Stephen Hawking’s “Brief History of Time” was one of the first popular science books I read, and I hated it. I hated it because I didn’t understand it. My frustration with this book is a big part of the reason I’m a physicist today – at least I know who to blame.

I don’t hate the book any more – admittedly Hawking did a remarkable job of sparking public interest in the fundamental questions raised by black hole physics. But every once in a while I still want to punch the damned book. Not because I didn’t understand it, but because it convinced so many other people they did understand it.In his book, Hawking painted a neat picture for black hole evaporation that is now widely used. According to this picture, black holes evaporate because pairs of virtual particles nearby the horizon are ripped apart by tidal forces. One of the particles gets caught behind the horizon and falls in, the other escapes. The result is a steady emission of particles from the black hole horizon. It’s simple, it’s intuitive, and it’s wrong.

Hawking’s is an illustrative picture, but nothing more than that. In reality – you will not be surprised to hear – the situation is more complicated.

The pairs of particles – to the extent that it makes sense to speak of particles at all – are not sharply localized. They are instead blurred out over a distance comparable to the black hole radius. The pairs do not start out as points, but as diffuse clouds smeared all around the black hole, and they only begin to separate when the escapee has retreated from the horizon a distance comparable to the black hole’s radius. This simple image that Hawking provided for the non-specialist is not backed up by the mathematics. It contains an element of the truth, but take it too seriously and it becomes highly misleading.

That this image isn’t accurate is not a new insight – it’s been known since the late 1970s that Hawking radiation is not produced in the immediate vicinity of the horizon. Already in Birrell and Davies’ textbook it is clearly spelled out that taking the particles from the far vicinity of the black hole and tracing them back to the horizon – thereby increasing (“blueshifting”) their frequency – does not deliver the accurate description in the horizon area. The two parts of the Hawking-pairs blur into each other in the horizon area, and to meaningfully speak of particles one should instead use a different, local, notion of particles. Better even, one should stick to calculating actually observable quantities like the stress-energy tensor.

That the particle pairs are not created in the immediate vicinity of the horizon was necessary to solve a conundrum that bothered physicists back then. The temperature of the black hole radiation is very small, but this is in the far distance to the black hole. For this radiation to have been able to escape, it must have started out with an enormous energy close by the black hole horizon. But if such an enormous energy was located there, then an infalling observer should notice and burn to ashes. This however violates the equivalence principle, according to which the infalling observer shouldn’t notice anything unusual upon crossing the horizon.

This problem is resolved by taking into account that tracing back the outgoing radiation to the horizon does not give a physically meaningful result. If one instead calculates the stress-energy in the vicinity of the horizon, one finds that it is small and remains small even upon horizon crossing. It is so small that an observer would only be able to tell the difference to flat space on distances comparable to the black hole radius (which is also the curvature scale). Everything fits nicely, and no disagreement with the equivalence principle comes about.

[I know this sounds very similar to the firewall problem that has been discussed more recently but it’s a different issue. The firewall problem comes about because if one requires the outgoing particles to carry information, then the correlation with the ingoing particles gets destroyed. This prevents a suitable cancellation in the near-horizon area. Again however one can criticize this conclusion by complaining that in the original “firewall paper” the stress-energy wasn’t calculated. I don’t think this is the origin of the problem, but other people do.]

The actual reason that black holes emit particles, the one that is backed up by mathematics, is that different observers have different notions of particles.

We are used to a particle either being there or not being there, but this is only the case so long as we move relative to each other at constant velocity. If an observer is accelerated, his definition of what a particle is changes. What looks like empty space for an observer at constant velocity suddenly seems to contain particles for an accelerated observer. This effect, named after Bill Unruh – who discovered it almost simultaneously with Hawking’s finding that black holes emit radiation – is exceedingly tiny for accelerations we experience in daily life, thus we never notice it.

The Unruh effect is very closely related to the Hawking effect by which black holes evaporate. Matter that collapses to a black hole creates a dynamical space-time that gives rise to an acceleration between observers in the past and in the future. The result is that the space-time around the collapsing matter, that did not contain particles before the black hole was formed, contains thermal radiation in the late stages of collapse. This Hawking-radiation that is emitted from the black hole is the same as the vacuum that initially surrounded the collapsing matter.

That, really, is the origin of particle emission from black holes: what is a “particle” depends on the observer. Not quite as simple, but dramatically more accurate.

