Falling into a black hole

[martix]

If, from an outsider's perspective it takes an infinite amount of time to cross the event horizon, but a finite amount of time for the black hole to evaporate, how does that look like from the perspective of the falling observer. For that matter, how does it look like from the perspective of the outside observer too? And how do black holes even grow from an outside observer's perspective anyway?

I feel like there's a major disconnect somewhere here, but none of the pop-sci I've seen on youtube has ever prepared me for this conundrum.

 

[Dale]

If, from an outsider's perspective it takes an infinite amount of time to cross the event horizon, but a finite amount of time for the black hole to evaporate

This isn’t true. Since the “if” is false none of the “then” applies.

I feel like there's a major disconnect somewhere here

A “takes forever to cross” spacetime is different from an “evaporates” spacetime. The “takes forever to cross” statement comes from some wonky coordinates that don’t mean much even in the spacetime where it fits.

 

[Ibix]

If, from an outsider's perspective it takes an infinite amount of time to cross the event horizon,

This isn't really correct. It's a result of taking the Schwarzschild time coordinate literally, which is a mistake similar to deciding you can't walk through the North Pole because that's the edge of the map. It's perfectly possible to construct definitions of "simultaneity" so that the infaller crosses in finite time, but it requires a bit more maths.

how does that look like from the perspective of the falling observer.

They fall in in finite time and, classically, hit the singularity very shortly after. The latter part may not be correct (it probably isn't, actually), but since we don't have a working theory of quantum gravity we don't actually know what happens.

 

[PeroK]

If, from an outsider's perspective it takes an infinite amount of time to cross the event horizon, but a finite amount of time for the black hole to evaporate, how does that look like from the perspective of the falling observer. For that matter, how does it look like from the perspective of the outside observer too? And how do black holes even grow from an outside observer's perspective anyway?

I feel like there's a major disconnect somewhere here, but none of the pop-sci I've seen on youtube has ever prepared me for this conundrum.

The first problem comes with the imprecise concept of "perspective". What does that even mean? It's true that light from a particular event may never reach a distant observer, but how much physical significance does that have? And, is that the only way to define your "perspective"?

The short answer is that using light signals from an event to determine the time of the event is not the only way to do things. And, indeed, in this case is a deficient approach. Your conundrum highlights this: a person falls into a black hole (event A) and distant observer (O) using his preferred coordinate system cannot assign a time to that event. Physics isn't broken. Instead, the coordinate system is not fully global, as it cannot be used to describe all events.

If you study GR, you would learn about other coordinate systems that can assign a time for event A. And the physics goes on.

 

[Nugatory]

I feel like there's a major disconnect somewhere here, but none of the pop-sci I've seen on youtube has ever prepared me for this conundrum.

Looks like a whole bunch of other people got answers in while I was typing my response (so that is what we're all doing on Saturday morning?) so the only thing I have to add is:

Take a look at a Kruskal diagram. You don't need the somewhat daunting math to understand the qualitative features of the diagram: what the path through spacetime of someone hovering outside the black hole looks, like what the path of someone falling in looks ;like, the path taken by light signals emitted by infaller.

 

[martix]

I have been lied to. Many times, but that one's just the latest instance.

> how does that look like from the perspective of the falling observer.

What I was considering behind that question was, if the black hole evaporates, and we have absurd levels of time dilation, could the falling observer get cooked by radiation or something. Kind of like Unruh radiation, which I learned about recently and coincidentally was reminded of because there's literally a thread right next to mine about it and black holes.

@Nugatory Could you link to one. Without knowing how to read one I can't seem to find one simple enough to parse.

 

[Vanadium 50]

I have been lied to.

Well, the quality and accuracy of YouTube videos is highly variable. That's why we recommend reliable sources like textbooks. YouTube was invented so people could share videos of the cute thing their cat did.

 

[Nugatory]

Could you link to one. Without knowing how to read one I can't seem to find one simple enough to parse.

You might try this one: https://en.m.wikipedia.org/wiki/File:Kruskal_diagram_of_Schwarzschild_chart.svg

All of spacetime outside of the black hole is in quadrant I. At any point in that quadrant, moving up the page is moving forward in time, moving sideways is moving either towards or away from the black hole (with only two dimensions in our diagram there’s no way of showing east/west or north/south motion so we’ll consider only radial motion).

