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In the standard model of particles it is understood that besides characteristics like momentum, spin, etc., two electrons are indistinguishable.

Are two black holes, in the same sense, indistinguishable given they have the same mass, momentum, etc.?

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    $\begingroup$ Location, location, location ! On a macroscopic scale (i.e. away from quantum level effects) location distinguishes two black holes. You can't generally say that about e.g. electrons in an atom. $\endgroup$ – StephenG Sep 27 '18 at 17:51
  • $\begingroup$ Perfect! But i meant the nature of matter and the resulting geometry of black hole $\endgroup$ – Marco Sep 27 '18 at 17:59
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    $\begingroup$ @StephenG, what do you mean? (The answer was fun but maybe leaves many people in the dark? : ) $\endgroup$ – Helen Sep 27 '18 at 18:15
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    $\begingroup$ @StephenG, the same can be said for two sufficiently well-separated elementary particles. Could one consider two black holes of equal mass and charge as two identical "elementary" particles? :-P $\endgroup$ – LLlAMnYP Sep 28 '18 at 13:01
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    $\begingroup$ @StephenG : But does the universe pick up a minus sign when I exchange to black holes differing only in position (which does happen when exchanging electrons)? $\endgroup$ – Eric Towers Sep 29 '18 at 19:01
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The answer to this question is not technically known. The theorem that applies to this question is the "No-Hair Theorem" which states that a black hole is described by only 3 externally observable properties - mass, charge, and angular momentum - and that's it. The No Hair Theorem implies then that two black holes which have the same mass, charge, and angular momentum are identical to each other no matter the actual matter that was used to create them. E.g. if you create one black hole using a bunch of atoms vs you create another black hole using neutrinos only - the no hair theorem says as long as the two black holes end up with the same mass, charge, and angular momentum, one could not tell the two apart. One could not say which one was the one created by neutrinos and which one was the one created by ordinary atomic matter.

The problem though is that the No Hair theorem is not technically a theorem in that it hasn't been proven yet. It's more of a conjecture or hypothesis at this point. There are motivating factors which seem to imply the No Hair Theorem is true, but alas there is no clear proof using GR that it is.

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    $\begingroup$ There are also some special cases in which it fails: en.wikipedia.org/wiki/No-hair_theorem#Counterexamples $\endgroup$ – J.G. Sep 27 '18 at 17:34
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    $\begingroup$ There is a more serious problem though. The No Hair Theorem applies to the exact solution of a rotating, charged black hole. In practice, black holes are often surrounded by matter in the form of accretion disks, and that exact solution is only an approximation. $\endgroup$ – Andrea Sep 27 '18 at 21:19
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    $\begingroup$ No-hair theorem also wrecks chaos in quantum mechanics in the form of the information paradox. $\endgroup$ – John Dvorak Sep 27 '18 at 21:48
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    $\begingroup$ Is there any proven theorem in Physics? I thought all we had were "good enough" models. $\endgroup$ – Eric Duminil Sep 28 '18 at 6:32
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    $\begingroup$ @EricDuminil You can certainly prove theorems within a particular model, because that's just mathematics, but you can't prove that the model corresponds to the real world. $\endgroup$ – David Richerby Sep 28 '18 at 11:07
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To expand on the answer of enumaris, there are four types of black holes based on their mass, charge, and angular momentum. Uncharged non-rotating black holes are called Schwarzschild black holes. These can be different only by mass. Rotating uncharged black holes are called Kerr black holes. Charged non-rotating black holes are called Reissner-Nordstrom black holes. And finally rotating charged black holes are called Kerr-Newman black holes. The physics of different types of black holes is quite different. While all of them contain a singularity, they may have a different number of event horizons of different types and shapes. For example, a charged black hole has a Cauchy horizon inside the Schwarzschild horizon.

The no-hair conjecture was proven for the Schwarzschild black holes for the simplified case of the uniqueness in 1967. The result since has been expanded to charged and rotating black holes. The general uncharged case has been partially resolved under the additional hypothesis of non-degenerate event horizons and the assumption of real analyticity of the space-time continuum. However, there still is no rigorous proof of the general case.

