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I feel the obvious answer should be "no" but help me think this through. It came from the previous Q on blackholes and am posting here for more visibility.

So considering two blackholes rotating about each other and eventually combining. It's in this situation that we get gravitational waves which we can detect (LIGO experiments). But what happens in the closing moments when the blackholes are within each others event horizon but not yet combined (and so still rotating rapidly about each other). Do the gravitational waves abruptly stop? Or are we privy to this "information" about what's going on inside an event horizon.

Thinking more generally, if the distribution of mass inside an event horizon can affect spacetime outside of the horizon then what happens in the following situation:

imagine a gigantic blackhole, one that allows a long time between passing the horizon and being crushed. You approach the horizon in a giant spacecraft and hover at a safe distance. You release a supermassive probe to descend past the horizon. The probe is supermassive in the way a mountain is supermassive. The intention is to be able to detect it's location via perturbation in the gravity field alone. Similar to how an actual mountain causes a pendulum to hang a miniscule yet measurable distance off the vertical.

Say the probe now descends down past the horizon, at some distance off the normal. Say a quarter mile to the 'left' if you consider the direction of the blackholes gravitational pull.

Let's say you had set the probes computer to perform some experiment, and a simple "yay/nay" indicated by it either staying on its current course down (yay) or it firing it's rockets laterally so that it approaches the direct line been you and the singularity and ends up about a quarter mile 'right' (to indicate nay).

The question is, is the relative position of the mass of this probe detectable by examining the resultant gravitational force exerted on your spaceship? Had it remained just off of centre minutely to the 'left' where it started to indicate the probe communicating 'yay' to you, or has it now deflected minutely to the right indicating 'nay'?

Whether the answer to this is yes or no, I'm confused what would happen in real life?

If the probes relative location is not detectable via gravity once it crosses the horizon, what happens as it approaches? Your very sensitive gravity equipment originally had a slight deviation to the left when both you and probe were outside the horizon. Does it abruptly disappear when it crosses the horizon? If so where does it go? The mass of the probe will eventually join with the mass of the singularity to make the blackhole slightly more massive. But does the gravitational pull of its mass instantly change from the location in the horizon where it crossed (about a quarter mile to the 'left') to now being at the singularity directly below. Anything "instant" doesn't seem right.

Or.. it's relative position within the horizon is detectable based on you examining the very slight deviations of your super sensitive pendulum equipment on board your space craft. And you're able to track it's relative position as it descends, until it's minute contribution to gravity has coalesced with the main blackhole.

But if this is the case then aren't we now getting information from within the horizon? Couldn't you set your probe to do experiments and then pass information back to you by it performing some rudimentary dance of manoeuvres? Which also seems crazy?

So both options seem crazy? Which is it?

(Note, this is a thought experiment. The probe is supermassive using some sort of future tech that's imaginable but far from possible by today's standards. Think a small planet with fusion powered engines or whatever. The point is, in principle, mass is detectable, and mass is moveable. Is this a way to peek inside a blackhole??)

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[–] vithigar@lemmy.ca 35 points 3 weeks ago (3 children)

Everything that approaches the event horizon appears to slow down to outside observers asymptotically approaching zero velocity at the moment it reaches the event horizon. At the same time it also red-shifts asymptotically toward infinite wavelengths, becoming undetectable.

An outside observer never sees it cross the event horizon.

[–] TachyonTele@lemm.ee 20 points 3 weeks ago* (last edited 3 weeks ago) (2 children)

To add to this, anything that crosses the "horizon" will appear to stay there, from an outside perspective, like a still photograph. Eventually that image of it will fade away as time goes on. Meanwhile that object has long since been violently destroyed and sucked into the center of the black hole. Nothing but particles survive this.

The only thing that comes out of a black hole is Hawking Radiation Which is basically the black hole spewing out radiation from the crushed mass.

Here's a good introduction to this idea by PBS Spacetime They're a great entry point into a lot of very complicated science, described in an easy to understand way.

Edit: The ending of Interstellar is just story telling. Cooper would have been stretched apart instantly and destroyed at a fundamental level when he got anywhere near the black hole due to the insanely extreme gravitational force.

[–] TheBat@lemmy.world 6 points 3 weeks ago

Edit: The ending of Interstellar is just story telling. Cooper would have been stretched apart instantly and destroyed at a fundamental level when he got anywhere near the black hole due to the insanely extreme gravitational force.