The image provided by Hawking with the virtual particle pairs close by the horizon has been so stunningly successful that now even some physicists believe it is what really happens. The knowledge that blueshifting the radiation from infinity back to the horizon gives a grossly wrong stress-energy seems to have gotten buried in the literature. Unfortunately, misunderstanding the relation between the flux of Hawking-particles in the far distance and in the vicinity of the black hole leads one to erroneously conclude that the flux is much larger than it is. Getting this relation wrong is for example the reason why Mersini-Houghton came to falsely conclude that black holes don’t exist.

It seems about time someone reminds the community of this. And here comes Steve Giddings.

Steve Giddings is the nonlocal hero of George Musser’s new book “Spooky Action at a Distance.” For the past two decades or so he’s been on a mission to convince his colleagues that nonlocality is necessary to resolve the black hole information loss problem. I spent a year in Santa Barbara a few doors down the corridor from Steve, but I liked his papers better when we was still on the idea that black hole remnants keep the information. Be that as it may, Steve knows black holes inside and out, and he has a new note on the arxiv that discusses the question where Hawking radiation originates.

In his paper, Steve collects the existing arguments why we know the pairs of the Hawking radiation are not created in the vicinity of the horizon, and he adds some new arguments. He estimates the effective area from which Hawking-radiation is emitted and finds it to be a sphere with a radius considerably larger than the black hole. He also estimates the width of wave-packets of Hawking radiation and shows that it is much larger than the separation of the wave-packet’s center from the horizon. This nicely fits with some earlier work of his that demonstrated that the partner particles do not separate from each other until after they have left the vicinity of the black hole.

All this supports the conclusion that Hawking particles are not created in the near vicinity of the horizon, but instead come from a region surrounding the black hole with a few times the black hole’s radius.

Steve’s paper has an amusing acknowledgement in which he thanks Don Marolf for confirming that some of their colleagues indeed believe that Hawking radiation is created close by the horizon. I can understand this. When I first noticed this misunderstanding I also couldn’t quite believe it. I kept pointing towards Birrell-Davies but nobody was listening. In the end I almost thought I was the one who got it wrong. So, I for sure am very glad about Steve’s paper because now, rather than citing a 40 year old textbook, I can just cite his paper.

If Hawking’s book taught me one thing, it’s that sticky visual metaphors that can be a curse as much as they can be a blessing.

ReplyDelete"If Hawking’s book taught me one thing, it’s that sticky visual metaphors that can be a curse as much as they can be a blessing."

Just like Feynman diagrams have made numerous physicists to believe that virtual particles (the internal lines of the diagrams) are real. Someone write a paper on that as well? :)

Nice post!

Is there a connection between the speed of light and the singularity in a black hole? If spacetime collapses at the speed of light then there must be some very deep relationship between the two. One can imagine that every photon is embedded in an event horizon.

ReplyDeleteSabine,

ReplyDeleteAnother place where this point is made is in Bill Unruh's 2007 paper "Where are the particles created in black hole evaporation"

http://inspirehep.net/record/775859?ln=en

It can be found on the PoS server

http://pos.sissa.it/cgi-bin/reader/conf.cgi?confid=43

Best,

John Donoghue

Poor Sabine. You are so very confused and picking buzz words here and there. Don't try to sell your personal confusion as community's opinion. Instead of moaning online just try to reproduce the calculations. I highly recommend it. Then you will understand what is going on...well, maybe...

ReplyDeleteMy advice: stop trying to associate yourself with respected names in the community to try and draw attention to yourself. It is cheap. Others are busy working, they cant waste time with your online whining. I have heard this comment from almost everyone you ever mentioned in your 'blogs'. If indeed your goal is to understand physics just sit and calculate. Good luck!

Anonymous? So you haven't the bollocks to reveal yourself, yet whine about Ms. Hossenfelder and her astute observations? I find you to be lacking in any credibility and your claims are just a pedantic attempt to discredit a better thinker and scientist than you pretend to be.

DeleteI do agree but Dr H will probably refrain from making this correction.

DeleteExcellent blog post! This reminds me of a very very old post by John Baez (1994, revised in 1997 by someone else):

ReplyDeletehttp://www.math.ucr.edu/home/baez/physics/Relativity/BlackHoles/hawking.html

although he doesn't discuss the additional subtleties addressed by Giddings that you clarified further. Baez write,

" How does this work? Well, you'll find Hawking radiation explained this way in a lot of "pop-science" treatments:

Virtual particle pairs are constantly being created near the horizon of the black hole, as they are everywhere. Normally, they are created as a particle-antiparticle pair and they quickly annihilate each other. But near the horizon of a black hole, it's possible for one to fall in before the annihilation can happen, in which case the other one escapes as Hawking radiation.

