We were wrong about how Super Massive Blackhole get bigger!

[Ziggurat]

At least in classical mechanics does it make sense to put a shell around the entire universe?

I'm not putting a shell around the entire universe. I'm cutting up the universe into shells.

 

[Mike Helland]

I'm not putting a shell around the entire universe.

I didn't say you were.

 

[Ziggurat]

I didn't say you were.

Then what the hell is your point?

 

[Mike Helland]

Then what the hell is your point?

Stemmed from Hans:

"Now go far enough back in time when the same mass was packed in a tighter radius, and it definitely should have just gone black hole."

So, if the whole universe was a black hole, but everything was inside it... does that actually mean anything to us?

 

[Ziggurat]

Stemmed from Hans:

"Now go far enough back in time when the same mass was packed in a tighter radius, and it definitely should have just gone black hole."

So, if the whole universe was a black hole, but everything was inside it... does that actually mean anything to us?

The point is that the naive view would predict even part of the universe had enough mass in a small enough radius to form a Schwarzschild black hole.

But the naive view is wrong, for the reasons I detailed.

 

[Mike Helland]

The point is that the naive view would predict even part of the universe had enough mass in a small enough radius to form a Schwarzschild black hole.

But the naive view is wrong, for the reasons I detailed.

Ok.

So about the SMBH's themselves.

If they don't develop through mergers, and develop through accretion, the universe isn't old enough for them to develop into their size.

Do I have that right?

 

[HansMustermann]

@Ziggurat
I'll refer you to my message #18.

Going Newtonian for a moment, like you do, it doesn't say that the shell theorem doesn't apply. It's a theorem. Even in an infinite continuous universe, if I carve an arbitrary sphere out of it, yes, the component of the vector caused by that sphere will be the same as if the rest of the universe didn't exist. It doesn't require anything like assuming the rest of the universe to be a void. It just requires me to be aware that it's just one component, not the final total vector. And it just happens that it cancels out with the attraction of the whole universe minus that sphere.

However that effectively flattens the universe instead of making it too curved for the shell theorem to apply.

 

[Ziggurat]

@Ziggurat
I'll refer you to my message #18.

Your post #18 was trying to get around discarding Birkhoff. But we have to. Birkhoff explicitly doesn't apply, because we're explicitly not talking about a vacuum solution.

One of the ways that we can see this conflict most obviously is in the fact that the Birkhoff solution must be static, but our universe is not.

Going Newtonian for a moment, like you do, it doesn't say that the shell theorem doesn't apply.

Sure, the shell theorem still applies to any particular shell. That's not where the problem is. The problem is in how you add them all up.

It's a theorem. Even in an infinite continuous universe, if I carve an arbitrary sphere out of it, yes, the component of the vector caused by that sphere will be the same as if the rest of the universe didn't exist. It doesn't require anything like assuming the rest of the universe to be a void. It just requires me to be aware that it's just one component, not the final total vector.

Yes, I know all that. I was counting on all of that.

And it just happens that it cancels out with the attraction of the whole universe minus that sphere.

Does it? How do you know? How are you calculating the attraction of the entire universe around you except this one sphere?

I described a way of calculating the attraction of the whole universe minus that sphere which gives an answer of zero, ie, the entire universe minus that sphere doesn't have a net attraction at all so that the attraction of the sphere is NOT cancelled.

 

[HansMustermann]

Easy. I draw another sphere centered on the object, and which includes the first sphere. It's trivial to integrate the attraction of that minus the first sphere, although I don't even actually have to, if I remember the shell theorem. Because it's literally just that: zero minus the attraction of the first sphere. The rest integrates further by radius just like before. Quite trivially in fact (letting aside for a moment the fact that the shell theorem already says I don't have to), because it's the integral of zero by radius. Each dr shell exerts a zero vector on the object, once we're past the radius that includes the first sphere. And technically I don't even need to integrate it to infinity (much as it would make no difference for the literal integral of zero), since realistically speaking, once you're past the radius of the visible universe at that point, nothing can affect you any more without violating causality.

 

[Mike Helland]

Easy. I draw another sphere centered on the object, and which includes the first sphere.

But the sphere can only include symmetrical objects where the center of gravity is equal for all points on its surface.

