[neilep (OP)]

Dearest Blackhologists,

If nothing can go faster than the speed of light in the universe, then  in contrast ,surely the 'gravity' of a black hole must do eh ? Is there an explanation for the gravitational pull/speed being so strong that light can not escape ? Is gravity...travelling then ?

whajafink ?











"no officer, I'm not driving I'm travelling'......sheeesh !!!!!!

[Halc]

Is gravity...travelling then ?

Gravity is not something that travels. It is a distortion of spacetime.
What does travel is gravitational waves, which carry information about the changes to the field. A Schwarzschild black hole doesn't emit any gravitational waves because it isn't changing, but say two black holes orbiting each other emit an incredible amount of energy in the form of gravitational waves. These are generated outside the black holes and travel at light speed.

Changes to masses inside a black hole emit gravitational waves that cannot leave the black hole for the same reason light cannot.

[neilep (OP)]

Thank you Halc,

I'm still a tad perplexed. Something must be holding light back faster than light itself travels. I understand gravitational waves propagating outside the black hole and so it's the propagation of internal gravity waves that stops light ?...does light even exist inside a black hole ?

[alancalverd]

Swimming through treacle is quite difficult, but (indeed because) the treacle doesn't move. Imagine a photon swimming through treacle, then increase the viscosity of the fluid until it becomes solid.

[Eternal Student]

Hi.

Something must be holding light back faster than light itself travels.

    Not really.    The easier way to imagine what is happening is to assume space itself is being pulled in toward the black hole singularity.   So light is travelling as fast as it can through space but it's not good enough, space itself is being pulled into the singularity faster than that.
    This is only an image or conceptual representation but some people have presented the idea as water flowing and people trying to travel through the water.   There's a reasonable animation on YouTube that I'll try and find and add to the post later.

LATE EDITING:   I can't find it in isolation.   You can see it in this video by Brian Greene,   "your daily equation #31", available on YouTube.      Aim for the time   between   4:20   and  6:00.

so it's the propagation of internal gravity waves that stops light ?

   No.

does light even exist inside a black hole ?

   According to theory,  yes it can   (for a while before it hits the singularity).   No one has actually been in there to see it.

Best Wishes.

[geordief]

Changes to masses inside a black hole emit gravitational waves that cannot leave the black hole for the same reason light cannot

What ,then, is the effect of changes to the distribution of mass inside a BH?  Anything?   Do we know?

A change in the distribution of mass would be a form of mass itself wouldn't it?

[Halc]

Something must be holding light back faster than light itself travels.

Nothing 'holds back' anything. Relative to anything inside a black hole, all future events are also inside. Trying to send light 'outside' is like trying to shine a light onto 2021 from here. Light doesn't travel into the past no matter how hard you attempt it.

I understand gravitational waves propagating outside the black hole

Gravitational waves generated outside the black hole propage outward, yes.

so it's the propagation of internal gravity waves that stops light ?

They have nothing to do with it. Gravitational waves are just another thing that moves at light speed, but also do not move into the past.

does light even exist inside a black hole ?

Of course. If you jump into a big one with a set of lights (say in a room full of glow sticks), you'd not notice anything different as you crossed the event horizon. Light from the glow sticks would still reach you from every direction.

What ,then, is the effect of changes to the distribution of mass inside a BH?  Anything?   Do we know?

Per the no-hair theorem, there is zero external effect of changes to internal mass distribution. Nobody outside could measure it.
A black hole has externally measurable (total) mass, angular momentum, and charge. That's it.

[paul cotter]

Pardon me for butting in but the easiest way to understand a black hole is in terms of escape velocity. As gravity gets stronger and stronger the escape velocity increases. When it reaches or exceeds c, the speed of light in vacuo, nothing including light can escape. I realise most of you understand this explicitly, this is for the op, neilep.

[Eternal Student]

Hi again.

    I'm sure that will be appreciated, @paul cotter .     @neilep  is a bit of a mystery.   My personal theory is that they are actually an expert in some area of science who appreciates that a forum and its members need something to do, some questions to answer etc.

   Can I throw in another minor question here, please?  It's vaguely on the same topic

If you throw a rock into* a black hole does it's mass parameter ever increase?   (does it "get bigger"?)
*into --->  perhaps I should have said towards the black hole, it hasn't actually gone in yet.