In fact this argument also does not correspond in any clear way to the actual computation. Or at least I've never seen how the standard computation can be transmuted into one involving virtual particles sneaking over the horizon, and in the last talk I was at on this it was emphasized that nobody has ever worked out a "local" description of Hawking radiation in terms of stuff like this happening at the horizon. I'd gladly be corrected by any experts out there... Note: I wouldn't be surprised if this heuristic picture turned out to be accurate, but I don't see how you get that picture from the usual computation."

In passing, I also recall that the book by J. Hartle "Gravity: an Introduction to Einstein's General Relativity", makes use of the same misconception for simplicity. Anyway, I like his explanation using the Killing vector.

Isn't it true, though, that an observer close to the black hole but accelerating strongly enough to not fall in, would still see a huge radiation flux? Would that just be the Unruh radiation she would see if not BH were involved?

ReplyDeleteRe the similarity between Unruh radiation and Hawking radiation, this leaves me confused about something: from a massive object that is not a black-hole, is there in principle some sort of Unruh or Hawking radiation due to the gravitational acceleration?

ReplyDeleteIf so, something wierd would have to happen to preserve conservation of energy - e.g., the gravitational interactions would have to allow normal matter to decay, presumably the process not preserving lepton & baryon numbers - but no more so than in the case of black hole formation + evaporation?

We often think of the event horizon as a knife edge, but the math and your description say that isn't so. The event horizon inhabits curved space. Where does it become so strong that an unfortunate astronaut would not be able to escape it?

ReplyDeleteOh finally, some sanity prevails. Hawking radiation has had far too much airtime, needs to be retired! Thanks, Sabine.

ReplyDeleteGravitation's indented stretched rubber membrane model has each of its triangles' three interior angles summing to fewer than 180°. Lofting a laser and two reflectors into solar orbit, the three interior angles sum to more than 180°. Other problems, arXiv:1312.3893, doi:10.1119/1.4848635

ReplyDeleteThe pendulum equation has no bob. Equivalence Principle (EP) violation consistent with all observations of measurable observables suggests contrasting an unmeasurable discontinuous (evade Noether's theorems) property. One property qualifies, removed from gravitation theory by postulate. Where could an EP violation otherwise hide? arXiv:1505.01774, doi:10.1103/PhysRevD.91.121101; FIG 1,2 - not there.

A facile model may be coincidental but not relevant to the phenomenon.

If delta(x) is comparable to R and there is wide disagreement for what is inside the black hole near singularity, how believable are such arguments without a good theory of quantum gravity?

ReplyDeleteGiannis,

ReplyDeleteYes, exactly! I was about to mention Feynman diagrams too, but then I thought long enough already :)

John,

ReplyDeleteThanks for the reference!

The comment that was posted under the pseudonym "D Biber" is from Laura Mersini-Houghton.

ReplyDeleteDear Laura:

ReplyDeleteYour comment is a pure ad hominem attack, it's void of scientific content. We have had a long email exchange over your paper so we both know that I did in fact try to reproduce your calculation. Unfortunately, I couldn't. Do I have to tell you one more time that the stress-energy close by the horizon is of the order of the curvature scale?

I am seriously shocked how personal you seem to take what is in my perspective an entirely scientific question.

CIP,

ReplyDeleteYes, that's right, an observer hovering close by the horizon will essentially see Unruh radiation with a temperature that is set by the acceleration necessary to remain stationary. Since the black hole is spherical symmetry though it's not the exact same radiation, the Unruh radiation has a different radial dependence in 3+1 dimensions.

Oh great, now Lubos is going to reply with a loooong rant about this.

ReplyDeleteGoogle [...]

ReplyDeleteThe Hawking radiation is really a result of the collapse - of the dynamical background. If you have an object that collapses without forming a black hole it does also, in principle, create Hawking radiation. But the temperature is even smaller than that of black holes already. In fact it drops exponentially with the ratio of the Schwarzschild-radius over the radius of the compact object. So there is no problem for static objects. (Gravitational interactions however lead in general to the emission of gravitational waves which do carry away energy.)

kashyap,

ReplyDeleteSorry, I don't know what you mean. What do you mean with 'delta(x)' and why would that be comparable to R?

