Given a black hole with mass of the Milky Way (1.6x10^42 kg), I built a frame that 1 m above the event horizon. I am jumping from this frame my legs down (since I am afraid to dive into a dangerous object my head down). As I understand, the tidal forces at this place are not yet dangerously pulling me apart.

As similar question about a rigid rod supposes that the end of the rod that approaches the horizon can "stuck" at this position and prevent the rod from moving in any direction. Here I have my legs instead of the rod and I don't quite understand how can I pass the even horizon if I jumped.

  1. Some other answers say that passing the event horizon would not be a big deal for me (yet), but how to connect this situation with the rod question? Would my legs get stuck near event horizon because they will stop in time?
  2. I have a flashlight with me. If I passed the event horizon, can I use it? If I turn it on and direct to my legs, will I see the light reflecting from them?
  3. Nearby the event horizon (where I was initially standing), is the gravitational formula still applicable? According to my calculations, gravitational pull will be GM/(R^2) = ~ 18.5 m/s^2 (numbers here), which is slightly less than double of that on Earth, and makes me think that I can bear that for some time.
  4. Will I be able to think and understand what is going on (supposing I am still sane and healthy)? Assuming thinking process as a combination of some chemical and physical processes happening in time, is there an obstacle for them to happen beyond the event horizon?
  5. Will I be able to move my body parts?

I have only one observer here (myself) and the reference frame is bound to my eyes (looking toward the BH along my body).

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    $\begingroup$ If you are standing in a magical platform 1m above a supermassive black holes, I am sorry to say your legs are gone. You will be a microscopically thin layer of neutron-degenerate matter splattered on the platform. $\endgroup$
    – lvella
    Mar 2, 2021 at 13:54

2 Answers 2


This answer is based on general relativity, which is currently the best foundation we have for understanding these things.

For some round numbers, let's use $M\sim 10^{12}\,M_\odot$ for the mass of our galaxy, where $M_\odot$ is the mass of the Sun. Consider a nonrotating black hole with mass $M$. The circumference of its event horizon is $4\pi GM/c^2$, where $G$ is Newton's constant and $c$ is the speed of light. Numerically, the circumference is $$ \frac{4\pi GM}{c^2} \sim 18\text{ trillion km}, $$ which is roughly five hundred times the circumference of Pluto's orbit.

Near the event horizon of such a large black hole, tidal effects are completely negligible. In other words, when you fall through the horizon, you are falling in a spacetime that is locally indistinguishable from flat spacetime. The experience of falling through the event horizon will be just like the experience of falling toward the earth (with no air resistance). If you want a brief taste of what it would feel like, go bungee jumping.

Regarding the numbered questions:

  1. You will not get stuck. (See benrg's answer.)

  2. Your flashlight will still work fine.

  3. You will feel weightless, just like when falling toward the earth or toward any other planet. (The planet's mass is irrelevant, as long as tidal effects are negligible, as they are for this particular black hole.)

  4. You will be able to think and understand, if the feeling of weightlessness isn't too distracting. You could bring a good book about general relativity and use your flashlight to read it while you're falling.

  5. And yes, you would still be able to move your body parts normally, including using your hands to turn the pages of the book.

Think of it as a really long bungee jump, but with no bungee to bring you back. Nothing can bring you back. It's a one-way trip, and the friends you left behind will never see you cross the horizon, but otherwise it's just like a bungee jump.

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    $\begingroup$ Is it possible to calculate how much time I have to read the book? Assuming my speed at 1 m before entering the event horizon was 0. $\endgroup$
    – greatvovan
    Mar 2, 2021 at 17:10
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    $\begingroup$ @greatvovan If umping from 1 m to such a huge black hole, your time inside (assuming this region exists) is practically $\tau=\pi M=\pi/2\cdot r_s$ directly proportional to the mass or radius. Sun's Schwarzschild radius is 3km, so you get only 16 microseconds inside. For a hypothetical trillion times heavier black hole you get a trillion times longer time to live or about 6 months. It also may be worth mentioning that the largest known black hole powering the quasar TON 618 is 66 billion times heavier than the Sun. There you get 12 days. And in our own galaxy's Sagittarius A* only 1 minute. $\endgroup$
    – safesphere
    Mar 3, 2021 at 6:37
  • $\begingroup$ @safesphere it is the time before what? Hitting singularity? I think an alive being will be terminated earlier, when spaghettification begins. $\endgroup$
    – greatvovan
    Mar 3, 2021 at 7:36
  • $\begingroup$ @greatvovan Yes, before the singularity. Please feel free to ask a separate question about tidal forces inside a black hole. $\endgroup$
    – safesphere
    Mar 3, 2021 at 8:47

Nothing special happens at the event horizon.

Time does not actually freeze at event horizons. There are two things that lead to this misconception:

  1. Things appear to freeze at event horizons, from outside. We don't perceive the world directly, we see light that enters our eyes (or cameras), and that light has to travel from the object we're seeing back to us. It can't escape from inside the horizon, and as the source approaches the horizon from the outside, the time the light takes to escape goes to infinity.

  2. $r=r_s$ in Schwarzschild coordinates is not the event horizon; it's a coordinate singularity. If you plot the worldline of an infalling object in Schwarzschild coordinates, it doesn't intersect the $r=r_s$ line, and people mistakenly think that means it doesn't cross the event horizon. It does cross the event horizon, but it has to leave the Schwarzschild chart to do it because the Schwarzschild chart doesn't include the event horizon. If you plot the same worldline in coordinates that do cover the event horizon (Kruskal-Szekeres or Eddington-Finkelstein or almost anything other than Schwarzschild coordinates) then it will look the same near the event horizon as anywhere else.

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    $\begingroup$ If I shot a laser directly at the black hole, the Shapiro time delay equations tell me that the photons will indeed freeze at the event horizon. This has nothing to do with direct observation, only with frames of reference. $\endgroup$
    – lvella
    Mar 2, 2021 at 13:52
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    $\begingroup$ @Ivella frames of reference and determining an observation are essentially the same thing $\endgroup$
    – OrangeDog
    Mar 2, 2021 at 15:46
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    $\begingroup$ @benrg I agree for the most part, but in point 2. you say $r=r_s$ is not the event horizon, which I don't think is right. The surface $r=r_s$ is where the Killing vector $\mathcal{K}_{t}$ associated with the Schwarzschild time $t$ becomes null and so defines the event horizon. Having said this, it is certainly true that there is no physical singularity there though and no observer would notice anything special while they pass through $r=r_s$. $\endgroup$ Mar 2, 2021 at 15:47
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    $\begingroup$ "𝑟=𝑟𝑠 in Schwarzschild coordinates is not the event horizon; it's a coordinate singularity". Could you tell more on this? In Wikipedia I read: "The surface at the Schwarzschild radius acts as an event horizon in a non-rotating body that fits inside this radius". $\endgroup$
    – greatvovan
    Mar 2, 2021 at 17:05
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    $\begingroup$ @benrg Can you expand on your statement that $g_{tt}=0$ does not mean the surface $r=r_s$ is null? This seems wrong. It is definitely true that for $r>r_s$ the coordinate $t$ is timelike, and for $r<r_s$ it is spacelike, what else can be the case at $r=r_s$? Also in your Kruskal-Szekeres link, it says that the event horizon is given by the surface $T^2 - X^2 = 0$. By the formula $T^2 - X^2 = (1 - r/r_s) e^{r/r_s}$ given there (valid for any $T^2 - X^2 < 1$) you can see that taking $r \to r_s$ puts you exactly on the event horizon so I'm not understanding your point. $\endgroup$ Mar 2, 2021 at 19:59

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