Redshifting Through Hole in Earth

[A.T.]

But once you go below the surface, the potential depends on the details of the density and pressure inside; change those details and you change the potential.

What about the classical Newtonian potential? Is it the same at the center of a uniform sphere vs. spherical shell of same mass & outer size?

 

[WannabeNewton]

What about the classical Newtonian potential? Is it the same at the center of a uniform sphere vs. spherical shell of same mass & outer size?

No. I already answered this above.

 

[ClamShell]

No. I already answered this above.

Don't be such a "NO" person. Show me how the mathematics gets
modified due to non-uniform radial distributions of mass.

 

[PeterDonis]

I'm thinking that what you have taught me is that, as more and more atoms and molecules are moved to the surface, the redshift of a photon traveling from the center to the surface would lessen as more and more matter was moved.

No, what we've taught you (or at least tried to) is that the redshift doesn't just depend on "where the matter is", i.e., the density--it also depends on the stresses within the matter (such as pressure).

Secondly, am I being too skeptical in the judging of theories that have inconsistant assumptions and/or predictions?

A theory can be perfectly consistent and still make bad predictions. (Also see below.)

I'm thinking that bad predictions mean bad theories

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.

Hawking has falsified Schwarzschild's prediction.

First of all, Hawking's prediction hasn't been experimentally verified at all, let alone to fourteen decimal places. Most physicists believe that it *would* be verified if we could run the appropriate experiments, but we won't be able to do that any time soon, and anyway believing it would be verified is not the same as it actually being verified. As far as experimental verification goes, Schwarzschild's prediction is the best we have.

 

[PeterDonis]

Show me how the mathematics gets modified due to non-uniform radial distributions of mass.

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.

 

[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.