Talk:Hawking radiation/Archive 2

Archive 1

Archive 2

Vapourisation of black holes, multi universes, worm holes and tachyons?

--Ganeshsashank (talk) 16:15, 2 October 2009 (UTC)Hello every one! My name is GANESH SASHANK. I'm just a student who has just completed 12th class. I'm of age 17 and if something wrong is there in my observations, kindly excuse me!. Acc. to hawking radiation, some micro sized black holes emitt more matter than they absorb. . In that process, they finally emitt all the matter in them and finally vanish. But a black hole is considered to be a region where there is infinite density and the matter absorbed is compressed to zero volume. But zero volume and infinite density, though supports the big bang singularity, then at that singularity, every mass should be converted to energy in accordance with EINSTEIN'S mass energy equivalence. Then, I say that every black hole must be able to emit mass by converting this energy to mass. And so, even a micro sized black hole should not vapourise. Now, based on this observation, one can easily support the concept of multiple universes and worm holes.. There are many universes, every universe present in a sea of infinite energy! Each universe rotates with its own frequency in the infinite energy medium.And any how, if a black hole emits matter...it is said that even light cannot escape the event horizon of a black hole... then how can mass emitted by a black hole come out of its event horizon? If hawking radiation is true, then the mass coming out of the event horizon should be coming with a velocity more tha that of c. Hence, there can be tachyons possibly. But when this happens, the einsteins equation of virtual mass converts to imaginary plane instead of real plane.. which suggests the presence of infinite energy beyond the black hole! Now if a question is raised as follows 'How can mass come out with speeds greater than speed of light out of the event horizon?', I would suggest a hypothesis--Energy flows from high energy region to low energy region. In this process, energy escapes out of event horizon as tachyonic pulses. After coming out of the event horizon, the tachyonic pulses slowly break down owing to the uniform distribution of energy. Finally some of it may convert to mass, some may remain as light or energy forms. Even if hawking's theory doesnt hold good, then just the tachyonic theory is ruled out. But still the concept of multiple universes can be accepted. This is because the presence of multiple universes can be explained on the basis of Doppler effect. Each universes rotates with its own characteristic frequency. And what can be the possible proof to this characteristic frequency thing?!! Guess what, I got the idea from comics--- Human brain is a radio which may receive many wavelengths. When the waves of the other universe enter our universe, one may consider them simply as empty cosmic waves. Another thing, here we must take into consideration, De Broglie's matter wave theory. Each incident and each mass is associated with its characteristic waves. So one may accidentally receive the waves of other universe, accept it as an idea and thus bring them out as comics!! like Batman, superman. And if hawking radiation is true, then worm hole concept can also be accepted. Black holes emit mass and energy of, probably, the other universe. This can be a truth because when black holes absorb matter, a black hole in the other universe may also absorb its matter. In this process, such masses accumulate in the infinite energy medium. And in due course of time, these masses may get exchanged and the masses of the other universe may come into ours along with the event wave energy of the other universe. Thus, a black hole might probably be emiting mass just by mutual exchange process!!

No, it just means Wikipedia isn't here to verify orginal research. ^Khukri 16:13, 11 October 2009 (UTC)

Ok, i understand now...i'm sorry because I thought that discussions were meant to share things and views!! —Preceding unsigned comment added by Ganeshsashank (talk • contribs) 06:31, 30 November 2009 (UTC)

Probable proof for multiple universes?

--Ganeshsashank (talk) 11:51, 11 October 2009 (UTC)Also, recently, I've studied the book as a part of my reading. This text book was prescribed to us in our university. As I mentioned, I'm a B.E first year student pursuing mechanical engineering. In that book, when I was reading Thermodynamics, I saw the lines defining an isolated system--No perfect isolated system exists in practice. However, our universe can be taken as a perfectly isolated system because it has no surroundings to exchange energy with. But in my intemediate text book, I saw a point in the same chapter that-entropy of the universe is increasing constantly. My point is- If the universe is isolated system and cannot exchange energy since it hasn't got any surroundings, then its energy-either in the form of entropy or enthalpy or any form should remain constant! But why should entropy of the universe increase??!!. In my view, this may be the scientific explanation for the existence of multiple universes!!!

Please post your comments if my thinking is wrong. I'm just an amateur among amateurs regarding these topics. So any one is welcome to find flaws in my thoughts

Rabinowitz formula

I don't believe the sentence about the Rabinowitz formula belongs in this article. According to the log, it was added by an anonymous user at 69.107.113.130, whose only other contribution is to the page on Rabinowitz himself. Incidentally, that IP is located in the SF Bay Area, which also happens to be where Rabinowitz resides.

The formula is at best ambiguous and you will find no references to it in any of the standard works on Hawking radiation. I would strongly suggest that it be removed, especially since this article will be distributed to schools.

Jbrnd (talk) 21:43, 13 March 2010 (UTC)

No answer in a week. Deleting it myself. Jbrnd (talk) 01:26, 22 March 2010 (UTC)

Emitting particles

http://en.wikipedia.org/wiki/Future_of_an_expanding_universe#Black_Hole_Era

"During the last stages of its evaporation, a black hole will emit not only massless particles but also heavier particles such as electrons, positrons, protons and antiprotons.^{, pp. 148–150.} "

Why isn't this mentioned? At what point does it occur? —Preceding unsigned comment added by 71.167.69.120 (talk) 17:38, 20 May 2010 (UTC)

Graph of Evaporation time versus size or mass

I would like to see a graph or illustration that coupled the evaporation time with the size or mass of a black hole. It would be interesting to see, for example, how long time it would take for a black hole with the mass of en electron, or 1 kg, or our sun, or the universe to evaporate. Perhaps not of much use to you people that can handle complicated formulas, but for the rest of us, it would be nice. --HelgeStenstrom (talk) 20:30, 15 August 2010 (UTC)

Future thresholds?

Does the Cosmic Background Radiation not become asymptotically fainter as the Universe continues to age? If so, it would appear that while black holes more massive than Earth's Moon grow at present, that threshold will rise as the Cosmic Background provides less (yet such that it would never reach 0) energy for mass increase to black holes in any given area of space. Would anyone like to assist in looking for literature on this?

