Redshifting Through Hole in Earth

[Bill_K]

What counts as a "bad" prediction? What you are calling "Schwarzschild's prediction", which I take to mean the prediction of classical GR with no quantum effects included, is accurate to fourteen decimal places in some cases.

What prediction of Schwarzschild's is accurate to fourteen decimal places??

 

[ClamShell]

There is no known closed-form analytical solution for a non-uniform spherically symmetric mass distribution; you have to integrate the equations numerically. I'm sure that's been done somewhere in the literature, but I don't have a quick reference handy; perhaps someone here does.

I feel that you're just trying to inform me that including radial pressure
as well as radial density make the problem too difficult for such a "trivial"
question. All I ask is a direction to expect if we move density(and pressure)
to the surface of the Earth. I'm thinking that pondering the work should
reveal a direction and ballpark estimate.

I can wait...

 

[ClamShell]

Perhaps there are three cases that could yield
the analytic solutions we are looking for.
Case 1. Redshift for uniform average density of the Earth,
from the center to R2(the surface).
Case 2. Redshift for uniform average density increased to
the density that puts the Earth's true mass between
R1 and R2 and includes a core between 0 and R1
that has the same density as between R1 and R2.
Case 3. Redshift of case 2's core between 0 and R1.

Then subtract case 3 from case 2 and compare that to
case 1. No con-sarned, new-fangled numerical simulation
would be necessary for this simplified analysis, but I'd be
happy with it. And not just happy, fascinated ! If it turned
out that the redshifts are essentially identical for a solid
Earth and a hollow core Earth, I'd buy a round for you at
my local pub...just kiddin', I don't even touch the stuff.

 

[ClamShell]

in fact, all of the quantum treatments of black holes assume that Schwarzschild's analysis gives the correct classical behavior of black holes.

I'm thinking that the Schwarzschild black-hole is more like a boundary
condition, complete with event-horizon and singularity at the center of
this "classical"(I thought classical meant Newtonian) and voracious
degenerate object. Then Hawking transforms this boundary condition
using some quantum chicanery and POOF!, the event-horizon is gone
in the "classical" sense. I'm getting ready for the coop-d-grass
possibilities that there is also no singularity, and gosh-forbid, no
gobbling either when Hawking manages to apply QMs to the inside
of a black-hole in addition to the previously real event-horizon.
IE, the assumption of the real event-horizon introduced into Hawking's
quantum analysis was an error, and "not-in-the-classical sense", a
euphemism and courtesy to the involved theorists.

 

[PeterDonis]

What prediction of Schwarzschild's is accurate to fourteen decimal places??

I was taking "Schwarzschild predictions" to refer to any prediction of GR for Schwarzschild spacetime, not predictions that Schwarzschild himself made. I was thinking of test of the equivalence principle, but I just looked up the Roll-Krotkov-Dicke experiment and realized the accuracy was 1 part in 100 billion, not one part in 100 trillion, so I should have said eleven decimal places, not fourteen.

 

[PeterDonis]

I feel that you're just trying to inform me that including radial pressure as well as radial density make the problem too difficult for such a "trivial" question. All I ask is a direction to expect if we move density(and pressure) to the surface of the Earth.

And I don't think there's a simple intuitive answer to that question. See below.

Case 1. Redshift for uniform average density of the Earth, from the center to R2(the surface).

This case has an exact analytical solution.

Case 2. Redshift for uniform average density increased to the density that puts the Earth's true mass between R1 and R2 and includes a core between 0 and R1 that has the same density as between R1 and R2.

This would just be re-computing the first case for an increased uniform density, so ok so far.

Case 3. Redshift of case 2's core between 0 and R1. Then subtract case 3 from case 2 and compare that to case 1.

However, this won't answer your question, because the presence of the core in case 2 changes the solution from what it would be if there were a spherical shell between R1 and R2, with the same total mass as the Earth, but vacuum inside. The internal stresses of the latter case will be different from the former case, because in the latter case, the shell has to support itself against its own gravity, whereas in the former case, the "shell" from R1 to R2 is being supported by the core from r = 0 to R1.

 

[ClamShell]

I'm for throwing out the pressure with the wash water and
hoping(against all hope) that pressure is not the baby.
If crazy stuff develops, then I can resort to goin' outside
to look for the baby in the weeds. Would circumferential
compression pressure be the same as radial compression
pressure? Dang-it, let's just ignore pressure; the Earth
is a pretty small object...let's look forward to putting
pressure back in when we try to model the Universe.

