# Shouldn't some stars behave as black holes?

Some of the "smaller" black holes have a mass of 4-15 suns. But still, they are black holes. Thus their gravity is so big, even light cannot escape.

Shouldn't this happen to some stars, that are even more massive? (mass of around 100 suns) If their mass is so much bigger, shouldn't their gravity be also bigger? (So they would behave like a black hole). Or does gravity depend on the density of the object as well?

• just a thought: galaxy has much higher mass than sun. Should not galaxy behave like a black hole? – Umaxo Nov 26 '20 at 12:26
• You need an energy gradient. The early universe had an energy density much larger than e.g. a star just before turning into a black hole - but it was nearly the same everywhere, so the gradient was nearly zero, and the spacetime curvature was also nearly flat. – Luaan Nov 27 '20 at 9:16
• @Luaan While you definitely need a gradient for black hole behavior, I think the early universe comment is misleading. The (idealized) early universe did not have a flat spacetime, even though it had essentially no gradients. It was spatially flat in the sense that the Riemannian constant-time slices were flat (even this is not immediate from homogeneity and isotropy, i.e. the lack of gradients-- it is inferred from observation), but the spacetime was very much curved, as manifest in the growth of the scale factor, which encodes all of the curvature of the "flat" FLRW metrics. – jawheele Nov 29 '20 at 17:59
• Comments made by Vilenkin about the Borde-Guth-Vilenkin Theorem (which specifies that universes that are on average expanding cannot be eternal to the past) suggest that contraction is considered to precede expansion, in the deSitter spacetime on which inflationary cosmologies are generally based, simply because the contracting phase would leave no evidence of its having happened, and would consequently leave no observable phenomena allowing scientific verification: Consequently, I'd agree with jawheele that "early" is only an idealization. – Edouard Dec 1 '20 at 20:30
• The OP may have an impression that black holes originate only from the collapse of individual stars: One clear instance of them having formed from the collapse of material scattered through vastly larger regions has been observed, in Sagitarrius A, and is discussed at astronomy.stackexchange.com/questions/25466/… . "Dust" is sometimes used in a very inclusive sense in discussions of such observations, just as stars are, in some contexts, referred to as "particles". – Edouard Dec 1 '20 at 20:39

The true answer lies in General Relativity, but we can make a simple Newtonian argument.

From the outside, a uniform sphere attracts test masses exactly as if all of its mass was concentrated in the center (part of the famous Shell theorem).

Gravitational attraction also increases the closer you are to the source of gravitation, but if you go inside the sphere, some of the mass of the sphere will form a shell surrounding you, hence you will experience no gravitational attraction from it, again because of the Shell theorem. This is because while the near side of the shell is pulling you towards it, so is the far side, and the forces cancel out, and the only gravitational forces remaining are from the smaller sphere in front of you.

Once you get near the center of the sphere, you will experience almost no gravitational pull at all, as pretty much all of the mass is pulling you radially away from the center.

This means that if you can get very close to the center of the sphere without going inside the sphere, you will experience much stronger gravitational attraction, as there is no exterior shell of mass to compensate the center of mass attraction. Hence, density plays a role: a relatively small mass concentrated in a very small radius will allow you to get incredibly close to the center and experience incredible gravitational forces, while if the same mass occupies a larger space, to get very close to the center you will have to get inside the mass, and some of the attraction will cancel out.

The conclusion is that a small mass can be a black hole if it is concentrated inside a small enough radius. The largest such radius is called the Schwarzschild radius. As a matter of fact our own Sun would be a black hole if it had a radius of less than $$3$$ km and the same mass, and the Earth would be a black hole if it had a radius of less than $$9$$ mm and the same mass.