 

[HansMustermann]

But the sphere can only include symmetrical objects where the center of gravity is equal for all points on its surface.

Only if the result were to be zero. But the result of this one is the opposite vector of what the first one was producing.

 

[Mike Helland]

Only if the result were to be zero. But the result of this one is the opposite vector of what the first one was producing.

Are you thinking of the electric field where the sphere contains a net zero charge?

If the target charge moves in relation to the sphere, that "shortcut" may no longer apply.

 

[Ziggurat]

Easy. I draw another sphere centered on the object, and which includes the first sphere. It's trivial to integrate the attraction of that minus the first sphere, although I don't even actually have to, if I remember the shell theorem. Because it's literally just that: zero minus the attraction of the first sphere. The rest integrates further by radius just like before. Quite trivially in fact (letting aside for a moment the fact that the shell theorem already says I don't have to), because it's the integral of zero by radius. Each dr shell exerts a zero vector on the object, once we're past the radius that includes the first sphere. And technically I don't even need to integrate it to infinity (much as it would make no difference for the literal integral of zero), since realistically speaking, once you're past the radius of the visible universe at that point, nothing can affect you any more without violating causality.

I don’t think you actually understand the scenario I described. And causality isn’t at play here. Newtonian gravity propagates infinitely fast.

 

[HansMustermann]

Let me explain. First we do this:

The original sphere is #1, the sphere I drew around the whole thing is #2, and the object it's exerting any attraction on is in the centre of that coordinate system. Sphere 2 is centered on it.

NOW... the attraction of sphere 1 is some value g. The attraction of the whole sphere 2 (including sphere 1) is zero. Because, quite trivially, if we have spherical symmetry, every point in sphere 2 that can exert a force on the object, has a mirror point that exerts the exact opposite vector.

And I could really write QED at this point and leave it at that. Whatever net force sphere 1 exerts, it is a part of sphere 2, and the total force from sphere 2 is zero. So the attraction of the difference between sphere 2 and sphere 1 (i.e., if we hollowed out sphere 1 inside sphere 2) has got to be the opposite vector of whatever sphere 1 was exerting.

But if anyone needs any further explanation, let's use a mirror of sphere 1, which I'll call number 3.

Whatever attraction is left after we take out sphere 1 is the attraction from sphere 3, which is the polar opposite of of sphere 1, plus whatever attraction is exerted by sphere 2 minus both spheres 1 and 3. But that part is symmetrical with respect of the origin again. Every infinitesimal element I want to take there, has a mirror element, so the attraction cancels out.

So basically I don't have to think that the shell theorem somehow doesn't apply. Of course it applies. It's a theorem. It's just that the rest exerts a force that counters the one from sphere 1.

Now, let's deal with whatever is outside sphere 2. Let's say sphere 2 has a radius R. To find the total contribution of the medium outside that, I have to integrate by r the contribution of concentric spheres from R to infinity. But each hollow shell from r to r+dr in there is exerting a force equal to zero, because it's symmetrical around our object. So we're literally taking the integral of zero from R to infinity, which of course it's zero.

 

[Ziggurat]

Let me explain.

I understand your scenario. But you don't understand mine.

Let me try with some more specifications. We are concerned with the gravitational field at the position (1,0,0). We position a sphere of radius 1 at the origin (0,0,0). Our point of interest is at the outer edge of this sphere. The gravitational field of this sphere at (1,0,0) points in the negative x direction, yes?

OK, now we have to add up the gravitational effect of everything else. Now consider a spherical shell around our original sphere, of inner radius 1 and outer radius 2, still centered at the origin. Our point of interest is on the inside of this sphere. Shell theorem applies, the net field at (1,0,0) from this surrounding shell is zero, yes?

So we keep going. Now we make a shell of inner radius 2 and outer radius 3, still centered at (0,0,0). No field at (1,0,0). And so on, and so on. Of course we need an infinite sum of such shells, but that's OK, because the contribution of each shell is always zero. And so I've proven that the gravitational field at (1,0,0) is pointing in the -x direction.

