The forums are so silent these days, so here's a challenge question, something I certainly struggle to wrap my head around.  For a distant observer it will take an infinite time to see an astronaut fall into a black hole.  But it will take a finite time for the same observer to see the black hole evaporate due to Hawking radiation, i.e. before the astronaut crosses the event horizon.  Black hole information paradox solved?  But how then explain the paradox that the astronaut will experience Hawking radiation at a normal rate and not experience that the event horizon slips away?

Below is what Chat-GPT had to say on the matter.  But it didn't really give an answer to my question.  It rather simply agreed to my assertions.

The black hole information paradox is one of the most fascinating and challenging problems in modern physics. One of the proposed solutions to this paradox involves the idea of black hole evaporation due to Hawking radiation.

According to this idea, a black hole will gradually evaporate over time, emitting particles and radiation due to quantum effects near its event horizon. This process is known as Hawking radiation, and it implies that information that falls into a black hole might not be lost forever.

One of the intriguing consequences of Hawking radiation is that, from the perspective of a distant observer, a black hole will appear to evaporate over a finite amount of time. However, the same observer would see an astronaut falling into the black hole taking an infinite amount of time to cross the event horizon. This apparent contradiction is known as the "black hole firewall paradox."

One possible resolution to this paradox involves the idea that the information of the infalling astronaut is encoded on the event horizon of the black hole, and that this information is preserved in the Hawking radiation emitted by the black hole. This means that, even though the distant observer sees the black hole evaporating before the astronaut crosses the event horizon, the information of the astronaut is still preserved and is not lost.

Regarding the paradox of the astronaut's experience of Hawking radiation, it is essential to note that this effect is due to the strong gravitational field near the black hole. From the perspective of the astronaut, the event horizon will appear to be stationary, and the rate of Hawking radiation will be normal. This is because the astronaut is falling towards the black hole at a constant rate, and their trajectory is determined by the curvature of spacetime near the black hole.

In summary, the black hole information paradox is a complex problem that involves fundamental concepts in quantum mechanics and general relativity. While there are several proposed solutions to this paradox, the role of Hawking radiation and its implications for the experience of infalling observers remain an active area of research and debate.

If pushed a bit further on the matter, ChatGPT just repeats that the paradox arises from different reference frames and different gravitational forces.

The forums are so silent these days, so here's a challenge question, something I certainly struggle to wrap my head around.  For a distant observer it will take an infinite time to see an astronaut fall into a black hole.  But it will take a finite time for the same observer to see the black hole evaporate due to Hawking radiation, i.e. before the astronaut crosses the event horizon.  Black hole information paradox solved?  But how then explain the paradox that the astronaut will experience Hawking radiation at a normal rate and not experience that the event horizon slips away?

That's really something to ask a university professor, rather than an AI that just regurgitates information (and is also terrible at maths).

The forums are so silent these days, so here's a challenge question, something I certainly struggle to wrap my head around.  For a distant observer it will take an infinite time to see an astronaut fall into a black hole.  But it will take a finite time for the same observer to see the black hole evaporate due to Hawking radiation, i.e. before the astronaut crosses the event horizon.  Black hole information paradox solved?  But how then explain the paradox that the astronaut will experience Hawking radiation at a normal rate and not experience that the event horizon slips away?

This is an excellent question. And I apologize this is a long answer, but it's hard to figure out how to make it shorter without skipping too much. Much of this will be stuff you already know, but it all plays together. I hope it helps!

Also, it is a very common misconception that the infalling astronaut must somehow be saved from falling through the horizon because of time dilation and the black hole's evaporation, so I'm glad we can get past that part right away!   For anyone who may be puzzled by that, the key is to consider which signals from the distant universe are able to catch up to the infalling astronaut before they meet the singularity. Because the proper time of their fall to the center is very short, only few of those signals can meet them. Imagine they have a friend orbiting the black hole at a safe distance. Their friend watches them plunge, then continues watching the hole evaporate. When the hole is half the size it was originally, they send in a signal saying "Hey! The black hole is now half as big!" That signal never reaches the astronaut. Basically, because the light has to make the same trip that they made. It cannot be that they are stuck on the horizon and the light just keeps falling onto them.

