# What's inside a frozen star?

Consider a specific black hole whose whole mass is located at the event horizon. How such a black hole can form is beyond he scope of this question. Perhaps a spherical shell made of the neutron star material has collapsed or whatever. For a remote observer, the matter at the horizon is frozen in time, so the black hole appears empty with no singularity inside.

My first question is if this concept is mathematically viable. Again, I am not concerned with the physics of how such a black hole can form. I only wonder if it can exist in principle with all its matter located at the horizon in the view of a remote observer.

According to the Schwarzschild solution, the gravity outside this black hole would seem to be the same as of a black hole with a singularity inside. However, obviously, the conditions inside would be very different. What would they be?

Normally, the gravitational time dilation inside an empty shell is the same as at the (inner surface of the) shell. If the same logic works for this black hole, time inside would not move and the entire volume would be lightlike. However, this intuition may be an oversimplification.

I would appreciate any insight, thoughts, comments, references, and constructive criticism of this concept.

• the entire volume would be lightlike What do you mean by that? You cannot have a 4D (spacetime) that is lightlike, metric must be nondegenerate. – A.V.S. Jul 2 '18 at 15:12
• @A.V.S. As a disclaimer and clarification: I am not stating what is inside - this is precisely what I am asking instead. So the "lightlike volume" is just a guess or an example of what may or may not be inside. It is perfectly fine with me if it is not. What I mean by it is that the properties of the volume being the same as the properties of the shell - any interval inside is lightlike (a thick event horizon). You seem to mean this is not possible and it is perfectly fine with me, it was just a guess anyway, but if you explain it in an answer that is convincing, I'd be grateful by 50 points. – safesphere Jul 2 '18 at 15:36
• @A.V.S. For example, why must the metric not be degenerate? Why not let it be degenerate inside? What is an overriding argument against a degenerate metric in this case? Sorry if I make no sense or if this argument is silly. I am not a mathematician but I have enough background in differential geometry to understand an expert answer. – safesphere Jul 2 '18 at 15:42

Ordinary matter (with realistic equation of state) cannot be static at (or near) the black hole horizon.

If we assume that matter is concentrated in a thin spherical shell, then the metric across this shell must satisfy certain junction conditions, (first obtained by W. Israel in 1966).

We are "gluing" together two spacetimes along a hypersurface of codimension one (this would be the worldvolume of a thin shell). First, the metric induced by both spacetimes on this hypersurface must be the same. Second, the discontinuity in extrinsic curvature across the hypersurface is connected with the stress-energy tensor of the shell integrated over its thickness.

Schematically: $g_{\mu\nu}$ is continuous, with kink, $\Gamma^\mu_{\nu\lambda}$ has step discontinuities, $G_{\mu\nu}$ has $\delta$-like singularities.

If we know the hypersurface beforehand, then junction conditions give us stress-energy tensor of this shell. We then might wish to check if this tensor satisfies appropriate energy conditions. If however, we only know the type of matter that constitutes the thin shell, then junction conditions together with the equation of state for the shell matter would provide the equation of evolution for the shell. A simple overview of the formalism with explicit formulas for the spherically symmetric shells could be found in:

• Goldwirth, D. S., & Katz, J. (1995). A comment on junction and energy conditions in thin shells. Classical and Quantum Gravity, 12(3), 769, web, arXiv.

• Sato, H. (1986). Motion of a shell at metric junction. Progress of Theoretical Physics, 76(6), 1250-1259, open access web.

Now, one of the spacetimes we want to glue is a Schwarzschild metric. One can glue together Schwarzschild metric for the exterior with flat Minkowski space for the interior and obtain a (generally) nonstationary spacetime describing collapse of a shell into a black hole. For the dustlike matter this has been first done by Oppenheimer & Snyder in 1939 (free to read).

Let us outline such solution in more detail. Let us assume that the motion of the shell if given by its radial coordinate $R(t)$ as a function of Schwarzschild time (time of outside observer, even though Schwarzschild coordinates are ill suited for analysis of the horizon crossing). For the metric we use: $${ds }^{2}=\begin{cases}\left(1-{\frac {2M}{r}}\right)\,dt^{2}-\left(1-{\frac {2M}{r}}\right)^{-1}\,dr^{2}-r^{2}d\Omega^2,& r > R(t) \\ f(t)^2\,dt^2-dr^{2}-r^{2}d\Omega^2, & r < R(t)\end{cases},$$ That the interior metric is Minkowski space is easy to see if we note that the proper time of static observers inside the shell is given by $dT=f(t) dt$. The function $f(t)$ is yet unknown and is given by the requirement that the proper time of the shell induced by interior metric is the same as proper time induced by exterior metric: $$\left(1-\frac{2M}{R}\right)-\frac{{\dot R}^2}{1-\frac{2M}{R}}=f(t)^2-{\dot R}^2,$$ where $\dot R \equiv \frac {dR}{dt}$ .

The remaining equation from junction conditions allows a simple physical interpretation: total mass $M$ is a sum of relativistic kinetic energy of the shell (including rest mass) and gravitational binding energy: $$M=\frac{m}{\sqrt{1-(\frac{dR}{dT})^2}}-\frac{m^2}{2 R},$$ where it is written in a most simple form for flat Minkowski space of the interior. Integrating this equation gives us the function $R(t)$ in terms of the Schwarzschild time $t$. This function asymptotically approaches Schwarzschild radius $r=2 M$ as $t$ approaches infinity, so to describe what happens with the shell inside the horizon we would need another set of coordinates. As one can expect, proper time measured by observers moving with the shell itself, as well as time $T$ measured by static interior observers remains finite both for horizon crossing and for singularity formation (when $R\to 0$).

