Post by: neilep on 14/06/2022 16:46:06
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 !!!!!!
Post by: Halc on 14/06/2022 17:38:15
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.
Post by: neilep on 14/06/2022 18:14:22
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 ?
Post by: alancalverd on 14/06/2022 18:18:25
Post by: Eternal Student on 14/06/2022 18:26:02
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.
Post by: geordief on 14/06/2022 18:30:05
Changes to masses inside a black hole emit gravitational waves that cannot leave the black hole for the same reason light cannotWhat ,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?
Post by: Halc on 14/06/2022 18:54:34
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.
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I understand gravitational waves propagating outside the black holeGravitational waves generated outside the black hole propage outward, yes.
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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.
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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.
Post by: paul cotter on 14/06/2022 19:47:17
Post by: Eternal Student on 14/06/2022 20:07:51
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.
Post by: paul cotter on 14/06/2022 20:38:59
Post by: Halc on 14/06/2022 21:52:56
If you throw a rock into* a black hole does it's mass parameter ever increase? (does it "get bigger"?)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.
*into ---> perhaps I should have said towards the black hole, it hasn't actually gone in yet.
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.
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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.
Post by: Eternal Student on 14/06/2022 23:02:15
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 existI 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.
Post by: evan_au on 15/06/2022 00:30:18
Quote from:
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.
Post by: evan_au on 15/06/2022 00:48:42
Quote from: OP
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
- 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
- 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
Post by: Eternal Student on 15/06/2022 02:02:07
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.
Post by: Halc on 15/06/2022 03:41:49
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.
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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.
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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.
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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.
Post by: evan_au on 15/06/2022 10:24:53
Quote from: NASA
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
Post by: neilep on 15/06/2022 12:15:21
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 '?
Post by: Halc on 15/06/2022 13:37:00
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.
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...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.
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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.
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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).
Post by: Eternal Student on 15/06/2022 19:54:34
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.
Post by: Halc on 15/06/2022 23:19:34
There's no disagreement about frames of reference and co-ordinate systems.There is if I'm choosing one that isn't singular at the event in question, and you are choosing a different one.
However, the Schwarzschild co-ordinates (r, θ, φ, t) used by the distant observer aren't just "abstract"[/quote]??? They very much are abstract. That coordinate system is an abstract assignment of those four values to events, and nothing more. The exact same physical spacetime can be assigned different values using a different choice of abstract coordinate system.
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(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).r is not spatial inside the event horizon, and events separated only by t are not time-like separated.
Not when the question is about what happens for the distant observer.Relativity of simultaneity says that the local time (at the distant observer) at which a distant event (object crosses EH) is coordinate system dependent. So there is no correct answer to the question unless you're an absolutist, in which case you should use the absolute foliation and no other. The coordinates you've chosen certainly do not qualify as an absolute foliation, but then I'm not suggesting one that serves that particular purpose any better.
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For the distant observer (holding constant r,θ,φ ), an infinite amount of time must pass before the rock reaches the horizon.Yes, if you choose the coordinate system you indicate, then the infalling thing never gets to the EH. That's probably a problem since I can think of a few contradictions that result from that, but the absolutists do actually posit something like that, denying the existence of black holes altogether. Coordinates of r <= rs are not valid coordinates of any real event. Maybe you can get around this by suggesting the EH reaches out due to a 2nd thing dropped in later, but there'd need to be an infalling metric describing it to be sure if this works or not. As I said, the absolutists deny this effect and say the material moves outward as the BH gains mass. I don't know if they've produced a metric satisfying the field equations that supports this, but they're already in denial of relativity, so they probably don't think they have to.
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The rock will reach the horizon when co-ordinate t =∞Not if the BH evaporates before then.
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No. You're deliberately trying to slip something past people here by tacitly switching to the time experienced by the rock.I'm not. Everybody already knows what the rock-time will be. That's a physical thing, not frame dependent. The frame of the rock is also a poor choice. I was thinking something like Kruskal–Szekeres coordinates where events for neither the distant observer nor the infalling object are ever singular along their worldlines. OK, the infalling object eventually reaches the central singularity (which might be a line or a plane), but it does so in finite time for everybody. No infinities.
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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 randomYes. I also assume the rock had a light beacon on it, perhaps a well aimed laser. Lots of photons, but there's always a last one.
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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.Totally agree. The Kruskal–Szekeres picture of the exact same geometry doesn't suggest otherwise. Said 'last photon' is a physical thing, not an abstraction after all.
Regards ES. Always an interesting conversation. Thank you for that.
Post by: Eternal Student on 16/06/2022 23:38:15
@neilep , I think you've had reasonable answers to your questions already .
