On the nature of the infinite fall toward the EH

[.Scott]

However, soon (I'm not sure how soon),

Pretty soon; for a BH of 100,000 solar masses, Alice's proper time to fall from the horizon to the singularity is about 1 second.

I don't know if it's that quick. Neither Alice nor Bob are really dropping "straight" down as if dropping through a sphere. They're in a TARDIS-like contraption - much bigger on the inside than on the outside. So I think they're free-fall is a kind of speed-of-light slide towards the center.

unseen by the outside world, tidal forces become extreme and Alice dies from spaghettification.

Ok here, but note that this assumes a classical model of the BH. Quantum effects might change this (and in fact many physicists expect them to). I mention that because you bring in quantum effects later on in your post, but if quantum effects are noticeable at the horizon they should certainly be noticeable near the singularity, yet you haven't taken them into account here. (My understanding of the OP is that it was assuming the classical model, no quantum effects.)

Quantum effects in her immediate world would not be seen by Alice as she crossed the event horizon. In fact, if Alice is the only object dropping into the BH, nothing really special should happen to Alice's immediate space until tidal forces start getting uncomfortable.

You mean towards Alice, correct? (I'll assume so and correct that in what follows.)

Thanks for the correction. I've edited my original post.

Bob drops through the Event Horizon just as his watch reaches noon time of that second day. At that point, he will still see Alice moving down at an unreachable and increasing distance below him.

No, he won't; the only light from Alice that he will see at the horizon is the light she emitted when she was at the horizon. To see light she emitted further in, he has to fall further in.

When Bob reaches the event horizon, he will not find Alice there. From Bob's point of view, she will have moved further down - and will be continuing to move further down. I think the main issue here is how much of that he can really "see".

However, as I noted above, you left out the question the OP actually asked, which is: when Alice falls in, is there a finite time by Bob's and Carol's clock at which they are no longer able to fly down and rescue Alice before she crosses the horizon? The answer is yes, there is. ... Again, the answer is yes, which means that there is some time by Bob's clock at which any light signal he sends inward towards Alice will not reach her until after she has crossed the horizon.

That last sentence is the key. If Bob can get a signal to Alice before Alice crosses the EH, then, in theory, something can be done to save her.

 

[PeterDonis]

I don't know if it's that quick.

I do; I calculated it. Most GR textbooks describe the calculation.

Neither Alice nor Bob are really dropping "straight" down as if dropping through a sphere. They're in a TARDIS-like contraption - much bigger on the inside than on the outside. So I think they're free-fall is a kind of speed-of-light slide towards the center.

I'm not sure this is a very good description. I would recommend studying a good GR textbook's treatment of free-fall into a black hole.

Quantum effects in her immediate world would not be seen by Alice as she crossed the event horizon. In fact, if Alice is the only object dropping into the BH, nothing really special should happen to Alice's immediate space until tidal forces start getting uncomfortable.

We don't know for sure that this is true because we don't have a good quantum theory of gravity. Many physicists think it's true, but not all; there are possible quantum models in which it isn't.

When Bob reaches the event horizon, he will not find Alice there.

Yes; I didn't say he would.

From Bob's point of view, she will have moved further down - and will be continuing to move further down.

Yes.

I think the main issue here is how much of that he can really "see".

Which depends on what light signals, emitted outward by Alice, will be received by Bob at what point along Bob's worldline. That's how I determined that Bob won't see Alice being spaghettified before he is spaghettified himself.

If Bob can get a signal to Alice before Alice crosses the EH, then, in theory, something can be done to save her.

As long as it's done within a finite time, by Bob's clock, after Alice starts falling. After that finite time has passed, nothing Bob does can save Alice.

 

[.Scott]

I'm not sure this is a very good description. I would recommend studying a good GR textbook's treatment of free-fall into a black hole.

I read several descriptions. I haven't seen one where the actual computation was done - using the reference frame of the projectile (Alice).
The problem I have is that gravity is going to pull you towards all the mass that entered the black hole before you got there. And all of that mass is a really long distance away. I realize that there is serious time dilation, but I can't see how that dilation will be enough to restore your transit time to seconds.

 

[pervect]

I read several descriptions. I haven't seen one where the actual computation was done - using the reference frame of the projectile (Alice).
The problem I have is that gravity is going to pull you towards all the mass that entered the black hole before you got there. And all of that mass is a really long distance away. I realize that there is serious time dilation, but I can't see how that dilation will be enough to restore your transit time to seconds.

