# Falling into a black hole "paradox"

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1 hour ago, Halc said:

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.

I agree, that would be better. Wouldn't the shape and orientation of the light cones be independent of the motion of the light source?, while the shape of the source's world line would be entirely dependent on it. So not all world lines would be pointed locally along the axis of the cone. I think the image is a fair representation of the world lines of inertial particles entering the BH at near the speed of light, which is not at all what I wanted to show.

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12 minutes ago, md65536 said:

Wouldn't the shape and orientation of the light cones be independent of the motion of the light source?, while the shape of the source's world line would be entirely dependent on it. So not all world lines would be pointed locally along the axis of the cone. I think the image is a fair representation of the world lines of inertial particles entering the BH at near the speed of light, which is not at all what I wanted to show.

Correct. You could go in spinning for instance, so feet/head worldlines switch sides repeatedly, taking turns being 'in the lead' so to speak. You could have two things orbiting the BH and slowly spinning in with high but opposing relative velocity.

The line defining the center of the light cone is frame dependent, so it's always in the center relative to the frame of the falling thing I think. There's not one unique worldline that the one at the center of a given event's light cone, but the BH itself sort of defines an obvious symmetrical frame where the object drops straight in instead of the more likely spiral.

Thus there's no worldline of a particle near the speed of light since in the frame of any falling thing, it is stationary with a worldline centered on its (local at least) light cone.

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6 hours ago, md65536 said:

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.

OK, let's first understand that I am absolutely no expert on BH's and SR/GR, but have read a fair bit. I still see some disagreement here as to what exactly happens, but I have heard and seen enough reputable sources to still insist as per

17 hours ago, MigL said:

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.

2 hours ago, J.C.MacSwell said:

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.

I see it as per those members, but again, essentially I'm here to learn, and if I have misunderstood something, then I would like it cleared up.

6 hours ago, md65536 said:

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.

I understand and agree with that. And I would add, that while a distant observer may never actually see anyone or anything actually cross the EH, ( just red shifted to infinity eventually fading from view) according to that person or thing doing the crossing, , it will cross the EH and head towards its eventual doom, depending on the size of the BH and the accompanied tidal forces. But also, both the observer and the person falling into the BH frames of reference, are valid.

6 hours ago, md65536 said:

The path of light that goes from feet to head also leads to the singularity.

Can you explain this bit further in simple language?

6 hours ago, md65536 said:

"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.

OK, I think I understand that...from your link:  "Light beams aimed directly outward from just outside the horizon don't escape to large distances until late values of t.  For someone at a large distance from the black hole and approximately at rest with respect to it, the coordinate t does correspond well to proper time".

But I must ask why? for some clarification. Please go easy with me...treat me like a virgin if you will. 😊 I'm only a poor old retired maintenance Fitter/Machinist/Welder.

Edited by beecee
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Maybe the confusion stems from the fact that we can't imagine not being able to see our own feet ...

Light, like any nformation, cannot travel to the exterior of the EH of a BH.
Gravity, or rather, the pre-existing gravitational field, does.
When we consider very large BHs, it is the differential gravity which leads to tidal forces, and since the differential gravity is small for very large radius, there is no spaghettification, and things are normal ( GRAVITATIONALLY ) for a laarge BH.
For a small BH, however, the differential gravity is very large because of the small radius, and the tidal forces will spaghettify.
That does not mean that everything else is also 'normal' on crossing the EH of a large BH.

Could a transmitter that is further inside the EH transmit a signal to your feet, which could then tranxmit it to your head ?
Could you have a chain of transmitters, all sending information to the next, all in close proximity, lowered all the way down to the center, and transmit information, in small steps, to your head, and out to distant observers ?

If you answer yes to any of the above, I congratulate you.
You have just 'discovered' a way to probe the singularity/center of a BH.

( i'm being sarcastic 🙂 )

the light cone of your feet is not the same as the light cone of your head.
Your feet have their own proper time, and will cross the EH, while your head is still an external observer.
Before reaching the EH, the light from your feet ) moving towards your head ) will be red shifted, and seemingly stop ( or rather, fade away ) at the EH.
It will be followed by the disappearance/stopping of your knees, thighs, hips , waist chest and shoulders; all fading away to infrared and nothingness before your head ( and eyes ) cross; and you see nothing else ahead of you.

