Aki Posted November 20, 2004 Report Posted November 20, 2004 Originally posted by: Tim_Loui read some article online before, it says that the way we detect black hole is some "vitrual photons" that are created and travelled faster than the speed of light and ultimately reached us... is that right or? I think you might be thinking about the anti-particles. Hawking said that every particle has an anti-particle, which once both come in contact, they will annihilate. He said that while the particle gets sucked into the black hole, its anti-particle might escape, and therefore, it looks like as if the black hole is emitting these ani-particles.
BlameTheEx Posted November 20, 2004 Report Posted November 20, 2004 Tormod You caught me using ambiguous language again. Ok. It is doubtful that a Black hole has a surface, but Singularity is a slightly different term. It defines an area of space outside normal laws, or so I have always taken the term. As applied to Black Holes, it defines the point of no return. In effect it is a sphere, and so it has a surface, although not the sort you walk on. It is more a concept than an actuality. If such a thing is possible, NOTHING can escape from a body with an escape velocity of C. You might argue that a photon could, as it has a velocity of C, but ALL its energy would be lost in the escape. A photon with no energy simply does not exist. The argument for Black Holes goes that although particles are apparently accelerated to greater than C within the Singularity, it does not count, as the inside of a Singularity is not part of the universe.
Aki Posted November 21, 2004 Report Posted November 21, 2004 Originally posted by: BlameTheEx If such a thing is possible, NOTHING can escape from a body with an escape velocity of C. You might argue that a photon could, as it has a velocity of C, but ALL its energy would be lost in the escape. wouldn't the escape velocity be greater than C?
Tormod Posted November 21, 2004 Report Posted November 21, 2004 Originally posted by: BlameTheExThe argument for Black Holes goes that although particles are apparently accelerated to greater than C within the Singularity, it does not count, as the inside of a Singularity is not part of the universe. No offense, Blame, but I think you are confusing the terms black hole and singularity. The black hole in itself has a dimension and size. If our own sun were to collapse and form a black hole it would be just a few kilometers across. The singularity resides at the heart of the black hole and has no dimension and no size, although it has infinite density (which is a rather meaningless term). So while technically the black hole has an "inside" and a "surface", the singularity does not. As applied to Black Holes, it defines the point of no return. That is called the event horizon.
BlameTheEx Posted November 21, 2004 Report Posted November 21, 2004 Tormod Hmmm. You may be right. I guess I should do more checking. Aki. _______________________________________________wouldn't the escape velocity be greater than C? ________________________________________________ Err. You are asking for more details on something I don't actually believe in, but I will do my best. First, nothing can escape from a body with an Escape velocity of C or above. C is sufficient, but perhaps you are asking if it can get higher? Tricky question. Best I can figure it the answer is no, even if I am wrong, and Black Holes exist. The reason is rather interesting. At the point which I think Tormod is accurately calling the event horizon, time dilation is total. Nothing can go further, or faster because there is no time for it to happen in.
Aki Posted November 22, 2004 Report Posted November 22, 2004 Well this is what I'm thinking, and I'm probably wrong:if the escape velocity is C, then wouldn't light be able to escape? So shouldn't the escape velocity be greater than C?
BlameTheEx Posted November 22, 2004 Report Posted November 22, 2004 Aki Well, light would escape, and it wouldn't. An escape velocity of C would result in a red shift of infinity. The light would escape, but with a frequency of zero hertz, and an energy of zero electron volts! In short it will disappear entirely. Nothing gets out except the idea of a photon.
Freethinker Posted November 22, 2004 Report Posted November 22, 2004 Originally posted by: TormodSo while technically the black hole has an "inside" and a "surface", the singularity does not. As applied to Black Holes, it defines the point of no return. That is called the event horizon.I think it might be more accurate to refer to the Event Horizon as a "Boundary", not a "surface".
Aki Posted November 23, 2004 Report Posted November 23, 2004 Originally posted by: BlameTheExAki Well, light would escape, and it wouldn't. An escape velocity of C would result in a red shift of infinity. The light would escape, but with a frequency of zero hertz, and an energy of zero electron volts! In short it will disappear entirely. Nothing gets out except the idea of a photon. Ah okay, I get it now
BlameTheEx Posted November 23, 2004 Report Posted November 23, 2004 Tormod. Done my checking. You are right. Event horizon is correct, and Singularity isn't. Thanks.
