Why you'll never escape from a black hole

Why You'll Never Escape From A Black Hole
Category: Gravity * black holes * relativity
Posted on: May 10, 2012 8:10 PM, by Ethan Siegel
"They say 'A flat ocean is an ocean of trouble. And an ocean of waves... can also be trouble.'
So, it's like, that balance. You know, it's that great Oriental way of thinking, you know, they
think they've tricked you, and then, they have." -Nigel Tufnel
Black holes* are some of the most perplexing objects in the entire Universe. Objects so dense,
where gravitation is so strong, that nothing, not even light, can escape from it.
(Image credit: Artist's Impression from MIT.)
But there are a number of very counterintuitive things that happen as you get near a black hole's
event horizon, and a very, very good reason why once you cross it, you can never get out! No
matter what type of black hole you had, not even if you had a spaceship capable of accelerating
in any direction at an arbitrarily large rate.
It turns out that General Relativity is a very harsh mistress, particularly when it comes to black
holes. It goes even deeper than that, mind you, and it's all because of how a black hole bends
spacetime.
(Image credit: Adam Apollo.)
When you're very far away from a black hole, spacetime is less curved. In fact, when you're
very far away from a black hole, its gravity is indistinguishable from any other mass, whether
it's a neutron star, a regular star, or just a diffuse cloud of gas.
The only difference is that instead of a gas cloud, star or neutron star, there will be a completely
black sphere in the center, from which no light will be visible. (Hence the "black" in the
moniker "black holes.")
(Image credit: Astronomy/Roen Kelly, retrieved from David Darling.)
This sphere, known as the event horizon, is not a physical entity, but rather a region of space --
of a certain size -- from which no light can escape. From very far away, it appears to be the size
that it actually is, as you'd expect.
(Image credit: Cornell University.)
For a black hole the mass of the Earth, it'd be a sphere about 1 cm in radius, while for a black
hole the mass of the Sun, the sphere would be closer to 3 km in radius, all the way up to a
supermassive black hole -- like the one at our galaxy's center -- that would be more like the size

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of a planetary orbit!
From a great distance away, geometry works just like you'd expect. But as you travel, in your
perfectly equipped, indestructible spacecraft, you start noticing something strange as you
approach this black hole. Unlike all the other objects you're used to, where they appear to get
visually larger in proportion to the distance you are away from them, this black hole appears to
grow much more quickly than you were expecting.
(Image credit: Ute Kraus, Physics education group Kraus, Universitat Hildesheim.)
By time the event horizon should be the size of the full Moon on the sky, it's actually more than
four times as large as that! The reason, of course, is that spacetime curves more and more
severely as you get close to the black hole, and so the "lines-of-light" that you can see from the
stars in the Universe that surround you are bent disastrously out of shape.
Conversely, the apparent area of the black hole appears to grow and grow dramatically; by time
you're just a few (maybe 10) Schwarzschild radii away from it, the black hole has grown to
such an apparent size that it blocks off nearly the entire front view of your spaceship.
(Image credit: Andrew Hamilton, who has some great visuals here.)
As you start to come closer and closer to the event horizon, you notice that the front-view from
your spaceship becomes entirely black, and that even the rear direction, which faces away
from the black hole, begins to be subsumed by darkness. The entirety of the Universe that's
visible to you begins to close off in a shrinking circle behind you.
Again, this is because of how the light-paths from various points travel in this highly bent
spacetime. For those of you (physics buffs) who want a qualitative analogy, it begins to look
very much like the lines of electric field when you bring a point charge close to a conducting
sphere.
(Image credit: J. Belcher at MIT.)
At this point, having not yet crossed the event horizon, you can still get out. If you provide
enough acceleration away from the event horizon, you could escape its gravity and have the
Universe go back to your safely (asymptotically) flat spacetime. Your gravitational sensors can
tell you that there's a definite downhill gradient towards the center of the blackness and away
from the regions where you can still see starlight.
But if you continue your fall towards the event horizon, you'll eventually see the starlight
compress down into a tiny dot behind you, changing color into the blue due to gravitational
blueshifting. At the last moment before you cross over into the event horizon, that dot will
become red, white, and then blue, as the cosmic microwave and radio backgrounds get shifted
into the visible part of the spectrum for your last, final glimpse of the outside Universe.
(Image credit: ZetaPrints.com.)