I will grant you that, depending on your speed of radial infall, those frozen, red-shifted images will experience some blue-shift because of your infall speed, so you may 'crash' into a sort of frozen image at the EH ( but I would think not, as the images have their wavelength stretched to infinity at the EH, and you cannot be moving fast enough to blue-shift them back ).

Edited by MigL
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17 minutes ago, MigL said:

Maybe the confusion stems from the fact that we can't imagine not being able to see our own feet ...

Light, like any nformation, cannot travel to the exterior of the EH of a BH.
Gravity, or rather, the pre-existing gravitational field, does.
When we consider very large BHs, it is the differential gravity which leads to tidal forces, and since the differential gravity is small for very large radius, there is no spaghettification, and things are normal ( GRAVITATIONALLY ) for a laarge BH.
For a small BH, however, the differential gravity is very large because of the small radius, and the tidal forces will spaghettify.
That does not mean that everything else is also 'normal' on crossing the EH of a large BH.

Could a transmitter that is further inside the EH transmit a signal to your feet, which could then tranxmit it to your head ?
Could you have a chain of transmitters, all sending information to the next, all in close proximity, lowered all the way down to the center, and transmit information, in small steps, to your head, and out to distant observers ?

If you answer yes to any of the above, I congratulate you.
You have just 'discovered' a way to probe the singularity/center of a BH.

( i'm being sarcastic 🙂 )

the light cone of your feet is not the same as the light cone of your head.
Your feet have their own proper time, and will cross the EH, while your head is still an external observer.
The light from your feet will be red shifted, and seemingly stop ( or rather, fade away ) at the EH.
It will be followed by the disappearance/stopping of your knees, thighs, hips , waist chest and shoulders; all fading away to infrared and nothingness before your head ( and eyes ) cross; and you see nothing else ahead of you.

I will grant you that, depending on your speed of radial infall, those frozen, red-shifted images will experience some blue-shift because of your infall speed, so you may 'crash' into a sort of frozen image at the EH ( but I would think not, as the images have their wavelength stretched to infinity at the EH, and you cannot be moving fast enough to blue-shift them back ).

At the EH light can still go 90 degrees to the radial. Up to that point it can do a little better., but light can essentially "orbit" at the EH. You can still "catch up" to light emitted from your feet, even if it is already past the EH. Big enough BH and it's barely redshifted for you as you speed toward the ultimate demise of whatever your atomic particles that made you up well after your normal lifetime, oblivious to the fact you entered a BH.

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If it went 900, it wouldn't hit your eyes.

( so you still wouldn't see it )

Edited by MigL
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Just now, MigL said:

If it went 900, it wouldn't hit your eyes.

But if it's capable of 90 degrees...it's capable of "waiting" for you...with an imperceptible redshift in your accelerating frame if the BH is big enough.

My point wrt 90 degrees was your claim light at (or inside) the EH could only go straight to...(Hello to the singularity?)

😀

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40 minutes ago, MigL said:

Light, like any nformation, cannot travel to the exterior of the EH of a BH.

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 sigularity. md65536's picture shows all the light paths and none crosses from in to outside the EH.

11 minutes ago, J.C.MacSwell said:

but light can essentially "orbit" at the EH

Typo? Perhaps you meant to say that it cannot.  Light cannot orbit (a Schwarzschild BH) any closer than the photon-sphere which is 1.5 times the radius of the EH. It gets more complicated for something like a Kerr-Newman BH which is a more realistic model of actual black holes.

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Just now, Halc said:

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 sigularity. md65536's picture shows all the light paths and none crosses from in to outside the EH.

Typo? Perhaps you meant to say that it cannot.  Light cannot orbit (a Schwarzschild BH) any closer than the photon-sphere which is 1.5 times the radius of the EH. It gets more complicated for something like a Kerr-Newman BH which is a more realistic model of actual black holes.

Not a typo, but I could be conflating the EH with the photon sphere.

Thinking about it I obviously am. +1. I don't think that affects the rest of my argument. Light can still spiral inward at the EH,correct?

It doesn't need to track radially.

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I don't understand, JC.
The photon has to hit your eyes  (radially ) for you to see it.
If it moves tangentially, at the speed of light, it will not be there anymore when your eye gets there.

As for red shift, the wavelength of radially outgoing light is redshifted to infinity at the EH, as measured at any distance outside the EH.
That means it disappears; there is nothing imperceptible about it.
( or should that be perceptible ? )

14 minutes ago, Halc said:

Nobody is suggesting light travelling from the interior to the exterior,

Except the OP, where he says ...

