Redshifting Through Hole in Earth

[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?

 

[PeterDonis]

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.

Assuming that "grow" means "increase in radius", then yes. (I use radius instead of volume because a BH doesn't have a well-defined "volume". But you can define its horizon radius--more precisely, you can define the surface area of the horizon, and then define the radius $R$ of the horizon such that the horizon's surface area is $4 \pi R^2$.) In terms of mass, all three will of course "grow" (increase in mass) as they collect dust. (And of course this is all assuming that no other processes are involved, which at least for the Sun is not a good assumption, as I discussed in a previous post. [Edit--also see my next post, the Sun isn't even close to being "cold, dead matter" so the analysis I was talking about doesn't apply to the Sun anyway.)

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

Yes.

 

[PeterDonis]

(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.)

Oops, on re-reading this I realized I left something very important out: the analysis I was talking about doesn't apply to the Sun anyway, because the Sun isn't even close to being "cold, dead matter"; the thermonuclear reactions going on inside the Sun make its internal structure very different from that of a ball of cold, dead matter with the same mass--for example, a one-solar-mass white dwarf or neutron star. So I'm not even sure that the Sun's volume as a function of mass would behave the way I was assuming it would in the quote above.

(The Earth isn't technically cold, dead matter either, but it's a lot closer to it than the Sun is, and the analysis I was talking about is at least a fairly good approximation for the Earth.)

 

[ClamShell]

Did I get this right?

For ideal spheres, (R2^3 - R1^3)/RDUST^3 = 1

And for ideal BHs, deltaM/deltaR = 1 = density * (Pi * 4/3) * R^2

Something looks wrong...

 

[PeterDonis]

For ideal spheres, (R2^3 - R1^3)/RDUST^3 = 1

I'm not sure what you mean by "ideal spheres". If you mean "perfectly spherical gravitating bodies in static equilibrium", then no, this is not correct, for two reasons.

First, the actual volume of a spherical gravitating body is not directly proportional to $R^3$; there is an additional factor due to the non-Euclidean geometry of space inside the body, which varies with both the mass and radius (more precisely, you have to express the actual volume of the body as an integral, not a closed-form formula).

Second, as we've discussed already, the change in volume of the body is not equal to the volume of the dust that falls onto it.

for ideal BHs, deltaM/deltaR = 1

This part is fine, assuming that you are using "geometric units" in which the speed of light and Newton's gravitational constant are both equal to 1. In conventional units, we would have $\Delta M / \Delta R = c^2 / 2 G$.

= density * (Pi * 4/3) * R^2

No, this is not correct, but now for a different reason than the above. As I said before, a BH does not have a well-defined "volume", so it doesn't have a well-defined "density" either.

 

[ClamShell]

I don't mean "actual spherical gravitating bodies in static equilibrium",
I mean old fashioned geometrical(Euclidian?, Pythagorean?) spheres.
I'm hoping its:

deltaVsphere/deltaVdust = (R2^3 - R1^3)/(Rdust^3) = 1.

and, for a big gas-filled balloon,

(density * (Pi * 4/3) * (R2^3 - R1^3 )/(R2-R1)/ (c^2)/(2*G) = 1.

Please bare with me, I'm trying to see how GR changes the trivial 3D
geometry case, especially for the BH.

And as customary, I suspect that I don't have it right yet. Just ask
yourself, "How would Pythagoras do it?" And how would he work it out
if someone told him that a "sphere" existed such that deltaM/deltaR
is a constant; "Weight?", "Density?", "I think Archimedes is working
on that."

 

[PeterDonis]

I don't mean "actual spherical gravitating bodies in static equilibrium",
I mean old fashioned geometrical(Euclidian?, Pythagorean?) spheres.

Ah, ok.

deltaVsphere/deltaVdust = (R2^3 - R1^3)/(Rdust^3) = 1.

Yes, for the straight Euclidean case, this is true.

for a big gas-filled balloon,

(density * (Pi * 4/3) * (R2^3 - R1^3 )/(R2-R1)/ (c^2)/(2*G) = 1.

I'm not clear on what you're trying to evaluate here. Are you trying to evaluate how the density of the balloon changes if its radius changes while its mass is held constant? Or how its volume changes if additional gas is added to the balloon, increasing its mass, but holding its density constant? Or something else? (The formula you give doesn't appear to fit either of the possibilities I just mentioned.)