But I could redo the same proof starting with a sphere and concentric shells centered at (2,0,0), and I would prove that the gravitational field at (1,0,0) is pointing in the +x direction. In fact, I can pick spheres to come up with any magnitude answer and pointing in any direction I want. This is a paradox. You cannot resolve it by appealing to the shell theorem, since that's part of how we got in this mess. You can't resolve it by saying the shell theorem is invalid, because it's mathematically rigorous. And you can't resolve it with your calculation method, because that's just one of many equally valid ways to calculate the answer. There is an explanation for why this paradox arises, but you have to go beyond the shell theorem to understand it.

ETA: and spheres and shells aren't even the only way to do this. You can also use infinite slabs with finite thickness (which is easy to solve analytically and produce finite fields), and the same thing applies: you can arrange the problem to get whatever answer you want.

 

[HansMustermann]

I suppose it's one of the things that you get when you plug infinity into most stuff. Seems rather trivially solvable, though, by remembering (again) that we have such a thing as the radius of the observable universe at the time as a limit. Which is centered on the object. It may not be strictly at 17'th century level, but since we know it exists, I have no problem with using it.

 

[Ziggurat]

I suppose it's one of the things that you get when you plug infinity into most stuff. Seems rather trivially solvable, though, by remembering (again) that we have such a thing as the radius of the observable universe at the time as a limit. Which is centered on the object. It may not be strictly at 17'th century level, but since we know it exists, I have no problem with using it.

Sure. Trying to do cosmology with Newtonian physics, where gravity travels instantly and so there is no limit to observation distance, is bound to cause problems. This is just one of the more abstract ones.

 

[HansMustermann]

Sure. Trying to do cosmology with Newtonian physics, where gravity travels instantly and so there is no limit to observation distance, is bound to cause problems. This is just one of the more abstract ones.

Guess there's a reason we had to come up with GR, huh?

 

[Roger Ramjets]

Ok.

So about the SMBH's themselves.

If they don't develop through mergers, and develop through accretion, the universe isn't old enough for them to develop into their size.

Do I have that right?

Apparently yes. The logical conclusion is that the Universe is older than we thought. Which conflicts with the Big Bang theory. Which is based on observation of the CMB. So we have two observations that appear to be in conflict.

But what if the CMB was not caused by a Big Bang? What if it is actually the remnants of events occurring over trillions of years that individually don't have much effect, but over a vast time frame built up to make the CMB we see today? Or what if it is actually caused some other effect that we are not aware of?

I think the Big Bang theory is on shaky ground. Sure the math works out if you make certain assumptions, but there are 'anomalies' it doesn't account for - and now we have another one.

The basic idea of a black hole is simple. If its gravitational field is strong enough to prevent light escaping then anything falling into it cannot get out. But when you look at the mechanics it gets a lot more complicated. Things don't just 'fall into' a black hole and collect in the middle, they spiral around it like water going down a drain hole, ripping matter apart and releasing vast amounts of energy as it accelerates close to the speed of light. We think of a space ship sedately wandering over the Schwarzschild radius like a ship crossing the equator (an invisible line that separates the Northern and Southern hemispheres) when it is actually more like a satellite burning up as it falls into the Earth's atmosphere at 17,000mph.

And when the encircling matter finally enters the black hole, what will it find? Imagine a black hole large enough that tidal forces are small and matter can exist in a 'normal' state. What would the event horizon look like to an observer inside? With all that stuff spinning around it you wouldn't be able to see the rest of the universe beyond, and any light beams you sent out to it would simply bend around back into the black hole. So the event horizon wouldn't just be invisible or opaque, it would be the edge of your 'universe' - an edge that you wouldn't even know was there. But matter is falling into it on a regular basis, enlarging your 'universe' and leaving behind a signature as it radiates energy. What would that energy signature look like?

Now imagine a black hole so large that it sucks in all the matter in the Universe, including all other blacks holes. Now the Universe is a black hole, but it has the entire Universe in it so it's basically the same as the current universe. IOW, the Universe is its own black hole. But it still has an event horizon, one we cannot see but which produces an energy signature - the CMB. Perhaps our calculations purporting to show the 'age of the Universe (ie. timing of the Big Bang) are actually telling us something quite different.

This is all idle speculation and no doubt someone will show me the math that proves it cannot be. You can prove anything with math if you make certain assumptions. But if the assumptions are invalid then the conclusions may be too. It wouldn't be the first time that a niggling anomaly blew the whole theory to bits.

 

[HansMustermann]

Yes, there is one hypothesis that we are inside the mother of all black holes.