The singularity is usual the infinite density of pure energy. But if normal matter like protons and electrons falls beyond the event horizon does this háve to turn into this infinite density of pure energy of the singularity or can it stay in a form of matter (perhaps neutron matter?) just near this singularity but don't annihilating the black hole?

Sometimes is stated that even our own universe is a black hole with it own event horizon, so this implies that normal matter could be inside a black hole. But I'm not sure if these things could be compared....

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    $\begingroup$ Hi Marijn The singularity is usual the infinite density of pure energy What is your source for this statement and what is pure energy? I ask because I think you might get d/v...... $\endgroup$ – user167453 Sep 4 '17 at 18:39
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    $\begingroup$ What is pure energy? $\endgroup$ – Kyle Kanos Sep 4 '17 at 18:45
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    $\begingroup$ physics.stackexchange.com/q/246061 $\endgroup$ – Marijn Sep 4 '17 at 18:56
  • $\begingroup$ We don't have a nailed down definition for energy, look at the article en.wikipedia.org/wiki/Energy , it's an historical record of changing definitions, so pure energy is very iffy as a concept. I read that article regarding energy density, but I honestly don't think that clears up energy, in general terms, which is what you have in your post. $\endgroup$ – user167453 Sep 4 '17 at 18:59
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    $\begingroup$ Sometimes is stated that even our own universe is a black hole with it own event horizon No, this is simply a misconception. physics.stackexchange.com/q/3294 $\endgroup$ – Ben Crowell Sep 4 '17 at 21:50

If you drop a normal matter particle into a non-rotating black hole it will quickly end up in the singularity (which is a region of infinite spacetime curvature rather than anything related to energy in mainstream general relativity (GR): attempts at quantum gravity has various other suggestions, but lets keep to straight GR). At least briefly there can be matter in the black hole separate from the singularity. However, the matter cannot stay apart from the singularity and will meet with it.

However, the meaning of "briefly" and "in" are slightly special here, since time and space are exactly what get bent here. In GR a black hole is described by the Schwartzschild solution, which is the solution of the field equations that correspond to an empty spacetime with a spherically symmetric mass at the centre. It describes how space and time curves, as well as the topology of manifold on which particles move.

The field equations are invariant under coordinate transformations: you can describe the same spacetime in very different ways. This is useful for untangling what is actually going on.

The standard Schwartzschild metric is a polar coordinate system that looks fairly "normal" at a distance from the black hole, but as you approach the centre it has a singularity. This is the real singularity: the manifold is indeed infinitely curved there. The metric also has a singularity at $r=2M$, which corresponds to the event horizon. This is not a true curvature singularity, it is just an awkward property of the coordinate system (a bit like the north and south pole in a latitude-longitude system on Earth: those spots are still normal parts of the surface, just with badly defined longitudes).

By choosing another coordinate system, Lemaître coordinates, that describe exactly the same situation but without the trouble at the horizon we can see what happens to freely falling particles.

[The metric is $ds^2=d\tau^2-(r_g/r)d\rho^2-r^2(d\theta^2+\sin^2\theta d\phi^2)$, with $r=[(3/2)(\rho-\tau)]^{2/3}r_g^{1/3}$]

Freely falling particles very conveniently keep a constant "radial" coordinate $\rho$ and the "time" $\tau$ corresponds to what would be measured by an observer following the particle. So a particle starting with (say) $\rho=10$ at $\tau=0$ will keep to $\rho=10$ as $\tau$ increases. Measured in Schwartzschild coordinates it will start out at infinite distance from the black hole in the far past, and as $\tau$ increases slowly fall towards the hole, moving ever faster.

When $(3/2)(\rho-\tau)=r_g$ it passes the event horizon, but from its perspective nothing unusual happens. In the Schwartzschild coordinates it seems to stop and shoot off to infinite future time: outside observers see its image being forever frozen and redshifted at the horizon, but this is the light from the particle, not the particle itself.

When $\tau=\rho$ it hits the singularity. So from time $\tau=\rho-(2/3)(r_g)$ when it passed the horizon it was "inside" the black hole, and it experienced $(2/3)r_g$ units of time inside. Note that the singularity is approached in a timelike direction: it is not a location an observer inside the hole can point at, but rather like an inevitable future point in time. This is the answer to the second part of your question: the matter cannot just "hang around" indefinitely inside, but will quickly meet the singularity just as an object existing right now will inevitably meet the next minute in a short while.

The timelike nature of the interior of the black hole (and hence the singularity) can be seen in the Schwartzschild metric too: when $r<2M$ the signs in front of the time term and the radial terms switch. So the brief proper time experienced between crossing the horizon and meeting the singularity corresponds to a time direction that is somewhat different from the outside world, and being inside doesn't look to the observer like being inside a normal 3D container.

The universe in relativistic cosmology has a very different spacetime from the black hole, the FLRW metric is the most popular model. It doesn't have the same kind of singularity, and the horizons that exist work a bit differently. This means that matter doesn't have to meet a singularity in a finite time.

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    $\begingroup$ The timelike nature of the interior of the black hole I'm not sure if this is a typo, or if you're using terms in a nonstandard way. In standard terminology, the singularity is spacelike, which means that it acts like a point in time. Is that what you meant? $\endgroup$ – Ben Crowell Sep 4 '17 at 21:42

A black hole can be observed to be made out of ordinary matter by a distant observer. The Schwarzschild metric is $$ ds^2~=~\left(1~-~\frac{2m}{r}\right)dt^2~-~\left(1~-~\frac{2m}{r}\right)^{-1}dr^2~-~r^2d\Omega^2 $$ Here of course $m~=~GM/c^2$. Now let us compute the path of a photon which has the null condition $ds~=~0$ We then integrate for a radial moving photon $t~=~\int dt$ so that $$ t~=~r^*/c~=~\int^r\left(1~-~\frac{2m}{r'}\right)^{-1}dr' $$ so that $r^*~=~r~+~2m~ln|r~-~2m|$. We then think of photon emitted from points ever closer to the horizon, so $r~\rightarrow~2m$. It is then clear that $r^*$ is going to be negative and of larger magnitude. This means that if you receive this photon it was emitted further back in the past. Of course as $r~=~2m$ it is infinite.

This means a black hole from the perspective of a distant observer is made of all the stuff that went into it, and it is in a sense piled up just above the event horizon. This material is for the most part ordinary matter. So from the perspective of the outside world black holes are made of ordinary stuff.

From the perspective of an observer in the interior things are maybe quite different. I wrote a post here on Stack Exchange on the Sen and Vafa model of a black hole singularity according to tachyons. I received some umbrage for this, even though it is not my construction.

There is then maybe a sort of complementary situation depending upon whether one is an exterior or interior observer of a black hole. It could be said though that because the black hole is made of ordinary matter from the exterior perspective that ultimately the same hold for the interior.

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    $\begingroup$ This doesn't answer the question. The OP asked whether infalling matter can arrive at some kind of equilibrium state inside the horizon, consisting of something like neutron matter, rather than accreting onto the singularity. $\endgroup$ – Ben Crowell Sep 5 '17 at 0:25
  • $\begingroup$ It answers the question completely. If with the tortoise coordinate matter of a black hole is just quantum modes on the horizon then it is like ordinary matter. It is also not in a state of complex compression or singularity. $\endgroup$ – Lawrence B. Crowell Sep 5 '17 at 0:54

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