Johnny5 Posted March 8, 2005 Share Posted March 8, 2005 I have a question. Is a black hole really hot, or really cold? Thank you Link to comment Share on other sites More sharing options...
Martin Posted March 8, 2005 Share Posted March 8, 2005 I have a question. Is a black hole really hot' date=' or really cold? Thank you [/quote'] depends on the mass, Johnny5. massive BH extremely cold BH with tiny mass can be very hot temp is proportional to RECIPROCAL of mass Link to comment Share on other sites More sharing options...
Martin Posted March 8, 2005 Share Posted March 8, 2005 in natural units the formula for the Bekenstein Hawking temperature of a BH is the simplest expressed in nat. units of type |8piG|=|hbar|= |c| = 1 (which is the system of units one typically sees used in QG research) the temp is simply equal to the recip of the mass well, in those units room temperature is about E-29 (tentominus29) so a black hole radiating with that temperature from surface would have mass (again expressed in nat. units) of E29 and that is E21 pounds mass, to put it in some more familiar language, in other words the mass is trillion billion pounds (E12 x E9 pounds) if it is LESS than that it is going to be hotter than room temp if it is MORE massive then it is going to be colder solar mass holes are very very cold because mass much more than E29 Link to comment Share on other sites More sharing options...
jdurg Posted March 8, 2005 Share Posted March 8, 2005 That would actually seem to make sense, since the more massive the black hole is, the more 'stuff' is packed into the smaller area. As a result, the molecules/atoms have very little space to move around, therefore their temperature would be cooler since they don't have the ability to move like molecules around us can (Since temperature is a measurement of molecular motion). Now while I'm probably WAAAAAAAAAAY off on that explanation, it does sound logical. Link to comment Share on other sites More sharing options...
Johnny5 Posted March 8, 2005 Author Share Posted March 8, 2005 Ok, go slowly. You say that the temperature of a black hole is inversely proportional to the mass. As the mass increases, the temperature decreases. According to the formula, what mass is required for a black hole to have a temperature of absolute zero? Infinite mass? I would think some quantum formula would sort of prevent that. Link to comment Share on other sites More sharing options...
Martin Posted March 8, 2005 Share Posted March 8, 2005 ...According to the formula' date=' what mass is required for a black hole to have a temperature of absolute zero? Infinite mass? I would think some quantum formula would sort of prevent that.[/quote'] even in pre-quantum, or classical, thermodynamics it is considered that the temperature of absolute zero is not physically achievable. no real physical system can have that temp that temp can only be approached closer and closer but never attained. that is probably some law of thermodynamics Link to comment Share on other sites More sharing options...
5614 Posted March 8, 2005 Share Posted March 8, 2005 Yes its the 3rd law of thermodynamics. Link to comment Share on other sites More sharing options...
Johnny5 Posted March 9, 2005 Author Share Posted March 9, 2005 Black holes aren't supposed to emit radiation. So if you put your hand close up to one, like you do a light bulb, you wont feel the heat. So you might suspect that the black hole is cold to the touch. Is it? Could it be, that past a certain mass, the black hole actually reaches absolute zero temperature? And if you are going to site the third law, are you certain it is true? Anyone care to answer? Thank you PS: I suppose, if there is internal motion, you could say that it's internal temperature isn't zero, but what about the surface? I read somewhere, that its easier to visualize internal thermal energy by focusing on the idea that the temperature inside is related to the speed of the parts in the 'gas' (Maxwellian speed distribution), but that this is too difficult to use for practical applications, and that is why the concept of entropy is used. [math] \frac{1}{T} = \frac{\partial S}{\partial U} [/math] But entropy is such a complicated variable, and I think there are problems with the mathematics which statistical mechanics is based on. Maybe you could think of this post more as, I want to understand U better. Link to comment Share on other sites More sharing options...
Martin Posted March 9, 2005 Share Posted March 9, 2005 Black holes aren't supposed to emit radiation... first let's both notice that this is purely theoretical, humans have not yet observed a black hole up close enough to detect its Hawking temperature and its Hawking radiation. black holes have been observed (or what reputable mainstream astronomers interpret as black holes) and some of them are Xray sources. the Xrays come from stuff spiraling into the hole which gets very hot before it finally reaches the eventhorizon and disappears. OK so there is a lot of observational evidence that holes exist and we see stuff orbiting them even including stars orbiting the massive one at the center of our own galaxy and we take photographs etc etc, and we detect the Xrays, and they are the only explanation for quasars that has survived scrutiny etc etc. so black holes are real. But nobody has ever detected the hawking radiation from one, yet. so hawking radiation is only theoretical, but it is a pretty good prediction of theory (I personally have no problem with it) and you can say "nothing can come from within the event horizon! how can hawking radiation escape from the hole?" the answer is that it comes from JUST OUTSIDE the event horizon. that is where virtual particleantiparticle pairs form and one falls in before they can mutually annihilate and the other turns into radiation and escapes. the hawking radiation comes from microscopically just outside. so it is not a contradiction after all. Link to comment Share on other sites More sharing options...
