How come the gravitational field produced by black hole is independent of mass distribution within the black hole? Suppose, for example, you have a specific massive object close to the surface of black hole , then wouldn't the gravitational field on that side of black hole somehow be different?(adsbygoogle = window.adsbygoogle || []).push({});

I guess I can PARTLY answer my own question. Namely, the very nature of black hole tells us that the object inside the black hole has no influence on what is outside. Since the influence includes gravitational influence, this means that the location of that object has no bearing on gravitational field outside.

However, this raises yet another question. If none of the objects inside the black hole have any gravitational influence on anything outside, wouldn't this imply that we shouldn't feel gravitational field of the black hole at all? If, on the other hand, we DO feel gravitational field it means that "by some mystery" the objects inside the black hole "mysteriously overcome" the "impossible" and do have the effect on the outside? If so, wouldn't that very "mystery" also imply that the gravitational field outside of the black hole will reflect the mass distribution inside the black hole as well?

I have one possible answer to the question of "gravity escaping black hole". Namely, the world path of an object falling inside the black hole goes to $t=\infty$ on its way to horizon and then it goes from $t=\infty$ BACK to finite $t$ once it is inside. This means that $t= \const$ hypersurface intersects the world path TWICE: one intersection is inside the black hole and one is outside. In light of this, it makes sense to say that we perceive the gravitational field coming from the intersection OUTSIDE of the black hole and not the one inside. Thus, any given object sends us gravitational signal exactly once, just like would be expected in the absence of black holes.

If that be the case, then the gravitational field outside the black hole is not really produced by black hole itself, but rather by all that dust that falls inside the black hole, which is still outside of its horizon. In other words these dust particles OUTSIDE of black hole horizon are attracting each other and thus "conspiring" in creating a black hole "in the middle" while none of them are in the black hole YET.

Now, if the above is, indeed, what happens, then the gravitational field outside of black hole would reflect the distribution of the dust that falls in. BUT the moment we write down Schwartzchild solution we are making an ASSUMPTION that the space is empty. Now, I realize just like any other person that making toy models is okay and even beneficial in many instances. But the catch is that toy model should not lead to LOGICAL contradiction. In case of black hole it does. The very existence of black hole LOGICALLY implies that there should be dust outside of it, and the "toy model" that there is no such dust is LOGICALLY impossible.

Now if we rule out this toy model on logical basis, we can no longer write down Schwartzchild solution and, instead, we would have to admit that the gravitational field of a black hole might violate spherical symmetry in response to possible violation of such symmetry by the mass distribution of the dust. This would result in violation of no-hair theorem. And you can't defend yourself by separating the gravitational field of a dust from the one of a black hole. After all, as argued earlier, the VERY gravitational field of a black hole IS in fact identical to the field of the dust OUTSIDE the black hole -- after all, nothing on its interior is capable of sending gravitational signals to the outside.

I guess the question I am trying to ask you guys is this. Is it possible to make a LOGICALLY CONSISTENT model of a black hole in such a way that

1. It, in fact, posesses gravitational field

2. It obeys no hair theorem

When I am trying to think about "logically consistent" pictures, I can not think of the one simultaneously satisfying 1 and 2 above, as explained earlier. But if you guys can, that would really help.

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# No hair theorem

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