Okay, what is going on from the orbiting friend's perspective?
This is stuff you already know, but it's important to review: The orbiting friend watches the astronaut fall, for a while getting faster and faster. Then, because of time dilation, their inward velocity slows and they gently come to rest on the horizon (asymptotically approaching it.) As you also know, their image is redshifted from view. What's important however, is how quickly this happens: it is about the same timescale as the proper time from horizon to singularity, which is extremely short compared to the lifetime of a black hole. The intensity of their light as seen from far away diminishes exponentially: by a factor of e (2.718) every ~10 microseconds per solar mass of the black hole. For a 4 million solar mass black hole like Sgr A*, that means their light is reduced by a factor of ten about every 90 seconds. Whereas the remaining lifetime of the black hole if we ignore accretion is about 10^87 years...

The image that the distant observer sees is very quickly fading away, both by wavelength and by brightness. 
Over a far shorter time than the black hole is evaporating, their image becomes indistinguishable from the horizon itself. In fact, we can't even tell where they are on the horizon anymore! Their redshifted light eventually goes to wavelengths comparable in size to the entire black hole, meaning we detect their scant remaining photons from all over the horizon. (More precisely, their radiation is redshifted down to the same effective temperature as the Hawking radiation, which is a thermal spectrum with a peak wavelength about 8 times larger than the diameter of the event horizon. A black hole imaged by its own Hawking radiation would look very fuzzy.)
 
From our external point of view, it just looks like they vanish onto the black hole and become part of it.
Their light becomes indistinguishable from Hawking radiation from the entire black hole. But contrary to our understanding that they simply get frozen in time at the event horizon, it is more that their image reddens, fades, and blurs out very rapidly and they take on the appearance of the black hole itself. Whereas the astronaut experiences no slowdown, and crosses the horizon extremely quickly. We can tell because the amount of light we get from them diminishes away so fast. That's how these wildly different perspectives of the trip are consistent.

Information paradox?
The astronaut experiences events within the interior of the black hole. Information from those events is swept into the singularity with them. We never know about it, nor is it even scrambled into the Hawking radiation for us to receive later. This is actually not the information paradox. Those events in the interior may as well be in another universe, or take place "after infinite time" in ours. The paradox is not about why we can't get that information out.

The paradox is instead about what happened to the information about the matter which originally created the black hole, or which fell into it after the horizon formed but before we were around to watch it. If one black hole formed from the collapse of a star, and another from a Kugelblitz, how do we know which is which? By the no hair theorem, there is no way to tell. 

If the astronaut dives into the black hole, but nobody shows up to that black hole until a few years later, how could they figure out that part of the mass of the black hole is due to the astronaut? The light by then is redshifted to oblivion, scrambled into Hawking radiation. The paradox was that this radiation was thought to be purely thermal, which is in a sense random and contains no information about the matter itself. (It is fully described by a single parameter: temperature.) 

The solution to the paradox is that the Hawking radiation actually does encode that information, in some statistical correlations between different photons. If you could intercept them all very carefully, it is theoretically possible (but incredibly impractical) to decode what went into the black hole.

Thanks, Wats, this is excellently put.  I think the point you're making that to the distant observer the astronaut actually disappears quickly both due to intensity loss and redshift, and is so utterly lost in Hawking radiation much much sooner than evaporation becoming a factor, is quite important.  This makes the usual picture that the astronaut freezes at the horizon quite misleading.  It would be better to say that the astronaut does quickly disappear, but not really into the black hole by crossing the event horizon, but rather disappearing into the extreme spacetime warping at the event horizon.  Is that a reasonable way to put it?