If so, this would mean that all black holes will eventually evaporate and terminate, even if they do so far enough in the future that the Cosmic Background Radiation will be appreciably less intense. The Mysterious El Willstro (talk) 06:37, 18 September 2010 (UTC)

A different point of view about this Zeldovich-Starobinsky-Bekenstein-Hawking stuff

Hi, my name is Mart Vabar and last week I got an idea about how to simplify this somewhat heavy construction:

Physical insight on the process may be gained by imagining that particle-antiparticle radiation is emitted from just beyond the event horizon. This radiation does not come directly from the black hole itself, but rather is a result of virtual particles being "boosted" by the black hole's gravitation into becoming real particles.

A slightly more precise, but still much simplified, view of the process is that vacuum fluctuations cause a particle-antiparticle pair to appear close to the event horizon of a black hole. One of the pair falls into the black hole whilst the other escapes. In order to preserve total energy, the particle that fell into the black hole must have had a negative energy (with respect to an observer far away from the black hole).

Here we have 3 different particles: 2 of these appear near the event horizon; the 3rd one is a part of the black hole and it disappears, when it meets the particle with negative energy.

Could it make sense to see all these 3 as one single paricle:

1)

once upon a time it fell into the black hole,

2)

but for some good reason it cannot fit a "possible place" there anymore.

3)

So, from a viewpoint of an outsider, it seemingly moves "back" in time and when doing so, it looks like an anti-particle (to have some fun with it: like outer planets sometimes move "back")...

4)

...and when seemingly "going back" in time, the particle also leaves the black hole,

5)

where it looks like a normal radiation with positive energy.

Comments anybody? —Preceding unsigned comment added by 194.150.65.25 (talk) 20:05, 20 September 2010 (UTC)

Hi, this falls squarely into wikipedia's guidelines on original research, these talk pages are for discussing verifiable and referenced improvements to the articles. Though the wikipedia help desk might be able to help at WP:QUESTIONS. Regards ^Khukri 21:36, 20 September 2010 (UTC)

Hawking radiation created in the lab

I'm a geek, but not a physics geek, so I don't know about this latest finding by Franco Belgiorno at the University of Milan who says he created and observed Hawking radiation in the lab. Should it be included in the article or is still too speculative? — Frεcklεfσσt | Talk 10:59, 27 September 2010 (UTC)

Dudes, I saw this information was added to the article, but why did you leave out a reference? — Frεcklεfσσt | Talk 22:42, 27 September 2010 (UTC)

---

Guys this article needs pics. —Preceding unsigned comment added by 78.56.217.206 (talk) 22:09, 27 October 2010 (UTC)

Effect of capture impact parameter on Hawking radiation

A black hole of temperature T should be in thermal equilibrium with a radiation bath at T. But the effective area of a black hole for absorption of radiation from a distance is not 4*pi*R^2 but larger than that by a factor 27/4 (MTW), due to the capture of deflected light rays that would miss the hole were they not deflected. For equilibrium the effective emission area must then also be larger by the same factor than the usually assumed (as in this article) 4*pi*R^2. R is the Schwartzschild radius.

No? Shell Ridge (talk) 00:22, 3 December 2010 (UTC) Gravitation, by Misner, Thorne and Wheeler (MTW), page 677.

Unprecedented evaporation observation

Need someone to clarify this point for laymen: if human were able to observe exotic events in the universe like hypernova, quasars, etc. then how come black hole evaporation is yet to be observed anywhere? Is it simply because they are black? If that's the case, what is the difference between observing an "alive" black hole and observing a black hole at its last phase?Mastertek (talk) 08:59, 25 October 2011 (UTC)

Strength of Evaporation

The article mentions the strength of a black hole evaporation lasting one second. Where can I find calculations to determine this? What is the strength of other sized black holes evaporating? Like say a black hole the mass of a car compared to one the mass of earth? — Preceding unsigned comment added by Nostawk (talk • contribs) 00:22, 29 June 2011 (UTC)

Why would only particles with positive energy escape?

Here is something I've often wondered, and don't see addressed in the article (although I find it rather dense and hard to follow, so maybe I missed it), so perhaps somebody could add this to the main article: if I understand it correctly, Hawking radiation means (according to one explanation) that matter/anti-matter particle pairs spontaneously come into existence right at the event horizon, such that the anti-matter particle falls into the black hole, reducing its energy, and the matter particle escapes into the universe, so that it looks like it has been radiated out by the black hole. Right?

But why would it always be the anti-matter particle which falls into the black hole? Why would it not be completely random which particle falls into the black hole, so that the net effect of many such events would be zero? I'm sure my question is due to a misunderstanding of the theory, but perhaps it would be a good thing to address in the article for others with the same question as me.83.87.133.123 (talk) 13:20, 19 July 2011 (UTC)

When particle-antiparticle pairs are created from the vacuum, they have the same mass (beacuse a particle and its antiparticle always do), and hence the same energy (by E=mc^2). Normally, they recombine within a time limit governed by the uncertainty principle, releasing the energy, and nature continues as if nothing happened. However, if such an event happens near the event horizon of a black hole, then one particle falls beyond the horizon in a finite proper time which is less than the limit imposed by the uncertainty principle, and the other escapes, so the two never recombine. Now, a distant observer sees a particle of finte mass and energy being ejected from the black hole; he can't see the other particle because it's beyond the event horizon. But by the conservation of mass-energy he reasons that, all other things being equal, the black hole has lost the same amount of mass-energy, and this is consistent because the observer can't actually look inside the event horizon and measure the masses of all the individual particles (an observer inside the horizon, however, would measure an increase in the black hole's mass). — Preceding unsigned comment added by 131.111.185.74 (talk) 14:16, 30 November 2011 (UTC)

Going back to the first question above and the answer above (with which I have difficulty). (1) If the particle & anti-particle both have the same mass, how does the black hole evaporate ie reduce its mass? I thought that the mass of the particle falling into the black hole had to have negative mass (or energy from E=mc2) but read elsewhere that antimatter has positive mass. (2) I thought that a particle & anti-particle pair had opposite charges. Does that mean that the same proportion of +ve and -ve charged particles are emitted by Hawking Radiation? — Preceding unsigned comment added by 78.143.210.190 (talk) 12:32, 12 April 2012 (UTC)

I think I'm a bit out of my depth here, and to understand it all thoroughly, you need to do calculations using quantum field theory in curved spacetime, which probably requires you to spend your whole life doing physics. But the main point is that mass is conserved, ALWAYS, and that means if a particle with mass comes out of a black hole then the black hole MUST lose mass. And that's the ONLY way of working out what's happened to the black hole because you can't look inside the black hole itself.

