There's something which I don't quite get about black holes and event horizons (feel free to tell me if I got something completely backwards at any step along the way):  Say we have a singularity and we launch a particle from the singularity's Schwarzchild radius R at the speed of light directed outwards from the black hole - wouldn't that particle be able to escape beyond R, slowly decelerating in a parabolic path to a halt somewhere outside it (at a radius of, say, R1)?  If so - wouldn't an observer A between R and R1 be able to see that particle as it decelerates? And if so, we couldn't define the event horizon to be at R, because light from R can affect A who's outside it. We couldn't define the event horizon to be at R1 either, because light from the observer could escape R1 entirely.  So where would we define the event horizon in the case presented above? How can there be an event horizon if light can be launched slightly below it and take an parabolic path outside it, even if only for a moment?
2284
In  (ignoring Hawking radiation), why is a  black? Why nothing, not even light, can escape from inside a black hole? To make the question simpler, say, why is a  black hole black?
66497
According to NASA      a black hole is anything but empty space. Rather, it is a great amount of matter packed into a very small area   According to the documentary  about Black Holes     Gravity here is so strong even light cannot escape   Which leads me to believe that Black Holes basically are masses so compact but great which gives them very strong gravitational influence (Newton's Gravitational Law) that light is not allowed to go out of, hence a "Black" hole.  If my above understanding of Black Holes is correct, then why is those Spiral Galaxies pictured using Hubble Telescope is showing a big shiny ball instead of a "Black" hole ? What is the obvious clue that I am missing here ?  Here is the image of Andromeda Galaxy image taken using Hubble.
16595
This question was prompted by . Notoriously, if you calculate the Schwarzschild coordinate time for anything, matter or light, to reach the event horizon the result is infinite. This implies that the universe ages by an infinite time before someone falling into the black hole reaches the event horizon, so could that person see the universe age by an infinite time?  To be more precise, suppose the observer starts falling from rest at time $t = 0$ and some initial distance $r &gt; r_s$. If we wait for some time $T$ then shine a light ray at the falling observer. Will the light ray always reach the falling observer before they cross the event horizon? If not, what is the formula for the longest time $T$ that we can wait and still be sure the ray will catch the observer? If $T$ is not bounded it implies that observer could indeed see the end of the universe.  I can think of a qualitative argument for an upper limit on $T$, but I'm not sure how sound my argument is. The proper time for the observer to fall to the event horizon is finite - call this $\tau$. The proper time for the light ray to reach the horizon is zero, therefore the light ray will reach the observer before they cross the event horizon only if $T &lt; \tau$. Hence $T$ is bounded and the observer won't see the end of the universe.  I think a more rigorous approach would be to determine the equations of motion (in the Schwarzschild coordinates) for the falling observer and the light ray, and then find the condition for the light to reach the falling observer at some distance $\epsilon$ from the event horizon. Then take the limit as $\epsilon \rightarrow 0$. In principle this seems straightforward, but in practice the algebra rapidly defeated me. Even for a light ray the radial distance:time equation isn't closed form (Wolfram claims it needs the $W$ function) and for the falling observer the calculation is even harder.
49922
I have a couple of questions about light here, and sorry of they are silly..  So since anything that goes beyond the event horizon can't go out, so what if a light beam was pointed somewhere behind the event horizon to the outside, would light have a slower speed then get sucked back? and what if the light particles were somehow exactly on the event horizon, would the light stand still?  And why can't light stand still? why does it always move at that really high speed?
97656
In quantum mechanics and gravity theories one always meets the word "singularity" in connection with talks about black holes and the big bang, etc. Now in mathematics a singularity is well defined – mostly as a pole of $n$-th order that makes the function value approach infinity near the singular point...  But I cannot imagine that a singularity can actually exist in the real world. It is simply "un-physical" and any equations having physical meaningful singularities must be wrong or the singularity somehow excluded just like saying $v/c$ is always less than 1.  Hence, a black hole can not in it self "be" a singularity but might behave as such near that point, which, however, nothing ever reaches. Thus it can not be observed and is physically non-existent as a point in time and space but only an abstraction – similar to what I would be tempted to say about the photon.