You can find the calculation of proper time to collapse in MTW's "Gravitation", pg 851, for the collapse of an observer on the surface of a sphere of presureless dust that is collapsing into a black hole.

eq 32.10c gives the proper time, in geometric units as $\pi ( R_i^3 / 8M)^ {\frac{1}{2}}$

It might not be done in the way you specify ("using the reference frame of the projectile"), but that doesn't make the computation invalid. One of the points of physics is that you can calculate observations in any convenient coordinate system.

It's a bit unclear to me exactly what you mean when you say "the reference frame of the projectile", or why you think you need to do it in that manner.

 

[PeterDonis]

I read several descriptions. I haven't seen one where the actual computation was done - using the reference frame of the projectile (Alice).

The simplest computation I know of is done in Painleve coordinates, which is about the best you can do for "the reference frame of the projectile" if you need to cover more than a small local patch of spacetime.

And all of that mass is a really long distance away.

Not when you reach the horizon. The time I gave you was the time to fall from the horizon to the singularity. That's because the statement of yours that I was responding to was about how long it takes Alice to fall from the horizon to the singularity (or at least close enough to the singularity to be spaghettified). The time to fall from a long distance away (meaning a long distance above the horizon) will be longer, of course (how much longer depends on how long a distance away--pervect gave the general formula).

 

[.Scott]

I guess it's that singularity that's bothering me.
At any point as you are descending into the BH, you could compute the difference in the circumference of the black hole at your location 1 microsecond earlier and the circumference of the black hole at your current position. Since you are descending towards the mass of the black hole, space curvature should be getting worse and worse so if you weren't accelerating that circumference difference should be getting smaller and smaller. But you are accelerating, and being almost at the speed of light anyway, you would be seeing distances down getting smaller and smaller - allowing the circumference difference to increase.

I'm sure that your equations are right, but when I try to visualize it, I see everything that has ever fallen into the black hole before me accelerating away from me making it harder and harder for me to further reduce the circumference of the black hole at my elevation.

 

[PeterDonis]

At any point as you are descending into the BH, you could compute the difference in the circumference of the black hole at your location 1 microsecond earlier and the circumference of the black hole at your current position.

I don't understand what you mean by this. Are you trying to figure out how much the mass of the BH increases when you yourself fall in? That would be the obvious interpretation of what you are saying here, but you go on to say...

Since you are descending towards the mass of the black hole, space curvature should be getting worse and worse so if you weren't accelerating that circumference difference should be getting smaller and smaller.

...which makes it seem like you mean something else by "circumference of the black hole", something to do with space curvature--which, btw, is different from *spacetime* curvature, so you need to clarify which one you mean.

If you mean space curvature, that is not an invariant; it depends on how you "slice up" the spacetime into space and time. For the most natural way of doing that for an observer free-falling into the hole, the way that corresponds to Painleve coordinates, space curvature is zero--the "space" that the observer finds himself falling through is flat! (Space is curved for the most natural way of slicing spacetime into space and time for a *static* observer--one who "hovers" at a constant altitude above the horizon--and there is a sense in which the "circumference" of this space curvature gets smaller for static observers closer to the horizon. But there are no static observers at or below the horizon.)

If you mean spacetime curvature, that is tidal gravity, and tidal gravity does get stronger as you get closer to the singularity at the center of the hole. But I don't know of any way of describing tidal gravity by a "circumference".

But you are accelerating, and being almost at the speed of light anyway, you would be seeing distances down getting smaller and smaller

No, you wouldn't; at least, not with the most natural slicing of spacetime into space and time for an infalling observer. Space is flat in that slicing, as I noted above, and spatial distances remain the same all the way down. The curved spacetime of a black hole does not work the same way as flat Minkowski spacetime, and you can't carry over all of your intuitions from special relativity.

I'm sure that your equations are right, but when I try to visualize it, I see everything that has ever fallen into the black hole before me accelerating away from me

There is a sense in which this is true: tidal gravity along the radial direction does make objects below you that are also falling inward radially appear to accelerate away from you. However:

making it harder and harder for me to further reduce the circumference of the black hole at my elevation.

Again, I don't understand what this means. See above.

 

[.Scott]

In a normal Euclidean sphere, with no space curvature, the circumference is w=2πr. The ratio 2πr/w = 1.
As space curvature increases, that ratio gets much larger 2πr/w>1.