ReplyDelete"Steve’s paper has an amusing acknowledgement in which he thanks Don Marolf for confirming that some of their colleagues indeed believe that Hawking radiation is created close by the horizon. I can understand this. When I first noticed this misunderstanding I also couldn’t quite believe it. I kept pointing towards Birrell-Davies but nobody was listening. In the end I almost thought I was the one who got it wrong. So, I for sure am very glad about Steve’s paper because now, rather than citing a 40 year old textbook, I can just cite his paper.

If Hawking’s book taught me one thing, it’s that sticky visual metaphors that can be a curse as much as they can be a blessing."

This isn't the only case when arguing too far from analogy leads to wrong conclusions, even among professional physicists. Maybe someone should write a review article about such misconceptions.

ReplyDelete"Gravitation's indented stretched rubber membrane model has each of its triangles' three interior angles summing to fewer than 180°. Lofting a laser and two reflectors into solar orbit, the three interior angles sum to more than 180°. Other problems, arXiv:1312.3893, doi:10.1119/1.4848635"Thanks for bringing this up. Should be required reading! But why do you claim that it is more than 180° in practice?

Here, I'll mention it again: http://arxiv.org/abs/1312.3893 and dig the title: "Circular orbits on a warped spandex fabric". No, it's not about groupies of heavy-metal bands back in the 1980s! :-)

This is another example of arguing too far from analogy. In practice, this occurs less often among professional physicists than is the case for the Hawking-radiation description, which many (perhaps without realizing it) get wrong. Hawking radiation doesn't come from pair-production on a knife-edge horizon, gravity is not just like a two-dimensional rubber sheet, and the universe is not really like a tightrope walker.

As usual, XKCD nailed it: https://xkcd.com/895/

Helbig, I owe you a tenner. No mbh at LHC. I pay you. Can provide postal address to dnikbin@yahoo.co.uk if necessary. I send you cash.

DeleteNikbin

Sabine, show me the calculation and I can tell you what you have done wrong.I doubt very much you even read the equations never mind solve them. Remember I have you recorded in the Hawking Radiation conference I organized, when for 10 minutes you ask what temperature I take at the horizon and I patiently show you the

ReplyDelete6 equations and explain that luminosity is a variable you Solve for in those equations, not something you give a value at the horizon. You would think this is trivial. If you still didn't get it, well I cant help you...But don't pass your confusion as someone else's. Better to be humble and accept you don't understand something instead of misleading people who are not experts to vent anger.

Bee,

ReplyDeleteSorry. My question was too brief. My understanding of virtual particles is that they are field effects due to uncertainty principle.Delta(x) I was talking about is delta(x) in Heisenberg's uncertainty principle. I thought you were talking about this delta(x) comparable to the size of the black hole which would mean that one is applying uncertainty principle to the heart of BH.This would take it to the interior and near singularity.Since this is precisely the region near singularity which is not understood, I was questioning this conclusion in absence of theory of quantum gravity. What am I missing?

Dear Sabine,

ReplyDeleteI really share your frustration! The image is particularly unfortunate. Having said that, I would be really interested to know your thoughts on the following work: http://arxiv.org/abs/1011.2994 (and the related work http://arxiv.org/abs/1302.5253). Particularly since those authors take the issues with particle interpretation in curved spacetime extremely seriously.

kashyap,

ReplyDeleteThe particles eventually end up in the singularity. I don't understand why you think this is a problem. Just because something runs into a singularity eventually doesn't mean you can't say anything about it previously. Maybe the issue is one of visualization - You should think of the particles as blurry, spherical shells, not as if they have a maximum in the center. Best,

B.

Umberto,

ReplyDeleteI don't know these works in particular, but my understanding is roughly that it's a different technical way to express the same thing. I really don't think one should interpret too much into mathematical expressions unless these describe observables.

Laura,

ReplyDeleteYes, I was very much hoping the recording would go online, where is it? It demonstrates that I asked you the same question in several different formulations - and you failed to respond to it each single time. Like you failed to reply in our email exchange.

Did you even notice that other people in the audience tried to tell you the same thing? I don't know who it was, sorry. I couldn't see very well from the doggy corner where you insisted that I had to sit.

The question Laura, to repeat it one more time: What is the luminosity in the vicinity of the horizon? It's not in your paper, I have turned it upside down. Do you reproduce the expression that is in Giddings' paper and that is in Birrell-Davies, yes or no? Write it down, by order of magnitude: What is the luminosity that you use?

In your paper 1409.1837, eq (3) is, as you state yourself, the tensor at infinity. The luminosity appears there. In eq (11c) you integrate that same expression. Where do you calculate the flux in the horizon area to get the backreaction? Where is this equation?