I don't know how you would really determine the early history of Sgr-A (or any other black hole) from what you can measure today. As someone previously mentioned, Black holes "have no hair" which basically means they are all quite featureless things. They have a mass parameter, angular momentum and some charge and that's all. Old conventional theory suggests that all other information about whatever fell into them has just gone (although some later theories suggest otherwise). So there's very little way to know what it has eaten by measuring or observing anything about it today.
So you would have thought the best hope of determining how Sgr-A started is just to run simulations based on what we think should have been around in earlier times. That will give you our best guess about what happened. @Halc seems to have already presented some reasonable ideas.
I can only apologise for taking the discussion off on a sidetrack, by the way. That is very likely to happen again because I've probably got some issues with some of the stuff Halc said and I'm assuming you won't mind. If you do, that's fine, ask me (or whoever else) to start a new thread of their own.
Best Wishes.
Post by: evan_au on 17/06/2022 09:36:38
Quote from: Halc
the time when the last photon is receivedIf photons are emitted in every direction from a beacon, they can take quite circuitous routes to reach the observer:
- Direct path
- One orbit around the black hole
- Two orbits around the black hole
etc.
So the photons arriving by the circuitous path arrive after a photon emitted at the same instant on the direct path
- But the observed source of these dizzy photons is spread thinly around the event horizon, rather than being localized at the place where the rock entered the event horizon, so its considerably harder to see.
Post by: Eternal Student on 17/06/2022 13:37:49
I hadn't even thought about photons going around the black hole with almost the right velocity (direction) to be trapped in the photosphere. Thanks for mentioning it, @evan_au .
About co-ordinate systems and Halc's comments:
They (co-ordinate systems) very much are abstract.Yes, OK, on one hand they are.
Suppose one person is using the co-ordinate system (x,y,z,t) which just turns out to be a set of co-ordinates that behave much as you'd expect. Specifically, their space is locally Minkowski space in those co-ordinates.
Another person can choose to use different co-ordinates with this transformation between the co-ordinate systems:
a =x ; b = y ; c = z
T = x + t
So that their co-ordinate system will be written as (a,b,c,T) with a,b,c exactly the same as x,y,z. It's only their T co-ordinate which was chosen to be a combination of some of what was previously considered space and time.
That's perfectly fine, the transformation from (x,y,z,t) → (a,b,c,T) is invertible, so the new system isn't degenerate, all events in spacetime can be written as distinct events in the new co-ordinate system etc.
However space isn't Mnkowski space in those new co-ordinates.
Proof: da = dx, db = dy, dc = dz
dT = dx + dt => dt = dT - dx = dT - da
Set c = 1 unit for convenience.
Hence ds2 = dx2 + dy2 + dz2 - dt2 [Minkowski metric in (x,y,z,t) co-ords ]
=> ds2 = da2 + db2 + dc2 - (dT - da)2 = db2 + dc2 - dT2 + 2 dT.da.
So that the metric is certainly not Minkowski in (a,b,c,T) co-ordinates (for example it has a mixed spatial and temporal term dT.da).
- - - - - - - -
Consider dropping a scientist and well stocked lab into some arbitrary place and time in the Universe. We know that space is always locally flat Minkowski space. Specifically, we can always find a set of co-ordinates (x,y,z,t) such that the Minkowski metric holds in those co-ordinates (at least locally). The scientist should have no difficulty identifying a suitable, natural set of co-ordinates because space won't be Minkowski space in very many co-ordinates. Specifically, they can choose to use some arbitrary co-ordinates but they will know and can tell that the metric isn't Minkowski in those co-ordinates - it it will only take them a few experiments to determine that.
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A more practical approach: Having just been dropped into some arbitrary space and time the first order of business for a scientist is obviously to sort out what co-ordinates they can use (not finding food and shelter). They will pick up a stop watch and set it going, they will say to themselves "that is time flowing in the positive direction". They will then rush to pick up three 1 metre rulers and a setsquare and construct a rigid frame they will call an x,y,z axis. Now they do have a little bit of freedom here. They know they can rotate that frame, pick it up and translate it a little so that the origin is somewhere else, or have a reflection of the axis to obtain a left-handed frame if they prefer - but that's about all they can do. Indeed the translation has to be small because we know that space is only guaranteed to be locally Minkowski, it might go wonky if they translate the frame too much and they'll end up constructing a suitable co-ordinate system for somewhere else rather than where they are now. They can't do very much with the time axis but they can put the origin t=0 wherever they wish (well, yes, OK they can if they start moving and have a Lorentz boost to be applied but that wouldn't be an arbitrary choice they can make - the space is locally Minkowski space and it will respond in a fixed way to their movement as determined in the frame of reference they have just established while trying to remain stationary at the event they were dropped into). Anyway, that's it... their co-ordinate system (x,y,z,t) while they remain at the event they were dropped into, is essentially determined up to some minor adjustments (e.g. rotations of the spatial axis etc.) They can now go off and start finding food and shelter.
