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54

This is a great question, because it's a subtle variation on the usual question about spaghettification and supermassive black holes, and shows somewhat deeper thinking. So let's assume the black hole is supermassive -- or more specifically that you are really tiny compared to the black hole -- so that we can ignore tidal effects. If your head were somehow ...


11

Let's start with what we mean by a horizon: The event horizon of an asymptotically-flat spacetime is the boundary between those events from which a future-pointing null geodesic can reach future null infinity and those events from which no such geodesic exists. A null geodesic is the path followed by a light ray, so the horizon marks the surface at ...


9

A falling observer does not experience passing through an event horizon as you describe. Instead a free-falling observer would see space-time as locally flat as long as the tidal forces were manageable. Your head and your feet are (nearly) sharing the same frame of reference. The falling observer always sees the apparent horizon in front of them until they ...


8

I think the problem in understanding this is the idea of "space being sucked into a black hole." The reality is matter is "sucked" into a black hole. Space is warped around the black whole, but space is not "sucked" into anything. Here's the issue. What is space? You can't touch space (or better, the space-time continuum). So, one view is that space is ...


5

This is really Kip Thorne's interpretation of the black-hole collapse. Many other scientists and theorists would disagree and in many quantum gravities the interior of the black hole is not empty and the central singularity is somehow regularized. In any case, what you really feel pulling on you is not the mass very far away, what you really feel is the ...


5

The geometry of spacetime is described by an equation called the metric. This is analogous to Pythagoras' theorem but with some key differences. Start with a 2d plane, where we identify positions of points by their $(x, y)$ coordinates. Suppose you move a distance $dx$ then a distance $dy$, then the distance from your starting point, $ds$, is given by ...


4

The answer must be closely related to my answer in Thought experiment - would you notice if you fell into a black hole? You can certainly use a similar Eddington-Finkelstein coordinate diagram to consider it (the E-F coordinates transform away the coordinate singularity at the event horizon). NB: This considers only GR and a non-rotating black hole (and ...


4

This is largely the same answer as Rob's, though rather than use Eddington-Finkelstein coordinates I'm going to use Kruskal-Szekeres coordinates because I think this makes the argument easier to understand. This is what the situation looks like in Kruskal-Szekeres coordinates: For the non-nerd the Kruskal-Szekeres coordinates seem formidably complicated, ...


4

Suppose you are moving toward the event horizon at .99999999 c. Your feet cross the horizon. No signal can leave your feet and reach your head, if your head stays outside the horizon. You are defeeted. Or are you? In a fraction of a fraction of a second, your head has crossed the horizon as well. The photons that your feet sent are passed by the head, ...


4

Look at these animations of the galactic center. You can see the stars orbiting some central object, but the central object is not a star, because it is not visible. But it is obviously very massive, because the other stars are orbiting it witout causing any wobbles on it. Our understanding is that the central object is a supermassive black hole. It is not ...


4

General Relativity is a purely geometrical theory of gravitation. Quantum effects have no place within GR, and more generally there is no scale to GR itself. For example, if you look at the Schwarzschild solution, you can set the mass $M$ to be whatever you want. But if you change $M$, you can also scale the time coordinate $t$ and the radial coordinate ...


3

Black holes are inferred through the confluence of two things. First, some observations are made of a dynamical phenomenon which tells us that there is a certain amount of mass present in a system and that this mass must be concentrated into a relatively small region. Usually, the limits on this region are far bigger than the Schwarzschild radius for a ...


3

If we take the Milky Way as an example, the black hole at the centre, Sagittarius A$^*$, has a mass of about 4 million times the Sun. However the mass of the Milky Way is somewhere around a trillion Suns. So the central black hole makes up 0.0004% of the total mass. So even if our central black hole was pure anti-matter it wouldn't come close to accounting ...


3

The answer you your question is no. There is a radius, larger than the Schwarzschild radius at which a neutron star, quark matter, whatever its equation of state, cannot be supported against collapse. There are limits imposed by causality and General Relativity on the structure of compact stars. In "Black Holes, White Dwarfs and Neutron Stars" by Shapiro ...


3

It is worth stressing that black holes are predictions of classical General Relativity models. Our experimental data have established that the underlying level of nature is quantum mechanical. There is a large body of research on quantizing gravity and unifying the three forces studied with particle physics experiments with the gravitational force. String ...