But if there is no space curvature without acceleration, then Alice will drop until she hits something or begins stretching out.

 

[PeterDonis]

In a normal Euclidean sphere, with no space curvature, the circumference is w=2πr. The ratio 2πr/w = 1.
As space curvature increases, that ratio gets much larger 2πr/w>1.

This kind of space curvature is the kind that depends on how you slice spacetime into space and time. The natural slicing of the infalling observer has no space curvature in this sense; see my previous post.

But if there is no space curvature without acceleration

What does acceleration have to do with it?

then Alice will drop until she hits something or begins stretching out.

A black hole spacetime is vacuum; there's nothing to hit. But I don't see what this has to do with space curvature.

 

[.Scott]

This kind of space curvature is the kind that depends on how you slice spacetime into space and time. The natural slicing of the infalling observer has no space curvature in this sense; see my previous post.

What does acceleration have to do with it?

If not acceleration (for example, resistance to falling), what does create a space curvature?

 

[PeterDonis]

If not acceleration (for example, resistance to falling), what does create a space curvature?

Space curvature is just an aspect of spacetime curvature, which is created by the presence of mass (or more generally, stress-energy). It has nothing to do with "resistance to falling"; I'm not even sure what you mean by that, or how it's equivalent to acceleration, but it doesn't seem like it has anything to do with mass.

 

[.Scott]

Having slept on it, let me try this approach. I'm sure it's a loosing argument, but I don't see the flaw.

First, the event horizon is all that matters. If I doubled the mass of the sun, it would take 9 minutes for the effects to reach Earth, but if I double the mass of a black hole singularity, the effects will never reach the event horizon. From the point of view of someone outside the black hole, the entire mass of the black hole is concentrated on or very near its event horizon.

Giving Alice a 24 hour head start into the 100,000 solar mass black hole means that Bob will see Alice on the surface of a black hole when he starts his drop. But he will never reach Alice. In fact Alice along with the entire mass of the black hole will retreat from him at an ever increasing rate.

With tidal forces being inversely proportional to the square of the distance, spaghettification is fighting an uphill battle - so to speak. My guess is that it would approach a limit - perhaps a survivable limit - leaving Alice and Bob in a never-ending fall. But I wouldn't bet my life on it.

 

[PAllen]

Having slept on it, let me try this approach. I'm sure it's a loosing argument, but I don't see the flaw.

First, the event horizon is all that matters. If I doubled the mass of the sun, it would take 9 minutes for the effects to reach Earth, but if I double the mass of a black hole singularity, the effects will never reach the event horizon. From the point of view of someone outside the black hole, the entire mass of the black hole is concentrated on or very near its event horizon.

Really? When the star was more like a neutron star, matter was spread throughout. Then it collapsed further. You think the matter from the center jumped to near the horizon? Any way in GR of asking what happened to this interior matter concludes it formed a singularity surrounded by an event horizon.

Giving Alice a 24 hour head start into the 100,000 solar mass black hole means that Bob will see Alice on the surface of a black hole when he starts his drop. But he will never reach Alice. In fact Alice along with the entire mass of the black hole will retreat from him at an ever increasing rate.

With tidal forces being inversely proportional to the square of the distance, spaghettification is fighting an uphill battle - so to speak. My guess is that it would approach a limit - perhaps a survivable limit - leaving Alice and Bob in a never-ending fall. But I wouldn't bet my life on it.

Forget 24 hours, but if Alice was hovering just above the horizon, and dropped, while Bob has been falling from far away, timed to cross the horizon just after Alice, the following is definitely possible: Bob will see Alice crossing the horizon when Bob crosses, then Bob will actually catch and pass Alice, reaching the singularity first.

 

[PeterDonis]

If I doubled the mass of the sun

How? You can't just magically double the Sun's mass; that violates the Einstein Field Equation. The added mass has to come from somewhere. The simplest case to analyze mathematically is the case where it falls in as a spherically symmetric shell of matter. For that case, this...

it would take 9 minutes for the effects to reach Earth

...is wrong; Earth will feel the effects as soon as the infalling spherically symmetric shell of matter passes Earth's orbit on the way in. You can construct other scenarios where there will be some time delay between the Sun feeling effects and the Earth feeling effects, but the scenario has to be consistent with the EFE, so mass can't just appear from nowhere.

Much the same point applies to the black hole case:

but if I double the mass of a black hole singularity, the effects will never reach the event horizon.

