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64

I'm not a quantum cosmologist, but I am an early-universe cosmologist, so I can give you my opinion after having read this paper. The article claims that Bohmian trajectories is a valid replacement for geodesics. This was claimed in the very beginning of the paper and not much is offered in the way of defense for this assumption. That's not to say that it's ...


27

Short answer is yes. But if you want to nit pick, I could argue that when a star collapses to form a BH, it first forms a horizon before the singularity forms (cannot form a "naked singularity"). And since time inside the horizon is essentially frozen with respect to that of an observer outside, the singularity NEVER forms. Yet from the point of view of the ...


25

While this work certainly investigates an interesting point, I think simply replacing geodesics in GR with similarly looking quantum trajectories does not solve the issues here. Finding the Friedmann equations while assuming large-scale homogeneity and isotropy is no surprise to me. There are a number of people working on so-called Big-Bounce Cosmologies. ...


17

There are only four known stable black hole geometries: Schwarzschild, Reissner-Nordstrom, Kerr and Kerr-Newman. We expect that any random assemblage of matter dense enough to form a black hole will relax into one of these four geometries by emission of gravitational waves. None of these geometries has two distinct singularities, so (as far as we know) it is ...


16

A popular assumption about black holes is that their gravity grows beyond any limit so it beats all repulsive forces and the matter collapses into a singularity. [...] Is there any evidence for this assumption? It's not an assumption, it's a calculation plus a theorem, the Penrose singularity theorem. The calculation is the Tolman-Oppenheimer-Volkoff ...


16

If something is infinitely dense, must it not also be infinitely massive? Nope. The singularity is a point where volume goes to zero, not where mass goes to infinity. It is a point with zero volume, but which still holds mass, due to the extreme stretching of space by gravity. The density is $\frac{mass}{volume}$, so we say that in the limit ...


14

Indeed you made one mistake: the infalling observer does not see the outside universe "speed up". Look at what happens in a space-time diagram. At the spacetime point where your astronaut passes the horizon, he can only see what's in his past light cone, and that's the universe at early times only. It's the signals that he sends back (or tries to) that reach ...


14

I think the closest model to what you're talking about would be two colliding black holes, during the intermediate period where their horizons had merged, but the central objects had not yet collided. These systems are very different from isolated black holes, as they give off significant gravitational radiation, and have horizons that rapidly change in ...


13

I) The substitution $f=r\psi$ is the standard substitution to get a radial 3D problem to resemble a 1D problem, see e.g. Ref. 1. II) From the perspective of the normalization of the wavefunction $\psi(r)$, a $1/r$ singularity of $\psi(r)$ at $r=0$ is fine because $|\psi(r)|^2$ is suppressed by a Jacobian factor $r^2$ coming from the measure in 3D spherical ...


12

In classical General Relativity, once an event horizon forms, every particle inside that event horizon will inevitably travel in the direction of the center of the black hole. This is what is meant by "gravitational collapse" and how matter comes to form a singularity in the center- no matter how small it is, or how close to the center it is, nothing can ...


12

Suppose you have some collection of matter that is so dense it has an event horizon where the escape velocity is greater than the speed of light. The escape velocity is obviously due to the strong gravitational field of the matter inside the event horizon, and equally obviously that matter is also pulled by its own gravity towards its centre of mass. Also ...


12

This is actually a common question. Many websites have been setup to try to explain this. I like this one for instance. I shall attempt to do my own layman explanation. First of all, in order to have a black hole, you need to have a place for it to be in. Since there was no such thing as a universe, there isn't a place for the black hole to actually ...


12

The many comments have covered the main points about the question, but I thought it would be worthwhile explaining how the behaviour is calculated. If we solve the Einstein equation for a point mass we get the Schwarzschild metric: $$ds^2 = -\left(1-\frac{2M}{r}\right)dt^2 + \left(1-\frac{2M}{r}\right)^{-1}dr^2 + r^2 d\Omega^2$$ All equations look scary to ...


11

It's important to understand the context in which statements like "there must be a singularity in a black hole" are made. This context is provided by the model used to derive the results. In this case, it was classical (meaning "non quantum") general relativity theory that was used to predict the existence of singularities in spacetime. Hawking and ...