They can decide to use a different co-ordinate system if they wish, they could choose to use the transformation to (a,b,c,T) co-ordinates as described earlier. However, what they can't do is change the behaviour of their local space and all the things in it just because they have chosen to use this other co-ordinate system. For example they can pick up the stop watch again and try it but it doesn't record the passage of T, it only records the passage of t. Recall that we had T = x + t but if the scientist quickly moves the stop watch 1 unit along the old x-axis (which is identical to the new a axis they are using) then it doesn't suddenly jump up 1 unit of reported time elapsed, instead it just continues to report the passage of the old co-ordinate time, t. Now they could build a new time (T) keeping device if they wanted, they can have a thing with wheels on the ground and gears inside that will push the hands of a time clock further on when they slide it along the x-axis BUT you've got to ask why? It is abundantly clear that ordinary stuff (chemical reactions, mechanical devices like stop watches etc.) are not behaving in a natural way and using T as if it was time just because you've told it that this T is the new temporal co-ordinate you would like to use.
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This post is getting too long, it's time to close and summarise: The choice of co-ordinates that a scientist uses is arbitrary, however the natural choice of co-ordinates so that space behaves (locally) like flat Minkowski space or so that objects behave naturally with respect to those co-ordinates is NOT arbitrary. There are only a few choices (like rotation of the spatial axis etc.) that can be made.
A scientist at a distance from a black hole (where the rock was heading into) can choose to use Kruskall co-ordinates but that doesn't mean that the scientist won't experience an infinite amount of time pass before the rock reaches the EH. I don't see that there needs to be an objective reality here. Similar to Unruh radiation which we have discussed before on some other thread (where an observer declared to be at rest doesn't see any particles but someone else who is accelerated w.r.t. the first observer can see a thermal bath of particles), there is no objective reality, reality is frame dependant. I think that for the observer that is distant from the black hole their universe is one where the rock will never reach the EH. The distant observer can change that (they can effectively change the universe around them) by moving (both their motion and also that they are now at a different location on a background of curved space will contribute to that). I know you (Halc) like to consider scientific facts or statements as something which is frame independent but I don't think you can have that here - to phrase that another way, are you sure that you (Halc) aren't trying to be the absolutist and suggesting that there would be an objective reality.
Best Wishes.
Post by: Halc on 18/06/2022 02:49:18
Suppose one person is using the co-ordinate system (x,y,z,t) which just turns out to be a set of co-ordinates that behave much as you'd expect. Specifically, their space is locally Minkowski space in those co-ordinates.OK, so you've assigned different coordinates to those same events using a system with non-orthogonal axes in which light moves at infinite speed in one (and only one) direction (which makes for an interesting sync convention). It indeed doesn't conform to the Minkowski metric, but the space itself is no different, just different abstract coordinates assigned to the physical events. The mathematics got more awkward, but not impossible.
Another person can choose to use different co-ordinates with this transformation between the co-ordinate systems:
a =x ; b = y ; c = z
T = x + t
So that their co-ordinate system will be written as (a,b,c,T) with a,b,c exactly the same as x,y,z.
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The scientist should have no difficulty identifying a suitable, natural set of co-ordinates because space won't be Minkowski space in very many co-ordinates.I didn't suggest anything not 'suitable'. In fact, I chose one far more more suitable. There's questions (about objective events) that cannot be asked using those coordinates, but which can be asked using different ones. That makes the 'different ones' a better choice for this scenario.
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Specifically, they can choose to use some arbitrary co-ordinates but they will know and can tell that the metric isn't Minkowski in those co-ordinates - it it will only take them a few experiments to determine that.This is an interesting assertion: that one can empirically test for an abstract choice of coordinate system without first begging the choice.
I can define a meter to be the length of my toaster, but I'm not sure if I've proven anything by measuring that light speed is now about a billion meters per second (assuming the second hasn't been redefined as well), since I need to beg the new definition of the meter to empirically make this measurement.
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They will pick up a stop watch and set it going, they will say to themselves "that is time flowing in the positive direction".Ah. Presentist scientists. The stopwatch demonstrates no such thing. Sorry. Just pointing out that they're begging a philosophical conclusion and have not demonstrated anything scientific yet. You can't discuss a black hole using a model where time is something that flows. Any such model denies the existence of the thing.
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For example they can pick up the stop watch again and try it but it doesn't record the passage of T, it only records the passage of t.If they hold it still relative to the x axis, it measures T. The Minkowskian guys don't get an accurate measurement of t either if the watch is moving. Funny that watches don't measure coordinate time.
