# Solid objects inside the event horizon - can they remain “solid”?

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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This is covered well in "spaghettificiation" near a neutral nonspinning black hole singularity. – Ron Maimon Aug 25 '11 at 17:07
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. – AdamRedwine Aug 25 '11 at 17:29
@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. – David Zaslavsky Aug 25 '11 at 22:11
@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. – Ron Maimon Aug 26 '11 at 1:24

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.

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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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this is completely wrong – Terminus Feb 25 '12 at 0:23
Terminus, can you please explain what you do not agree with? – Anixx Feb 25 '12 at 0:47

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.

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 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. – Anixx Feb 25 '12 at 0:49