 

[ClamShell]

Can anybody tell me how to quote, sub-quotes?
I tried to highlight what I want to quote, but it
didn't work.

 

[DrGreg]

Can anybody tell me how to quote, sub-quotes?
I tried to highlight what I want to quote, but it
didn't work.

When you press the Quote button, it quotes the whole message. Then delete the bits you don't want.

 

[dauto]

When you press the Quote button, it quotes the whole message. Then delete the bits you don't want.

That's not correct. I pressed the quote button and ClamShel's subquote is gone

 

[PeterDonis]

I'm thinking that the Schwarzschild black-hole is more like a boundary
condition, complete with event-horizon and singularity at the center of this "classical"(I thought classical meant Newtonian) and voracious degenerate object.

I don't know what you mean by "boundary condition" here.

(Also, "classical" means "non-quantum". GR is classical just as much as Newtonian physics is.)

Then Hawking transforms this boundary condition using some quantum chicanery

I don't know what "boundary condition" means here either, but whatever it means, I don't see any connection between this description and what Hawking's model actually does.

and POOF!, the event-horizon is gone in the "classical" sense.

In Hawking's original quantum model of an evaporating black hole, there is still an event horizon and a singularity; they just don't exist forever into the future. They only exist until the black hole evaporates away completely.

There are other proposed quantum models in which there is no event horizon and no singularity. However, all of them (including Hawking's original model) are speculative at this point, since we have no way of testing any of them.

 

[PeterDonis]

I'm for throwing out the pressure with the wash water

I'm not sure how you propose to do that, since the math says pressure has to be included, and its presence is crucial for having a static equilibrium (see further comments below). You can't just arbitrarily "throw it out". If you want to try to come up with your own theory of gravity that doesn't include pressure but still makes correct predictions, good luck.

Would circumferential compression pressure be the same as radial compression pressure?

"Compression pressure" is redundant, isn't it? Pressure *is* compression. As far as whether it's the same (by which I assume you mean "works the same", not "is of the same magnitude", since we've been talking about the case where it isn't of the same magnitude), yes, pressure "works the same" regardless of which direction it's in.

Dang-it, let's just ignore pressure; the Earth is a pretty small object

As far as the actual magnitude of the pressure inside the Earth, it is in fact very, very small compared to the Earth's energy density (which is the relevant comparison). However, that doesn't mean you can neglect the pressure's contribution to determining the redshift; you can't. The pressure contributes to determining the redshift by determining the static equilibrium of the matter inside the Earth; if there's no pressure there's no static equilibrium at all--the Earth would implode.

 

[A.T.]

Can anybody tell me how to quote, sub-quotes?
I tried to highlight what I want to quote, but it
didn't work.

I think this is disabled to avoid to much nested quotes. It annoys me too, because to show the context of a quote at least 2 nesting levels should be included by default, not just 1 as it is now. I use the multi quote function instead, but it's less convenient, because you have to jump to all the quoted posts and delete much stuff.

 

[ClamShell]

However, this won't answer your question, because the presence of the core in case 2 changes the solution from what it would be if there were a spherical shell between R1 and R2, with the same total mass as the Earth, but vacuum inside. The internal stresses of the latter case will be different from the former case, because in the latter case, the shell has to support itself against its own gravity, whereas in the former case, the "shell" from R1 to R2 is being supported by the core from r = 0 to R1.

Well how about comparing case 1 from r = 0 to R2, to case 2 from r = R1 to R2? And
leave case 2's core intact, so it won't collapse.

 

[PeterDonis]

Well how about comparing case 1 from r = 0 to R2, to case 2 from r = R1 to R2? And leave case 2's core intact, so it won't collapse.

What would this show? The case 2 core is still there, so it still isn't the same as having a spherical shell with vacuum inside.

 

[ClamShell]

What would this show? The case 2 core is still there, so it still isn't the same as having a spherical shell with vacuum inside.

Yah, what WOULD it show? That's my new question...would the redshift in case 2
(from R1 to R2) be less, more, or equal to the redshift of case 1? Maybe we
could even intuitively guess the result.

 

[PeterDonis]

Yah, what WOULD it show? That's my new question...would the redshift in case 2 (from R1 to R2) be less, more, or equal to the redshift of case 1? Maybe we could even intuitively guess the result.