"At some point, the astronaut sends a message basically saying "My feet are now inside the event horizon", and Earth eventually receives that message."

IOW, he 'sees' ( gets information from ) his feet, inside the EH, and relays that information arbitrarily far away, to Earth.

For simplicity's sake, I recommend considering only radial movement, and Schwarzschild BHs

Edited by MigL
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23 hours ago, Halc said:

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.

Ok, thanks for the reply! I'll read through the rest of the posts to see where my understanding is wrong.

23 hours ago, Halc said:

It's not like the head is hovering on the outside.

Hovering is not what I meant, but watching the feet continuously, including the brief moment while feet has passed event horizon and eyes have not.

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5 minutes ago, MigL said:

I don't understand, JC.
The photon has to hit your eyes  (radially ) for you to see it.
If it moves tangentially, at the speed of light, it will not be there anymore when your eye gets there.

As for red shift, the wavelength of radially outgoing light is redshifted to infinity at the EH, as measured at any distance outside the EH.
That means it disappears; there is nothing imperceptible about it.

Except the OP, where he says ...

"At some point, the astronaut sends a message basically saying "My feet are now inside the event horizon", and Earth eventually receives that message."

IOW, he 'sees' ( gets information from ) his feet, inside the EH, and relays that information arbitrarily far away, to Earth.

For simplicity's sake, I recommend considering only radial movement, and Schwarzschild BHs

Yeah, not with the Earth remaining outside the EH.

38 minutes ago, MigL said:

The photon has to hit your eyes  (radially ) for you to see it.
If it moves tangentially, at the speed of light, it will not be there anymore when your eye gets there.

Agree with both.

41 minutes ago, MigL said:

As for red shift, the wavelength of radially outgoing light is redshifted to infinity at the EH, as measured at any distance outside the EH.
That means it disappears

But it doesn't disappear immediately. It takes time, and the bigger the BH the longer it takes. In the mean time you move toward it, and overtake it.

Compare your reference frame to that of the BH centered on the singularity/future singularity. Neither are perfect inertial frames.

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My limited understanding has me agreeing with MigL, for whatever that's worth.

1 hour ago, Halc said:

Typo? Perhaps you meant to say that it cannot.  Light cannot orbit (a Schwarzschild BH) any closer than the photon-sphere which is 1.5 times the radius of the EH. It gets more complicated for something like a Kerr-Newman BH which is a more realistic model of actual black holes.

Agree with that particularly getting more complicated with the Kerr metric, where we have two EH's and two photon spheres each orbiting in opposite directions.

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15 minutes ago, beecee said:

My limited understanding has me agreeing with MigL, for whatever that's worth.

I'm tending to agree with Halc on this one...I don't think either of us is in bad company...

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re. "The path of light that goes from feet to head also leads to the singularity."

3 hours ago, beecee said:

Can you explain this bit further in simple language?

Sure, I'll use another diagram:

Here, the purple line could represent a path of light that goes from an event at the feet, and intersects the head's green world line, and ends up at the singularity.

The pink line represents light from an event, that is aimed 'directly' toward the singularity. It has a slope of 45 degrees, representing a coordinate speed of c. No object anywhere on this graph can have a slope shallower than 45 degrees.

Purple represents light from the same event, directed as close as you can get to away from the singularity. As you can see, it still reaches the singularity.

Blue represents all the locations at which light from the one event can reach the singularity, at different times.

Red represents an impossible object that "catches up to light" directed away from it. This is impossible, as has been stated a few times in this thread. No path can go from the right side of the pink line, to the left side of it.

Yellow represents some impossible objects travelling faster than light in the "opposite" direction. No path can go from the left side of the purple line to the right.

Green represents a possible world line for an object, that is traveling slower than light, and sees the event (when it intersects the purple line). Green could be the world line of the head. Even though it entered the BH after the feet, light from the feet can still reach it. Notice that at no point is there light moving away from the singularity. Notice that the possible paths for light from the event are not all the same, or same length. Their different slopes represents different coordinate speeds (in the graphed coordinates) of light. Inside the EH, all velocities in these coordinates are toward the singularity. The vertical alignment of light cones on the EH represents a coordinate speed of light equal to 0 for light directed away from the BH.

There is no "catching up to light" needed, because the light from the feet that intersect the head's world line is aimed *toward* the head, as well as toward the singularity.

