I've heard many scientists, when giving interviews and the like, state that if one were falling into a black hole massive enough that the tidal forces at the event horizon weren't too extreme, that you wouldn't "notice" or "feel" anything, and so forth.

Thinking about this for a few minutes, it seems to be quite wrong. If you're falling feet first for example, as your feet cross the horizon, your brain can no longer receive any information from them, as the information would have to travel faster than light. Once you are entirely within the horizon, no part of your body closer to the singularity can send any sort of signal to any part of your body that is further away, for the same reason.

Even bloodflow would stop, as blood that is pumped downward towards your feet could never be pumped back up again.

In other words, inside the event horizon is a series of even more event horizons, like the layers of an onion, infinitely thin.

Am I missing something important?

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    $\begingroup$ Fantastic question! $\endgroup$
    – Mike
    Commented Jun 5, 2015 at 22:04
  • $\begingroup$ Just came here to ask exactly the same question. I doubt I could ask it better) $\endgroup$ Commented Aug 8, 2015 at 2:04
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    $\begingroup$ You've argued correctly that you would definitely notice if you were hovering stationary right at an even horizon. You haven't argued that you'd notice if you fell through a horizon. $\endgroup$
    – tparker
    Commented Jul 26, 2017 at 8:22
  • $\begingroup$ You cross the event horizon at the infinite coordinate time. No BH exists infinite time. So, you do not cross. $\endgroup$
    – Anixx
    Commented Apr 12, 2021 at 0:31

8 Answers 8


This is a great question, because it's a subtle variation on the usual question about spaghettification and supermassive black holes, and shows somewhat deeper thinking.

So let's assume the black hole is supermassive -- or more specifically that you are really tiny compared to the black hole -- so that we can ignore tidal effects. Tidal effects are the difference in gravitational "force" on two different parts of an object. In this case, I mean the difference between the acceleration of your feet and your head. Your feet are slightly closer to the center of the black hole, so they will experience a slightly greater acceleration than your head. You would feel this as a slight tug on your feet. The bigger the hole or the farther you are from it, the smaller the difference will be. At some point, it will be so small that it's "in the noise" and you don't even notice it. We're assuming that.

If your head were somehow stuck just outside the horizon, you would be right. I don't think anyone would claim you wouldn't feel anything if your head were attached to a rocket keeping you out, while your feet dangled inside the black hole. :) But those aren't tidal effects; they're acceleration effects.

On the other hand, if you are falling into the supermassive black hole (even if you jumped off this crazy rocket just an instant earlier), things are very different. Your head and feet are being "accelerated" at basically the same rate (relative to some stationary coordinate system, let's say) because you are so small compared to the black hole. So your head is moving at roughly the same speed as your feet, which means that the signal doesn't have to actually move outward relative to these stationary coordinates (it can't). Instead, it just needs to move inward more slowly than your head. And that's entirely allowed everywhere, even well inside the black hole.

You'll typically see this sort of thing represented by a graph of the light cones. And inside the horizon, those light cones "tip over" towards the singularity. This means that even light pointed outward can't actually move outward; the outward-pointing light ray will still be moving toward the singularity. But your head (and your feet) are moving toward the singularity faster, so your head enters into the light cone of your feet. Which means that relative to your head light can still move outward, as can a nerve impulse. Basically, think of two light rays given off by your feet: one directed toward the singularity, and the other directed away from it. You'll probably believe that they have different speeds. The speed of your feet is somewhere between those two, as is the speed of your head.

So all that needs to happen is for your head to enter the future light cone of your feet before your head hits the singularity. Not a problem, since the black hole is so large and you've still got a while to go. Now, you might be concerned that your feet will hit the singularity before your head gets that first signal, which would seem weird. But then you remember that the concept of simultaneity is relative. Your head and feet are in the same reference frame -- at least far from the singularity -- so they experience things at basically the same rate, and nearly the same time as judged in their own reference frame.

Just as a side note, you should try to distinguish between an event horizon and an apparent horizon. Technically, you're talking about the latter, which is the local surface where light rays that are directed outward can't actually move outward. An event (or absolute) horizon, on the other hand, has nothing to do with local effects -- at least not directly. You can only know if something is an event horizon if you know the entire future history of the universe. Unfortunately, the term "event horizon" is thrown around in popular descriptions of black holes when it shouldn't be. They happen to be the same for certain special black holes, but they really are different concepts, and the right way to think about a horizon is different in the two cases. I just use the term "horizon", and anyone who knows the difference will figure it out. A good (and accurate) popular reference for all such things is Thorne's "Black holes and time warps". The standard technical reference is Hawking & Ellis's "The large-scale structure of space-time".

