# When a neutron star accumulates enough mass, does a black hole start at its center?

With less dense bodies, such as the earth and the sun, the center has less gravity/density (since there's an equal amount of mass surrounding the center, pulling out on it from all directions).

Does this also apply to neutron stars? That is, is gravity/density highest in a shell near the center, but not at the center?

If so, wouldn't countless black holes form in said shell and start merging until one large black hole exists?

More precisely, wouldn't groups of quarks fuse beyond the critical density beginning in the most dense part of the neutron star (the shell surrounding its core), resulting in a shell of merging microscopic black holes surrounding a core of quark-gluon soup technically still outside any event horizon until the merging black holes reach it?

• In most objects, the center is generally the densest part. Also, if an object did collapse into a black hole, the entire thing would collapse, not just part of it. There would be no merge. But yes, neutron stars can collapse into black holes if they are massive enough. Commented Sep 2, 2015 at 18:10
• @Paul You're mistaken on your hypothesis that the center has less density: d2vlcm61l7u1fs.cloudfront.net/… (Earth. Note the axis is "depth") aanda.org/articles/aa/full/2002/12/aa1471/img21.gif (neutron star, note the axis is "radius")
– user12029
Commented Sep 2, 2015 at 18:14
• @HDE 226868 and NeuroFuzzy That makes sense to me. I wrongfully assumed that the drop in gravity at the center reduced the density by some degree.
– Paul
Commented Sep 2, 2015 at 18:52
• Here is a paper discussing simulations of the formation of a black hole from a collapsing neutron star: Richers, "Equation of State Effects on Gravitational Waves from Rotating Core Collapse," arxiv.org/abs/1701.02752 . There are a lot of unknowns. It is not even really established whether real-world astrophysical gravitational collapse of a star leads to a black hole. It may lead to a naked singularity: Joshi and Malafarina, "All black holes in Lemaitre-Tolman-Bondi inhomogeneous dust collapse," arxiv.org/abs/1405.1146
– user4552
Commented Sep 12, 2017 at 22:06
• The OP makes an argument based on density, which is clearly not right for the reasons explained by Timaeus. But I also don't think Timaeus's answer is valid, for the reasons given in comments.
– user4552
Commented Sep 12, 2017 at 22:31

Black holes (and event horizons) are not caused by density. And yes, they always start at the center, even when 100% of the mass and energy is located on the shell far away from the center.

First thing to remember is that you don't "feel" an event horizon. In fact, if aliens were headed towards the Earth from all directions, and they were hoarders (or just preferred to bring their planets with them rather than building space ships), then we might be inside an event horizon right now and not even know it! You don't feel an event horizon, and it does not form because of large density.

You can even have an infinite density and still not form an event horizon.

So let's ask what an event horizon actually is. Firstly, it's a surface that has an inside and an outside. Secondly, it's a one way surface. Things can go from the outside to the inside but not vice versa. Thirdly, the observers that stay outside never see the inside, not because you have a Faraday cage or something that blocks light, but because of the geometry not allowing even the speed of light to cross the horizon.

You can think of the horizon as something that's itself passing past any event at the speed of light going outwards.

For instance if you have an observer that's accelerating at a constant rate (and direction) so that in Minkowski space they would have a hyperbolic notion, then that observer has a horizon, the boundary between events they see and those that they never see. And it is a perfectly normal surface moving along a fixed direction at the speed of light.

It's a "last call", like if you send a light signal now you can reach that observer eventually but if you wait past this event then you will never reach them. It doesn't feel like a "thing" per se.

So of a giant shell of matter were collapsing around you and you were regular matter and the shell is slowly contracting there might be a point where it gets too concentrated (too much on the inside compared to the surface area). Then from the outside it would seem like a black hole. The black hole is an event horizon.

It is basically grouping all the observer that stayed outside and marking a point when you could contact them and when you couldn't.

For instance in the above example if you didn't want the aliens contracting on you from all directions and trap you inside a black hole, you could send a them message and if you send it early enough you could tell them to stop contracting (whether they'd oblige is of course, up to them).

At some point if you hadn't had sent the message, it would've been too late now. Your message wouldn't get to them in time. You waited too long. That is the event where the event horizon forms. In the center, at the moment you waited too long. It expands at the speed of light in all directions. It might merge with other event horizons that also started at events and then started expanding at the speed of light. Eventually it expands and hits the surface.

If the surface were far from the center it might be a very long time before a singularity forms. But the event horizon started as soon as it was too late (depending on the location) to get to the outside.

So you can even have a completely empty spherical ball and then have an infinitely dense shell of mass on the outside. As long as you don't have too much mass per surface area, you are fine. So a mass that's infinite per unit volume but still finite per unit area, given that the area is bigger than $4\pi (2m)^2$ for the mass $m$, things would still be fine. But if the shell contracts to have too little surface area then it gets engulfed by an event horizon.

