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No hair theorem

  1. Mar 19, 2012 #1
    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?

    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.
     
    Last edited: Mar 19, 2012
  2. jcsd
  3. Mar 19, 2012 #2
    I'm not sure what the problem here is. The Schwarzschild solution is just an empty spacetime which is curved and which has a singularity at the center, and which solves the Einstein equations. There is no logical contradiction here.
     
  4. Mar 19, 2012 #3
    If the distribution of mass inside the BH had a measurable effect outside the BH, then someone inside could manipulate the position of a mass to send coded messages to the outside measurer - yes/no answers to pre-formulated questions by moving the mass left/right... or using this method to send Morse code (slowly)... or in theory any digital format (voice, photo/video, data...) and if this did work then one would want to test that it worked in the reverse direction from outside to within (an internal measurer detecting changes in position of external masses - leading to a two way communication...

    Pretty sure the theory is that this is forbidden for the within to without direction, not so sure on the reverse direction...?

    I sent a note to Perlmutter years ago asking about how the external gravitational influence comes to escape or "gets out" of the BH and he wrote that it doesn't, it originates from the outside, but not significantly from mass outside, just curvature... what curvature was there at the time of initiation of the BH just stays there and maintains...
     
  5. Mar 19, 2012 #4
    Okay in order to say what you just did, one has to assume that all of the mass of the black hole is concentrated in the center. After all, if you substitute Schwatzchild solution into Einstein's equation you will see that the density is zero both inside the horizon AND outside. However, according to the ''standard'' understanding, only OUTSIDE the horizon the density is zero while inside it might be non-zero. After all, we are told that different density distributions on the interior of black hole are all equivalent as far as outside observer is concerned. And this statement is what creates the problem. Any interior distribution ''other than'' a point would logically DEMAND the presence of matter outside as well -- after all whatever is inside ''used to be'' on the outside at some point. Thus if we are sticking to the toy model that NOTHING is outside then we have to assume that interior mass is all at the center -- and this is somethign we are told not to assume.

    The other problem is the following. If we are to say that the black hole was ''formed'' at some point in time, we would again be forced to claim that, at any given time slice, there is some matter outside the black hole ''still'' falling in -- after all the point of crossing horizon is $t= \infty$. Thus, in order to naively say that we just have simple matter-free solution to Einstein's equation, we have to assume that black hole never formed, but rather that it is a simple ''topological'' structure that ''existed from very beginning''. Now in principle it is possible to make this claim. But that is not what most physicists believe. Most physicists DO believe that the black hole has, in fact, been ''formed''. And this is exactly what makes it incompatible with the toy model of empty space outside of it.
     
  6. Mar 19, 2012 #5
    I follow this argument, and this is precisely what I was thinking. But the problem is that it raises the following question: if you are not allowed to know ''where'' the mass is located inside the black hole, perhaps you are not allowed to know if the mass is there to begin with? Yet you do know whether the mass is there, and you know how much mass, by analyzing the gravitational field outside. So this implies apparent contradiction. On the one hand, any given particle inside the black hole can send gravitational signal and tell you it ''exists'', but on the other hand it can't tell you ''where'' it exists.
     
  7. Mar 19, 2012 #6
    Okay, maybe I can try to take the issue of $t= \infty$ out of the way by considering two blackholes of equal mass falling into each other and forming ''mutual horizon''. In this case, if each black hole has mass M and radius R=2M, they would form mutual horizon when their centers are separated by R'=4M=2R away from each other. Thus, they can reach this configuraiton within finite t (after all, you only need infinite t to reach R=2M, not R'=4M)

    Now in this case we can ask the question in the following way. Suppose we take for granted the gravitational field of each of the ''small'' black holes, and our task is to see whether the gravitational field of a ''big'' black hole will be what we expect. Before the mutual horizon forms, gravitational field is NOT spherically symmetric. If gravitational field is to ''become'' spherically symmetric ''immediately'' after horizon forms, then Einstein's equation would imply singularity.

    If, on the other hand, a ''short time'' after horizon forms the spherical symmetry continues to be violated, then we can simply say that we are NOT sensing the gravitational field from the interior of ''big'' black hole; rather, we are sensing ''lingering'' gravitational field from the time the ''big'' black hole did not exist. This, in fact, is pretty logical to say. After all, the fact that spherical symmetry ''immediately after'' the formation of ''big'' black hole will lead to singularity is ''logically related'' to the gravitational field ''before'' the ''big'' black hole was formed, which, in turn, makes it logical to call it ''lingering gravitational effect from the past''.

    Now, our claim, therefore, is that the gravitational field will get ''closer and closer'' to the Schwardschild solution in the ''distant future''. This means that ''lingering gravitational field'' does not dissipate; but instead it simply redistributes itself into spherically symmetric form. Now, any linear field would be expected to dissipate. Thus, the only possible explanation why gravitational field does not is its non-linearity. Physically, it means that gravitons get ''gravitationally attracted'' towards the ''big'' black hole, which prevents them from flying away, which, in turn, causes other gravitons to be ''physically attracted'' and prevents them from flying away, and so forth. In other words, the ''big'' black hole never emits a single graviton; we simply having ''lingering'' gravitons from the past that are keeping each other from flying away!

    Now, we can go ahead and make similar claim regardnig the ''small'' black holes as well. In other words, the gravitons that were surrounding ''small'' black holes were, likewise, ''lingering gravitons from the past'', namely from the time that small black holes were forming. So if we are to trace down these ''lingering gravitons'' to the time they were ''actually'' emitted, that would be the time before ANY black hole has formed. What do you guys think? Is this the answer to my question or do you have some other answer?

    Here is the question I still have Were there any studies that were looking at the dynamics of the formation of mutual horison and subsequent evolution of gravitational field? I mean, what I described above was just a ''logical'' guess, devoit of any numeric verification; so I would like to know whether or not simulation of gravitational field shows the picture resembling what I was guessing.
     
  8. Mar 20, 2012 #7

    PAllen

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    See:

    http://www.black-holes.org/explore2.html

    and look down the page for merging event horizons. The whole site should be of interest to your questions.

    Note also, the no hair theorems only state that a black hole that doesn't grow will eventually settle into a state described only by mass, charge, and angular momentum. It is allowed and predicted that this can take a while, especially for a rotating black hole.
     
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