# Why is there no black hole-like activities inside a massive star?

Black holes are very dense objects in our universe, where a star that is several times more massive than our sun is compressed within a few kilometers. And here, the intense gravity of the black hole is not getting "added" after a star's collapse. Instead it is created by the same remnant of the star that exploded.

In simple terms, Newton's law of gravitation is enough to explain it. Just like the surface gravity of Betelgeuse is less compared to our sun's (if I am right), the gravity deep inside the star, let's say a few hundred kilometers from its center will be so intense, by Newton's law, we can say that the distance is so small. Then why is there no black hole-like activities there?

• I think your confusion might go away if you consider that the gravity force inside an homogeneous sphere is lower than at the surface and it will go to 0 at the center of the sphere. Commented Mar 7, 2022 at 16:46
• The gravitational attraction of the star is the sum of all the gravitational attractions of each particle in the star. If you are inside the star, some of the mass is pulling you away from the center. (And if you are at the center, all of it is pulling you away.) It's only when you are outside the surface of the star that you can begin to treat the star as a point mass at the center of the star. Commented Mar 7, 2022 at 21:59

The answer is that the cores of massive stars never become dense enough or centrally concentrated enough to form a black hole during all but the final few seconds of their lives.

The criterion for a (Schwarzschild) black hole is that its mass must be contained within the Schwarzschild radius. The fact that the mass is at the centre of a star is not relevant - the mass surrounding it has no effect.

The Schwarzschild radius is $$r_s = 2GM/c^2$$. If we write the mass as $$M = 4\pi r_s^3\rho/3$$, where $$\rho$$ is a density, then a black hole would be formed if the mass inside $$r$$ $$M(r) > \frac{c^2r}{2G}$$ $$\frac{4\pi}{3}r^3 \rho > \frac{c^2r}{2G}$$ $$\rho > \frac{3c^2}{8\pi G}\left(\frac{1}{r^2}\right)\ .$$ If we put this in sensible stellar units $$\rho > 3.3\times 10^8 \left(\frac{r}{R_\odot}\right)^{-2}\ {\rm kg}/{\rm m}^3.$$

Thus a black hole would form if the average density within a solar radius was more than 330 million kg/m$$^{3}$$.

If you want to consider the core of a star (say the inner $$0.1R_\odot$$), then the density threshold is 100 times higher.

Clearly, the density of the interior of a star cannot grow faster than, or even as fast as $$r^{-2}$$, otherwise the density would become infinite at the centre.

Thus the answer to your question is that the centres of stars never become dense enough or centrally concentrated enough, except in the late stages of a massive supernova when the core can collapse to $$\sim 10^{-5}R_\odot$$ amd the density does exceed $$3\times 10^{18}$$ kg/m$$^3$$.

Shell Theorem, proved by Isaac Newton:

• A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its center.
• If the body is a spherically symmetric shell (i.e., a hollow ball), no net gravitational force is exerted by the shell on any object inside, regardless of the object's location within the shell.

So in the center of a massive star, say 100 km from the center, the gravity of everything further than 100km from the center can be disregarded, and the net gravity becomes weaker with decreasing distance, in fact dropping to 0 in the center.

• It's the gravitational potential energy rather than the gravitational acceleration that is key. In particular, is the escape velocity at some point from the center so great that even light cannot escape? If it is, the object is a black hole. If it's not, the object is not a black hole. ProfRob's answer addresses this issue. Commented Mar 9, 2022 at 0:28

So can we say that the star that is hundred times massive than our sun, has a "black hole" in it?

The answer is no if the star is still undergoing fusion. It's only until fusion reaches its conclusion that even the most massive of stars can collapse into a black hole. Until then, the high temperatures that result from fusion works to counteract gravitational collapse.

• I can understand that the internal pressure is enough to prevent gravitational collapse, but just like the surface gravity of Betelgeuse is less compared to our sun's (if I am right), the gravity deep inside the star, let's say a few hundred Kms fromits center will be so intense, by newton's law, we can say that the r is so small . Then why there is no black hole like activities there ? Commented Mar 7, 2022 at 15:48
• No, the gravity force is lower near the center and highest at the surface. Commented Mar 7, 2022 at 16:49
• @pygri What you wrote is true assuming a uniform density. At any point inside an object with non-uniform density, gravitational acceleration increases with depth if the local density is less than 2/3 of the mean density of all the stuff at greater depth. This means that, for example, gravitational acceleration inside the Earth is greater than gravitational acceleration at the Earth's surface all the way down to the core-mantle boundary. Density variation is even stronger in a star. Gravitational acceleration increases with depth well down into a star's core. Commented Mar 7, 2022 at 17:34
• This should be the accepted answer. “the high temperatures...works to counteract gravitational collapse” is the crucial point. Commented Mar 8, 2022 at 22:34
• @leftaroundabout I disagree that this should be the accepted answer. ProfRob's answer is better. I however do not understand the downvote. Commented Mar 9, 2022 at 0:24

In the gravitational formula, there is a term $$r$$ that refers to the distance between two objects. This assumes that there is one single distance. But given two objects, such as the Sun and Earth, there are lots of different distances. For instance, the distance from the part of the Earth closest to the Sun to the part of the Sun closest to the Earth is smaller than the distance from the part of the Earth farthest from the Sun to the part of the Sun farthest from the Earth. So the former contributes more gravitational force than the latter. The distance between the Sun and the Earth is so much larger than the radii of the Earth and Sun that this difference is negligible, and we can get a reasonable approximation of the force by using one distance. Furthermore, it turns out that the gravitational field outside a spherical symmetrical object is the same as if all of the mass were at the center. This is known as the shell theorem.

If you go close to the center of the Sun, the radius of the Sun is no longer negligible. Furthermore, the shell theorem does not apply, or at least does not apply in the same way: the shell theorem is about how much gravity there is outside of an object, and you're no longer outside the Sun. Now that you're inside the Sun, if you want to use the shell theorem, you need to use it on the mass of the Sun that's closer to the center than you are, rather than on the total mass of the Sun.