The scale is chosen so that a flash of light will follow a path along a 45-degree angle: one light-second sideways for every second upwards. (If you’re familiar with the Minkowski spacetime diagrams of special relativity, this is the same concept). Anything moving at less than the speed of light (which is to say everything, including someone falling into the black hole) will follow a path steeper than 45 degrees: less than one light-second sideways for every second forward in time.

The hyperbolas labeled $r=1.2$, $r=1.4$ and so forth are the paths of objects hovering at a constant height above the black hole. They are hyperbolas instead of straight vertical lines because of the curvature of spacetime, analogous to the way that the straight line path of an airliner appears as a great circle on a two-dimensional map of the earth.
The dashed line is the event horizon, so everything in quadrant II is inside the black hole. The blue hyperbola in that quadrant represents the singularity at $r=0$.
The squiggly black line with the triangles is the path of someone falling into the black hole.

Once inside the event horizon, everything (including light following 45 degree paths) must eventually reach the singularity.
Light emitted at the horizon stays there forever; anything else on a path at a steeper angle must eventually reach the singularity.

 

[PeterDonis]

@martix, you might try reading this Insights article and comment thread:

https://www.physicsforums.com/threads/do-black-holes-really-exist-comments.849716/

 

[PeterDonis]

I have been lied to.

YouTube videos are not good ways to learn actual science. Yes, you should expect to be "lied to" if you watch them expecting to learn actual science.

What I was considering behind that question was, if the black hole evaporates, and we have absurd levels of time dilation, could the falling observer get cooked by radiation or something. Kind of like Unruh radiation, which I learned about recently and coincidentally was reminded of because there's literally a thread right next to mine about it and black holes.

Neither Unruh radiation nor Hawking radiation are detectable by free-falling observers. At least, that's what Hawking's original model for an evaporating black hole implies for Hawking radiation (for Unruh radiation it is a simple consequence of Unruh's derivation). You will find various speculations about "firewalls" and other different models in which an observer free-falling into a black hole does encounter very high temperature radiation (though none of these models imply that a free-falling observer in flat spacetime will see radiation in a vacuum), but those are just speculations at this point.

It is also possible that we will eventually find that the objects we now call "black holes" actually don't have any event horizons or singularities--instead they have something else inside them that just makes them look like "black holes" for a very long period of time, on the order of the Hawking evaporation time. An example is the "Bardeen black hole", which has been discussed in previous PF threads. But this is all just speculation as well; we simply don't have enough evidence or enough theoretical understanding at this point to do more than speculate in this regime.

 

[Grinkle]

@Nugatory Thanks for linking this.

In quadrant II, what is the interpretation of the constant 'r' curves? The infalling particle path looks to my eye close to perpendicular to these curves. Are these curves just for reference (as opposed to potential allowed paths for a particle)?

It looks like the triangle path crosses the EH at t = infinity and then counts down - is that how one reads this? If so, what are the implications of time starting at infinity and counting down in quadrant II, after having counted up to infinity on its way to the EH in quadrant I?

 

[Tomas Vencl]

Many

If, from an outsider's perspective it takes an infinite amount of time to cross the event horizon, but a finite amount of time for the black hole to evaporate, ………

Many of you responsed in sense that this is not true, but….
Although I know and do not dispute the possibility of using comoving coordinate systems, I think it is somewhat escape from the problem. If we consider a distant static observer, then I think the only physical coordinate system for them is the Schwarzschild system (or its derivative with the coordinate singularity). No other system (which removes the coordinate singularity) has any physical consequences for them. So while a hovering observer might say (based on calculations) ‘now he has fallen,’ it has the same physical significance (none) as if they had made up the time. I mean to say that for a static observer, the Schwarzschild coordinate system still holds a privileged position because I can tie the reception of signals to some physical process.

So I think that this is relevant guestion, which is not anwered in this thread “If, from an outsider's perspective it takes an infinite amount of time to cross the event horizon, but a finite amount of time for the black hole to evaporate, ………how does it look like from the perspective of the outside observer too?”

 

[Dale]

If we consider a distant static observer, then I think the only physical coordinate system for them is the Schwarzschild system

This is wrong on a couple of levels.

First, it is wrong because in a basic sense there are no physical coordinate systems. Coordinate systems are just mathematical mappings between events in spacetime and points in $\mathbb R^4$.

Second, it is wrong because there are many different coordinates that have some straightforward mapping to something or other for a distant observer.