In which case the event horizon would start at the center, and just in time that it needs to expand to become the boundary around the shell before the critical mass per area were reached (versus after and thus being trapped inside forever).

And the exact critical moment actually just depends on the geometry. The whole thing is just geometry! The moment geometry stops letting escape, horizon forms. It expands at the speed of light (otherwise you could cross it). And it does so for an inertial frame. Eventually it gets to a point where expanding at the speed of light means staying at a place that has the same local geometry. When it gets to the critical surface it can expand at the speed of light and have that same geometry outside and inside. And that trapped surface is really what makes the event horizon "stick."

So in essence the horizon starts on the inside, and keeps expanding until it just reaches the "outside", then it just "sticks" onto there.

• In the case of spherical symmetry, you're certainly right. It's pretty trivial to see this on a Penrose diagram, where the event horizon is a null surface that originates at the center of symmetry. But you're just assuming spherical symmetry, and that's basically assuming what you're trying to prove. In the scenario the OP has in mind, there would not be spherical symmetry in the first place. It seems perfectly possible that the event horizon would initially consist of disconnected pieces.
– user4552
Commented Sep 12, 2017 at 22:29
• If you look at realistic simulations of gravitatational collapse of a neutron star, such as Richers, arxiv.org/abs/1701.02752 , they totally lack spherical symmetry. There is an initial collapse, at least one bounce, and complicated flows and shock waves.
– user4552
Commented Sep 12, 2017 at 22:34

Your question implicitly assumes that above some critical density, a region of space will collapse into a black hole. Well, let's find that density:

The Schwarzschild radius is:

$$r_s = \frac{2 GM}{c^2},$$

and density is given by:

$$\rho = \frac{M}{V} \approx \frac{4 M}{3 r^3}$$

Which means that

$$\rho_s = \frac{3M}{4 \pi\left(\frac{2 GM}{c^2} \right )^3} = \frac{3 c^6}{32 \pi G^3 M^2}.$$

That's problematic because the density depends on the mass... But from this relationship, we can see that as the mass gets bigger, the critical density actually gets smaller! Let's try finding the density in terms of radius instead:

$$M = \frac{4 \pi r_s^3 \rho_s}{3} = \frac{r_s c^2}{2 G}$$ $$\rho_s = \frac{3 c^2}{8 \pi G r_s^2}$$

As you can see, we have a similar relationship: As the radius gets bigger, the critical density still shrinks.

This is not a full answer because HDE's comments cover your main point (I think), it's just to deal with your first question:

With less dense bodies, such as the earth and the sun, the center has less gravity/density (since there's an equal amount of mass surrounding the center, pulling out on it from all directions).

An estimate of the densities of the various layers of the Earth

Density of Earth Layers

Estimates vary, but some approximate values should be as follows (in grams per cubic centimeter):

Continental Crust:    2.7 to 3.0
Oceanic Crust:        3.0 to 3.3
Mantle (silicates):   3.3 to 5.7 (increasing with depth?)
Outer Core (liquid):  9.9 to 12.2
Inner Core (solid):  12.6 to 13.0


The rest of your question:

More precisely, wouldn't groups of quarks fuse beyond the critical density beginning in the most dense part of the neutron star (the shell surrounding its core), resulting in a shell of merging microscopic black holes surrounding a core of quark-gluon soup technically still outside any event horizon until the merging black holes reach it?

seems to be based on the idea that density is less at the centre.

Pressure at Earth's core might be related to your question.

• Thanks Acid Jazz, as stated in the link you provided "pressure is dominated by the matter far from the centre where the gravity is not zero". I'm not a physicist, but after thinking about it that does make sense, and if true, negates my conclusion.
– Paul
Commented Sep 2, 2015 at 18:47
• Hi Paul, I think you might be getting confused with the idea that, if someone dug a hole in the middle on the Earth, would you be weightless? Yes, you would, as gravity is cancelled out all around you. If you look up Newton's Sphere/Shell it's an interesting version of the same sort of idea.
– user81619
Commented Sep 2, 2015 at 18:53

Just for fun, though I think this was largely covered already. The Earth's inner core is by far the most dense part of the earth (lets say 13 grams per CM), but it's also quite small by comparison. It's only about 20% of the Earth by radius (and less than 1% by volume, less than 2% of the mass).

So, on the surface of the Earth's Iron core, ignoring the outer layers, you'd be 5 times closer to the center of gravity (corresponding to 25 times greater gravity given equal mass), but the mass is less than 1/50th - so the gravity is actually lower using the Earth as an example.

Sources: Source_1 and Source_2

Though, curiously, I think you may actually be right that the black hole may form inside first. I'm not positive and it could only happen if the inside collapsed to much greater density than the outer layers, essentially undergoing collapse as the Degeneracy or degeneracies are overcome by the pressure.

As the neutron star undergoes collapse, it's entirely possible that the black hole first forms inside and expands outside as the outer matter falls in. But I'm just theorizing. I could be wrong.