5614 Posted March 9, 2005 Share Posted March 9, 2005 Black holes aren't supposed to emit radiation. So if you put your hand close up to one, like you do a light bulb, you wont feel the heat. As Martin said... hawking radiation. What do you mean you can't feel a light bulb emitting EMR? Or are you referring to when the light bulb is off? That has no link to the temperature of a black hole anyway, but I suppose it wasn't meant like that. Furthermore obviously this is all theoretical, no one has been up to a black hole and stuck a thermometer next to it!!! So it's theoretical, just because you haven't seen it doesn't mean it's not true though... To answer your question we must use theories, realistitc probably correct theories, just because you cannot prove it now doesn't mean its wrong... of course it doesn't mean it's right. The fact is that you (Johnny5) like to prove things and you cannot ask a totaly theoretical question, get a theoretical answer and then try to prove it! I mean it's a good way to go about physics, proving it all as you go, but with things which are totaly theoretical such as hawking radiation you really need to study it in depth and then conclude whether or not you agree with it. Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted March 9, 2005 Share Posted March 9, 2005 first let's both notice that this is purely theoretical' date=' [b']humans have not yet observed a black hole up close[/b] enough to detect its Hawking temperature and its Hawking radiation. Except Geoffrey, and we all know what happened to him! Link to comment Share on other sites More sharing options...
Johnny5 Posted March 9, 2005 Author Share Posted March 9, 2005 As Martin said... hawking radiation. What do you mean you can't feel a light bulb emitting EMR? Or are you referring to when the light bulb is off? That has no link to the temperature of a black hole anyway' date=' but I suppose it wasn't meant like that.[/quote'] Sorry, I didn't say what I mean there. I meant... When you do put your hand near a light bulb, you do feel heat. The explanation is that the light bulb is the source of the heat, and the 'heat' is electromagnetic radiation/photons. But, photons must leave a body at speed c=299792458 m/s. A black hole is defined as a body with a large enough mass, so that its escape velocity is greater than or equal to c. So my question is here... since light can only leave things at speed c, and the mass of a black hole is so great that light cannot escape, then the black hole can't be emitting light, and so it cannot be 'hot'. So is there an upper bound to a black holes mass (or more appropriately density) at which its temperature reaches absolute zero? (Hawking radiation aside) But I guess more centrally, I am wondering about the meaning of temperature here. A measure of how hot something is, or a measure of internal energy. In most cases they are related, but I'm not so sure about the case of a black hole. I don't see how anyone could intelligently talk about the entropy of a black hole, though I am sure some have tried. Link to comment Share on other sites More sharing options...
5614 Posted March 9, 2005 Share Posted March 9, 2005 What gives you the idea that the surface of a black hole will be cold? I mean, so it gives off no heat, that doesn't mean the surface cannot have thermal energy, no conduction/convection (vacum) and no radiation can espace, but if you touched it conduction could occur. Think of the energy that goes in, just because no energy escapes doesn't mean it can't have thermal energy... in fact as a guess based on logic I'd say that surely a black hole would be hot? Link to comment Share on other sites More sharing options...
ed84c Posted March 9, 2005 Share Posted March 9, 2005 No. The photons are craeted at the edge of the shwarzchild radius NOT IN IT, and therefore can escape as 'heat' They are created because 'virtual photons' are made all the time (heisenberg uncertanty principle) and cancelled out and give the energy back to the universe on an EXTREEMLY small time scale. The tidal forces neer the edge of the black holes make a posative and negative energy one. The posative energy one escapes ( described as heat) and the negative falls into the black hole. Black holes therefore indirectly give it away energy. Imagine Mr A Mr B and Mr C Mr A gives Mr B £5 and therefore is £5 down. So he gives and IOU to Mr C for Himself to break even. The IOU however is special. It takes away £5 out of Mr Cs pocket without it actually leaving him (i.e. negative energy). So a black hole does not directly evaporate it receives negative energy as a cancelling out of the posative energy photons that its tidal forces create, therefore conserving the energy. I think. Link to comment Share on other sites More sharing options...