My point was the poor utility of said Minkowski coordinates or perhaps Schwarzschild coordinates when considering what's going on with our infalling observer. You seem to be explicitly avoiding that point, concentrating on nice little local experiments instead. Your choice of coordinate system is far more suitable for those purposes, but those purposes are not the topic of this thread.
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A scientist at a distance from a black hole (where the rock was heading into) can choose to use Kruskall co-ordinates but that doesn't mean that the scientist won't experience an infinite amount of time pass before the rock reaches the EH. I don't see that there needs to be an objective reality here.Hey, this comment was on point. Yay!
No, there's never an objective answer to 'what time is it here simultaneous with remote event there?'. Relativity of simultaneity doesn't allow that. But you can ask for instance if an observer is falling into a black hole and looking back at the distant lab with a bright clock showing, what time does he see on that clock when crossing the EH? What is the very last time he sees period? Both coordinate systems listed above cannot answer that, thus giving rise to the misconception that the infalling observer sees billions of years of the outside universe as he falls in (sees distant clocks speed up) which of course is backwards. He sees them slow down, but never stop.
As for our distant scientist using Kruskall co-ordinates, it very much does mean that he'll experience finite time until the rock crosses the EH. That's what those coordinates are for. He'll of course not see that event ever, but seeing it is an objective measurement, not something coordinate dependent.
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I know you (Halc)Nit: Everyone knows who you're talking to, it being implicit in the quotes to which you're replying. You don't need to do that every time you use a pronoun. Sorry, I'm sounding grumpy now. Don't take this as hostility. :D
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to phrase that another way, are you sure that you (Halc) aren't trying to be the absolutist and suggesting that there would be an objective reality.Classically, there are objective events. Its only getting down to the quantum level where I would suggests a lack of objective reality. A rock/laser/observer falling into a black hole is a classic scenario. Unruh radiation is not.
Post by: Eternal Student on 18/06/2022 20:03:33
Kindly accept my apologies below where I seem to be objecting to things that are perhaps not important here.I'd get bored if there were no replies. While an agreement is satisfying for 1 minute, a disagreement is a discussion waiting to happen which is obviously better. However, I have taken on board the implication that the discussion is drifting off the OP a bit and won't continue writing much.
There's probably just one or two things I can't allow to pass without a comment.
As for our distant scientist using Kruskal co-ordinates, it very much does mean that he'll experience finite time until the rock crosses the EH. That's what those coordinates are for.No.
As you have implied in several earlier posts, the distant scientist cannot change the way her local space behaves or the laws of physics in her local space just by changing her co-ordinates.
They can use Kruskal co-ordinates (T,R, Ω) and they will see on a diagram that the rock reaches the event horizon at an event specified by (T',R', Ω') with T' = some value and R' = +T'. However they can't escape the fact that R=T is a surface where the Schwarzschild co-ordinate time, t is specified by t = +∞ and Schwarschild r = +2GM. That Schwarzschild time, t, isn't unimportant or arbitrary to the scientist. That co-ordinate t is what they will experience as local time (if they hold still). Assuming the scientist stays still (holds constant r and angles in Ω) the scientist's hair turns grey and they die according to that value of that time t, in the usual way. It doesn't necessarily matter too much, it's only interesting in that nothing that scientist will ever experience has been the result of the rock reaching the event horizon. For example, the mass parameter of that black hole will not change for the scientist until t > ∞ (assuming the rock was only a test mass and didn't alter the solution of the EFE at all as it fell toward the EH, it's a bit different if the rock does change the solution and the EH can come out to meet the rock and engulf it).
I'm not ignoring or oblivious to the tendancy to switch to notions of presentism by the way - but this post is already too long. A similar argument can be made just by drawing a Kruskal diagram:
Kruskall.png (31.38 kB . 1197x777 - viewed 1852 times)The event where the rock crosses the EH never falls inside a past light cone for an observer on the blue line of constant Schwarzschild radial co-ordinate r shown. So that event is never something than can cause an effect for the observer.
Classically, there are objective events.That's fine. Everyone agrees that there is an event with the rock on the event horizon. The Schwarzschild co-ordinates are not good for providing a co-ordinate description of that event and fail completely when attempting to describe events within the event horizon. Some other co-ordinates like Kruskal co-ordinates are a better choice to describe that event.
There's no need to reply and a lack of response will not be taken as implied agreement.
Thanks for your time and best wishes.
Post by: Halc on 19/06/2022 16:19:32
As you have implied in several earlier posts, the distant scientist cannot change the way her local space behaves or the laws of physics in her local space just by changing her co-ordinates.But you are doing this in your prior posts, implying that 'the way space behaves' is a function of your frame dependent abstraction, and not a function of the physical geometry of the spacetime.
