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Do Schwarzschild black holes exist in reality? I have searched answer for this question but am not fully satisfied. Everything in the universe, including planets, stars, and galaxies, is spinning. How can something nonrotating/non-spinning exist in the universe? From this link, it says that

A non-rotating black hole is extremely unlikely … even if one existed, it would only take one photon to hit the event horizon off-centre to give it angular momentum, ie start it rotating.

Reference:

If they don't exist in reality then why do we study them?

Glorfindel
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apk
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    For the same reason you study frictionless perfect inclined planes in Physics class; The simplest model provides a starting point for developing more complex models. – notovny Jul 19 '22 at 12:11
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    Wait, it would only take one photon to give the black hole spin? I find this unlikely. – White Prime Jul 19 '22 at 23:22
  • @WhitePrime wouldn't the quantization of angular momentum play a role here?something analogous to how particles can only have discrete energy states? – eps Jul 19 '22 at 23:57
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    @eps That question is a bit too advanced for me, man. I'm not a galaxy brain like many on here; I'm just extremely interested in cosmology. – White Prime Jul 20 '22 at 00:01
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    A single photon doesn't carry much momentum, so its effect on the spin of the BH is infinitesimal. Anything falling into a BH will contribute to the linear & angular momentum of the BH, and although, on average, all the stray particles & radiation falling into a BH will have almost no net angular momentum, it's unlikely to be exactly zero. OTOH, a Kerr BH can lose angular momentum via the Penrose process. – PM 2Ring Jul 20 '22 at 00:43

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No, Schwarzschild black holes probably do not exist. We expect astrophysical black holes to be Kerr black holes, and we expect that most of them have a lot of spin. As the diagram at the end of this answer shows, supermassive black holes generally spin at relativistic speeds.

Stellar mass black holes are formed in core-collapse supernovae. They can also form when a neutron star collides with its companion (which could be another neutron star or a normal star); neutron stars are also formed in core-collapse supernova events. Collapses and collisions (of course) conserve angular momentum, and many young neutron stars are pulsars, spinning many times per second. However, some of that angular momentum may be carried away by the ejecta of the supernova explosion. It appears that core collapse can be highly asymmetric, which can give the remnant considerable proper motion, a phenomenon known as a pulsar kick:

A pulsar kick is the name of the phenomenon that often causes a neutron star to move with a different, usually substantially greater, velocity than its progenitor star.

The cause of pulsar kicks is unknown, but many astrophysicists believe that it must be due to an asymmetry in the way a supernova explodes. If true, this would give information about the supernova mechanism.

It's not easy to detect an isolated inactive black hole, or to determine its angular momentum. And if the black hole is active, the accretion disk will have high angular momentum simply due to its orbital speed, even if the spin of the black hole itself is relatively slow.

So why do we study Schwarzschild black holes? For the same reason we study Special Relativity even though spacetime is generally not flat. You need to thoroughly understand flat spacetime before you try to learn General Relativity. And you need to understand the Schwarzschild metric before you add the extra complexity that spin brings to the picture.

Besides, the Schwarzschild solution is a useful model for any spherical body with relatively low spin, it doesn't only apply to black holes. Thus you can use it (for example) to calculate the gravitational time dilation on the surface of the Earth.

As jawheele mentions in the comments, real black holes aren't exactly Kerr black holes either. The Kerr solution, like the Schwarzschild solution, is an eternal vacuum solution to the Einstein Field Equations. And we don't expect real black holes that formed through astrophysical processes to be the multi-universe gateways that the Penrose diagram indicates.

Bear in mind that we need a quantum gravity theory to properly talk about what happens at the core of a black hole, and even with such a theory we cannot empirically validate its predictions directly because we cannot extract information from the other side of the event horizon(s).