1 hour ago, beecee said:

My limited understanding has me agreeing with MigL, for whatever that's worth.

MigL is simply wrong, using arguments that apply to a stationary or distant observer, applied incorrectly to an infalling observer.

We can all agree at least that it's possible to survive falling into a large enough black hole without tidal effects ripping you apart, correct? Writeups about this can be found everywhere with a simple google search. Is anyone arguing that those are all wrong?

What does it mean if you can survive falling through an EH intact, but you lose sight of your feet? You lose feeling from your feet, and your heart for that matter. You lose blood from your heart. You lose the half of your head underneath your eyeballs for that matter! What does it mean to survive intact, not be ripped apart by tidal forces, and yet be literally cut off from your body because light has suddenly taken on a completely different local behaviour?

2 hours ago, MigL said:

"At some point, the astronaut sends a message basically saying "My feet are now inside the event horizon", and Earth eventually receives that message."

IOW, he 'sees' ( gets information from ) his feet, inside the EH, and relays that information arbitrarily far away, to Earth.

No, you're purposefully neglecting that light takes time to travel from the feet to the head, about the same (in this scenario) time that it takes the EH to travel from the from the feet to the head.

This all implies the astronaut has a meaning for the use of the word "now". If the astronaut can say that the feet cross the EH "before" the head, then there is a meaningful point where it can say the feet are now inside and the head is not. If what the astronaut "sees" is also what the astronaut calls "now", then by necessity the head and feet passed the EH at the same time, and Earth does not receive that message. Either way, there is no paradox, there is no path of light from inside the EH to outside, and there is no point where the feet suddenly disappear because delay of light ceases to be a thing.

Edited by md65536
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3 hours ago, MigL said:

Except the OP, where he says ...

"At some point, the astronaut sends a message basically saying "My feet are now inside the event horizon", and Earth eventually receives that message."

IOW, he 'sees' ( gets information from ) his feet, inside the EH, and relays that information arbitrarily far away, to Earth.

That was the question in the OP, yes, and it has been answered. The feet that he sees are the feet from a few nanoseconds ago (falling frame), same as the feet you see now are not the present feet, but the feet in the past.  Those feet were outside the EH when the light was emitted from them.  I've said as much since my first reply in this thread.  A body does not experience itself in the present since it takes time for any information to travel from say feet to head.

So there's no anomaly in spacetime from the PoV of the falling guy. It's all normal physics to him, and he has no awareness of having crossed over unless he has a computer with him that informed him of the situation, since it is a calculable thing.  The view outside the window also doesn't change appreciably.  There is a section of the normal universe that occupies a percentage of his view, with the rest of the view black except for other sources of light falling in with him, such as his feet.  If 50 guys all fell in sort of with each other, not all at the same time or place, they'd all still be able to see each other at all times, but tidal effects would eventually alter their distribution.

5 hours ago, MigL said:

Before reaching the EH, the light from your feet ) moving towards your head ) will be red shifted, and seemingly stop ( or rather, fade away ) at the EH.

It is red shifted only from the perspective of something hovering outside. Our infalling guy is in a completely different frame of reference in which no red-shift is observed at all, at least not until the spaghettification begins to occur. Per the equivalence principle, he cannot perform a test that demonstrates when the EH is crossed. GR rests on that principle and would collapse if the descriptions you paint actually occurred.  My feet appear to me to be more red shifted here at my desk than would be seen by our guy falling in. This is because the falling guy is inertial and I am not.

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21 hours ago, md65536 said:

Isn't that the same thing?

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, the tangent space will fall to the interior of the light cone; if the surface is null, the tangent space will coincide with the surface of the light cone. So this isn’t the same - you can (at least in principle) escape from a light-like surface to infinity, but you can’t escape from a null surface.

21 hours ago, md65536 said:

Yes, but if both ships fell in, they could remain close enough to relative rest to not notice, for some time.

For some time, yes.

21 hours ago, md65536 said:

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).

This is true. I’d just like to point out that the non-existence of stationary frames below the horizon is not a consequence of tidal forces, but is due to the causal structure of spacetime; but you are right in that, for very massive BHs and just below the horizon, one could remain very nearly stationary for some time.

21 hours ago, md65536 said:

I can't say what would be observed on the inside of a BH

General relativistic optics is a notoriously tricky subject, so I won’t speculate on this too much, also because it would in some ways depend on how exactly you move once you are below the horizon. In principle though, for very massive BHs and just below the horizon, there shouldn’t be any extraordinary visual effects, other than some blue-shifting of distant stars.