    I'll be an observer outside the black hole maintaining a constant distance  (constant radial coordinate, r)  from the black hole.
    I'll pull everyones attention to how much of my time passes before the rock reaches the event horizon.
    So how do black holes get bigger - other than through black hole mergers?

Best Wishes.

[paul cotter]

Yes indeed, that has always puzzled me. All matter should, as viewed from a safe distance, get stuck at the event horizon.

[Halc]

If you throw a rock into* a black hole does it's mass parameter ever increase?   (does it "get bigger"?)
*into --->  perhaps I should have said towards the black hole, it hasn't actually gone in yet.

This mass parameter is frame dependent, but from your distant viewpoint, the mass/energy it gains from KE is balanced by the PE mass/energy lost from it being at an ever lower potential. So no. A 1 kg rock dropped into a black hole increases the BH mass by 1 kg.

If there is somebody falling in locally with the rock who is in possession of some kind of mass-measuring device, the rock won't change mass along the way. This is a very different frame, but same answer.

So how do black holes get bigger - other than through black hole mergers?

Relative to a distant frame, they grow. A rock never falls into a Schwarzschild black hole, but a BH with a rock dropping into it doesn't conform to the Schwarzschild metric. I'm unfamiliar with the name of a metric describing a mass falling in. Surely somebody must have worked it out.

Your implication that black holes can grow only through mergers suggests that none exist, since it takes two to make one. I do think there are absolutists that suggest the non-existence of black holes since they very much do contradict descriptions in absolute terms.  In order to do this, I think they must suggest that matter stuck on the 'surface' of nothing must actually move outward despite lack of force pushing it that way. Not sure if the people who actually know their physics are on board with that. I've never seen a formal absolutist theory presented as a replacement for GR.
Violation of conservation of baryon number is also a contradiction with such a theory.

[Eternal Student]

Hi and thanks @Halc

   It wasn't really the mass parameter of the rock I was interested in.   However, your statements explain what happens there well.   It's the mass parameter of the Black Hole that I was more interested in.
Anyway,  I am inclined to agree with the later comments you made.    The rock never reaches the black hole event horizon IF the space in that region retains its Schwarzschild geometry.   However, the geometry of spacetime isn't going to be exactly Schwarzschild with a rock close to the event horizon.   

    Overall, I can't really say what happens - which is why I was asking.    The PopSci explanations usually gloss over this issue completely.  One reasonable idea is that there is a significantly different, if temporary, solution to the EFE when the rock is near the (former) event horizon of the black hole, the event horizon extends outward to meet the rock falling in and does engulf it.   Having engulfed it, spacetime settles back down to a typical Schwarzschild solution again but now with a slightly greater mass parameter.  That may all happen in a finite amount of time for an observer well away from the black hole.   
     Obviously that narrative is very different from the usual version of what happens when something falls towards a black hole and is said to never reach the event horizon - as far a distant observer is concerned.   I've always wondered if that is only true for a "test mass" dropped in toward a black hole,  i.e. one that does not change the curvature of spacetime itself.

Your implication that black holes can grow only through mergers suggests that none exist

    I wasn't suggesting that black hole mergers are the ONLY way that a black hole can grow.   I was only stating that this is at least one way that it can happen in a finite amount of time.  LIGO have strong evidence for black hole mergers and they do seem to happen in a finite amount of time.   It seems that, for certain, even if a rock doesn't ever reach the event horizon of a black hole,  another event horizon from a different black hole can.
     It's probably just slightly sloppy use of language to say that two black holes merged.   They were black holes (well more or less) when they were well separated and they are one big black hole after the merging -   but during the merger they were "unusual sources of gravitation" that don't conform to the Schwarzschild solution well enough to be called black holes.

Best Wishes.

[evan_au]

matter stuck on the 'surface' (of the event horizon)

Time dilation is so extreme within millimeters of the event horizon that the image of an infalling rock would be very quickly red-shifted into oblivion.
- The speed of a rock falling into a black hole from a great distance approaches a significant fraction of the speed of light. So, from its frame of reference, it would spend very little time in close proximity to where the event horizon "seemed" to be (when they were far away). So it would emit very few photons in that time.

So: Very few photons, severely red-shifted: The rock would not "float" near the event horizon, it would just disappear.

[evan_au]

Is gravity...travelling then ?