Laura Mersini-Houghton champions the multiverse. If it interacted with this universe, thermodynamic degrees of freedom would be greater than classical or stat mech. They aren't, so there isn't.

ReplyDelete@Phillip Helbig. Simple spacetime curvature is elliptic geometry, yes? Spacetime torsion is chiral (e.g., Lorentz force). Measure the differential enthalpy of fusion of chemically identical enantiomorphic space group single crystals. Opposite shoes embed within a spacetime left foot with different energies. Maximum signal at 45° latitude melting them to identical socks (benzil, mp = 95°C) gives the energy difference: insertion plus 24-hour sinusoidal Equivalence Principle violation.

http://thewinnower.s3.amazonaws.com/papers/95/v1/sources/image006.png

Only crystal lattice benzil is a homochiral helix.

http://thewinnower.s3.amazonaws.com/papers/95/v1/sources/image008.png

static insertion difference

http://www.mazepath.com/uncleal/orbit.png

dynamic EP violation

Id love to watch a live debate between Sabine Hossenfelder and Laura Mersini-Houghton over their disagreement.

ReplyDeleteAny kind of force causes acceleration which in turn is equivalent to gravity appearing in the accelerating frame. How can this link of all types of forces to gravity be understood in physical terms? Why would all forces cause gravity or Unruh radiation?

ReplyDeleteTheories and speculations concerning event horizons are not necessarily immune to challenges that might be successful, or partially successful.

ReplyDelete"Dark Energy Stars" by G. Chapline, 2005

I'm only interested in the recording as that is as close to an accurate account on recent events and...as a reaction back;

ReplyDeletewhat the heck did that just happen??

Won't a quantum object many kilometers across and subject to interactions with matter infalling into the black hole quickly decohere?

ReplyDelete"Matter that collapses to a black hole creates a dynamical space-time that gives rise to an acceleration between observers in the past and in the future. The result is that the space-time around the collapsing matter, that did not contain particles before the black hole was formed, contains thermal radiation in the late stages of collapse."

ReplyDeleteEven eternal black holes produce hawking radiation. The key concept behind hawking radiation of stationary BHs is the horizon, no horizon no Hawking radiation.

Giotis,

ReplyDeleteEternal black holes only "produce" Hawking radiation because one can set up their environment so that they do, by balancing the outgoing flux with a suitable ingoing flux. The horizon does not produce Hawking radiation in any meaninful way - the horizon is what makes the radiation thermal, by separating it into two disconnected regions. If that wasn't so, black holes with apparent horizons couldn't produce any radiation. It's the dynamical background that makes for non-trivial Bogoliubov transformations between different regions (as opposed to between different observers, as you have with the Unruh effect). The easiest way to see this is this is to use the model of a collapsing sphere and let the collapse velocity go to zero. The radiation goes to zero with it.

Best,

B.

sabine:

ReplyDeleteyou said in a previous blog posting that you had heard stephen's wolfram talk about of spacetime and thought it was rather interesting. here's a new blog that stephen just wrote on the subject. i'd like to hear your opinion on it.

http://blog.stephenwolfram.com/2015/12/what-is-spacetime-really/

richard

The infalling observer would note that light coming from space is Blue Shifted, at the event horizon it is infinitely blue shifted and that would burn our intrepid traveller to a crisp.

ReplyDeleteBlue shift of incoming light is inconvenient so an oft used solution to this problem is evoked: the blue shift of incoming light

is ignored.Yet if outgoing light is redshifted then incoming light is blue shifted. Why does anyone even doubt this??

You don't have to collapse anything (although that is how it was originally derived by Hawking). The whole point of the BH radiation is that a static gravitational field can produce particles due to the fact that there is a horizon.

ReplyDeleteJust take a scalar field and quantize it in the background of an existing BH already formed.

Giotis,

ReplyDeleteA static black hole can only emit particles if you feed in the same amount of particles, otherwise it wouldn't be static. Can we agree on this? I think you've spent too much time in an AdS box ;) Yes, I take the scalar field and quantize it in the static Schwarzschild background and.... what's the boundary condition again? Best,

B.

naivetheorist,

ReplyDeleteI only met Wolfram once and that was almost 10 years ago. Yes, it was interesting. But it wasn't very convincing.