I went to Brian Cox' talk about black holes the day before, but this clarification was missing.  Brian is an excellent presenter, though.  But I kind of found Kip Thorne's talk more interesting when he visited Oslo some years ago since we where closer to 30 people in the room allowing interaction rather than 3000 in Brian's one-way event.

Thanks, Wats, this is excellently put.  I think the point you're making that to the distant observer the astronaut actually disappears quickly both due to intensity loss and redshift, and is so utterly lost in Hawking radiation much much sooner than evaporation becoming a factor, is quite important.  This makes the usual picture that the astronaut freezes at the horizon quite misleading.  It would be better to say that the astronaut does quickly disappear, but not really into the black hole by crossing the event horizon, but rather disappearing into the extreme spacetime warping at the event horizon.  Is that a reasonable way to put it?

The usual description of the astronaut freezing at the horizon is definitely misleading. I think part of the problem is that in special relativity, a lot of emphasis is placed on the idea that different frames of reference are equally valid, even if seemingly paradoxical. For instance, two ships passing each other at relativistic speeds each see the clocks on the other ship ticking more slowly, and both views are correct. With the black hole, the astronaut experiences one thing, while the distant observer sees something else entirely. In this case the astronaut's description is the more correct one. It is the observer who is contending with the effects of time dilation and redshift as they watch the astronaut, and not so much the astronaut contending with those effects as they fall.

I don't think saying that they vanish into extreme spacetime warping is quite the right way to describe it, because the space time curvature at the event horizon is finite, and actually quite weak for a supermassive black hole. The bigger the hole, the weaker the curvature at its horizon, which is why the tidal forces are survivable. A better way might be to say that the astronaut really does fall into the black hole extremely quickly, while the distant observer can only see the light they emitted from outside the horizon. The astronaut is long gone, but portions of the light they emitted from when they were closer to crossing the horizon take longer and longer to reach us. 

An analogy would be to try to listen to someone as they accelerated away from you and broke the sound barrier. You would hear a slowdown of their speech, rapidly shifting down in both pitch and loudness. Some of their words will never reach you -- they were said behind an acoustic horizon, of sorts. But you know they really did break the sound barrier, and you can be sure they aren't hearing your speech sped up. 

(This analogy isn't perfect because when falling into a black hole, you still receive some light that was sent in after you, while if breaking the sound barrier you are cut off from any further sound signals from behind you. A better analogy would be with an actual acoustic black hole.)

I don't think saying that they vanish into extreme spacetime warping is quite the right way to describe it, because the space time curvature at the event horizon is finite, and actually quite weak for a supermassive black hole. The bigger the hole, the weaker the curvature at its horizon, which is why the tidal forces are survivable.

Yes, but the spacetime curvature between the observer and the astronaut is extreme, which is what the observer is seeing.  Anyeay, it's not easy to find a way to explain what is happening in a way easy to picture..

Yes, but the spacetime curvature between the observer and the astronaut is extreme, which is what the observer is seeing.

For clarity, let's call this gravitational redshift and gravitational time dilation (actually these are the same thing), not spacetime curvature. Spacetime curvature relates to tidal forces, which become extreme near the central singularity, but are finite and may be arbitrarily weak at the event horizon.

The time dilation and redshift that the observer sees does go to infinity at the horizon. To the observer, the astronaut seems to freeze in place on the horizon, though as we saw this is more illusory than real -- an artifact of the light taking a very long time to climb out of the gravity well to reach us, whereas the astronaut has really fallen through and experienced no slowdown whatsoever.

Anyeay, it's not easy to find a way to explain what is happening in a way easy to picture..

For sure... 

The best visual I know of is to imagine an analogue of a black hole with water flowing toward a drain. Water waves take the role of light waves. Replace the infalling astronaut with a fishing bob floating on the surface, which bounces up and down to create circular water waves at regular time intervals. Replace the distant observer with a sensor that measures water depth. Then think about what these waves look like to a "global observer" who can see everything, vs. what the wave sensor measures, as the bob gets drawn toward and then sucked down the drain. 