• @StephenG Thanks for the comment, I'm not trying to explain why large stars don't form black holes, just why a small, light object can be a black hole while a large, heavy object isn't, the point being that the entirety of the object must be inside the Schwarzschild radius. I'm not commenting on the stability of very massive very large object, all the statement I made are about a test mass at various points. – user2723984 Nov 26 '20 at 15:47
• A wrong but amusing argument: if you set $\tfrac12mc^2=\tfrac{GMm}{r}$ in Newtonian physics to discover the radius of a spherical mass $M$ with surface escape velocity $c$, you get $r=\tfrac{2GM}{c^2}$, which is just the Schwarzschild radius. – J.G. Nov 26 '20 at 21:06
• @user2723984 Well, it wouldn't be surprising if it were wrong by a factor, but thanks to dimensional analysis that's the worst that could happen. I'm trying to remember the name of an effect in atomic physics that's famously twice as large when you take special relativity into account. – J.G. Nov 27 '20 at 8:31
• @J.G. Are you thinking of the gyromagnetic ratio, which is 1 for a classical electron, 2 (Dirac) for relativistic, and 2.00........ when QED has thrown all its corrections at it. – Neil_UK Nov 27 '20 at 11:48
• @RobJeffries There's no gravitational field inside a spherical shell because the gravity from mass behind you and the mass in front of you cancel out. Which is what the answer says. – Carmeister Nov 29 '20 at 0:42

Stars generate a great deal of energy through fusion at the core. Basically the more massive a star is, the more pressure the core is under (due to the star's own gravity) and the more energy it can generate (somewhat simplified).

That energy of course radiates outward and heats everything outside the core making it a something like a pressure cooker, with heat creating pressure and the outer regions of the star being kept in place by it's own gravity. Stars would collapse into more dense objects (like white dwarfs and neutron stars and black holes) if this outward heat driven pressure did not exist.

Black holes are created when the fusion process can no longer generate enough energy to produce that pressure to prevent collapse and the star is massive enough so that it's gravitational field can compress itself so far it becomes dense enough to be a black hole.

• Also, when the pressure is high enough, it cannot prevent collapse because pressure is a component of the stress-energy tensor which is the source of spacetime curvature. So increased pressure leads to increased gravity, which leads to further increased pressure, etc, and that feedback loop causes runaway collapse of the star core. – PM 2Ring Nov 26 '20 at 17:13

Roughly speaking, for a star to become a black hole, its physical radius has to become smaller than its Schwarzschild radius. So even the Earth could be a black hole if it shrinks to below 9 milimiters. It is not precise to say that a black hole depends on the density of the object, since a Schwarzschild metric is a vacuum solution of Einstein's field equations.

• Exactly. Those stars simply occupy too much space to be black holes. You can weight the mass inside any sphere within a star, and you will never have enough mass in there for the sphere to be an even horizon. It's only when the stars compress down their core later in their life that the core may become smaller than its Schwarzschild radius, causing it to become a black hole. – cmaster - reinstate monica Nov 29 '20 at 14:51
• The extremely complex Kerr metric is also a vacuum solution, but the Kerr-Newman metric is not, I guess because it contains electrons, which are material. – Edouard Dec 1 '20 at 21:02

Or does gravity depend on the density of the object as well?

The problem with this question is that it's rather ambiguous as to what you mean by "gravity". An object doesn't have a single number that is its "gravity". If a ship is near a star, the gravitational force that the ship feels depends on the mass of the star, the mass of the ship, and the distance between them. If we consider the acceleration, rather than the force, then we can divide out by the mass of the ship. So rather than saying "gravity", I will talk about the gravitational acceleration. We can take the mass of the star as being fixed, but that still leaves the variable of the distance between them.

So the question is whether this distance is measured from the center of the object, or from the surface of the object. If the distance is measured from the center, then gravitational acceleration does not depend on the density of the object. If the Sun were to contract and become more dense, the orbit of the Earth would not be affected.

However, the less dense the object is (for a fixed mass), the further the surface will be from the center. So decreasing the density of an object decreases its surface gravitational acceleration. If the Earth were to expand in volume, but not increase in mass, then the gravitational acceleration at its new surface would be lower.

Also, it's more the escape velocity, rather than the gravitational acceleration, that determines whether something is a black hole. However, the escape velocity follows the same pattern as gravitational acceleration: the escape velocity relative to the center of an object does not depend on the density, but the surface escape velocity does. As a star collapses, its surface escape velocity increases, and once the surface escape velocity reaches the speed of light, it is a black hole.

If the visible matter became enough dense to be concentrated inside its Schwarzschild radius, it becomes a BH. Until their inner pressure withstand the gravitation they stay being stars.