# Working of black holes

Why does a black hole attract matter with such a huge amount of force? Does its mass increase on becoming a black hole? Is it due to its volume decreasing? In the formula for gravitational force, $\frac{Gm_1 m_2}{r^2}$, there is no mention of the volume of the bodies, just their masses. The amount of matter present must be same as that present before the collapse of the star, so why does gravity increase so much?

Black Holes exert no more gravitational force than other matter. They are just so massive that light can't escape. When they collapse, they typically lose some matter which means their mass actually decreases during the collapse. You are correct to note that gravity is independent of the volume of the object. Nearby to a black hole, Newtonian physics starts to break down which is where General Relativity comes in, but from afar, $\frac{GMm}{r^2}$ continues to hold true.

And as far as i think the amount of matter present must be same as that present before the collapse of star so why does it gravity increase manifold?

It doesn't. If you replaced our sun with a black hole of the same mass, it wouldn't change the Earth's orbit. Everything would stay the same - except for the noticeable lack of sunlight, of course.

You're right that the Newtonian formula for the gravitational force $$F = G\frac{Mm}{R^2}$$ does not say anything about the size of the attractor - the distance $R$ is measured from the center of mass of the attracting body. So one might ask what happens when $R\rightarrow 0$? The force appears to grow without limit. But we've forgotten that day-to-day attractors, like the Earth and the sun, have a nonzero size.

If we fly toward the sun, we will plunge into the surface (where $R=R_0$ is the radius of the sun) long before $R=0$. Once we're inside, the gravitational force becomes roughly

$$F = \frac{GMm}{R_0^2} \cdot \frac{R}{R_0}$$ which goes smoothly to zero as we reach the center.

On the other hand, a black hole is special because there is no surface to plunge into. Rather than emanating from a volume (which we could get inside, at which point the gravity would start to decrease), the force from a black hole appears to come from a single point. It therefore becomes possible to get closer and closer to the source of gravity, at which point the gravitational force grows without bound.

The weird stuff starts to happen when we get to a radial distance on the order of the Schwarzschild radius $$R_s = \frac{2GM}{c^2}$$

Plugging in the values for the sun, we find that $R_s \approx 2$ miles - but of course, the sun's radius is about 200,000 times that distance.

• I've had the impression that, although all orbits initiated under its effect would continue due to inertia, the passage of the last gravitational energy emitted by the star before its collapse would leave orbits initiated anywhere in the zone already passed by that energy unaffected by its gravity (although they could still enter the BH if their orbital trajectory would take them through it). I'm wondering if this conclusion appears correct to you, as you seem to have the math, whereas my impression is more from the philosophical idea of causal separation at event horizons. – Edouard Jan 23 '18 at 18:00

Approaching in a Newtonian sense, if you equate the escape velocity from a body to the speed of light, you will get yourself a black hole by definition.

$$v_{escape} = \sqrt { \frac{2GM} {R} } = v_{light} = 299 \ 792 \ 458 \ m / s$$

Note while the term $\frac {M}{R}$ is not density itself, it is obviously proportional to it; so this is why when any body is dense enough, it can become a black hole in some sense.

You are talking about black holes in a Newtonian / astrophysical sense. But, there black holes are a purely general relativistic phenomena. In particular, if we just consider non-rotating black holes, then, the existence of such solutions is by Birkhoff's theorem, namely: every spherically symmetric vacuum spacetime is static. The resulting spacetime is the Schwarzschild solution, which is the original "black hole" solution of Einstein's field equations. The black hole is a singularity that occurs at $r=0$, which is the result of computing the Kretschmann scalar: $K = R^{abcd}R_{abcd} \sim r^{-6}$.

In this context, one cannot talk about such solutions of Einstein's equations with terms like "force", "volume", "mass".

A separate question is that of how a star / astrophysical object can collapse into a black hole, and for that you must use the TOV equation: https://en.wikipedia.org/wiki/Tolman%E2%80%93Oppenheimer%E2%80%93Volkoff_equation

Integrating this equation / solving the ODE, one obtains for the pressure of the spherically symmetric object at some distance $r$, :

$p(r) = \mu \left[ \frac{\sqrt{1-r_s / R} - \sqrt{1 - r_s r^2 / R^3}}{ \sqrt{1 - r_s r^2 / R^3} - 3 \sqrt{1 - r_s / R}}\right]$

where $\mu$ is the mass density of the object, and $r_s$ I have denoted as the Schwarzschild radius: $r_s = 2G M$.

Now, look at this result: it is only valid for when $r \leq R$. The pressure at $r=0$ which is what you're concerned with is:

$p(0) = \mu \left[ \frac{\sqrt{1 - r_s / R} - 1}{1 - 3 \sqrt{1 - r_s /R}} \right]$.

This becomes negative, $p(0) < 0$, when the denominator of this expression becomes negative, since $\mu > 0$ by assumption. So, this becomes negative when:

$1 - 3 \sqrt{1 - r_s / R} < 0$.

This inequality is essentially the collapse condition. So, your object will collapse into a "black hole" as long as this inequality is satisfied.

You essentially use G.R. to see when the pressure of the object you are trying to "collapse" becomes negative. If your object is spherically symmetric, you can use the TOV equation to show that this collapse into a "black hole" occurs for $R>9/8 r_s$, where $r_s$ is the Schwarzschild radius. The spherically symmetric case is nice because of symmetry, but for general geometries, it is slightly more difficult. This hopefully answers your question. (On a side note, one can always have black holes without any such notions of things collapsing: just take a Minkowski spacetime and cut out a hole, you'll get the Schwarzschild metric!)