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110

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 ...


20

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 ...


17

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 ...


13

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 ...


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 ...


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 ...


9

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 ...


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

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

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

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 ...


8

"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 ...


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

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 ...


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

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 ...


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

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

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

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

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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