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And then... blackness. Nothing. From inside the event horizon, no light from the outside
Universe hits your spaceship. You now think about your fabulous spaceship engines, and how
to get out. You recall which direction the singularity was towards, and sure enough, there's a
gravitational gradient downhill towards that direction.
But your sensors tell you something even more bizarre: there's a gravitational gradient that's
downhill, towards a singularity, in all directions! The gradient even appears to go downhill
towards the singularity directly behind you, in the direction that you knew is opposite to the
singularity! How is this possible?
(Image credit: Cetin Bal.)
Because you're inside the event horizon, and even any light beam (which you could never
catch) you now emitted would end up falling towards the singularity; you are too deep in the
black hole's throat! What's worse is that any acceleration you make will take you closer to the
singularity at a faster rate; the way to maximize your survival time at this point is to not even try
to escape! The singularity is there in all directions, and no matter where you look, it's all
downhill from here.
Like I said, General Relativity is a harsh mistress, particularly when it comes to black holes.
(* -- This is all done for a non-rotating, or Schwarzschild black hole. Other forms of black
holes are similar, but slightly different, and much more complicated, quantitatively.)
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Comments
1
I can see the headline now

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Scientist announces existence of 'perfectly equipped, indestructible spacecraft', itself to be used
in planned manned mission to a blackhole
Posted by: Justicar | May 10, 2012 9:25 PM
2
Nice account, but I can't help trying to pick some nits . . .
changing color into the blue due to gravitational blueshifting.
Are you sure there will be blue shifting even if I'm in free fall as I cross the horizon? If I'm
accelerating to remain a fiducial observer, then I'll see blue shift -- but if I'm freely falling in,
then the only thing I'd notice would be tidal effects (which can be arbitrarily small for a
sufficiently large black hole).
From inside the event horizon, no light from the outside Universe hits your spaceship.
Unless I'm missing something, this claim isn't quite right either. Look at a Penrose conformal
diagram of a black hole. It's obvious that the null rays still intersect the world line of an
infalling observer inside the event horizon. Seems to me that we still get to see what's going on
in (some of) the outside world even when we're inside (until we go squish, of course).
And while I assume that you're right that accelerating will reduce my proper time, it would
allow me to see more of the outside world (blue-shifted, of course) before I go squish.
Posted by: Physicalist | May 10, 2012 9:55 PM
3
Ethan, you say "For a black hole the mass of the Earth, it'd be a sphere about 1 cm in radius,
while for a black hole the mass of the Sun, the sphere would be closer to 3 km in radius..." Is
there a minimum mass needed to form a singularity? Can 1 cm black holes be formed in nature?
Posted by: Wesley Dodson | May 10, 2012 11:02 PM
4
Wesley, what you're asking about is called a quantum black hole, and theoretically they could
have been made during the Big Bang, when the titanic forces involved could have compressed
any size mass into a black hole. However, as black holes slowly "evaporate" through Hawking
radiation, they eventually shrink, and the smaller ones shrink faster. So a small black hole
would probably be gone by now, 14+ billion years later. Otherwise, I think it takes a star a
couple times larger than our sun to make a black hole.
Posted by: Artor | May 10, 2012 11:50 PM
5
It might deserve mention that it's only totally black, as you say, if you are the only thing falling
into the black hole. If anything emitting light is falling in ahead of you, you'll see that, at least
until you get very near the central singularity.
Something that I didn't really appreciate until recently is, when you start on this journey, the
event horizon is an infinite proper distance away. You reach it in finite time because as you fall
in, you gain momentum, and the fiducial distances are Lorentz-contracted in such way that you

