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132

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 ...


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 ...


23

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 ...


20

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 ...


18

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 ...


17

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 ...


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 ...


15

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 ...


14

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 black holes are oblivious. Now he admits (like some other people do) black holes have perfect memory, just ...


13

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

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 ...


12

The bright parts around the black hole are the accretion disk, which is in reality just a flat disk in the equatorial plane similar to the rings of Saturn, but is distorted visually by gravitational lensing. You can see a page here that gives some code for creating images using ray-tracing of light rays in curved spacetime, which offers a more schematic ...


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

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 ...


11

It is a truly unfortunate coincidence that the "escape velocity" at the Schwarzschild radius $R_\mathrm{S}$ is $c$. This leads people to think Newtonian mechanics can predict black holes. It can't. Note that a Newtonian escape velocity is the minimum velocity needed to coast to infinity against gravity. If you have less speed, you still travel away from the ...


10

This is explained thoroughly in Thorne's book "The Science of Interstellar". There were two scientific papers based on the simulations: One in physics and one in computer rendering. The two circles are caused by gravitational lensing by a very rapidly spinning black hole. The radius of this black hole is 150 million kilometers with a mass of 100 million ...


10

Assume the object falling in is a blue laser that you launched directly (radially) towards the Schwarzchild (non-rotating) black hole that is aimed directly at you and that you are far from the black hole. The massive object is the laser itself, the light that you are watching is your way to "see" the object as it approaches the event horizon. First of all ...


10

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

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

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 ...


9

"According to Newton's law the negative mass should be repelled" -- Nope, in both Newtonian physics and in general relativity, negative mass would be attracted gravitationally to positive mass, although negative mass would exert a repulsive gravitational effect on positive mass (but if the negative mass is small compared to the mass of the black hole this ...


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

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 ...


8

The reason has to do with time dilation, and specifically, with the resulting red shift. A black hole forms from a collapsing star, which is of course made of brightly glowing matter. The event horizon forms in the centre and moves outwards while the star-matter falls towards it. Because of gravitational time dilation, the infalling matter never crosses the ...


8

Hawking radiation is such a miniscule effect we can be sure we'll never detect it for a real astrophysical black hole. The Wikipedia article gives some numbers: For a black hole of the mass of the Sun, the power emitted in Hawking radiation amounts to $9\times10^{-29}\ \mathrm{W}$. Even if all this energy were converted to visible-light photons ...


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. ...


8

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

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 ...



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