However space isn't Mnkowski space in those new co-ordinates.Here you changed your coordinate system and suggests that somehow the spacetime is different, but when I do the same and you say it hasn't changed. You need to be consistent. Is spacetime being locally Minkowskian an abstract choice, or are you referring to the fact that the physical spacetime is locally flat such that a Minkowskian metric can be meaningfully mapped to it?
So I agree that spacetime hasn't physically changed, but my choice of abstract coordinate system does assign simultaneity to different events, and it is that simultaneity which is under scrutiny here.
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They can use Kruskal co-ordinates (T,R, Ω)I was using (T,X). There is an r and t that corresponds to Schwarzschild coordinates, but those are different coordinates. The rock reaches the event horizon in finite time T as illustratred in your picture. The singularity has been omitted from your picture, but the rock also reaches that in finite time.
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However they can't escape the fact that R=T is a surface where the Schwarzschild co-ordinate time, t is specified by t = +∞ and Schwarschild r = +2GM.That's right. Different coordinates are singular there, which is why I didn't choose them.
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That Schwarzschild time, t, isn't unimportant or arbitrary to the scientist. That co-ordinate t is what they will experience as local time (if they hold still).This is wrong. How does one 'experience' any kind of abstract time? One experiences proper time. That's the only time that's physical. One does not 'experience' the time for some worldline not in one's presence.
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The event where the rock crosses the EH never falls inside a past light cone for an observer on the blue line of constant Schwarzschild radial co-ordinate r shown.Of course not. It's a physical (coordinate independent) fact that and event on the event horizon cannot causally effect one outside that horizon. Light cones (physical) do not define simultaneity (abstract).
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That's fine. Everyone agrees that there is an event with the rock on the event horizon.Not the people using the x,y,z,t or say the cosmic coordinates. There are people that very much disagree that the rock physically crosses the EH, and that the experience of falling in would be cessation of existence right there. That's a crock of course, it leading to inconsistencies.
Post by: Eternal Student on 20/06/2022 19:04:33
Here's a rushed attempt to respond to some of the comments you (Halc) raised (and yes I do need to put Halc in brackets there, I haven't referred to them anywhere earlier yet). None of it's urgent, or important for that matter.
But you are doing this in your prior posts, implying that 'the way space behaves' is a function of your frame dependent abstraction, and not a function of the physical geometry of the spacetime.That was not my intention. Space and the way things behave in space follows the physical laws of science. Changing co-ordinates can't change that.
However, some co-ordinate systems make things seem unnatural when expressed in those co-ordinates. E.g. Objects move around in circles in some some co-ordinates but physically they are always obeying Newton's laws, it's just that the chosen co-ordinates don't describe an inertial frame.
Is spacetime being locally Minkowskian an abstract choice, or are you referring to the fact that the physical spacetime is locally flat such that a Minkowskian metric can be meaningfully mapped to it?Yes to the last point. These are all ways of saying the same thing:
Spacetime is locally flat around any given event.
<=> Centred at any event, there should always be a LIF (Local Inertial Frame).
<=> There is a co-ordinate system centred on the event that reduces the metric to the Minkowski metric (provided you stay local to that event).
And critically, no to the first point. You don't have much choice about the co-ordinate system if you do want to describe a LIF around an event (equivalent to having the Minkowski metric in those co-ordinates, equivalent to saying you do want space to be flat, ... centred around that event).
There are very few co-ordinate systems you could choose to use. If S is a LIF centred at an event and S' is another LIF centred at the same event then the transformations that map S to S' are of one small class. I don't think anyone wants to see the Mathematics - but I think the Conformal group transformations are all you could choose from. (Note: I spent about 30 minutes trying to prove that before deciding it doesn't add much to the discussion, it seems right but shouldn't be taken as gospel). You can further reduce that choice if you did as described in my earlier posts - drop a scientist with a well stocked lab into an arbitrary event.
If you drop a scientist and their lab into an arbitrary event in spacetime, then Nature has already determined what happens to a 1 metre ruler and similar pieces of equipment.