While by the clock of an outside observers the shell never crosses the sphere $r=2M$, nevertheless there is a horizon at a finite times $t$ in this spacetime. It starts at origin at some moment $t_0$ and asymptotically approaches $r=2M$ from the inside as $t\to\infty$. So all stationary observers strictly inside the sphere $r=2M$ would enter the black hole horizon at a finite time $t$.

We may also consider a static shell with the radius $R$ larger than $2M$ (in GR units). For the interior solution we could simply choose Minkowski space (spatial ball of a radius $R$ and time dilated just like the outer shell surface). The stress energy tensor would have positive energy density and positive pressure, needed to keep the shell in equilibrium. Of course, no horizon is ever formed for this spacetime, and signals from the inside could reach infinity.

However, if we try to glue exterior Schwarzschild metric along the event horizon, we would encounter problems. This hypersurface has null Killing vector field as well as Killing vector fields generating rotations, so the interior spacetime also must have Killing horizon with a spherical topology. This requirement excludes Minkowski spacetime as well as many other simple spacetimes which do not have the horizon. What is left (if we restrict ourselves to interior solutions without matter and with spherical symmetry) is another outside Schwarzschild metric or the De Sitter space.

The first case (two exterior Schwarzschild metrics glued together along the horizon) is known as Einstein–Rosen bridge and is historically the first (1916, 1935) example of a wormhole. However, such solution is unstable, it decays into a pair of black holes under small perturbations.

One could even go further, and identify the mirror points on two sheets of the Einstein-Rosen solution and obtain a spacetime without any interior: the spacetime simply ends at the null boundary. Such manifold would be geodesically incomplete, but if we have the appropriate rules for the boundary, we would be able to describe what an outside observer sees, how the boundary responds to perturbations. etc. This is provided by black hole membrane paradigm.

The second option, de Sitter space in the interior, Schwarzschild for the exterior, would be a type of gravastar (gravitational vacuum star), a relatively recently proposed alternative to black hole. While original proposal by Mazur and Mottola has a shell of finite thickness, thin shell simplification could be found here:

• Visser, M., & Wiltshire, D. L. (2004). Stable gravastars — an alternative to black holes?. Classical and Quantum Gravity, 21(4), 1135, web, arXiv.

Aside: Manifolds with boundary are not the same as regions with degenerate metric, those a simply are not allowed in classical GR, since GR is a study of Lorentzian manifolds, however some versions of quantum gravity are expected to have phases with degenerate metric. For example, strings above Hagedorn temperature are expected to form more symmetric, topological phase with zero metric.

If we leave confines of the classical general relativity and look for a proposed modifications/extensions then there are quite a few alternatives to the usual descriptions of black holes, which have some form of structure/matter at the horizon (or where the horizon should be if it does not forms): gravastars, AdS bubbles, 2-2 holes, superspinars, fuzzballs, collapsed polymers, quantum bounces. For a sample (with an eye toward observational signatures of such Ultra Compact Objects — UCOs) have a look at the following paper:

• Cardoso, V., & Pani, P. (2017). The observational evidence for horizons: from echoes to precision GW physics. arXiv:1707.03021.
• The question asks about a shell at the event horizon as viewed by a "remote" observer. This isn't a static shell it is spherically symmetric infalling matter. – Rob Jeffries Jul 5 '18 at 3:05
• @RobJeffries Hi Rob and thank you for your comment! I'd like very much to understand this matter better and would appreciate your help. It seems to me that the shell is static for every observer outside of it. Is this not true? – safesphere Jul 5 '18 at 11:43
• @safesphere No, what you are asking about is not static. Do you think if something falls into a black hole, that that is treated as a static situation? Whereas this answer appears to be about a static shell. Please be clear about whether you are talking about spherically symmetric infall or a static spherical shell. – Rob Jeffries Jul 5 '18 at 12:13
• I am glad you've pointed this out, because it is exactly the point of this question that there may be no such distinction. The shell is asymptotically static for all external observers. Matter outside of the EH is falling with a deceleration. Once a quantum distance to the EH is reached, the shell becomes static for all outside observers. Is it not? (I would be happy to assign 50 points for a convincing answer). Thanks! – safesphere Jul 5 '18 at 13:24
• @safesphere BTW, I think this answer is fine in describing solutions to GR for shells, but as you can see in paragraphs 6 and 7, the collapsing Oppenheimer &Snyder solution is not static and a static shell must have a radius larger than $r_s$ and would not be described as a black hole. – Rob Jeffries Jul 5 '18 at 13:25

If all the mass is located on a spherical shell, the object would just be a classical ball, not a black hole. There would be no event horizon or any sizable time dilation.

• not true, when you build a Kugelblitz, see youtube.com/watch?v=v3hd3AI2CAA&t=4m10s the Kugelblitz only converges to r=rs, but never reaches it, in the external observers frame of reference, and therefore stays a hollow shell. Only in the system of the internal and infalling observers the collapse happens in a finite time, for the external observer it takes an infinite time. – Yukterez Jul 3 '18 at 0:59
• @СимонТыран The video says, we wouldn't see the he Kugelblitz light coming until it hits us. They fail to mention that long before it hits us we would see the uninverse blueshift until it fries us with gamma rays. – safesphere Jul 5 '18 at 22:07
• We would not see the universe blueshift before the Kugelblitz hits us, why should we – Yukterez Jul 7 '18 at 1:40
• @СимонТыран Sorry I did not see your question before, because you did not address it to me with the @ sign so I was not notified. I see your point. We would see the universe blueshift, if the shell is collapsing slower than light. However, if the shell moves with the speed of light, then it comes before the blueshifted light from the universe. The light that is already inside the shell will not blueshift. So yes, you are correct, we would be hit before we know. – safesphere Aug 24 '18 at 21:52