3

Yes, a sufficiently massive neutron star can prevent collapse to a black hole, up to a point. The most massive neutron stars are thought to be spinning very rapidly, and they exceed the mass of the theoretical limit of a non-rotating neutron star. However, most models predict that a neutron star couldn't spin much faster than 1 kHz because they would break ...


2

You are right that the $(\Omega_H-\omega)^2$ term doesn't contribute. This is because it is the square of something that vanishes at the horizon: when you take the derivative, there remains a vanishing factor. As for the other term, since $\Delta$ vanishes at the horizon, this term vanishes except when the derivative hits $\Delta$. This yields the last ...


2

The Schwarzschild metric as you've written it is only one particular coordinate system and the fact that $r$ and $t$ switch roles at the evnt horizon is an artifact of that coordinate system. There are other coordinate systems which make certain properties of the metric more intuitive. The ones that might be most useful for you are ones which can be drawn as ...


2

Really, to answer this carefully, we have to really think through what a horizon is. And for a general spacetime, there are several different notions of horizon, and "event horizon" is probably the most difficult of them to work with. The formal definition of "event horizon" says "Let's go to the distant future, take every freely-falling path that ...


2

The maximal analytic continuation of the Kerr metric extends the $r$ coordinate from the $r=0$ "ellipsoid" to negative values of $r$ which involves literally making another copy of your coordinates and switching from one to another when you pass through the disk $r=0.$ But there is another sense where there is another universe for the Kerr solution. As long ...


2

For a non-rotating uncharged black hole (the Schwarzschild metric) once you have crossed the event horizon there is no timelike trajectory that will increase the distance between you and the singularity. By this I mean that to increase your distance from the singularity would require you to be moving faster than light, which is of course impossible. So all ...


2

I just want to add to the answers that for some time it was believed (maybe considered is a better word), that galactic centers were actually stars. It was believed that the Milky Way center could be a very compact cluster of stars to dim to see them, or a collection of very heavy giant stars called warmers. So yes, your theory was seriously considered for ...


2

You're confusing yourself. A black hole state has nothing to do with the lack of energy in a system. A star doesn't become a black hole because it can't fuse stuff anymore (i.e. the end of its energy), it becomes one because of what happens to the star after its fusion process ends**. Black holes form when a high amount of energy is concentrated at one ...


1

I think your considerations are led by the daily experience that what you see actually happens as you see it. However, this is an illusion. All you perceive are photons, but they can (1) take their time and (2) can be redshifted or absorbed. Of course, the object quickly passes the event horizon (in its own frame it doesn't even notice the event horizon), ...


1

There are a lot of review articles and books on these topics. But most of them require some knowledge of basic General Relativity and QFT. For Black hole thermodynamics I found the reviews by Jacobson and Ross very useful. On EE from vondensed matter (QFT) point of view there are papers by Cardy and others. But for those you need CFT background. Here is a ...


1

I think I found a coordinate transformation that shines more light on this. Instead of looking for a conical singularity in $g_{tt}$ alone I must look at conical singularities in any part of the metric (apart from $g_{rr}$) First of all the metric can be expressed in terms of $r_\pm$ and Wick rotated $t\to it_E$ such that $$ ds_{E}^2 = \frac{ ...


1

I agree that you would not "notice" or "feel" anything as you "fall" into a black hole. As far as matter is concerned, the event horizon is not a line of demarcation were "strange" things happen on one side but not on the other. As far as you are concerned, all your molecules would be accelerated the same before and after crossing the event horizon, and ...


1

The smallest possible black hole that could be observed would be one with a mass on the scale of the Planck mass (~ 22 µg) and a radius on the scale of the Planck length (really small). Thermodynamics makes it impractical to pack multiple particles into such a small space, so the best bet would be to accelerate elementary particles to have relativistic mass ...


1

There is already a good answer from John Rennie. Here we will just mention that if the spherically symmetric metric (2) is supposed to be a vacuum solution to Einstein field equations with $\Lambda=0$, then Birkhoff's theorem shows that the metric (2) [after a possible reparametrization of the time coordinate $t$] is exactly the Schwarzschild metric. See ...


1

What the author means is as follows. Consider the (un-normalized) vector field $\partial_r$ where $r$ is the radial coordinate; $\partial_r$ is thus just the vector field orthogonal to the level sets $r = \text{const.}$ or, equivalently, it is the vector field foliating said level sets. As an aside, note that Schwarzschild coordinates are perfectly valid ...



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