Here's an analogy. There are two rooms, A and B, with a doorway between them. The two rooms are the black hole and the rest of the universe, and the door is the event horizon. These rooms have a certain number of little balls that are like particles. Every so often a pair of balls appears but then disappears very quickly, so most of the time there are, say, 100 balls in each room. Balls can pass through the door from A to B but not B to A.

Now let's say Alice is room A and Bob in room B. A pair of balls appears close to the doorway; so close in fact that one appears on each side. The ball on Bob's side is sucked deep into the room before it can can cancel with the one on Alice's side.

Alice counts the balls in her room, and finds there are 101. She knows that there are always 200 balls overall, so she reasons there are 99 in Bob's room. Bob does the same, and reckons that HE has 101 balls to Alice's 99.

Alice then goes to the doorway. (And this is the weird bit I can't explain in laymen's terms, you'll just have to accept it). It turns out that, although she saw the ball come through the doorway, she can never do so herself. Time slows down for her as she approaches, and when she gets to the door, time stops completely. This is really what happens for black holes, you can see something else enter a black hole but you can never enter yourself! So Alice can't ask Bob how many balls are in his room.

So who is right, Alice or Bob? Well, since we are in room A (outside the black hole), it makes sense for us to say Alice is right, that her room has gained a ball and Bob's has lost one. We can't ever get into Bob's room and find out whether we are right! However what we can't say (as you may think is obvious) is that there are 101 balls in each room. It is a rule that the number of balls is ALWAYS 200, except for the occasional production of a pair of balls that doesn't last very long (Actually, the rule would say that the number of balls is approximately 200, and how close you get to this depends on how long you spend counting. See uncertainty principle for more details).

As for charge, yes you are right, antiparticles have opposite charge. So if a black hole emits, say, an electron (-ve), then a positron (+ve) falls in and charge is conserved. In this case, actually, there is no paradox, since the total charge of the pair is zero, unlike their total mass.

When virtual particles form they are not particle and antiparticle, they are positive and negative energy (or mass) pairs, if they were particle and antiparticle when they combine they would add energy to the universe and violate the conservation of energy. The reason the negative and not the positive energy or mass particle enters the black hole is because gravity is caused by the attraction of negative to positive energy. Seeing as a black hole has a large amount of positive matter only the negative particles fall in. — Preceding unsigned comment added by 174.131.116.139 (talk) 17:32, 9 June 2012 (UTC)

Actually, my understanding is that virtual particles are indeed created as particle/antiparticle pairs for this case (they're a bit like excitons that way). All quantum numbers are conserved (so total charge, spin, etc. is zero). The general idea is that energy is conserved too, but is violated in the short-term (with a handwavy way of describing it being that a "debt" is associated with the total energy of the pair that must be repaid within the time described by the uncertainty principle). A more accurate description is that the system as a whole (black hole and its environment) is allowed to evolve from any valid state to any other valid state, as long as any rule-breaking that happens during the transitions doesn't happen for long enough to be measurable. Virtual particles are a way of doing the bookkeeping for the analysis of that evolution between states.

Disclaimer: I am not a physicist. --Christopher Thomas (talk) 21:29, 9 June 2012 (UTC)

So where does the negative energy come from that falls in the black hole? It’s not good enough to just say that it “is” negative, because the particle that escapes is positive. Again if both particles had positive mass, like a typical particle anti particle pair the black hole would gain mass and you would have a violation of the conservation of energy. If there is no real negative energy there is no black hole radiation. Where else can it come from except the virtual particle pair? Were else can it come from unless the virtual particle pair has both a negative and positive energy component? — Preceding unsigned comment added by 174.131.116.139 (talk) 23:11, 9 June 2012 (UTC)

Actually, it is good enough to say that. Per my description above, "virtual particles" aren't things that exist per se - they're a convenient way for setting up the math describing how a system transitions from one state (black hole and nothing else) to another (black hole with less mass, and a particle escaping). They show up in a lot of other places, too; Feynman diagrams make use of them when setting up the math for calculating the forces between charged particles (described as mediated by an exchange of virtual particles which are allowed to travel faster than light for short distances).

For more information about what virtual particles are and why they show up with certain types of analysis (ones based on perturbation theory, mostly), see the full description at virtual particle. --Christopher Thomas (talk) 00:16, 10 June 2012 (UTC)

I realize there are other circumstances where it is not necessary for virtual particle pairs to be negative and positive energy but for black hole radiation I believe it must be true. — Preceding unsigned comment added by 174.131.116.139 (talk) 19:41, 10 June 2012 (UTC)

Now I'm confused too. The problem I have with the virtual particle explanation is that these particles persist indefintiely after being created. How can they go from being virtual to real? — Preceding unsigned comment added by 86.26.13.2 (talk) 10:04, 1 October 2012 (UTC)

The handwavy, not-quite-correct explanation is that a virtual particle has an "energy debt" associated with it (which is why it vanishes after a short time), and that if that "energy debt" is paid, it becomes real. This can happen anywhere that you have a strong potential energy gradient - it happens near black holes due to the steep gravity well, but it also happens in strong electric fields (which is one way of thinking about how pair production works).

The actual explanation has to do with the fact that this is a mathematical tool, not actual particles. When you construct a Feynman diagram showing a possible set of particle interactions, only the starting and ending state have to be situations that are actually possible. Intermediate states - lines representing particles in the middle of the diagram - can do unrealistic things like moving faster than light or other tricks, subject to certain constraints. Particles that break the rules in this manner are called "virtual particles". The probability distribution of possible (realistic) outcomes from a given (realistic) input is found by looking at contributions from all possible Feynman diagrams you could draw connecting a given input to an output. You _can_ interpret this in terms of "diagram X happens, where these virtual particles come into being and vanish again", but it's not necessarily meaningful to do so. In the case of Hawking radiation, virtual particles are one description of how you get from state A (black hole with no emitted particles) to state B (slightly lighter black hole and one or more emitted particles). Other descriptions exist; virtual particles are just a description that makes it easier to work the math for it. --Christopher Thomas (talk) 20:23, 1 October 2012 (UTC)

Hawking Temperature

Recently, an edit doubled the boxed temperature formula. It left no citation and did not change the other appearances of the same expression. There are two reasons that this could have happened.

