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106

This is really a footnote to Adobe's answer. Light cannot escape from an event horizon. But how can you check that light can never escape? You can watch the surface for some time $T$, but all you have proved is that light can't escape in the time $T$. This is what we mean by an apparent horizon, i.e. it is a surface from which light can't escape within a ...


16

Light cannot move outwards inside the event horizon. I would guess you're thinking that an outgoing light ray might leave you in the outgoing direction, then slow to a halt and return - hence you would see yourself. However this doesn't happen. The light leaving you moves inwards not outwards, but since you fall inwards faster than the light does, the light ...


15

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 ...


14

Well, it can't (float), since a Black Hole is not a solid object that has any kind of surface. When someone says that a super massive black hole has less density than water, one probably means that since the density goes like $\frac{M}{R^3}$ where M is the mass and R is the typical size of the object, then for a black hole the typical size is the ...


13

It is true that, from an outside perspective, nothing can ever pass the event horizon. I will attempt to describe the situation as best I can, to the best of my knowledge. First, let's imagine a classical black hole. By "classical" I mean a black-hole solution to Einstein's equations, which we imagine not to emit Hawking radiation (for now). Such an ...


12

Let me start with your question about stability: Any astrophysical object is subject to a battle between two forces: gravity (which will try to collapse the object) and whatever force prevents that collapse. A regular star uses heat (generated by thermonuclear fusion) to counteract gravity. When it runs out of fuel, gravity begins to compress the star ...


11

Craig Feinstein asked: Does Stephen Hawking believe that General Relativity is wrong? Here is my answer (I will shift my answer there if some one reopen that question): Stephen Hawking did NOT say that black holes do not exist. Hawking used to think balckholes are oblivious. Now he admits (like some other people do) balckholes have perfect memory , just ...


11

Indeed, nothing can get under the horizon. The stuff close to the event horizon does move outwards as the BH radius increases. Even more with any BH deformations such as waves on its surface, the tidal deformations or the change of the rotation speed, all the oblects close enough to the horizon remain "sticked" to it and follow all the changes of the BH ...


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

First note that this is a fictional movie and the image is an artist's impression, not a detailed simulation. The public seems to think the movie is some sort of fictionalized documentary, which it never claimed to be. That said, the image is qualitatively conveying some of what happens near a black hole. The diagonal disk is the accretion disk -- this is ...


8

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

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 it's own gravity towards it's centre of mass. Also ...


8

The distance where light has a circular orbit is actually $1.5r_s$ not the event horizon. This distance is known as the photon sphere. In principle a shell observer hovering at this distance could indeed see their own back. The proper distance is indeed just $2\pi r$, however the object would look bigger than expected because the curvature of spacetime has ...


7

As Chris White points out, this is a subtle issue, so I'm eager to see some more answers - perhaps someone can some up with a good car analogy ;) In the meantime, here's my best shot at an explanation: First, accept that the existence of a preferred spatial slicing does not make FLRW spacetime into Minkowski spacetime: Proper distance at constant ...


7

Whether it's a black hole or some other more ordinary mass pulling on your rope isn't actually that interesting. Let's think about a cable unrolling above Earth to start with. What we have is a pulley with a rope hanging off one side. The weight of the rope exerts some force on the edge of the pulley, causing it to undergo angular acceleration (starts to ...


7

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. ...


7

To add to John's answer: black hole with nonzero angular momentum is represented by Kerr metric. It's horizon is a spherical surface, but it also has a special surface: ergosphere that is oblate spheroid touching horizon at two 'poles'. The no-hair theorem of black hole physics precludes them from having more complicated shapes, because such shape would have ...


7

No, there's no detectable dispersion in gravitational lensing, at least not when the wavelength is much shorter than the curvature radius. The reason is simple to see: one may approximate the light by rays propagating along geodesics. They have to be null geodesics because the photons are massless. And given the location of the source and initial ...


6

We are already living in a nearly empty de Sitter space - the cosmological constant already represents 73% of the energy density in the Universe - and the Universe won't experience any qualitative change in the future: the percentage will just approach 100%. However, once the space may be approximated as an empty de Sitter space, all moments of time are ...


6

This the classic "hurling a stone into a black hole" problem. It's described in detail in sample problem 3 in chapter 3 of Exploring Black Holes by Edwin F.Taylor and John Archibald Wheeler. Incidentally I strongly recommend this book if you're interested in learning about black holes. It does require some maths, so it's not a book for the general public, ...


6

The final stages of star collapse include various stages, but three common ones to consider are white dwarfs, neutron stars, and black holes. White Dwarfs are formed when gravitational forces of the mass of the remnants of the star cannot overcome the repulsion of the electron degeneracy pressure. So think of gravity competing with the electromagnetic ...


6

Spacetime is made out of events. An event simply means a moment in time plus a point in space. Events can cause other events, e.g., if a spaceship flies from event A to event B, or a radio signal travels from A to B. It's also possible to have events that can't be causally linked, e.g., if B is 10 light-years away from A and 5 years in A's future, then A ...


6

There are actually several types of black-hole horizons with different definitions, each of which are sometimes called the event horizon. But the one that scientists typically mean when they use this phrase is more precisely called the "absolute horizon". Wikipedia's page is accurate, but not very complete. There is also a nice overview given by the ...


6

I suspect you're not asking the question you're really interested in, because the answer to your question is really boring. If you jump into a black hole you'll see the event horizon retreating before you, and you'll never cross it. The distance you've travelled is an ambiguous quantity since of course in your frame you're stationary and have travelled no ...


6

What would happen if I were to allow one end of a rope to fall past the event horizon of a black hole while I held the other end? As usual, this is in the context of a Schwarzschild black hole. First, outside the horizon, a object with constant radial coordinate 'feels' a constant proper acceleration, i.e., an accelerometer (think of a weight scale) ...


6

Basically, the reason is that if the matter is to keep a positive mass, the amount of force required to keep the matter distribution stable tends to infinity once the matter is fit within the schwarzschild radius for that given mass. This was proven in a very strong sense, without much in terms of assumptions about the particular form of the metric, by ...


6

Dark matter as far as gravitational forces go , has the same behavior as normal matter. That is how it was discovered and defined. By balancing gravitational forces in the motion of galaxies etc, it was found that more matter was needed than the matter estimated from the luminosity of the bodies. It was observed that the trajectories would not fit the ...


5

If you stick to the theory of general relativity then what happens to the matter is fairly straightforward. As the matter falls inwards it experiences increasing tidal forces. The matter reaches the singularity in a finite (short!) time, and at the singularity it is compressed into a point with zero size and infinite density. Note that nothing special is ...


5

The reference is Pulsar's answer, see detailed explanations and graphics. @Christoph 's answer is correct, so it is just an other way to express the answer. First, there is a notion of instantaneous proper distance (between us and some galaxy), where we may imagine that the expansion is stopped at time $t$, and that we measure the distance between us and ...



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