# Falling into a black hole "paradox"

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An astronaut falls into an extremely large Schwarzschild black hole, so large that they don't notice any spaghettification-like effects. Their head can see their feet the entire time, and their helmet is constantly sending information to Earth, say. At some point, the astronaut sends a message basically saying "My feet are now inside the event horizon", and Earth eventually receives that message.

Is there a mistake with this scenario? What prevents the astronaut's head outside the event horizon from seeing their feet inside the horizon, and how does that look to them if they're watching their feet the entire time?

I have an answer in mind but I'm curious if there's any room for experts to disagree on a resolution, and also curious if fellow amateurs can resolve it with about a pop-sci level understanding. I'm interested in how different the answers might be, so please share if you have an opinion!

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You're still not reading the posts. Nobody is suggesting light travelling from the interior to the exterior, or for that matter, in any direction that doesn't head steadily in the direction of the sig

Consider an arbitrary event located directly on the surface in question, and attach a light cone to that event. Now look at the tangent space to the surface at that event. If the surface is like-like,

One of the easier visualisations is, IMO, a Kruskal diagram. Null paths in it are 45 degree lines, and the event horizon is one such. The axes of local inertial frames are (very small) x shapes, squis

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Posted (edited)

The head cannot have measured the feet having entered the black hole since the head always measures the feet in the past.  So sure, he can send that message, meaning the feet are in there 'now' in the falling frame, but he is sensing the part of the feet worldline still outside the horizon.

Edited by Halc
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Posted (edited)

Taking the PoV of the feet first, infalling astronaut ( proper, where the clock is attached and moves along  his worldline ), I would think, as he approaches the EH, he would see it grow disproportionately larger, as it rises up like a 'cup' to envelop him, eventually 'closing off' behind him, due to the strongly curved spacetime.
This is all before any part of himself crosses the EH. Just before he is fully enveloped by the EH, and most certainly after, he would lose sight of anything ahead of himself, such as his feet, as light has no way of coming 'back' to his eyes.
It would be a dark journey to his future doom.

Edited by MigL
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49 minutes ago, MigL said:

I would think, as he approaches the EH, he would see it grow disproportionately larger, as it rises up like a 'cup' to envelop him, eventually 'closing off' behind him, due to the strongly curved spacetime.

They have videos of this point of view, and the most notable part of it is that nothing particularly weird happens at the event horizon. You can't tell when you cross over.

Just before he is fully enveloped by the EH, and most certainly after, he would lose sight of anything ahead of himself, such as his feet, as light has no way of coming 'back' to his eyes.

No, you never lose sight of your feet. Everything is pretty normal until the tidal forces start to rip you apart, and for a very large BH, that's well after you cross over. You can still see in all directions in 3D, and you can see the stars in the universe you've left behind, however much you can no longer reach them.

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44 minutes ago, Halc said:

No, you never lose sight of your feet.

If you enter a black hole feet first being able to see your feet would imply that photons move against gravity from beyond the event horizon to reach the eyes of the individual falling in. Unless I misunderstand the thought experiment that does not sound correct. Photons beyond event horizon can't travel in the direction away from the black hole's centre.

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Posted (edited)
2 hours ago, Ghideon said:

If you enter a black hole feet first being able to see your feet would imply that photons move against gravity from beyond the event horizon to reach the eyes of the individual falling in.

No it does not. By the time the photon from the feet meets your eyes (a few nanoseconds later, eye-time), those eyes are long since beyond the horizon.  It's not like the head is hovering on the outside. That certainly would kill you, but at least prevent you from seeing your severed feet.

Edited by Halc
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Light has no path ( geodesic ) that leads away from the center of the EH, or outside of the EH.
We are not discussing tidal forces as md65536 specified an extremely large BH where tidal forces at the EH are trivial; and they have nothing to do with 'seeing' your feet ahead of you.

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This is straight up Einstein's equivalence principle which says that in a small enough box (effectively one where tidal effects are not significant), one cannot tell free-fall in a gravitational field from inertial motion in no gravity. That means there's no obvious effects as you cross the event horizon. You still see your feet, and can go about your business just like before just like anybody falling such as on say the ISS.

Nobody is suggesting light traveling from inside the EH to the outside. I never suggested it. Read my first post above and understand what I'm saying. The guy could be broadcasting a video from a head-cam so everybody can watch, and sure enough, the feet are visible in that broadcast, but the feet seen are not yet inside the EH.  The light was emitted outside the EH horizon and headed towards the camera. You don't see your feet now, you see them about 6 nanoseconds ago, and nobody seems to realize that, or they're assuming that the head is somehow hovering outside the EH, which would require enough proper acceleration to indeed make your feet invisible, not to mention detached. No, the guy is falling and notices absolutely nothing change as he crosses over.