A Gravitational Field can be viewed as a distortion in spacetime (thanks to Einstein).

Stellar-mass black holes typically start off as a massive star.
- This star bends spacetime around the star
- At the end of its life, after it has burnt all the fuel in the core to iron, the star explodes/implodes as a supernova, forming a black hole, with almost the same the same order of magnitude as the mass of the star before it imploded.
- Before the supernova, the distortion of spacetime outside the surface of the star emulates the distortion as if all the mass of the star existed at a single point at its center (thanks to Newton's shell theorem)
- After the supernova, the distortion of spacetime outside the (original) surface of the star shell of ejected material emulates the distortion as if all the mass of the star existed at a single point at its center (thanks to Newton's shell theorem). This point is now at the center of the newly-formed black hole.
- So the gravitational field outside of the star does not really change before and after the supernova, so the gravitational field does not need to "travel" for light-years.
- There are major changes in the gravitational field between (the original surface of the star) and (the surface of the new black hole). Changes in the shape of spacetime/Gravitational Field would propagate within this zone, but this is typically within a radius of a few light-seconds.

If the implosion of the supernova were completely symmetrical, no gravitational waves would be emitted outside the star
- However, computer simulations (and some recent observations) suggest that a supernova implosion is a very chaotic process, and often very asymmetrical, meaning that gravitational waves may be detectable from a nearby supernova (and neutrinos too - but astronomers have been waiting since 1987).

https://en.wikipedia.org/wiki/SN_1987A#Neutrino_emissions

Update: Corrected to account for visible supernovae ejecting 75% of their mass

[Eternal Student]

Hi.
   Thanks for the extra information, @evan_au .  There's stuff I didn't know there and I may look into Supernova again to see what the latest is.   
   The rest of this post may look like a criticism, sorry.  It isn't meant to be.

Time dilation is so extreme within millimeters of the event horizon that the image of an infalling rock would be very quickly red-shifted into oblivion.

    I think we need to establish where the observer is and assume they are using a frame of reference where they are at rest.   The distant observer = An observer at a fixed radial co-ordinate, r >> Schwarzschild radius.      The rock = the rock or anyone close to the rock and falling in with it.

      We all agree that the rock doesn't spend long outside the event horizon, it just falls in within a finite amount of time as far as its concerned.   However, for the distant observer the rock takes an infinite amount of time to reach the event horizon   (well, certainly if the rock is treated as such a small mass that it doesn't affect the spacetime geometry - a previous post discussed that and mentioned my uncertainty about it).
     As far as the distant observer is concerned, light from the rock is progressively red-shifted and total luminosity from it reduces.   This isn't a quick process.

So: Very few photons, severely red-shifted: The rock would not "float" near the event horizon, it would just disappear.

   Bits of this are OK.   However, it doesn't "just disappear", it fades away.  It's like having a studio engineer with the slowest hands in the world, turning the fader knob so slowly that the universe will end before the stage actually goes dark. 

...forming a black hole, with almost the same mass as the star before it imploded...

    Is that right?  I know stellar collapse varies quite a lot and I'm not aware of the latest ideas for the typical behaviour.   Old texts used to suggest that typically there is a supernova explosion where the outer layers of the star making up about 20% of the original mass of the star is blown away.    Some sources put the amount of matter ejected far higher than that...
About 75% of the mass of the star is ejected into space in the supernova.   

https://imagine.gsfc.nasa.gov/science/objects/stars1.html

Best Wishes.

[Halc]

The rock never reaches the black hole event horizon IF the space in that region retains its Schwarzschild geometry.

I didn't say that at all. It not reaching the EH is an artifact of an abstract labeling of the crossing event in the 'frame of the distant observer' which just happens to be singular at the point of contention, making it a very poor choice to answer the question. That abstract coordinate system happens to assign infinity to the EH events, and thus no event outside is after the crossing. But that's just an abstraction, not a physical barrier to the object going in. Choose a coordinate system that isn't singular at the EH and the object goes in without any fuss, in finite time according to everybody.
Schwarzschild geometry has nothing to do with any of that. The metric simply fails to describe a black hole with material infalling, but it gets really close to describing a tiny infalling thing.   

 

Obviously that narrative is very different from the usual version of what happens when something falls towards a black hole and is said to never reach the event horizon - as far a distant observer is concerned.