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.
87877
The photon sphere is a spherical region in space where photons are forced to travel in an orbit at $r = \frac{3GM}{c^{2}}$.   Is it possible to detect these spheres?  What happens if I fall through it? What happens to a photon that tangentially hits the event horizon, wouldn't this also lead to a photon orbiting the black hole? EDIT Since neutron stars also might have a , do we know neutron stars that should have such a sphere?
9328
I guess nobody really knows the true nature of black holes, however, based on everything I know about black holes, there is a "singularity" at their center, which has finite mass but is infinitely small and thus infinitely dense.  So a black hole is really just a point mass that dramatically warps the space-time around it so that an event horizon is achieved.  What doesn't make sense to me is how a true mathematical "point" can rotate, or at least how it is even possible to determine if a point is rotating (or if it should even make a difference about whether or not it is rotating, for the same reason it would be impossible to tell if a perfect, featureless sphere were rotating (I know it would have higher rotational energy but lets assume you can't measure that because a true "point" can't attain rotational energy since it has no radius)).  The only conceptual solution I have to describe a rotating black hole would be to compare it to a rotating whirlpool, where the surface of the water is the fabric of space-time... but this is different because there is no "singularity" in a whirlpool that causes the water around it to rotate, whirlpools form from flow of water around a point (put into motion perhaps by some kid swirling water in a 2-liter bottle... it's due to an external cause), but the only thing that could cause rotation around a black hole would be the singularity itself, which doesn't make sense because, once again, how can we know if a true point is actually rotating?  Aside from everything I've said, here is what I am actually asking:  Why/how do some black holes rotate... what is the mechanism of their rotation (is analogous to a spinning point or a 3-d whirlpool, or something different)?
68645
Since the final Black Hole (BH) had 3 solar masses less of mass than the original binary BH, it seems the 2 BHs lost mass, and with it event surface area, entropy, and information. If that came from inside the BHs it seems that it would be a contradiction of what a BH is --'nothing escapes its event horizon'. Some may say that came from the rotational energy of the pair, but the new BH now seems to have less than the two original ones summed. So, did some of that escape the horizons?
35600
Observations:   I have read that for a free-falling observer within the event horizon of a black hole that all lines of sight will end at the singularity which is black. I also look up and see that the sky is mostly black. I also know that by measuring the red-shifts of the galaxies that they are all accelelerating towards that blackness. (rather than accelerating away from a big-bang event, which makes less intuitive sense) The red-shifts are greatest for those galaxies closest to the singularity (i.e. furthest away from us).   I therefore conclude from these consistent observations that all the matter in at least the finite observable region of the universe in which we are located, is accelerating due to gravity towards the singularity of a super-massive black hole.  In the spirit of good science I will happy accept being wrong, but I need to understand why I am wrong.  Predictions from the theory:  The light from the universe outside the event horizon of the black hole should be visible as a small disc or point light source in the direction facing away from the singularity. Of course, if this happens to be inconveniently located on the far side of the milky way's core from us, then we will be unable to observe it.  Further Speculations:  It may be that in regions of the universe far ouside the influence of this black hole (e.g. >50 billon light years away) that the sky is white hot due to Olber's paradox.  The black hole may have been formed from a large quantity of anti-matter, which might explain the local perponderance of ordinary matter.  The microwave background radiation may be a result of Hawking radiation at the event horizon (which we are viewing fromn the inside), or, depending upon the non-euclidian geometry, we may be observing the result of matter being destroyed at the singularity.