21 hours ago, md65536 said:

like suddenly being unable to communicate with or see "nearby" infalling particles.

It depends what you mean by “nearby”. Since below the horizon the r-coordinate becomes time-like in nature, all light cones will be tilted inwards - meaning you cannot see anything that is below you, since it is impossible for a photon to increase its r-coordinate, irrespective of how it is emitted. You can still communicate with particles co-moving along with you at the same radial distance. A particle higher up than you can send you messages, but your reply won’t ever reach that particle.

So locally in your own frame nothing special happens, but once you start interacting with other local frames, I think you can always deduce that you are below a horizon.

20 hours ago, JohnSSM said:

Why exactly do people believe that going into a black hole is like entering stretched space?

It’s a direct consequence of the geometry of this kind of spacetime, and you can fairly straightforwardly calculate the tidal effects that occur. For a radial in-fall into a Schwarzschild black hole, what you’ll find is that the test body gets stretched along the radial direction, and compressed perpendicular to it (this effect is hence called “spaghettification”). The magnitude of these effects follows an inverse cube law, and also depends on the mass of the black hole. This is linked to, but not necessarily dependent on, gravitational time dilation - you can have time dilation without there being spatial tidal effects (but not vice versa).

17 hours ago, SergUpstart said:

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????

For an observer who is stationary just outside the horizon, the astronaut will fall past him at nearly the speed of light, so the actual time it takes for a human body to cross the horizon is so short that no adverse effects could occur. The astronaut himself will never notice anything special as he falls through the horizon.

14 hours ago, md65536 said:

Also, if you still think the head can't see the feet inside a BH, please show how this would look with light cones.

If you look at the diagram you posted, you will notice that below the horizon all light cones are tilted inwards, towards the singularity. Photons “live” on the surface of light cones, meaning no photon could ever increase its radial coordinate, i.e. move away from the singularity. This is not a tidal effect, but due to the causal structure of spacetime. Once emitted, a photon can only decrease its radial position wrt the singularity. Hence, for an astronaut falling feet-first, a photon emitted from his foot should not be able to travel “upwards” to his eyes.
On the other hand though the astronaut himself is of cause also falling - so the real question is whether it is possible to set up the scenario such that the relative motion between photon and eyes can be made such that the falling astronaut can somehow “catch up” with the (also falling) light.
I reserve final judgement here, as I think this is one of those situations where one would really have to go and work through the maths.

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1 hour ago, Markus Hanke said:

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, the tangent space will fall to the interior of the light cone; if the surface is null, the tangent space will coincide with the surface of the light cone. So this isn’t the same - you can (at least in principle) escape from a light-like surface to infinity, but you can’t escape from a null surface.

But the future light cone interior is in the direction of the BH interior, that doesn't matter? Is it that the EH is a null surface, but also a light-like surface, yet the latter does not give you enough information to define the horizon?

1 hour ago, Markus Hanke said:

It depends what you mean by “nearby”. Since below the horizon the r-coordinate becomes time-like in nature, all light cones will be tilted inwards - meaning you cannot see anything that is below you, since it is impossible for a photon to increase its r-coordinate, irrespective of how it is emitted. You can still communicate with particles co-moving along with you at the same radial distance. A particle higher up than you can send you messages, but your reply won’t ever reach that particle.

My main argument in this thread is basically that you can see what is below you, because it doesn't involve a photon increasing its r-coordinate, but rather the observer decreasing its r-coordinate to reach that light. Is that argument wrong?

Inside the horizon, the r-coordinate of all photons must decrease over time. Near the horizon, the photons aimed "upward" decrease r very slowly. By necessity, an infalling object must decrease in r faster than some of its photons could---the object can't leave the future light cone of a past event on its own world line, and fall slower into the singularity than all of its light. This implies that an infalling particle can decrease in r faster than the "upward"-directed photons from events on the world line of an object below it. Thus, the infalling observer must be able to see objects that are always below where it is now, but were above where it is now when they emitted the light the observer is now seeing, which appears to come from below. In Schwarszchild coordinates (I think???) those photons are decreasing r very slowly, but in the observer's local coordinates they are moving upward at the speed of light. Just like the EH has a fixed r coordinate but locally moves past the infalling observer at the speed of light.

If this is wrong, I don't see how the astronaut could see their own feet, because their feet are at a lower r coordinate.