Third, it is misguided because distant observers aren’t uniquely important or valid.

Finally, and most importantly, the evaporating spacetime is physically different from the Schwarzschild spacetime. The usual things you think you know about Schwarzschild coordinates don’t apply in an evaporating spacetime. Specifically, in an evaporating spacetime it does not take an infinite amount of coordinate time to cross the horizon in Schwarzschild-like coordinates

No other system (which removes the coordinate singularity) has any physical consequences for them.

No coordinate system has any physical consequences for anyone, including Schwarzschild.

I mean to say that for a static observer, the Schwarzschild coordinate system still holds a privileged position because I can tie the reception of signals to some physical process

No coordinate system holds a priveliged position. That is essentially the principle of relativity. All physical processes, including signals, can be tied to any coordinate system.

“If, from an outsider's perspective it takes an infinite amount of time to cross the event horizon, but a finite amount of time for the black hole to evaporate, ………how does it look like from the perspective of the outside observer too?”

The answer to this question was given above. In post 2

 

[PeterDonis]

In quadrant II, what is the interpretation of the constant 'r' curves?

They are best viewed as moments of time that the infalling particle passes through between the horizon and the singularity.

 

[PeterDonis]

I think that this is relevant guestion, which is not anwered in this thread “If, from an outsider's perspective it takes an infinite amount of time to cross the event horizon

As @Dale has already pointed out, in a spacetime with an evaporating black hole, it is not true that "from an outsider's perspective it takes an infinite amount of time to cross the event horizon".

Please read the Insights article referenced in post #9.

 

[Nugatory]

In quadrant II, what is the interpretation of the constant 'r' curves? The infalling particle path looks to my eye close to perpendicular to these curves. Are these curves just for reference (as opposed to potential allowed paths for a particle)?

In quadrant II the $r$ coordinate is timelike, so curves of constant $r$ are at less than a 45-degree angle and as @PeterDonis says should be interpreted as moments in time instead of points in space. Meanwhile curves of constant Schwarzchild $t$ are straight lines with slope greater than 45 degrees so are possible paths for an infalling object. (This is the basis for the common misstatement that “inside the event horizon time and space switch places”).
What’s really going on is that we cannot smoothly extend the Schwarzschild coordinates across the horizon because of the coordinate singularity there, so the $r$ and $t$ in quadrant II are different from the $r$ and $t$ in quadrant I, even if they’re represented by the same letters. It would make more sense to use different letters; see for example “Remark 1” on page 5 of https://arxiv.org/abs/0804.3619.

 

[Tomas Vencl]

Specifically, in an evaporating spacetime it does not take an infinite amount of coordinate time to cross the horizon in Schwarzschild-like coordinates

This is not necessarily true. As you yourself mention, the metric of an evaporating black hole is not Schwarzschild, and the exact solution for such a metric is unknown. Recently, there was a very interesting discussion in another thread about an article
https://arxiv.org/abs/1102.2609
which claims (see fig.3) that infaling do not cross the horizon at all.
Another possibility is (my favorite) that an evaporating “black” hole does not have a horizon at all.

 

[Dale]

This is not necessarily true. As you yourself mention, the metric of an evaporating black hole is not Schwarzschild, and the exact solution for such a metric is unknown. Recently, there was a very interesting discussion in another thread about an article
https://arxiv.org/abs/1102.2609
which claims (see fig.3) that infaling do not cross the horizon at all.
Another possibility is (my favorite) that an evaporating “black” hole does not have a horizon at all.

That metric winds up being an evaporating white hole. So I was excluding it because the OP specified a black hole.

 

[Tomas Vencl]

That metric winds up being an evaporating white hole. So I was excluding it because the OP specified a black hole.

May be, but fact is that exact metric for evaporating “black hole” is unknown, so claiming that “in an evaporating spacetime it does not take an infinite amount of coordinate time to cross the horizon in Schwarzschild-like coordinates” is at least based on aproximations and rather speculative.

 

[Grinkle]

are possible paths for an infalling object.

Is the following question purely a choice of co-ordinates, or is there some physics involved? Sorry to get so "meta", asking a question about my question, but after a few attempts that were much more wordy this was the shortest way I could think to ask.

Is it possible for the depicted object with some physically allowed acceleration to take, for instance, the t=2 path directly to the singularity after intersecting it? So I mean fire rockets at t=2 and given strong enough rockets experience no further 'time' (aging?) before encountering the singularity?