{We need a brief aside to justify or explain this BUT also the post is long and I've got to rush: nature isn't personified and it isn't choosing, it's just following rules, all that matters is that it has done this and the scientist can't change that. We will take for granted that well formulated laws of physics hold in all co-ordinate systems. A "well formulated" law of physics is one that was expressed in a co-ordinate independent way, for example it might be given in covariant form such as the geodesic equation from General Relativity. As such the atoms in a 1 metre ruler behave a certain way irrespective of the co-ordinate system. More-over using any LIF centred on the ruler (and in which the ruler is stationary) the length of that many million molecules of the ruler will be 1 metre. You could obtain similar fixed lengths by using other pieces of equipment in the lab like a light source and a stop watch and allowing light to travel for 1/300 000 000 th. of a second as reported by that stop watch.... there are various ways in which the scientist can fix the numerical values of some important spacetime intervals. Re-scaling space and time or choosing arbitrary numerical values for spacetime intervals isn't something they can do - if they want the 1 metre ruler to be 1 metre and space to be locally flat etc.}
Returning from that brief aside.... The scientist has very limited choices about choosing a co-ordinate system now. Not only must the metric remain Minkowski, the numerical value of a spacetime interval between any two (local) but otherwise arbitrary events must be the same. So the spacetime interval is an invariant under any choice of co-ordinates the scientist can make (if space is to be flat and the lab equipment behaves normally and measures the correct spacetime intervals etc.) It shouldn't be a surprise then that the scientist's choice of co-ordinates follows the ideas presented in most good textbooks about Special relativity - we could only have Poincare transformations between any two LIF's. We can continue to eliminate more possibilities, we know space is only locally flat so we can disallow the selection of translations of the frame, also we can insist the LIF should have the object of interest remain at rest at the origin - which eliminates Lorentz boosts etc. Anyway..... overall what we obtain is the following:
The scientist with their well stocked laboratory, dropped into an arbitrary event in spacetime, actually has very few choices about the co-ordinate system they can choose to use to define the LIF that exists around them and in which they remain stationary. They can rotate the spatial axis and have some reflections but that's about it.
That Schwarzschild time, t, isn't unimportant or arbitrary to the scientist. That co-ordinate t is what they will experience as local time (if they hold still).There's no disagreement here. The original sentence had the phrase "if they hold still" in it and the distant scientist is located on a surface of constant radial co-ordinate r, their entire worldline is on that surface. For the distant scientist, the proper time interval they experience (between two events in their worldline) = the difference in the Schwarzschild co-ordinate time, t, between those two events.
This is wrong. How does one 'experience' any kind of abstract time? One experiences proper time. That's the only time that's physical. One does not 'experience' the time for some worldline not in one's presence.
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Now, some of that was interesting but I'm struggling to recall why it was relevant and which previous posts it connects with. This is what I think mattered:
1. Selecting co-ordinates that describe a LIF around an object (or a scientist) isn't arbitrary. Modulo spatial rotations and reflections, there's only one co-ordinate frame you could choose. This was mentioned many posts earlier in the context that the Schwarzschild co-ordinates aren't unimportant to the distant scientist. Now technically they aren't a Cartesian frame, they are more like a spherical co-ord system BUT you can obtain a corresponding Cartesian reference frame and co-ordinate system from them, with that tacit understanding we can say that the Schwarzschild coordinates are (up to some minor changes like rotation of the spatial axis), the unique co-ordinates describing a LIF centred around the distant scientist.
The scientist can choose to use other co-ordinates but it's impossible that everything seems natural in the local vicinity of the scientist when described with those co-ordinates (local space won't be Euclidean in the presumed spatial co-ordinates or stop watches won't report the passage of the presumed temporal co-ordinate etc. - at least some things will be abnormal when expressed in those co-ordinates just because they won't be matching the Minkowski metric in those other co-ordinates or having the same numerical spacetime intervals reported that their lab equipment demands). There are many ways for the scientist to establish co-ordinates that describe a LIF centred around them and having done that they then know that locally this is identical to the co-ordinate system they would have obtained by using the Schwarzschild co-ordinates (up to some minor changes like spatial rotations etc.). This is just because the scientist was distant from the Black hole and we know that the Schwarzschild metric is well approximated by the Minkowski metric at a distance r >> rs. So the Schwarzschild co-ordinates are one set of co-ordinates that will (approximately) be suitable to describe locally flat Minkowski space around the distant scientist. That approximation becomes exact if the scientist is infinitely far from the black hole but the approximation is good enough for r >> rs. Anyway, since those Schwarzschild co-ordinates will describe a LIF around the scientist, we know that they are the only co-ordinate system that will do so (up-to minor adjustments like spatial rotations etc.).
2. As shown on the Kruskal diagram (which was produced in paintbrush and took what seemed like hours before you criticize it again for not showing irrelevant details like the singularity). Where the notation (T, R, Ω) was just based on what Sean Carroll used in his book Spacetime and Geometry. The Schwarzschild co-ordinates were always kept in lower case r, t . Meanwhile Ω is just shorthand symbolism for the angle co-ordinates like θ, φ because they are technically still there and still required although they are not important or adjusted between Schwarz. and Kruskal co-ordinate systems. Anyway, that T and R on the diagram are the Kruskal co-ordinates.
I'm fairly sure I started with an "As shown by..." but the sentence then went off explaining other things. Anyway, the event with the rock crossing over the EH is never in the past light cone of the distant scientist (whose entire worldline is on the blue line of constant Schwarz. radial co-ordinate r). We both seem to agree on that. So that event never causes an effect for the distant scientist.