Yeah except

Tap for spoilerit's not actually a black hole, it's a supermassive structure specifically built to allow him to communicate with his past self.

[–] metaStatic@kbin.earth 3 points 3 weeks ago (1 children)

Meanwhile that object has long since been destroyed

people fundamentally misunderstand time dilatation, the observer will be dust long before the object passes the event horizon, if anything can pass it at all.

[–] TachyonTele@lemm.ee 5 points 3 weeks ago

Either way the object doesn't survive.
The event horizon isn't a line around the blackhole, it's a horizon. Of course matter passes through it.

[–] FourPacketsOfPeanuts@lemmy.world 2 points 3 weeks ago* (last edited 3 weeks ago) (1 children)

So.. Let's think that through.. say a planet was falling into the blackhole. With respect to visible light, yes, makes sense to me it's progressively red shifted and you never see it cross the horizon. But if this applies to the gravity of that planet too, then its gravity is forever detectable at that point where it appears to be crossing. The blackhole therefore appears to have a 'bobbled' structure. The main mass of the blackhole, and a minor mass 'glued' to its side. It means the gravity of mass falling into the blackhole never joins it at the centre but is forever detectable at its surface? And if mass fell into the blackhole in an uneven way then gravitationally the blackhole would forever appear 'uneven' rather than perfectly smooth?

Now let's do the same for two blackholes colliding. They each get progressively 'red shifted' and never cross they others event horizon

So are we saying if we were close enough to two black holes that merged, we'd detect, gravitationally, that they actually have not merged, but are sort of frozen at each others horizon??

I mean, this may well be the case, but as far as data LIGO detected goes (there are places where you can hear a sound representation of merging blackholes circling closer and closer until they merge). The impression is that their orbit round each other gets faster and faster until there's a single abrupt moment they collide.

But this seems at odds with them 'red shifting' and getting stuck at the event horizon. Surely that signal would look like gravitational oscillations (of their mutual orbit) getting slower and slower the closer to the horizon they got? Doesn't seem to add up

[–] vithigar@lemmy.ca 6 points 3 weeks ago (1 children)

Gravity is a vector field without distinguishing differences between one source and another. The gravity of the falling mass always was and always will be "joined" with that of the black hole, and every other piece of surrounding matter. It's not like light where two nearby sources remain distinguishable. There's no "bobbled" structure, just a very very slight shift in the location of the center of mass which gets smaller as the falling object gets closer.

As for the faster rotation of colliding black holes, event horizons aren't objects, they're regions of spacetime, and larger than the actual "surface" of a black hole. They combine into a single event horizon long before they ever actually "touch" each other.

[–] FourPacketsOfPeanuts@lemmy.world 0 points 3 weeks ago* (last edited 3 weeks ago) (1 children)

just a very very slight shift in the location of the center of mass which gets smaller as the falling object gets closer.

Sure. But if we're observing this "slight shift in the centre of mass" outside the event horizon then that suggests one could message from inside the horizon to outside by moving a suitably heavy mass side to side.

[–] vithigar@lemmy.ca 5 points 3 weeks ago* (last edited 3 weeks ago) (1 children)

Well, no. Because the "event" of that movement never makes it outside. Hence event horizon. Causality itself cannot traverse it.

edit: Put differently, the propagation of gravity still happens at the local speed of light and is constrained by curved spacetime, just like everything else. Gravity itself is, in a sense, affected by gravity.

[–] FourPacketsOfPeanuts@lemmy.world 2 points 3 weeks ago (1 children)

So when LIGO is detecting gravitational waves of two blackholes rapidly orbiting each other until they collide. Do the gravitational waves gradually slow and then stop as the blackholes approach each others event horizon? Are we ever actually observing the blackholes merge? Because if we are then doesn't that suggest we're observing something via gravitational waves inside the event horizon? And if we're aren't, doesn't that suggest the black holes never actually merge (from our frame of reference) they just get infinitely close to each others event horizon without us observing them crossing it?

[–] vithigar@lemmy.ca 3 points 3 weeks ago (1 children)

I already addressed this in an earlier comment. Event horizons aren't objects, they're regions. Your final suggestion here is roughly correct, the black holes proper never merge from our frame of reference but their event horizons do long before that would happen.

[–] FourPacketsOfPeanuts@lemmy.world 1 points 3 weeks ago* (last edited 3 weeks ago) (1 children)

Yes, I was confusing the picture there a bit.. I meant would we ever actually see the result of one black hole crossing where the others event horizon would have been.. but that was a bit of a mouthful.