Syd Posted March 9, 2005 Share Posted March 9, 2005 Hawking Radiaton doesn't come from the black hole itself, but from the Horyzont Zdarzeń (don't know how it is in english but it's a distance from black hole from were You can still escape, one more step and Your trapped). Hawking theory also says that consequence of radiaton is that black hole disappears after some 1000... years. So existance of hawking rad. can be easyli check(if You've got 10000000 years free time ps. soon i'll finish my theory connected with black holes.I hope that it will be quite interestin for You Link to comment Share on other sites More sharing options...
BlackHole Posted March 9, 2005 Share Posted March 9, 2005 Black holes were deduced from Einstein's theory of gravitation. Basically the geometrical structure of a black hole is calculated from GR. A “singularity” of zero volume and infinite density (probably also infinite temperature) pulls in all matter and energy that comes within an event horizon, defined by the Schwarzschild radius, around it. Here is the solution: [math]ds^2 = - (1 - \frac{2M}{r})dt^2 + (1 - \frac{2M}{r})^{-1} dr^2 + r^2d\Omega^2[/math], [math]c=G=1[/math] and [math]d\Omega^2 = d\theta + \sin^2\theta; d\phi^2[/math]. In 1965, Roger Penrose proved the singularity theorem, which says that a singularity must reside inside every imploding star, and therefore every black hole. At the singularity, though, the laws of physics, including GR, break down. We don't know what happens at and beyond the singularity. The problem is that GR incompletely describes gravity. Inflationary models are becomming the leading candidates now. Like GR, cosmic inflation also predicts that a characteristic pattern of long-wavelength gravitational waves would have been created in the early universe. These waves are literally gravitons - the hypothetical particles that carry the gravitational force - that have been stretched to macroscopic lengths by the cosmic expansion. The detection of these waves would provide a unique signature of inflation. Link to comment Share on other sites More sharing options...
The Rebel Posted March 9, 2005 Share Posted March 9, 2005 Lots of comments about the outside of a black hole and what happens just beyond the event horizon. What happens when the black hole is smaller than the Schwarzchild radius? Is anything transmitted, energy or otherwise between the black hole and its event horizon? What would be its surface temperature then? Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted March 9, 2005 Share Posted March 9, 2005 What happens when the black hole is smaller than the Schwarzchild radius? ? Aren't they identical? Link to comment Share on other sites More sharing options...
ed84c Posted March 9, 2005 Share Posted March 9, 2005 Yes they are. Shwarzchild radius= Event Horizon= Edge of 'hole' Link to comment Share on other sites More sharing options...
The Rebel Posted March 9, 2005 Share Posted March 9, 2005 Ok then, take an object of mass 7e10kg that has a radius of 10km. There is a good 400m between the object and the schwarzchild radius. That's the area of interest. Link to comment Share on other sites More sharing options...
swansont Posted March 9, 2005 Share Posted March 9, 2005 Ok then' date=' take an object of mass 7e10kg that has a radius of 10km. There is a good 400m between the object and the schwarzchild radius. That's the area of interest.[/quote'] How can you say that the object has that size? Link to comment Share on other sites More sharing options...
Martin Posted March 9, 2005 Share Posted March 9, 2005 Hawking Radiaton doesn't come from the black hole itself, but from the Horyzont Zdarze? (don't know how it is in english but it's a distance from black hole from were You can still escape, one more step and Your trapped). Horyzont Zdarzen must mean "Event Horizon" for a nonrotating black hole this is spherical and the radius is called the "Schwarzschild radius" You are right, the radiation comes from just slightly outside the event horizon. So it can escape. so the hole can glow like a hot piece of metal, if it is hot enough, and the radiation is something you could feel (but only if the hole is hot enough) Hawking theory also says that consequence of radiaton is that black hole disappears after some 1000... years. So existance of hawking rad. can be easyli check(if You've got 10000000 years free time that is right and it is easy to calculate if you use Quantum Gravity units (c = hbar = 8piG = 1) in that case the evaporation time is proportional to the cube of the mass. the time a hole of mass M will take to evaporate (radiate away all its mass-energy) is equal to (80/pi) M3 so if the black hole has mass E33 (approx the mass of the earth) then the evaporation time is (80/pi) E99. this is about 25E99 to put it in familiar terms, E50 quantum gravity time units is an earth year. so the evaporation time is about 25E49 years. Obviously black holes which are much less massive can evaporate rapidly. If the mass is roughly a million pounds, which means E14 natural units, then the evaporation time is 25E42, which is about 7 seconds if you're lucky, just time to get out of the building Link to comment Share on other sites More sharing options...
AtomicMX Posted March 10, 2005 Share Posted March 10, 2005 any x<0 k Link to comment Share on other sites More sharing options...
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