This is getting to the crux of matter: We orbit around Sagittarius-A* which seems to be a big black hole, so we are that distant scientist, following a worldline that lies (more or less) at constant Schwarzschild radius r. Is it possible for that black hole to engulf a rock and grow, so that it's mass parameter is now larger, during a finite amount of time for us scientists? Will the mass parameter of Sgr-A ever change in my lifetime? (Assuming that I do not ever get off planet earth and do something like travel fast or travel toward the black hole etc). It makes little practical difference if the gravity we experience from the centre of the galaxy is always caused by a black hole of Mass parameter M plus a small rock close to the event horizon with mass m, or if eventually we just experience the gravity from a Black hole with mass parameter M+m. However, there is a small difference, one is symmetric, the other is not. When the galaxy eventually spins around and we get to the far side of Sgr-A the rock could be further away from us and then there is a difference. There's no such difference if the black hole has grown and has Mass parameter is M+m.
That's too long, sorry and already too much to think about. Anyway, it's not urgent and it's always a pleasure to discuss something. The last question (will the Mass parameter of Sgr-A*) change in my lifetime is something I genuinely don't know. As mumbled about in much earlier posts, if the rock isn't treated as a test mass, so that it does influence the metric in it's locality then the Schwarzschild solution doesn't apply well, the black hole EH might actually come out to meet and engulf the rock in a finite amount of time for me on planet earth.
Best Wishes.
Post by: Deecart on 20/06/2022 22:59:44
You can add english subtitles using the parameter button if you dont understand french langage.
It is very long but it is worth to be seen if you want to become a real physicist.
Post by: Eternal Student on 21/06/2022 01:01:07
Thanks @Deecart .
That is five and a half hours in French. At the very least, it's going to take me a long time to watch that. The subtitles might be automatically generated and therefore full of errors. To be quite honest, it's unlikely that I would finish watching it - but thank you for your time and effort. Someone else who speaks better French may very well enjoy that.
Best Wishes.
Post by: geordief on 21/06/2022 01:22:07
https://en.m.wikipedia.org/wiki/Jean-Pierre_Petit
"In the 1990s, he stated on various French TV shows that some of his main scientific ideas were directly from his analysis of the Ummo case and documents, questioning their terrestrial origin.[citation needed] He said in 2018 that he experienced personal contacts with unidentified entities that may or not be related to the Ummo case, but that he believes are aliens.[5][non-primary source needed]
The Ummo affair is generally believed to have been a sophisticated hoax elaborated by a person without technical knowledge.[6
Have we been blessed with a visitation of emissaries from the stars?
Post by: Deecart on 21/06/2022 18:45:15
Quote from: Eternal Student
That is five and a half hours in French. At the very least, it's going to take me a long time to watch that. The subtitles might be automatically generated and therefore full of errors. To be quite honest, it's unlikely that I would finish watching it - but thank you for your time and effort. Someone else who speaks better French may very well enjoy that.
Yes you are right.
I have tried to let youtube automaticaly translate ... and this is awfull (i couldnt really understand anything, its about tomato and so on :D).
But good news.
Someone have translated the short version of the video (around 1h21 long) in english.
The link here : https://obliviousphysicist.wordpress.com/2022/05/07/are-black-hole-models-mathematically-consistent/
I have viewed this video too and i recommend it.
Post by: Halc on 22/06/2022 02:00:15
Space and the way things behave in space follows the physical laws of science. Changing co-ordinates can't change that.Agree, but this contradicts what you said before. I had needed (and got) some clarification before knowing which one was the contradictory one. It concerns your alternate metric with T = x + t.
Consider dropping a scientist and well stocked lab into some arbitrary place and time in the Universe.This suggests that the 'way things behave in space' can be changed by a coordinate change. They're apparently performing experiments to empirically demonstrate an abstraction (their alternate choice). No experiment will show that, because as you say, the choice of abstraction can't change the way things physically (empirically) behave. One can tell the metric isn't Minkowskian simply with a pencil and paper. The experiments will all be unaltered by the choice.
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Specifically, they can choose to use some arbitrary co-ordinates but they will know and can tell that the metric isn't Minkowski in those co-ordinates - it it will only take them a few experiments to determine that.
We seem to have a fundamental disagreement about the line between arbitrary abstraction and objective (and classical) physical fact.
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However, some co-ordinate systems make things seem unnatural when expressed in those co-ordinates. E.g. Objects move around in circles in some some co-ordinates but physically they are always obeying Newton's laws, it's just that the chosen co-ordinates don't describe an inertial frame.Newton's laws are local simplifications and what might be a natural coordinate system for local description will be inevitably entirely unnatural for one's larger purpose. Yet again, we're not discussing local physics here, so choosing a nice neat local CS is inappropriate (not a natural choice). Most of your post focused on this 'LIF', but the 'L' there makes it unnatural for a non-'L' description unless spacetime remains effectively flat between observer and measured event, which in this scenario is not at all the case. Your scientist with the well stocked lab isn't considering anything in the lab, and he isn't even taking any actual measurements. The question wasn't 'what will the distant observer measure?'.
