in this case there is not even the "in principle" possibility of setting up sight lines in this way.

[Q-reeus]

Originally Posted by Q-reeus:
"If I read that right, you are perhaps inadvertently agreeing with my notion of 'spatial gradients'. That length measured with a microscopic ruler will not correspond exactly with length measured with a large ruler, even 'locally', because of local gradients."

The only thing that would affect a large ruler, as opposed to a very small ruler, is tidal gravity. That's the only kind of "gradient" in the field that can cause actual physical stress on a ruler and thus change its physical length. The K factor does *not* do this.

Remember I said the K factor is *not* the same as curvature; it's related but it's not the same. Spacetime curvature is tidal gravity, not the K factor.
We agreed on first part earlier, no need to repeat. But the last bit 'spacetime curvature is tidal gravity' is surely too restrictive a definition. It would leave out redshift for one. I mean, you earlier agreed curvature is what allows K to exceed unity, but if curvature = tidal gravity (grad(grad(potential))), and tidal forces don't effect K, something is missing here. I use the term curvature a bit loosely maybe, but we had a mutual understanding I thought that K is a manifestation of that curvature, in a distinctly different way to tidal forces. And that gradient of K is not tidal forces at work. Unless restraints are imposed, no stresses from varying K - always understood ('solid sphere' argument excepted).

Where did this start? Oh yes - your argument in #126 that varying K, just over the radius of a hoop, throws out the ability to accurately determine local curvature via marble count:
"Yes, but as I pointed out, to calculate the count, you need to know the K factor for a whole range of "r" values, from r = 0 out to the "r" of the hoop, which is its circumference divided by 2 pi. This brings in additional complications which are not present if you consider the area between two nearby circles of circumference C and C + dC."

Are we yet again misunderstanding each other's words? I took the above to mean, since it was not specified any more clearly, that 'varying r' was from the center of the hoop to it's periphery. And you meant something different? What exactly? If not, you are saying gradient of K locally matters re count - just as I thought. If not, how should one take it to mean? By the time that bit was written, you were quite aware of what I meant by hoop.
Originally Posted by Q-reeus: "Well gaps must open before one can pop in a new marble, yes? But I could have made that statement a bit clearer, sure."

What do you mean, "pop in a new marble"? We are not talking about that kind of experiment; we are talking about packing marbles into various pre-existing spaces. If you're envisioning a "gap" opening up, you're envisioning something like this: we take a circular "hoop" of circumference C and place it on a sphere, centered on the North Pole. Then we take a second circular "hoop" made of some elastic material, so it can stretch; we place it on the sphere starting with circumference C + dC(0), and then slowly move it away from the first circle, so dC gradually gets larger compared to dC(0). As we do that, yes, open space will appear that we now need to pack with more marbles. But also, as we do that, the K factor will vary between the two circles, so things become more complicated. It would be really good if you would stick to the purely "local" case first, to avoid confounding factors, like for example tidal gravity--see my comments elsewhere in this post.

Completely wrong at the start - more misunderstanding. Once you got the right idea of 'hoop' as a circular perimeter, why go bringing in this business of adding another one? Go right back to #108 where the hoop thing was introduced:
"Let's take your analogy of north pole - or anywhere on a curved spherical surface. Instead of concentric circles, just take a hoop, fill it with tiny marbles. We know that non-euclidean surface curvature means being able to fit more marbles inside the hoop than would be the case on a flat surface. But the analogy is flawed - we can move the hoop anywhere over a spherical surface and marbles fit the same. The proper analogy is more like a surface in the shape of an egg - with pointy end corresponding to the source of gravity in 'real' case. We note now that our hoop, despite having a fixed locally measured perimeter, fits more and more marbles within upon approach to the pointy end. Do you still say there will be no observable 'delta K factor'?"

Notice - one hoop, sampling a varying surface curvature. As it does so, the marble packing density alters - gaps will open - and to maintain packing density, one every now and then 'pops an extra one in'. Hope this part at least is perfectly bedded down. Sheesh.:grumpy:
Originally Posted by Q-reeus: "Stop right there...As you argued re hoop - there are gradient effects here."

Not if we're ignoring tidal gravity. See above. The only "gradient" effect that will actually stretch a container and change its physical size is tidal gravity. The K factor will *not* do this.

This is why I keep saying we should stick to the local case first; it avoids introducing confounding factors like tidal gravity that are *not* the same as the K factor. If you keep muddling these things together, you will keep on being confused. Once again: the K factor does *not* cause any stress on objects. Therefore, the K factor *cannot* change the physical size of a container; that would require causing stress on the container's walls. This is a basic point of the physics involved, and if it's not clear, we need to stick to the local case until it is.
Unless you can prove that the marble filled hoop will *not* experience changed packing density (restraint = fixed marble count) in heading towards pointy end, you have to face the fact that locally observed effects are present. And one possible *interpretation* by a local flat-land observer, who can't discern curvature directly, is varying hoop size, or alternately, shrinking marbles. Stresses can't explain it, but effects normally put down to changing container size and/or marble size are there. All you have to do to end that argument, is what I asked above - can the hoop packing density/number be independent of surface curvature? Around and around it all goes. Where it ends, nobody knows!:cry:

Originally Posted by Q-reeus:
"To put it more shortly - this means a simple subdivision of shell volume into 'containers' implies a radial stretching/stressing must occur after gravity is 'switched on'."

This implies that you're subdividing the shell volume by *coordinate* r, *not* by physical size of a very small object. You are still insisting on "labeling" every small point in the volume by its *coordinate*, r, which can be thought of as its "Euclidean coordinate", because it's derived from the area of the associated 2-sphere via the Euclidean geometry formula. But that labeling is *not* physical--it is coordinate-dependent. Physically, as I keep saying, Euclidean geometry is *not* privileged, and the "Euclidean coordinate" of a particular point has *no* physical meaning.
In order to properly match the s sided cube to spherical shells, which fails to fit re your scenario used earlier, one must stretch it further. That's what I meant. That stretch factor - what one would need to do, was my way of understanding your point about the misfit. But note my square bracketed comments in #128. :zzz:

[PAllen]

But the last bit 'spacetime curvature is tidal gravity' is surely too restrictive a definition. It would leave out redshift for one.

Gravitational redshift can occur in flat spacetime (e.g. uniform acceleration in SR produces it, yet the spacetime has no curvature).

Peter is correct that if there is curvature of spacetime in a region, then there is tidal gravity. There may also be non-euclidean effects on space alone, these being much harder to detect.

[PeterDonis]

But the last bit 'spacetime curvature is tidal gravity' is surely too restrictive a definition.

No, it's the exact definition in GR. Tidal gravity is spacetime curvature. There's a geometric way to express the meaning of this in more detail, but I won't go into it unless you want me to, as it is rather a tangent relative to the topic of this thread. But it's worth seeing how some of the other things you cite relate to the definition.

It would leave out redshift for one.

Redshift can be observed between accelerating observers in flat spacetime; curvature is not required. It can also be caused by curvature, of course, as it is around a gravitating body. But it is not the same thing as curvature.

I mean, you earlier agreed curvature is what allows K to exceed unity

Not quite; I said (or meant to say; I don't think I've gone into any detail about this yet, so it may be that you are reading too much into something I said rather quickly) that if K is not unity, there must be curvature present. The converse is *not* true; it is possible for there to be curvature present but still have K = 1. For example, "Painleve observers" who are free-falling towards a black hole from rest "at infinity" see K = 1, even though the spacetime is curved.

but if curvature = tidal gravity (grad(grad(potential))), and tidal forces don't effect K, something is missing here.

I didn't say tidal forces don't affect K. See above for how the two are related. With regard to the marbles and containers and so forth, I said that K does not cause stress in objects, whereas tidal gravity does.

I use the term curvature a bit loosely maybe, but we had a mutual understanding I thought

Apparently not. See above for clarification.

And that gradient of K is not tidal forces at work. Unless restraints are imposed, no stresses from varying K - always understood ('solid sphere' argument excepted).

This is true; K and variation in K does not cause stresses in objects, whereas tidal gravity does.

Where did this start? Oh yes - your argument in #126 that varying K, just over the radius of a hoop, throws out the ability to accurately determine local curvature via marble count:

That's not what I said. You even quoted what I said, but apparently failed to notice that I was talking about determining the marble count itself, *not* determining local curvature via marble count. Actually, you can't determine local curvature from the marble count, even if we restrict ourselves to a small enough range that K can be considered constant. K is not curvature, and all that the marble count allows us to measure is K. Measuring curvature is more complicated, as others' posts have illustrated.

Are we yet again misunderstanding each other's words? I took the above to mean, since it was not specified any more clearly, that 'varying r' was from the center of the hoop to it's periphery. And you meant something different? What exactly? If not, you are saying gradient of K locally matters re count - just as I thought. If not, how should one take it to mean? By the time that bit was written, you were quite aware of what I meant by hoop.

You've lost me here; I don't understand how this relates to the part of my post that you quoted. But again, all this about what happens when K varies is PREMATURE. Sorry for shouting, but I made a point of repeating this several times in my last post. You *need* to get the "local" case, with constant K, figured out *first*, before even *thinking* about gradients of any kind. Most of the rest of your post is the same thing. Let's get the constant K, local case agreed first. As far as I know, you still believe that even in that case, there is some effect that causes objects to "change size" somehow, when there isn't. We need to get that cleared up first. Until we do, I can't respond to any talk about what happens when there is a gradient in K, because we don't have a common base to start from.

[PeterDonis]

Peter is correct that if there is curvature of spacetime in a region, then there is tidal gravity.

And it's also true that if there is tidal gravity, then there is curvature of spacetime. So the two are equivalent, which is why I made the stronger claim I did in my post.

I should note that by "tidal gravity" I include *any* effect that causes initially parallel geodesics to converge or diverge. That's exactly what the Riemann curvature tensor captures.

[PAllen]

Unless you can prove that the marble filled hoop will *not* experience changed packing density (restraint = fixed marble count) in heading towards pointy end, you have to face the fact that locally observed effects are present. And one possible *interpretation* by a local flat-land observer, who can't discern curvature directly, is varying hoop size, or alternately, shrinking marbles. Stresses can't explain it, but effects normally put down to changing container size and/or marble size are there. All you have to do to end that argument, is what I asked above - can the hoop packing density/number be independent of surface curvature? Around and around it all goes. Where it ends, nobody knows!:cry:


Forget hoops, filled or otherwise. You cannot detect curvature in a 4-manifold with anything restricted to a 2-surface, in any orientation (anything you think you might detect this way will be a an embedding feature, similar to embedding a curved 2-sphere in flat 3-space). You need lots of measurements of a substantial spatial region, as in the examples Peter and I have been discussing (his volume examples, and Synge's many point, many measurement examples).

[PeterDonis]

In order to properly match the s sided cube to spherical shells, which fails to fit re your scenario used earlier, one must stretch it further. That's what I meant. That stretch factor - what one would need to do, was my way of understanding your point about the misfit.

I'm responding to this separately because it actually *is* about the local issue, when K is constant. Why would you want to stretch the cube? That invalidates the property that makes the cube useful in the first place: that it has sides of a known length. There is no force due to the K factor that *makes* the cube stretch, so if you think of it as being stretched, you're bringing in some external force, *not* due to the K factor, to stretch it. Why do that? It just complicates things without any benefit in understanding what we're trying to understand, which is the physical effects of the K factor.

However, the last part of that paragraph is *not* about the local issue:

But note my square bracketed comments in #128. :zzz:

Your square bracketed comments in #128 were:

[Hardening my stance on this. Comments in #123 apply: "While your concentric circles around north pole analogy in #99 talked in terms of perimeter-to-radius ratio, one could equally talk in terms of an enclosed surface area-to-perimeter ratio of a dished annulus (numerically different, but having in common dependence on surface curvature). Once you see it the latter way, the hoop thing springs out as a more evident manifestation that local phenomena will exist, which is why I used it." There is a marble count excess for dished annulus sitting on 2-sphere (2D analogue of spherical shells experiencing 3-curvature). Same general effect must apply to a hoop, and likewise 3D container]

The enclosed surface area to perimeter ratio is *not* local; it requires sampling K over a range of perimeters (circumferences in my terminology), over which K varies. The "annulus" scenario I used does not; that's why I used it. So this part is out of bounds: I won't discuss it until we've got the local, constant K part figured out.

[pervect]

Forget hoops, filled or otherwise. You cannot detect curvature in a 4-manifold with anything restricted to a 2-surface, in any orientation (anything you think you might detect this way will be a an embedding feature, similar to embedding a curved 2-sphere in flat 3-space). You need lots of measurements of a substantial spatial region, as in the examples Peter and I have been discussing (his volume examples, and Synge's many point, many measurement examples).

You could, I suppose, also replace the hoops with spheres, which seems more in spirit with the discussion.

[PAllen]

You could, I suppose, also replace the hoops with spheres, which seems more in spirit with the discussion.

Q-reeus introduced hoops (to simplify? to detect orientation? not quite sure). So, I want to emphasize it will never work.

The simplest thing you can do with spheres to detect curvature is up for discussion. I don't think anyone here so far claims to know what the simplest construction involving spheres or parts of spheres that would detect curvature is.

[The hard part of detecting curvature from purely spatial measurements is avoiding embedding artifacts. Anything dependent on a particular foliation doesn't cut it. So, if one hypothesized that volume to surface area ration differed from 1/3, you would have to show that there does not exist any foliation in which the ratio is 1/3.]

[pervect]

Taking a large, hollow sphere, and counting the number of smaller spheres you can pack into it, to measure it's volume, would (at least in principle) give you a measure of spatial curvature. But it wouldn't give a measure of space-time curvature, it would measure the spatial curvature of some particular spatial slice.

I think thats what was wanted, though I haven't been following in detail and the thread is too long to try and catch up.

Another minor issue is that the Riemann of a plane only has 1 component, but the Riemann of a three-space should have 3. So the circle-packing tells us as much as we can know about the curvature of a plane, but sphere-packing doesn't tell us everything about the curvature of some particular spatial slice.

[PeterDonis]

Taking a large, hollow sphere, and counting the number of smaller spheres you can pack into it, to measure it's volume, would (at least in principle) give you a measure of spatial curvature. But it wouldn't give a measure of space-time curvature, it would measure the spatial curvature of some particular spatial slice.

I think thats what was wanted, though I haven't been following in detail and the thread is too long to try and catch up.

Kinda sorta. You bring up a good point, we've been using the word "curvature" without always being clear about what kind.

When I said that tidal gravity is the same as curvature, I meant specifically *spacetime* curvature. (I said so explicitly at least once.)