The feet inside the EH also emit light, and that light also reaches the camera, but not before the camera is inside the EH.  The broadcast from the head-cam never reaches the outside observer. So the guy falling in notices absolutely nothing amiss as he crosses the EH. The distortion of spacetime will eventually get noticed (arms getting ripped out of shoulders, etc), but we're positing a large BH here with a still reasonably uniform gravitational field at the EH.

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52 minutes ago, MigL said:

Light has no path ( geodesic ) that leads away from the center of the EH, or outside of the EH.
We are not discussing tidal forces as md65536 specified an extremely large BH where tidal forces at the EH are trivial; and they have nothing to do with 'seeing' your feet ahead of you.

Assuming the black hole is big enough, It's not necessary. You see your feet consistently the whole time, simply by catching up. When your feet enter the black hole you don't see that version of the feet until you get there.

We could be inside the event horizon of a large enough black hole right now, so large there is nothing to tip us off. So large the gradient is insignificant to our lives, and the Universe as we know it.

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I agree with Halc. When the feet cross the EH, the photons making up the image of that, and directed outward, forever remain at the EH, which is a lightlike surface. It's stationary in the very distant observer's coordinates, but locally moves at the speed of light. The astronaut notices nothing unusual because the EH and image move past its head at the speed of light, just like photons from the feet do in usual circumstances.

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Posted (edited)

I will have to disagree with you gentlemen.
Light does not stand still, even inside the EH, so there is no 'catching up' to the light emitted by your feet.
As soon as your feet cross the EH, light emitted inexorably moves towards its future, the center of the BH.
There is no path backwards to your eyes, and you cannot catch up to it.

As for md's assertion, any light that is 'frozen at the EH, is red -shifted to infinite wavelength long before it can reachthe head which is still outside the EH, even for such a miniscule distance.

So if no light can possibly reach your eyes, what do you see ???

Sorry JC, we cannot be inside the EH of a BH right now.
The geodesics that light ( or anything else ) must follow inside the EH, have only one direction, towards the future singularity.
That is what GR predicts, f this happens to conflict with observation, I will stand corrected.

Edited by MigL
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Posted (edited)

Is the tiny distance between the head and the feet a misleading example?

Would whatever is seen,(or not seen) be  essentialy the same as if the distance was the same as that between any object and an observer  say 1,2 or 1000 light seconds away?

So the object crosses the EH in its own timeframe  and the observer sees it up until it crosses the EH in the observer's timeframe.

And the timeframes differ.

Edited by geordief
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Posted (edited)
1 hour ago, MigL said:

Light does not stand still, even inside the EH, so there is no 'catching up' to the light emitted by your feet.
As soon as your feet cross the EH, light emitted inexorably moves towards its future, the center of the BH.
There is no path backwards to your eyes, and you cannot catch up to it.
So what do you see ???

Regardless of what happens inside the BH, the original scenario doesn't involve any light from inside the EH whatsoever. It doesn't say anything about what happens after some moment when the astronaut's head is less than a body's length from the EH, still outside, still only receiving light from outside the EH.

But, I think you're also wrong about what happens inside the Schwarzschild BH. Light inside can still move in different directions. Draw some light cones on a diagram. Yes, all light cones inside the BH will be tilted so that all light paths head toward the singularity, but the light cone from an event inside the EH can still intersect the worldline of an infalling particle. The lightcones do not instantly tilt an extra 45 degrees or narrow and degenerate into a single line, the instant you cross the EH.

56 minutes ago, geordief said:

Is the tiny distance between the head and the feet a misleading example?

Would whatever is seen,(or not seen) be  essentialy the same as if the distance was the same as that between any object and an observer  say 1,2 or 1000 light seconds away?

So the object crosses the EH in its own timeframe  and the observer sees it up until it crosses the EH in the observer's timeframe.

Basically. Any observer outside the EH will see the object approach the EH but never appear to cross it. The exact measurements will be different for different observers. My example was basically two observers who both fall in to the BH.

But sure, you could make the head and feet two ships arbitrarily far apart, and just make the BH big enough. Actually this thought experiment started from trying to figure out what would happen if two ships at relative rest were flying far from a BH so large that they couldn't detect it, such as J.C.MacSwell suggested, but if one ship ended up inside the EH and one was outside. I think either both ships enter the BH and don't notice, or the ships necessarily separate (head rips off), as the outside ship has to escape the inside ship if it is to escape the EH.

It causes problems thinking of the BH as a giant sphere whose event horizon can hover harmlessly between the two ships, or between head and feet. It is a lightlike surface. If the head and feet see each other, or the ships are in communication, the EH between them is moving as fast as they can communicate.