25687
Imagine a black hole that is fast-approaching its final exponential throws of Hawking evaporation. Presumably, at all points in this end process there will remain a region that identifiably remains &quot;the black hole&quot; until the the very end, as opposed to huge swarm of fundamental particles that is being radiated out from it. As the mass of the black hole descends to that of individual particles, it would seem entirely feasible that the very last fermionic Hawking radiation event available to the almost-deceased black hole could leave it with an unbalanced charge, e.g. -1, and an unbalanced spin, say 1/2. It would also have some kind of mass of course, but that aspect of the final residue could be fine-tuned to any specific value by photon emissions of arbitrary frequencies. After photon emission mass trimming, the resulting black hole residuum would reach a point where it is no longer be able to evaporate into any known particle, because there is no longer any lower-mass option available to it for removing the -1 charge and 1/2 spin. The black hole residuum will at that point be stuck, so to speak, stuck with exact charge, spin, and mass features of an electron. And so my question: Is it an electron? And if so, by equivalence, is every electrons in the universe really just a particular type of black hole that cannot evaporate any further due to the constraints of charge and spin conservation? And if so, why are charge and spin so uniquely combined in such black hole remnants, so that e.g. a remnant of -1 charge and zero spin is not permitted, at least not commonly, and the mass is forced to a very specific associated level? Is there anything in the current understanding of general relativity that would explain such a curious set of restrictions on evaporation? The full generalization of this idea would of course be that all forms of black hole evaporation are ultimately constrained in ways that correspond exactly to the Standard Model, with free fundamental particles like electrons being the only stable end states of the evaporation process. The proton would be a fascinating example of an evaporation that remains incomplete in a more profound way, with the three quarks remaining incapable of isolated existence within spacetime. The strong force, from that perspective, would in some odd sense have to be a curious unbalanced remnant of those same deeper constraints on the overall gravitational evaporation process. This may all be tautological, too! That is, since Hawking radiation is guided by the particles possible, the constraints I just mentioned may be built-in and thus entirely trivial in nature. However, something deeper in the way they work together would seem... plausible, at least? If an electron is an unbalanced black hole, then the particles given off would also be black holes, and the overall process would be not one of just particle emission, but of how black holes split at low masses. Splitting with constraints imposed by the structure of spacetime itself would be a rather different way of looking at black hole evaporation, I suspect. (final note: This is just a passing thought that I've mulled over now and then through the years. Asking it was inspired by  by Ben Crowell. I should add that I doubt very seriously that that my wild speculations above have anything to do with Wheeler's concept of geons, though.)
95948
Often this is asked if it was pointed towards a black hole. so I thought, if a scientist decided to sacrifice himself and jumped in a black hole, released some dry-ice mist and pointed a laser pointer away from him, what would he see.   (The scientist is just to help you understand. Maybe this experiment is impossible because he'd get spaghettified first)  And the laser pointed is pointed directly away from a black hole, not slightly to the left or right, where I'd assume it would curve and fall back.
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$$ A $$  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?
112806
Consider a photon coming from the infinity in a unbounded orbit to a Schwarzschild black hole (Schwarzschild radius $r_{s}$) (see  for illustration). Its impact parameter is $b$ and its distance of closest approach is $r_{0}$ with $$b^2=\frac{r_{0}^{3}}{r_{0}-r_{s}}$$.  Then its trajectory in polar coordinates is defined by :  $$\frac{d\varphi}{dr} = \frac{1}{r^2\sqrt{\frac{1}{b^2}-\left(1-\frac{r_s}{r}\right)\frac{1}{r^2}}}$$  Consequently : $$\varphi\left(r\right) = \int_{r_{0}}^{r} \frac{dp}{p^2\sqrt{\frac{1}{b^2}-\left(1-\frac{r_s}{p}\right)\frac{1}{p^2}}}$$  and one can compute the total deviation using :  $\Delta\varphi = 2\times\left(\lim_{r\to+\infty}\varphi\left(r\right)-\frac{\pi}{2}\right)$  But my question is : how can I plot/draw the trajectory using the integral expression of $\varphi\left(r\right)$ ?    Because if I compute : $$f\left(r\right) = 2\times\left(\int_{r_{0}}^{r} \frac{dp}{p^2\sqrt{\frac{1}{b^2}-\left(1-\frac{r_s}{p}\right)\frac{1}{p^2}}}-\frac{\pi}{2}\right)$$ I obtain $f\left(r_{0}\right) = -\pi$, and then $f$ increases up to zero, crosses zero, and tends to its positive value at infinity $\Delta\varphi$. It does not make sense for me and I do not understand how to compute the trajectory from that.    If I compute : $$g\left(r\right) = \int_{r_{0}}^{r} \frac{dp}{p^2\sqrt{\frac{1}{b^2}-\left(1-\frac{r_s}{p}\right)\frac{1}{p^2}}}$$ it starts from $0$, and increase up to $\frac{\pi}{2}+\frac{\Delta\varphi}{2}$.    I would like to compute the trajectory in the $\left(x, y\right)$ plane, so how to use the values of $f\left(r\right)$ or $g\left(r\right)$ to compute the function $y\left(x\right)$ ?