There is also, as you say, the curvature of a spatial slice. That, of course, depends on how you cut the slice, so to speak, out of spacetime. Also, as you note, the sphere packing, which measures what I've been calling the K factor, is not a complete measure even of the spatial curvature. (Also, as I've noted, the measurement you describe samples the K factor over a range of radial coordinates, or sphere areas, so it's more complicated than just measuring the K factor between two spheres that are very close together. I'm trying to stick to the "local" case, where K is effectively constant, until we get that sorted out, before bringing in variation in K.)

(Another minor point is that what you described is the *intrinsic* curvature of the spatial slice; there is also the extrinsic curvature of the slice, which is something else again.)

[PAllen]

Taking a large, hollow sphere, and counting the number of smaller spheres you can pack into it, to measure it's volume, would (at least in principle) give you a measure of spatial curvature. But it wouldn't give a measure of space-time curvature, it would measure the spatial curvature of some particular spatial slice.

I think thats what was wanted, though I haven't been following in detail and the thread is too long to try and catch up.

Another minor issue is that the Riemann of a plane only has 1 component, but the Riemann of a three-space should have 3. So the circle-packing tells us as much as we can know about the curvature of a plane, but sphere-packing doesn't tell us everything about the curvature of some particular spatial slice.

Well I was interested in something intrinsic. I see no reason you can't construct a non-euclidean spacelike 3-surface in Minkowski flat spacetime. What would be the physical significance of that? Whereas, with spacetime curvature present, while you can generally find a flat 2-surface, you cannot find a flat 3-surface (a while back I opened a thread on embedding like this, and determined this based on number of coordinate conditions that can be imposed on a metric). So, to have real meaning, the condition to look for isn't ability to find a curved spatial slice; instead, it is inability to find a flat one.

Separately, I don't know if every 3-surface with non-vanishing Riemann tensor must deviate from the Euclidean sphere area/volume ratio. Have you determined that this is true?

[pervect]

Kinda sorta. You bring up a good point, we've been using the word "curvature" without always being clear about what kind.

When I said that tidal gravity is the same as curvature, I meant specifically *spacetime* curvature. (I said so explicitly at least once.)

There is also, as you say, the curvature of a spatial slice. That, of course, depends on how you cut the slice, so to speak, out of spacetime. Also, as you note, the sphere packing, which measures what I've been calling the K factor, is not a complete measure even of the spatial curvature. (Also, as I've noted, the measurement you describe samples the K factor over a range of radial coordinates, or sphere areas, so it's more complicated than just measuring the K factor between two spheres that are very close together. I'm trying to stick to the "local" case, where K is effectively constant, until we get that sorted out, before bringing in variation in K.)

(Another minor point is that what you described is the *intrinsic* curvature of the spatial slice; there is also the extrinsic curvature of the slice, which is something else again.)

I agree with everything you wrote, though I'm still not sure which one of the various aspects of curvature is of interest. I suspect that the idea is just to overall describe curvature.

I don't regard extrinsic curvature as being physically very interesting, because we'd have to stand outside of space-time to do define it. So I'm mostly interested in intrinsic curvature. I suppose that the extrinsic curvature might be of some use if you're doing ADM stuff, but it's outside the scope of my current interests.

As regards intrinsic space-time curvature, I'd go with the perhaps overly mathematical approach that says that it's completely defined by the Riemann tensor, and that tidal forces are described by one part of the Riemann tensor, the part that's sometimes called the electro-gravitic part in the Bel decomposition.

There are two other parts of the Bel decomposition in the 4d spacetime of GR. One of them is the topo-gravitic part. This describes the purely spatial part of the curvature.

The remaining part is the magneto-gravitic part, that describes frame dragging effects.

So my take is that tidal gravity is part of the mathematical entity (the Riemann) that completely describes all the aspects of space-time, curvature, but it's not the complete story.

Though I think that if you have the tidal forces for observers in all state of motion (rather than just the tidal forces for one observer), you can recover the Riemann, just as you can do it from a set of multiple sectional curvatures of planar slices, though I couldn't write down exactly how to perform either operation.

If we focus on the deviation between a reference geodesic and nearby geodesics (via the geodesic deviation equation), we can neatly categorize the various parts of the Bel decompositon as follows:

The geodesic deviation (the relative acceleration between nearby geodesics) will depend on both the spatial separation (and be proportional to it), and will also depend on the relative velocity (said velocity being measured in the fermi-normal frame of the reference geodesic).

The deviation turns out to be quadratic with respect to the velocity. The terms independent of velocity, presnet at zero velocity, will give rise to the electro-gravitic part of the tensor, and are described by the tidal forces.

The parts that are proportional to velocity describe the magnetic part.

The parts that are proportional to velocity squared are due to the spatial curvature (i.e. the topogravitic part of the tensor). They're rather analogous to the v^2/R type forces that an object moving in a circular path of radius R with velocity v experiences.

[PeterDonis]

Whereas, with spacetime curvature present, while you can generally find a flat 2-surface, you cannot find a flat 3-surface (a while back I opened a thread on embedding like this, and determined this based on number of coordinate conditions that can be imposed on a metric).

You can in certain special cases. One has been mentioned in this thread: Painleve coordinates in Schwarzschild spacetime; the slices of constant Painleve time are flat. Another is FRW spacetime with k = 0; spacetime as a whole is curved but the spatial slices of constant "comoving" time are flat. I don't know of any other such cases off the top of my head.

So, to have real meaning, the condition to look for isn't ability to find a curved spatial slice; instead, it is inability to find a flat one.

Provided it isn't one of the special cases.

[PeterDonis]

I don't regard extrinsic curvature as being physically very interesting, because we'd have to stand outside of space-time to do define it.

Not the extrinsic 3-curvature of a spatial slice taken out of a 4-d spacetime. But I agree extrinsic curvature isn't very interesting compared to intrinsic; certainly not for the topic of this thread.

As regards intrinsic space-time curvature, I'd go with the perhaps overly mathematical approach that says that it's completely defined by the Riemann tensor,

Agreed.

and that tidal forces are described by one part of the Riemann tensor, the part that's sometimes called the electro-gravitic part in the Bel decomposition.

There are two other parts of the Bel decomposition in the 4d spacetime of GR. One of them is the topo-gravitic part. This describes the purely spatial part of the curvature.

The remaining part is the magneto-gravitic part, that describes frame dragging effects.

So my take is that tidal gravity is part of the mathematical entity (the Riemann) that completely describes all the aspects of space-time, curvature, but it's not the complete story.


Well, the term "tidal gravity" may be a bit ambiguous. I posted earlier the definition I was using: *anything* that causes geodesic deviation. All parts of the Riemann tensor describe this in some form; you describe how later in your post.

I recognize now that my terminology may be non-standard. I first saw it in Kip Thorne's Black Holes and Time Warps, and it made sense to me. But my arguments in this thread are essentially unchanged if "tidal gravity" is taken to describe just the electrogravitic part of the full curvature tensor, as you say. It's still true that what I am calling the K factor is not the same as tidal gravity, and is not equivalent to it, nor is it equivalent to curvature in full.

Though I think that if you have the tidal forces for observers in all state of motion (rather than just the tidal forces for one observer), you can recover the Riemann, just as you can do it from a set of multiple sectional curvatures of planar slices, though I couldn't write down exactly how to perform either operation.

I think this is true, yes.

[PAllen]

You can in certain special cases. One has been mentioned in this thread: Painleve coordinates in Schwarzschild spacetime; the slices of constant Painleve time are flat. Another is FRW spacetime with k = 0; spacetime as a whole is curved but the spatial slices of constant "comoving" time are flat. I don't know of any other such cases off the top of my head.



Provided it isn't one of the special cases.

Ok, interesting. I only looked at the general case. I see you are right. So I would interpret that as saying that your K can be considered as an artifact of the 'natural' way stationary (not inertial) observers set up simultaneity.

Also interesting, is that G-P coordinates have the feature (just like SC) that (t,t) metric component becomes spacelike inside the horizon. It looks to me, then, that inside the horizon, where the metric is not static, you do not have flat spatial slices. I might guess that you can't, in this region.

[Passionflower]

You can in certain special cases. One has been mentioned in this thread: Painleve coordinates in Schwarzschild spacetime; the slices of constant Painleve time are flat.
Yes an observer traveling radially at escape velocity observes his space as flat in that the distance between two r-coordinate values is exactly the difference. Other observers, who are also on a radial path or stationary, observe this distance differently. One can obtain this distance by applying the Lorentz factor (local speed wrt a free falling at escape velocity observer) before integration between these two r-coordinates.

[PeterDonis]

Ok, interesting. I only looked at the general case. I see you are right. So I would interpret that as saying that your K can be considered as an artifact of the 'natural' way stationary (not inertial) observers set up simultaneity.

Yes, I agree.

Also interesting, is that G-P coordinates have the feature (just like SC) that (t,t) metric component becomes spacelike inside the horizon. It looks to me, then, that inside the horizon, where the metric is not static, you do not have flat spatial slices. I might guess that you can't, in this region.

Actually, you still can. G-P coordinates are very weird inside the horizon; *all four* coordinates are spacelike! That means that, even though the time coordinate T is spacelike, the surfaces of constant T are as well. I agree this is extremely odd, but it's true.

To be more precise in stating what this means: inside the horizon, the vector \frac{\partial}{\partial T} corresponding to Painleve time T *and* the vector \frac{\partial}{\partial r} corresponding to the radial coordinate r (which is defined the same for Painleve as for Schwarzschild coordinates) are spacelike. Since the angular vectors are also spacelike, you can form a spacelike 3-surface using the three "spatial" coordinates inside the horizon just like you can outside, and these 3-surfaces will be orthogonal to the worldline of an infalling "Painleve" observer. (Note that they are *not* orthogonal to the vector \frac{\partial}{\partial T}, obviously, because that vector is spacelike inside the horizon. But the 3-surfaces I've just defined aren't orthogonal to \frac{\partial}{\partial T} *outside* the horizon either, because of the dTdr term in the Painleve metric.)

[PAllen]

Yes, I agree.



Actually, you still can. G-P coordinates are very weird inside the horizon; *all four* coordinates are spacelike! That means that, even though the time coordinate T is spacelike, the surfaces of constant T are as well. I agree this is extremely odd, but it's true.

To be more precise in stating what this means: inside the horizon, the vector \frac{\partial}{\partial T} corresponding to Painleve time T *and* the vector \frac{\partial}{\partial r} corresponding to the radial coordinate r (which is defined the same for Painleve as for Schwarzschild coordinates) are spacelike. Since the angular vectors are also spacelike, you can form a spacelike 3-surface using the three "spatial" coordinates inside the horizon just like you can outside, and these 3-surfaces will be orthogonal to the worldline of an infalling "Painleve" observer. (Note that they are *not* orthogonal to the vector \frac{\partial}{\partial T}, obviously, because that vector is spacelike inside the horizon. But the 3-surfaces I've just defined aren't orthogonal to \frac{\partial}{\partial T} *outside* the horizon either, because of the dTdr term in the Painleve metric.)

Fascinating. After a little thought I agree. So I guess the SC geometry everywhere has too many symmetries to preclude construction of flat spatial slices.

[Q-reeus]

Originally Posted by Q-reeus:
"Unless you can prove that the marble filled hoop will *not* experience changed packing density (restraint = fixed marble count) in heading towards pointy end, you have to face the fact that locally observed effects are present. And one possible *interpretation* by a local flat-land observer, who can't discern curvature directly, is varying hoop size, or alternately, shrinking marbles. Stresses can't explain it, but effects normally put down to changing container size and/or marble size are there. All you have to do to end that argument, is what I asked above - can the hoop packing density/number be independent of surface curvature?"

Forget hoops, filled or otherwise. You cannot detect curvature in a 4-manifold with anything restricted to a 2-surface, in any orientation (anything you think you might detect this way will be a an embedding feature, similar to embedding a curved 2-sphere in flat 3-space). You need lots of measurements of a substantial spatial region, as in the examples Peter and I have been discussing (his volume examples, and Synge's many point, many measurement examples).
I'm putting it down to this: both you and Peter 'know' I probably have everything wrong, so what I actually argue tends to go in one ear and out the other, to be replaced with an image of what 'I probably meant', and then proceed to tear down that straw man. Point to anywhere, not just above but any previous entry, where I claimed what you think I did. I'm not so stupid as to imagine that running a hoop over an 'egg' allows one to detect 3-curvature at all. Have endlessly now referred to this as 2D analogue of 3D situation. Having said that, there *is* I think an in principle 2D manifestation of 3-curvature, which will be discussed in another posting.

[Q-reeus]

You could, I suppose, also replace the hoops with spheres, which seems more in spirit with the discussion.
And in #139:
"Taking a large, hollow sphere, and counting the number of smaller spheres you can pack into it, to measure it's volume, would (at least in principle) give you a measure of spatial curvature. But it wouldn't give a measure of space-time curvature, it would measure the spatial curvature of some particular spatial slice.
I think thats what was wanted, though I haven't been following in detail and the thread is too long to try and catch up."
Full marks from me for carefully reading what I have been arguing - yes got the drift admirably.

[Q-reeus]

Rather than spend time arguing over every matter of who said and meant what in recent posts, may I propose to look at this from a slightly different angle - literally. Back in #113 angles of triangles in positively curved spacetime was mentioned. Let's take it a bit further. In flat spacetime we have a flat equilateral triangular surface formed from a fine tiling of much smaller uniform equilateral triangles. There are no gaps - and all internal angles = 600. Now move this composite triangle to a region of 'uniformly' positively curved spacetime. Do we all accept as a given that, no matter the orientation, included angles now add to more than 600, -provided that is the sides of all triangles are geodesically 'straight' in that curved spacetime? And with that proviso that the angular departure is larger the larger the triangle? I will take it there is unanimously a yes and yes to the above. So here's the thing. If one constrains the outer triangle to have straight sides and therefore vertex angles significantly > 600, it follows gaps must appear in the tiling, since the much smaller triangular tiles will have vertex angles insignificantly > 600. This straight sides constraint necessarily means lowered surface packing density, or alternately filling with micro tiles between the mini tiles to maintain surface density.

Alternately, imposing the constraint that tiles pack uniformly, we must have that sides are no longer geodesically staight. In particular, the outer triangle must become puckered - inwardly bowing sides. 2D manifestation of 3-curvature. Now go the next step to 3D. Instead of flat tiles, build an outer tetrahedron from much smaller ones. Anyone doubt the same issues will manifest as for 2D, only proportionately greater in effect? Now head back to the s-sided cube arrangement between concentric shells. It didn't fit. But then it was supposed to be a perfect cube. Still wouldn't exactly fit given the above, but depending on assumed constraints, that certified in flat spacetime perfectly cubic cube has either puckered like a pin-cushion, or one notices a 'strange' addition to the vertex angles that 'shouldn't' be there. And 'strangely', one can fit more counting spheres inside said cube given the latter constraint (geodesically straight edges and flat faces).

Now all my yokel arguing here is apparently imposible, but seems inevitable to me. So where is this all falling apart?

