I'm going to ask/explain this as best I can; I'm sure I have some fundamentals wrong here.

Spaghettification is a phenomenon which occurs only in stellar-mass black holes owing to the immense gravity gradients experienced when approaching the event horizon. I have read elsewhere that in the case of supermassive blackholes, one could cross the event horizon and travel a measurable amount of distance toward the singularity without knowing until it is too late, whilst remaining in one piece.

This previously asked question sprung to mind - Fighting a black hole: Could a strong spherical shell inside an event horizon resist falling in to the singularity?

I understand clearly, from the answers to that question, that forces holding matter together cannot propagate faster than light, so, my question is:-

How can someone/thing travel past and beyond the event horizon of a supermassive blackhole and travel an arbitrary distance when, in my understanding, they should effectively vapourise once the event horizon has been crossed?


3 Answers 3


I suspect that what has confused you is the difference between remaining a fixed distance from the black hole and falling freely into it. Let me attempt an analogy to illustrate what I mean.

Suppose you are carrying a large and heavy backpack. You can feel the gravitational force of the backpack weighing you down. However this only happens because you're staying a fixed distance from the centre of the Earth i.e. you're standing stationary on the Earth's surface. If you and the backpack were to leap from a cliff then (ignoring air resistance) you would feel no gravity as you plummeted downwards and the backpack wouldn't weigh anything.

If we now switch our attention to the black hole, if you attempt to stay a fixed distance from the black hole (presumably by firing the rocket motors on your spaceship) you'd feel the weight of the backpack, and the weight would get bigger and bigger as you approach the event horizon. In fact the weight is given by:

$$ F=\frac{GMm}{r^2}\frac{1}{\sqrt{1-\frac{r_s}{r}}} \tag{1} $$

where $m$ is the mass of the backpack, $M$ is the mass of the black hole, $r_s$ is the event horizn radius and $r$ is your distance from the centre of the black hole. As you approach the event horizon, i.e. as $r \rightarrow r_s$, equation (1) tells us that the force goes to infinity. That's why once you reach the event horizon it is impossible to resist falling inwards.

But you only feel this force because you're trying to resist the gravity of the black hole. If you just fling yourself off your spaceship towards the black hole then you will feel no weight at all. You would fall through the event horizon without noticing anything special. In fact you would see an apparent event horizon retreating before you and you would never actually cross anything that looks like a horizon to you.

But there is another phenomenon that can cause you problems, and this is related to the phenomenon of spaghettification that you mention. At any moment some parts of you will be nearer the centre of the black hole than others. For example if you're falling feet first your feet will be nearer the centre than your head. That means your feet will be accelerating slightly faster than your head, and the end result is that you get slightly stretched. This is called a tidal force, and it happens with all sources of gravity, not just black holes. Even on the Earth the gravitational force on your feet it slightly higher than on your head, though the difference is so small that you'd never notice it.

The thing about a black hole is that because its gravity is so strong the tidal forces can get very strong indeed. In fact they can get so strog that they'd pull you out into a long thin strip like a piece of spaghetti - hence the term spaghettification.

But the tidal forces only become infinite right at the centre of the black hole. They are not infinite at the event horizon, and in fact for large enough black holes the tidal forces at the event horizon can be negligably small. The equation for the variation of gravitational acceleration with distance is:

$$ \frac{\Delta a}{\Delta r} = \frac{c^6}{(2GM)^2} \tag{2} $$

If we take a black hole with the mass of the Sun and use equation (2) to calculate the tidal force we get $\Delta a/\Delta r \approx 10^{9}g$/m. So if you're two metres tall the difference between the acceleration of your head and feet would be $2 \times 10^9g$, where $g$ is the gravitational acceleration at the Earth's surface. This would spaghettify you very effectively. However at the event horizon of a supermassive black hole with the mass of a million Suns the difference between your head and feet would be only 0.001$g$ and you'd struggle to feel it.

  • 1
    $\begingroup$ I interpreted the question to be based on the idea that since "forces holding matter together cannot propagate faster than light", an atom at just a slightly smaller radius in Schwarzschild coordinates would not be able to transmit forces to an atom at a slightly larger Schwarzschild radius once both were inside the horizon, so the object made of these atoms would fall apart...but this is based on erroneously treating the radial coordinate as a spatial direction inside the horizon, a conformal diagram makes clear signals can still travel back and forth between nearby atoms inside the horizon. $\endgroup$
    – Hypnosifl
    Jan 7, 2015 at 17:48
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    $\begingroup$ @Hypnosifl: presumably something like this. I confess I assumed this would be far too complicated an answer. $\endgroup$ Jan 7, 2015 at 18:07
  • $\begingroup$ Of course the event horizon on a stellar-mass black hole is only about 3km from the singularity, so I guess that much spaghettification is understandable. $\endgroup$
    – Gabe
    Jan 8, 2015 at 5:18
  • $\begingroup$ Thanks for this guys. John's answer coupled with the addition from @Hypnosifl cleared this up for me. Wise men. $\endgroup$
    – Phizzy
    Jan 8, 2015 at 10:18

The forces holding matter together wouldn't need to propagate any faster than light to hold things together inside the horizon--it's easiest to see what's going on inside the horizon if you use a "conformal" diagram which represents all light rays as diagonals 45 degrees from the vertical, with all timelike worldlines having a slope closer to vertical than 45 degrees at all points, so that the light cone structure works just like in a Minkowski diagram in special relativity (see here for a basic introduction to spacetime diagrams in special relativity if you're not familiar). For a non-rotating black hole you can use Kruskal-Szekeres coordinates (see the "qualitative features" section of the wiki article for an introduction), and for both rotating and non-rotating black holes you can use a Penrose diagram (Penrose diagrams for different types of black hole spacetimes can be seen on this page from the inside black holes site, though note that they all represent idealized 'eternal' black hole spacetimes, which are simpler mathematically than realistic spacetimes where a black hole forms from collapsing matter). In such a diagram, the reason it's impossible to escape the event horizon once you've entered it is just that the event horizon itself is moving outward at the speed of light, so it's impossible to exit for the same reason it's impossible to exit some event's future light cone once you've entered it. But you are still free to travel in any spatial direction you like once inside the horizon, including the "outward" radial direction (this would not be true in Schwarzschild coordinates, a non-conformal coordinate system where the "radial" coordinate actually becomes timelike inside the horizon).