PM 2Ring
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  • There is no particular reason for typical stellar mass black holes to have lots of spin. LIGO observations currently pointing to most stellar black holes having relatively little spin. Larger black holes on the other hand are expected to have quite a bit of spin, because it is very hard to grow a black hole without adding spin. – TimRias Jul 19 '22 at 18:33
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    +1 for the second paragraph – Ethan Bolker Jul 19 '22 at 22:33
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    @TimRias Interesting! Could you give me a reference for that? In the meantime, I'll modify my answer. – PM 2Ring Jul 19 '22 at 23:47
  • @PM2Ring According to that paper published in nature which you mentioned in your answer, supermassive black holes rotate extremely fast. But all real black holes are supermassive only? What if the black hole is not supermassive? then can it rotate slowly? – apk Jul 20 '22 at 07:17
  • @apk No, most real black holes are not supermassive (SMBH). They are usually much smaller, formed by the processes I described in the 2nd paragraph. Most galaxies, even dwarf galaxies, appear to have a large BH at their core, although there are exceptions. Presumably, galaxies without a central big BH lost their BH in a galactic collision. We're still not exactly sure how SMBHs form: they seem to be present in the early universe, but it's not clear how they managed to get so big so early. – PM 2Ring Jul 20 '22 at 07:35
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    (cont) Yes, a small BH could rotate slowly, but as I mention in my answer we expect stellar BHs to have a lot of spin because of conservation of angular momentum. – PM 2Ring Jul 20 '22 at 07:35
  • @apk This article talks about the difficulties involved in the merger of large black holes: https://en.wikipedia.org/wiki/Binary_black_hole#Final_parsec_problem – PM 2Ring Jul 20 '22 at 07:40
  • @PM2Ring The last paragraph of the answer - "Besides, the Schwarzschild solution is a useful model for any spherical body with relatively low spin" is unclear to me. What I know is Schwarzschild metric describes gravitational field outside a spherical mass which is static nonrotating then how can we apply it to earth or any other star, as earth and any other star will be rotating? Also, Schwarzschild is the vacuum solution to EFE, and outside of earth or any star will not be vacuum so how is it applicable to that? Please clarify my confusion. – apk Jul 20 '22 at 08:04
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    @apk In physics, we almost always have to make approximations. We can't include the whole universe in our calculations! So a simple model of the Earth's gravity treats the Earth as a uniform sphere & ignores other effects. Then gradually we improve the model, adding in other effects, as necessary. Eg, the centripetal acceleration at the equator due to the Earth's rotation is only ~0.0337 m/s^2, which is very small compared to g, and we get a reasonable result if we just combine them linearly. – PM 2Ring Jul 20 '22 at 08:40
  • (cont) Of course, JPL use a much more sophisticated model of the Earth's gravitation in their calculations, using precision measurements of the Earth's rotation & orientation from the IERS, etc. There are further details in The JPL Planetary and Lunar Ephemerides DE440 and DE441 , Ryan S. Park et al 2021 AJ 161 105. – PM 2Ring Jul 20 '22 at 08:41
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    For example look at the right panel of Figure 11 in https://iopscience.iop.org/article/10.3847/2041-8213/abe949/pdf. This is from the first two LIGO-Virgo observation runs. The distribution of inferred effective spin values peaks very close to zero. Of course, the population observed in GW mergers may not representative for the BH population as a whole. Creating a pair of black holes that merger within a Hubble time requires specific formation channels. – TimRias Jul 20 '22 at 11:47
  • For O3 see: https://dcc.ligo.org/public/0177/P2100239/009/o3_population.pdf Figure 21. – TimRias Jul 20 '22 at 11:51
  • Thanks for providings those references, @TimRias. – PM 2Ring Jul 20 '22 at 12:41
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    This answer might be improved by noting that exactly Kerr black holes don't exist either, and that GR theorists generally expect that realistically formed black holes have some features more in common with Schwarzschild than with Kerr (in particular, the structure of their Penrose diagram) whether they have large spin or not. This is essentially the content of the Strong Cosmic Censorship Conjecture. – jawheele Jul 20 '22 at 15:16
  • @jawheele Fair point. I'll add a short note. I originally intended this answer to be short & sweet, and I've already expanded it once. ;) – PM 2Ring Jul 20 '22 at 15:23
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The Schwarzschild solution is static. There are certain technical intricacies to explain what does it mean exactly, starting just from the fact that the t coordinate is not defined everywhere on what the “Schwarzschild metric” today refers to. But it’s sufficient to remember that the external observer sees the Schwarzschild black hole to have constant mass. Such presumed hole has no beginning and never decays.

Indeed, this staticity doesn’t go well with our cosmology. Can the hole be permanent while all the Universe was formed with the Big Bang? Ī̲ doubt it. As a side note, we theorize known black holes to form via gravitational collapse. Moreover, the provision to exist forever conflicts with the Hawking radiation and all plausible scenarios of the ultimate fate of our universe.

Well, we study the Schwarzschild metric because it’s explicit and simple yet has such defining features as the event horizon and singularity.

Incnis Mrsi
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Not only would a single photon falling into the black hole make it no longer a Schwarzschild black hole, but even in an otherwise empty universe you would have Hawking radiation that would cause an infinitesimal amount of spin. Thus they can't actually exist even if a K3 civilization tried to make one.

Loren Pechtel
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    Why would Hawking radiation cause spin? Shouldn't it be spherically symmetric for a BH with no spin? – PM 2Ring Jul 21 '22 at 16:48
  • @PM2Ring Any given photon can't be symmetric. It's the same situation as an infalling photon but since it's self-generated you can't possibly stop it from happening. – Loren Pechtel Jul 21 '22 at 22:21
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    Photons can be symmetric. If the photon emitter is perfectly spherically symmetric, then the photons it emits are also spherically symmetric. See https://physics.stackexchange.com/a/189776/123208 – PM 2Ring Jul 22 '22 at 04:17
  • @PM2Ring But the photon is emitted when half a virtual pair gets gobbled up--it's going to appear on one side of the black hole. – Loren Pechtel Jul 22 '22 at 14:06
  • The virtual particle picture of Hawking radiation (HR) is just a heuristic, you shouldn't take it too seriously. OTOH, the formal derivation of HR involves rather heavy mathematics, and I don't follow it fully. But you can't really localise the region where HR is produced. There's some good info here: https://physics.stackexchange.com/a/252236/123208 And a more in-depth treatment here: https://physics.stackexchange.com/q/634972/123208 Also, https://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/hawking.html – PM 2Ring Jul 22 '22 at 15:38
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    Indeed. The "virtual pair" description is just an analogy and not how the calculation is actually done. Hawking himself said of the virtual pair description: It should be emphasized that these pictures of the mechanism responsible for the thermal emission and area decrease are heuristic only and should not be taken too literally. – John Rennie Jul 22 '22 at 15:55
  • @PM2Ring But the photon is going to have a direction. Unless that's exactly normal to the black hole you imparted spin. – Loren Pechtel Jul 23 '22 at 19:39
  • @LorenPechtel No, it doesn't have a direction, as explained in the 1st Physics.SE link I posted. In a comment on that answer, Emilio Pisanty, a quantum optics researcher, gives a little more detail. But maybe you should ask a question about this on Physics.SE if the existing questions & answers on this important topic in QM don't convince you. – PM 2Ring Jul 23 '22 at 20:02