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    $\begingroup$ Sources, please? $\endgroup$ – N. Steinle Sep 27 '18 at 18:23
  • $\begingroup$ @N.Steinle The source for the information on the No-Hair theorem is in the link in the answer by enumaris (or here: en.wikipedia.org/wiki/No-hair_theorem). The source for the types of black holes is Wiki: en.wikipedia.org/wiki/Charged_black_hole $\endgroup$ – safesphere Sep 27 '18 at 18:27
  • $\begingroup$ Doesn't seem like mass factors into the "type" of black hole, although it's certainly a distinguishing factor. There are two choices for charge (charged or uncharged) and two for rotation (rotating or non-rotating), which makes for 2x2=4 types of blackholes. Meanwhile mass is a continuous variable. $\endgroup$ – BallpointBen Sep 27 '18 at 20:30
  • $\begingroup$ @BallpointBen Yes indeed. I can't think of any even hypothetical possibility for a any type black hole not to have mass. For example, a collapsing spere of light does have a rest mass even before it forms a black hole despite each photon being massless. $\endgroup$ – safesphere Sep 27 '18 at 20:39
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    $\begingroup$ I think these four types of BH are just the four mathematical possibilities. In the real world, matter is entering the BH at random off-axis vectors so it will acquire angular momentum in the same way as every other star, planet, moon, and heavenly body. Therefore, all BHs will be rotating. As for charge; this is more likely to be zero, averaged over time (just like stars, planets, moons, etc.). Interestingly, if you ever wanted to move a BH, you could get a handle on it by charging it up (e.g., with an electron beam) and then moving it around with an electric field. $\endgroup$ – Oscar Bravo Sep 28 '18 at 6:13
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Always be careful when your question involves gravity and quantum physics together. We have only very partial knowledge of that combination. In particular, note that the No Hair Theorem is a statement within classical general relativity, and as soon as you bring in quantum physics then things are complicated by the black hole entropy. As I understand it, this entropy is sufficiently well established that we would be very surprised if it turned out to be not there, and the place where it is found is on the horizon, and it typically has a large value. In this sense, black holes have a lot of "hair", i.e. physical properties that can distinguish one of them from another. The entropy is often huge: they have a huge amount of "hair", in the sense of available microstates consistent with their macrostate.

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I'm going to take a different approach to this :

Macroscopic vs. Quantum worlds.

In standard model of particles it is understood that besides characteristics like momentum, spin, etc two electrons are indistinguable.

Let's remember that an elementary particle is, by definition, identified by these characteristics. The (rest) mass is a fixed value. The spin is fixed, the various charge values are fixed. If they deviate from specific values then you do not have one of these particles at all.

But we say that two e.g. electrons are indistinguishable because in any system we cannot label them and track them in any way. We can make a measurement that says there is an electron at position one and two, but the instant we make those measurements we (in general) cease to know anything about the actual positions of the electrons.

We can't say that when we next measure those electron positions which electron is which - we can't track them. The electron at position one could be at position three or four next, and the electron that was at position two could now be at position three or four. We just know the next measurements produce two distinct measurements for positions of electrons.

Are in the same sense two black holes indistinguable given they have same mass, momentum, etc?

So not in the same sense as elementary particles.

If I have two black holes they have a macroscopic position. Unlike my electrons I can track them easily and there is no mystery between measurements as to which is which. Regardless of the no hair theorem, regardless of their size, rotation, charge, etc. they have distinct locations which can be tracked.

So the black hole that started at position one, I can say with essentially perfect confidence is the same one I measured at position three later. Likewise there's no confusion about whether the other black hole could be at position three and the one I thought was there is actually at position four.

So for macroscopic objects location is a property that labels them uniquely.

But for elementary particles this is not generally the case.

The No Hair Theorem

I think I should point out that the No Hair Theorem does not say we cannot distinguish black holes from each other, even if they are identical in external characteristics. It says that the only information we can determine about the black hole's interior (on the other side of that event horizon) are these "statistical" values.

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    $\begingroup$ The critical distinction you bring is the ability to track. This ability is indeed different in the quantum and classical cases, but it has more to do with the environment than with the objects themselves. You can put two electrons in two chambers isolated enough to make tunneling improbable (like tunnelling of two black holes into each other). Then you can track each electron separately. On the other hand, two black holes of the same parameters (e.g. spinning around each other) are different only by your prior knowledge. If you lose your records, you would not be able to tell them apart. $\endgroup$ – safesphere Sep 27 '18 at 21:08
  • $\begingroup$ If you are allowed to consider the location to be a distinguishing characteristic, why not the accretion disks also? Those must certainly have significant differences. $\endgroup$ – D. Halsey Sep 27 '18 at 21:32
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    $\begingroup$ You can also say with absolute confidence that the electrons at A and B are different if A and B are space-like separated $\endgroup$ – John Dvorak Sep 27 '18 at 21:52
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    $\begingroup$ I don't think that what this answers is what the question was about. If black holes are indistinguishable, then you can "lose" a black hole in the same way as an electron, just by looking away for long enough. When you find it again, you don't know if it's the same one, or another with the same characteristics. But if they retain information, then it is conceivable that you could "tag" it in some way by feeding it a particular hunk of matter, and find evidence of that later on. $\endgroup$ – hobbs Sep 28 '18 at 18:14
  • $\begingroup$ I think that this answer misses by a lot the original question. The question was about identity of characteristics. $\endgroup$ – Helen Sep 30 '18 at 8:13
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Maybe you can use the light bending properties of mass as hypothesized in general relativity and verified by experiment in 1918 - that gravity actually bends space. They did not look at a black hole, but rather a star, (maybe the sun, I don't remember), and the angular distortion created by the mass should make you able to calculate the mass, angular position and distance to object, and this mass you could calculate would make you able to distinguish black holes from less massive objects.

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    $\begingroup$ The OP was not about detecting different masses, but distinguishing between black holes with identical masses. The 'light bending properties' do cause extreme delays for photons from objects crossing the event horizon to escape, so in a way we can see the entire history of each black hole and thus distinguish them. $\endgroup$ – amI Sep 30 '18 at 4:52

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