External observer will never see any mass crossing event horizon - any object near it would appear frozen, red-shifted and then disappearing. In this case - how black hole can possibly get any mass from the point of view of external observer?
14125
The event horizon of a black hole is where gravity is such that not even light can escape. This is also the point I understand that according to Einstein time dilation will be infinite for a far-away-observer.  If this is the case how can anything ever fall into a black hole.  In my thought experiment I am in a spaceship with a powerful telescope that can detect light at a wide range of wavelengths.  I have it focused on the black hole and watch as a large rock approaches the event horizon.  Am I correct in saying that from my far-away-position the rock would freeze outside the event horizon and would never pass it? If this is the case how can a black hole ever consume any material, let alone grow to millions of solar masses.  If I was able to train the telescope onto the black hole for millions of years would I still see the rock at the edge of the event horizon?  I am getting ready for the response of the object would slowly fade.  Why would it slowly fade and if it would how long would this fading take?  If it is going to red shift at some point would the red shifting not slow down to a standstill? This question has been bugging me for years!  OK - just an edit based on responses so far.  Again, please keep thinking from an observers point of view.  If observers see objects slowly fade and slowly disappear as they approach the event horizon would that mean that over time the event horizon would be "lumpy" with objects invisible, but not passed through?  We should be able to detect the "lumpiness" should we not through?
111772
This question appeared quite a time ago and was inspired, of course, by all the fuss around "LHC will destroy the Earth".   Consider a small black hole, that is somehow got inside the Earth. Under "small" I mean small enough to not to destroy Earth instantaneously, but large enough to not to evaporate due to the . I need this because I want the black hole to "consume" the Earth. I think reasonable values for the mass would be $10^{15} - 10^{20}$ kilograms.   Also let us suppose that the black hole is at rest relative to the Earth.  The question is: How can one estimate the speed at which the matter would be consumed by the black hole in this circumstances?
119367
The first-ever image of a black hole released this week was taken with radio telescopes, suggesting that it is a  image and doesn't represent what a person would see in reality, or visible light. This seems to be confirmed by this , but I am perplexed as to why this is not widely mentioned, including the announcement on the telescope's  and Wikipedia (, ). Perhaps not mentioning it makes for better headlines, but it feels wrong.  What's going on here?
91280
Does the mass of matter falling into a black hole affect the size of an event horizon the moment it passes through it, or when it has been incorporated into the singularity?
90380
If a large star goes supernova, but not enough mass collapses to form a black hole, it often forms a neutron star. My understanding is that this is the densest object that can exist because of the Pauli exclusion principle: It's made entirely of degenerate matter, each particle of which cannot occupy the same quantum state of any other.  So these objects are so massive that they gravitationally lens light. If you make them more massive, they bend the light more. Keep going and going until they bend the light so much that light passing near the surface can barely escape. It's still a neutron star. Add a bit more mass, just enough that light passing just over the surface cannot escape. Now it's a black hole with an event horizon (I think?). Does this mean the neutron star has become a singularity? Isn't it still just a neutron star just beneath the event horizon?  Why are black holes treated as having a singularity instead of just an incredibly massive neutron star at its center? Does something happen when an event horizon is "created?"
76662
The well-known fable of an astronaut sending signals out to an external observer while falling toward an event horizon states that the time lapse between such signals becomes greater even if in the astronaut sends them out periodically (as judged in his inertial frame). When viewed from earth and weighing time-dilation due to the gravitational field of the collapsing star, how is it possible that a black hole can form in finite time (for any external observer) if it takes an infinite amount of time to "see" events occurring at the event horizon?
49057
I understand that the event horizon of a black hole forms at the radius from the singularity where the escape velocity is $c$. But it's also true that you don't have to go escape velocity to escape an object if you can maintain some kind of thrust. You could escape the earth at 1 km/h if you could maintain the proper amount of thrust for enough time. So if you pass just beneath the event horizon, shouldn't you be able to thrust your way back out, despite the $&gt;c$ escape velocity? Or does this restriction have to do solely with relativistic effects (time stopping for outside observers at the event horizon)?
79549
If time stops at the event horizon, can we ever detect two black holes merging? In other words, if you are a short distance away, would you encounter a spherically symmetric gravitational field, or a dipole field?
77184
Photons do not have (rest) mass (that's why they can move at speed of "light").   So, my question is how the gravity of classical$^1$ black hole can stop light from escaping?  --  $^1$ We ignore quantum mechanical effects, such as, Hawking radiation.