We can walk through this analogy in more detail later if you like. We can even bring in the apparent paradox you started off with -- what exactly is going on with the astronaut and his image as the black hole evaporates? Imagine the bob being swept down the drain, while the drain gets closed extremely slowly (closing up long after the bob went down it). What does this look like to the bob, the wave sensor, and the "global observer" who has the ability to see everything?

Imagine the bob being swept down the drain, while the drain gets closed extremely slowly (closing up long after the bob went down it). What does this look like to the bob, the wave sensor, and the "global observer" who has the ability to see everything?

This analogy implies that the sensor will detect the final wave that could escape long after the bob got swallowed, even after the drain was closed (as seen by the global observer).  Transferred to the block hole what the observer would finally see is the frozen image of the astronaut in the act of crossing the event horizon.  Thanks to Hawking radiation it will actually be the exact moment of crossing, not only infinitely close to the moment before, since Hawking radiation will release the photons captured at the horizon as the astronaut crossed.

Yep, that's the right intuition! There's just one catch, which is what we encountered before: even though hypothetically the distant observer should continue catching signals for the remaining lifetime of the hole, in practice there won't be much of an image or a detection at such late times after the plunge happened. The signals fade away too quickly. We also won't really see anything from the event of the horizon crossing itself.  

Think of it this way: the horizon is an unstable position for a wave to be. It's a powerful separator of nearby waves. Waves emitted outward from just above the horizon move away and escape, while those emitted from just below get dragged inward. The speed at which they move away from the horizon is proportional to their distance from the horizon. This leads to an exponentially rapid separation of waves from the vicinity of the horizon. (Which combined with the extremely finite amount of waves emitted near the horizon, because it was crossed so quickly, is why the distant observer sees such a rapidly fading image.)

Yep, that's the right intuition! There's just one catch, which is what we encountered before: even though hypothetically the distant observer should continue catching signals for the remaining lifetime of the hole, in practice there won't be much of an image or a detection at such late times after the plunge happened. The signals fade away too quickly. We also won't really see anything from the event of the horizon crossing itself.  

Yes, that is understood, but it was a new insight that the effect of the Hawking radiation is that the photons of the exact moment of crossing will be released, though this "image" is attenuated by orders of orders of magnitude.
Since Hawking radiation is assumed to be real, I suppose that would also imply that some information from inside the black hole actually can escape, say an electron from the falling astronaut if the electron happened to quantum tunnel from the inside to the outside of the black hole in the moment of crossing?

Since Hawking radiation is assumed to be real, I suppose that would also imply that some information from inside the black hole actually can escape

Sorry for the delay! I've been thinking more about this, but I wanted to create a few visuals to go with the explanation. 

With our analogue black hole model using waves in a fluid, we get the intuition that some waves should be able to escape from on or even slightly below the horizon, if the hole is shrinking. It's a neat thing to consider. Essentially we must make a distinction between defining the horizon as "boundary from within which nothing can escape to infinity", and "boundary where flow speed exceeds signal speed", or "the Schwarzschild radius". We usually consider these all to be the same, (though the former is the closest to the technical definition used in general relativity). With evaporation, that first definition becomes slightly smaller. However, when we apply this to a real black hole, we'll find the distinction can be 100% ignored. The math behind that claim is astonishing, and we'll go over that, too. Basically, black hole evaporation is just way too slow. But it's fun to see what this looks like if we exaggerate the effect a whole lot, and I think it helps with understanding the apparent paradox from time dilation as something falls into a black hole, too.

First to help build visual intuition, I coded up a simple model using waves in a fluid which is flowing radially towards a drain. A particle is released into the flow, and it emits wave pulses at regular time intervals. (Whose time? That measured by the infalling particle, which is what we build our "global observer" map from.) The speed of the waves relative to the flow is set to unity, while the fluid flow speed goes as -1/r, so there is an event horizon at a radius of r=1. Waves are also bent by the flow, which creates an analogue to the photon orbit at 1.5 radii. A particle moving tangent to the hole at that radius can make multiple orbits. In these respects this is a quantitatively correct analogy for a black hole in general relativity.