 

[ClamShell]

I should have said: An incompressible liquid-filled balloon analogy of a BH.

deltaM/deltaR = (c^2)/(2 * G) = density * (Pi * 4/3) * (R2^3 - R1^3)/(R2 - R1) (equ. 2)

and if R2 = 2 * R1, then R2^3 = 8 * (R1^3),
or,
deltaM/deltaR = density * (Pi * 4/3) * (8 * R1^3 - R1^3)/R1

= density * (Pi * 4/3) * (7 * R1^3)/R1

= density * (Pi * 4/3) * (7 * R1^2) = (c^2)/(2 * G)
or,
density * (Pi * 4/3) * (7 * R1^2)/(c^2)/(2 * G) = 1 (equ. 2a)

Can you see where I'm going with this? Even if it's going off a cliff.
(equ 2) should give what's constant for a BH in sharp contrast to
what's constant for the Euclidean:

deltaVsphere/deltaVdust = (R2^3 - R1^3)/(Rdust^3) = 1. (equ. 1)
where if R2 = 2* R1 and Rdust = R1,

deltaVsphere/deltaVdust = (7 * R1^3)/(R1^3) = 7 (equ. 1a)

 

[PeterDonis]

An incompressible liquid-filled balloon analogy of a BH.

Which is not a good analogy. The mass of an ordinary object made of an incompressible substance (i.e., one whose density is constant) will vary as the cube of its radius, not linearly with the radius. Your equation is fine for an ordinary incompressible substance, but it doesn't work for a BH at all, and setting your deltaM/deltaR for an ordinary object equal to the deltaM/deltaR we derived earlier for a BH is physically meaningless.

(equ 2) should give what's constant for a BH

No, it gives nothing meaningful at all. A BH doesn't have a well-defined density or a well-defined volume. Patching formulas together won't change that.

 

[ClamShell]

I don't see where I say that the mass will vary linearly with radius;
just looks like I say constant density mass varies as the cube of the
radius.
Consider two water balloons with radius of 2 inches; each holds
Pi * (4/3) * 8 cubic inches of water. Both together total
Pi * (8/3) * 8 cubic inches of water. And if we transfer both to
a third balloon, the relation for the the third balloon would be

Pi * (8/3) * 8 = Pi * (4/3) * R^3
or,
2 * 8 = R^3
or,
R = cube root of 16 = 2.51984209979 inches

That's how water balloons should add.

But two 2 inch radius BHs add to get 4 inches instead of
2.51984209979 inches.

I'm really not after an explanation for this right now, I
would only like to describe this algebraically to distinguish
between the addition of water balloons and the addition of
BHs. Got any ideas?

 

[PeterDonis]

I don't see where I say that the mass will vary linearly with radius

You didn't; I was pointing out that for an actual BH, mass *does* vary linearly with radius. But I'm also pointing out that "radius" for a BH doesn't mean what it means for an ordinary object. See below.

two 2 inch radius BHs add to get 4 inches instead of 2.51984209979 inches.

When you "add" two 2-inch radius water balloons to get a water balloon with a radius of 2.51984209979 inches, that's because the water balloons have a well-defined volume which varies as the cube of the radius, and the radius is just the distance from the center.

When you add two 2-inch "radius" black holes to get a black hole with a 4-inch "radius", you're not "adding" objects with a well-defined volume, and the "radius" does not correspond to "distance from the center", which also isn't well-defined for a BH.

If you really want to compare apples to apples, I think the best you can do would be to look at surface area, since a black hole's horizon does have a well-defined surface area. The comparison would look like this:

* Take two water balloons, each with a surface area of $4 \pi$. Then "add" them together to make one water balloon, without changing the total mass or density of the water. You will get one water balloon with surface area $2^{8/3} \pi \approx 6.3496 \pi$.

* Take two black holes, each with a horizon surface area of $4 \pi$. Then "add" them together to make one black hole. The one black hole will have a horizon surface area of $16 \pi = 2^4 \pi$.

 

[ClamShell]

I'm now wondering if another relativistic object, the observable universe,
with a horizon area of 4*Pi, would end up with a horizon area of 16*Pi
if something happened to cause its mass to double.

 

[PeterDonis]

I'm now wondering if another relativistic object, the observable universe, with a horizon area of 4*Pi

The boundary of our observable universe is not a "horizon" in any meaningful sense; there's nothing preventing objects beyond that boundary from eventually sending light signals to us, they just haven't had time to do so yet. A "horizon" in the usual sense is a causal boundary--objects behind it can't send light signals outside forever.

If the expansion of the universe continues to accelerate (which, according to our present theories, it will), then there is a "cosmological horizon" such that objects beyond it will never be able to send us light signals at all. There are some analogies which are being explored between this kind of horizon and an ordinary black hole horizon. However, none of them have been experimentally tested or are likely to be any time soon.

Also, there is one key respect in which a cosmological horizon is *not* the same as a black hole horizon; it is a two-way boundary, not a one-way boundary. That is, if there is some galaxy which is beyond our cosmological horizon, so it can't send light signals to us, then we are also beyond *its* cosmological horizon, so we can't send light signals to it either. So there's no way for a cosmological horizon to "gain mass" the way a black hole can.