[pervect]

Rather than spend time arguing over every matter of who said and meant what in recent posts, may I propose to look at this from a slightly different angle - literally. Back in #113 angles of triangles in curved spacetime was mentioned. Let's take it a bit further. In flat spacetime we have a flat equilateral triangular surface formed from a fine tiling of much smaller uniform equilateral triangles. There are no gaps - and all internal angles = 600. Now move this composite triangle to a region of 'uniformly' curved spacetime. Do we all accept as a given that, no matter the orientation, included angles now add to more than 600, -provided that is the sides of all triangles are geodesically 'straight' in that curved spacetime?


Well, to use one of the examples already mentioned, if you consider a FRW space-time is uniformly curved (offhand, I'm not sure of the definition of uniformly curved), the spatial part of the curvature, which is what you're measuring (assuming also that you use the cosmological time-slice) could be positive, negative, or zero, and hence the sum of the angles could be greater, equal, or less than 360.

[Q-reeus]

Well, to use one of the examples already mentioned, if you consider a FRW space-time is uniformly curved (offhand, I'm not sure of the definition of uniformly curved), the spatial part of the curvature, which is what you're measuring (assuming also that you use the cosmological time-slice) could be positive, negative, or zero, and hence the sum of the angles could be greater, equal, or less than 360.
Fair comment. I'm assuming we are relating this to the case of something similar to the original setup - stationary spherical mass and stationary observers. That would be positively curved spacetime I presume? In which angles add to more than 1800. [have since edited and added word 'positively' to curved - thanks]

[PAllen]

I'm putting it down to this: both you and Peter 'know' I probably have everything wrong, so what I actually argue tends to go in one ear and out the other, to be replaced with an image of what 'I probably meant', and then proceed to tear down that straw man. Point to anywhere, not just above but any previous entry, where I claimed what you think I did. I'm not so stupid as to imagine that running a hoop over an 'egg' allows one to detect 3-curvature at all. Have endlessly now referred to this as 2D analogue of 3D situation. Having said that, there *is* I think an in principle 2D manifestation of 3-curvature, which will be discussed in another posting.

Then it is not a straw men. Your last sentence is a mathematical falsehood, with no possible valid interpretation. That is, anything you can detect restricted to a 2-surface (or thin 3-analog of it) will tell you nothing about presence of absence of spatial curvature.

[PAllen]

Rather than spend time arguing over every matter of who said and meant what in recent posts, may I propose to look at this from a slightly different angle - literally. Back in #113 angles of triangles in positively curved spacetime was mentioned. Let's take it a bit further. In flat spacetime we have a flat equilateral triangular surface formed from a fine tiling of much smaller uniform equilateral triangles. There are no gaps - and all internal angles = 600. Now move this composite triangle to a region of 'uniformly' positively curved spacetime. Do we all accept as a given that, no matter the orientation, included angles now add to more than 600,

No, this is false. As long as you restrict to a 2-surface, and use physical procedures (which naturally pick out the flattest possible interpretation of local reality), you will detect no deviation. To make this plausible, go right back to my argument that you accused of being strawman - 2-sphere embedded in Euclidean 3-space. The 2-surface is curved, the 3-space is flat. As a result, anything you find restricted to a 2-surface cannot be distinguished from an embedding artifact; further, in seeking straightness, you will pick out a flat 2-surface and find no deviation.

The discussion I had with Peter went further than this. That if spacetime curvature is simple enough, then even fully general measures of 3-space curvature (spanning large regions) can show exact flatness. In each case Peter mentioned it is the 'most natural observer' - the inertial one, or the one who sees CMB radiation as isotropic - that detects no spatial at curvature at all, by any means. His K, in SC geometry, only applies to stationary observers, who are non-inertial. One can detect similar things for non-inertial observers in flat spacetime, so even K is related more to non-inertial motion than intrinsic spatial curvature.

[PeterDonis]

Now move this composite triangle to a region of 'uniformly' positively curved spacetime.

What is the triangle made of? The "triangle" is not a physical object; it's an abstraction. To "move" it, you have to realize the abstraction somehow. How?

Do we all accept as a given that, no matter the orientation, included angles now add to more than 600, -provided that is the sides of all triangles are geodesically 'straight' in that curved spacetime?

How are the sides going to be kept geodesically straight? As an abstraction, yes, a "triangle" in a curved spacetime, with geodesically straight sides, will have its angles add to something different from the Euclidean sum--more if the spacetime is positively curved, less if it's negatively curved. That much is simple geometry. But to move from that to the actual physical behavior of a physical object, you have to bring in, well, physics.

Btw, in the examples we've been discussing, both in 2-D and 3-D, the "hoops" and 2-spheres are *not* necessarily composed of geodesics. In the 2-D case, the "hoops" are lines of latitude, which are not geodesics except for the equator. In the 3-D case, the 2-spheres themselves have geodesic tangent vectors, but the static objects sitting between them do not if you include the time dimension; static objects hovering above a gravitating body are accelerated.

If one constrains the outer triangle to have straight sides and therefore vertex angles significantly > 600, it follows gaps must appear in the tiling, since the much smaller triangular tiles will have vertex angles insignificantly > 600. This straight sides constraint necessarily means lowered surface packing density, or alternately filling with micro tiles between the mini tiles to maintain surface density.

How will you constrain the triangle to have straight sides (where I assume "straight" means "geodesic in the curved surface"--again, precise terminology really helps in these cases)? If you work it out, you will see that, if you imagine taking a triangle of actual, physical material and moving it from flat to curved space this way, constraining the sides as you suggest will necessarily cause stresses in the material. As soon as that happens, you can no longer use the material as a standard of distance to compare things with flat space, because in flat space the material was unstressed.

Alternately, imposing the constraint that tiles pack uniformly, we must have that sides are no longer geodesically staight. In particular, the outer triangle must become puckered - inwardly bowing sides. 2D manifestation of 3-curvature.

In the sense that you can't "wrap" a flat surface onto a curved surface without distorting it, yes--so if you don't distort the flat surface, it won't fit properly onto the curved surface. That's a quick way of summarizing what you've been saying. But the distortion is a *physical change* in the flat object, so once that object is distorted you can't use it to compare the flat with the curved surface. That's what you appear to be missing.

I won't bother commenting on the 3D case because it works out exactly the same.

Now all my yokel arguing here is apparently imposible, but seems inevitable to me. So where is this all falling apart?

None of what you say is incorrect. It just doesn't mean what you appear to think it means.

Added in edit, after seeing PAllen's post: Here's another way to say what I just said. I say none of what you have said is incorrect; but at the same time, PAllen is right when he says the comment he quoted is false. That's because I am saying you are correct as a matter of abstract geometry only, while he is saying you are incorrect in how you are trying to relate the geometry to the physics. *Both* of our comments are valid. Can you see why?

[Q-reeus]

Originally Posted by Q-reeus: "Now move this composite triangle to a region of 'uniformly' positively curved spacetime."

What is the triangle made of? The "triangle" is not a physical object; it's an abstraction. To "move" it, you have to realize the abstraction somehow. How?
From #151 "In flat spacetime we have a flat equilateral triangular surface formed from a fine tiling of much smaller uniform equilateral triangles."
Miss that bit - tiling, as in tiles? A composite, physical object.
Originally Posted by Q-reeus:
"Do we all accept as a given that, no matter the orientation, included angles now add to more than 600, -provided that is the sides of all triangles are geodesically 'straight' in that curved spacetime?"

How are the sides going to be kept geodesically straight? As an abstraction, yes, a "triangle" in a curved spacetime, with geodesically straight sides, will have its angles add to something different from the Euclidean sum--more if the spacetime is positively curved, less if it's negatively curved. That much is simple geometry. But to move from that to the actual physical behavior of a physical object, you have to bring in, well, physics.
Like gaps appearing for instance. I gave two restraint examples - gaps appear, or sides curve. Not physics? Assuming some kind of elastic glue joining the triangle tiles, naturally, forcing the outer triangle's to be straight requires imposing stresses - evidence of living in curved spacetime. Gaps or curved sides or stresses. Not there, or not needed in flat spacetime. Can't see any physics at work; no makings of an in principle 'detector'?
Btw, in the examples we've been discussing, both in 2-D and 3-D, the "hoops" and 2-spheres are *not* necessarily composed of geodesics. In the 2-D case, the "hoops" are lines of latitude, which are not geodesics except for the equator. In the 3-D case, the 2-spheres themselves have geodesic tangent vectors, but the static objects sitting between them do not if you include the time dimension; static objects hovering above a gravitating body are accelerated.
Fine, but in present scenario, whether triangles have perfectly 'straight' sides is just a convenience - we are only really interested in detecting change. Straight sides just makes the job easier. Point is, there is detectable change - choose a restraint - see the effect.
Originally Posted by Q-reeus:
"If one constrains the outer triangle to have straight sides and therefore vertex angles significantly > 600, it follows gaps must appear in the tiling, since the much smaller triangular tiles will have vertex angles insignificantly > 600. This straight sides constraint necessarily means lowered surface packing density, or alternately filling with micro tiles between the mini tiles to maintain surface density."

How will you constrain the triangle to have straight sides (where I assume "straight" means "geodesic in the curved surface"--again, precise terminology really helps in these cases)? If you work it out, you will see that, if you imagine taking a triangle of actual, physical material and moving it from flat to curved space this way, constraining the sides as you suggest will necessarily cause stresses in the material. As soon as that happens, you can no longer use the material as a standard of distance to compare things with flat space, because in flat space the material was unstressed.
Gone over that above. But another angle on it - what assumptions are you making in saying one must force the outer triangle to have straight sides? That assumes, as I guessed in answering above, that the tiles are glued together somehow. That's one configuration. Another could be tiles just sitting together snuggly but loosely on a plane. Moved en masse to curved space, why would they not simply drift apart slightly owing to curvature? There are any number of possible constraints - everyone I have considered results in detectable change - my idea of physics at work.
Originally Posted by Q-reeus:
"Alternately, imposing the constraint that tiles pack uniformly, we must have that sides are no longer geodesically staight. In particular, the outer triangle must become puckered - inwardly bowing sides. 2D manifestation of 3-curvature."

In the sense that you can't "wrap" a flat surface onto a curved surface without distorting it, yes--so if you don't distort the flat surface, it won't fit properly onto the curved surface. That's a quick way of summarizing what you've been saying. But the distortion is a *physical change* in the flat object, so once that object is distorted you can't use it to compare the flat with the curved surface. That's what you appear to be missing.
Huh!? We *want* physical change - that's the detection of being in curved spacetime. Things change. Please, no endless circling here.

[PeterDonis]

Warning: long post! :redface:

Huh!? We *want* physical change - that's the detection of being in curved spacetime. Things change. Please, no endless circling here.

Yes, things change. And the change means you can't use the things as standards of distance.

Here's another way of stating what I just said: the things you are saying change as a result of curvature, change because you are specifying that the objects have forces exerted on them in order to match the curvature. Those forces are external forces; they are not due to the curvature itself, at least not in the scenarios you are describing.

At the risk of piling on scenario after scenario, consider the following:

Take a piece of paper and try to wrap it around a globe. You can't; the paper will buckle. That's a physical realization of one scenario you were describing, the one where the sides of the triangles remain straight in the Euclidean sense; if we imagine the paper tesselated with tiny triangles, the paper behaves in such a way as to keep the triangles as flat Euclidean triangles. The paper may buckle along the edges between triangles, but the triangles themselves maintain their shape. So the paper can't possibly conform to any curved surface.

Now take a piece of flat rubber and wrap it around the same globe. You can do it, but only by exerting force on the rubber to deform it into the shape of the globe. This is a physical realization of another scenario you described, where we allow the triangles to adjust to the curvature of the surface; if we imagine the rubber tesselated by tiny triangles, the deformation of the rubber will deform the triangles too. But the deformation is not somehow magically caused by the globe's curvature; it's caused by us, exerting external force on the rubber to change its shape.

We can use the rubber to measure the K factor as follows: suppose that, when sitting on a flat Euclidean plane, the paper and the rubber have identical shape and area. We cut the paper into little tiny squares, and use them to tile the rubber. They fit exactly. Now we take the rubber and stretch it over the globe, and it deforms; we can see the deformation by watching the little squares and seeing that they no longer exactly tile the rubber; we might, as you say, see little gaps open up between the squares, assuming that the rubber is being deformed appropriately. But in order to know exactly how the squares will tile the rubber once it's wrapped around the globe, we have to specify *how*, exactly, the rubber is being wrapped.

Here's one way of specifying the wrapping. Take a circular disk of rubber instead of a square, and a circular piece of paper that, on a flat Euclidean plane, is exactly the same size. We cut the paper up into tiny shapes (triangles, squares, whatever you like) and verify that on a flat Euclidean plane, they tile the rubber exactly. Now take a sphere and draw a circle on it, centered on the North Pole, such that the circumference of the circle is exactly the same as the circumference of the rubber circle when the latter is sitting on the flat Euclidean plane. If we wrap the rubber circle onto the sphere in such a way that its circumference lies exactly on the circle we drew on the sphere, we will need to stretch the rubber; its area will now be larger than it was on the flat plane, and we can verify this by trying to tile it with the little pieces we cut out of the paper and seeing that there is still area left over. This is a manifestation of the K factor being greater than 1. But again, we had to physically stretch the rubber--exert external force on it--to get it to fit the designated area of the sphere.

(As I've noted before, the K factor will vary over the portion of the sphere being used for this experiment, but the average will be greater than 1; if we wanted to limit ourselves to a single value for K, we could cut a small annular ring of rubber on the flat plane, draw two circles on the sphere that matched its outer and inner circumferences, and stretch the rubber between them, and verify that the area increased by tiling with little pieces cut out of a paper annulus that exactly overlapped the rubber on a flat plane. We'll need this version for the 3-D case, as we'll see in a moment.)

All of the above is consistent with what you've been saying about how curvature of a surface manifests itself, and nobody has been disputing it. Now carry the analogy forward to the 3-D case. Here, since we're going to have a gravitating body in the center, we need to use the "annular" version of the scenario, as I just noted. We go to a region far away from all gravitating bodies, so spacetime in the vicinity is flat, and we cut ourselves a hollow sphere of rubber, with inner surface area A and outer surface area A + dA. We also cut a hollow sphere of metal (we use this instead of paper as our standard for material that will maintain its shape instead of deforming) with the same inner and outer surface area. Then we cut the metal up into little tiny pieces and verify that the pieces exactly fill the volume of the rubber. (This is a thought experiment, so assume that we can do this "tiling" in the 3-D case as we did in the 2-D case.)