1. Confusion between the Mass of a black hole and the Radius of the black hole. In geometrical units $R=2M$ or in SI units $R={\frac {2GM}{c^{2}}}$
2. Confusion from some published sources which purport to calculate a black hole temperature distinct from the Hawking temperature. ( Mitra, P. (2007). "Hawking temperature from tunnelling formalism". Physics Letters B. 648 (2–3): 240–242. arXiv:hep-th/0611265. Bibcode:2007PhLB..648..240M. doi:10.1016/j.physletb.2007.03.002. ) corrects one such example and gives the Hawking temperature in geometrical units as $T={\frac {\hbar }{4\pi R}}={\frac {\hbar }{8\pi M}}$

In SI units, we insert factors of Boltzman's constant, Newton's constant (G) and the speed of light to recover the standard Hawking Temperature $T_{H}={\frac {\hbar c}{4\pi k_{B}R}}={\frac {\hbar c^{3}}{8\pi k_{B}GM}}$ which is what I restored - 209.131.48.66 (talk) 18:07, 2 August 2011 (UTC)

Black Holes in Thermal Equilibrium

I wish to edit the last paragraph in the overview section regarding black holes of various masses being in thermal equilibrium with their environment. However, I've never edited before and don't know all the etiquette.

A black hole, since its temperature is inversely proportion to its mass, has a negative specific heat. If it accepts energy/mass, it becomes colder. Consequently, it cannot come into thermal equilibrium with an infinite (or at least very large) CMB photon bath at 2.76 K. Confer any basic text or in particular the excellent review article arXiv:gr-qc/9707012v1 . The maximum volume with which a black hole could reach thermal equilibrium is V = (2^20)(pi^4)(E^5)/(5^5 * sigma) where E is the total energy of the reservoir and black hole and sigma is the Steffan-Boltzman constant. Dirac1234 (talk) 00:06, 30 November 2011 (UTC)

See WP:BOLD for instructions on editing articles. Be aware that arxiv isn't a reliable source, I suggest you source to the final and published review article if you add it to the article (which you may already have done I haven't checked). IRWolfie- (talk) 16:21, 8 January 2012 (UTC)

False claims in article

The calculational section of this article is not correct.

1. The article claims black holes are perfect blackbodies, with emissivity = 1. This is false. There are "greybody" factors that make epsilon < 1, and reduce the power output of black holes.

2. The article assumes there is only one massless particle, the photon. In fact, black holes will also emit gravitons, and, if (as is conceivable) there is a massless neutrino, it will emit that too. When then black hole gets small enough (and it's temperature therefore large enough) it will start emitting electrons, protons, etc. too.

The first mistake makes the lifetime longer than is quoted, the second effect makes the lifetime shorter. Every equation after the one that says "epsilon =1" is wrong. — Preceding unsigned comment added by RoadMap (talk • contribs) 19:33, 13 March 2012 (UTC)

Evaporation time, verifying calculations

In Black_hole_evaporation#Black_hole_evaporation, the following 2 values are given for $t_{\operatorname {ev} }$:

The evaporation time of a black hole is proportional to the cube of its mass:

$t_{\operatorname {ev} }={\frac {c^{2}M_{0}^{3}}{3K_{\operatorname {ev} }}}=\left({\frac {c^{2}M_{0}^{3}}{3}}\right)\left({\frac {15360\pi G^{2}}{\hbar c^{6}}}\right)={\frac {5120\pi G^{2}M_{0}^{3}}{\hbar c^{4}}}=8.407\times 10^{-17}\left[{\frac {M_{0}}{\mathrm {kg} }}\right]^{3}\mathrm {s} \;$

The time that the black hole takes to dissipate is:

$t_{\operatorname {ev} }={\frac {5120\pi G^{2}M_{0}^{3}}{\hbar c^{4}}}\;$

$\approx 1.33*10^{-17}M_{0}^{3}{\frac {s}{kg^{3}}}\;$ Where $M_{0}$ is the mass of the black hole.

Are these equations talking about the same quantity? If so, shouldn't their values be the same? Instead, the first line is it is 8.407x10^-17 and in another it is 1.33x10^-17. They seem to differ by a factor of 2*pi. I suspect that in the first line the listed Reduced_Planck_constant was used, and the second line the Planck_constant was used instead by mistake. But I don't understand the physics well enough to be sure. If you do understand, would you please recheck these calculations? Mynameisnoted (talk) 05:03, 18 October 2011 (UTC)

The second one is wrong (more likely mistake is that they used the value of "h" rather than "hbar". Though both are incorrect (see below). RoadMap (talk) 20:21, 13 March 2012 (UTC)

Decoupling of physics and mathematics

This problem arises over a whole series of articles stretching from Hawking radiation to Rindler coordinates and possibly beyond. I encountered it when I tried to use Wikipedia to grok (understand) Hawking radiation.

The problem is that: (1) all articles reference a new physical process or mathematical transformation that is not natively explained by the article. (2) following the hyperlink to the article that was supposed to explain the unexplained produces only a partial explanation and repeats the problem (1) for the rest of the explanation. (3) the number of questions and unknowns increases during the following of this chain instead of decreasing.

To illustrate: I want to know what is the process by which Hawking radiation comes about, but the explanatory paragraph references the Unruh effect without much explanation (really, without any explanation). Then I go to the Unruh effect article and it references Minkowski spacetime and Rindler coordinates. By the time I get to the Rindler coordinates article, I am not reading about Hawking radiation anymore, I am studying hard-core mathematical topology.

The core problem then is that the amount of (human) energy required to parse the Hawking radiation is at least an order of magnitude larger than what the article size and apparent complexity would lead you to believe. Because you first have to parse topology itself and then work your way up to what you actually need.

What should be done is that that energy expenditure for understanding Hawking radiation should be decreased, and namely this should be done by not requiring the reader to understand the whole scientific field just so he/she can understand a small part of the field. The concrete solution appears to be decoupling of various topics. The explanations should be converging and if a hapless reader needs to follow a hyperlink to a different article, he/she should do so with a clear picture in his/her head about the questions that need to be answered to complete the understanding.

I am willing to do the legwork for this, however I can not do this in a vacuum. I am thinking about, first, explaining all the equations right there where they appear ("explain" means that all variables and constants are identified) and, second, by adding "in-a-nutshell" explanations of lower-level phenomena (for example: Unruh effect) inline with the main explanation.

Since I am not a physicist, someone will have to make sure I did not misinterpret stuff. Triklod (talk) 04:07, 11 August 2013 (UTC)

Deletion of BICEP2 experiment info?

The following edit was deleted as an "inaccuracy." I quoted a credible source, but don't know enough about the topic to say if it should stay or go. Anyway, this is what it was...

The BICEP2 experiment detected the early universe's horizon's Hawking radiation, in the form of gravitational waves.