  • $\begingroup$ His question is about event horizons, i.e. the former, isn't it? I'm not entirely convinced right now that e.g. your head "catches up" with the information transmitted from your feet $\endgroup$
    – innisfree
    Commented Jun 5, 2015 at 22:49
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    $\begingroup$ No, an event horizon is defined by what the light rays eventually do in the infinite future. The apparent horizon is technically what he's talking about. Though in the case of an (eternally) static black hole, they're the same thing. $\endgroup$
    – Mike
    Commented Jun 5, 2015 at 22:50
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    $\begingroup$ a black hole is usually thought of as e.g. Schwarzschild metric with an event horizon isn't it? this is a thought experiment about about black holes with event horizons. the detail about apparent horizons is a red herring. $\endgroup$
    – innisfree
    Commented Jun 5, 2015 at 22:57
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    $\begingroup$ It's not a red herring; it's a definition. This is a physics site, so we need to be precise about our definitions. Event horizons may or may not coincide with apparent horizons, but the concept the OP is talking about is definitely an apparent horizon, and not necessarily an event horizon. The OP never said anything about Schwarzschild, say you can't say if they're the same. $\endgroup$
    – Mike
    Commented Jun 5, 2015 at 23:04
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    $\begingroup$ Thanks Mike! For some reason it didn't occur to me that the freefalling head would be approaching the singularity faster than the signals from the feet. I didn't mean to imply that your head would be tied to the rocket while the feet dangle! $\endgroup$
    – alzee
    Commented Jun 6, 2015 at 1:36

A falling observer does not experience passing through an event horizon as you describe.

Instead a free-falling observer would see space-time as locally flat as long as the tidal forces were manageable. Your head and your feet are (nearly) sharing the same frame of reference.

The falling observer always sees the apparent horizon in front of them until they reach the singularity in a finite proper time.

Up to that point, as Mike correctly describes, because no stationary "shell" observers are allowed, and space itself moves towards the singularity, it is possible for both the following statements to be true.

  1. Light directed outwards still travels at the speed of light outwards according to the falling observer.

  2. That light will always end up at the singularity in the future, but after the observer gets there.

I find the following diagram helpful. It shows the world lines of head and feet in Eddington Finkelstein coordinates and I obtained it here. In this diagram the singularity and the event horizon are shown as vertical lines. The curved solid lines are the world lines of your head and feet respectively. Light cones are shown and these are limited by the trajectories of light directed radially inwards or outwards. Far from the black hole these would just be lines at $\pm 45^{\circ}$. At the event horizon the outgoing side of the light cone is vertical. Inside the event horizon we see that the future light cone is directed inwards and that nothing can escape to outside the event horizon.

Now follow what happens when the "feet" signal to the "head" by following the world line of an outging photon (the right hand side of the light cone). You can see that a signal from the feet is always able to reach the head right up until the head meets the singularity. But of course that outgoing signal never makes it out of the black hole, it too reaches the singularity sometime after the "head" does.

Head and feet in Eddington-Finkelstein coordinates

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    $\begingroup$ This is succinct enough to have a sort of ring of truth to it, but it seems to imply (for example) that the observer would be able to see their feet when they are inside the horizon while their head is not, which doesn't seem right, as they would be receiving information from beyond the EH. $\endgroup$
    – alzee
    Commented Jun 5, 2015 at 22:17
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    $\begingroup$ What does it mean for your feet to be inside the horizon "while" (i.e. at the same time as) your head is not? Some observers might say that there is a time when your feet are inside and your head is outside, while others might say that there is no such time. $\endgroup$
    – WillO
    Commented Jun 5, 2015 at 22:27
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    $\begingroup$ @shawnhcorey You are completely wrong (and it's not my diagram). Instead of downvoting things you are not familiar with, do some reading about Eddington-Finkelstein coordinates. $\endgroup$
    – ProfRob
    Commented Jun 7, 2015 at 18:56
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    $\begingroup$ @shawnhcorey You should have said that you were pursuing some pet theory. Well, the accepted answer relies on what I've said being correct too. Why don't you put forward your ideas as an answer so that we can all appreciate them. I mean, that Eddington chap was a bit of an amateur wasn't he... $\endgroup$
    – ProfRob
    Commented Jun 7, 2015 at 23:37
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    $\begingroup$ @shawnhcorey Coordinates can do basically anything you want them to do -- which is one of the main lessons of General Relativity. You really shouldn't downvote a good answer like this just because you don't understand what's going on. Feel free to ask questions, instead. $\endgroup$
    – Mike
    Commented Jun 8, 2015 at 3:38

Suppose you are moving toward the event horizon at .99999999 c.

