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Tags astrophysics , Becky Smethurst , black holes

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Old 15th August 2021, 02:07 PM   #1
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We were wrong about how Super Massive Blackhole get bigger!

Dr. Rebecca Smethurst, aka Dr. Becky on YouTube, has released her latest paper on her current research into Supermassive Black Holes and how they get bigger. Previously it was thought that most of their mass comes from the combining of the Black Holes when galaxies collide, but her team has discovered that this is incorrect and that the Black Holes in galaxies that have never collided are of similar size to those that have, meaning there was another mechanism. This paper discusses what they found and opens up some new theories about galaxy formation and evolution. If you are into astrophysics, have a look.

https://arxiv.org/pdf/2108.05361.pdf
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Old 15th August 2021, 08:12 PM   #2
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I always assumed that they got bigger by swallowing up nearby stars, other black holes, neutron stars, and any other matter in the vicinity.

I'll go watch the video. I also wonder whether they formed in the Big Bang, or later. Did they begin as stellar-mass black holes that grew by swallowing up more matter, or did they somehow start out as supermassive black holes and only grow a little bit after that?
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Old 16th August 2021, 03:11 AM   #3
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Originally Posted by Puppycow View Post
I always assumed that they got bigger by swallowing up nearby stars, other black holes, neutron stars, and any other matter in the vicinity.

I'll go watch the video. I also wonder whether they formed in the Big Bang, or later. Did they begin as stellar-mass black holes that grew by swallowing up more matter, or did they somehow start out as supermassive black holes and only grow a little bit after that?
I'm not sure that I can summarize the paper in a way that actually does it justice, but it seems that galaxies structures such as spiral arms and bars are systems that channel material and gas down into the centre of a galaxy, and it is this material that the SMBH feeds on, while also throwing a lot of it out again.

Dr. Becky talks about the paper and its finding in this video at 16:12...

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More on Black holes with Dr Becky...
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Old 16th August 2021, 04:54 AM   #4
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If she is right then here is my prediction. If you look back in time the supermassive black holes would be much smaller than they are now.
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Old 16th August 2021, 05:41 AM   #5
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Originally Posted by rjh01 View Post
If she is right then here is my prediction. If you look back in time the supermassive black holes would be much smaller than they are now.
I predict they will get bigger in the future.
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Old 16th August 2021, 08:55 AM   #6
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Originally Posted by Puppycow View Post
I always assumed that they got bigger by swallowing up nearby stars, other black holes, neutron stars, and any other matter in the vicinity.

I'll go watch the video. I also wonder whether they formed in the Big Bang, or later. Did they begin as stellar-mass black holes that grew by swallowing up more matter, or did they somehow start out as supermassive black holes and only grow a little bit after that?
There is a limit to how fast a black hole can grow that way. The process of ďeatingĒ a star causes the a massive release of energy near the black hole as the start is torn apart, and this pushed nearby matter away from the black hole slowing down itís growth. Apparently 13 billion years isnít nearly long enough for supermassive black holes to get as big as they are just by eating stars and dust clouds.
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Old 16th August 2021, 04:50 PM   #7
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Originally Posted by lomiller View Post
There is a limit to how fast a black hole can grow that way. The process of ďeatingĒ a star causes the a massive release of energy near the black hole as the start is torn apart, and this pushed nearby matter away from the black hole slowing down itís growth. Apparently 13 billion years isnít nearly long enough for supermassive black holes to get as big as they are just by eating stars and dust clouds.
They need to be eating about an average of a sun's worth of material each year.
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Old 12th September 2021, 04:25 AM   #8
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Let us assume when all the matter was in one place and not in far flung places, it was simpler for black holes to form.
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Old 12th September 2021, 07:45 AM   #9
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Originally Posted by Samson View Post
Let us assume when all the matter was in one place and not in far flung places, it was simpler for black holes to form.
Well, in fact simpler than to explain why the heck isn't the whole universe a black hole. There's still some debate as to whether basically the Schwarzschild radius for the mass of the visible universe is actually the radius of the visible universe. 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.
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Old 12th September 2021, 09:45 AM   #10
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Mod WarningOff topic posts fed to our very own supermassive blackhole - AAH. It is estimated that within the next decade 99% of all posts will have disappeared into the black hole.
Responding to this mod box in thread will be off topic Posted By:Darat
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Old 12th September 2021, 01:00 PM   #11
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Originally Posted by lomiller View Post
There is a limit to how fast a black hole can grow that way. The process of ďeatingĒ a star causes the a massive release of energy near the black hole as the start is torn apart, and this pushed nearby matter away from the black hole slowing down itís growth. Apparently 13 billion years isnít nearly long enough for supermassive black holes to get as big as they are just by eating stars and dust clouds.
How long is long enough?
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Old 12th September 2021, 01:32 PM   #12
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Originally Posted by HansMustermann View Post
Well, in fact simpler than to explain why the heck isn't the whole universe a black hole. There's still some debate as to whether basically the Schwarzschild radius for the mass of the visible universe is actually the radius of the visible universe. 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.
The larger a black hole is, the less dense it is, if you calculate the density by dividing the mass by the volume inside the Schwarzchild radius. So at any given density, you can calculate a radius that would give you a black hole. If the universe has some particular density at some particular time, you can calculate a corresponding Schwarzschild radius, and see if that radius is smaller than the universe is. And if you take this approach, it may indeed look like the universe should have turned into a black hole.