Now take the hollow rubber sphere and place it around a gravitating body, in such a way that its inner and outer surface area are exactly the same as they were in the flat spacetime region. (Again, this is a thought experiment, so assume we can actually do this as we did in the 2-D case.) We will find that we need to physically stretch the rubber in order for the inner and outer surface areas to match up; there is "more volume" between the two surfaces than there was in the flat spacetime region. Again, we can verify this by trying to tile the volume of the rubber with the little metal pieces, and finding that we run short; there is volume left over when all the pieces have been used up. This excess volume is a physical measure of the K factor. (If dA is small enough compared to A, K is basically constant in this scenario, as it was in the "annular" 2-D case.)

The key fact, physically, is that we have to apply an external force to the rubber, stretch it, to make it fit when it's wrapped around a gravitating body, just as we had to apply an external force to the 2-D rubber circle to make it fit in the specified way around the sphere. So we *cannot* say that the rubber was stretched "by the K factor"--the K factor wasn't what applied the force. We did. Remember that we are assuming that tidal forces, and any other "forces", are negligible; the *only* effect we are considering is the K factor. (Again, this is easier to do because we are now dealing with a "local" scenario, where K is constant; if we covered a wider area, so K varied, we would also find it impossible, or at least a lot harder, to ignore or factor out the effects of tidal gravity. In the "local" case it's easy to set up the scenario so that tidal gravity is negligible, while the K factor itself is not; we just choose the mass M of the gravitating body and the radial coordinate r at which we evaluate K appropriately.)

You may ask, well, what about if we *don't* exert any force on the rubber? What then? In the 2-D case, the answer is that if we exert no force on the rubber, we can't make it conform to the curved surface at all. We specified one way of fitting the rubber to the sphere--so that its circumference remained the same (we checked that by drawing in advance a circle on the sphere centered on the North Pole, to use as a guide). We could specify another way--keep the total area of the rubber constant. But if we do that, then the rubber's circumference will be smaller, so again the rubber has to deform. There is *no* way to make the flat piece of rubber exactly fit to the curved surface without deforming it somehow.

For the spacetime case, it's a little more complicated, because if we take the hollow sphere of rubber we made in a flat spacetime region, and move it around a gravitating body, *something* will happen to it if we exert no force on it (other than the force needed to move it into place). But the general point still applies: there is no way to "fit" the object into a curved spacetime region without deforming it somehow. Since we can't actually wrap a full hollow sphere around a gravitating body without breaking it, consider a portion of the sphere covering a given solid angle, as PAllen proposed in a previous post. Suppose we have done all the work described earlier in a flat spacetime region: we have the inner surface area A and outer surface area A + dA measured (now not covering the total solid angle 4 pi of the sphere, but something less), and we have a piece of metal of the exact same shape and size in flat spacetime cut up into little pieces, which tile the volume of the rubber exactly in flat spacetime.

Now we slowly lower the rubber into place "hovering" over a gravitating body. We have to specify *some* constraint for where it ends up. Suppose we specify that the inner surface lies in a portion of a 2-sphere such that the solid angle it covers is exactly the same as what it covered in flat spacetime--i.e., the ratio of the inner surface area A of the rubber piece to the total area of the 2-sphere is the same as it was for the 2-sphere that the inner surface was cut out of in flat spacetime. That tells us that the inner surface of the 2-sphere is not deformed; but now we have to decide what to do with the outer surface. We have the same problem as we did in the 2-D case: if we specify that the outer surface area remains the same, relative to the inner surface area (i.e., the outer surface "fits" into a corresponding 2-sphere the same way the inner surface does--the outer surface is not deformed), then the rubber will have to stretch to match up, and there will be extra volume in the rubber, according to the K factor (which we can verify by trying to tile with the little pieces of metal and seeing that there is volume left over). Or, if we specify that we want the volume of the rubber to remain the same (i.e., we want to be able to tile it exactly with the little pieces of metal), then the outer surface area will "fit" to a smaller 2-sphere, so the shape of the outer surface will deform. And, of course, if we relax the constraint on the inner surface, there are even more ways we could deform the rubber; but there is *no* way to fit it without *some* kind of deformation.

As far as how the rubber would deform if we carefully exerted *no* force on it except to lower it into place, I think it would end up sitting with the inner surface not deformed and with the volume the same (meaning the outer surface would be deformed, as above). But even in this "constant volume" case, the shape of the rubber would still not be exactly the same. This is a manifestation of curvature, but I don't think you can attribute it to the K factor, because the K factor is measured by "volume excess" when I impose a particular constraint (having the inner and outer surfaces both the same shape as in flat spacetime, as above), and we're now talking about a *different* constraint, one where we keep the volume the same, since, as I said, that's the "natural" constraint that I think would hold if we carefully exerted no "extra" forces on the rubber.

Sorry for the long post, but this is not a simple question, and I wanted to try to get as much as possible out on the table for discussion. To summarize: the K factor is measured by "area excess" or "volume excess" under a particular physical constraint; with that constraint, K is a physical observable, but you have to be careful about interpreting what it means. More generally, curvature manifests itself as the inability to "fit" an object with a defined shape in a flat region, onto a curved region without changing its shape somehow. That change in shape requires external forces to be exerted on the object, which distort the object and make it unusable as a standard of "size" or "shape"; this is always true in the 2-D case we were discussing, and in the spacetime case, it is "almost always" true; there is *some* way to place the object without exerting any "extra" forces, and even in this case, the object's shape will not be the same as it was in flat spacetime, but its volume will be the same, as long as we can ignore tidal gravity.

[Q-reeus]

Warning: long post! :redface:
You are I deduce a speed typist - massive effort!
[QUOTE]...Here's another way of stating what I just said: the things you are saying change as a result of curvature, change because you are specifying that the objects have forces exerted on them in order to match the curvature. Those forces are external forces; they are not due to the curvature itself, at least not in the scenarios you are describing.
Understand and agree with that almost completely. Sure the curvature is not exerting any direct forces, so in that sense yes is not physics. Had thought though we could use the 'deformity correction' externally applied forces as a gauge of 'geometry'. More on that below.
As far as how the rubber would deform if we carefully exerted *no* force on it except to lower it into place, I think it would end up sitting with the inner surface not deformed and with the volume the same (meaning the outer surface would be deformed, as above). But even in this "constant volume" case, the shape of the rubber would still not be exactly the same. This is a manifestation of curvature, but I don't think you can attribute it to the K factor, because the K factor is measured by "volume excess" when I impose a particular constraint (having the inner and outer surfaces both the same shape as in flat spacetime, as above), and we're now talking about a *different* constraint, one where we keep the volume the same, since, as I said, that's the "natural" constraint that I think would hold if we carefully exerted no "extra" forces on the rubber.
This is where something, perhaps much different than straight K, should still allow measurement of a kind. Take an equalateral triangle composed of rigid tubes joined by free-hinging joints. In this configuration, one should expect vertex angles will exceed 60 degrees as discussed before. If small triangular gussetts were initially glued to the apexes, owing to their reduced susceptibility to angular change, ought to partially restrain angular expansion of the much larger triangle tube joints. In other words, there should be some internal stresses - miniscule but in principle measurable. I think.
Sorry for the long post, but this is not a simple question, and I wanted to try to get as much as possible out on the table for discussion. To summarize: the K factor is measured by "area excess" or "volume excess" under a particular physical constraint; with that constraint, K is a physical observable, but you have to be careful about interpreting what it means. More generally, curvature manifests itself as the inability to "fit" an object with a defined shape in a flat region, onto a curved region without changing its shape somehow.
Many thanks Peter for going to all that fuss - I do now appreciate more just why K factor, relating to a global geometry restraint, requires an implied stretching to make sense. So an isolated, stress-free container (or sheet in 2D case) cannot exhibit that unless forced into an equivalent geometry - like the solid angle segment mentioned. OK then agree that my earlier subdividing of concentric shells argument is not appropriate. And the hoop/marbles thing is presumably analogous. To explain a null result, we must assume there is no effective sampling of curvature, which is still hard for me to fathom re curved vs flat surface. Or is it a fundamental difference betwen 2D vs 3D , which may have been talked about elsewhere.
That change in shape requires external forces to be exerted on the object, which distort the object and make it unusable as a standard of "size" or "shape"; this is always true in the 2-D case we were discussing, and in the spacetime case, it is "almost always" true; there is *some* way to place the object without exerting any "extra" forces, and even in this case, the object's shape will not be the same as it was in flat spacetime, but its volume will be the same, as long as we can ignore tidal gravity.
So how to interpret goings on in the case of say a very rigid tubular frame in the shape of a perfect tetrahedron in flat spacetime. It is the simplest polyhedron that is fully constrained re shape when all sides are connected via freely hinging ball joints. A 3D arrangement of four triangles. If we accept the vertex angles will be greater in +ve curved spacetime, there is a sense that the enclosed volume should be greater than in flat spacetime. Can we say just how that would not be true? S'pose it's down to fierce math proof - yes? Scratching head over that one.:confused:
Thanks again, Cheers :wink::zzz:

[PeterDonis]

You are I deduce a speed typist

Thanks to my Dad forcing me to learn to type one summer in high school, yes. I learned on the old Olympic typewriter that he had used to type his master's thesis, which was older than I was. So my fingers learned how to squeeze a decent word rate out of those old, half-rusted keys that took about ten pounds of effort each to strike, and now they just laugh at a typical computer keyboard, which takes practically no effort by comparison. (It probably helps that I play keyboards too.)

This is where something, perhaps much different than straight K, should still allow measurement of a kind. Take an equalateral triangle composed of rigid tubes joined by free-hinging joints. In this configuration, one should expect vertex angles will exceed 60 degrees as discussed before.

I think this is right, provided that the tetrahedron is large enough that spatial curvature is measurable over its size, and that we deform the tetrahedron to conform to the curvature. Consider the analogous case on a 2-D surface, where we make a triangle out of rigid tubes joined by free-hinging joints, and then measure the joint angles after carefully wrapping the triangle onto a sphere so it conforms to the sphere's curvature. This case is exactly like that of the circular rubber disk or annulus; to see any change in shape, we need to exert external force on the rubber to make it conform to the shape of the sphere, and we have to decide how it is going to conform. We also have to make the shape we are trying to wrap around the sphere large enough that it "sees" the curvature; if it's too small we will be unable to use it to detect any difference from a flat plane (like the tiny pieces of paper we cut out of the paper disk or annulus in my previous example).

Given that these conditions are met, yes, we can in principle use the triangle as described (or the tetrahedron in the 3-D case) to measure differences between the space we are interested in and flat space; but there is still the question of how we decide to constrain the deformation. In the case quoted above, you are essentially constraining the side lengths of the tetrahedron (or triangle) to be constant, and letting the angles vary. Consider the 2-D case first (the triangle), and note that to physically realize this, we not only need to let the joint angles expand; we also need to bend the sides of the triangle since they are no longer Euclidean straight lines, but geodesics of the sphere, i.e., segments of great circles. (At least, I assume this was your intention in specifying side lengths constant and letting angles vary.) For thought experiment purposes, we can stipulate that this can be done while keeping the length of the sides constant, and similarly, in 3-D, the sides of the tetrahedron will have to bend slightly, but we can stipulate that their lengths are still held constant. In both cases, however, the sides will clearly undergo deformation, and we will have to exert external force on the tetrahedron to effect this deformation.

If we accept the vertex angles will be greater in +ve curved spacetime, there is a sense that the enclosed volume should be greater than in flat spacetime.

With the constraint as described above, yes; I would expect the volume inside the tetrahedron (or the area of the triangle, in the analogous 2-D case) to be larger if the side lengths are stipulated to be held constant. But as I just noted, to realize this case we have to exert external force to deform the tetrahedron (or triangle), and this external force is what causes the volume to expand.

OTOH, we could stipulate a different constraint, that the tetrahedron should be "unstressed"--or more precisely, that we should not *impose* any stress on it by exerting external force. In this case, I'm not sure what would happen, other than that I would expect the tetrahedron's volume to remain constant--always assuming that we can neglect tidal forces (as I noted previously, we can always choose the mass M of the central body and the radius r appropriately to make K measurably different from 1 while still having tidal gravity negligible). I *think* that the angles would still expand, but the sides would shorten, keeping the resulting volume constant but changing the shape.

[JDoolin]

Hi, I apologize if this seems off topic. You guys are talking about a stationary shell, but another interesting idea to consider is an expanding shell.

http://www.physicsforums.com/attachment.php?attachmentid=40428&stc=1&d=1319836650

The particles making up the shell need not be attached, but are simply Lorentz-contracted and time-dilated to an extremely high density.

This is a question I've been pondering for some time; precisely what would the effect of gravity be on the internal particles? It's quite possible that you would have large regions where the gravitational potential would be extremely great, but fairly constant, so there is no net force in any direction.

Since everything would be so uniform, would there be any observable effect at all?

[PeterDonis]

The particles making up the shell need not be attached, but are simply Lorentz-contracted and time-dilated to an extremely high density.

Lorentz-contrated and time-dilated relative to what?

This is a question I've been pondering for some time; precisely what would the effect of gravity be on the internal particles?

It would depend on what the shell was made of and how you specified the initial conditions of the expansion.

For a shell made of "normal" matter, i.e., with pressure no greater than 1/3 its energy density, the gravity of the shell would cause the expansion to decelerate, similar to a matter-dominated expanding FRW model. Depending on the initial conditions (velocity of expansion vs. shell energy density and pressure), the shell might stop expanding altogether and re-contract, or it might go on expanding forever, continually slowing down but never quite stopping.

It's quite possible that you would have large regions where the gravitational potential would be extremely great, but fairly constant, so there is no net force in any direction.

Since everything would be so uniform, would there be any observable effect at all?

If the shell's density is uniform (i.e., uniform throughout the shell at any particular "time" in the shell's comoving frame--the density could still change with time, as long as it remained uniform spatially within the shell), then I think you are right to guess that there would be no observable effect from the "potential" within the shell itself. There might still be an effect relative to the potential in the spacetime exterior to the shell. (The potential interior to the shell would be the same as the potential on the shell's inner surface, just as for a static shell.)

[Q-reeus]

Originally Posted by Q-reeus:
"This is where something, perhaps much different than straight K, should still allow measurement of a kind. Take an equalateral triangle composed of rigid tubes joined by free-hinging joints. In this configuration, one should expect vertex angles will exceed 60 degrees as discussed before."

I think this is right, provided that the tetrahedron is large enough that spatial curvature is measurable over its size, and that we deform the tetrahedron to conform to the curvature. Consider the analogous case on a 2-D surface, where we make a triangle out of rigid tubes joined by free-hinging joints, and then measure the joint angles after carefully wrapping the triangle onto a sphere so it conforms to the sphere's curvature. This case is exactly like that of the circular rubber disk or annulus; to see any change in shape, we need to exert external force on the rubber to make it conform to the shape of the sphere, and we have to decide how it is going to conform.
Having accepted your explanation of why K cannot apply to a simple container that does not enclose the source of curvature, there remains a point of disagreement here. My understanding of triangles adding to more than 180 degrees in +ve curved 3-space is that the weirdness here is precisely due to that, as measured by say laser theodolite, the tubular sides of said triangle will be *exactly* straight and thus entirely unstressed (assuming 'gussetts' are absent). It is understood here measurements are taken in the curved environment - not some distant coordinate reference. Hence the specification of free-hinging pinned joints, and rigid tubes that are not 'floppy'. Otherwise, what is implied is surely an intrinsic, locally measurable curvature of just one straight rod, in going from flat spacetime to curved. But in 3-curvature, how will the 'straight' rod/tube 'know' which way to bend?