This is the relevent reference from the S&T article, which I did not include in full due to copyright concerns:

This is the first detection of Hawking radiation. Hawking radiation is usually associated with the slow evaporation of black holes, as photons emitted from the event horizon. But the observable universe also has a horizon. Hawking radiation should be coming from this horizon, and also from every horizon in the universe — in other words, from every point in the universe, says cosmologist Max Tegmark (MIT). Today the cosmic horizons are huge and their Hawking radiation is utterly insignificant. But in the universe’s first fraction of a second, the horizons were tiny and sharply curved. The gravitational waves announced today are these horizons’ Hawking radiation.

— Eoghanacht ^talk 16:48, 31 March 2014 (UTC)

far from the black hole

What exactly is meant by "far from the black hole" in this section of the article?

"... nothing, not even electromagnetic radiation, can escape from the black hole. It is yet unknown how gravity can be incorporated into quantum mechanics. Nevertheless, far from the black hole the gravitational effects can be weak enough for calculations to be reliably performed in the framework of quantum field theory in curved spacetime."

Can Hawking radiation be derived using calculations that only "look at" locations far outside the event horizon, and so can be calculated even for objects that are not yet black holes?

If so, I suggest that section be rephrased to something like

"Nevertheless, far outside a black hole's event horizon, the gravitational effects can be weak enough ..."

Or does Hawking radiation, like spaghettification inside a supermassive black hole, require assumptions about stuff that happens at the event horizon? I.e., assumptions that may be perfectly reasonable and perhaps can be indirectly confirmed, but no one can directly confirm those assumptions and return to tell about it? Then I suggest that section be rephrased to something like

"Nevertheless, far from a black hole's central singularity, the gravitational effects at the event horizon can be weak enough ..."

--DavidCary (talk) 16:23, 21 April 2015 (UTC)

Common errors in calculating radiated Hawking power.

Perhaps every discussion of the power radiated by a black hole that I have seen makes the following error: The total radiating area of the (Schwartzschild) black hole is taken to be the area of the event horizon, 4 pi Rs^2. But this is not the effective radiating area as seen from a distance. As stated on p 679 of Misner, Thorne and Wheeler, the capture cross section of a black hole is 27 pi M^2 (gravitational units), larger than pi Rs^2 because of the gravitational bending in of light rays. This makes the effective absorbing, and hence radiating, area larger than the area of the horizon by a factor of 6.75. This is pretty piddling compared to the tiny and huge numbers coming out of the power and lifetime calculations, but deserves to be considered if factors like pi are not also dropped.

There is also a physical optics effect that I have never seen mentioned, and that in fact will probably heavily modify what I have said above in a way that I am not prepared to estimate. This is the fact that the black hole source of the Hawking radiation is very small compared to the dominant wavelength of thermal radiation at the Hawking temperature. When the radiation must squeeze out of such a small hole it is probably not correct to use the geometric area in the way it is usually done, even as corrected above. And finally of course, as HAS been pointed out by others, a black hole is black for all forms of radiation, not just electromagnetic.

Sprite82 (talk) 22:30, 11 April 2013 (UTC)

The 1977 Page analysis, which I just added, appears to get take into account all those elements you mentioned. I couldn't find a later analysis factoring in that neutrinos have mass; maybe Misner et al has it but I don't have access to that text. Rolf H Nelson (talk) 03:48, 22 December 2015 (UTC)

External links modified

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Hawking Radiation not visible to an observer falling into a black hole

I'm not an expert on this at all myself. But in other accounts of this, they say that the particle count depends on your rest frame. If you are in an accelerating frame falling into a black hole, then you won't see any particles at the event horizon. But if you could be suspended just above the event horizon you'd see the Hawking radiation because you count particles differently.

Calling it "an accelerating frame" is confusing, but yes, you're correct that your particle detector clicks if you're suspended above the black hole while your free-falling neighbor's detector does not. If you have a specific source we can add it in. Rolf H Nelson (talk) 02:27, 27 January 2016 (UTC)

Also - there's a question, what type of radiation is produced? The particle / anti-particle explanation would seem to suggest all the radiation would be protons, anti protons, electrons, anti-electrons and such like - and probably equal numbers of each as there is no particular reason for one or other of the pair to fall into the black hole. But Hawking radiation is said to be mainly photons, and when a black hole evaporates it's almost entirely gamma rays at that point. So that suggests something wrong with this simple picture.

Photons are their own anti-particles, so there's no contradiction. If we believe this is a common source of confusion, we should definitely clarify it in the article. Rolf H Nelson (talk) 02:27, 27 January 2016 (UTC)

So, I think it would be good if someone expert on this was to write something about this. Either a clear explanation if it is well understood but not widely known, or if nobody knows, to explain what is known and what is uncertain about it. It's possible that the technical section of this paper makes it clear for theorists who are expert already as that is too technical for me to read, but it's not clear from the intro anyway.

Here are a couple of academic web pages on these points:

• Hawking Radiation by John Baez
• Hawking Radiation poses new questions on the official string theory site.

Thanks Robert Walker (talk) 17:35, 26 January 2016 (UTC)

FWIW Mr. Walker, I agree with everything you're saying. I came here to understand exactly that question: what type of radiation is Hawking Radiation? If that info is implied in the article, it's certainly not clear to me as a layman. Thanks! 148.87.19.210 (talk) 20:40, 14 June 2016 (UTC)

What would people think of mostly removing the pop-science "virtual pair production" content? According to John Baez [1], it's not really how the physicists who work with it seem to frame it. I'm not a physicists, but to me it doesn't seem to me very insightful for analyzing the hypothetical phenomenons of Hawking and Unruh radiation at a deeper level, and it probably confuses people about how virtual particles normally behave. The main scholarly articles don't seem to lean heavily on the virtual pair production analogy; the closest I found was Kiefer 1998, which says only as an aside "It turns out that the vacuum expectation value of the energy-momentum tensor of the quantum field is negative near the horizon, corresponding to a flux of negative energy into the hole (this is the basis for the pictorial interpretation of the Hawking effect, where one partner of a pair of virtual particles can fall into the hole, thus enabling the other partner to become real and escape to infinity, where it can be observed as Hawking radiation)". Rolf H Nelson (talk) 00:27, 19 June 2016 (UTC)

Why Hawking?