We can see the source is drawn down to the singularity extremely quickly according to its own clock, along with all waves it emitted from beneath the horizon. Whereas if we imagine an observer stationed at the location x=4, y=0, what do they see? At first a regular set of wave pulses, but then the delay between each successive wave grows longer. At first slowly, but then suddenly. This is of course the redshift and time dilation effect.

Aside: it's interesting to see how at the moment the particle crosses the horizon, portions of every wave pulse it emitted during its plunge toward the horizon are also simultaneously crossing the horizon, in a large shell. That shell collapses down into the singularity with them. This is a neat visual effect that isn't mentioned often. It is often said that an astronaut hovering at the photon sphere could see the back of their head. (It would be severely lensed and hard to see in practice.) But here, we're finding an astronaut falling into the hole would see an image of their own fall, the entire history seen all at once, and from everywhere on the sky, right as they met the singularity. The saying "my life flashed before my eyes" comes to mind.   However, this is not the same as saying the astronaut sees the entire history of the universe flash before them. That is a misconception of the effect of time dilation. If we think of regular light pulses from arbitrary points everywhere in the universe falling in with the astronaut, they only see a tiny subset of them, and not all at once.


Okay, let's add in the effect of black hole evaporation. For simplicity, we'll make the evaporation rate constant, and for the effect to be visible, decently fast compared to the time needed to reach the center. We'll also zoom in a little, have the source emit waves much more frequently, and only after it crosses the photon orbit.



Some observations:

1) The effective event horizon, defined as "the boundary from within which no signals can escape" is slightly smaller than "the boundary where flow speed exceeds signal speed" (the latter is the one rendered as the black dashed circle). We can see one wave pulse emitted beneath the latter boundary where parts of it manage to escape. There is even a dot on the wave pulse after that, which was very nearly stuck on the effective horizon, but fell into the singularity shortly before the end. Some dots also wrap around much of the inside of the black hole and then get saved by the evaporation.

2) The difference is hard to tell for a distant observer. They still see an extremely rapid redshifting and fadeout of the signals, just as if the horizon had the same behavior, but was slightly smaller than they expected.

What is the size of the horizon, as defined by from where no signals can escape?
It is the distance where the inward speed (due to the flow) of a radially outward-moving wave pulse is equal to the speed that the horizon is shrinking. If there is no evaporation, then this distance is where the flow speed equals the wave speed. With evaporation, if the current Schwarzschild radius is R and is changing at the rate R' (with R' < 0 for evaporation), and the wave speed is c, then this effective horizon lies at r/R = 1/(c-R'). For example, if the horizon is shrinking at half the wave speed, then the effective horizon is 2/3 as large as the Schwarzschild radius.


What does this mean for real, astrophysical black holes?

The rate a black hole evaporates is inversely proportional to the square of its mass. The power emitted in Hawking radiation goes as 



Power is energy per time, and mass is energy divided by the speed of light squared, so the rate of mass loss is the Hawking radiation power divided by c^2. Finally, the horizon radius is 2GM/c^2, so the rate it shrinks is equal to the power times 2G/c^4, or



Ignoring growth by accretion (in fact all astrophysical-sized black holes are currently growing even just by absorbing CMB photons), how fast is this for a one solar mass black hole? It comes out to 1.5x10^-72 m/s, or 5x10^-81 times the speed of light. 

For the SgrA* black hole at the center of our galaxy, weighing 4 million solar masses, it works out to 9.3x10^-86 m/s, or 3.1x10^-94 of c. These are exceptionally small numbers. Plugging into our formula for the corrected event horizon size above, then for the one solar mass black hole, the difference is about 10^-77 of a meter. For bigger black holes the difference is smaller. For SgrA* it is about 10^-84 of a meter.