I think the proper analogy here in going from flat to curved is not trying to wrap a flat object onto a curved surface. Rather, think of drawing an equalateral triangle on the surface of a large balloon (low surface curvature). 2D flat-landers living on the balloon surface cannot directly detect the surface curvature, but with their 2D confined 'laser theodolites' will confirm the triangle sides are straight, and the vertex angles are 'near enough' to 60 degrees. Now deflate the balloon to a much smaller radius. Flat-landers now attempt to construct another equalateral triangle of the same side lengths as before (meaning triangle occupies a much larger portion of the balloon surface than before). Their theodolites continue to say the sides are perfectly straight, but are puzzled to find the vertex angles now significantly exceed 60 degrees. That's how I got what curvature does here - there's a faint whiff of sanity to Dr Who's 'Tardis' if you like.
We also have to make the shape we are trying to wrap around the sphere large enough that it "sees" the curvature; if it's too small we will be unable to use it to detect any difference from a flat plane (like the tiny pieces of paper we cut out of the paper disk or annulus in my previous example).
Yes, and is it not just this size related differential that allows flat-landers to detect curvature induced angular changes by means of a small, 'standard' protractor that minimally 'feels' curvature. Thoughts?

[PAllen]

Having accepted your explanation of why K cannot apply to a simple container that does not enclose the source of curvature, there remains a point of disagreement here. My understanding of triangles adding to more than 180 degrees in +ve curved 3-space is that the weirdness here is precisely due to that, as measured by say laser theodolite, the tubular sides of said triangle will be *exactly* straight and thus entirely unstressed (assuming 'gussetts' are absent). It is understood here measurements are taken in the curved environment - not some distant coordinate reference. Hence the specification of free-hinging pinned joints, and rigid tubes that are not 'floppy'. Otherwise, what is implied is surely an intrinsic, locally measurable curvature of just one straight rod, in going from flat spacetime to curved. But in 3-curvature, how will the 'straight' rod/tube 'know' which way to bend?

I think the proper analogy here in going from flat to curved is not trying to wrap a flat object onto a curved surface. Rather, think of drawing an equalateral triangle on the surface of a large balloon (low surface curvature). 2D flat-landers living on the balloon surface cannot directly detect the surface curvature, but with their 2D confined 'laser theodolites' will confirm the triangle sides are straight, and the vertex angles are 'near enough' to 60 degrees. Now deflate the balloon to a much smaller radius. Flat-landers now attempt to construct another equalateral triangle of the same side lengths as before (meaning triangle occupies a much larger portion of the balloon surface than before). Their theodolites continue to say the sides are perfectly straight, but are puzzled to find the vertex angles now significantly exceed 60 degrees. That's how I got what curvature does here - there's a faint whiff of sanity to Dr Who's 'Tardis' if you like.

Yes, and is it not just this size related differential that allows flat-landers to detect curvature induced angular changes by means of a small, 'standard' protractor that minimally 'feels' curvature. Thoughts?

One part you don't get is the issue of embedding. Please think about how a curved spherical surface is embedded in flat 3-space without telling you anything about the curvature of the 3-space. Similarly, in curved spacetime you can embed flat planes, and any procedure looking for straight lines will pick out this embedded flat plane. Thus no procedure limited to a plane can detect curvature of a 4-manifold.

I have explained this multiple times and you have continued to ignore it.

[Q-reeus]

I have explained this multiple times and you have continued to ignore it.
Ignore is not perhaps the right word, as all I could pick up were assertions of how it is - may well be true, but to me they were just assertions.
One part you don't get is the issue of embedding. Please think about how a curved spherical surface is embedded in flat 3-space without telling you anything about the curvature of the 3-space. Similarly, in curved spacetime you can embed flat planes, and any procedure looking for straight lines will pick out this embedded flat plane. Thus no procedure limited to a plane can detect curvature of a 4-manifold.

I have quoted a proof by J.L. Synge that five vertices are the minimum to detect curvature of spacetime (that is, even a tetrahedron can always be constructed to conform to Euclidean expectations).
Allright, given that is so, what sense does one make of the 'popular' statement that the internal angles of a triangle do not generally add to 180 degrees in curved spacetime? Having not studied the subject, I took my que from such as:
"In other words, in non-Euclidean geometry, the relation between the sides of a triangle must necessarily take a non-Pythagorean form." at http://en.wikipedia.org/wiki/Pythagorean_theorem#Non-Euclidean_geometry
[Another grab, from: http://en.wikipedia.org/wiki/Curved_space#Open.2C_flat.2C_closed
"Triangles which lie on the surface of an open space will have a sum of angles which is less than 180°. Triangles which lie on the surface of a closed space will have a sum of angles which is greater than 180°. The volume, however, is not (4 / 3)πr3" In context this appears to me to be a generalized statement applicable to higher than 2D curvature. As I say , haven't studied this subject at all.]

[PAllen]

Ignore is not perhaps the right word, as all I could pick up were assertions of how it is - may well be true, but to me they were just assertions.

Allright, given that is so, what sense does one make of the 'popular' statement that the internal angles of a triangle do not generally add to 180 degrees in curved spacetime? Having not studied the subject, I took my que from such as:
"In other words, in non-Euclidean geometry, the relation between the sides of a triangle must necessarily take a non-Pythagorean form." at http://en.wikipedia.org/wiki/Pythagorean_theorem#Non-Euclidean_geometry

If you look at that link, the context is geometry on a 2-surface. Popular statements don't get into the issue of embedding - they are over-simplifed. I am not just making assertions, I am asking you to think, as I'll do again. Apply your idea to 3-space "if a triangle is non-pythagorean, the *space* is non-euclidean' to flat 3-space. On a 2-sphere in 3-space, you conclude, correctly, that the 2-sphere is curved (triangle is non-pythagorean). What does that tell you about the 3-space: nothing. The 3-space is still flat.

One specific argument that flat planes are embeddable in a 4-manifold is simply to note that for an arbitrary symmetric metric with 10 components, you can apply 4 coordinate conditions. This is sufficient to make e.g. the x,y components of the metric [[1,0],[0,1]], that is a Euclidean plane. As a result, any non-Euclidean behavior of a plane is just a function of coordinate choice, and is not telling you anything intrinsic about the manifold.

[Q-reeus]

If you look at that link, the context is geometry on a 2-surface.
More inclusive quote from that same passage:
"The Pythagorean theorem is derived from the axioms of Euclidean geometry, and in fact, the Pythagorean theorem given above does not hold in a non-Euclidean geometry.[51] (The Pythagorean theorem has been shown, in fact, to be equivalent to Euclid's Parallel (Fifth) Postulate.[52][53]) In other words, in non-Euclidean geometry, the relation between the sides of a triangle must necessarily take a non-Pythagorean form. For example, in spherical geometry, all three sides of the right triangle (say a, b, and c) bounding an octant of the unit sphere have length equal to π/2, and all its angles are right angles, which violates the Pythagorean theorem because a2 + b2 ≠ c2."

I can only take that one way - in *any* non-Euclidean geometry. Just how that applies to triangle in 3-curvature is the question.
One specific argument that flat planes are embeddable in a 4-manifold is simply to note that for an arbitrary symmetric metric with 10 components, you can apply 4 coordinate conditions. This is sufficient to make e.g. the x,y components of the metric [[1,0],[0,1]], that is a Euclidean plane. As a result, any non-Euclidean behavior of a plane is just a function of coordinate choice, and is not telling you anything intrinsic about the manifold.
Not familiar with this to argue what you say here, other than to ask you to explain the full passage I quoted above. Put it simply please - are you saying that angles will add to 180 degrees in a generally 3-curved space, or not? :zzz:

[PAllen]

More inclusive quote from that same passage:
"The Pythagorean theorem is derived from the axioms of Euclidean geometry, and in fact, the Pythagorean theorem given above does not hold in a non-Euclidean geometry.[51] (The Pythagorean theorem has been shown, in fact, to be equivalent to Euclid's Parallel (Fifth) Postulate.[52][53]) In other words, in non-Euclidean geometry, the relation between the sides of a triangle must necessarily take a non-Pythagorean form. For example, in spherical geometry, all three sides of the right triangle (say a, b, and c) bounding an octant of the unit sphere have length equal to π/2, and all its angles are right angles, which violates the Pythagorean theorem because a2 + b2 ≠ c2."

I can only take that one way - in *any* non-Euclidean geometry. Just how that applies to triangle in 3-curvature is the question.

Not familiar with this to argue what you say here, other than to ask you to explain the full passage I quoted above. Put it simply please - are you saying that angles will add to 180 degrees in a generally 3-curved space, or not? :zzz:

The whole passage refers to geometry of a 2-surface. Euclid's Parrallel postulate is a postulate about plane geometry. Spherical geometry is the geometry of a 2-sphere - the *surface* of a sphere. And I ask you again to think about my simple embedding example. The surface of a globe is non-euclidean. The globe is sitting in a flat euclidean 3-space. Contradiction? No. The embedded space is curved, the space embedded in happens to be flat.

[PAllen]

More inclusive quote from that same passage:
"The Pythagorean theorem is derived from the axioms of Euclidean geometry, and in fact, the Pythagorean theorem given above does not hold in a non-Euclidean geometry.[51] (The Pythagorean theorem has been shown, in fact, to be equivalent to Euclid's Parallel (Fifth) Postulate.[52][53]) In other words, in non-Euclidean geometry, the relation between the sides of a triangle must necessarily take a non-Pythagorean form. For example, in spherical geometry, all three sides of the right triangle (say a, b, and c) bounding an octant of the unit sphere have length equal to π/2, and all its angles are right angles, which violates the Pythagorean theorem because a2 + b2 ≠ c2."

I can only take that one way - in *any* non-Euclidean geometry. Just how that applies to triangle in 3-curvature is the question.

Not familiar with this to argue what you say here, other than to ask you to explain the full passage I quoted above. Put it simply please - are you saying that angles will add to 180 degrees in a generally 3-curved space, or not? :zzz:

I am saying if you try to make Euclidean triangles in a general 4-d semi-riemannian manifold, you will succeed. For 3-space, I'm not sure whether you can *always* do it. There are fewer degrees of freedom, and the simple counting argument I used for 4-manifold does not work. It is certainly true that not all 2-manifolds can be embedded in flat 3-space, so it is plausible that some 3-manifolds will not embed a section of flat 2-surface.

Also recall my discussion with Peter: the SC geometry allows embedding of completely flat 3-space regions. The K factor is actually a feature of a particular class of observers (static observers, which are non-inertial observers), not something intrinsic to the geometry. It is analogous to distortions seen in an accelerating rocket in flat spacetime - a feature of the observer, not the intrinsic spacetime geometry. GP observers in the same SC geometry, experience absolutely flat 3-space.

[JDoolin]

Lorentz-contrated and time-dilated relative to what?



If you look carefully, there is one particle in the diagram that doesn't move. All motion in this diagram, then, is relative to that stationary particle.

This deserves further animations; I'd rather show you, if I can, rather than tell you. But suffice it to say, the animation above could be Lorentz transformed so as to place any particle in the center of the circle. However, because there are only a finite number of particles in the picture, there would be a gravitational asymmetry for every particle, except for one.


It would depend on what the shell was made of and how you specified the initial conditions of the expansion.



There are a couple different ways I might specify the initial conditions; (1) starting with an equipartition of rapidity (2) starting with an equipartition of momentum.

In this animation I assumed equal masses, and used equipartition of momentum.

Let

q\equiv \left \|\frac{\vec p}{m c} \right \| = \left \|\frac{(\vec v/c)}{\sqrt{1-v^2/c^2}} \right \|<3


\frac{v}{c}=\sqrt{\frac{q^2}{q^2+1}}

Once I calculated the velocities, I animated, finding position, by simply multiplying the velocities by t.



The animation renders about 79,000 dots of equal mass with random q between 0 and 3 in random directions. The particle at the center of this distribution should experience no net acceleration; however, closer to the edges, there is more and more gravitational force.

There's no limit to what q might be. If I set q=10^100, then all the particles that are in this animation would be so close to the center that they would experience essentially no net force.


For a shell made of "normal" matter, i.e., with pressure no greater than 1/3 its energy density, the gravity of the shell would cause the expansion to decelerate, similar to a matter-dominated expanding FRW model. Depending on the initial conditions (velocity of expansion vs. shell energy density and pressure), the shell might stop expanding altogether and re-contract, or it might go on expanding forever, continually slowing down but never quite stopping.

If the shell's density is uniform (i.e., uniform throughout the shell at any particular "time" in the shell's comoving frame--the density could still change with time, as long as it remained uniform spatially within the shell), then I think you are right to guess that there would be no observable effect from the "potential" within the shell itself. There might still be an effect relative to the potential in the spacetime exterior to the shell. (The potential interior to the shell would be the same as the potential on the shell's inner surface, just as for a static shell.)

I have a plan in mind to present the animation from the reference frame of a particle on the edge of the mass by Lorentz Transforming all of the velocities. Definitely, as long as the total mass of the distribution is finite, then there would be particles on the edge that experience extreme accelerations toward the center. If the mass were infinite (and thus no edge), you could invoke symmetry, and there would be no net force in any direction for any particle.

But then you'd have the problem, being inside a shell of infinite mass, that at any given point inside, you are at an infinite negative gravitational potential

On the other hand, with a finite mass, the particles at the edge could experience, at least for a time, acceleration equivalent to a black hole. I'm not at all sure what theoretical ramifications that would have.

Anyway, I'll work on the other animations, and hope that makes my meaning clearer.

[PeterDonis]

Having accepted your explanation of why K cannot apply to a simple container that does not enclose the source of curvature, there remains a point of disagreement here. My understanding of triangles adding to more than 180 degrees in +ve curved 3-space is that the weirdness here is precisely due to that, as measured by say laser theodolite, the tubular sides of said triangle will be *exactly* straight and thus entirely unstressed (assuming 'gussetts' are absent).

If the triangle is *assembled* in the curved space, in the right way, this will be true. But you specified that the triangle (or tetrahedron) was assembled in flat space, and then *moved* to curved space. That means the "natural" configuration of the sides is the flat configuration, Euclidean straight lines, and they must be deformed to get into the curved configuration, geodesics on a curved surface.