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cover infinite proper distance in finite proper time. (From the perspective of a distant observer,
you would still have to travel infinite proper distance, but your increasing speed grants you the
benefit of a diverging time dilation factor, which is how she understand that you will reach the
horizon in finite proper time on your clock even as it takes infinite time on hers.) I say this
because I think the imagery is vivid yet I've never heard it used in a popular description.
Posted by: Matty | May 11, 2012 12:30 AM
6
Wouldn't the light falling in behind you become red shifted? The light once you're past the
event horizon from an object nearer than infinity would still, for a while, be visible to you, it's
just that the universe would get SMALLER (rather than 15 bn light years, it would drop to
5Bn, then 1Bn, then 1 light year, then 1 meter, since nothing further away would be "visible" to
you, and likewise visible in return).
Although the geodesics of highly curved surfaces may ensure that this "obvious" flat-world
view doesn't hold. When entirely within the horizon, where all geodesics lead toward the
singularity, there would be nowhere for light to come from except from something else inside
that volume. And so that, if emitting, would be receding quickly and therefore redshifted.
But there would be nothing available from outside the range of the (now closed) geodesics,
therefore "black".
Posted by: Wow | May 11, 2012 7:12 AM
7
The gradient even appears to go downhill towards the singularity directly behind you, in the
direction that you knew is opposite to the singularity! How is this possible?
Because you're inside the event horizon..."
:D Care to elaborate?
Posted by: Uli | May 11, 2012 7:21 AM
8
I too am disturbed by the blithe assertion that the Universe behind you is blueshiftef to infinity.
If you are freely-falling, there should be no shift at all, either way (I think). Could you
elaborate?
Posted by: Bernard Gilroy | May 11, 2012 7:26 AM
9
":D Care to elaborate?"
He already did :-P
"Because you're inside the event horizon, and even any light beam (which you could never
catch) you now emitted would end up falling towards the singularity; you are too deep in the
black hole's throat! What's worse is that any acceleration you make will take you closer to the
singularity at a faster rate; the way to maximize your survival time at this point is to not even try

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to escape! The singularity is there in all directions, and no matter where you look, it's all
downhill from here."
Posted by: Wow | May 11, 2012 8:39 AM
10
I don't understand how time-dilation would behave upon close approach to the singularity. Is it
not true that a body would never be able to reach the singularity, due to the nature of time
changing in relation to light speed? In that case, nothing around you would change as you fell,
since time would change as well, making things seem rather normal.
On another note, I believe (correct me please) Hawking's theory on the evaporation of black
holes has been revised. See Leonard Susskind's writings on the subject, namely The Black
Hole Wars.
Posted by: jim | May 11, 2012 9:09 AM
11
"He already did :-P"
I read the rest of the paragraph, but i don't see, how this could affect gravity gradients ....
Posted by: Uli | May 11, 2012 9:57 AM
12
Is it not true that a body would never be able to reach the singularity, due to the nature of time
changing in relation to light speed?
To an observer remaining far from the event horizon, it takes an infalling body an infinite
amount of time to reach the event horizon. In the frame of the infalling body, it takes a finite
amount of time. Mathematical explanation: the coordinate transformation (specifically the
timelike coordinate) between the two frames blows up at the event horizon.
I'm less sure about the blueshifting which others have commented on. I understand the
argument that, since your speed is approaching c (assuming you have made no attempt to slow
down) as you approach the event horizon, this should produce a redshift. OTOH, these
photons are also gaining energy as they fall inward, which would be the source of the blueshift
mentioned in the original post. I haven't done the math, so I'm not sure whether one effect wins
out over the other or whether they cancel exactly.
Posted by: Eric Lund | May 11, 2012 10:16 AM
13
"I don't understand how time-dilation would behave upon close approach to the singularity"
From the POV of the fallee (yes, I made a cromulent word!), there is no time dilation.
From the outside, past the event horizon, there can be no singularity seen.
When talking about extreme science, you have to be VERY precise in your definitions!