Your feet cross the horizon. No signal can leave your feet and reach your head, if your head stays outside the horizon. You are defeeted. Or are you?

In a fraction of a fraction of a second, your head has crossed the horizon as well. The photons that your feet sent are passed by the head, you notice nothing.

But wait! Why are we crossing the horizon so fast? Why not just mosey on over it, instead?

The horizon we are approaching is pulling space inward at a ridiculous pace. For a black hole to be flat at the event horizon, it has to be huge, and if it is huge, the rate at which it "pulls space in" gets ridiculous. (well, more like the radius over which its "pull space in at a ridiculous rate" is large: all black holes "pull space in" ridiculously fast if you are near enough to the event horizon).

In order to prevent ourselves from crossing the horizon at a ridiculous pace, we have to accelerate away from the black hole. But acceleration itself generates an apparent horizon. If we accelerate fast enough to "hover" crossing the edge of the black hole, we'll end up putting an apparent horizon between our head and our feet: you will be torn apart, but you'd be torn apart if you did it in empty space.

Space near the event horizon, if viewed to be stationary, has an event horizon sweeping through it at the speed of light. In order to keep ahead of it, you need to accelerate fast enough that events "near" the horizon never reach you -- generate an apparent horizon between you and it. Because if events "near" the horizon reach you, so does the horizon! If your feet are danging over this apparent horizon, no force will allow them to communicate with you. Go to empty space, accelerate in the same manner, and your feet still get disconnected from you causally. (the difference being, if you stop accelerating in empty space, you are now next to your torn-off feet. If you stop accelerating near the black hole, you cross the event horizon, and are now next to your torn-off feet.)

If you "stand still" near the event horizon, you don't feel anything as it sweeps over you. Signals from the part of your body that cross over first are sent, don't cross the event horizon, but they do reach the other side of your body -- after that part of your body crosses the event horizon.

The reduction in tidal forces (how hard it is to notice the black hole's gravity) and to the rate of acceleration required to keep ahead of the event horizon when near it are both functions of the size and mass of the black hole. And in the limit, an infinite mass black hole looks like the future: coming at you at the speed of light, uniform over all of space and time, and no way to go the other way. Your feet cannot send messages to your head in the present time, but they can send messages to your head in the future, both plummet futureward.


Event horizons are everywhere

The top-rated answers both talk about "tipping light cones" as though that were a special phenomenon that happens in black holes. In reality, nothing special happens at an event horizon, and you see your feet there for the same reason you see your feet anywhere else. In fact, event horizons (one-way surfaces in spacetime) are literally everywhere. Your feet are falling through one right now.

Forget about black holes for a moment. In fact, forget about general relativity, and consider a 1+1 dimensional special-relativistic world, with a couple of astronauts in it (Alice and Bob), and a diagonal line drawn arbitrarily through it:

       |      /.
       |     /.       ^
       |    /.        |
       |   /.      future
       |  / .
       | / .
       |/  .
       /  .
      /|  .         past
     / |  .           |
    /  |  .           v
   /   |  .
  E    A  B

E is the worldline of a point (or, in 3+1 dimensions, a plane) moving rightward at the speed of light. Alice crosses that worldline (feet first—i.e., her feet are on the left), while Bob fires his thrusters and goes into hyperbolic motion which avoids crossing E indefinitely. (Again, E is not a physical object, just a line I've drawn, but Bob can fire his thrusters whenever he wants, and chooses to fire them here for whatever reason.)

E is an event horizon. It should be easy to see that once Alice has crossed it, she can never go back, even if she can travel at the speed of light. Likewise (and for the same reason), if Bob never crosses the horizon then he will never see any light (or anything else) from Alice after she crosses the horizon.