If, instead, you assembled the triangle carefully in a curved region, so that the "natural" configuration of its sides was as geodesics on the curved surface, and the "natural" angles at each vertex summed to more than 180 degrees, then the triangle would deform if you tried to make it conform to a flat Euclidean plane. Similarly, if you assembled a tetrahedron in a region of curved space, hovering over a gravitating body, so that it was unstressed in that configuration, and then took it far away from gravitating bodies where space was flat, it would undergo stress and deformation in the course of conforming to the flat space.

I think the proper analogy here in going from flat to curved is not trying to wrap a flat object onto a curved surface. Rather, think of drawing an equalateral triangle on the surface of a large balloon (low surface curvature). 2D flat-landers living on the balloon surface cannot directly detect the surface curvature, but with their 2D confined 'laser theodolites' will confirm the triangle sides are straight, and the vertex angles are 'near enough' to 60 degrees. Now deflate the balloon to a much smaller radius. Flat-landers now attempt to construct another equalateral triangle of the same side lengths as before (meaning triangle occupies a much larger portion of the balloon surface than before). Their theodolites continue to say the sides are perfectly straight, but are puzzled to find the vertex angles now significantly exceed 60 degrees. That's how I got what curvature does here - there's a faint whiff of sanity to Dr Who's 'Tardis' if you like.

It depends on what you want to make an analogy to. If we are trying to construct an analogy to what happens when we take an object constructed far away from gravitating bodies and move it into a gravity well, the above is *not* a good analogy for that, because deflating the balloon corresponds to contracting the spacetime as a whole. That would do as an analogy for a collapsing universe, but *not* for moving into a gravity well. A better analogy for that would be to consider a surface like the Flamm paraboloid...

http://en.wikipedia.org/wiki/File:Flamm%27s_paraboloid.svg

...and think of moving a triangle from the flat region to the curved region.

[PeterDonis]

If you look carefully, there is one particle in the diagram that doesn't move. All motion in this diagram, then, is relative to that stationary particle.

Then I'm not sure I would call this a "shell". "Shell" implies a thin region of matter with complete vacuum inside and outside it, at least in the context of this thread.

The animation renders about 79,000 dots of equal mass with random q between 0 and 3 in random directions. The particle at the center of this distribution should experience no net acceleration; however, closer to the edges, there is more and more gravitational force.

How are you calculating the force?

[JDoolin]

Then I'm not sure I would call this a "shell". "Shell" implies a thin region of matter with complete vacuum inside and outside it, at least in the context of this thread.



How are you calculating the force?

This is a good question, or more specifically, how should I (or how should we) calculate the force? So far, essentially, I'm not calculating the force. Invoking symmetry, there is no net force. There's a gravitational potential, but no net force on a particle at or near the center of the distribution.

But let's look at the other extreme now, from the very edge of the distribution.
http://www.physicsforums.com/attachment.php?attachmentid=40459&stc=1&d=1319912254

Here, obviously we don't have symmetry. The stationary particle should experience a net force to the right.

Let me give you a couple of premises of how I would go about calculating the force, and see if you agree with these premises, or if I am hopelessly naive.

#1 The net gravitational force on a particle is F = G m \sum_i \frac{M_i}{r_i^2} \vec u_i
#2 The force on a particle should be calculated based on the reference frame of the particle that is undergoing the force.
#3 The mass of each particle affecting gravitation is the rest-mass of the particle. i.e. the Lorentz factor affects momentum, but not gravitational attraction.
#4 The location of the particle is not the "simultaneous" location, but rather the speed-of-light delayed location of the particle. i.e. the speed of gravity is the same as the speed of light, so we must find an intersection of the past-light-cone of our particle of interest with the world-lines of the particles involved.

Another peculiarity of relativity, (whether using Galilean Transformation or Lorentz Transformation) is that when one accelerates toward a future event, it leans toward him, becoming directly in his future, but when one accelerates toward a past event, it leans away. We may find that even if the particles at the edge accelerate "toward" the center, that in fact, the end result is not at all what common-sense would suggest. By accelerating toward the center, the particle continually enters reference frames where the center is further and further away.

[PeterDonis]

Here, obviously we don't have symmetry. The stationary particle should experience a net force to the right.

Well, the problem as a whole has spherical symmetry--or at least, you can impose that condition as a reasonable idealization to make the problem tractable. If the problem has spherical symmetry, then the "force" on any particle (a) must point in the radial direction, and (b) must be a function only of its radial coordinate. If the matter is all ordinary matter, as I described in my last post, then (a) can be further refined: the force on any particle must point radially *inward*, i.e., the expansion of the shell must be decelerating. As I note below, viewing this deceleration as caused by a "force" may not be the simplest way to view this problem.

#1 The net gravitational force on a particle is F = G m \sum_i \frac{M_i}{r_i^2} \vec u_i

This requires the problem to be non-relativistic. That is not consistent with your #4. In fact, the inconsistency between #1 and #4 was one of the stumbling blocks on the road to General Relativity, back in the early 1900's. See further note below.

#2 The force on a particle should be calculated based on the reference frame of the particle that is undergoing the force.

A better way to state this would be: the 4-force on a particle is a covariant geometric object, it is the particle's rest mass times the particle's 4-acceleration, which is the covariant derivative of its 4-velocity with respect to its proper time.

#3 The mass of each particle affecting gravitation is the rest-mass of the particle. i.e. the Lorentz factor affects momentum, but not gravitational attraction.

A better way to state this would be: the "gravitational field" is determined by the stress-energy tensor of the matter, which is a covariant geometric object. The SET determines the field via the Einstein Field Equation. It also correctly accounts for the fact that "rest mass" is what affects gravity, not momentum due purely to kinematics.

#4 The location of the particle is not the "simultaneous" location, but rather the speed-of-light delayed location of the particle. i.e. the speed of gravity is the same as the speed of light, so we must find an intersection of the past-light-cone of our particle of interest with the world-lines of the particles involved.

If you are assuming this, then you can't use #1 as your force equation. Consider a simple example: the force on the Earth at a given instant does *not* point towards where the Sun was 8 minutes ago by the Earth's clock. It points towards where the Sun is "now" by the Earth's clock. (Technically, there are some small correction factors, but they can be ignored for this discussion.) So if you use your #1, you have to plug in the position of the Earth relative to the Sun "now", not the "retarded" position, or you'll get the wrong answer. (Steve Carlip wrote an excellent paper some time ago that explains all this: it's at http://arxiv.org/abs/gr-qc/9909087.)

As I noted above, using #1 requires the problem to be non-relativistic, and you don't seem to be imposing that limitation. For the relativistic case, you can't really use a "force" equation for this problem, or at least it does not seem to be the easiest way to approach it. A better model would be an expanding matter-dominated FRW-type model, as I mentioned in a previous post, especially since it doesn't seem like your particles are a "shell", since there's no interior vacuum region, as far as I can see. This type of model does not view gravity as a "force"; it just solves for the dynamics of a curved spacetime using the EFE and an expression for the stress-energy tensor of the matter. You can still view individual particles as being subject to a "force", but that force can be more easily calculated *after* you have constructed the model from the EFE.

If the spatial extent of the expanding "shell" is limited, then at the surface of the shell, the FRW-type solution would be matched to an exterior Schwarzschild vacuum solution; this would basically be the time reverse of the Oppenheimer-Snyder solution for the gravitational collapse of a star.

Another peculiarity of relativity, (whether using Galilean Transformation or Lorentz Transformation) is that when one accelerates toward a future event, it leans toward him, becoming directly in his future, but when one accelerates toward a past event, it leans away. We may find that even if the particles at the edge accelerate "toward" the center, that in fact, the end result is not at all what common-sense would suggest. By accelerating toward the center, the particle continually enters reference frames where the center is further and further away.

Huh? I don't understand what you're getting at here at all. How can you accelerate toward a past event? For that matter, how can you accelerate toward a future event? You accelerate toward a point in space, not an event in spacetime. Also, what kind of "reference frames" are you talking about? If you're talking about ordinary Lorentz frames, they are only valid locally; you can't use them to correctly evaluate distances to faraway objects.

It looks like this discussion might be better moved to a new thread, since it appears to be getting further from the topic of this one.

[JDoolin]

This requires the problem to be non-relativistic. That is not consistent with your #4. In fact, the inconsistency between #1 and #4 was one of the stumbling blocks on the road to General Relativity, back in the early 1900's. See further note below.



Okay, I see where I had one mistake.


http://en.wikipedia.org/wiki/Gauss's_law#Relation_to_Coulomb.27s_law

Strictly speaking, Coulomb's law cannot be derived from Gauss's law alone, since Gauss's law does not give any information regarding the curl of E (see Helmholtz decomposition and Faraday's law). However, Coulomb's law can be proven from Gauss's law if it is assumed, in addition, that the electric field from a point charge is spherically-symmetric (this assumption, like Coulomb's law itself, is exactly true if the charge is stationary, and approximately true if the charge is in motion).

I was thinking that coulomb's law and gauss law were equivalent. But they are only equivalent when the charge is not moving. I'll try to read more of the article to see what I've missed. May take me a bit of time to catch up so I don't think I'm ready to start another thread at this time.

If indeed, the gravitational field from a receding object has some predictable variation based on its relative velocity, then we should, of course, use that modification. But if you want to claim that there is no predictable variation; no predictable "meaningful" position we can use for distant objects, I beg to differ.

On another topic, if possible, can you support your argument that "ordinary Lorentz frames...are valid only locally." I've heard this time and time again, but no one has ever explained what it means. Are you saying that when I point at a distant galaxy, that that direction that I am pointing only exists locally? Are you saying that the very concept of direction is only a local phenomenon?

[PeterDonis]

If indeed, the gravitational field from a receding object has some predictable variation based on its relative velocity, then we should, of course, use that modification. But if you want to claim that there is no predictable variation; no predictable "meaningful" position we can use for distant objects, I beg to differ.

The answer to this will be easier to understand if I answer your other question first. See below.

On another topic, if possible, can you support your argument that "ordinary Lorentz frames...are valid only locally." I've heard this time and time again, but no one has ever explained what it means.

It means that you can set up a Lorentz frame centered around any event you like in spacetime, but the frame will only obey the laws of special relativity in a restricted local region of spacetime around that event. Strictly speaking, it will only obey the laws of special relativity exactly *at* that event; but in practice, there will be a finite region around the event where the deviations from the laws of special relativity due to the curvature of spacetime are too small to detect. The size of that region depends on how accurate your measurements are and how curved the spacetime is.

Here's a simple example: take an object that is freely falling towards the Earth from far away (say halfway between the Earth and the Moon), and set up a Lorentz frame using its worldline as the t axis. Since the object is in free fall, i.e., inertial motion, its worldline can be used this way according to the laws of SR. Now take a second object which is also in free fall towards the Earth, but which is slightly lower than the first object. Suppose there is an instant of time at which the second object is at rest relative to the first; we take this instant of time to define t = 0 in our Lorentz frame, and the position of the first object at this instant to define the spatial origin, so the frame is centered on that event on the first object's worldline.

It should be evident that, for a given accuracy in measuring the relative velocity of the two objects, there will be some time t > 0 at which it becomes apparent that the second object is no longer at rest relative to the first. (This is because it is slightly closer to the Earth and therefore sees a slightly higher acceleration due to gravity.) But SR says that' can't happen: both objects are moving inertially, both were at rest relative to each other at one instant, and SR says that they therefore should remain at rest relative to each other forever. They don't. So we can only set up a Lorentz frame around the first object's worldline for a short enough period of time that the effects of tidal gravity (which is what causes the difference in acceleration of the two objects) can't be measured. Similarly, if the second object were further away from the first, its relative acceleration would become evident sooner; so there is a limit to how far we can extend a Lorentz frame around the first object in space as well, before the effects of spacetime curvature become evident.

Are you saying that when I point at a distant galaxy, that that direction that I am pointing only exists locally? Are you saying that the very concept of direction is only a local phenomenon?

The direction you are pointing is the direction from which light rays emitted by the distant galaxy are entering your eyes. It is certainly reasonable to call that "the direction of the galaxy", but only if you are aware of the limitations of that way of thinking. Light paths are bent by gravity, so the direction you are seeing the light come from may not be "the" direction the galaxy is "actually" in. This is another manifestation of spacetime curvature, and it means you can only set up a Lorentz frame locally with regard to directions and positions as well as relative motion of objects.

This also applies to what you were saying about assigning a "distance" to distant objects. There is no unique definition of "distance" in a curved spacetime; there are at least three different ones that are used in cosmology. You can uniquely define distance in a local Lorentz frame, but that only works locally; once you are out of the local region where the laws of SR can be applied to the desired accuracy, your unique definition of distance no longer works.

[pervect]

There's a well known solution for an expanding (or collapsing) sphere of pressureless dust. Which isn't quite what was asked for, but you might be able to graft onto it to get the solution. The expanding pressureles dust sphere solution is just the FRW solution of cosmology, by the way.

As far as forces go, they're pretty much not used in GR. You can calculate them as an afterthought when you have the metric by evaluating u^a \nabla_a u^a, where u^a is your velocity vector.

[add]Conceptually, what the above expression does is calculate what an accelerometer following the worldline would read.

The reason forces aren't used much in GR that they transform in a complex manner - i.e. they don't transform as a tensor.

In SR, you can still deal with 4-forces as a tensor under Lorentz boosts. In GR, with accelerated coordinate systems, forces obviously cannot transform as a tensor. For instance, if you have an unaccelerated system with zero force, an accelerated system will have a nonzero force, but a tensor that is zero in one coordinate system must be zero in all.

[Q-reeus]

And I ask you again to think about my simple embedding example. The surface of a globe is non-euclidean. The globe is sitting in a flat euclidean 3-space. Contradiction? No. The embedded space is curved, the space embedded in happens to be flat.
No problem in agreeing with that situation.
Also recall my discussion with Peter: the SC geometry allows embedding of completely flat 3-space regions. The K factor is actually a feature of a particular class of observers (static observers, which are non-inertial observers), not something intrinsic to the geometry. It is analogous to distortions seen in an accelerating rocket in flat spacetime - a feature of the observer, not the intrinsic spacetime geometry. GP observers in the same SC geometry, experience absolutely flat 3-space.
In that more generaized context I see the point, no problems.

[Q-reeus]

A better analogy for that would be to consider a surface like the Flamm paraboloid...
http://en.wikipedia.org/wiki/File:Fl...paraboloid.svg
Handy reminder; while I think DrGreg had posted on it much earlier, upon finding the original article at http://en.wikipedia.org/wiki/Flamm%27s_paraboloid#Flamm.27s_paraboloid , studied that piece with some more attention. Had never bothered to understand the significance of the w ordinate, but now appreciate how it gives a useful handle on visualising spatial part of SM. 'Inverting' w ordinate gives the sense of radial contraction as I had envisaged. Matching that to shell transition is the interesting exercise. Brings it back to on topic, and having come to appreciate the general view of the difficulties of applying a simple reading of SC's, conclude the original posting was an inappropriate vehicle for finding problems with SM. Thanks for all the inputs.