Edit: I am not ignoring the discussions around r being the timelike co-ordinate on this side of the EH, I am making the assumption that a traveler continues to experience time and aging for as long as they remain alive even after crossing the EH.

 

[Ibix]

Is it possible for the depicted object with some physically allowed acceleration to take, for instance, the t=2 path directly to the singularity after intersecting it? So I mean fire rockets at t=2 and given strong enough rockets experience no further 'time' (aging?) before encountering the singularity?

The Kruskal diagram deliberately has the same property as the Minkowski diagram that light travels on 45° lines. All timelike paths are closer to vertical. Inside the black hole you can travel on lines of constant Schwarzschild $t$, but that's only because $t$ is a misnomer there - it's not a timelike coordinate inside the horizon. You continue to age as normal.

You can accelerate inside the hole. The strategy that maximises your survival time is to match speeds with an object that fell in from rest at infinity (IIRC), then freefall. Any other approach has a shorter proper time to the singularity.

 

[martix]

You might try this one: https://en.m.wikipedia.org/wiki/File:Kruskal_diagram_of_Schwarzschild_chart.svg

All of spacetime outside of the black hole is in quadrant I. At any point in that quadrant, moving up the page is moving forward in time, moving sideways is moving either towards or away from the black hole (with only two dimensions in our diagram there’s no way of showing east/west or north/south motion so we’ll consider only radial motion).

The scale is chosen so that a flash of light will follow a path along a 45-degree angle: one light-second sideways for every second upwards. (If you’re familiar with the Minkowski spacetime diagrams of special relativity, this is the same concept). Anything moving at less than the speed of light (which is to say everything, including someone falling into the black hole) will follow a path steeper than 45 degrees: less than one light-second sideways for every second forward in time.

The hyperbolas labeled $r=1.2$, $r=1.4$ and so forth are the paths of objects hovering at a constant height above the black hole. They are hyperbolas instead of straight vertical lines because of the curvature of spacetime, analogous to the way that the straight line path of an airliner appears as a great circle on a two-dimensional map of the earth.
The dashed line is the event horizon, so everything in quadrant II is inside the black hole. The blue hyperbola in that quadrant represents the singularity at $r=0$.
The squiggly black line with the triangles is the path of someone falling into the black hole.

Once inside the event horizon, everything (including light following 45 degree paths) must eventually reach the singularity.
Light emitted at the horizon stays there forever; anything else on a path at a steeper angle must eventually reach the singularity.

Thank you for this explanation. I have yet to parse it fully, but let's start somewhere. I am wondering whether this diagram can finally make me understand the nature of gravity as spacetime curvature rather than a force.

So, in Minkowski spacetime (aka flat spacetime?) an unmoving object's worldline is just a straight line going up - only changes in time, not in space.
But let's put a gravitating object in the vicinity, which will give us a space-time like you see in a Kruskal diagram. Then the same straight worldline will instead take us closer to the gravitating object, because of the curved spacetime, which is effectively what gravity is.

Is that reasonable?
(You say the squiggly triangle line is an infalling object, and it's curved, so I'm not sure how that fits in, nor am I sure what an inertial or non-inertial frame of reference looks in this context.)

P.S.

YouTube videos are not good ways to learn actual science. Yes, you should expect to be "lied to" if you watch them expecting to learn actual science.

I do. Though it's sad to see how many people don't and I understand the need to issue such warnings.

 

[Grinkle]

that's only because is a misnomer there

Sorry if the answer is in the paper that @Nugatory linked, I have it printed but haven't read it yet.

In what sense is it less confusing to call that co-ordinate the 't' co-ordinate? What is the motivation for that naming convention? Does it make equations on either side of the EH look more similar to each other?

I'm sure it makes sense to folks who learn the math, and the primary motivation for GR math in general is certainly not to make the topic convenient for parlor discussion.

 

[Nugatory]

I am wondering whether this diagram can finally make me understand the nature of gravity as spacetime curvature rather than a force.

For that particular question the video below by our member @A.T. may be even more helpful.

(You say the squiggly triangle line is an infalling object, and it's curved, so I'm not sure how that fits in, nor am I sure what an inertial or non-inertial frame of reference looks in this context.)