But this approach is wrong.

The thing about the Schwarzschild radius is that it's part of the Schwarzschild solution, and the Schwarzschild solution is for a spherically symmetric mass distribution in an asymptotically flat universe. And this solution works pretty well for black holes we know of because at the scale of individual stars and even galaxies, the universe is reasonably close to flat.

But the universe is not actually asymptotically flat, especially the early universe. The fact that the universe is expanding makes the Schwarzschild solution completely invalid at the kind of scales you're describing. So you cannot use the Schwarzschild solution when evaluating the universe as a whole. It doesn't work.
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Old 12th September 2021, 07:33 PM   #13
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Well, the thing is, if you go back in time far enough, the whole mass was concentrated in something the size of a galaxy or indeed of a (really big) star. So if calculating the Schwarzschild radius works at that scale, we still have a problem: the mass in the universe at that point is in fact creating a Schwarzschild radius larger than the actual radius of the universe. Again, at that point in time.

"But that radius is very big, compared to the size in which the mass is" doesn't really work, because it doesn't work for the singularity in a black hole.

It seems to me like Birkhoff pretty explicitly says that the solution of the gravity of that sphere WILL be asymptotically flat, and that whatever growth or shrinking happens in the size of that sphere, is not going to matter. Well, at least as long as it's reasonably spherical, non-rotating, etc.

So SOMETHING must have still pushed hard outwards for it to grow instead of just implode right back.
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Old 12th September 2021, 07:48 PM   #14
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Originally Posted by HansMustermann View Post
Well, the thing is, if you go back in time far enough, the whole mass was concentrated in something the size of a galaxy or indeed of a (really big) star. So if calculating the Schwarzschild radius works at that scale, we still have a problem: the mass in the universe at that point is in fact creating a Schwarzschild radius larger than the actual radius of the universe. Again, at that point in time.
No, that's not how it works. Right now, space is close enough to asymptotically flat for the approximation to work for galaxies. But in the early universe, when huge amounts of matter were within the volume of a galaxy, the curvature of the universe was also larger. On the scale of a galaxy-sized object, the universe was not asymptotically flat. The scale at which the flatness approximation breaks down isn't fixed. It's smaller the farther back we go. So yes, everything was denser in the early universe, but everything was also more curved.

Quote:
It seems to me like Birkhoff pretty explicitly says that the solution of the gravity of that sphere WILL be asymptotically flat, and that whatever growth or shrinking happens in the size of that sphere, is not going to matter. Well, at least as long as it's reasonably spherical, non-rotating, etc.
No. Asymptotic flatness isn't a result of the solution, it's a required input. Without that input, you cannot get that solution.
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Old 12th September 2021, 07:54 PM   #15
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Originally Posted by Ziggurat View Post
No, that's not how it works. Right now, space is close enough to asymptotically flat for the approximation to work for galaxies. But in the early universe, when huge amounts of matter were within the volume of a galaxy, the curvature of the universe was also larger. On the scale of a galaxy-sized object, the universe was not asymptotically flat. The scale at which the flatness approximation breaks down isn't fixed. It's smaller the farther back we go. So yes, everything was denser in the early universe, but everything was also more curved.
Isn't that how it works for a black hole, though? If it's curved enough, you get an event horizon. I'm not sure how something can be too curved to be a black hole, if we're talking mass alone. (But of course it can be if some other effect curves it the other way around.)

Originally Posted by Ziggurat View Post
No. Asymptotic flatness isn't a result of the solution, it's a required input. Without that input, you cannot get that solution.
It may be my misunderstanding of Birkhoff, but it seems to me like nowhere does it require asymptotic anything inside the sphere. The whole point is that the space-time distortion of any spherical and non-rotating distribution of mass can be calculated as being a point mass in the centre. It doesn't matter if the distribution by radius is uniform, asymptotic from the centre, all below surface of the sphere, or whatever.

Also, exactly what does Birkhoff say, if asymptotic flatness
1. outside the sphere,
2. of the effect of the mass inside the sphere,
apparently isn't it?
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Old 12th September 2021, 08:35 PM   #16
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Originally Posted by HansMustermann View Post
Isn't that how it works for a black hole, though? If it's curved enough, you get an event horizon. I'm not sure how something can be too curved to be a black hole, if we're talking mass alone.
The Schwarzschild solution is a solution for a black hole sitting in a flat spacetime. The spacetime far away from the black hole must be flat, the spacetime near the black hole is obviously not.