Most of the "phenomenology" of black hole event horizons applies also to this horizon. Bob will see Alice "frozen" on the horizon at the moment of falling through, redshifted into the indefinite future. (If you don't see why, just look at the path of the light.) Black hole horizons behave like electromagnetic conductors; so does this horizon. Black hole horizons emit Hawking radiation; so does this horizon (it's called Unruh radiation, but it's just the infinite-radius limit of Hawking radiation). For the most part, if you want to know what happens at a black hole horizon, you can work it out in this special-relativistic analogue. (The exception is when a large part of the area of the horizon is involved, so that the spherical shape is significant. E is the infinite-mass (infinite-radius) limit of a black hole horizon.)

Obviously, Alice sees her feet at all times. That's not because of tilting light cones or because she's moving very quickly. It's because her feet are radiating light at all times, and it's never blocked by anything. Just before (resp. after) her head crosses E, she sees her feet just before (resp. after) they cross E. Bob never sees the light from after she crosses E, but that's only because he's moving in such a way that it never reaches him, not because it was absorbed or otherwise blocked.

What makes black hole horizons special is the (generally unknowable) future

Black hole event horizons (unlike E) are not at arbitrary locations, but that isn't because anything happens at the location of the horizon. It's because there is a future singularity where everything is destroyed.

This also has a special-relativistic analogue. Imagine that a region of space is peppered with undisarmable time bombs with synchronized countdown timers. The spacetime diagram looks like this:

   **********        ^
    \      /         |
     \    /       future
      \  /

Each asterisk is an explosion. The diagonal lines bound the region of spacetime from which you can't avoid being blown up, even if you can travel at the speed of light. Those lines are meaningful—their location is fixed by a real physical event—but there is nothing there. Nothing detectable happens when you cross them. Nor is there anything to detect anywhere inside the certain-death region. The causal future (future light cone) of the explosions is entirely outside that region, so no one inside it can detect the explosions in any way. They might infer that they are going to happen if they can see the bombs and countdown timers (which is analogous to noticing the rapidly increasing tidal forces), but they will never see (or otherwise detect) their feet blowing up before their head does. Because the explosions are spacelike separated, there is no sense in which their feet do blow up before their head, even if they crossed the certain-death horizon feet first.


Whether an in-falling observer can see (or more generally be able to receive sensory input from) their feet as they pass through the event horizon is an interesting question. The answer is yes - you will be able to see your feet the entire time, putting aside the whole spaghettification issue.

This question provides an excellent opportunity to examine what coordinates actually mean in GR, and how we define and interpret them. If you want the punchline, skip to the end of the answer.

The reason can best be understood by adopting the Gullstrand-Painlevé coordinates, also called raindrop coordinates. In this coordinate system, we utilize the same $r,\theta,\phi$ coordinates as the standard Schwarzschild metric along with a new time coordinate $$T \equiv t - 2M\left(-2y + \log\left[\frac{y+1}{y-1}\right]\right), \qquad y\equiv \sqrt{\frac{2M}{r}}$$ in units such that $c=G=1$.

Before proceeding, it is worth taking a moment to review the precise physical interpretation of the Schwarzschild coordinates.

  • Because the metric is static, we may slice the spacetime into space-like surfaces $\Sigma_t$ labeled by a timelike coordinate $t$ in such a way that

    1. the components of the spatial metric $g_\Sigma$ on $\Sigma_t$ are independent of $t$, and
    2. the vector field $\partial_t$ is orthogonal to the surfaces $\Sigma_t$.

    With this choice of time coordinate, we manifest the idea that the spacetime is not changing with time; any two slices $\Sigma_t$ and $\Sigma_{t'}$ are identical (more precisely, isometric). Furthermore, if an object is at rest on the slice $\Sigma_t$ (by which I mean that $\mathrm dx^i/dt = 0$ for each of the spatial coordinates $x^i$), then it will have the same spatial coordinates on the slice $\Sigma_{t+\delta t}$ (to first order in $\delta t$).

  • Because the metric is spherically symmetric, each $\Sigma_t$ can be decomposed into a family of concentric $2$-spheres. More precisely, given a point $p\in \Sigma_t$, the set of points which can be reached from $p$ by performing a rotation about the center of the black hole constitutes a 2-sphere, which can be coordinatized with the standard angles $(\theta,\phi)$. Spherical symmetry implies that the induced metric on each sphere is given by $g_{\mathrm S^2} = K\big(\mathrm d\theta^2 + \sin^2(\theta) \mathrm d\phi^2\big)$, where $K\in(0,\infty)$ does not depend on $\theta$ or $\phi$.

