# How can black hole increase its mass? [duplicate]

From observer point of view an object, which falls into black hole never crosses its horizon. Then how does black hole appears and grows its mass?

Or does any black hole looks (and feels by all other information sources) for us like a void sphere with all mass around its horizon?

## 3 Answers

The Schwarzschild coordinates (which seem to suggest that no object ever crosses the event horizon when viewed from far outside) were derived for stationary case: no matter flows onto the black hole, the black hole has constant mass. In fact, Schwarzschild was assuming zero stress-energy tensor (vacuum solution).

However if you start adding a lot of mass to the black hole, the situation changes. Imagine you throw a little object towards the event horizon. It "seems" to freeze on the surface of the horizon (it actually visually disappears due to the red shift). Later on, there is a huge amount of material streaming to the black hole. It is thousands of times more mass than the original mass of the black hole. At this point the conditions under which Schwarzschild found his solution no longer stand, because the stress-energy tensor is far from being zero. The event horizon will grow, since it forms wherever the gravitational potential reaches certain value. By adding more mass you unavoidably enlarge the volume where the potential has the required value to form the event horizon.

The case of non-constant mass is described by the Vaidya metric. Mathematically this is described on pages 133-134 of this book.

• I believe this answer doesn't solve the question posed as it is right now. Could you elaborate on how the Vaidya metric shows that an (very small) mass falling into a black hole won’t take an infinite amount of time to reach it for a distant observer?. The problem is that it isn’t intuitive from the link provided how that metric behaves differently than Schwarzschild’s on this particular aspect, as it is “just” a change of $M$ (constant) into a $M(v,u)$ (changing). Nov 16 '21 at 9:50
• Without a more explicit explanation on how that changes the problem, one is tempted to think that at moment $u_0,v_0$ the black hole has a mass $M_0$ and a Schwarzschild solution with mass ($M_0$) and therefore infinite time. At moment $u,v$ the mass will be $M(u,v)$ and at this instant we “just” have another Schwarzschild solution with mass ($M(u,v)$) and therefore infinite time again. So why is it not so? Does the explicit expression of $t$ (or $u$ and $v$) from solving the Vaidya metric shows anything different? Thanks. Nov 16 '21 at 9:50

Mass still falls into a black hole, it's just that from the point of view of the mass, its last instant outside of the hole is infinitely long. The mass falls into the hole, but just can't perceive that it has.

We detect black holes because they emit x-rays, have accretion discs, and have a very strong gravitational field. We cannot "see" the singularity because it is hidden beyond the event horizon - we only detect its presence indirectly.

That's more of an optical illusion than anything else. There is nothing that can physically prevent entering mass from continuing onto the center of the blackhole, so that's what it must do.

Theoretically, one could tell the difference between a true singularity with a concentrated central mass from a thin sphere with all its mass at the event horizon, but I don't we'll prove that anytime soon.

TTFN

• "Theoretically, one could tell the difference" how? Jun 8 '14 at 20:41
• What difference between illusion and not illusion? You can't preserve it using any force, even gravity (the information can't reach you by definition). Jun 8 '14 at 20:42