I don't have an intuitive guess off the top of my head. But that wasn't my point anyway: my point was, even if we had an answer to this question, what would it show?

 

[ClamShell]

I don't have an intuitive guess off the top of my head. But that wasn't my point anyway: my point was, even if we had an answer to this question, what would it show?

Ummm, maybe the solution of case 2 as R1 approaches R2 would
mimic the evolution of an Earth-sized black hole. Or maybe it
would just be "fun" in the sense often written by the late Dr. Feynman.

 

[pervect]

Some general comments from skimming the thread

1) The idea that pressure causes gravity is part of GR, like it or not. The least technical explanation I've seen of this is in Baez's "The Meaning of Einstein's Equation", You can find it online at http://math.ucr.edu/home/baez/einstein/ , you can find it in print in the American Journal of Physics, Am. J. Phys. 73, 653 (2005); http://dx.doi.org/10.1119/1.1852541

Another remarkable feature of Einstein's equation is the pressure term: it says that not only energy density but also pressure causes gravitational attraction. This may seem to violate our intuition that pressure makes matter want to expand! Here, however, we are talking about gravitational effects of pressure, which are undetectably small in everyday circumstances.

So we have very direct statements in the published literature that pressure causes gravity.

2) I'm not sure what calculation is being proposed, but I saw a notice of subtraction. You can't, in general "subtract" solutions in GR, the field equations aren't linear. And there isn't anything personal about noting that certain problems in GR requiring numerical integration, it happens a lot. If you're lucky it's just numerical integration of a few integrals.

3) It's not the work done by the pressure that causes the extra gravity - it's the pressure itself. (I don't know if this was asked, but this is a common confusion).

4) Negative pressure exists, we call it tension. But it's not too important for stellar models :-). Negative pressure is important in some thought experiments like the "box of light", and (I think) a few wormhole solutions.

 

[PeterDonis]

Ummm, maybe the solution of case 2 as R1 approaches R2 would mimic the evolution of an Earth-sized black hole.

No, the solutions that describe the formation of black holes are known, and are quite different from what we've been discussing. For the idealized case of a spherically symmetric collapse with zero pressure, the solution was discovered by Oppenheimer and Snyder in 1939, and is briefly described here:

http://grwiki.physics.ncsu.edu/wiki/Oppenheimer-Snyder_Collapse

There is a much more complete description in MTW. For nonzero pressure, or non-spherically-symmetric collapse, there is no known exact analytical solution, but these solutions have been extensively studied numerically.

The key differences between these solutions and the ones we've been discussing are (1) the black hole collapse process is highly non-static, whereas the solutions we've been discussing are static; and (2) R2 in the solutions we've been discussing is much, much larger than the Schwarzschild radius for a black hole with the mass of the Earth, whereas in the black hole collapse solutions, the event horizon does not form until the collapsing matter falls inside the Schwarzschild radius corresponding to its total mass.

 

[ClamShell]

R2 in the solutions we've been discussing is much, much larger than the Schwarzschild radius for a black hole with the mass of the Earth

Of course...a BH the size of the Earth would contain the mass of maybe 25,000
of our sun's. I'm not correcting you, just reminding you that as R1 approaches
R2 for case 2, the density and pressure of this hypothetical "earth sized" object
would go to infinity. In my imagination...R1 even appears to be a bit
"event-horizonish" .

Makes me even wonder if dust falling on the Earth increases the volume of
the Earth, or more imaginatively, the dust does not increase the volume of
the Earth, but instead, changes the radial density and pressure. Collapse
may be how supernova's produce BH's, but maybe dust comglomerating
can evolve a BH too. And maybe case 2 as R1 approaches R2, is closer
to how dust conglomerates.

 

[ClamShell]

So we have very direct statements in the published literature that pressure causes gravity.

Strength or potential, or both?

Do you mean to imply that ALL gravity is caused by pressure?
That would seem to be a fascinating breakthrough.

 

[PeterDonis]

Of course...a BH the size of the Earth would contain the mass of maybe 25,000 of our sun's.

If you mean a black hole with a Schwarzschild radius equal to the Earth's radius, it would have a mass of about 2,200 suns. (Earth radius 6,378 km; Sun's Schwarzschild radius about 2.9 km.)

I'm not correcting you, just reminding you that as R1 approaches R2 for case 2, the density and pressure of this hypothetical "earth sized" object would go to infinity.