Edited by md65536
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I'm still not happy with the descrepancy in answers given. 🤔

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2 hours ago, md65536 said:

But the future light cone interior is in the direction of the BH interior, that doesn't matter?

Correct, it is indeed, but that isn’t how such a surface is defined (that would be difficult, since all light cones have a light-like interior). The simplest formal definition I know of for any kind of boundary surface like this is by way of what kind of normal vector with respect to the local metric they admit. In the case of an event horizon, wrt to the local Lorentzian metric, the unit normal vector at all points is a null vector, so this is a null (hyper-)surface. In fact, in can be shown that all event horizons are always null surfaces.

If I remember correctly, Wald (General Relativity) formalises this by using the pullback of the metric, but tbh I don’t remember the details exactly. I’d have to find that in my notes first.

2 hours ago, md65536 said:

My main argument in this thread is basically that you can see what is below you, because it doesn't involve a photon increasing its r-coordinate, but rather the observer decreasing its r-coordinate to reach that light. Is that argument wrong?

I won’t claim that this is wrong, because I am honestly not sure how this would play out. I had similar thoughts actually, which is why I mentioned the relative motion between photon and falling astronaut. Your general line of thought is not wrong, since both photon and astronaut are falling, so neither is increasing its r. Nonetheless, wrt to the astronaut the photon must of course still propagate at exactly c, so I am unsure what form the relative motion between the two would need to take. I don’t see how the eyes of the astronaut could possibly “catch up” with the photon, while still preserving the usual local laws of SR. Perhaps the answer is obvious (lol), I just don’t see it right now.

24 minutes ago, beecee said:

I'm still not happy with the descrepancy in answers given. 🤔

This is one of those questions that seem trivial at first glance, but if you really think about them, you’ll find a lot of little devils in the details. Both the photon emitted from his boots, as well as the astronaut, can only fall along allowed geodesics in this region of spacetime, which means they both can only decrease their r-coordinates as they age into the future, wrt to the central singularity. However, this does not necessarily preclude a relative motion between photon and helmet (and photon and boot) such that the astronaut might see something, so long as this relative motion is in accordance with the usual laws of SR - so the astronaut must determine the photon’s propagation velocity to be exactly c in his own frame. It should be possible to set this up accordingly - after all, a freely falling test particle in a region where tidal forces are negligible is locally inertial, i.e. it finds itself in a small local patch of Minkowski spacetime, irrespective of whether this is above or below a horizon surface. So in principle, so long as the BH is massive enough, the astronaut should be able to see his boots for a while at least, because otherwise he couldn’t be considered to be in inertial motion within a Minkowski patch.

That being said, if my years of looking into GR have taught me anything, then it is to be suspicious of what seems “intuitively obvious” - I’ve fallen on my nose often enough through this mistake. So perhaps I’m overlooking something here.

Edited by Markus Hanke
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1 hour ago, Markus Hanke said:

This is one of those questions that seem trivial at first glance, but if you really think about them, you’ll find a lot of little devils in the details. Both the photon emitted from his boots, as well as the astronaut, can only fall along allowed geodesics in this region of spacetime, which means they both can only decrease their r-coordinates as they age into the future, wrt to the central singularity. However, this does not necessarily preclude a relative motion between photon and helmet (and photon and boot) such that the astronaut might see something, so long as this relative motion is in accordance with the usual laws of SR - so the astronaut must determine the photon’s propagation velocity to be exactly c in his own frame. It should be possible to set this up accordingly - after all, a freely falling test particle in a region where tidal forces are negligible is locally inertial, i.e. it finds itself in a small local patch of Minkowski spacetime, irrespective of whether this is above or below a horizon surface. So in principle, so long as the BH is massive enough, the astronaut should be able to see his boots for a while at least, because otherwise he couldn’t be considered to be in inertial motion within a Minkowski patch.

That being said, if my years of looking into GR have taught me anything, then it is to be suspicious of what seems “intuitively obvious” - I’ve fallen on my nose often enough through this mistake. So perhaps I’m overlooking something here.