79828
A thought experiment:   Take an object that is massive enough so we can detect it's position in space by measuring its gravitation field (for example, by sending almost tangential laser beams and measure how they deviate in the object's gravitation field).  Take a simple (non-rotating, non-charged) black hole. Send this object to the black hole. While the object falls into the black hole, measure the object's gravitation field (therefore, the object's position).   I'm wondering what will we see when the object:   Is getting closer to the event horizon. Crossing the event horizon. Fails to the singularity under the event horizon.   My intuition tells me following:   As time around the black hole slows down for an external observer, we may see that the object (i.e. its gravitation field) is always in the place it touched the black hole. (EDIT: According to "", this will not happen.) If my previous assumption is wrong, we may see how the object falls down to the singularity behind the event horizon. Also, the shape of the event horizon will deviate from an ideal sphere and by measuring these deviations we will see a position of the object under the horizon, which seems to contradict to the definition of the event horizon.   My questions are:   In any generally recognized theory that describes this situation? What will happen to the object and its gravitation field according to this theory (theories)? How confident we can be in the description given by this theory?
10927
I understand that an  is the boundary of trapped codimension-2 surfaces in the spacetime (surfaces from which both ingoing and outgoing light rays must be converging). Meanwhile, an  is the boundary of the past of future null infinity (boundary of points from which a light ray can escape to infinity).  I have a few questions about this:   I've read that apparent horizons are always inside or coincident with event horizons. If it's inside then doesn't that allow for points from which null rays can travel outwards i.e. points inside the event horizon that are causally connected to infinity? Apparent horizons are observer dependent. Why? Shouldn't all observers agree on light converging or diverging? If apparent horizons are observer dependent, why are they at all useful in GR which is famously diffeomorphism invariant (observer independent)? Both definitions seem to me to be a boundary between points that can and points that can't send light to infinity. Can someone explain the difference between them please? I know there are several posts on this but I'm afraid it has not helped clear it up in my head. Does solving $g^{rr}(r)=0$ locate the apparent horizon? Wouldn't we need to find this by solving e.g. the Raychauduri equation or something?
50637
If an answer does exist, I'd love to hear it. I'm trying to incorporate a doomed earth story in something I'm writing, and the end of the world I'm going for is a black hole.  Let's say the black hole was created in the Philippines (I chose it because it's on the equator). Here's  for reference.  If I was standing 90 degrees of that, in Kenya, or if I was standing 180 degrees of that, in Mexico; what would I see in those two separate instances? What if I was standing in Japan above it, or in Australia below it?  Let's also say I was riding a boat towards the source of the black hole. Unlike most imagined scenarios where I am orbiting the black hole and slowly falling into it, this time I'm no longer orbiting it - I'm getting sucked into it instead. I'm imagining that the black hole will "half" be inside Earth and "half" will be in the atmosphere. But as it pulls me in, what will it look like to me (if I was massless, so no tidal forces)? It's probably not going to be a black hemisphere sticking out of the ground. Light will be lensed around it in some way so as to appear like a hole in the air.  Also, if the black hole was very small, it probably won't have a strong gravitational pull, and assuming it doesn't disappear, you'd be able to look at it (or at least, its effect on the light in its surroundings). Of course, as you approach it, it would still have an event horizon, so spacetime would still be warped if you're near enough and the surroundings will still experience tidal forces.  As the black hole gains mass, it will get larger, and so its event horizon will increase. If I was standing in Kenya and the event horizon "washes over me," (as in a water wave) what will I see?  I've so far imagined that, at least when seen from outer space, there will be a hole on the ground and light will be lensed around it. When you are on the ground, though, it becomes a bit harder to imagine.  I do hope someone can lend me a hand!  Edit: Someone had previously answered this question, but I'm not sure why it's been deleted. It was pointed out to me, though, that a black hole cannot stay in place. So an additional question for me is, does that mean it is impossible to be standing still, and the event horizon "washes over you"?  Edit2: Here's a guiding thought experiment that could answer the question. Imagine that you are in outer space directly above the black hole, looking at the black hole that is on the earth's surface. Can you imagine what it would look like? It would simply be a hole with the surrounding light gravitationally lensed around it (possibly lensing the entire earth if it's big enough - but let's say it's not).  Now imagine that you are on the direct opposite of where you were above - you're right behind the black hole this time, looking directly at the black hole but unable to see it because the earth is blocking your view.  Now try to imagine, from the second image, rotate the earth into the first image. I find it a bit hard to do. Just as the black hole is appearing as you rotate the earth, how would the light be lensed?