The squiggles just indicate that the infalling object is occasionally accelerating/decelerating, which is to say changing its speed, which is to say changing the angle of its worldline. If the infaller was free-falling, its path would be perpendicular to curves of constant $r$ everywhere and would look like a smooth curve instead of squiggles. I expect that the original author of this diagram added the squiggles to reinforce the point that no matter what the infaller does, even firing arbitrarily powerful thrusters to resist their fall, there is no escaping the singularity in their future.

Video:

 

[Ibix]

In what sense is it less confusing to call that co-ordinate the 't' co-ordinate?

What happens is that when $r<R_S$ the $1-R_S/r$ terms flip sign, and one of the spacelike coordinates inside the horizon has the same prefactor in the metric as the timelike coordinate outside, and the timelike coordinate has the same prefactor as one of the spacelike coordinates outside. So sometimes people carry the labels across. The $r$ still labels the size of nested spheres, albeit they now decrease in size into the future instead of radially inwards, but $t$ isn't really defensible except in the sense that any letter will do and one should always take care to check one's preconceptions.

Personally, I don't think it is less confusing to keep the $r$/$t$ notation, and some sources pick other labels.

 

[Nugatory]

In what sense is it less confusing to call that co-ordinate the 't' co-ordinate? What is the motivation for that naming convention? Does it make equations on either side of the EH look more similar to each other?

I expect that it is largely historical. When we're solving the EFE for a spherically symmetrical spacetime, it's just another differential equation: We write down an ansatz in $t$ and $r$, do differential equation solving stuff with it, and we end up with the Schwarzschild metric written as a function $t$ and $r$ and applicable both outside and inside the horizon.

So it's natural enough to use the same letters for the coordinates on both sides of the horizon, and anyone who has gotten that far into the math will comfortable with the essential arbitrariness of the labels, won't be confused by $r$ and $t$ being timelike on one side of the horizon and spacelike on the other, and won't feel any particular need to relabel anything. It's only a pitfall for people starting with the worked solution and making the (natiural enough) assumption that $t$ always means "time" and $r$ always means “distance”.

 

[martix]

For that particular question the video below by our member @A.T. may be even more helpful.

The video shows a different example, but that's exactly what I was imagining.

The squiggles just indicate that the infalling object is occasionally accelerating/decelerating, which is to say changing its speed, which is to say changing the angle of its worldline. If the infaller was free-falling, its path would be perpendicular to curves of constant $r$ everywhere and would look like a smooth curve instead of squiggles. I expect that the original author of this diagram added the squiggles to reinforce the point that no matter what the infaller does, even firing arbitrarily powerful thrusters to resist their fall, there is no escaping the singularity in their future.

I'm confused now. Which squiggle are we talking about now? The one with the triangles? If the dashed diagonal line is the event horizon arbitrarily powerful thrusters can always save you, right? As long as you're still in quadrant I.
It's curved towards the black hole, so to me it looks like the infaller is firing their thrusters to help, rather than counteract gravity. I imagine doing nothing would be a straight line up (i.e. similar idea to the video) and trying to counteract the gravity of the black hole would be a line curving towards the right.

P.S. t=∞ again rears its head, making the Kruskal diagram, if anything, reinforce what started all this, about never crossing the event horizon.

 

[Ibix]

If the dashed diagonal line is the event horizon arbitrarily powerful thrusters can always save you, right? As long as you're still in quadrant I.

Well, if you don't mind being in a very thin layer across the back wall of your ship, yes.

The black line on the Wikipedia image looks like a freefall path to me, although you'd need to do a calculation to be sure. The hyperbolae in region I are the paths of objects hovering at constant $r$. Anything that isn't curving to the right at least that much is falling in.

 

[Nugatory]

If the dashed diagonal line is the event horizon arbitrarily powerful thrusters can always save you, right? As long as you're still in quadrant I.

Sorry, yes, I was talking about quadrant II.

 

[martix]

I suddenly feel like I'm wasting everyone's time. There's literally a similar thread from a week ago... I guess we might have seen the same video. (Well thank you everyone for humoring me.)

Anyway

If the infaller was free-falling, its path would be perpendicular to curves of constant $r$ everywhere and would look like a smooth curve instead of squiggles.

The black squiggle looks smooth to me.
And does that mean a free-falling object would have an even greater curve to the left? Honestly, I'm having difficulty visualizing what a free-fall curve would look like on that diagram (I thought it was a straight line up - like in the linked video).