Robert Stonjek:

ReplyDeleteThe misconception that you demonstrate exactly the reason I wrote this blogpost. Maybe you should now go and read what I wrote.

http://arxiv.org/abs/hep-th/9907001

ReplyDeleteMight be of interest.

stor, yes see comment of Umberto and my reply above.

ReplyDeleteVery interesting post. Thanks, Bee.

ReplyDeleteThat was very interesting.

ReplyDeleteI always found Hawking's pop science explanation rather unconvincing - the idea that "one virtual particle falls into the black hole and the other escapes" would seem to depend on all sorts of details about the position and velocity of such particles.

However, I had a question. As per the Unruh effect, someone hovering just above the event horizon would see thermal radiation.

But as per the Hawking effect, even an observer at infinity - who is presumably not accelerating - sees radiation. What is the connection (if any) ?

I also found this paper which seems interesting..

http://xxx.lanl.gov/pdf/gr-qc/0304042v1.pdf

senanindya:

ReplyDeleteThat's what I meant with the rather vague statement that the black hole background 'freezes in' the acceleration. In the black hole background the two observers (early and late stages of collapse) don't agree on what is a vacuum, even though none of them is accelerated. Mathematically, the calculation is literally the same as with the Unruh effect. (Not taking into account that the one situation is spherically symmetric and the other isn't.) Best,

B.

In static black holes BH radiation is due to two different vacua one of the free infalling observer (kruskal coordinates) and one of observer at some constant distance.

ReplyDeleteI 'm not sure why you bring the thermal bath into the discussion, this concerns the thermodynamic stability.

As I see it you can perfectly examines BH radiation without referring to the BH formation during collapse.

Also don't forget complementarity. Where is the membrane (which absorbs thermalize and re-emits)in the complementarity argument? It is at a Planck length distance from the mathematical horizon.

Listen I'm not saying that this intuitive picture is the correct one and indeed you shouldn't take it too seriously but it doesn't do any harm either. It has qualitatively good characteristics and that's why it is widely used even in many text books.

To tell you the truth these BH discussions usually turn into some kind of mythopoeia. It's hard to distinguish the reality from the myth and one myth from the other. In fact they are boring.

that's why

Wow this D. Biber person is feeling CATTY!

ReplyDeleteI can accept that a thermometer far from a black hole says that the temperature of the black hole is 3 K.

ReplyDeleteAnd I can accept that a thermometer falling very near that black hole says that the temperature of that black hole is 0.1 K.

But I can not accept that a thermometer falling very near that black hole would say that the temperature of that black hole is 0 K.

Do I need to come up with some argument? Doesn't 0 K sound improbable enough?

Jari,

ReplyDeleteI don't know what you mean. A 'thermometer' falling through the black hole horizon wouldn't say that the temperature is zero. Provided it is large enough, it would detect Hawking quanta. The point is that these are so large they're on the order of the curvature radius, and that is compatible with the equivalence principle. (Ie, you'd need a huge thermometer, thus the scare quotes.)

@Sabine: I realize that I'm much too late for this discussion, but wouldn't a free-falling thermometer be in the vicinity of the horizon for much too short a time to interact with Hawking quanta? Thus, it would only be able to provide an upper bound for the temperature (one that is quite a bit higher than the real temperature). Although I suppose you could send $10^{10}$ "thermometers" through the black hole horizon one at a time, and use the statistics from the radiation they observe to calculate the temperature.

DeleteHi Peter,

DeleteYes, instead of taking one large thermometer, you can take a lot of small ones. This decreases the chances of it becoming excited, but then you can rely on statistics.

Consider a burst of outward-aimed light emitted by an infalling object at the precise moment that it passes through r=2M.

DeleteWith a Wheeler back hole, that light is supposed to remain frozen into the horizon, so if you are outside r=2M, you don't see it. Temperature=0.

However, if you let yourself fall into the hole, and pass inwards at speed through r=2M, then you are passing through that frozen wavefront, and should register a non-zero temperature, regardless of Hawking radiation issues.

Well I guess I have to actually start thinking why the 0 K thing is impossible. The reason is some kind of thermodynamical argument, if there is a reason, I'm not so sure anymore.

ReplyDeleteNow I'm wondering: If only a black hole size thermometer with the help of Doppler shift can absorb some quanta, then how can a black hole size black hole emit those quanta? I mean, in the black hole's rest frame those quanta are really huge.