Why isn't it named Starobinsky-Zel'dovich? Or: did the argument Hawking provided transcend in any important way what the Soviets said? (The cited website is down now.) — Preceding unsigned comment added by 189.6.253.80 (talk) 01:34, 24 May 2014 (UTC)

And what's about Gribov?

He told Zel'dovich about black holes' emission. — Preceding unsigned comment added by 46.10.176.6 (talk) 10:14, 14 March 2018 (UTC)

We just call it what our sources call it; feel free to provide sources for additional things it's called and based on their quality we can weigh adding them to the article. Rolf H Nelson (talk) 06:03, 16 March 2018 (UTC)

Hawking Radiation really is Dis-proven

Lets stick to NPOV policy and not cover up what another scientist showed.2601:447:4101:41F9:C0C1:6F90:38C9:B404 (talk) 01:13, 6 July 2018 (UTC)

Are you talking about Susskind? He agrees with Hawking that black holes radiate. Waleswatcher (talk) 11:44, 6 July 2018 (UTC)

Evaporation Times are Inconsistent

We give three very different answers to the question of how long a solar-mass black hole would take to decay: 6.617E+74 years (formula), 6.81E+73 years (Page), and 2.098E+67 years (unknown). They can't all three be right.

The article derives a formula for computing the time required for a black hole of a particular mass to evaporate. I've found the same formula derived in the slides for college classes on relativity, so I'm pretty sure it's right. However, the article says that for a 1-solar-mass black hole, "we get an evaporation time of 2.098 × 10^67 years." This is not correct, but then it immediately gives 6.617E+74, which is correct. (I actually get 6.6131.)

Then it says "But for a black hole of 10^11 kg, the evaporation time is 2.667 billion years," even though using the formula, you'd get 8.4109E+16 years.

Since the whole section is titled "A crude analytic estimate," it's possible that the whole mathematical workup is wrong, except that, as I said, it's easy to find online just searching for "black hole evaporation times" and I don't think professors are just mindlessly copying from Wikipedia.

The other numbers seem to be coming from texts published in the 1980s, but, even so, it's hard to believe they'd be off by such a huge amount. Conversely, if the "crude estimate" is this crude, what good is it?

There's also a section called the "Page Numerical Analysis," but this gives numbers that differ by almost exactly a factor of 10 from the formula derived in the article.

Someone who understands this better than I do should fix it. Or, if somehow I've just utterly misunderstood it, make it a little clearer. Greg (talk) 20:11, 27 June 2019 (UTC)

Radiation power

Any sources on radiation power? I did some crude estimate based on other values, like 1 solar mass black hole having temperature of 62 nK, and about 3000 meters radius, thus 1.13e8 m^2 surface area. Plug that into Stefan-Boltzman equation, and I got less than 94x10^{-30} W, I just got lost at the SI prefixes. I was expecting it to be very small, but would be nice for some confirmation. Also how small would the BH need to be to lets say produce 1W or 1TW of power? 81.6.34.246 (talk) 22:19, 11 December 2019 (UTC)

The Xaonon Hawking Radiation Calculator was a popular go-to, but it seems to be down these past few months. If you use Firefox with you-know-what, please stop by my user page for a link to something along the same lines (my apologies for the not-so-great hosting service). DWIII (talk) 09:11, 22 December 2019 (UTC)

BTW, here's a slew of online math/physics calculators provided by PhysicsForums, which includes another Hawking radiation application. DWIII (talk) 11:21, 22 December 2019 (UTC)

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Quest for help

What is the reference of emission process in blackhole particularly. ????? RASHMI RANJAN ROUTARAY (talk) 15:43, 25 February 2020 (UTC)

Unsourced and dense, unreadable, unhelpful content

Hello. I am moving this content here for further study and possible re-use if references can be found to explain where this comes from. Jehochman ^Talk 12:55, 9 December 2019 (UTC)

While I'm not able to comment on the correctness of the text in question, I always found it to be very helpful: It gives readers an idea of the magnitude of the explosion that happens when a black hole evaporates. Especially the weight, initial power output, and TNT equivalent of a 1s black hole are extremely instructive. So, I would vote to put this kind of information back into the article.

If the entire text is too broad for keeping, at least the core info about a 1s black hole should be retained imho: Weight, radius, initial power, and TNT equivalent. 77.8.226.179 (talk) 23:14, 1 January 2020 (UTC)

Most of this text violates WP:NOTTEXTBOOK and does not belong in the article. XOR'easter (talk) 14:48, 17 May 2020 (UTC)

A crude analytic estimate

The power emitted by a black hole in the form of Hawking radiation can easily be estimated for the simplest case of a nonrotating, non-charged Schwarzschild black hole of mass M. Combining the formulas for the Schwarzschild radius of the black hole, the Stefan–Boltzmann law of blackbody radiation, the above formula for the temperature of the radiation, and the formula for the surface area of a sphere (the black hole's event horizon), several equations can be derived:

Stefan–Boltzmann constant:

$\sigma ={\frac {\pi ^{2}k_{\mathrm {B} }^{4}}{60\hbar ^{3}c^{2}}}$

Schwarzschild radius:

$r_{\mathrm {s} }={\frac {2GM}{c^{2}}}$

Acceleration due to gravity at the event horizon:

$g={\frac {GM}{r_{s}^{2}}}={\frac {c^{4}}{4GM}}$

Hawking radiation has a blackbody (Planck) spectrum with a temperature T given by:

$E=k_{\mathrm {B} }T={\frac {\hbar g}{2\pi c}}={\frac {\hbar }{2\pi c}}\left({\frac {c^{4}}{4GM}}\right)={\frac {\hbar c^{3}}{8\pi GM}}$

Hawking radiation temperature:

$T_{\mathrm {H} }={\frac {\hbar c^{3}}{8\pi GMk_{\mathrm {B} }}}$

For a one solar mass black hole, the peak Hawking radiation temperature is:

$T_{\mathrm {H} }={\frac {\hbar c^{3}}{8\pi GM_{\odot }k_{\mathrm {B} }}}=6.170\times 10^{-8}\;{\text{K}}\,.$

The peak wavelength of this radiation is nearly 16 times the Schwarzschild radius of the black hole. Using Wien's displacement constant b = hc/4.9651 k_B = 2.8978×10⁻³ m K:

$\lambda _{\mathrm {max} }={\frac {b}{T_{\mathrm {H} }}}={\frac {8\pi ^{2}}{4.9651}}\,r_{\mathrm {s} }=15.902\,r_{\mathrm {s} }$

Schwarzschild sphere surface area of Schwarzschild radius r_s:

$A_{\mathrm {s} }=4\pi r_{\mathrm {s} }^{2}=4\pi \left({\frac {2GM}{c^{2}}}\right)^{2}={\frac {16\pi G^{2}M^{2}}{c^{4}}}$

Stefan–Boltzmann power law:

$P=A_{\mathrm {s} }j^{\star }=A_{s}\varepsilon \sigma T^{4}$

For simplicity, assume a black hole is a perfect blackbody (ε = 1).