1453
How much extra distance would I have to travel through space to get from Earth to a stellar mass event horizon? (compared to the same point in space without a black hole)
55353
Last week, I went to Switzerland and visited the LHC. I took a tour, and the guide told us that it is possible for black holes to appear in the LHC, but they will be so small, that they would evaporate away due to Hawking Radiation and there would be virtually no time for the black hole to "suck up" anything. I suddenly remembered something I read on  that said a "a 100 tonne black hole would evaporate in 8.4×10−2 s, emitting approximately E=Mc2=9×1021 joules of energy as it does so – equivalent to more than a million megatons of TNT. I guess you could call this an explosion!"  If Mount Everest was condensed into a black hole, it would be smaller than a hydrogen atom. Mount Everest is 357 trillion pounds. 357 trillion pounds converted into energy would be a mind-boggling amount.  If a black hole formed in the LHC, wouldn't it completely annihilate it due to the spontaneous energy release due to Hawking Radiation? I'm new to the topic of Hawking Radiation, so I do not know if I have been misinformed in any way.
9332
Are black holes in a sense transparent? Meaning, they don't 'hide' any stars that happen to be 'behind' them?  A regular object obviously blocks a point source of light emanating from 'behind' it. Meaning it creates an area of space photons cannot get to us from. But a massive object curves space, and so via gravitational lensing we can sometimes see distant stars that are 'behind' other stars (by behind I mean that if the star in the middle would suddenly vanish, the trajectory of photons traveling from the distant star to Earth would have passed through the volume now occupied by the middle star).  My thinking, is that black holes curve space so much, that from ANY source of light outside it, light has a trajectory to reach us on Earth (assuming of course no other object blocks it on its way).  The reasoning for this is: Suppose we draw the straight line between us and the light source (as if the black hole wasn't there at all). Photons from our light source crossing the event horizon would obviously 'fall' inside. A photon in theory traveling exactly tangent to the event horizon would in a sense go 'into orbit' around the black hole (obviously there is no such thing, because the black hole isn't perfectly spherical, and there are quantum effects and so forth. But I'm focusing on the geometry), crossing our straight line exactly at the intersection of our line and the event horizon. Any photon passing near the event horizon, but a bit farther, would follow a curve that intersects our line a little farther from the event horizon. Continuing this, there would be at last be a photon that would exactly curve our way.
82686
$$ A $$  Light is travelling away from the absolute center of the Schwarzchild black hole (i.e. none of that spinning/stretching field) at $\ C $, and so shouldn't $\ \Delta C$ need to be greater than $\ C $ in order for the light to collapse into the black hole, therefore violating the relativistic limit of $\ C $?    It seems like light would just 'hover' right in the center of a black hole then. Because if it goes away from it, the event horizon pulls it back, and if it goes into it, the velocity change must be greater than $\ C $ which violates general relativity; so technically there's no way for the light to move away, and so it would just be still.  The thing is that the light must move at the speed of light or general relativity doesn't work. It's massless... But in the center of a black hole, if the above is true (it can't move or escape), then it is not travelling at the speed of light. So it cannot be still.  It cannot be spinning in a constant orbit either, because that would require velocity in an outward angle. And it cannot move in an outwardly fashion, only in an inwardly one.  Is it spinning in an infinitely decaying spiral towards the center?  If so what happens to it if and when it reaches the direct and complete center, i.e. The case of it being 'shot' from the center.    $$ B $$  If you shot light at an angle towards a black hole, but not directly to the center of it  then that light from the outside will eventually become the same as the light that I said was shot from the center to the outside; because in both cases, they end up in an infinitely decaying spiral.    If so it would end up exactly like the light in the above paragraph, so then we need to figure out what happens to that light to figure this one out, or vice versa.    $$ C $$  If you shot light directly towards the center of a black hole it would have no spiral to decay into, so therefore it must stop at some point in the center of a black hole. If it stops though, it loses it's momentum and violates general relativity again because it's energy "disappears".    If so then it violates conservation laws. So I'm assuming this simply is not true.    Is the light travelling into a black hole somehow still travelling at the speed of light, in every direction possible where that direction keeps it facing towards the center?  If so, why is there not a point at which the only thing left for the light to do is stay still, since all other paths would increase the distance between it and the black hole's center, and therefore become Case A. And if there isn't, what would happen if there was?  Also, if it's a time and observation answer where it supposedly takes and infinite amount of time for C to hit the center, wouldn't it be the case that the black hole is 'eating' all of the available light in the universe and keeping it in stasis?  Also, what happens in a Kerr black hole then? Does it just rip the field of time/gravity at an angle such that the light would never get closer than the center?
49611
In the simple explanation that a black hole appears when a big star collapses under missing internal pressure and huge gravity, I can't see  any need to invoke relativity. Is this correct?