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The famous "No-Hair Conjecture" states that a blackhole can have only 3 hairs: Mass, Angular Momentum, and Electric Charge. It occurred to me that the basic underlying reason behind this might be that the interior of a blackhole can't causally influence the exterior. For the interior to not affect the exterior, it appears to be a necessary condition that the observations made in the exterior should only depend on the conserved quantities of the interior. Otherwise, a cosmohiker who has passed the horizon may change a non-conserved quantity (say, the number of particles) and (through the dependence of exterior observations on this quantity) send a message to the exterior. But no matter what he does, a conserved quantity is not going to change and thus, the exterior observations can possibly depend on only such conserved quantities.

I realize that it just puts an upper limit on how many hairs the blackhole can have and there certainly is a greater number of conserved quantities than the three included in the "No-Hair Conjecture" and the rest of them are not considered as hairs of the blackhole, e.g., the baryon number. So, this reasoning can not explain the whole picture but can a reasoning based on the causal arguments explain the rest of the picture? More importantly, is this reasoning appropriate at least for what it seems to explain (that a hair of a blckhole must be a conserved quantity)? Also, if the blackhole hairs are related to conservation laws then one should be able to relate them to symmetries. Is there any such interesting link that is known?

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    $\begingroup$ Maybe it's worth noting that the No-Hair Theorem is a theorem, not a conjecture. It says that in GR + EM, the stationary black hole solutions have a certain form, the Kerr-Newman metric. $\endgroup$
    – user1504
    Commented Jun 29, 2017 at 11:17
  • $\begingroup$ @user1504 Yes, the statement you asserted is certainly a theorem. But I think by the no-hair statement, we mean more than that. I mean we expect it to hold for non-classical (i.e. real) and non-eternal black holes as well. $\endgroup$
    – user87745
    Commented Jun 29, 2017 at 12:00
  • $\begingroup$ Related: physics.stackexchange.com/questions/335410/… $\endgroup$
    – gj255
    Commented Jul 4, 2017 at 18:56

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Your line of argument cannot be fleshed out to make a proof of the no-hair theorems because it omits two necessary assumptions, stationarity and electrovac. If you omit stationarity, you have counterexamples such as the one in AGML's answer. If you omit electrovac, then there are counterexamples where the black hole is coupled to other fields besides the electromagnetic field.

More importantly, is this reasoning appropriate at least for what it seems to explain (that a hair of a blckhole must be a conserved quantity)

No-hair theorems (unlike Birkhoff's theorem) take stationarity as an assumption. So it's trivially true that if you can write down a definition of some property of a stationary black hole, then that property is conserved. For example, the maximum value of the Kretschmann scalar outside the horizon is guaranteed to be conserved. There are infinitely many such (automatically) conserved quantities. A no-hair theorem has to do something very different: it has to prove that out of all these conserved quantities, only three are independent parameters, and all the others can be determined from them.

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  • $\begingroup$ Tangentially related to my question: If we consider a spherical shell concentric to a black hole currently outside the black hole which - under the effect of the gravity of black hole is (spherically) symmetrically shrinks towards the black hole and gets absorbed - making no gravitational waves thanks to the spherical symmetry of the motion. In such a case, the moment the shell crosses the horizon, the black hole should be described by only 3 parameters and would be indistinguishable from an eternal black hole. Is this claim correct? $\endgroup$
    – user87745
    Commented Jul 5, 2017 at 21:43
  • $\begingroup$ Thanks for your answer and elucidating the importance of the stationary assumption in the no-hair theorem. The part you told about other conserved quantities in such cases make obvious sense and thus, I only tried to make a partial statement that this (my) reasoning should put only a qualification on a possible hair of the black hole that it must be a quantity that can't be affected by an insider (which I think should be a locally conserved quantity). Is this partial statement appropriate? $\endgroup$
    – user87745
    Commented Jul 5, 2017 at 21:47
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Well, in the case of an isolated black hole, you certainly have some symmetries. Considering a horizon as a null 3-surface, with null tangent vector $\ell^{a}$ and (degenerate) 3-metric $q_{ab}$, the isolated horizon condition can be $£_{\ell}q_{ab} = 0$, which automatically makes $\ell$ a killing vector to the surface, and induces a conserved quantity, which we call the mass.

Furthermore, you have, normal to $\ell$, a spacelike 2-surface at each point on the horizon. Gauss's law works perfectly well here, which gives you the charge conservation law. And, if this surface has a rotational killing vector $\phi^{a}$, then you have an induced angular momentum.

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  • $\begingroup$ This can all be made more rigorous than what I said above (and extended to non-isolated horizons), but to first approximation, this is the idea. $\endgroup$ Commented Jun 28, 2017 at 21:57
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    $\begingroup$ This doesn't seem to answer the question. The OP outlines an (incorrect) line of attack to attempt to prove a no-hair theorem based on the conservation of the three quantities that are referred to in no-hair theorems. You outline an argument that these three quantities are well defined and conserved (in an asymptotically flat spacetime?), but that has nothing to do with no-hair and doesn't answer the OP's question about whether his line of attack is correct (which it isn't). Your outline doesn't get at no-hair at all (and can't, because it doesn't invoke necessary assumptions like electrovac). $\endgroup$
    – user4552
    Commented Jul 4, 2017 at 18:23
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Restricting to vacuum, the conserved mass and angular momentum are respectively conjugate to the timelike and rotational symmetries of the black hole. No-hair may roughly be equivalently stated as "isolated black holes do not radiate and are axisymmetric".

This is obviously not true of all black holes. For example the members of a black hole binary tidally deform one another and emit gravitational radiation, so you will need more than these parameters to describe them. Similarly, if you throw something really big into a black hole, it will wobble for a bit, which will break the symmetries. No-hair says rather these transient behaviours will be exponentially damped (i.e. a black hole has no stable ringing modes).

No-hair can be sort of thought of as an amped-up version of Birkhoff's theorem. The latter says that the Schwarzschild black hole is the unique spherically symmetric vacuum solution to general relativity. This doesn't quite put limitations on what "cosmohikers" could or could not do beneath the event horizon, because the theorem gives no reason to expect black holes to in fact be spherically symmetric. But it does imply that if you ever found a spherically symmetric black hole, it would have a conserved mass: spherically symmetric gravitational radiation is impossible, because that would constitute a spherically symmetric solution different from the Schwarzschild black hole.

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    $\begingroup$ Although you never come out and say it directly, I think your answer boils down to a "no," i.e., the OP's proposed justification for no-hair is simply wrong. If the OP's outline of an argument had been correct, then it would have proved a stronger version of no-hair, to which your example of the binary black holes is a counterexample. Another way to produce counterexamples is that there are known examples in which no-hair is false if there is a field present other than the electromagnetic field. $\endgroup$
    – user4552
    Commented Jul 4, 2017 at 18:11
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    $\begingroup$ Thanks for your answer. I probably wasn't clear enough but I never meant to talk about putting restrictions on what a cosmohiker can do inside the black hole. Rather, I was saying that since she could almost anything she would like to, we better describe the black hole by those quantities only who can not be possibly affected by whatever the cosmohiker chooses to do inside the black hole. But I think the rest of your answer stands on its own without much reference to this point. $\endgroup$
    – user87745
    Commented Jul 5, 2017 at 21:53

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