41356
I'm not very knowledgeable about physics generally, but know that nothing can escape a black hole's gravitational pull, not even light (making them nearly invisible?).   My question is: What has been obtained from direct observation of a black hole to prove their existence? Please note that I'm not questioning the existence of black holes, but am just curious as to how we have come to realize it.
107143
Under Newton, the escape velocity is  $$v_{esc} = \rm c \ \sqrt{r_s/r}$$  where $\rm r_s=2 \ GM/c^2$. In the nonrotating relativistic case (the Schwarzschild case) the radial escape velocity is the same:  $$v^{\parallel}_{esc} = \rm c \ \sqrt{r_s/r}$$  which is the speed of light, $v={\rm c}$, at the event horizon at $\rm r=r_s$. Since the photon sphere, where an object with $v={\rm c}$ would not escape, but be captured into orbit, is at $r=3 \ {\rm GM/c^2}$, the transverse escape velocity is  $$v^{\perp}_{esc} = \rm c \ \sqrt{r_s/r}\div \sqrt{1-r_s/r}$$  and for the observed (shapiro delayed) velocity ${\rm v}$ the transformation is  $${\rm v}^{\perp}=v^{\perp} \ \sqrt{\rm 1-r_s/r} \ , \ \ {\rm v}^{\parallel}=v^{\parallel} \ (\rm 1-r_s/r)$$  so the observed angular motion is the same as with Newton, yet locally higher, while the observed radial motion is slowed down because of the radial length contraction and time dilation.  But what about the Kerr case where the rotation is significant and the event horizon is at $\rm r=(1+\sqrt{1-a^2}) \ {\rm GM/c^2}$ instead of $\rm r=r_s$?  A testparticle which is locally at rest and therefore corotating with the angular velocity of the frame dragging  $${\Omega = \rm d \phi/ {\rm d t} = {\rm 2 \ a \ r/((a^2+r^2)^2-a^2 (a^2+(r-2) \ r) \sin ^2(\theta ))}}$$  should experience the time dilation factor  $$\rm \dot t = {\rm d t}/ {\rm d}\tau = \sqrt{{\rm ((a^2+(r-2) \ r) (a^2 \cos ^2(\theta )+r^2))/((a^2+r^2)^2-a^2 (a^2+(r-2) \ r) \sin ^2(\theta ))}}$$  relatively to the far away observer, so multiplying $\Omega$ with $\rm \dot t$ should give the local frame dragging angular velocity, which, multiplied with the $\rm \sqrt{x^2+y^2}$ value of the Boyer-Lindquist to cartesian transformation should give the transverse velocity - this is slightly below $\rm c$ for a maximaly rotating black hole (with $\rm a = 1$ - at the edge of the outer equatorial ergosphere where $\rm r=1$ and $\rm R=\sqrt{r^2+a^2}$).  But how do I find the radial and total escape velocity, and what are the rules to convert the shapiro-delayed coordinate velocities $\rm d r/d t \ , \ d \phi/d t \ , d \theta/d t$ to the local 3-velocity components $v^{\parallel}$ and $v^{\perp}$ of a test particle (local meaning relatively to a frame dragged probe with contant $\rm r$)? The radial length contraction seems a bit different than in the Schwarzschild-case, where it is the same as the gravitational time dilation factor...
82528
The classical explanation of a black hole says that if you get to close, you reach a point - the event horizon radius - from which you cannot escape even travelling at the speed of light. Then they normally talk about spaghetti.  But here's a thought experiment. What if I have a BH with event horizon radius R such that the gravitational gradient at the event horizon is far too weak for creating pasta. I build a Ring with radius R+x around the BH. Then I lower a pole of length x+d from my ring towards the BH, so that the tip passes beyond the event horizon.  Now what happens when I try to pull the pole back?