Edited by md65536
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2 hours ago, md65536 said:

It is a lightlike surface.

It’s a null surface, actually.

2 hours ago, md65536 said:

what would happen if two ships at relative rest were flying far from a BH so large that they couldn't detect it, such as J.C.MacSwell suggested, but if one ship ended up inside the EH and one was outside.

The geometry of spacetime below the horizon is such that no stationary frames exist - in other words, no matter how much radial thrust the engines of the unfortunate ship put out, it will continue to experience radial decay as it ages into the future. So the two ships couldn’t remain at relative rest.

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Posted (edited)
1 hour ago, Markus Hanke said:

It’s a null surface, actually.

Isn't that the same thing?

1 hour ago, Markus Hanke said:

The geometry of spacetime below the horizon is such that no stationary frames exist - in other words, no matter how much radial thrust the engines of the unfortunate ship put out, it will continue to experience radial decay as it ages into the future. So the two ships couldn’t remain at relative rest.

Yes, but if both ships fell in, they could remain close enough to relative rest to not notice, for some time. Just like a person falling into a large enough black hole doesn't feel themselves being pulled apart for some time, even if tidal forces are always present (though negligible here). I'm talking about a black hole so immense that the gravitational effects are very weak at r_s, like an r_s of tens of billions of lightyears. If that's too small, maybe a billion times that! If you make it big enough, you should be able to continue falling toward the singularity for a lifetime without ever noticing you're trapped in a black hole. All the rules still apply, you can't communicate with anyone outside the EH, but the EH has expanded beyond you at the speed of light, and anyone you try to communicate with who hasn't escaped, is inside with you.

I can't say what would be observed on the inside of a BH, but my understanding is that accepted theory does not predict any strange local effects of crossing the event horizon, like suddenly being unable to communicate with or see "nearby" infalling particles.

By "paradox" I meant it's not a real paradox, just a conceptual problem to resolve.

Edited by md65536
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Why exactly do people believe that going into a black hole is like entering stretched space?  Time dilation dictates that time runs slower near massive objects.  With space and time being linked, how does slowing down time, also create an effect of a hole that is stretching space as you fall into it?   DO you imagine there being very much space, because it takes so long to get though it?  Are you imagining an inverse relationship between time and space or spacetime?  I dont believe the continuum runs with an inverse.  So if you slowed down time, wouldn't you also need to make less space in keeping with non inverse relativity?  Or is time now running slower, because space is stretched?  IS that where im getting confused?  I know that time runs slower near massive objects.  How does that mix with this reality of falling into a black hole and stretching in space.  A dip in the fabric creates a hole we fall into?  No, a concentration of energy, defined by the amount of compression exerted upon it, changes spacetime and the effect is called gravity and it seems to cause attraction.  But if that massive energy is truly just taking the value away from time, by slowing it down, as it loses strength as it extends outwards,  this bending space time all around it will curve light and even changes the definition of what we experience as a straight line.  It is literally bending spacetime, in concentric rings of compression, which become less compressed the farther away from the massive energy that you get.  And then time speeds up.  Less compressed the farther away you get.  Less compressed is stretched, and it happens the farther away you get from a black hole.  So a black hole must create a compression the closer you get, which is why time does run slower, and they all think space gets bigger?  Or do you just look stretched, to someone in a less compressed reality.

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5 hours ago, MigL said:

I will have to disagree with you gentlemen.
Light does not stand still, even inside the EH,

So if no light can possibly reach your eyes, what do you see ???

2 hours ago, Markus Hanke said:

It’s a null surface, actually.

The geometry of spacetime below the horizon is such that no stationary frames exist -

My understanding is as the two members above. Once inside the BH's EH, all paths lead to the singularity, so a person falling in feet first would not get to see his feet.

I'll throw something else in for debate...light/photons that are emitted just fractionally this side of the EH? All that light/photons not emitted directly radially away from the EH, will eventually arc back to fall into the BH. But any photons emitted directly radially away will always remain just outside the EH, never falling in and never quite getting away.

Perhaps some info here may help.....https://jila.colorado.edu/~ajsh/insidebh/index.html

In actual fact, things probably get more complicated as realistcally any BH would be a Kerr type.

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19 hours ago, md65536 said:

An astronaut falls into an extremely large Schwarzschild black hole, so large that they don't notice any spaghettification-like effects. Their head can see their feet the entire time, and their helmet is constantly sending information to Earth, say. At some point, the astronaut sends a message basically saying "My feet are now inside the event horizon", and Earth eventually receives that message.