If the spacetime far away from the black hole is NOT flat, then the Schwarzschild solution is not valid.

Quote:
It may be my misunderstanding of Birkhoff, but it seems to me like nowhere does it require asymptotic anything inside the sphere.
I'm not talking about flatness inside the sphere. Asymptotic flatness means that spacetime approaches perfect flatness as you go out infinitely far away. That's the whole point. The exterior solution is what must be asymptotically flat. But the universe isn't.

I think what may be going on is that you're misunderstanding the Wikipedia entry, or some other source that's copied from it. So let's take a look at it for a moment.
In general relativity, Birkhoff's theorem states that any spherically symmetric solution of the vacuum field equations must be static and asymptotically flat.
This makes it sound like asymptotic flatness is an output of the theory, not an input. But that's not quite true. And the reason it's not really true is hiding in this very sentence, though it's easy to miss: the vacuum field equations. What does vacuum mean here? It means no mass.

So what we're looking at with this theorem is a spherically symmetric solution in which there's no mass in our region of interest. There can be mass somewhere, but (1) that mass must be spherically symmetric so our solution is as well, and (2) we're only looking at the solution outside of that mass, and there is no mass in this exterior region.

But that lack of mass anywhere else is what leads to asymptotic flatness. By saying that there's basically no mass anywhere except in this specific region, we've basically made asymptotic flatness an input, just under a different name.

And when we're talking about cosmology and what's happening at very large scales, you're obviously NOT talking about the universe having no mass outside of a particular region. There's mass everywhere. You can't get away from it. So even ignoring asymptotic flatness as such and using the phrasing as presented by Wikipedia, Birkoff's theorem is invalid from the start for examining cosmology because we can't use a vacuum solution when the universe isn't a vacuum.
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Old 12th September 2021, 08:55 PM   #17
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Still not sure how that's supposed to work. Whether it's GR or even newtonian mechanics, sure, mass outside can add its own distortion. Like, we can calculate the gravity of the Sun at 150 million km to a value, and yes, it's calculated as if there was a void outside the Sun, but if you're on Earth chances are it's not the dominant component. Doesn't prevent us from calculating the gravity of the Sun at that distance that way anyway. In fact, at least for classic gravity, it's kind of the whole point of the shell theorem.
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Old 12th September 2021, 08:58 PM   #18
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That said, it seems to me like an easier case could be made for it just being an effect of the universe being actually infinite or a closed manifold, rather than trying to make Birkhoff not apply, innit? But that goes through making the space flatter in every point (at least as far as mass contribution to that curvature goes), rather than too curved.
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Old 12th September 2021, 09:25 PM   #19
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Originally Posted by HansMustermann View Post
Still not sure how that's supposed to work. Whether it's GR or even newtonian mechanics, sure, mass outside can add its own distortion. Like, we can calculate the gravity of the Sun at 150 million km to a value, and yes, it's calculated as if there was a void outside the Sun, but if you're on Earth chances are it's not the dominant component. Doesn't prevent us from calculating the gravity of the Sun at that distance that way anyway. In fact, at least for classic gravity, it's kind of the whole point of the shell theorem.
Well, let's take a look at the classical Newtonian gravity case for a moment.

Let's suppose we're in a Newtonian universe, where space is infinite and mass is distributed uniformly throughout that universe.

Now let's take a sphere of radius R, with us positioned on the right side of that sphere (so the sphere is to our left). The mass within that sphere will pull on us, and so we should feel a gravitational pull to the left from that sphere. OK, but what about the mass from the rest of the universe? Well, for that, we can break the rest of the universe into concentric shells of mass. And we're inside each of those shells. So we should feel no gravitational attraction. That means we should feel a net force to the left.

But what if, instead of starting with a sphere on to our left, we start with a sphere to our right? We go through the same process, and determine that we should feel a net force to the right.

There's a conflict here. How can we resolve it? What the hell is going on?

What's going on is that the shell theorem is essentially giving us an infinite series, and the terms of the infinite series aren't actually getting smaller. The series can appear to converge on different answers, depending on how you order your terms. This is a strong indicator that that's not the right approach.

The same will be true in general relativity. Each unit of farther away mass will have a smaller effect, but there's also more mass farther away. In Newtonian mechanics, the terms of any one shell will always cancel, and you're OK as long as you don't go infinite. But Newtonian gravity is a linear theory. General relativity is not linear. And while general relativity does still have its equivalent of a shell theorem, it isn't the same shell theorem. Things don't simply cancel.

