As we know that light photon cannot escape the gravity of a black hole so I was thinking that if that is the surface of the black hole would be bright as all the photons would be there only. Am I right or wrong?
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$\begingroup$ What you're describing is similar to the photon sphere, the radius at which a photon can orbit a black-hole. $\endgroup$– DilithiumMatrixApr 9, 2013 at 17:23
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2$\begingroup$ @akash Perhaps a change of title is in order? Are you interested in the "surface" (aka event horizon) or the "center" (aka singularity)? $\endgroup$– user10851Apr 9, 2013 at 23:10
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$\begingroup$ sorry that was surface of the black hole $\endgroup$– DeiknymiApr 10, 2013 at 6:26
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7 Answers
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Well, a black hole doesn't have a solid surface, it has an event horizon that, as you might already know, represents a mathematical boundary beyond which no matter/energy can escape, which includes photons. Any light that the "star" may have released is no longer visible because it cannot escape, but incoming light may either fall into the hole, or "orbit" the hole just outside of the Schwarzschild radius. This light would be visible only to an indestructible observer, otherwise he would be destroyed by the sum of all forces at this stage. But the center is not "bright" because light would have to emitted and reach our eyes for us to see it.
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$\begingroup$ I didn't understood what do you mean by it doesn't have solid surface as it is made-up from mass and matter $\endgroup$– DeiknymiApr 9, 2013 at 18:10
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1$\begingroup$ Well, yes. But the mass and matter has virtually no volume, it has collapsed into a singularity, a point. And that point has no surface. This is the theory, but theoretically, a singularity could also be an object of extremely high density and infinitely small volume. $\endgroup$ Apr 9, 2013 at 20:36
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A photon emitted outside the event horizon can escape outwards, while a photon emitted inside the event horizon can only move inwards. The only way for a photon to be trapped at the event horizon would be for it to be emitted (by some infalling body) exactly at the event horizon.
Although this is possible, there are two problems. The first problem is that the event horizon is a 2D surface so the probability of a photon being emitted exactly at the event horizon is zero. Since photons emitted a tiny bit inside or a tiny bit outside move away from the event horizon the photon density is going to fall to zero at the surface itself. The second problem is that a photon emitted exactly at the surface would be infinitely red shifted for any external observers so it would appear dark not bright.
So the surface of the black hole, i.e. the event horizon, isn't bright.
answered Apr 9, 2013 at 17:18
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$\begingroup$ You wrote that because the event horizon is a 2d surface the probability of a photon being released there is zero. By that reasoning the probability of a photon being released at any particular point (which must lie on some 2d surface) would also be zero. That can't be right. $\endgroup$ Apr 10, 2013 at 2:54
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1$\begingroup$ @Richardbernstein: It may seem paradoxical, but it's right. If you pick a random real number uniformly distributed between 0 and 1, the probability of any particular number is 0. If it weren't zero, what would it be? $\endgroup$– user4552Apr 10, 2013 at 4:59
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$\begingroup$ No the point is not seeing it from a far away point I was saying if we reach the surface or as you say event horizon of the black hole will it be bright there $\endgroup$– DeiknymiApr 10, 2013 at 6:29
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$\begingroup$ @Akash: No, if you leap into a black hole you will notice nothing special as you pass the position where far away observers see the event horizon. In fact there would be no way to tell when you pass through the event horizon except by calculating where it should be. $\endgroup$ Apr 10, 2013 at 7:05
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$\begingroup$ Ben Crowell is correct. There are an infinite number of points in space at which the particle could be emitted, but infinity in three dimensions (not including time). A plane only has two degrees of freedom, so the chances of a particle being in any specific point are one in infinity, or zero. $\endgroup$ Apr 29, 2013 at 20:45
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A photon can escape the black-hole if it doesn't fall into the event horizon. If the photon hits the surface, it just orbits around the black-hole through the horizon. The curvature is so strong that it can't escape through the exit light cone. Here is an image from "A Brief History of Time". This makes sense that only photons that trace above the boundary can be perceived.
My question is: "How can we perceive something when the element of perception (light or any other EM radiation - photons) can't reach us?" (It can't even escape...)
answered Apr 9, 2013 at 17:02
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http://www.aoc.nrao.edu/~gcnews/gcnews/Vol.11/
The link above is one example of a serious scientific simulation. It shows more light comes from the sides than the center, which looks more or less darker than the outside. This is called "seeing the black hole's shadow", and some serious astronomers expect to do it in less than ten years.
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Firstly there are a lot of post with “we know” and “Yes” and “No” The simple fact is we don’t know, every bit of knowledge we have concerning black holes is theoretical, so keep an open mind, keep asking questions and don’t accept anyone answers as final.
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$\begingroup$ The basis for black holes is theoretical. Observations are just now starting to point to black hole like phenomena as unequivocal fact. The theory makes very specific predictions about black holes and every definitive answer is always within the context of the theory. $\endgroup$ Apr 9, 2013 at 22:51
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Yes the singularity of a black hole would be incredibly bright once you get to the center because the light and energy can't escape the black so it is all compressed into and infinity dense small point and even if it is just normal matter going inside the black hole it would even make it more bright because the black hole tidal forces would rip the atoms apart because it's gravity gets so strong not just that at the center of a black hole it is so dense that the atoms would heat up a lot and turn back into energy and since it is infinitely dense with a lot of energy and photons inside there all compact in a single point it will stay burning hot and bright its just we can't see the center because the gravity sucks in the light before it reaches our eyes.
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4$\begingroup$ I'm just gonna leave this here:
,,.,;use them to your heart's content $\endgroup$ May 12, 2016 at 20:01
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There are alot of theories about black holes.
It can take you to another universe on our multiverse, so you can say yes if you exit near a star or something.
It can give more energy to the photons that enter so you will not see it.
It can even turn the light in something else. Small chances.
But these are just some theories. By normal physics yes, you should see light, but... we know that nothing acts normaly inside a black hole so the answer is undefined. We may found it just experimentaly, and this would be a suicide mission. I think that only a type III civilization could enter a black hole and we are a milion years away from that stage of the civilization so... you can only hope that immortality and other things can occur in your lifespan. But again by normal physics the answer is YES! PS: Sorry for my bad english, I am a 12 year old boy from Romania.
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1$\begingroup$ John: although black holes are pretty weird they're probably less weird than you think. It's going to be a bit hard to get a real idea of how they work because you won't have the maths background at age 12, but there are good articles around. For example see physics.stackexchange.com/questions/19636/… for why a black hole doesn't really act as a gateway to another universe. $\endgroup$ Apr 9, 2013 at 17:34