As for whether forces would need to travel faster than light to hold things together, imagine drawing two timelike curves which are right alongside each other as they traverse the inside of the horizon (just draw a second blue curve alongside one of the blue curves seen on the page of Penrose diagrams I mentioned before), curves which could represent the worldlines of two nearby atoms in a solid. In that case, if you draw a diagonal light ray emitted from a point on one curve (John Rennie linked in a comment to this answer he'd given to another question which includes diagrams showing light rays emitted in both directions from a timelike curve in Kruskal-Szekeres coordinates), the ray will reach the other one after only a short interval on the vertical time axis, regardless of which direction it's going. So, light signals can repeatedly travel back and forth between the two worldlines up until the point they hit the singularity, so the inter-atomic forces can still hold the atoms together (at least until the tidal forces overwhelm them).

  • $\begingroup$ Only radial light rays travel at $45^\circ$ in the usual Penrose diagram for a Schwarzschild black hole. I agree you are free to travel in any spatial direction, even inside the event horizon. But this is true for Schwarzschild coordinates too! For $r<2M$, an object is free to move in both positive and negative directions of $t$. Increasing $t$ is more "outward" motion than decreasing $t$, but outward is a relative term as $r$ will inevitably decrease. More precisely, the spatial radial direction is a combination of the vectors $\partial_t$ and $\partial_r$, depending on the observer. $\endgroup$ Feb 8, 2018 at 6:53

There aren't big tidal forces at the event horizon of a big black hole, but it is still pulling you in, and pulling you in hard. Just like astronauts orbiting the earth don't notice huge tidal forces but there isa big effect of being pulled closer to the earth, they just don't feel that because they let themselves fall. When you jump up you don't feel yourself falling, but on the way down you do see the earth rushing up, gravity is real, but the tidal forces are what you feel when you are falling.

Nothing special happens locally when you freely fall through an event horizon, and locally means when everything around you is falling with you. If the gravitational source is small then long before you get to the event horizon you'll experience strong tidal forces. For a spherical body, once you cross the event horizon you will inevitably and eventually get to regions of huge tidal forces, because you will inevitably and eventually get really close to the dense and compact gravitational source.

Getting close to a gravitational source is what causes tidal forces, because when you are close to a small body then different parts of you are differently close to it (percentagewise) and differently directed (non small angles).

The point isn't that you can go past the horizon and then for as far as you want and then stop. The point is that if you are falling towards the black hole and it is large then that point of no return happens so far away that there still aren't any significant tidal forces and even if you keep falling there still aren't big tidal forces for awhile.

You don't get vaporized when you fall past an event horizon, you are simply simply falling but all the parts of you are falling almost equally (since parts of you are percentagewise at almost the same distance from the source and the different directions parts of you are to the body are very small angles different). But if you want to run away from the black hole, you have to start accelerating away and it takes time for your rockets to start and during that time you fall even closer so now your rockets have more falling to overcome. The point of no return is the event horizon and that's when it is too late, not because of tidal forces, but because of the falling.

  • $\begingroup$ "But if you want to run away from the black hole, you have to start accelerating away and it takes time for your rockets to start and during that time you fall even closer so now your rockets have more falling to overcome." -- I don't think this explanation really makes sense, the reason you can't increase your radius in Schwarzschild coordinates has nothing to do with the time for your rockets to start (even if your rockets could instantaneously start and cause your proper acceleration to jump discontinuously to a large value, it wouldn't help), it's that the "radial" coordinate is timelike. $\endgroup$
    – Hypnosifl
    Jan 7, 2015 at 18:36
  • $\begingroup$ (continued) Whereas in Kruskal-Szekeres coordinates or a Penrose diagram, you can increase the value of your radial coordinate by firing your rockets, the issue in this case is just that your radial velocity outwards is always slower than that of the event horizon, which moves outward at the speed of light in these coordinates. $\endgroup$
    – Hypnosifl
    Jan 7, 2015 at 18:38
  • $\begingroup$ Exactly at the event horizon if you can instantly change your velocity to speed c in the direction directly away, then you can escape (the singularity, though not get away). The radial direction being timelike is exactly what I was trying to describe, you fire your rockets as hard as you want, but you end up closer than before. Once you are within the horizon you can fire your rockets and travel for a finite interval (whether an impulse like you suggest or a slower burn) but at the end of your interval you end up closer than you were at the beginning of the interval. $\endgroup$
    – Timaeus
    Jan 8, 2015 at 16:08
  • $\begingroup$ I'm not sure how physical an instantaneous impulse is, I always imagine it as an idealization of a very very short burn, and during your very very short burn you fall farther from the horizon than you climb out, and I use that terminology merely to contrast the radial fall and radial climb contrasting against the tidal forces which are a completely separate issue. My goal is to make that distinction clear without obscuring things with a choice of (an arbitrary) coordinate system. $\endgroup$
    – Timaeus
    Jan 8, 2015 at 16:13

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