Yes, they will, because to double the mass of the hole's singularity, the mass has to fall in, which means it has to pass through the horizon first. And someone orbiting the hole will see the effects in much the same way as the Earth would see the effects of a large mass falling into the Sun, as above.

From the point of view of someone outside the black hole, the entire mass of the black hole is concentrated on or very near its event horizon.

No, from the point of view of someone outside the black hole, the mass of the hole is somewhere inside the horizon, but he has no way of telling where. Nor does it matter to him, because it makes no difference to any observations he can make.

Giving Alice a 24 hour head start into the 100,000 solar mass black hole means that Bob will see Alice on the surface of a black hole when he starts his drop.

Huh? The hole has no "surface". If by "surface" you mean "horizon", Bob can't see Alice on the horizon until he himself reaches the horizon; at that instant he will see light that Alice emitted when she crossed the horizon, no matter how long ago.

But he will never reach Alice.

If they are both freely falling in, this is correct; someone who free-falls in later can never catch up to someone who free-falls in earlier. (Things get more complicated if one or both of them can fire rockets to accelerate inward or outward, but I don't think we need to go into that here.)

In fact Alice along with the entire mass of the black hole will retreat from him at an ever increasing rate.

True for Alice, false for the mass of the hole. The mass of the hole never changes at all from Bob's (or Alice's) point of view; and the "location" of the mass--the singularity--isn't a spatial location anyway; it's an instant of time. So it makes no sense for Bob or Alice to ask "how far away" the singularity is; it's in their future. The only question that makes sense is how long it will take them, by their own clock, to reach the singularity. Asking how far away it is would be like asking how far away, in the sense of spatial distance, next Tuesday is.

With tidal forces being inversely proportional to the square of the distance, spaghettification is fighting an uphill battle - so to speak. My guess is that it would approach a limit - perhaps a survivable limit - leaving Alice and Bob in a never-ending fall.

Nope. Tidal forces increase without bound as the singularity is approached, and both Bob and Alice reach the singularity in a finite time by their clocks, meaning that they will be spaghettified in a (slightly shorter) finite time by their clocks.

But I wouldn't bet my life on it.

Good call.

 

[.Scott]

Really? When the star was more like a neutron star, matter was spread throughout. Then it collapsed further. You think the matter from the center jumped to near the horizon? Any way in GR of asking what happened to this interior matter concludes it formed a singularity surrounded by an event horizon.

I was only talking about an existing black hole. I wasn't trying to address exactly what happens when one forms.

Forget 24 hours, but if Alice was hovering just above the horizon, and dropped, while Bob has been falling from far away, timed to cross the horizon just after Alice, the following is definitely possible: Bob will see Alice crossing the horizon when Bob crosses, then Bob will actually catch and pass Alice, reaching the singularity first.

In the scenario created a week or two ago, Alice drops into the black hole first. Bob remains outside the black hole and sees Alice approach the event horizon.
For as long as Bob remains outside the black hole, he can never see Alice cross the event horizon.

Then Bob drops. In that scenario, Bob will never cross paths with Alice. Instead, Alice and everything else that makes up the black hole will fall further and further away. One notion is that Alice will stop when she reaches the singularity, but that singularity doesn't seem reachable. PeterDonis stated there was nothing for Alice to his - just the vacuum of space. If that's true, everyone just keeps falling and the tidal forces should never get extreme.

 

[PeterDonis]

Alice and everything else that makes up the black hole will fall further and further away.

True for Alice, false for everything else. See my previous post.

One notion is that Alice will stop when she reaches the singularity

It's not a "notion"; it's an inescapable consequence of classical GR.

but that singularity doesn't seem reachable. PeterDonis stated there was nothing for Alice to his - just the vacuum of space. If that's true, everyone just keeps falling and the tidal forces should never get extreme.

Incorrect; see my previous post.

 

[PAllen]

I was only talking about an existing black hole. I wasn't trying to address exactly what happens when one forms.

But you said "the entire mass of of the black hole is concentrated on or very near its event horizon". This is pure and simply false.

In the scenario created a week or two ago, Alice drops into the black hole first. Bob remains outside the black hole and sees Alice approach the event horizon.
For as long as Bob remains outside the black hole, he can never see Alice cross the event horizon.

Then Bob drops. In that scenario, Bob will never cross paths with Alice. Instead, Alice and everything else that makes up the black hole will fall further and further away. One notion is that Alice will stop when she reaches the singularity, but that singularity doesn't seem reachable. PeterDonis stated there was nothing for Alice to his - just the vacuum of space. If that's true, everyone just keeps falling and the tidal forces should never get extreme.