11

A high enough energy density is a necessary condition but not a sufficient condition for black holes to form: one needs to have a center which will ultimately become the center of the black holes; one needs the matter that collapses to the black hole to have a low enough velocity so that gravity may squeeze it before the matter manages to fly away and dilute ...


11

If the Universe is spatially infinite, it always had to be spatially infinite, even though the distances were shortened by an arbitrary factor right after the Big Bang. In the case of a spatially infinite Universe, one has to be careful that the singularity doesn't necessarily mean a single point in space. It is a place - the whole Universe - where ...


11

Although we don't have a quantum theory of gravity, we think we have some reliable knowledge about the properties of black holes from general relativity. One thing we think we know is the so-called "No-hair conjecture", which says that black holes can be described by just three numbers: mass, charge, and angular momentum (i.e. how much they are spinning). ...


11

You are correct when you concluded that two classical point electrons could never touch each other. It would take infinite energy.


10

There are a number of models for the universe over the years. The Big Bang as you show it in the figure has become the "standard model" for the creation of the observed universe as we know it because it fits observations, i.e. data, using known theories and behaviors from elementary particle theories. This model has been evolving as data are added in our ...


10

No. Firstly, weak cosmic censorship can only hold in the generic sense, as there are known examples of nakedly singular space-times. (See, e.g. Christodoulou 1993, and Christodoulou 1999.) Observe in particular that the nakedly singular space-time constructed in the 1993 paper is spherically symmetric with a central axis, and the initial data is ...


10

We don't know what will happen when a photon or any other particle hits a singularity of a black hole. The singularity is a phenomenon of classical general relativity and the singularity is really is an indication that classical general relativity breaks down there. To really understand what happens near a singularity we need a full quantum mechanical ...


9

While there is a general consensus aligned with the Big Bang theory's historical and current stages of the universe. To note, there are three theories with focus on this topic regarding the future, namely: the open universe, flat universe and closed universe theories. Ultimately, the fate of the universe depends on the outcome of the competition between the ...


9

Black holes and "anti"-black holes are the same objects. A black hole resulting from the collapse of normal matter, and a black hole resulting from the collapse of antimatter, are indistinguishable. Recall that black holes only have charge, mass, and spin and there is no way to tell that a black hole originally was matter or not (e.g., we can't measure B or ...


9

Nothing is unusual to an observer falling into a black hole at the event horizon. He does not "hit" it. It is crossed with no fuss or bother. As he falls farther and farther into the black hole, though, tidal gravitational forces "spaghettifi" him. I do not know what you mean by saying that the external universe is "speeded up infinitely." The moment the ...


9

Good question! I often pondered that myself. As this website explains, the Big Bang wasn't a black hole basically because it couldn't be! A black hole is the mathematical solution to Einstein's equations of General Relativity that describes a pre-existing region of spacetime that has gravitationally collapsed and formed a singularity. Since there was no ...


9

As dmckee says in his comment, the answer is no, a stationary spherical shell isn't possible. This is because not even the interparticle forces in neutronium are strong enough to support it. The problem is that once inside the event horizon there is no way to travel away from the singularity, or even maintain your distance from it, without travelling faster ...


8

It's difficult to know what happens on the other side of a black hole, since no information can cross back through the event horizon (the radius at which light and therefore any information can no longer escape). The leading idea is that near the center of every black hole lies a singularity, or a point where the density (and therefore the curvature of ...


8

The only definitive answer is maybe. Observations now seem to indicate that the Universe is dominated by dark energy, which leads to the most likely conclusion that the Universe will expand faster and faster, eventually resulting in the disappearance of everything that isn't bound to whatever object in which you reside. In the case of humans, that would be ...


8

A stationary uncharged black hole is described the the Schwarzschild metric: $$ ds^2 = -\left(1-\frac{2GM}{c^2r}\right)dt^2 + \frac{dr^2}{\left(1-\frac{2GM}{c^2r}\right)} + r^2 (d\theta^2 + sin^2\theta d\phi^2) $$ The event horizon is at $r = 2GM/c^2$, where the $dr^2$ term goes to infinity, so it is a surface of constant $r$ i.e. it is indeed a sphere. ...



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