Thanks, yes, understand that the event horizon becomes one big region roughly like a figure 8 on its side as the blackholes approach each other

But it still feels like we detect things happening after this merging of event horizons? Is that not so?

Either we get gravitational waves of the two singularities continuing to orbit each other within this one big event horizon, or we don't. If we do that's information from inside an event horizon. And if we don't it suggests that that (from our reference) the two singularities remain apart, the two blackholes now having this one 8-shaped event horizon, which forever remains in this shape.

Neither seem right to me to be honest lol..

[–] vithigar@lemmy.ca 1 points 3 weeks ago

Either we get gravitational waves of the two singularities continuing to orbit each other within this one big event horizon, or we don’t.

We don't. That's why the indicator of it happening is a rapidly elevating frequency as they get closer that suddenly drops to nothing once the event horizons converge.

https://www.youtube.com/watch?v=1agm33iEAuo

[–] metaStatic@kbin.earth 1 points 3 weeks ago (1 children)

my theory is that nothing ever crosses the event horizon. as you approach a black hole time dilates so the event horizon should recede from you as you approach it until the singularity evaporates. the idea that reality decouples never made sense to me but I'm sure my idea has been disproved by someone with an above room temperature IQ.

[–] folkrav@lemmy.ca 2 points 3 weeks ago* (last edited 3 weeks ago) (1 children)

Reality never decouples, it's just always relative but our usual point of views are similar enoug that we typically ignore the difference. Time will fly slightly faster for an astronaut on the ISS, but we're talking nanosecond differences over long periods.

From your POV, you fall into the black hole and that's it. From an external observer's POV, they see you falling slower and slower until you appear to stop at the event horizon then redshift to disappearance. The EH is just the description of the boundary where gravity becomes great enough that nothing can escape it.

[–] metaStatic@kbin.earth 1 points 3 weeks ago

there's a theory that all information is preserved on the surface of the event horizon and that's basically where my idea comes from. for it to be preserved nothing would ever be able to pass the horizon so as you approach time dilates so the external universe is moving fast enough for the event horizon to evaporate before you ever reach it. you would of course be shredded by hawking radiation pretty quickly, relatively speaking, but a hypothetical you would basically travel to the end of time.

[–] AbouBenAdhem@lemmy.world 22 points 3 weeks ago* (last edited 3 weeks ago) (1 children)

An object that accelerates as it falls past the event horizon toward the singularity will cause gravity waves to propagate outward from the object at the speed of light. But those waves will never reach the horizon (or an outside observer) because everything leaving the object at any speed or direction will converge at the singularity; from its frame of reference the horizon only existed in the past. Anything that enters the horizon is no longer causally connected to the outside universe.

[–] FourPacketsOfPeanuts@lemmy.world 5 points 3 weeks ago (1 children)

So how does mass falling into a blackhole ever appear to add to its mass at the centre? If we're saying even gravitational waves can't escape. This sounds like two merged blackholes don't actually ever merge, they get forever 'stuck' at each others event horizon.

If that were the case it feels like the rapid oscillation of gravitational waves that LIGO detected should actually slow down as the blackholes horizons touch. But as far as I've seen this isn't the case. The waves increase rapidly in frequency until there's an abrupt moment the blackholes merge and the waves abruptly stop?

But that makes no sense if each blackhole is progressively 'red shifted' at the others horizon

[–] AbouBenAdhem@lemmy.world 3 points 3 weeks ago* (last edited 3 weeks ago) (1 children)

As a mass falls into a black hole, the radius of the event horizon increases, the black hole is pulled toward the mass enough to conserve their combined momentum, and their electric charge and angular momentum similarly combine. Those are the properties that change when a black hole merges with another black hole or any other infalling mass, and none of it depends on anything that happens inside the horizon.

And it’s true that events near the merging event horizon would appear to undergo time dilation to an outside observer, but I assume that rotational frame-dragging would balance out that effect on the gravitational waves produced (although the actual math is beyond me).

[–] FourPacketsOfPeanuts@lemmy.world 1 points 3 weeks ago (1 children)

When blackholes merge there must be a period of time between the singularities crossing each others event horizon and the singularities actually merging. During this brief period they are still rotating each other and therefore still doing the thing that's generating gravitational waves.

What I'm trying to figure out whether LIGO is detecting any of these gravitational waves produced after the blackholes have crossed each others event horizon?

If they are then that's information being conveyed out of the blackhole by gravity.