Meaning, the density and pressure at the center? Yes, if nothing else intervened, they would go to infinity; but something else does intervene: static equilibrium becomes impossible. And in the case under discussion, that happens *before* the total mass of the object (i.e., counting all the matter from r = 0 to r = R2) because equal to 2,200 suns. Static equilibrium in this case is only possible for an object whose surface radius is greater than 9/8 times the Schwarzschild radius for its mass; Einstein proved this as a theorem in the 1930's. So the equilibrium condition is R2 > 9/8 M_total, or M_total < 8/9 * 2,200 suns, or M_total < 1,955 suns. Such an object would not be a black hole.

In my imagination...R1 even appears to be a bit "event-horizonish".

No, it wouldn't be, because if the object is in static equilibrium, meaning that its total mass meets the above condition, then that condition will also hold at any radius inside the object; i.e., the total mass contained inside any given radius inside the object (including R1) will be less than 8/9 of the mass of a black hole with that Schwarzschild radius.

You appear to be thinking of a black hole as a static object. That's not really a good way to think of a black hole. The spacetime *outside* a black hole is static (at least in the idealized case where the hole is spherically symmetric, and quantum effects are ignored so the hole never evaporates), but spacetime at and inside the event horizon of the hole is not.

Makes me even wonder if dust falling on the Earth increases the volume of the Earth, or more imaginatively, the dust does not increase the volume of the Earth, but instead, changes the radial density and pressure.

Any mass falling into an object in static equilibrium will change the static equilibrium, yes. But that change will most likely involve changing the object's radius, not just the radial distribution of density and pressure. See below.

Collapse may be how supernova's produce BH's, but maybe dust comglomerating can evolve a BH too.

This is possible, but if it happens, it will involve the same kind of collapse that happens in a supernova; the object will no longer be able to support itself in static equilibrium and it will implode. Once it implodes inside the Schwarzschild radius for its mass, an event horizon forms, and that is the criterion for saying that a black hole has formed.

And maybe case 2 as R1 approaches R2, is closer to how dust conglomerates.

Not really, because in any real case of dust conglomeration, the surface radius of the object would change. In the case of R1 approaching R2, you are artificially holding R2 constant, but in a real case, there would be no physical constraint enforcing that, so it most likely would not happen that way.

 

[pervect]

Strength or potential, or both?

Both, though in GR one can't always define a potential.

Do you mean to imply that ALL gravity is caused by pressure?
That would seem to be a fascinating breakthrough.

No, I didn't say that, and neither did Baez. In fact, if you read the quote from the literature, it mentions energy density specifically as causing gravity along with pressure.

I'm a little dissapointed that you had to ask, since if you read what I wrote you'd have known better.

 

[ClamShell]

I'm a little dissapointed that you had to ask, since if you read what I wrote you'd have known better.

Yes, my imagination disappoints me much of the time, too.
But seriously, what if the pressure of spacetime on
neutrons, protons, electrons, etc. is what makes
them acquire gravity? And without this pressure they'd
be as big as houses and have no attraction for one
another at all.

 

[ClamShell]

The idea that pressure causes gravity is part of GR, like it or not

I like it, but am still having trouble visualizing the violation of the statement,

"This may seem to violate our intuition that pressure makes matter want to expand!"

I'm still hung-up thinking that pressure would cause a repulsion, not an attraction.
And if it was a repulsion...that a beam of light might even be blushifted as it
travels from the center of a neutron star to its surface; a preposterous thought
indeed.

 

[PeterDonis]

what if the pressure of spacetime on neutrons, protons, electrons, etc. is what makes them acquire gravity?

There's no such thing as "the pressure of spacetime". Pressure is a property of matter, not spacetime.

 

[PeterDonis]

I'm still hung-up thinking that pressure would cause a repulsion, not an attraction.

It causes both. If an object is in static equilibrium, then the pressure of the inner parts of the object pushes outward on the outer parts. However, the pressure also increases the strength of gravity, so it increases the weight of the outer parts pushing inward on the inner parts.

 

[ClamShell]

It causes both. If an object is in static equilibrium, then the pressure of the inner parts of the object pushes outward on the outer parts. However, the pressure also increases the strength of gravity, so it increases the weight of the outer parts pushing inward on the inner parts.