Stefan–Boltzmann–Schwarzschild–Hawking black hole radiation power law derivation:

$P=A_{s}\varepsilon \sigma T_{\mathrm {H} }^{4}=\left({\frac {16\pi G^{2}M^{2}}{c^{4}}}\right)\left({\frac {\pi ^{2}k_{\mathrm {B} }^{4}}{60\hbar ^{3}c^{2}}}\right)\left({\frac {\hbar c^{3}}{8\pi GMk_{\mathrm {B} }}}\right)^{4}={\frac {\hbar c^{6}}{15360\pi G^{2}M^{2}}}$

This yields the Bekenstein–Hawking luminosity of a black hole, under the assumption of pure photon emission (no other particles are emitted) and under the assumption that the horizon is the radiating surface:

$P={\frac {\hbar c^{6}}{15360\pi G^{2}M^{2}}}$

where P is the luminosity, i.e., the radiated power, ħ is the reduced Planck constant, c is the speed of light, G is the gravitational constant and M is the mass of the black hole. It is worth mentioning that the above formula has not yet been derived in the framework of semiclassical gravity.

Substituting the numerical values of the physical constants in the formula for luminosity we obtain P= 3.562×10³² W kg^2/M^2. The power of the Hawking radiation from a solar mass (M_☉) black hole turns out to be minuscule:

$P={\frac {\hbar c^{6}}{15360\pi G^{2}M_{\odot }^{2}}}=9.007\times 10^{-29}\;{\text{W}}\,.$

It is indeed an extremely good approximation to call such an object 'black'. Under the assumption of an otherwise empty universe, so that no matter, cosmic microwave background radiation, or other radiation falls into the black hole, it is possible to calculate how long it would take for the black hole to dissipate:

$K_{\mathrm {ev} }={\frac {\hbar c^{6}}{15360\pi G^{2}}}=3.562\times 10^{32}\;{\text{W}}\;{\text{kg}}^{2}\,.$

Given that the power of the Hawking radiation is the rate of evaporation energy loss of the black hole:

$P=-{\frac {dE}{dt}}={\frac {K_{\mathrm {ev} }}{M^{2}}}\,.$

Since the total energy E of the black hole is related to its mass M by Einstein's mass–energy formula E = Mc²:

$P=-{\frac {dE}{dt}}=-\left({\frac {d}{dt}}\right)Mc^{2}=-c^{2}{\frac {dM}{dt}}\,.$

We can then equate this to our above expression for the power:

$-c^{2}{\frac {dM}{dt}}={\frac {K_{\mathrm {ev} }}{M^{2}}}\,.$

This differential equation is separable, and we can write:

$M^{2}\,dM=-{\frac {K_{\mathrm {ev} }}{c^{2}}}\,dt\,.$

The black hole's mass is now a function M(t) of time t. Integrating over M from M₀ (the initial mass of the black hole) to zero (complete evaporation), and over t from zero to t_ev:

$\int _{M_{0}}^{0}M^{2}\,dM=-{\frac {K_{\mathrm {ev} }}{c^{2}}}\int _{0}^{t_{\mathrm {ev} }}\,dt\,.$

The evaporation time of a black hole is proportional to the cube of its mass:

$t_{\mathrm {ev} }={\frac {c^{2}M_{0}^{3}}{3K_{\mathrm {ev} }}}=\left({\frac {c^{2}M_{0}^{3}}{3}}\right)\left({\frac {15360\pi G^{2}}{\hbar c^{6}}}\right)={\frac {5120\pi G^{2}M_{0}^{3}}{\hbar c^{4}}}=8.4109\times 10^{-17}\;\left[{\frac {M_{0}}{\mathrm {kg} }}\right]^{3}\;\mathrm {s} \,.$

Therefore, the evaporation time of a black hole is also proportional to its volume (and the cube of its Schwarzschild radius):

$r_{s}={\frac {2GM_{0}}{c^{2}}}$

$V_{0}={\frac {4\pi r_{s}^{3}}{3}}={\frac {32\pi G^{3}M_{0}^{3}}{3c^{6}}}$

$t_{ev}={\frac {5120\pi G^{2}M_{0}^{3}}{\hbar c^{4}}}={\frac {480c^{2}V_{0}}{\hbar G}}=6.129\times 10^{63}{\frac {V_{0}}{m^{3}}}s$

The time that the black hole takes to dissipate is:

$t_{\mathrm {ev} }={\frac {5120\pi G^{2}M_{0}^{3}}{\hbar c^{4}}}={\frac {480c^{2}V_{0}}{\hbar G}}$

where M₀ and V₀ are the mass and (Schwarzschild) volume of the black hole.

The lower classical quantum limit for mass for this equation is equivalent to the Planck mass, m_P.

Hawking radiation evaporation time for a Planck mass quantum black hole:

$t_{\mathrm {ev} }={\frac {5120\pi G^{2}m_{\mathrm {P} }^{3}}{\hbar c^{4}}}=5120\pi t_{\mathrm {P} }=5120\pi {\sqrt {\frac {\hbar G}{c^{5}}}}=8.671\times 10^{-40}\;{\text{s}}$

$t_{\mathrm {ev} }=5120\pi {\sqrt {\frac {\hbar G}{c^{5}}}}$

where t_P is the Planck time.

For a black hole of one solar mass (M_☉ = 1.98892×10³⁰ kg), we get an evaporation time of 2.098×10⁶⁷ years—much longer than the current age of the universe at (13.799±0.021)×10⁹ years:

$t_{\mathrm {ev} }={\frac {5120\pi G^{2}M_{\odot }^{3}}{\hbar c^{4}}}=6.617\times 10^{74}\;{\text{s}}\,.$

But for a black hole of 10¹¹ kg, the evaporation time is 2.667 billion years. This is why some astronomers are searching for signs of exploding primordial black holes.

However, since the universe contains the cosmic microwave background radiation, in order for the black hole to dissipate, it must have a temperature greater than that of the present-day blackbody radiation of the universe of 2.7 K = 2.3×10⁻⁴ eV. This implies that M must be less than 0.8% of the mass of the Earth – approximately the mass of the Moon.

