Redshifting Through Hole in Earth

[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!