In that scenario, it is true that Bob will never cross paths with Alice. The rest of what you suggest is false. Bob sees Alice as of when Alice crossed the horizon at the moment Bob crossed the horizon. As Bob get close enough to the singularity to be seriously spaghettified, Bob will see Alice as of when she was spaghettified. Bob will see every stage of Alice's unfortunate fate, as Bob suffers the same.

 

[PAllen]

If they are both freely falling in, this is correct; someone who free-falls in later can never catch up to someone who free-falls in earlier. (Things get more complicated if one or both of them can fire rockets to accelerate inward or outward, but I don't think we need to go into that here.)

Yes, this is true in terms of time of starting free fall (assuming the events are causally connected). However, free faller (Alice) starting from a static position very near the horizon being caught up to inside the horizon by someone free falling from much further away (Bob) timed to cross the horizon just after Alice crossed, is possible. Of course Bob started falling earlier, but crossed the horizon later.

 

[PeterDonis]

Yes, this is true in terms of time of starting free fall (assuming the events are causally connected). However, free faller (Alice) starting from a static position very near the horizon being caught up to inside the horizon by someone free falling from much further away (Bob) timed to cross the horizon just after Alice crossed, is possible. Of course Bob started falling earlier, but crossed the horizon later.

Yes, you're right, I should have been clearer that I was talking about two free-fallers starting from the same altitude.

 

[.Scott]

My assumption was that both Alice and Bob dropped from the same altitude. Also, I imagine that giving Alice a 24-hour head start should be more than enough to keep them separated forever.

When I suggested that Alice might hit something, I got this response:

A black hole spacetime is vacuum; there's nothing to hit. But I don't see what this has to do with space curvature.

When something reaches the singularity is either stops or otherwise accelerates or it doesn't. If it doesn't, then it will continue to increase its distance from everything that falls in behind it.
It seems as though the singularity is a self-fulfilling creation. If you assume there is something there to crash into, then that abrupt deceleration allows objects that follow to catch up and feel the extreme tidal forces. Otherwise, everything flows freely with only moderate tidal forces.

 

[PeterDonis]

When something reaches the singularity is either stops or otherwise accelerates or it doesn't.

No, the correct answer is "none of the above". The singularity is an "edge" of spacetime; there is no spacetime beyond it, so any worldline that reaches the singularity ceases to exist once it reaches it.

If you assume there is something there to crash into

There isn't. The singularity is not a place in space, it's an instant of time. Can you crash into next Tuesday?

Just to clarify, all this is according to classical GR. The standard view among physicists seems to be that quantum gravity effects will take over before the singularity is reached, so there won't actually be an "edge" to spacetime where worldlines just stop and objects cease to exist. However, whatever takes the place of that will be something that probably can't be modeled using a classical spacetime at all; it certainly won't be anything normal like decelerating to a stop.

 

[WannabeNewton]

When something reaches the singularity is either stops or otherwise accelerates or it doesn't.

Do you actually know how a singularity is defined in classical GR? I would strongly suggest reading chapter 9 of Wald "General Relativity" for a basic introduction to the formalism of singularities in classical GR. You can't just hand-wave this stuff, there's a mountain of formalism that comes with the concept of a singularity.

 

[.Scott]

No, the correct answer is "none of the above". The singularity is an "edge" of spacetime; there is no spacetime beyond it, so any worldline that reaches the singularity ceases to exist once it reaches it.

Does it totally cease to exist or does its charge and mass get added into the singularity? If the mass gets added in, does that mass now appear to be a stationary mass? Also, is the singularity a point or a sphere? I don't think that once a black hole has been established, the radius of this "singularity" can ever reach zero.

 

[.Scott]

Do you actually know how a singularity is defined in classical GR? I would strongly suggest reading chapter 9 of Wald "General Relativity" for a basic introduction to the formalism of singularities in classical GR. You can't just hand-wave this stuff, there's a mountain of formalism that comes with the concept of a singularity.

I'm hoping there's some way of showing that the tidal forces would inevitably reach the most extreme conditions without attacking a "mountain of formalism".

 

[PeterDonis]

Does it totally cease to exist or does its charge and mass get added into the singularity?