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re: 11, then you should have said what you wanted elaborated. The reason for the "because"?
The reason for the gravity lines to all bend toward the singularity? What?
Posted by: Wow | May 11, 2012 11:00 AM
14
"this should produce a redshift. OTOH, these photons are also gaining energy as they fall
inward, which would be the source of the blueshift mentioned in the original post"
Energetically, a photon "falling down" toward you will gain energy from the gravitational field
and be blueshifted (since this is the only way a photon can express a higher kinetic energy).
However, you are accelerating away from the source, therefore redshifting.
Which wins?
Well, since Ethan can actually DO the maths, I'd take his word for it as long as he'd noted the
confusion over the tussle here. Blueshift for a non-accelerating body in/near the event horizon
would see blueshifting.
But if you're freefalling, you're accelerating.
So does that change things?
What about something "further in" the hole that is emitting? Is that redshifted too, or does this
discussion not discuss help signals from other doomed astronomers, falling to their doom? Is
that where we're getting confused here?
Posted by: Wow | May 11, 2012 11:05 AM
15
What I don't understand, or rather, what I really don't understand, is why is Hawking radiation
so special that it can "escape" the black hole, where nothing else can? Also, why can't we detect
Hawking radiation? Seems it would be a real easy way to find black holes.
Posted by: Marty Lovik | May 11, 2012 11:20 AM
16
Why should a black hole be a real point-like object, and not just be normal space, like the eye of
a hurricane that is a region of mostly calm weather, and around which all pressure and density
circulates. We aren't falling straight into the sun so why should it be the case for light and a
black hole? Just a regular fast rotating matter processing transit-zone, just like any other
swirling whirlpool / tornado.
http://upload.wikimedia.org/wikipedia/commons/5/52/Hurr_cross.jpg
Posted by: Chelle | May 11, 2012 11:32 AM
17
Marty @16: As I understand it, Hawking radiation is related to virtual particles. Energy and
time obey a similar uncertainty relation to that of position and momentum, so that one can
temporarily create a particle-antiparticle pair as long as they annihilate within a time interval

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inversely proportional to the particle energy. The trick is that near the event horizon of a black
hole, one of these virtual particles can escape while the other one is sucked back into the event
horizon. Result: the escaped virtual particle is now a real particle, and the mass of the black hole
is that much lower. The probability that a virtual particle will escape is inversely related to the
mass of the black hole, so stellar mass black holes won't disappear by this route (at least over
reasonable time scales).
Chelle @16: The reason your analogy does not work is because maintaining an orbit just
outside the event horizon requires a speed higher than c, which is not allowed. Once you fall
inside the radius of the innermost circular orbit (which turns out to be 1.5 Schwarzschild radii),
you are doomed to fall into the black hole if you do not have a sufficiently powerful propulsion
system. It's true that the singularity need not be a point (I believe that for rotating black holes
the solution is a ring rather than a point). There has also been speculation about black holes
being wormholes, but so far that is purely speculation.
Posted by: Eric Lund | May 11, 2012 12:03 PM
18
Mostly Correct- : -If photons (light) have no mass as many areas of physics believe then you
would not be able to see while traveling the speed of light, much less see light comming from
outside of the ship while entering the even horizon. (I suspect it would be nearly impossible or
difficult to see while traveling even near the speed of light.)
Because photons have no mass it will not be like dropping a baseball while moving in a car (or
even the spaceship) where the ball would 'appear' to fall straight down to the floor of the car or
ship. A ball has mass, and is moving in perspective to ones observation. Photon's have no
mass and would therefore not follow this same rule of physics. And because of this, simply put
and assuming that you could: if you move at or faster than light then light will not be able to
reach your eyes for you to see anything.
This would be especially true for light from coming from outside of the ship if it uses some
sort of "bubble warp" like in *cough* the Star Trek theory around ship. Sorry for the Star Trek
reference, but their theory is valid if a similar method is used. Meaning light produced from
within the ship would remain in observable perspective to anyone else in the ship. Therefore if
this be the case one could see only things from within the ship. However, if it were possible it
were possible to travel at or faster than light and no bubble "warp" method were used then
you'd not be able to see outside or inside of the ship either. It would be perfectly dark.
Posted by: Brian Moote | May 11, 2012 12:41 PM
19
Thanks Eric-
I can kind of grasp the Particle/Anti particle concept. Is that the particles that manage to escape
are so few that we can't detect the radiation? Sounds like what you are saying is that the larger
the balck hole, the smaller the amount of Hawking radiation it emits.
Posted by: Marty Lovik | May 11, 2012 1:03 PM
20