Does this tell us anything about how much of a small chunk of dust
added to the solid surface of the Earth would increase the volume
of the Earth? Say, given 1 cubic centimeter of dust, how much of
that 1 cubic centimeter contributes to the Earth's volume, and
how much of that 1 cubic centimeter is "absorbed" by the Earth?
And by "absorbed" I mean that part that translates into a higher
average density and higher average pressure with no change in
the volume of the Earth. Newton would probably tell us that the
Earth would increase its volume by 1 cubic centimeter. Whereas
Einstein would probably object to that...?

I'm thinking so-called "collapse" is happening all the time, but on
a much smaller scale than a supernova. I'm also thinking that
a BH supernova remnant, would grow as ejected matter returns
to be gobbled-up by the BH; and grow such that doubling the mass
would double the radius of the BH.

 

[PeterDonis]

Does this tell us anything about how much of a small chunk of dust
added to the solid surface of the Earth would increase the volume of the Earth?

Not by itself, but back in the 1950's John Wheeler and two of his students, Harrison and Wakano, did an analysis of the possible static equilibrium states of what they called "cold, dead matter", which basically means matter that can't undergo any further chemical or nuclear reactions and is at zero temperature, so the only significant factors determining the static equilibrium are gravity and the Pauli exclusion principle, which provides the pressure (because fermions repel each other when they are squeezed together). Their analysis is described in MTW, and also in Kip Thorne's book Black Holes and Time Warps, but unfortunately I haven't found a good discussion of it available online.

The key point here is that their analysis showed that for objects with a total mass smaller than a certain value (which with modern calculations is, IIRC, somewhere between the mass of Jupiter and the mass of the smallest dwarf star we've seen), adding mass to the object increases its outer radius; but once the total mass goes above that certain value, adding mass to the object *decreases* its outer radius. In other words, that certain value is the value at which pressure increasing the strength of gravity begins to outweigh pressure increasing the outward push resisting gravity. So, since the Earth's mass is much smaller than that threshold value, adding matter to the Earth would increase its radius, and hence its volume. But I don't have an easy quantitative answer to the question of how much.

I'm thinking so-called "collapse" is happening all the time, but on
a much smaller scale than a supernova.

What do you mean by "collapse" here? I don't understand.

I'm also thinking that a BH supernova remnant, would grow as ejected matter returns to be gobbled-up by the BH

Possibly; it depends on how fast the matter is ejected during the supernova explosion. AFAIK numerical simulations show that virtually all the ejected matter is moving outward fast enough to escape the gravity of the BH that remains behind.

and grow such that doubling the mass would double the radius of the BH.

This is true of any BH that accretes matter; the hole's Schwarzschild radius is a linear function of its mass.

 

[ClamShell]

So there might be a mass somewhere between Jupiter and
a smallest brown dwarf, that absorbs dust without changing
its radius? Interesting. Maybe it's somewhere else for dust
instead of cold, dark stuff.

Ummm, collapse means that part of the dust's volume that
gets lost from the equation that trivially predicts a one cubic
centimeter increase in volume for every cubic centimeter of
dust collected. Guess the dust density would to need to be
the same as where you put it...if dust lands on previous
dust, that wouldn't seem to matter. Newton would probably
predict that the trivial result always happened. That the analytical
solution even suggests that there is an inflection means that the
actual increase in volume is always less than the trivial case.

Maybe it's happening on the surface of our Sun(is its name Sol?),
as I sit here and type.

Maybe it's happening on the surface of our Earth, because the
surface of the Earth is pretty "spongy".

 

[PeterDonis]

So there might be a mass somewhere between Jupiter and a smallest brown dwarf, that absorbs dust without changing its radius?

Technically, the critical value is just a single value; a mass exactly at that value would decrease its radius slightly if it absorbed a small amount of dust, and a mass just short of that value would increase its radius slightly if it absorbed a small amount of dust. Practically speaking, yes, there will be a range of masses around the critical value for which the radius doesn't change appreciably (how wide the range is will depend on how accurately we can measure the radius).

Maybe it's somewhere else for dust instead of cold, dark stuff.

In standard terminology, "dust" just means "ordinary matter with zero pressure"; and such matter can't exist in static equilibrium. The analysis I mentioned only applies to objects in static equilibrium.

Ummm, collapse means that part of the dust's volume that gets lost from the equation that trivially predicts a one cubic centimeter increase in volume for every cubic centimeter of dust collected.