Yes. The infalling object ceases to exist at the singularity, but its mass and charge are added to the mass and charge of the hole. Strictly speaking, the mass and charge of the hole are not "at" the singularity, because the singularity is not a place in space. The mass and charge are really properties of the spacetime as a whole. As far as an observer outside the hole's horizon is concerned, the mass and charge of the infalling object are added to the hole's mass and charge, at the latest, when the object crosses the horizon. (I say "at the latest" because the details depend on where the object falls in, relative to where the observer outside the horizon is.)

If the mass gets added in, does that mass now appear to be a stationary mass?

This question isn't really well-defined as it stands. See above.

Also, is the singularity a point or a sphere?

Neither. It's an instant of time.

I don't think that once a black hole has been established, the radius of this "singularity" can ever reach zero.

The singularity doesn't "reach" a radius of zero; it *is* at a "radius" of zero--but "radius" isn't really the right term, because, once again, the singularity, $r = 0$, is not a place in space; it's an instant of time.

As far as how the singularity forms, it forms when the object that originally collapses to form the hole reaches zero size and infinite density. Up to that point, $r = 0$ is a (non-singular) place in space, at the center of the object (we're talking about an idealized, perfectly spherically symmetrical collapse here, to keep things simple; non-symmetrical collapses just make the math more complicated, so that it requires numerical simulations, without changing the key conclusions); but at the instant the collapsing object reaches zero size and infinite density, the singularity forms and $r = 0$ becomes an instant of time.

 

[.Scott]

Strictly speaking, the mass and charge of the hole are not "at" the singularity, because the singularity is not a place in space. The mass and charge are really properties of the spacetime as a whole. As far as an observer outside the hole's horizon is concerned, the mass and charge of the infalling object are added to the hole's mass and charge, at the latest, when the object crosses the horizon. (I say "at the latest" because the details depend on where the object falls in, relative to where the observer outside the horizon is.)

The singularity doesn't "reach" a radius of zero; it *is* at a "radius" of zero--but "radius" isn't really the right term, because, once again, the singularity, $r = 0$, is not a place in space; it's an instant of time.

As far as how the singularity forms, it forms when the object that originally collapses to form the hole reaches zero size and infinite density. Up to that point, $r = 0$ is a (non-singular) place in space, at the center of the object (we're talking about an idealized, perfectly spherically symmetrical collapse here, to keep things simple; non-symmetrical collapses just make the math more complicated, so that it requires numerical simulations, without changing the key conclusions); but at the instant the collapsing object reaches zero size and infinite density, the singularity forms and $r = 0$ becomes an instant of time.

So when material is said to fall into the singularity, it's not a movement through space but though time? ...To reach that "instant of time".

But is it possible to reach this singularity without also reaching an r=0 position? I suspect the answer is yes.

 

[.Scott]

This just in: Stephen Hawking has completely remodeled black holes, eliminating the event horizon and spaghettification and replacing it with a "wall of fire".

 

[Mike Holland]

HJi guys, I've been following this revival of this topic, and I have one query. Once Alice passes the EV, all light she emits will be directed inwards, towards the singularity. So how can Bob ever catch up with this light after he crosses the EH, and see her getting spaggettified?
Surely there is no light moving towards him after he enters the BH. except from behind him.
Mike

 

[.Scott]

HJi guys, I've been following this revival of this topic, and I have one query. Once Alice passes the EV, all light she emits will be directed inwards, towards the singularity. So how can Bob ever catch up with this light after he crosses the EH, and see her getting spaggettified?
Surely there is no light moving towards him after he enters the BH. except from behind him.
Mike

If Alice is pointing her flashlight outward, Bob can catch up to some of this light after he falls through the event horizon.

 

[WannabeNewton]

HJi guys, I've been following this revival of this topic, and I have one query. Once Alice passes the EV, all light she emits will be directed inwards, towards the singularity. So how can Bob ever catch up with this light after he crosses the EH, and see her getting spaggettified?

Your question is easily answered by means of a simple Kruskal diagram. In fact your question is almost identical to one of the homework problems I had in a GR class last semester. Unfortunately I can't pull up the solutions publicly from my own university website for the class but I did find the exact same problem set assigned by some other university-they use an Eddington-Finkelstein diagram but the idea is the same: http://dafix.uark.edu/~danielk/Relativity/HW8Soln.pdf

See section 11.1.5 of the following document for the general formalism of Kruskal diagrams: http://eagle.phys.utk.edu/guidry/astro490/lectures/lecture490_ch11.pdf