Dr Hossenfelder - I'm a layman who loves to try to understand a bit of these discussions. I'm aware of being very confused about one thing: if the escapee was formed out of the vacuum a few radii away from the BH, then isn’t it the vacuum that is losing energy, rather than the BH? If the escapee’s partner ends up falling into the black hole, then it would seem to me these events only increase the energy of the BH, not decrease it (the deficit being in the vacuum outside the hole). Obviously, I am laboring under some basic misunderstanding.

ReplyDeleteLinking Hawking radiation and the Unruh effect impresses me a lot more than the usual virtual particles getting trapped explanation. It is a rather amazing fact that there is a temperature associated with acceleration whether something is just getting jerked around by external forces or yanked by black hole levels of gravity. I'm sure I've embedded a few misunderstandings in there, but I can appreciate that this links quantum mechanical ideas about vacuum energy and virtual particles with general relativistic ideas about the importance of the observer's reference frame.

ReplyDeleteIt is articles like this that make the internet worthwhile.

Jari,

ReplyDeleteThey are not emitted at the horizon. That was the whole point of this post.

Steve,

ReplyDeleteThe particle pair takes energy from the background field. One of them falls into the black hole, the other one escapes. That leads to a net loss of mass (energy) from the black hole.

What happens technically is somewhat different because in the mathematical treatment the background is fixed and cannot lose energy. For this reason, mathematically one of the two particles actually has negative energy - that's the only way energy can remain conserved if the background remains fixed. It's the negative energy particle that falls into the black hole, leading to a net loss. The easiest way to understand that this is the same is to think of a background that is some kind of matter field. In these cases you can have 'holes' next to particles. That the mathematical treatment doesn't quite capture what actually happens is known as the 'backreaction problem' (origin of this blog's name). Best,

B.

I hated A Brief History of Time also. Not that I expect to understand everything I read, but I came away thinking it was REALLY badly written.

ReplyDeleteDr Hossenfelder, I like the description of a particle & hole pair forming out of the vacuum, analogous to similar descriptions for electronics phenomena. Is there some way to explain why it is always the hole that falls into the BH, and never the particle? By the way, I asked my original question on John Baez's blog and suggested that the particle falling into the hole might have negative mass (whatever that could mean) - I got some interesting replies to my suggestion.

ReplyDeletePersonally, I like the "acoustic metric" interpretation of Hawking radiation.

ReplyDeleteIn this, some statistically-unlikely combinations of particle collisions result in a few particles escaping from the hole. The forced physical accelerations are associated with geometry-change, and the associated geometry-changes near the horizon allow the “effective” horizon for a distant observer to briefly jump discontinuously to a position lower than the final accelerated particle, allowing escape.

This indirect escape mechanism doesn't work for single particles (so it doesn't appear in a "test particle" based description), and since it also depends on brief physical-acceleration-dependent distortions, it also doesn't appear if we assume that the geometry of the region is defined only by the black hole's central mass - assuming that a test particle's contribution to the geometry is minimal results in our predicting (in this case, wrongly) the absence of any such radiation.

Now let's place a detector far from the hole, which registers particles coming from the hole's direction. Since no particles can escape from within the horizon along a non-accelerated path, a misguided attempt to back-calculate a ballistic trajectory for these particles breaks down some distance outside the horizon. For a given escaped particle, its “fake” trajectory appears to start out, impossibly, at more than the speed of light, and since this earlier section of extrapolated path is tachyonic, this section of the particle's extrapolated path gets described in observerspace as appearing to be time-reversed.

Combine the two sections of path, and we get a description of a particle-pair apparently popping into existence outside the horizon, with one member of the pair moving away from the hole and escaping, and its reversed twin moving in the opposite direction, back into the hole. In this “artificial” description, we get weird results, like the height along the calculated path at which p-p is deemed to “really” happen being a function of a hovering observer's height, and radiation that is classifies as being “virtual” for the distant observer, appearing to be “real” for the nearby hovering observer ... but these effects are also present in the QM description.

In the "acoustic" explanation, pair-production is a projective mathematical artefact of an attempt to model a radiative effect that depends critically on acceleration, while ignoring how physical acceleration affects the shape of spacetime. The explanation seems (AFAIK) to be the most efficient way so far of explaining (and predicting!) discoveries in quantum mechanics that relate to event horizon behaviour.

Sabine,

ReplyDeletere the backreaction problem, is it accurate to say that it does not exist in background independent formulations, say those that utilize the geometric algebra of Hestenes et.al.?

P.