Is there a mistake with this scenario? What prevents the astronaut's head outside the event horizon from seeing their feet inside the horizon, and how does that look to them if they're watching their feet the entire time?

I have an answer in mind but I'm curious if there's any room for experts to disagree on a resolution, and also curious if fellow amateurs can resolve it with about a pop-sci level understanding. I'm interested in how different the answers might be, so please share if you have an opinion!

And if the astronaut falls forward head first, then when the neck is at the level of the event horizon, it becomes impossible to transmit nerve impulses from the head to the heart and lungs, it turns out that the event horizon will work like a guillotine or a cholinesterase inhibitor, so what????

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5 hours ago, md65536 said:

By "paradox" I meant it's not a real paradox, just a conceptual problem to resolve.

Isn't that the definition of a "real paradox"?

We can never know...

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5 hours ago, JohnSSM said:

Time dilation dictates that time runs slower near massive objects.  With space and time being linked, how does slowing down time, also create an effect of a hole that is stretching space as you fall into it?

The slowing of time is only relative. An external observer measures the infalling clock slowing relative to its own, but the infalling observer measures their own local clock's proper time ticking at a rate of 1s/s. Their own clock does not stop as they fall into a BH.

If an infalling astronaut would lose sight of their feet, and smaller distances, this would imply that a light clock would stop functioning, or at least would stop depending on its orientation. Theory doesn't predict that.

5 hours ago, beecee said:

My understanding is as the two members above. Once inside the BH's EH, all paths lead to the singularity, so a person falling in feet first would not get to see his feet.

I'll throw something else in for debate...light/photons that are emitted just fractionally this side of the EH? All that light/photons not emitted directly radially away from the EH, will eventually arc back to fall into the BH. But any photons emitted directly radially away will always remain just outside the EH, never falling in and never quite getting away.

I really didn't expect there to be debate after a resolution was posted.

You still have light cones inside the EH, outside of the singularity. The cones do not become lines, where there is only one path for light from an event to the singularity. The path of light that goes from feet to head also leads to the singularity.

"Light beams aimed directly outward from just outside the horizon don't escape to large distances until late values of t." -- https://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/fall_in.html but they do escape.

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12 minutes ago, md65536 said:

You still have light cones inside the EH, outside of the singularity.

Isn't a singularity the wrong word to use for a black hole?

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Posted (edited)

I've attached an image showing light cones inside the horizon. The labelled events are:

A - Feet pass the EH.

B - Head passes the EH.

C - Astronaut twitches her feet

D - Astronaut sees her feet twitch

E - Photons from twitched feet reach the singularity.

The scale of this is out, the astronaut twitches her feet in the short time before the head has entered the BH. More realistically, the head's world line would be very close to the feet's.

The light cones inside the BH show that light from a single event does not take the same amount of time to reach the singularity, in all directions.

Also... I think I made a mistake, the lines I'm using as the head and feet worldlines appear to be light-like. Can you see from the diagram as-is how it should look, or should I post a correction? Also, if you still think the head can't see the feet inside a BH, please show how this would look with light cones.

Edited by md65536
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much like Event horizone, is the wrong word to use for a black hole...

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2 hours ago, md65536 said:

Also... I think I made a mistake, the lines I'm using as the head and feet worldlines appear to be light-like.

Yes, the worldline would be through the center of a given light cone, not the edge, such as a line connecting A directly through D.  I think your head and feet need to be a little closer together to illustrate the light going from one to the other. There are no inertial worldlines in your picture.

2 hours ago, dimreepr said:

Isn't a singularity the wrong word to use for a black hole?

1 hour ago, dimreepr said:

much like Event horizone, is the wrong word to use for a black hole...

Neither term means the same thing as 'black hole'.  It seems like a black hole is the collection of events enclosed by the event horizon, the event horizon being a boundary between events that can have a causal effect on a distant wordline and events that cannot.

As for singularity, there is a coordinate singularity at the event horizon in some coordinate systems (not the one md65536 shows), and there is a physical singularity inside the black hole, the nature of which varies between say Schwarzschild BH, Kerr BH, or other metrics.

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15 hours ago, MigL said:

Sorry JC, we cannot be inside the EH of a BH right now.
The geodesics that light ( or anything else ) must follow inside the EH, have only one direction, towards the future singularity.
That is what GR predicts, f this happens to conflict with observation, I will stand corrected.

They have multiple directions, just none of them can include a vector in the opposite direction of the future singularity.

I don't think that GR conflicts with my take on this. The larger the BH the more normal things would seem as you cross the EH. Large enough and you can't observe any difference from what we observe right now, in which case neither of us would likely ever stand corrected. Nothing could substantially change in our lifetime.

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