 

[ClamShell]

$\Delta M / \Delta R = c^2 / 2 G$

So there is a dimensionless number,

$2GM / Rc^2$

In heat transfer, there is a dimensionless "Fourier Number"(Fo) that
seems analogous. It is the (diffusive transport rate) divided by
the (storage rate).

or,

$Fo = αt / L^2$

where α is the thermal diffusivity, t is time, and L is length.

First, is there possibly an analogy here?... like a "material diffusivity"?

Second, is there a name for this dimensionless number? Does anyone
call it a "Schwarzschild Number" or "Einstein Number"?

 

[ClamShell]

The boundary of our observable universe is not a "horizon" in any meaningful sense; there's nothing preventing objects beyond that boundary from eventually sending light signals to us, they just haven't had time to do so yet. A "horizon" in the usual sense is a causal boundary--objects behind it can't send light signals outside forever.

If the expansion of the universe continues to accelerate (which, according to our present theories, it will), then there is a "cosmological horizon" such that objects beyond it will never be able to send us light signals at all. There are some analogies which are being explored between this kind of horizon and an ordinary black hole horizon. However, none of them have been experimentally tested or are likely to be any time soon.

Also, there is one key respect in which a cosmological horizon is *not* the same as a black hole horizon; it is a two-way boundary, not a one-way boundary. That is, if there is some galaxy which is beyond our cosmological horizon, so it can't send light signals to us, then we are also beyond *its* cosmological horizon, so we can't send light signals to it either. So there's no way for a cosmological horizon to "gain mass" the way a black hole can.

Maybe Dr. Hawking is on to something and will help to reconcile the
differences between black holes and universes.

"The boundary of our observable universe is not a "horizon" in any
meaningful sense."

Isn't that what Dr. Hawking is saying about black holes too?

 

[PeterDonis]

First, is there possibly an analogy here?... like a "material diffusivity"?

Not really. There are lots of dimensionless numbers in physics, and there's no reason why there should be an analogy between any two of them.

Second, is there a name for this dimensionless number?

Yes, it's often called the "gravitational potential". More precisely, $- GM / r$ is often called the "gravitational potential", sometimes denoted $\phi$, and the metric coefficient $g_{tt}$ is written as $- \left( 1 + 2 \phi / c^2 \right)$. But in "natural" units, where $G = c = 1$, the "potential" $\phi$ is dimensionless.

 

[ClamShell]

Yes, it's often called the "gravitational potential". More precisely, $- GM / r$ is often called the "gravitational potential", sometimes denoted $\phi$, and the metric coefficient $g_{tt}$ is written as $- \left( 1 + 2 \phi / c^2 \right)$. But in "natural" units, where $G = c = 1$, the "potential" $\phi$ is dimensionless.

I guess it will not surprise you that I come from a background
in electronics and heat transfer, and there is much similarity
between these two disciplines. I'm now trying to find
analogous similarities with Relativity theory. Concerning
Special Relativity, I don't know what in the world to do with
that pesky Lorentz factor. But in General Relativity I do see
a similarity between gravity and charge; ie, $G / c^2$
is $k$ (Coulomb's constant) in electronics. In heat
transfer the Fourier Number is sometimes referred to as
dimensionless time. And I think "gravitational Potential"
is referred to as "Volts" in electronics.

Perhaps I should get right in there and take some electives
in classical(modern?) physics, but for now, all I know is
Newtonian physics where the orbit of the planet Mercury
wouldn't dare to "orbit" too. So instead, I have come here
to PF to see what's going on in the hope that previous
"knowledge" can help.

Do you know of any author who has attempted to explain
the similarities? I'm thinkin' it might be titled "Relativity
for Dummies".

 

[PeterDonis]

I'm now trying to find analogous similarities with Relativity theory.

...

Do you know of any author who has attempted to explain the similarities?

Not really, no. I'm also not sure that looking at relativity in terms of "similarities" (I think "heuristic analogies" would be a better term) with other disciplines is the best way to learn relativity. Relativity is highly counterintuitive, and trying to understand it in terms of intuitive analogies with other disciplines will at some point cause you to hit a wall, where the intuitive analogies break down and there is simply no substitute for discarding the analogies and learning the actual underlying physics from scratch. (From what you say about "that pesky Lorentz factor" in SR, it looks like you've already hit such a wall at least once.) Since you're going to have to do that anyway, why not do it sooner rather than later?

 

[ClamShell]

Relativity is highly counterintuitive

"Counterintuitive?"...I was hoping the "mysterious" appearance of magnetism
in electronics due to moving a charge, is as counterintuitive as it gets; I need
to go swallow another aspirin. :=) Thanks Peter, you're an excellent teacher.

 

[PeterDonis]

"Counterintuitive?"...I was hoping the "mysterious" appearance of magnetism in electronics due to moving a charge, is as counterintuitive as it gets

Unfortunately not.

I need to go swallow another aspirin. :=)

You might want to lay in a good stock.

Thanks Peter, you're an excellent teacher.

You're welcome, thanks for the kudos!