If they are not then that suggests that we never really observe blackholes merge. And the movement of one round the other becomes unobservable at the point they reach each other's event horizon. But if this is the case then the blackholes merging must appear 'frozen' at the event horizon in much the same way light is?

[–] AbouBenAdhem@lemmy.world 2 points 3 weeks ago (1 children)

Black holes (and their singularities) never cross each other’s event horizons—their event horizons just merge.

Maybe you’re imagining event horizons as being caused by the singularities inside them. That isn’t strictly true: when a star collapses into a black hole, the event horizon forms before the singularity does. And once the event horizon forms, nothing inside it (including the not-yet-formed singularity) is part of our universe any longer—as far as the outside universe is concerned, the event horizon itself is the black hole (and is its own cause).

Think of black holes merging like soap bubbles merging—it’s the surfaces themselves that merge, not anything inside that’s generating them.

[–] FourPacketsOfPeanuts@lemmy.world 1 points 3 weeks ago* (last edited 3 weeks ago)

Thanks that makes sense. I guess the consequence of that is that merged blackholes are "lumpy". As in there's an orientation of the two previous centres of mass frozen in time some distance from each other. Presumably if we were close enough we'd be able to detect if we're looking at the two singularities side on or in a line? The event horizon would be a sort of oblate 8 shape on its side?

Edit; forgot to add.. this doesn't sound like what we're hearing where LIGO has translated its signals into sound. That sounds like the ever increasing rate of the blackholes orbiting each other until there's an abrupt culmination and calm afterwards.

[–] remotelove@lemmy.ca 7 points 3 weeks ago (1 children)

My guess is that you are visualizing the event horizon as a gradient when it should be viewed as a hard-line barrier.

As anything approaches the event horizon, it still has a chance to escape. Once an object crosses that line, it's game over: All arrows point in.

Now, I have also heard that of you wait long enough for the black hole to completely evaporate and are able to collect every bit of the black hole as it does, you should be able to reassemble the data you desire. It would probably take a supercomputer more massive than the original black hole, but it's worth a shot. As a bonus, I believe you have to solve for an information duplication paradox that is tucked in there somewhere as well.

Yup, am visualising it as a distinct 'hard line' barrier.

Nothing that's crossed the horizon can casually affect anything outside the horizon. But that means when our super massive probe approaches the horizon and is gradually red shifted, never actually appearing to cross, then it's mass is also forever detectable at that point. That means it's mass never joins the main black hole at the centre. The blackhole has a 'bobbled' appearance, gravitationally speaking, with uneven blobs of mass falling into it forever stuck at its surface. But if this is the case then how do blackholes get bigger? If any mass that falls into them after they first form is stuck at the surface. Does that mean the singularity is forever the mass of the instant the blackhole first formed and ALL additional mass consists of mass frozen at the surface of the event horizon?

What does that mean for merging blackholes? They never actually merge? They're just forever frozen at each others even horizon?

That could well be the case, it just seems weird. And it doesn't seem to be what the LIGO gravitational wave signal seems to imply (from my completely unprofessional, uncredentialled, uninformed view)

[–] LostXOR@fedia.io 3 points 3 weeks ago (1 children)

As I understand it, a very massive object sent into a black hole would, after passing the event horizon, be undetectable; the black hole would simply appear to have more mass. The gravitational field of the object doesn't disappear suddenly, it just gradually merges with the black hole's gravitational field until they appear as one mass.

With a big enough object, you'd get some gravitational waves and warping of the event horizon as it enters, but you'd still be unable to send information out because the object's energy and momentum is conserved. So, regardless of what you do on the inside of the black hole, the resulting gravitational waves will look the same from outside. (Think about a spacecraft deflecting its trajectory; it has to throw mass in the opposite direction with an equal amount of momentum).

Yes, mass has to be thrown out as rocket fuel and momentum conserved. But this doesn't mean the distribution of mass has to be equal. Think of a spacecraft (single concentrated area of mass) and the gaseous trail of its spent fuel (indistinct tail of low density mass spread over a much wider area). It would seem in that circumstance the single area of mass of the spacecraft is detectable as having a different profile to the much wider spread out exhaust. (Or to make the comparison more extreme, it's the gravity of a mountain sized 'probe' versus a trail of propellant the consistency of cloud stretching thousands of miles. Seems like these would have different gravitational profiles)

[–] arthur@lemmy.eco.br 2 points 3 weeks ago (1 children)

No. The escape velocity from "inside" the black hole is higher than the light speed. So, even if something happens "inside" the black hole to generate gravity waves, those waves will not be fast enough to escape.

