So, once something is inside a black hole's event horizon, it can only move towards the center. This is fine for a point-object. But 3D solid objects rely on molecular forces to stay in one piece. These forces act in all directions inside the solid. But there could not be any interaction between atoms at different distances from the singularity, right?

So, what happens to a solid (let's assume a crystalline lattice, for simplicity) hypothetically placed inside the event horizon? Does it get cleaved in ultra-thin concentric layers, centered on the singularity?

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    $\begingroup$ This is covered well in "spaghettificiation" near a neutral nonspinning black hole singularity. $\endgroup$
    – Ron Maimon
    Aug 25, 2011 at 17:07
  • $\begingroup$ I have not taken a GR class and so am not really qualified to answer but I have heard Neil DeGrasse Tyson claim that an observer could potentially fall past the event horizon and not even notice provided that the radius was sufficiently large to avoid the spaghettification issue. I believe the event horizon is the maximum extent to which photons leaving the singularity (perpendicular to the horizon surface) may travel. That does not, however, limit particles traveling in that direction whose origin is not the singularity. In other words, the object could remain in tact. $\endgroup$ Aug 25, 2011 at 17:29
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    $\begingroup$ @Adam: actually no, nothing leaves the singularity. (As far as we know.) The event horizon is the surface from within which a particle (or light ray) will never be able to escape. Inside the event horizon, it is not possible for anything to avoid moving closer to the black hole. $\endgroup$
    – David Z
    Aug 25, 2011 at 22:11
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    $\begingroup$ @David: be careful to always say "uncharged, non-rotating" when making statements about the inner world of a black hole. Nothing is known for sure in the rotating/charged case. $\endgroup$
    – Ron Maimon
    Aug 26, 2011 at 1:24

4 Answers 4


Whatever happens, one should always remember that for any observer gravity manifests itself only through second order effecs in the distance to the observer. In other words, in the coordinates, comoving with any observer, metric is always flat along the observer's world line and is quadratic in spatial distance to the world line (see comoving Fermi coordinates, e.g. in the MTW book).

Hence, for a sufficiently small solid, crossing an event horizon of a black hole will have a vanishingly small effect. No concentric layers, no problems with the atoms interacting with each other.

However, sooner or later during the infall the second order terms in metric will reveal themselves through tidal forces, increasing infinitely, as one goes closer to the singularity. This will result in mechincal destruction of the solid through squeezing in one direction and stretching in the other.

Closer to singularity, as higher order effects grow stronger, the mechanical destruction will go on in a more sophisticated manner. As the singularity is reached, GR breaks down.

For an illustration, it might be interesting to imagine a supermassive black hole. Its mass amounts to between $M_{BH}=10^5$ and $10^9 M_\odot$, or $10^{11}$ to $10^{15} M_{Earth}$. Its Schwartzschild radius will be $R_{BH}=\dfrac{2 GM_{BH}}{c^2}\sim 3\cdot (10^5$-$10^9)$km$\sim 0.5(10^2-10^6)R_{Earth}$. The tidal forces, created by the black hole near its surface, as calculated in the comoving reference frame of the falling solid body, will be proportional to $F\sim \dfrac{M_{BH}}{R_{BH}^3}\sim F_{Earth}(10^5$-$10^{-3})$. Hence, for the most massive supermassive black holes the tidal forces are even much smaller than those we feel on the Earth!

Concerning the confusion with causality, it is not violated. One wouldn't be able to send signals between arbitrary points under the event horizon, if the points had a fixed spatial coordinate. However, crossing the event horizon means that it is impossible in principle to stop the body from falling into the singularity, hence it will never follow the worldines of any spatially fixed spacetime points, and hence the causality arguments do not lead to a contradiction.


As something falls towards the event-horizon Lorentz transformation on the object becomes apparent to an outside observer. Specifically, space contracts increasingly tighter as the object nears the event-horizon; time for the object slows, slowing to the point that it either appears stopped relative to the age of the universe, or is actually stopped; all entropy is lost to kinetic energy, thermal or Hawking radiation, and so the object cools to the point of becoming Bose-Einstein condensate; and the object's velocity increases, approaching the speed of light.

Relative to the outside observer, this event-horizon appears to be a zero-energy boundary that takes infinitely long to reach, with infinitely vast distances squished into infinitesimally small regions near the horizon itself. As the object approaches this boundary it will appear to be going near the speed of light (taking infinitely long to get there), and so cold it appears to contain zero entropy!

Accordingly, there's some doubt an object can actually fall through this boundary.

Nothing with mass can go the speed of light. Nothing can contain exactly zero entropy absolutely (you will sometimes see where this is implied but it is always relative to something else), or reach a point where increasingly vast distances are being squished into infinitesimally smaller regions. For the sake of your question, however, let's assume these restrictions don't exist. Let's also assume the problem of tidal forces wasn't an issue. From the perspective of the object falling towards the event-horizon, time wouldn't appear stopped, or space contracted, however the increase in velocity would be apparent, and so would the loss of energy. Nevertheless, the object would see itself falling for a very long time, approaching the speed of light, but never actually reaching that speed, and getting increasingly colder (stiller, with no vibration) There would come a point at which energy loss would be so great that even the strong and weak nuclear forces would break down and so nothing could or would hold the solid together.

Now to arrive at this we assumed that tidal forces played no role, but the fact of the matter is that these would actually be the first effects to rip a solid apart, at least long before the strong and weak nuclear forces stopped working. Let's assume for the sake of argument that a body did somehow make it past the event-horizon, gravity is still increasing as the body nears the centre of mass within the black-hole. So any effects within the event-horizon will be amplified exaggerations of effects seen at the horizon itself, that is, unless physics breaks down, or some other type of physics is in effect ... and that, we just don't know.


Inside the event horizon notion of 'center' ceases to exist. The 'distance from the center' has no meaning either. Actually, black hole has no 'center' at all! The singularity that exists in spacetimes containing black holes is space-like - so it can be interpreted as a certain moment of time for an infalling object. The objcet falling through the event horizon experiences tidial forces very similar (and by no means infinite) to those that create tides in the ocean because of moon's proximity. If those forces do not break the rigid object (e.g. it is made of really rigid material) the object survives intact. There can be perfectly normal interactions between objects that have falen beside the horison (if causality permits, of course, but this is no different that for objects outside BH). The singularity manifests itself in that an object that has fallen under the horizon meets it after some time regardless of what direction it chooses to travel to. So it reassembles death in a way - it gets you whatever you do and wherever you go :) it just needs some time. And, just like with death, noone really knows what happens after - the predictive power of general relativity breaks there.

  • $\begingroup$ No, predictive power of General Relativity breaks much earlier. Before anything crosses the horizon, the BH will evaporate, but GR does not account for this. $\endgroup$
    – Anixx
    Feb 25, 2012 at 0:49

No solid can be placed inside an event horizon because any solid matter under the horizon means information about it is lost.

Also a solid object inside the BH would allow to transfer information from inside, say, by changing its gravitational field by rotating it.

The both things are impossible thus no body containing information can be inside the BH.

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    $\begingroup$ this is completely wrong $\endgroup$
    – Terminus
    Feb 25, 2012 at 0:23
  • $\begingroup$ Terminus, can you please explain what you do not agree with? $\endgroup$
    – Anixx
    Feb 25, 2012 at 0:47

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