  • The surface area of each of the aforementioned 2-spheres is given by $$A = \int_0^{2\pi} \mathrm d\phi \int_0^\pi \mathrm d\theta \sqrt{\mathrm{det}(g_{\mathrm S^2})}=4\pi K$$ If we define the coordinate $r \equiv \sqrt{K} = \sqrt{A/4\pi}$, then the metric on $\Sigma_t$ takes the form $$g_{\Sigma} = a(r)\mathrm dr^2 + r^2\big(\mathrm d\theta^2 + \sin^2(\theta)\mathrm d\phi^2\big)$$ where $a$ is an unknown function.

Under these assumptions, the full spacetime metric $g$ is given by $$g = -b(r) \mathrm dt^2 + a(r) \mathrm dr^2 + r^2 \big(\mathrm d\theta^2 + \sin^2(\theta) \mathrm d\phi^2\big)$$ where $a$ and $b$ are unknown functions. Plugging them into the Einstein equations and applying the appropriate boundary conditions (specifically, that the spacetime approaches a Newtonian limit at $r\rightarrow \infty$), we find that $$a(r) = \frac{1}{b(r)} = 1- \frac{2M}{r}$$ where $M$ is the mass of the black hole. The full metric in Schwarzschild coordinates is then $$g = -\left(1-\frac{2M}{r}\right)\mathrm dt^2 + \left(1-\frac{2M}{r}\right)^{-1}\mathrm dr^2 + r^2\big(\mathrm d\theta^2 + \sin^2(\theta) \mathrm d\phi^2\big)\tag{$\star$}$$

So to conclude this review:

  1. The temporal component $t$ is chosen to make the metric components independent of $t$, and the spacelike surfaces of simultaneity $\Sigma_t$ orthogonal to the "time direction" $\partial_t$. If we would like to assign a physical meaning to the coordinate $t$, we might imagine a family of observers each of whom sits at fixed spatial coordinates $(r,\theta,\phi)$, and who all start their wristwatches when their worldlines intersect the surface $\Sigma_0$. When we say that an event has coordinates $(t,r,\theta,\phi)$, we mean that it occurs at spatial position $(r,\theta,\phi)\in \Sigma_t$ and that as we walk away from the black hole along the slice $\Sigma_t$, the time on the observers' wristwatches will approach $t$ as $r\rightarrow \infty$. As a crucial note, $t$ is not the time on the wristwatch of the observer fixed to the point $(r,\theta,\phi)$; their watch reads $t\sqrt{1-\frac{2M}{r}} < t$.
  2. The angular coordinates $(\theta,\phi)$ are chosen because we decompose each slice $\Sigma_t$ into a family of nested 2-spheres.
  3. We label each such 2-sphere by the areal radius $r\equiv \sqrt{A/4\pi}$. Note that this is not this distance from a point on the 2-sphere to the center of the black hole! Rather, it is the square root of the area of the 2-sphere divided by $4\pi$. This coincides with the radial distance if $\Sigma_t$ is flat - but that is not the case here.

With that review under our belts, we may now provide an interpretation to the raindrop time $T$. Rather than choosing our time coordinate to reflect the static nature of the metric as we did with $t$, we should like to choose a time coordinate which reflects the experience of a radially in-falling observer. To do this, we first draw the trajectories of in-falling observers in Schwarzschild coordinates, where each observer starts from some initial radius $r_0$. We will take $r_0$ sufficiently large that the gravitational time dilation is negligible, and so the time coordinate $t$ corresponds to the wristwatch time of a stationary observer fixed at $r_0$.

enter image description here

A generic event $E$ lies on exactly one of these trajectories. When the in-falling observer departs from $r_0$ at Schwarzschild time $t=t_0$, their wristwatch time matches the Schwarzschild time. When they arrive at $r_E$, however, their watch is no longer synchronized with the observer at $r_E$; this is due to a combination of gravitational and kinematic time dilation.

One can compute the discrepancy between the wristwatch time of our in-falling observer and the local Schwarzschild time at $t_E$. Between the time the observer departs from $r_0$ and the time they arrive at $r_E$, the time elapsed on their wristwatch $\Delta \tau$ differs from the elapsed Schwarzschild time $\delta t \equiv t_E-t_0$ by an amount

$$\Delta \tau - \Delta t = 2M \Delta\left(2y + \log\left[\frac{y+1}{y-1}\right]\right), \qquad y \equiv \sqrt{\frac{2M}{r}}$$ which is independent of $r_0$ except for the fact that we have sneakily used the assumption that $r_0 \gg 2M$ during the derivation (figuring out where is a good exercise).