[JDoolin]

The answer to this will be easier to understand if I answer your other question first. See below.



It means that you can set up a Lorentz frame centered around any event you like in spacetime, but the frame will only obey the laws of special relativity in a restricted local region of spacetime around that event. Strictly speaking, it will only obey the laws of special relativity exactly *at* that event; but in practice, there will be a finite region around the event where the deviations from the laws of special relativity due to the curvature of spacetime are too small to detect. The size of that region depends on how accurate your measurements are and how curved the spacetime is.

Here's a simple example: take an object that is freely falling towards the Earth from far away (say halfway between the Earth and the Moon), and set up a Lorentz frame using its worldline as the t axis. Since the object is in free fall, i.e., inertial motion, its worldline can be used this way according to the laws of SR. Now take a second object which is also in free fall towards the Earth, but which is slightly lower than the first object. Suppose there is an instant of time at which the second object is at rest relative to the first; we take this instant of time to define t = 0 in our Lorentz frame, and the position of the first object at this instant to define the spatial origin, so the frame is centered on that event on the first object's worldline.

It should be evident that, for a given accuracy in measuring the relative velocity of the two objects, there will be some time t > 0 at which it becomes apparent that the second object is no longer at rest relative to the first. (This is because it is slightly closer to the Earth and therefore sees a slightly higher acceleration due to gravity.) But SR says that' can't happen: both objects are moving inertially, both were at rest relative to each other at one instant, and SR says that they therefore should remain at rest relative to each other forever. They don't. So we can only set up a Lorentz frame around the first object's worldline for a short enough period of time that the effects of tidal gravity (which is what causes the difference in acceleration of the two objects) can't be measured. Similarly, if the second object were further away from the first, its relative acceleration would become evident sooner; so there is a limit to how far we can extend a Lorentz frame around the first object in space as well, before the effects of spacetime curvature become evident.



The direction you are pointing is the direction from which light rays emitted by the distant galaxy are entering your eyes. It is certainly reasonable to call that "the direction of the galaxy", but only if you are aware of the limitations of that way of thinking. Light paths are bent by gravity, so the direction you are seeing the light come from may not be "the" direction the galaxy is "actually" in. This is another manifestation of spacetime curvature, and it means you can only set up a Lorentz frame locally with regard to directions and positions as well as relative motion of objects.

This also applies to what you were saying about assigning a "distance" to distant objects. There is no unique definition of "distance" in a curved spacetime; there are at least three different ones that are used in cosmology. You can uniquely define distance in a local Lorentz frame, but that only works locally; once you are out of the local region where the laws of SR can be applied to the desired accuracy, your unique definition of distance no longer works.

I think my main objection to this is that if two objects are in free-fall, you cannot claim that they are both moving inertially.

Although, depending on what you mean by "set up a Lorentz frame using its worldline as the t axis," I may have an even greater objection:

Are you suggesting that geodesic paths in a gravitational field can represent straight lines in global inertial Lorentz Frames (in which case, I cannot agree; curved paths are not straight lines) or are you saying that a momentarily comoving reference frame with time axis tangent to the geodesic worldline at one particular event represents an inertial Lorentz Frame?

In any case, this no longer has anything to do with shells, so I can invite you to read my comments here: http://www.physicsforums.com/showthread.php?t=545002 or if you prefer, you or I could start a new thread regarding whether Lorentz Transformations are purely a local phenomenon.

[PeterDonis]

I think my main objection to this is that if two objects are in free-fall, you cannot claim that they are both moving inertially.

Um, excuse me? "Moving inertially" and "in free fall" are the same thing.

Are you suggesting that geodesic paths in a gravitational field can represent straight lines in global inertial Lorentz Frames

Obviously not; the whole point is that geodesic paths in a gravitational field are not straight lines in a global Lorentz frame. That's why SR cannot be applied on a large scale in a curved spacetime.

are you saying that a momentarily comoving reference frame with time axis tangent to the geodesic worldline at one particular event represents an inertial Lorentz Frame?

Yes, that's a more precise way of stating what I meant by "use the worldline of one freely falling body as the t axis of the Lorentz frame". So it seems that you understand that I can only set up a momentarily comoving reference frame tangent to a worldline in a curved spacetime at one particular event. That's what "Lorentz frames are only valid locally" means. So I'm having trouble understanding why you're not clear about the meaning of "Lorentz frames are only valid locally".

I'll look at what you posted in the other thread and make further comments on that if need be.

[PeterDonis]

I'll look at what you posted in the other thread and make further comments on that if need be.

Quick answers to the questions you posed in the other thread; I won't comment in that thread unless it seems warranted.

(1) my answer is (b)

(2) my answer could be (a), but only in the vacuous sense; in the presence of gravity there are no flat regions of spacetime, there is some curvature everywhere. (b) is the strictly correct answer, but as I said in an earlier post, for a given accuracy of measurement there will be some finite region where the deviations from flatness are not observable, and within that region a Lorentz transformation on a local coordinate patch of "flat" Minkowski coordinates will work

(3) my answer is (b), but "similar" has to be interpreted carefully; Fredrik's comments in an earlier post in that thread on the properties a coordinate chart has to have to admit these transformations are worth reading

(4) my answer is "it depends"; (a) is correct *only* if the coordinate chart meets the requirements for having a rotation transformation in the first place, per Fredrik's post; otherwise the answer is undefined because there is no well-defined assignment of coordinates to objects beyond a small local coordinate patch; (b) is never correct, if the transformation is valid at all when extended to distant objects it will change their coordinate positions

[Passionflower]

I think my main objection to this is that if two objects are in free-fall, you cannot claim that they are both moving inertially.

As Peter already wrote, inertial movement and free fall refers to the same thing.

One can have many objects in free fall all with lower and higher local velocities wrt a free falling object at escape velocity.

[JDoolin]

As Peter already wrote, inertial movement and free fall refers to the same thing.

One can have many objects in free fall all with lower and higher local velocities wrt a free falling object at escape velocity.

Newton's first law is often referred to as the law of inertia. The velocity of a body remains constant unless the body is acted upon by an external force.

An object in free fall is changing its velocity because it is acted upon by an external force. The force of gravity.

Inertial movement is movement in a straight line.

Free-fall is not inertial movement. It's accelerated movement.

This is a basic reality-check. You mean to tell me that if an object is in orbit, you actually consider that to be a straight line?

[PeterDonis]

Newton's first law is often referred to as the law of inertia. The velocity of a body remains constant unless the body is acted upon by an external force.

An object in free fall is changing its velocity because it is acted upon by an external force. The force of gravity.

Ah, ok. We have a terminology problem. See below.

Inertial movement is movement in a straight line.

Not in GR. In SR, yes, but that's only because SR assumes spacetime is flat. In GR, where spacetime can be curved, inertial motion has to be defined physically. An object is moving inertially if it feels no force, i.e., is weightless, i.e., is in free fall.

Free-fall is not inertial movement. It's accelerated movement.

In Newtonian terms, yes. But that's because Newtonian physics defines "force" and "acceleration" differently than GR does. In GR, an object in free fall, that is weightless, *feels* no force, and therefore there is no force on it in any coordinate-independent, invariant sense, and it is *not* accelerated. More precisely, it is not accelerated in the coordinate-independent sense of "acceleration": the covariant derivative of its 4-velocity with respect to its proper time is zero. This is sometimes called proper acceleration or 4-acceleration. And since "force" in GR is defined as the object's rest mass times its proper acceleration, a freely falling object is experiencing zero force in GR.

The object is "accelerated" with respect to the Earth, but that is coordinate acceleration, not proper acceleration; similarly, the "force" of gravity in Newtonian terms is not a "force" in GR, because it does not cause any 4-acceleration of the object. That's how I was using the terms, and how they are standardly used in GR.

You mean to tell me that if an object is in orbit, you actually consider that to be a straight line?

In GR, yes. More precisely, the object's *worldline* is a geodesic of the curved spacetime around the Earth, so it's as "straight" as any worldline can be in that spacetime. I was not saying that the object's path in *space* was straight; obviously it's not. But that's irrelevant to this discussion.

[Passionflower]

This is a basic reality-check. You mean to tell me that if an object is in orbit, you actually consider that to be a straight line?
It is not really relevant to the question what is free falling and inertial, every test observer that has no proper acceleration is free falling and inertial.

From one perspective one certainly could consider it a straight line as a test object in orbit differs only from a radially free falling test object by having a non-zero angular momentum. For instance do you feel yourself turned often when the seasons pass? The same for an astronaut who travels in orbit, as far as he is concerned he travels in a straight line as no forces act on him.

[Passionflower]

I was not saying that the object's path in *space* was straight; obviously it's not. But that's irrelevant to this discussion.
I disagree with that.

Whether a free falling object's path in space is straight depends solely on the chosen coordinates. In simple terms, the Sun just as much goes around the Earth as the Earth goes around the Sun it simply depends on the point of view.

[JDoolin]

Um, excuse me? "Moving inertially" and "in free fall" are the same thing.



Obviously not; the whole point is that geodesic paths in a gravitational field are not straight lines in a global Lorentz frame. That's why SR cannot be applied on a large scale in a curved spacetime.

Okay, basic reality check passed. RIGHT, geodesics are not straight lines in a global Lorentz Frame. Of course they aren't. But why would that be a failure of Special Relativity?

In fact, I would say that points to the sanity of Special Relativity. In fact, if Special Relativity somehow claimed that geodesic paths in spacetime were straight, I would find that to be much more alarming, and troubling.

Yes, that's a more precise way of stating what I meant by "use the worldline of one freely falling body as the t axis of the Lorentz frame".

You can't make the worldline of a freely falling body as the t-axis of an inertial frame. Because a freely falling body is changing velocity, and if there is a change in velocity, it's not the same inertial reference frame.

So it seems that you understand that I can only set up a momentarily comoving reference frame tangent to a worldline in a curved spacetime at one particular event. That's what "Lorentz frames are only valid locally" means.

Even if the object is only momentarily comoving with the reference frame at a single event, it's a GLOBAL LORENTZ frame. The Lorentz Frame extends for infinity into the past, the future, and in all directions.

In the very next instant, the body will be comoving with an entirely different GLOBAL Lorentz Frame.

So I'm having trouble understanding why you're not clear about the meaning of "Lorentz frames are only valid locally".

Because the Lorentz Transformation, just like the rotation transformation, takes as its input, the location of every event in space and time, and outputs the locations of every event in space and time.

It's like this. If I have a function whose domain is the real numbers, and its range is the real numbers, then I would say that function is valid globally. If I have a function whose domain is 0 to 1 and range is 0 to 1, then I would say the function is valid only locally.

The Lorentz Transformations take as input and output global inertial reference frames, representing every event that ever has and ever will happen in the entire universe.

[JDoolin]

It is not really relevant to the question what is free falling and inertial, every test observer that has no proper acceleration is free falling and inertial.

From one perspective one certainly could consider it a straight line as a test object in orbit differs only from a radially free falling test object by having a non-zero angular momentum. For instance do you feel yourself turned often when the seasons pass? The same for an astronaut who travels in orbit, as far as he is concerned he travels in a straight line as no forces act on him.

There are tidal forces in effect, if you have more than a point-observer. Even locally, free-fall is different from straight-line motion. The speed of light has a different geodesic from the falling box. Objects moving at different velocities have different geodesics, and with a careful bit of study, you might be able to figure out what you're orbiting around and how far away it is, even if you can't look outside your box.

The local phenomena within the area of a geodesic would be small, perhaps too small for your most sensitive equipment to detect, but traveling along a geodesic is different from traveling in a straight line.

[PeterDonis]

Okay, basic reality check passed. RIGHT, geodesics are not straight lines in a global Lorentz Frame. Of course they aren't. But why would that be a failure of Special Relativity?

Because SR requires initially parallel geodesics (freely falling worldlines) to remain parallel. Otherwise the whole mechanism for setting up Lorentz frames does not work.

In fact, I would say that points to the sanity of Special Relativity. In fact, if Special Relativity somehow claimed that geodesic paths in spacetime were straight, I would find that to be much more alarming, and troubling.

Then you should definitely be alarmed and troubled, because that's exactly what SR does claim. If you disagree, please provide an explicit counterexample: a geodesic path in a spacetime in which the laws of SR apply, that is not straight.

Of course it's easy to find geodesic paths in a curved spacetime that are not "straight" in the sense you're using the term, but the laws of SR don't apply in those spacetimes, precisely because they are curved. I've already given a specific example of such a law: SR requires initially parallel geodesics to remain parallel. (Or, in more ordinary language, SR requires that two objects, both in free fall and weightless, feeling no force, which are at rest relative to each other at one instant of time, must remain at rest relative to each other at all times.) In a curved spacetime, this law is violated, as my example of bodies falling towards Earth made clear.

You can't make the worldline of a freely falling body as the t-axis of an inertial frame. Because a freely falling body is changing velocity, and if there is a change in velocity, it's not the same inertial reference frame.

Exactly. And that violates the laws of SR. Therefore, SR only applies "locally" in a curved spacetime--in a small enough region that the changes in "velocity" you speak of are not observable within the accuracy of measurement being used.

Btw, it's also worth noting that you speak of "changing velocity" without defining what that means. The 4-velocity of a freely falling body does *not* change in the coordinate-independent sense I gave in my last post: its covariant derivative with respect to the body's proper time is zero. So if you think its velocity is changing, what is it changing relative to? Any such definition of velocity "changing" will be a coordinate-dependent definition. The GR definition is not; it's a genuine physical observable (whether or not the body feels acceleration).

Even if the object is only momentarily comoving with the reference frame at a single event, it's a GLOBAL LORENTZ frame. The Lorentz Frame extends for infinity into the past, the future, and in all directions.

And how are the coordinates in this supposed global Lorentz frame to be defined? Can you specify a way to do it that is both consistent with all the laws of SR, *and* works in a curved spacetime, where gravity is present? If so, please elucidate.

Because the Lorentz Transformation, just like the rotation transformation, takes as its input, the location of every event in space and time, and outputs the locations of every event in space and time.

If by "locations" you mean "coordinates", then yes, mathematically, this is what the LT does. That's the easy part. You have completely glossed over the hard part, *assigning* those coordinates in the first place in a way that is consistent with all the laws of SR. If you think you can do that in a curved spacetime, again, please elucidate.

[pervect]

There are tidal forces in effect, if you have more than a point-observer. Even locally, free-fall is different from straight-line motion. The speed of light has a different geodesic from the falling box. Objects moving at different velocities have different geodesics, and with a careful bit of study, you might be able to figure out what you're orbiting around and how far away it is, even if you can't look outside your box.