Tormod Posted November 23, 2004 Report Posted November 23, 2004 Originally posted by: FreethinkerOriginally posted by: TormodThat is called the event horizon.I think it might be more accurate to refer to the Event Horizon as a "Boundary", not a "surface". I did not say the event horizon has a surface. The black hole has a "surface", the event horizon is the boundary which Blame correctly labeled the "point of no return". I think however that for all practical purposes they are one and the same. The "point of no return" idea is true no matter what happens, even if we see particles escaping. When, say, a proton reaches the horizon it will decay into smaller parts, some of which will fall into the black hole, while some will "miss" it. The original particle is nevertheless destroyed by the interaction. this is, however, true for *all* events in quantum physics. I found an interesting (not too technical) read on black hole formation: The Formation and Growth of Black Holeshttp://www.mathpages.com/rr/s7-02/7-02.htm
paultrr Posted November 23, 2004 Report Posted November 23, 2004 Has anyone else ever noticed that the zero point out of quantum physics shares a lot in common with a blackhole state? Granted the gravity context is different. But zero time type background frame still applies.
paultrr Posted November 23, 2004 Report Posted November 23, 2004 Basically, backing up another post with the words, "this is, however, true for *all* events in quantum physics."
Freethinker Posted November 23, 2004 Report Posted November 23, 2004 Originally posted by: TormodOriginally posted by: FreethinkerI think it might be more accurate to refer to the Event Horizon as a "Boundary", not a "surface".I did not say the event horizon has a surface. The black hole has a "surface", the event horizon is the boundary which Blame correctly labeled the "point of no return". I think however that for all practical purposes they are one and the same.I just have a problem with using the word "surface" to symbolize the "boundary" point of the Event Horizon, just as I would to symbolize the Terminal Shock point of transition for the solar system. There is no physical layer to detect as one would connect to a "surface". It is a distance at which interactions are changed. e.g. the Earth has a "surface" which is easily definable, while a Country has a boundary which is not. You KNOW where the earth's surface is at any one point on it. While you may only know where a Country's Boundary is only know if an effect is experienced at some point which is different from the previous point. Perhaps by being shot at?
Tormod Posted November 23, 2004 Report Posted November 23, 2004 Yeah, I see your argument, and not having seen a black hole I am hard pressed to say you are wrong. But would you say that the sun has a surface? Where does the sun's gravity field end, and where is the "point of no return" for anything that falls towards it?
Freethinker Posted November 23, 2004 Report Posted November 23, 2004 Yes I would say the Sun does have a "surface". Being gaseous, it may not fit the "hard" def of Earth's. But it is obvious it has a "substance" of which it is made and defined. I am not aware of a PoNR for the Sun. Yes any substance we know of would be destroyed at some point before reaching the surface, but as far as I am aware some particles can still pass thru and light obviously is NOT captured by it.
paultrr Posted November 25, 2004 Report Posted November 25, 2004 The Sun has no event horizon in its present state. As for gravity its field drops off following the normal 1/r^2 rule. In essence, while getting weaker as one goes outward, its field extends towards infinity, so to speak, the same as an EM field does. The sun has a surface. The event horizon of a blackhole is not an actual surface at all. It's a point of no return beyond which something would have to exceed the speed of light to escape. Its not composed of matter in the common meaning of such and is simply a horizon as far as observation goes. The gravity field of a Blackhole like every other gravity field extends far further out than its own body itself. The Hawking radiation(Virtual particles) that escape from this event horizon do so by quantum tunneling which is a known shortcut path through spacetime itself. One virtual particle at the event horizon can borrow enough energy to become a real particle. That particle escapes while its twin does not. The timing of this escape via quantum effects is shorter in duration than light speed allows along normal paths. So is essence via a quantum path such a particle does exceed the escape velocity of the event horizon. But it does so by a quantum allowed path that at macroscopic scales is not allowed. Also, this type of event takes place right at the horizon itself. So in essence, the virtual particle made real that escapes has never really been inside that horizon. Its been at that horizon and the small path difference allowed under quantum theory is all the boost it needs to escape. The escape velocity right at the boundary of the event horizon itself is C, beyond that point its greater than C.
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