As a black hole evaporates away to nothing, does the effect ever become great enough to be important? Not really. Even for a black hole with a horizon the size of a proton, which weighs about 10^12 kg and is radiating so violently that nothing could even fall into it, the rate of shrinking is only about 10^-36 m/s, which still corrects the horizon size by much less than a Planck length.

It gets interesting to think of what happens when the mass gets down to about 10 micrograms, which is basically the final moment since the horizon radius is approximately the Planck length and the remaining lifetime is about the Planck time. The black hole at this point would be radiating photons with approximately the Planck temperature. I think it would be a neat coincidence if the rate of horizon shrinking was the speed of light. Alas, it's not. But it's getting there -- about 0.02% of c! I suppose it's reassuring that the rate the horizon can shrink is never faster than c, since that would raise concerns about causality.

One other fun thing I came across while reading articles about analogue black holes. Here's an analogue white hole that can exist in nature, created by whales.   It's of course also possible to make a white hole with a water faucet. The spreading of water away from the stream at the bottom is faster than the speed of surface waves out toward a distance where there is a hydraulic jump, and surface waves can exit but not enter that region.

As a black hole evaporates away to nothing, does the effect ever become great enough to be important? Not really. Even for a black hole with a horizon the size of a proton, which weighs about 10^12 kg and is radiating so violently that nothing could even fall into it, the rate of shrinking is only about 10^-36 m/s, which still corrects the horizon size by much less than a Planck length.

Much to digest.  Thank you!  These numbers are surprising.  A proton, almost 10^-15m in size, shrinking at a constant rate of 10^-36 m/s, would require 0.5 * 10^21s to disappear or about 1.5*10^12 years.  That is a very long time.  In reality, the shrinking accellerates, so it will evaporate much faster, but likely still taking several billion years.  If we then consider more familiar black holes like SgrA*, evaporation will take a ridiculously long time, even if we don't consider the time for it to exhaust the photons of the accretion disk.  Even the time it takes to shrink a Planck length is ridiculous.

These numbers are so small that the chance of something dodging a shrinking event horizon due to Hawking radiation appears to be comparable to the chance of dodging due to the expansion of the universe.  In an expanding universe a black hole should be "shrinking" relative to everything else in the universe.

Let's say that Alcubierre drive shipships are possible.  Could they go through a black hole by bending away the event horizon?

Speaking of which, in the case of merging black holes as illustrated in  - at 0:50 we see an extreme opposite warping.  Does this imply that it's possible to ride these waves for a split second to achieve faster-than-light travel as seen for a distant observer?

The slowness of black hole evaporation really is astonishing. Even though I already understood stood as much, I was still taken aback by how small the correction to the event horizon size because of it turns out to be. I figured it would at least be, like, a micrometer or something, which you could say is significant enough to be real but still small enough to ignore. Less than a Planck length is ridiculous.

Aside: a similar thing happens with how solar wind modifies planetary orbits. The wind constantly pushes outwards, so shouldn't the orbits expand? Again it's one of those things that turns out to be negligible. Since it pushes radially outward, it doesn't give the planet orbital energy. Instead, the pressure from solar wind acts like a constant centrifugal force, which is like slightly decreasing the Sun's effective gravitational force. This makes the circular orbit at a given orbital speed slightly larger than you'd expect, but it does not grow over time. For Earth, it raises the orbit by about 1 micrometer. 

Let's say that Alcubierre drive shipships are possible.  Could they go through a black hole by bending away the event horizon?

If the ship could go faster than light, absolutely. Not by bending the horizon away, but by following a spacelike path, which exits its own future light cone. At the event horizon, the outgoing edge of the light cone is parallel to the horizon, and tilted more inward the further in you go, so a greater FTL speed is needed to get closer to the center and emerge again. When would the ship emerge, according to an external observer, I'm not sure has an invariant answer. But this scenario of course introduces all sorts of paradoxes even without a black hole involved.