Perhaps my main problem is putting too much faith in the river analogy? https://aapt.scitation.org/doi/10.1119/1.2830526

With this analogy, the EH is "flowing" at "c" and increases inside. Like a fish [photon of light coming from an astronauts feet] swimming upstream at 5kms/hr, against a current [spacetime within EH] falling at 10kms/hr, it aint going to make any headway.  https://jila.colorado.edu/~ajsh/insidebh/waterfall.html

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Just now, beecee said:

it aint going to make any headway

No, but that doesn’t mean that within the fish blood cannot circulate.
Likewise, the composite system “astronaut + photon” cannot move away from the central singularity (only towards it) - but that doesn’t necessarily mean there can’t be ordinary (i.e. respecting the laws of SR) relative motion between the photon and the astronaut’s eyes on a small enough local scale.

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5 hours ago, md65536 said:

My main argument in this thread is basically that you can see what is below you, because it doesn't involve a photon increasing its r-coordinate, but rather the observer decreasing its r-coordinate to reach that light. Is that argument wrong?

I was going to jump on that. +1 for catching it. In my reply above your post, I mentioned the other companions all around you falling with you. They don't disappear from your sight. They're not 'below' you, they trace a worldline near your worldline and there is no way light from one nearby worldline doesn't reach the other, regardless of the direction in which this companion appears in relation to you.

3 hours ago, Markus Hanke said:

I won’t claim that this is wrong, because I am honestly not sure how this would play out. I had similar thoughts actually, which is why I mentioned the relative motion between photon and falling astronaut. Your general line of thought is not wrong, since both photon and astronaut are falling, so neither is increasing its r. Nonetheless, wrt to the astronaut the photon must of course still propagate at exactly c, so I am unsure what form the relative motion between the two would need to take. I don’t see how the eyes of the astronaut could possibly “catch up” with the photon, while still preserving the usual local laws of SR. Perhaps the answer is obvious (lol), I just don’t see it right now.

The usual local laws of SR very much do apply here per the equivalence principle.

7 hours ago, Markus Hanke said:

I’d just like to point out that the non-existence of stationary frames below the horizon is not a consequence of tidal forces, but is due to the causal structure of spacetime

The falling observer is very much stationary in his own reference frame. That frame simply doesn't correspond to any  inertial reference frame relative to a distant outside frame of reference, say the inertial frame in which the center of mass of the BH is stationary. This is not a contradiction, just an abstract consequence that no outside inertial coordinate system foliates the events inside a black hole. That in no way implies that the events inside a black hole cannot be locally mapped with an inertial coordinate system.

The falling observer is stationary in his own frame and sees the objects all around him in any direction. If not, he'd not see his feet or hands.

Edited by Halc
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46 minutes ago, Halc said:

I was going to jump on that. +1 for catching it. In my reply above your post, I mentioned the other companions all around you falling with you. They don't disappear from your sight. They're not 'below' you, they trace a worldline near your worldline and there is no way light from one nearby worldline doesn't reach the other, regardless of the direction in which this companion appears in relation to you.

The usual local laws of SR very much do apply here per the equivalence principle.

The falling observer is very much stationary in his own reference frame. That frame simply doesn't correspond to any  inertial reference frame relative to a distant outside frame of reference, say the inertial frame in which the center of mass of the BH is stationary. This is not a contradiction, just an abstract consequence that no outside inertial coordinate system foliates the events inside a black hole. That in no way implies that the events inside a black hole cannot be locally mapped with an inertial coordinate system.

The falling observer is stationary in his own frame and sees the objects all around him in any direction. If not, he'd not see his feet or hands.

How will we ever know?

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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, squished towards one or other of the light cone arms. If this sounds a bit like a Minkowski diagram, it should. But here the representation of spacetime is very distorted, so you can't extend those local axes infinitely as you can in Minkowski spacetime. And there's a lot of additional structure (a second exterior region and a white hole) that are artefacts of taking the Schwarzschild solution a bit too seriously.

Here's a Kruskal diagram, copied from Wikipedia, with toe and head worldlines drawn (badly hand-drawn black lines), and a few light signals exchanged (red from toe to head, green from head to toe).

I've only sketched the infaller's paths (badly), not calculated them. And they're quite a long way apart, so there isn't time for a light signal from the head to reach the toes inside the horizon. This would be possible if the lines were closer together.

The reason I like this diagram is the simplicity of the light cones. It's easy to see what happens as someone falls in.

• There is no invariant sense in which you can say that "your toes have now crossed" until your head also crosses, so you cannot report this. In your local inertial frame your toes do cross first, and you are free to report that your toes have now crossed (but others are free to disagree).
• Light signals can be exchanged both ways at all times (as long as both your head and toes cross), so you can never not see your toes.

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