So basically, by "collapse is taking place all the time" you mean "real massive objects in the real universe, like the Sun, always have some kind of matter falling in on them, and the resulting increase in volume, if there is any, is always less than the volume of the matter that falls in". Yes, that's true; for an object smaller than the critical mass (like the Earth), the volume will increase, but by an amount smaller than the volume of the infalling matter, although it will be hard to detect the difference unless the amount of matter falling in is significant in comparison with the mass of the object; and for an object larger than the critical mass (like the Sun), the volume will decrease, although again it will be hard to detect the difference unless the amount of matter falling in is significant in comparison with the mass of the object. In practical terms, although there is certainly matter falling into the Earth and the Sun, the amount is much too small for us to detect any effect on the volume of those objects.

(With the Sun, there are two additional factors to consider. First, the Sun is continuously *ejecting* matter, in the form of the solar wind, coronal mass ejections, etc. On balance, the amount of matter being ejected probably exceeds the amount of matter falling into the Sun, which means the Sun's volume would actually be very slowly increasing as its mass decreases due to this effect--though again the amount is much too small for us to measure. Also, since the Sun has nuclear reactions taking place in its core, converting mass to energy, and the energy eventually gets radiated away as visible light from the surface, the Sun's mass is slowly decreasing due to this as well.)

 

[pervect]

Actually, on thinking this over, I'm no longer sure it's true. The redshift factor depends on the pressure as well as the density, and changing the density distribution will change the pressure distribution as well. It's not clear to me from looking at the equations (which are in this post on my PF blog) that the changes will always work out to keep the redshift factor at the center the same.

I believe we had a thread on a related matter, in the thread "How does GR handle metric transition for a spherical mass shell?". The perma-link doesn't seem to be working properly, I have some remarks in post #202.

A sphere of highly pressurized gas, or liquid, or even a box of light can exist , but it must be enclosed in some pressure vessel, a sphere in tension, to keep the contents from escaping. Since pressure (and tension) cause gravity, one has to analyze the equilibrium system to get the correct equilibrium gravitational field, so one needs to be concerned with the details of the shell.

There are various ways to analyze it, I chose to have an enclosing sphere of exotic matter, whose density $\rho$ was zero, but whose tension was not. Tension is just negative pressure.

To recap the highlights:

Using a general spherically symmetric metric:

$$s = -f(r)\,dt^2 + h(r)\,dr^2 + r^2 \left(d \theta ^2 + \sin^2 \, \theta \: d\phi^2 \right)$$

f is the square of the "time dilation" factor.
h is another metric coefficient (closely related to spatial curvature as I recall).

I used eq 6.23 and 6.24 from Wald's "General Relativity", I recall you (Peter) had your own, the ones you mention in your blog, I haven't recast them in your form.

Setting ρ to zero and using eq 6.2.3 from Wald 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.

Adding together 6.2.3 and 6.2.4 from Wald, we get

$$8 \pi \left(\rho + P \right) \; = \; \frac{ \left(dh/dr \right) } {r h^2 }+ \frac{ \left( df/dr \right) } {r \, f \, h} \; = \; \left( \frac{1}{ r \, f \, h^2 } \right) \, \frac{d}{dr} \left[ f \, h \right]$$

With $\rho=0$ and P nonzero, we must have f*h vary through the exotic matter shell.

Re-reading this, with P negative, f*h must be decreasing through the shell, which means f is decreasing. (For a thin shell, in the limit f jumps discontinuously). The decrease in f is consistent with the expected negative "Komar mass" of the exotic matter shell.

I haven't analyzed an "ordinary matter shell", but I would expect that f wouldn't decrease.

In any event, the time dilation factor won't be the same inside and outside the shell. In the limit of the shell having zero mass (which requires exotic matter), the difference in the time dilation between the inside and outside can be entirely attributed to the gravitational effects of the tension in the shell.

 

[PeterDonis]

I haven't analyzed an "ordinary matter shell", but I would expect that f wouldn't decrease.

It obviously can't, because an ordinary matter shell must have $\rho + P > 0$, so the LHS of your equation must be positive, hence $f h$ must increase with increasing $r$, hence $f$ must increase from inside to outside the shell.

 

[ClamShell]

So the Earth would grow as it collects dust.
And the Sun would shrink as it collects dust.
And a BH would grow as it collects dust.

Have I got that right?

And the radius of the 4 million solar mass SMBH
at the center of the Milky Way galaxy would be
about 11 million kilometers?