I think you're asking though if a slower than light ship using a warp bubble could escape by having some effect on the horizon. My current thinking leans against it, for probably an unintuitive reason. The reason is that the curvature of spacetime near the horizon is small, and unrelated to the existence of the horizon. If we imagine an arbitrarily large black hole and a ship a small distance inside it, the local environment of the ship is no different than being in flat spacetime far from any black hole at all. But the horizon now behaves the same way as a receding flash of light. Even if you accelerate to arbitrarily large fractions of the speed of light, whether by using rockets or by a slower than light warp bubble, that horizon remains ahead of you (even getting further away), in the same way as if you were trying to catch up to a light flash.

That answer might be confusing, because isn't curvature the very thing that "causes gravity" by "telling matter how to move" in the first place? That's right, but in the sense of telling how nearby particles will accelerate relative to one another (tidal forces). Otherwise, the only thing any particle does at the local level is "move straight", i.e. along a geodesic, in spacetime. It is the global geometry rather than local curvature that constrains geodesics to move more inward.

at 0:50 we see an extreme opposite warping.  Does this imply that it's possible to ride these waves for a split second to achieve faster-than-light travel as seen for a distant observer?

This is pretty neat. The short answer is no, but there's a lot of information to digest here, so let's cover each part of what the video is showing.

The colors represent the magnitude of time dilation. Arrows represent the direction and magnitude of acceleration (as seen by a stationary observer at that location). The height of the sheet represents a measure of the warping of space. What measure, exactly? I'm pretty sure it is the spatial part of the metric, which is the most common way to create this visual (also called an embedding diagram) for a black hole. 

The Schwarzchild metric for a static, non-spinning black hole in the time and radial directions is 



The first term in parentheses gives you the time dilation, while the second term tells you how much the space has been distorted. This is not the curvature in the sense of tidal forces, but rather in the sense of how much the proper distance (which you would measure with a ruler) between two points in the radial direction is different from what you would expect if you were in flat spacetime. See my earlier post here for more explanation and a thought experiment you could imagine doing around the Earth, in order to interpret this type of spatial distortion.

For a single, non-spinning black hole, this distortion changes with distance in the same way as the time dilation does (those two metric components are just negative inverses of each other), and the spatial distortion is typically plotted with a minus sign so that it looks like a funnel. For a black hole merger, some regions get the spatial component distorted the other way (upwards in the diagram), even while the time dilation is still very strong so the colors are red. That upward distortion isn't like a region of antigravity or a white hole though -- we can see the arrows still point towards the black hole through this region, and have very high magnitude. A test particle released in this region would quickly fall into the merged black holes. But the warping of spacetime is obviously very different from normal. If the usual downward curve means that there is more distance between nearby points in space than you expected, an upward curve means there is less. Space is, in a sense, more compressed in the radial direction than you would have expected. (I expect it is more stretched out in the perpendicular directions.) The large magnitude also implies large tidal forces -- this is where the gravitational waves have the greatest effect.

It's fun to do a quick calculation to see how violent the gravitational waves are this close to a merger. The fractional amount something is stretched and squeezed by the wave (called the strain) is inversely proportional to the distance from the wave source (whereas the energy of the wave drops off with the inverse square law.) At Earth, the strain was about 1 part in 10^21, and the source was about 400 Mpc away (give or take about 150 Mpc.) Each black hole was about 30 solar masses, or about 180 km in diameter. 

If we ignore the fact that the regular tidal forces are already lethal this close to black holes of this size, the gravitational waves are probably lethal, as well. At a distance of 1000km (just outside of the final orbit they made), the waves carried a strain of about 1 part in 100. That would stretch and squeeze your body in alternating directions by several centimeters, multiple times in much less than a second. I'm not sure what that would feel like, but I doubt it would be pleasant, or even survivable!