This provides us with the desired interpretation - the coordinate $T$ of an event is the wristwatch time of an in-falling observer, whose clock was initially synchronized to the Schwarzschild time but has lost synchronicity due to kinematic and gravitational time dilation.

In these new raindrop coordinates, the spacetime metric takes the form

$$g = -\left(1-\frac{2M}{r}\right) \mathrm dT^2 - 2\sqrt{\frac{2M}{r}} \mathrm dT \mathrm dr + \mathrm dr^2 + r^2 \big(\mathrm d\theta^2 + \sin^2(\theta) \mathrm d\phi^2\big)$$

This differs from $(\star)$ in two major ways. First, this expression is not singular at $r=2M$, and so it covers the entire Schwarzschild spacetime (not just the region outside of the event horizon). Second, the spacelike hypersurfaces of simultaneity $\Sigma_T$ are now spatially flat (the spatial part of the metric looks like ordinary spherical coordinates); the price we pay for this is the cross term proportional to $\mathrm dT\mathrm dr$, which we can loosely think of as as a kind of drag - a natural "flow" through space towards $r=0$. See e.g. the river model of black holes for a more thorough description of this picture.

In these coordinates, the velocity of an in-falling observer (our namesake raindrop) who was initially at rest at $r_0\rightarrow \infty$ is found to be $$\frac{dr_{raindrop}}{dT} = - \sqrt{\frac{2M}{r}}$$ On the other hand, the trajectories of light beams are given by $$\frac{dr_{light}}{dT} = - \sqrt{\frac{2M}{r}} \pm 1$$ We see that inside the event horizon where $r<2M$, light beams must travel toward the center - $dr/dT <0$ - but can do so at two different rates, corresponding to us pointing our flashlights toward or away from $r=0$. Even in the latter case, the light travels only radially inward, but it does so more slowly than our freely-falling observer.

The Punchline

In the context of this question, once you get inside the event horizon the light emanating from the tops of your shoes travels inward toward $r=0$ at a rate of $dr/dT = -\sqrt{2M/r} + 1$, which is smaller (i.e. less negative) than the rate at which your head falls toward $r=0$, namely $dr/dT = -\sqrt{2M/r}$. The relative speed between the light and your eyes is then simply $1$ (or $c$, in SI units, as we might expect from the equivalence principle), and so you will indeed be able to see whether your shoes remain tied as you are swallowed by the gaping maw of infinity.


The thing you are missing is the change in perspective from a distant observer and a very nearby observer.

It's not terribly unreasonable for the distant observer to mark a fixed sphere on her maps and say "that's roughly the event horizon of the black hole".

However, this picture is not very good if she's very near the event horizon: a much more accurate picture would be that the event horizon is rushing towards her at the speed of light.

(maybe it's reasonable to say it looks like it's a little slower than that if she's outside, and a little faster than the speed of light if she's inside)

From this up-close picture, the reason signals from inside the black hole can't reach outside is now clear: they simply fail to outrun the event horizon.

If you're outside the horizon, then with enough thrust you can stay ahead of it, although the closer it gets, the more thrust you need. If you let your feet fall in, but applied enough thrust to your head to stay outside, the stress of that acceleration would rip your body apart. (although the G-forces involved in keeping your spacecraft hovering would have killed you long before you got close enough to try this experiment)


I agree that you would not "notice" or "feel" anything as you "fall" into a black hole. As far as matter is concerned, the event horizon is not a line of demarcation were "strange" things happen on one side but not on the other. As far as you are concerned, all your molecules would be accelerated the same before and after crossing the event horizon, and since you are in "free fall," the only thing you feel, is the acceleration. If for some reason some part of your body is accelerated more than another part, the parts will be separated, and your brain would feel it, because your nerve signals are operating in "local time" with respect to your body.


Actually around event horizon always have intense photon area that would burn anythings near event horizon so when you fell to that place you feel hot and dying with cremation before touching event horizon

If we could skip that then when you foot touch the event horizon, the nerve signal of your foot disintegrate may able to transfer upwards because the signal came from nerve cell over event horizon breaking from nerve cell touching it. So you may feel the pain of all your body disintegrate bottom up


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