The local phenomena within the area of a geodesic would be small, perhaps too small for your most sensitive equipment to detect, but traveling along a geodesic is different from traveling in a straight line.

Well, there's never (or at least not usually) a straight line in our actual universe, because space-time is in general not flat. So geodesics are as close as we come.

Geodesics are of course described by the geodesic equation.


\frac{d^2x^\lambda }{dt^2} + \Gamma^{\lambda}{}_{\mu \nu }\frac{dx^\mu }{dt}\frac{dx^\nu }{dt} = 0


Futhermore, if we multiply through by m, and replace the right hand side by a force, this is about as close as GR comes to Newton's equations of motion.

Specifically, if we have a test particle moving under the action of an external non-gravitational force, (for instance, an electric field), we can write in GR


m\,\frac{d^2x^\lambda }{ds^2} + m\,\Gamma^{\lambda}{}_{\mu \nu }\frac{dx^\mu }{ds}\frac{dx^\nu }{ds} = F


So, it shouldn't be too surprising that the Christoffel symbols act pretty much like forces. In particular, we can identify some of them as being equal to what we used called the "force" of gravity in Newtonian theory.

And in GR, we can replace solving F=ma with solving the geodesic equations (well, there are occasions where this doesn't work, and we have to worry about the Paperpatrou equations, but this is rare)

BUT

As I remarked earlier, the components of the Christoffel symbols transform in a complex way. From wiki http://en.wikipedia.org/w/index.php?title=Christoffel_symbols&oldid=455650120 they transform like:


\overline{\Gamma^k_{ij}} =
\frac{\partial x^p}{\partial y^i}\,
\frac{\partial x^q}{\partial y^j}\,
\Gamma^r_{pq}\,
\frac{\partial y^k}{\partial x^r}
+
\frac{\partial y^k}{\partial x^m}\,
\frac{\partial^2 x^m}{\partial y^i \partial y^j}


which is not the tensor transformation law. So it's convenient to think of Christoffel symbols as "forces" in any one particular coordinate system that you want to work in, but it's a mistake to think they'll transform in the same manner as the forces in flat space-time that you may be thinking of them as being analogous to.

[JDoolin]

A couple of notes that I wanted to throw out before I go to work:

http://www.physicsforums.com/attachment.php?attachmentid=40428&stc=1&d=1319836650


But let's look at the other extreme now, from the very edge of the distribution.
http://www.physicsforums.com/attachment.php?attachmentid=40459&stc=1&d=1319912254


(1) I wanted to point out that this distribution is NOT an equipartition of rapidity, but an equipartition of v/(c sqrt(1-1/v/c^2)

I think if I redo it with equipartition of rapidity, we'll have distribution that looks the same (density-wise) in the center after Lorentz Transformation.

(2) I want to make an analogy. Let's say I take a cone, lay it out flat, and draw a straight line on it, and then wrap it back up in the cone shape again. We live in a cartesian coordinate system. Has that Cartesian Coordinate system failed us because that line that we have defined as straight is now curved? No. The cartesian coordinate system is alive and well, but it has a cone in it, and it has curved lines in it.

I'm not trying to criticize the application of geometry to take a cone and wrap it, and say, YES, you can draw "straight" lines on a curved object. I have no problem with that. But when you go on to say that the overlying cartesian geometry has somehow been invalidated because you can shape a paper into a cone, that's where I have a disagreement.

(3) I'm not saying that straight lines are always easy to detect. Of course if I look at the moon on the horizon, it looks distorted, because the light has NOT moved in a straight line (mostly because of atmospheric rather than gravitational effects). But the fact that the light doesn't move in a straight line does NOT mean that straight lines don't exist. But on the whole, how often does that happen? The great majority of things in the universe are not significantly affected by this sort of phenomenon. You point at them, and that's the direction they (not are) WERE, when the light left them. You can't see where the object IS, but you can see where the object WAS when the light left it.

[PeterDonis]

(2) I want to make an analogy. Let's say I take a cone, lay it out flat, and draw a straight line on it, and then wrap it back up in the cone shape again. We live in a cartesian coordinate system. Has that Cartesian Coordinate system failed us because that line that we have defined as straight is now curved? No. The cartesian coordinate system is alive and well, but it has a cone in it, and it has curved lines in it.

I'm not trying to criticize the application of geometry to take a cone and wrap it, and say, YES, you can draw "straight" lines on a curved object. I have no problem with that. But when you go on to say that the overlying cartesian geometry has somehow been invalidated because you can shape a paper into a cone, that's where I have a disagreement.

The cone's surface (barring the singularity at the apex) is still "flat" in the sense that it has zero intrinsic curvature. It has nonzero *extrinsic* curvature--the way in which it is embedded in the surrounding 3-D space is curved--but that's not what we're talking about in this discussion. If you calculated the connection coefficients in the formulas that pervect was writing, working with the Cartesian coordinates you assigned to the cone when it was laid out flat and you drew a straight line on it, they were zero when the cone was laid out flat and they are still zero when the cone is wrapped back up into a cone. So the cone is still flat in the intrinsic sense. (Technically, you could construct coordinates on the cone where the connection coefficients were nonzero, but the point is that you don't *have* to--there will always be *some* coordinate chart where they are all zero, whether the cone is laid out flat or wrapped up. The same is *not* true for a sphere--see below.)

Now try the same thought experiment with a sphere. You can't do it. There is no way to "lay a sphere out flat" and draw straight lines on the flat version, and then wrap it all back up into a sphere again. It's impossible--any such operation will distort the surface and invalidate its geometric invariants. That's one way of expressing the fact that a sphere has nonzero *intrinsic* curvature. The connection coefficients in pervect's formulas are nonzero on a sphere (i.e., there is *no* coordinate chart on the sphere that makes them all zero). And that's the kind of curvature we're talking about when we say that gravity is curvature of spacetime.

(3) I'm not saying that straight lines are always easy to detect. Of course if I look at the moon on the horizon, it looks distorted, because the light has NOT moved in a straight line (mostly because of atmospheric rather than gravitational effects). But the fact that the light doesn't move in a straight line does NOT mean that straight lines don't exist. But on the whole, how often does that happen? The great majority of things in the universe are not significantly affected by this sort of phenomenon. You point at them, and that's the direction they (not are) WERE, when the light left them. You can't see where the object IS, but you can see where the object WAS when the light left it.

No, you are not even seeing that. Light is bent by gravity. This has been measured for light passing close by the Sun: take two stars that are a little further apart in the sky than the apparent diameter of the Sun, when the Sun is in a different part of the sky--say the are separated by an angle a. When the Sun is in between them in the sky, they are separated by a *larger* angle, a + da, because the Sun bends their light towards itself. And what's true for the Sun is true for any large gravitating body.

The spacetime we are living in has intrinsic curvature because of gravity; it is a spacetime analogue of something like a sphere (actually more like a saddle, but the same argument I gave for a sphere would apply to a saddle too). It is *not* the spacetime analogue of something like a cone or a cylinder that can be laid out flat and wrapped up again without changing its intrinsic geometry. And in the presence of intrinsic curvature, all your intuitions about how things work in flat Euclidean space, or flat Minkowskian spacetime, are simply wrong on any large scale; they only approximately work over very small patches. You can only treat the Earth's surface as flat over a small area, and you can only treat spacetime as flat over a small region.

[JDoolin]

I have made several further posts in the other thread.

http://www.physicsforums.com/showthread.php?t=545002

I think I have explained myself better there.

I'm trying to find the most concise description of our disagreement: You seem to take the attitude that infinite straight lines do not exist. And I take the attitude that infinite straight lines do exist, or at least are definable. In my thinking, regardless of the curvature of paths caused by gravity, it is always possible to imagine what would happen if matter wasn't there. In your thinking, regardless of how we try to imagine what would happen if matter wasn't there, there would always be more matter, screwing things up. Now I won't argue with that, but the question is not whether we can really map the straight lines with perfect accuracy. The question is whether those straight lines exist at all. I think they do. You think they don't.

I explain why I think they do in the other thread.

[PeterDonis]

I'll comment in the other thread.

[DaleSpam]

I will do that. It will take a few days.This came up in another thread. I worked on this for a few days, got stuck, and did not complete it. The matter distribution in question had some discontinuities which made it difficult for me to handle, also, the matter distribution didn't have a nice convenient form like a fluid.

The bottom result is that I was not able to conclusively prove or disprove either position. I still think that I am probably correct, but not so strongly as before.

[George Jones]

This came up in another thread. I worked on this for a few days, got stuck, and did not complete it. The matter distribution in question had some discontinuities which made it difficult for me to handle, also, the matter distribution didn't have a nice convenient form like a fluid.

The bottom result is that I was not able to conclusively prove or disprove either position. I still think that I am probably correct, but not so strongly as before.

I'm having trouble tracing this back. What, precisely (and succinctly), is the matter distribution in question.

[Q-reeus]

I'm having trouble tracing this back. What, precisely (and succinctly), is the matter distribution in question.
Try #17 where the specs were set out.

[George Jones]

I am looking for a new post, written in succinct scientific style (without a lot commentary), of the form

"For a matter distribution given by ..., calculate ..."

[JDoolin]

Here's an idea for a non-stationary matter distribution:

Assume you have 5000 particles, originating from an event (t=0,x=0) and each is assigned a random x and y rapidity between -3 and 3, traveling away from each other in the xy-plane. Assume that these particles move with constant velocity until time t=1.

http://www.spoonfedrelativity.com/pages/ClockExplosion/ClockExplosionBlue.gif

At time t=1, the particles spontaneously develop mass. Given one of the particles, calculate the force on this particle resulting from the other 4999 particles.

(The proper-time/coordinate-time of the spontaneous development of mass, and the delay before that mass is detected, would need to be more fully described to answer the question.)

[DaleSpam]

I'm having trouble tracing this back. What, precisely (and succinctly), is the matter distribution in question.The details are a little fuzzy, but if I recall correctly Q-reeus was claiming that GR was not self consistent because of the transition from the Schwarzschild metric outside a solid shell to the flat metric inside the solid shell.

Somehow the discussion became focused on the space-space components of the (Ricci) curvature tensor and the space-space components of the stress-energy tensor. Q-reeus though you could neglect the space-space components of the stress-energy tensor simply because the time-time component was larger (which somehow led to a contradiction, though I don't remember how).

So I was going to calculate the metric for a finite-thickness solid shell and show that the space-space components of the stress-energy tensor could not be neglected and that accounting for them resolved the supposed contradictions.

[pervect]

At the risk of hijacking the thread (which seems hopelessly confused anyway), I do have a specific problem along the lines of boundary conditions - one I think I solved correctly, that I presented earlier. I think there was some questions raised about it, but I didn't follow the questions.

If we consider one of the simplest possible forms for the interior metric of a photon star,

from http://arxiv.org/abs/gr-qc/9903044

eq (1)

\frac{7}{4}\, dr^2 + r^2 \,d \theta^2 + r^2 sin^2 \theta \, d\phi^2 - \sqrt{\frac{7}{3}}\,r\,dt^2

we might ask how do we go about enclosing said interior metric in a thin, massless shell, so we get the exterior Schwarzschild metric. I.e. how do we match up the exterior and interior Schwarzschild soultions at the boundary so that we have a solution for light in a spherical "box" by matching the interior solution given by (1) to some exterioor Schwarzschild solution.

I started with the line element from Wald for the spherically symmetric metric:

eq(2)


-f(r)\,dt^2 + h(r)\,dr^2 + r^2 \left(d \theta ^2 + sin \, \theta \: d\phi^2 \right)


Einsteins' equations give via equations 6.2.3 and 6.2.4 from Wald, General Relativity

8 \, \pi \rho = \frac{ \left( dh/dr \right) }{r \, h^2} + \frac{1}{r^2} \left( 1 - \frac{1}{h} \right) \; = \; \frac{1}{r^2} \frac{d}{dr} \left[r \, \left(1 - \frac{1}{h} \right) \right]


8 \, \pi \, P = \frac{ \left( df/dr \right) } {r \, f \, h} - \frac{1}{r^2} \left( 1 - \frac{1}{h} \right)

Here \rho and P are the density and pressure in the spherical shell.

Setting \rho to zero and using 6.2.3 immediately tells us that r (1 - 1/h) is constant through the shell. For a thin shell, this means that h is the same inside the shell and outside the shell, because r is the same at the interior of the shell and the exterior of the shell, so h-, h inside the shell, equals h+, h outside the shell.

We can add 6.2.3 and 6.2.4 together to get

8 \pi \left(\rho + P \right) \; = \; \frac{ \left(dh/dr \right) } {r h^2 }+ \frac{ \left( df/dr \right) } {r \, f \, h} \; = \; \left( \frac{1}{ r \, f \, h^2 } \right) \, \frac{d}{dr} \left[ f \, h \right]

So we can see that the product (f * h) can't be constant through the shell. So, we known that the right boundary conditions are that h is constant and f varies. In a shell of finite thickness, f will increase continuously throughout the shell. As we shrink it to zero width, f jumps discontinuously.

Simply put, for a _massless_ shell, we can say that the spatial curvature coefficient, h, is the same inside the shell and outside. This is a consequence of Einstein's equations. While h is constant, f, the time dilation metric coefficient, is NOT constant. This also follows from Einstein's equations.

We can do some more computation and find the exterior metric if we assume that the boundary of the shell is located at r=1. (It turns out we can place it wherever we like).

Then, the metric previously given in (1) is used for r<1, and for r> 1, we use


\frac{dr^2}{1-\frac{3}{7r}}+ r^2 \,d \theta^2 + r^2 sin^2 \theta \, d\phi^2 - \left( 1-\frac{3}{7\,r} \right) dt^2


We can do some more interesting stuff along the lines of comparing the Komar mass to the Schwarzschild mass parameter, but I think it suffices to say that the two agree for the total mass M as judged by the observer in asymptotically flat space-time, but are distrubuted differently in the interior.

[Q-reeus]

At the risk of hijacking the thread (which seems hopelessly confused anyway),...
Feel free to 'hijack' this confused thread, which evidently has kicked back into life, and hopefully end the confusion.
Won't comment on the specifics of your photon gas inside a containing shell model, other than to say that there shell self-gravitation as contribution to it's own stress seems to be, understandably, an entirely absent factor. In my scenario, it is the only contribution. If you go back to #1 hopefully my problem statement is made clear enough, and basically what DaleSpam said in #201 sums it up. Again, the specific model settled on was in #17, but that can be obviously generalized.
I walked away from this thread owing to a general failure to get agreement on being able to apply differential length, radial (dr) vs azimuthal (rdΩ), in coordinate measure, as suggested by the standard Schwarzschild coordinates. As expressed in earlier entries, I accept boundary matching from exterior to interior regions is always possible mathematically. Whether that math is properly based on a physical principle (and I was genuinely shocked when it was claimed shell stresses for an entirely self-gravitating shell would do that trick) is another matter!