But more important than that, the blackhole (IMHO) is an event that is happening, there is no "inside" yet, because the body that originated the blackhole is frozen in time, from an outside perspective, there's no singularity yet because as density goes up, the proper time from this object slows down infinitely close to zero. If you wanna know more about that perspective on blackholes, look for "frozen stars".

[–] FourPacketsOfPeanuts@lemmy.world 3 points 3 weeks ago (1 children)

Thanks, I'll look that up

I guess I'm getting used to the idea that the result of two blackholes merging is not spherical. If you imagine one mass falling into the other it never gets there, just slows. So although the event horizon has coalesced, the two centre masses have not physically met (from our point of view). So the merged blackholes event horizon should really be like a very oblate 8 shape?

[–] arthur@lemmy.eco.br 1 points 3 weeks ago (1 children)

That will demand a direct observation to answer

[–] FourPacketsOfPeanuts@lemmy.world 1 points 3 weeks ago (1 children)

To my completely unscientific gut instinct, the point at which movements of a singularity cease being detectable is when it gets to where the event horizon of the other blackhole was (prior to them starting the merge). Not that any of this makes any sense. But if the event horizon is where regular matter appears to freeze and red shift without ever seeming to cross. Then that is because light cannot escape the gravitational well. But then the same surely applies to gravitational waves travelling at the speed of light. They can't escape the gravitational well either. So the last we can detect of a merging black hole is when its centre has approached to the same distance at which other matter appears to freeze too - at the event horizon.

So for two equally sized blackholes that would leave them about an event horizons width away from each other. My point being, that would surely be distinct enough to be observable. Would could at least tell if such a system were spinning because the gravitational profiles would change with each rotation.

[–] arthur@lemmy.eco.br 1 points 3 weeks ago (1 children)

Yeah... But the singularity is hypothetical, we don't know if they exists. We know that blackholes exists, but how their inside's are is a different topic. And things don't just appear to be frozen when they approach the blackhole, they (from our perspectives) are frozen, they did not crossed it yet because of the time dilation.

[–] FourPacketsOfPeanuts@lemmy.world 1 points 3 weeks ago (1 children)

For 'singularity' really I'm just referring to 'mass at the centre of a blackhole'. It seems like this should behave like any other mass falling into a second blackhole. Ie frozen and 'redshifted' forever at the event horizon. Difference here being the first blackhole causes the event horizon to be around it too. It just doesn't feel right that we'd be getting gravitational waves of two blackholes circling each other until they physically combine. That would be gravitational waves coming out of the event horizon somehow.

[–] arthur@lemmy.eco.br 1 points 3 weeks ago (1 children)

The waves come from the circling, not the merge itself.

[–] FourPacketsOfPeanuts@lemmy.world 1 points 3 weeks ago (1 children)

Right? Which makes sense.

But if you see (or hear) the results at LIGO they appear to convey the frequency of rotation getting higher and higher to a crescendo.

This seems at odds with what's generally supposed to happen as a mass approaches a blackhole: its time appears to slow, any signals from it appear increasingly redshifted.

So why are the gravitational waves not "redshifted", that is, lower and lower frequency, petering out to nothing the same way light would. The opposite seems to happen.

It was because of the difference between the two that I read under the impression that there was something special about gravitational waves and that we were observing the blackholes continuing to circle each other inside their event horizons until their singularities merge. At which point it makes sense for the gravitational waves to abruptly stop.

If that's not the case (and it does make sense that it isn't), then why the almost opposite character of light and gravitational waves? Both propagate through spacetime at c. So why aren't both affected the same?

[–] arthur@lemmy.eco.br 1 points 3 weeks ago (1 children)

If the frequency does lower abruptly due the redshifting, could we tell the difference between that and it just stopping?

That's a good point, yes. It's just I've watched quite a lot of LIGO scientists talking about it and they all seem to use the same language of the black holes "merging" or "joining". And the quiet period afterwards being a "cool down". None of them mention the relativistic effects of one hole almost-but-not-quite reaching the other which I'd have thought was a fairly easy observation to make when explaining the distinct phases of the signal?

[–] dream_weasel@sh.itjust.works 2 points 3 weeks ago
[–] M137@lemmy.world 2 points 3 weeks ago

"a information"

I did two informations and am now a black hole, never do more than one information.