It all depends on the coordinate system chosen by said distant observer. I can similarly choose a coordinate system where I cannot reach the next room due to a singularity along the way, but I don't actually notice anything when I go there.

I was only stating that this is at least one way that it can happen in a finite amount of time.

How long it takes is a purely abstract duration, not a physical one. Physically, it falls in without fuss, but physically there are no objective time coordinates to events, so the question of how long it takes is essentially meaningless. The proper time is not meaningless.

LIGO have strong evidence for black hole mergers and they do seem to happen in a finite amount of time.

The gravitational waves quickly die down from its peak, well below the ability of LIGO to detect them. But it's the same as any light emitted near the event horizon: it never stops arriving, being red-shifted arbitrarily long. The merger, as observed by a perfect LIGO sensitive to all wavelengths, will be (classically) observed for nearly forever. But that's an observation, not what's actually going on. Observations are physically objective, and are not frame dependent.

So: Very few photons, severely red-shifted: The rock would not "float" near the event horizon, it would just disappear.

Yes, since the image isn't classic, but is quantum. At some point the last photon is emitted that will reach the observer in question. Ditto for the last graviton detected by the perfect LIGO.

[evan_au]

About 75% of the mass of the star is ejected into space in the supernova.

That is the case for visible supernovas.

However, the observed rate of supernovas in our galaxy is lower than calculated by astrophysicists
- This may be due to mundane reasons like the large amount of dust in the plane of the Milky Way
- But some speculate that there may be "dark" supernovas or "failed" supernovas, where the black hole eats the star from the inside, and doesn't blast most of the star's envelope into space.
- Astronomers are hopeful that  the next supernova in our galaxy will leave an imprint in neutrinos and/or gravitational waves, even if it is not visible in electromagnetic radiation.
See: https://en.wikipedia.org/wiki/Failed_supernova

[neilep (OP)]

WOW !!..Thank ewe all for your wonderful contributions and responses. The time and effort you put into these replies are greatly appreciated. I am finding this fascinating.


phew !!...


Can I also ask is there a  way to know how large Sagittarius A was when it formed ?  I know it currently has a mass circa 4 million sol.....


...and then.....ewe have TON 618....... 66 TRILLION sol mass !!..... 66000000000 !!!! how is this possible ? did it swallow a few galaxies ?


From what I understand it takes a long long time(millions of years ?).....just for one sol's worth of mass to be gobbled up by a black hole.........so what kind of commencement did TON 618 have and how large would it have been  when it was ' born '? 

[Halc]

is there a  way to know how large Sagittarius A was when it formed ?

It was probably one or more stellar black holes, most of which are born perhaps twice the mass of our sun. It gets larger by having other mass fall in (which is in abundance in the galactic center, especially in the early formation years.

Sgr-A might have formed a bit larger than a typical supernova event, maybe a stealth event like the dark event Evan describes where a mass just 'blinks out' rather than explodes. Suppose you have a really big star, and it's happily burning away and keeping itself un-holely through that combustion. But rather than quickly using up its fuel and going supernova, the new fuel dumps in far faster than the star can burn it, and eventually it gets so massive that even the fusion going on in its core cannot prevent the gravitational collapse. It blinks out in seconds, and probably produces a black hole massing maybe 20-100 solar masses. That's still way smaller than it is today. It got that big by consuming dust, whole stars, other black holes, and even the occasional galaxy. Andromeda's central black hole will definitely eat Sgr-A in about 20-30 billion years. It's about 10 times the mass of Sgr-A.

...and then.....ewe have TON 618....... 66 TRILLION sol mass !!..... 66000000000 !!!! how is this possible ? did it swallow a few galaxies ?

Lots of them I think. There are insanely large black holes at the centers of superclusters like (from small to large) Virgo, the Great Attractor, and the biggest one 'nearby', the Shapley attractor.

From what I understand it takes a long long time(millions of years ?).....just for one sol's worth of mass to be gobbled up by a black hole.