If you are working on a mass density scale where the universe as a whole has a similar density to the area you're interested in, you cannot ignore the gravitational effects of the rest of the universe. We can ignore the gravitational effects of the universe as a whole when looking at, say, the orbits of the planets in our solar system, because the mass density of the solar system is so much higher than the mass density of the universe as a whole. But that obviously won't work when you're talking about why the observable universe itself isn't a black hole, because the universe beyond the observable universe is going to have similar mass density to the observable universe.
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Old 12th September 2021, 09:43 PM   #20
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Originally Posted by Ziggurat View Post
Well, let's take a look at the classical Newtonian gravity case for a moment.

Let's suppose we're in a Newtonian universe, where space is infinite and mass is distributed uniformly throughout that universe.

Now let's take a sphere of radius R, with us positioned on the right side of that sphere (so the sphere is to our left). The mass within that sphere will pull on us, and so we should feel a gravitational pull to the left from that sphere. OK, but what about the mass from the rest of the universe? Well, for that, we can break the rest of the universe into concentric shells of mass. And we're inside each of those shells. So we should feel no gravitational attraction. That means we should feel a net force to the left.

But what if, instead of starting with a sphere on to our left, we start with a sphere to our right? We go through the same process, and determine that we should feel a net force to the right.

There's a conflict here. How can we resolve it? What the hell is going on?

What's going on is that the shell theorem is essentially giving us an infinite series, and the terms of the infinite series aren't actually getting smaller. The series can appear to converge on different answers, depending on how you order your terms. This is a strong indicator that that's not the right approach.

The same will be true in general relativity. Each unit of farther away mass will have a smaller effect, but there's also more mass farther away. In Newtonian mechanics, the terms of any one shell will always cancel, and you're OK as long as you don't go infinite. But Newtonian gravity is a linear theory. General relativity is not linear. And while general relativity does still have its equivalent of a shell theorem, it isn't the same shell theorem. Things don't simply cancel.

If you are working on a mass density scale where the universe as a whole has a similar density to the area you're interested in, you cannot ignore the gravitational effects of the rest of the universe. We can ignore the gravitational effects of the universe as a whole when looking at, say, the orbits of the planets in our solar system, because the mass density of the solar system is so much higher than the mass density of the universe as a whole. But that obviously won't work when you're talking about why the observable universe itself isn't a black hole, because the universe beyond the observable universe is going to have similar mass density to the observable universe.
At least in classical mechanics does it make sense to put a shell around the entire universe?

That would imply you're calculating the effects of gravity on something not in the universe. So... pretty dead end, aint it?
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Old 12th September 2021, 09:48 PM   #21
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Originally Posted by Mike Helland View Post
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.
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Old 12th September 2021, 09:53 PM   #22
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Originally Posted by Ziggurat View Post
I'm not putting a shell around the entire universe.
I didn't say you were.
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Old 12th September 2021, 10:01 PM   #23
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Originally Posted by Mike Helland View Post
I didn't say you were.
Then what the hell is your point?
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Old 12th September 2021, 10:05 PM   #24
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Originally Posted by Ziggurat View Post
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?
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Old 12th September 2021, 10:38 PM   #25
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Originally Posted by Mike Helland View Post
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.
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Old 12th September 2021, 11:06 PM   #26
Mike Helland
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Originally Posted by Ziggurat View Post
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?
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Old 12th September 2021, 11:20 PM   #27
HansMustermann
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@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.
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Old 13th September 2021, 02:10 AM   #28
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Originally Posted by HansMustermann View Post
@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.

Quote:
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.

Quote:
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.

Quote:
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.
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Old 13th September 2021, 03:15 AM   #29
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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.
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Old 13th September 2021, 03:25 AM   #30
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Originally Posted by HansMustermann View Post
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.
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Old 13th September 2021, 03:28 AM   #31
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Originally Posted by Mike Helland View Post
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.
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Old 13th September 2021, 03:31 AM   #32
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Originally Posted by HansMustermann View Post
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.
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Old 13th September 2021, 04:05 AM   #33
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Originally Posted by HansMustermann View Post
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 donít think you actually understand the scenario I described. And causality isnít at play here. Newtonian gravity propagates infinitely fast.
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Old 13th September 2021, 04:20 AM   #34
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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.
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Old 13th September 2021, 07:21 AM   #35
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Originally Posted by HansMustermann View Post
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.
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Old 13th September 2021, 07:30 AM   #36
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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.
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Old 13th September 2021, 10:26 AM   #37
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Originally Posted by HansMustermann View Post
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.
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Old 13th September 2021, 10:27 AM   #38
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Originally Posted by Ziggurat View Post
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?
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Old 19th September 2021, 07:35 PM   #39
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Originally Posted by Mike Helland View Post
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.
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Old 19th September 2021, 09:44 PM   #40
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Yes, there is one hypothesis that we are inside the mother of all black holes.
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