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

[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 modeled 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-r_s/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-r_s/r, whereas that should have been J = (1-r_s/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 gravitational 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 modeled 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 modeled 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-r_s/r, whereas that should have been J = (1-r_s/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 gravitational 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 ρ_t = ρ_m + 2p, where ρ_t, ρ_m are the total and matter only contributions. So dm = ρ_tdx³. 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 r_s = 2M, so for that p(0) expression, one gets as you said p(0) = ρr_s/r, which is exceedingly small. The fractional p modification to m(r) is then of the order 2r_s/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 spatial 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 modeled 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 spatial 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–Oppenheimer–Volkoff_equation

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–Oppenheimer–Volkoff_equation

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.

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. [STRIKE]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'?[/STRIKE]
[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.