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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?

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    $\begingroup$ 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. $\endgroup$
    – notovny
    Jul 19, 2022 at 12:11
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    $\begingroup$ Wait, it would only take one photon to give the black hole spin? I find this unlikely. $\endgroup$ Jul 19, 2022 at 23:22
  • $\begingroup$ @WhitePrime wouldn't the quantization of angular momentum play a role here?something analogous to how particles can only have discrete energy states? $\endgroup$
    – eps
    Jul 19, 2022 at 23:57
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    $\begingroup$ @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. $\endgroup$ Jul 20, 2022 at 0:01
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    $\begingroup$ 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. $\endgroup$
    – PM 2Ring
    Jul 20, 2022 at 0:43

3 Answers 3

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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).

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    $\begingroup$ +1 for the second paragraph $\endgroup$ Jul 19, 2022 at 22:33
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    $\begingroup$ @TimRias Interesting! Could you give me a reference for that? In the meantime, I'll modify my answer. $\endgroup$
    – PM 2Ring
    Jul 19, 2022 at 23:47
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    $\begingroup$ (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. $\endgroup$
    – PM 2Ring
    Jul 20, 2022 at 7:35
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    $\begingroup$ @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. $\endgroup$
    – PM 2Ring
    Jul 20, 2022 at 8:40
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    $\begingroup$ For example look at the right panel of Figure 11 in 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. $\endgroup$
    – TimRias
    Jul 20, 2022 at 11:47
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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.

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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.

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    $\begingroup$ Why would Hawking radiation cause spin? Shouldn't it be spherically symmetric for a BH with no spin? $\endgroup$
    – PM 2Ring
    Jul 21, 2022 at 16:48
  • $\begingroup$ @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. $\endgroup$ Jul 21, 2022 at 22:21
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    $\begingroup$ Photons can be symmetric. If the photon emitter is perfectly spherically symmetric, then the photons it emits are also spherically symmetric. See physics.stackexchange.com/a/189776/123208 $\endgroup$
    – PM 2Ring
    Jul 22, 2022 at 4:17
  • $\begingroup$ @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. $\endgroup$ Jul 22, 2022 at 14:06
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    $\begingroup$ 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. $\endgroup$ Jul 22, 2022 at 15:55

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