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"Sounds like what you are saying is that the larger the balck hole, the smaller the amount of
Hawking radiation it emits."
It emits more.
The "surface" is more curved, therefore there is more chance of capture of one of the pair
without the other.
Entropically, the surface area of the event horizon defines the entropy of the black hole
(inversely), therefore the smaller the sphere that it makes, the more entropy it has, the higher the
"temperature" it has and the more energy it radiates (Stephan's law).
Odd way to think, but it's an excellent display of using a consequence of a scientific theory to
"prove" that theory (prove in the old sense of "test").
Posted by: Wow | May 11, 2012 1:14 PM
21
"What I don't understand, or rather, what I really don't understand, is why is Hawking radiation
so special that it can "escape" the black hole, where nothing else can?"
Anything outside the event horizon can escape, anything within cannot.
Hawking radiation is no different.
It's a consequence of just that feature in virtual particles.
If they appear ON the event horizon (the centre of action is on that geometric plane), then one is
within the event horizon and one is outside the event horizon.
One cannot get out, the other can.
If the one that *can* get out goes in instead, then it doesn't radiate at all.
If the one that *can* get out goes a different direction, then the partner cannot get there and the
virtual pair cannot annihilate.
And therefore that lucky partner is real and can go anywhere.
It never was inside the event horizon.
Its partner was not so lucky.
Posted by: Wow | May 11, 2012 1:18 PM
22
Is that the particles that manage to escape are so few that we can't detect the radiation?
If by "few" you mean relative to other possible sources of such particles, then yes, that is why
we can't detect Hawking radiation. I haven't crunched the numbers to see what kind of
experiment would be needed to get good enough statistics on this (Ethan might have a better
idea), but for stellar mass black holes the resource requirements (in terms of detector area and

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time) are likely to be well beyond what is currently feasible.
Yes, the rate of Hawking radiation emission decreases as black hole mass increases. There is a
narrow range of quantum black holes we might be able to detect--the ones that are just finishing
their evaporation now--but anything smaller than that will be already gone, and anything much
larger than that won't be emitting enough to be detectable. And that assumes that such black
holes actually exist; at present we have no evidence either for or against this proposition.
Posted by: Eric Lund | May 11, 2012 1:32 PM
23
I can't even come close to understanding the physics but what he describes is eerily similar to
some of my relative's homes!
Posted by: Jeff Jones | May 11, 2012 2:25 PM
24
@17 - Eric Lund
The reason your analogy does not work is because maintaining an orbit just outside the event
horizon requires a speed higher than c, which is not allowed.
Who says that photons can't be just squashed into Gamma Rays at that pressure area just
outside the event horizon, generating those bubbles that are up and bellow our Milky Way.
There would be no need to have something within the event horizon, just regular empty space.
There is no logic in stuff falling straight into a gravity hole, everything always circulates
around...
http://tinyurl.com/gamma-ray-emitting-bubbles
Posted by: Chelle | May 11, 2012 3:03 PM
25
In an addition to my previous post: I'd like to add a comment for those who believe photons do
have mass:
Regardless to whether they do or do not. If you out-run or go at or near the same speed as
photons "light" then the photons will never hit your eyes (hit them properly at near/same
speeds) for you see any objects in your ship or the light from stars as well. This is true
regardless as to whether photons have mass or not and regardless to the ball being dropping
inside of a moving vessel analogy that I used.
Posted by: Brian Moote | May 11, 2012 4:27 PM
26
I am no physicist, so these questions may seem simple, but I am curious. Black Holes emit jets
of radiation from their centers. Are these jets fixed in their relation to the event horizon? If so,
does that mean that a black hole can be approached from different angles that would mean
differences in the expected effects on the observer? What if, for example, I approach a black
hole from behind it? If I had a starship that could bridge the distance between us and a black
hole, then certainly I could plot a course beyond it - not through, naturally - and then approach
it from that angle. What, if anything, would be different? Would the jet of radiation be pointed

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