Cosmic microwave background radiation universe temperature:

$T_{\mathrm {u} }=2.725\;{\text{K}}$

Hawking total black hole mass:

$M_{\mathrm {H} }\leq {\frac {\hbar c^{3}}{8\pi Gk_{\mathrm {B} }T_{\mathrm {u} }}}\leq 4.503\times 10^{22}\;{\text{kg}}$

$M_{\mathrm {H} }\leq {\frac {\hbar c^{3}}{8\pi Gk_{\mathrm {B} }T_{\mathrm {u} }}}$

${\frac {M_{\mathrm {H} }}{M_{\oplus }}}=7.539\times 10^{-3}=0.754\;\%$

where M_E is the total Earth mass.

In common units,

$P=3.563\,45\times 10^{32}\;\left[{\frac {\mathrm {kg} }{M}}\right]^{2}\;\mathrm {W}$

$t_{\mathrm {ev} }=8.410\,92\times 10^{-17}\;\left[{\frac {M_{0}}{\mathrm {kg} }}\right]^{3}\;\mathrm {s} \quad \approx \ 2.665\,32\times 10^{-24}\;\left[{\frac {M_{0}}{\mathrm {kg} }}\right]^{3}\;\mathrm {yr}$

$M_{0}=2.282\,71\times 10^{5}\;\left[{\frac {t_{\mathrm {ev} }}{\mathrm {s} }}\right]^{\frac {1}{3}}\;\mathrm {kg} \quad \approx \ 7.2\times 10^{7}\;\left[{\frac {t_{\mathrm {ev} }}{\mathrm {yr} }}\right]^{\frac {1}{3}}\;\mathrm {kg}$

So, for instance, a 1-second-life black hole has a mass of 2.28×10⁵ kg, equivalent to an energy of 2.05×10²² J that could be released by 5×10⁶ megatons of TNT. The initial power is 6.84×10²¹ W.

Black hole evaporation has several significant consequences:

• Black hole evaporation produces a more consistent view of black hole thermodynamics by showing how black holes interact thermally with the rest of the universe.
• Unlike most objects, a black hole's temperature increases as it radiates away mass. The rate of temperature increase is exponential, with the most likely endpoint being the dissolution of the black hole in a violent burst of gamma rays. A complete description of this dissolution requires a model of quantum gravity, however, as it occurs when the black hole approaches Planck mass and Planck radius.
• The simplest models of black hole evaporation lead to the black hole information paradox. The information content of a black hole appears to be lost when it dissipates, as under these models the Hawking radiation is random (it has no relation to the original information). A number of solutions to this problem have been proposed, including suggestions that Hawking radiation is perturbed to contain the missing information, that the Hawking evaporation leaves some form of remnant particle containing the missing information, and that information is allowed to be lost under these conditions.

References

New Article

Hello, I found this article today and I'm not sure if it can be incorporated into the wiki article. Both articles are so far above my head that I don't understand either of them.  :) Just FYI. Bill the Cat 7 (talk) 23:02, 10 July 2020 (UTC)

Target audience...

Are we trying to explain this to students already in physics, or to a wider audience? High level language, and writing is all well, and good, but it risks putting the article out of reach of people with even basic scientific understanding. Oceanic84 (talk) 21:12, 26 May 2021 (UTC)

Encyclopedias address multiple audiences. Some of the material will necessarily be above or below the level that individual readers are looking for. Readers should be able to find material that addresses their level of comprehension. -- Jibal (talk) 19:19, 8 July 2021 (UTC)

Do rotating black holes evaporate?

"... The process obeys the laws of black hole thermodynamics. A consequence of these laws is that if the process is performed repeatedly, the black hole can eventually lose all of its angular momentum, becoming non-rotating, i.e. a Schwarzschild black hole. In this case the theoretical maximum energy that can be extracted from an uncharged black hole is 29% of its original mass.[4] Larger efficiencies are possible for charged rotating black holes.[5]" From the Penrose Process Wikipedia link. If I read the text correctly, a rotating black hole emits radiation until not longer rotating, then radiation stops. The black hole does not evaporate, Yes? Wabarr52 (talk) 17:02, 19 July 2021 (UTC)

The quick answer is 'we don't know yet'. "Space and time reverse roles at the event horizon". We don't even know the timescale at which your very reasonable question applies, much less the timescale for Hawking radiation. I personally would search for other jets out of other galactic nuclei (like the jet out of M87*), using the Hubble or the Webb space telescopes. At least we could get an idea of the lifespan of these jets. --Ancheta Wis   (talk | contribs) 20:07, 19 July 2021 (UTC)

References

Hawking Radiation Missing Assumption or Explanation

"The amplification gives rise to a "partner wave", which carries negative energy and passes through the event horizon, where it remains trapped, reducing the total energy of the black hole." The statement assumes only negative energy remains trapped by the black hole. What the explanation does not describe yet crucial is the mechanism that pulls negative energy exclusively into the black hole. Without such an explanation an equal amount of positive and negative energy would remain in the black hole due to random process. What is the missing explanation professional physicists assume layman know? Wabarr52 (talk) 11:16, 18 July 2021 (UTC)

I found Penrose process to be helpful. --Ancheta Wis   (talk | contribs) 13:52, 18 July 2021 (UTC)

The article is actually quite explicit: the positive energy of the escaping photon has a "partner", a wave with negative energy the enters the BH. The only assumption is that energy is conserved. Everything captured by the BH "remains trapped". -- Jibal (talk) 01:21, 20 July 2021 (UTC)

Incorrect explanation of how Hawking radiation works conceptually

The explanation in the form of "virtual particles" is not really what happens. First of all virtual particles as the name suggests do not ever really exist or form. It's just an analogy for an energy transfer, so the explanation is actually rather circular.

I also don't think scholarpedia is really a good reliable source to cite within a Wikipedia article? Which seems to be the source used for this incorrect explanation. Freeze4576 (talk) 12:49, 10 September 2021 (UTC)

Since all reference to virtual particles has been removed from this article, probably we should also remove the redirect to this article from "virtual particle production"? ToxicPsychotic (talk) 04:11, 30 August 2022 (UTC)

Hawking temperature too small by a factor of 2

Pointed out here. If information is to be preserved then the temperature of the Hawking radiation must be twice the value computed by Hawking. Count Iblis (talk) 22:54, 29 November 2022 (UTC)