There's no disagreement here. The original sentence had the phrase "if they hold still" in it and the distant scientist is located on a surface of constant radial co-ordinate r, their entire worldline is on that surface. For the distant scientist, the proper time interval they experience (between two events in their worldline) = the difference in the Schwarzschild co-ordinate time, t, between those two events.[/quote]OK, I see what you mean. The same could be said worldline a meter above the event horizon, despite the objective massive dilation of the lower time relative to the distant time.Quote from: Eternal StudentThat Schwarzschild time, t, isn't unimportant or arbitrary to the scientist. That co-ordinate t is what they will experience as local time (if they hold still).This is wrong. How does one 'experience' any kind of abstract time? One experiences proper time. That's the only time that's physical. One does not 'experience' the time for some worldline not in one's presence.
Yes, in answering 'when does the rock cross the EH?', I was using time T (not t) to express the simultaneity since T is not singular. It may take some arithmetic, but one can very much compute distant-observer-t from a given T, even if T isn't something the guy's clock on the wall measures.
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As shown on the Kruskal diagram (which was produced in paintbrush and took what seemed like hours before you criticize it again for not showing irrelevant details like the singularity).Fantastic job then. I never managed reasonable curves with the primitive tools I have. I'd have just grabbed one from the web.
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Anyway, the event with the rock crossing over the EH is never in the past light cone of the distant scientistOf course not. It wouldn't be an EH if it was.
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So that event never causes an effect for the distant scientist.None claimed.
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This is getting to the crux of matter: We orbit around Sagittarius-A* which seems to be a big black hole, so we are that distant scientist, following a worldline that lies (more or less) at constant Schwarzschild radius r. Is it possible for that black hole to engulf a rock and grow, so that it's mass parameter is now larger, during a finite amount of time for us scientists?Hard to say, since the question is abstract, not physical. Your scientist might pick a metric that is singular at the EH, but that metric cannot actually describe the situation. The LIF doesn't work when there's gravity involved at all. The Schwarzschild metric doesn't work in anything but a static black hole. Even the distant orbiting thing violates that if it has any mass.
So I think I discussed this before. Absent a metric describing an infalling mass, one has to simply approximate and imagine it, possibly giving wrong answers. More below, but your comments are on point.
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Will the mass parameter of Sgr-A ever change in my lifetime?If it didn't, it wouldn't have a mass parameter in the first place. Based on that alone, you have only two choices, a singular infalling metric that either allows mass at all, or one that doesn't. The rock (and everything else in its history) goes in or it doesn't. Keep in mind that the question isn't physical. It is strictly an abstract one unless one asserts physicality to a particular abstraction.
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(Assuming that I do not ever get off planet earth and do something like travel fast or travel toward the black hole etc). It makes little practical difference if the gravity we experience from the centre of the galaxy is always caused by a black hole of Mass parameter M plus a small rock close to the event horizon with mass m, or if eventually we just experience the gravity from a Black hole with mass parameter M+m.A black hole with no mass at all, but a lot of crap almost in it is (must be) empirically indistinguishable from a black hole of mass <a lot of crap>. Thus we will very much experience M+m because m is there, inside or not. What we experience isn't abstract.
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However, there is a small difference, one is symmetric, the other is not.Yes! That's a huge problem with plan B above (it all stuck on the surface). Suppose we start with a solar mass black hole (about 3km). Now we take a concrete cylinder 100m in diameter and massing 100 stars. It's a super-long cylinder. We jam that thing into the small black hole and it all sticks to the non-rotating surface in one place. That puts all the mass off to one side, not centered at all. That would violate the whole no-hair thing. The black hole (after the bar thrown in) is still stationary in the frame of the system CoM, (which is nowhere near where the small black hole was at first). Where is the mass? All on the one side, or centered on the radius? It can't be the former since an off-center mass would be empirically detectable, not just an abstraction. Right? No? My logical seems a little naive/Newtonian, so maybe I'm just doing the mathematics wrong.
So maybe a tiny mass gets stuck, but the next tiny mass (on the same side??) grows the EH, swallowing the first. You drop in a big rock, and all but the trailing bit gets in, at least relative to this chosen metric.
Post by: Eternal Student on 22/06/2022 19:03:38
Thanks for your time. I've enjoyed reading the stuff. I can't think of much else that's relevant.
We seem to have a fundamental disagreement about the line between arbitrary abstraction and objective (and classical) physical fact.I'm not sure. I think a fair amount was misunderstanding and error in what was written and/or what was read.
Best Wishes.