Sgr-A is a known slow eater (at least currently), but nowhere near that slow. A well-aimed star will just fall straight in, so one can consume a star in moments. Most stars are not well aimed, so if they get too close they just get torn apart and distributed into the accretion disk, some of which is slowly consumed by the black hole below, but the energy released by the infalling stuff adds kinetic energy to the atoms left behind, so much of the material gets shot away at the polar jets.

what kind of commencement did TON 618 have and how large would it have been  when it was ' born '

Probably the same as any other. Probably the larger 'blink out' birth of 20-100 solar masses (a guess). It probably happened earlier than almost any other black hole since for it to get that big today, it had to be near the center of an obscene density of material where stars form quickly and grow too large before they can burn almost any of their fuel. There were probably many such large-but-infant black holes formed, all of which merged after not too long. Determining which one was the original TON 618 itself is like trying to figure out which exact puddle is the head of the Thames river (without a map showing which one they picked).

[Eternal Student]

Hi.

It not reaching the EH is an artifact of an abstract labeling of the crossing event in the 'frame of the distant observer'

   There's no disagreement about frames of reference and co-ordinate systems.   However, the Schwarzschild co-ordinates (r, θ, φ, t) used by the distant observer aren't just "abstract" and they weren't arbitrarily chosen by the distant observer.    They are natural co-ordinates to use for the distant observer:
1.   Provided they are quite distant then their space is locally flat Minkowski space in those co-ordinates.
2.   The intervals of time along the t co-ordinate is what their clocks will measure when they stay still (hold constant r, θ, φ ).   Similarly (r,θ,φ) will be a natural spherical spatial co-ordinate system for them (centred at the black hole admittedly rather than being centred on themselves which is a bit unusual but not totally bizarre).   Everything will make sense and seem natural for the distant observer.
     The distant observer would be deliberately making things counter-intuitve and abstract for themselves if they did choose some other co-ordinate system which was anything other than just a spatial translation or linear re-scaling of that co-ordinate system.  A spatial translation won't affect their time co-ordinate, a re-scaling won't matter either because that will only multiply time intervals (and we will be considering an infinite time interval, so 2∞ or 3.5∞ won't look any different).

making it (Schwazscild co-ordinates) a very poor choice to answer the question.

   Not when the question is about what happens for the distant observer.  For the distant observer (holding constant r,θ,φ ), an infinite amount of time must pass before the rock reaches the horizon.
    Let's rephrase this one more time:    The rock will reach the horizon when co-ordinate t =∞  and it would reach a region with r < 2GM    when  |t| > ∞   but that, of course, is nonsense -  the (r,θ,φ,t) co-ordinate system is just broken or inadequate for |t| ≥ ∞ or  r ≤ 2GM.   However, that co-ordinate system is the most natural one for the distant observer to use and it isn't broken or inadequate until t=∞ .

Choose a coordinate system that isn't singular at the EH and the object goes in without any fuss, in finite time according to everybody.

    No.   You're deliberately trying to slip something past people here by tacitly switching to the time experienced by the rock.   Yes, everyone agrees on the proper time interval or spacetime interval between the events where the rock was just outside the EH and precisely at the EH.  (And, as it happens, the distant observer can't use the Schwarzschild co-ordinates to perform that calculation, they would have to use some other co-ordinates to avoid having 0 .∞ and division by 0  appear in the calculation).   So the rock experiences a finite amount of time to reach the EH because a clock travelling with the rock would record that spacetime interval* as the time elapsed (it doesn't record what happens to the t co-ordinate).
(* For accuracy, the clock would record that time if the rock takes a straight line path connecting those events).
   However, the distant observer does not experience the passage of time like a clock travelling with the rock, a clock in the possesion of the distant observer records the proper time they experience and that is precisely the change in the t co-ordinate that occurs (provided the distant observer stays still, holds constant r,θ,φ ).  So the distant observer can and does experience an infinite amount of time passing in order to have the rock reach the horizon. 
 

Yes, since the image isn't classic, but is quantum. At some point the last photon is emitted that will reach the observer in question. Ditto for the last graviton detected by the perfect LIGO.

    I knew "the last photon" would be mentioned before your reply appeared,  it was just bound to be mentioned.   I think the usual model assumes the emission of individual photons from the rock is random (over a proper time interval Δτ for the rock, the expectation value is for n photons to be emitted but the precise number and precise times of an emission is random and given by a Poisson distribution).   Then the time when the last photon is received can't be predicted and the distant observer can't be sure that this was the last photon, there could always be one more.
   Also, as I'm sure you know, it's irrelevant if you use a classical model for e-m radiation and there's an argument you could make about GR being just a classical theory, so it can't be taken for granted that GR affects these quantum events in a simple way.

Best Wishes.