Sabine,

ReplyDeleteStaying with geometric algebra approach for one more post, how about the proof offered by those folks that gauge theory gravity in flat space is equivalent to GR in curved space? See for instance

http://geocalc.clas.asu.edu/pdf/GTG.w.GC.FP.pdf

It seems to me this plus holographic principle is sufficient to remove the paradox, however one describes the local effects. See for instance

https://www.osapublishing.org/abstract.cfm?URI=QIM-2013-W6.01

Which is not to say local effects should be ignored. Of particular interest is the Planck particle, the 'electromagnetic black hole', which by consideration of local (ie near field) effects should be stable, yet is claimed to be dramatically unstable from perspective of Hawking radiation.

P.

I heard Unruh twice this year, and in both the cases he said explicitly that the radiation coming from the horizon was a common wrong belief. Moreover, he underlined that actually most of the radiation is coming from quite far from the horizon, if I remember correctly, the biggest part is produced around r ~ 3M and you can see it also experimentally with analogue gravity. So the question "who did believe it?" is a good question that deserves acknowledgments to who answered :)

ReplyDeleteThe height at which the radiation is nominally created is supposed to be observer-dependent. If you are a very, very long way away, the radiation might indeed be calculated as apparently coming from ~r=3M, but if you are //at// r=3M, the radiation will seem to be coming from between r=3M and r=2M. If you are at r=2.1m, the radiation will appear to you to be be originating at somewhere between r=2.1M and r=2M.

DeleteThere's a simple sanity-check for this: Hawking radiation inherits the matter-antimatter bias that applied to the matter that previously fell into the hole. So in a matter-biased universe, the radiation is predominantly matter, and the supposed infalling antiparticles are described as being predominantly antimatter.

so, question: if you hover above a black hole horizon, should you get zapped by the infalling antimatter coming from above you, as the infalling partners of the escaping Hawking radiation? If not, why not? The answer would seem to be, "because the Hawking radiation always seems to you to be coming from somewhere between your own location and the horizon".

This is how things work with Hawking radiation in acoustic metrics, in am's the radiation actually migrates through the horizon along accelerated paths, but external observers catching the radiation and extrapolating non-accelerated paths, end up describing the same result as being due to particle-pair-production outside the horizon, but at a height somewhere lower than the observer.

Hmmm... Gerard 't Hooft repeats the claim about Hawking radiation being produced near the horizon in a November Arxiv paper. Link: http://arxiv.org/abs/1511.04427 On page 2 he states, " Hawking particles are now understood to be formed at

ReplyDeletethe horizon, not, as was originally thought, somewhere near the r = 0 singularity in its

past."

Thank you, Sabine (and Steve Giddings) for taking a stand for the

ReplyDeletetruth and writing such a good explanation to bring it to a wider

audience. I think the people you heard from the doggy corner

were Emil Mottola and I (and possibly Paul Davies and Philippe

Spindel at times). By the way, Bill Unruh wrote two papers with

the same title, "Origin of the particles in black hole

evaporation", one in 1977 (cited by Giddings) and one in 2008

(cited by an earlier commenter); IMHO the first one is more

relevant to the present discussion. The original paper on the 2D

black hole model, which seems to get forgotten, is

Davies, Fulling, and Unruh, Phys Rev D 13 (1976) 2720.

And while I'm at it, I wrote a paper on the physical

implications, Phys Rev D 15 (1977) 2411, agreeing with Unruh's

but from a slightly different point of view. All these 2D papers

may be informative to people who want to see explicit

calculations with a minimum of geometrical complications.

Giotis and Sabine,

ReplyDeleteWhat is the reason for researchers, particularly string theorists I've found, insisting that horizons create particles? I agree its true that if you take a static, eternal black hole spacetime you can pick certain states that give an outgoing flux of particles with a thermal distribution of their frequencies; you can also pick states that DO NOT give an outgoing flux of particles. However in reality, or in any more realistic calculation, the production of particles at a point outside of the black hole could not depend on the horizon because the horizon surface is not in the past light cone of the observer.

senanindya,

The relation between the Unruh effect and Hawking radiation is superficial in my opinion. It has been shown that the energy to produce particles in the Unruh effect comes from the acceleration of the observer while the energy to produce particles in the Hawking effect should come from the black hole.

As you correctly noted, an observer orbiting a black hole has zero acceleration and will not see particles according to the Unruh effect, where the temperature is proportional to acceleration. But this observer will see Hawking radiation. More so, any static, constant radial observer outside of a black hole should see a Hawking spectrum that is not in line with what one would expect from the Unruh effect, that is the temperature of radiation is not equal to hbar*a/(2*pi*c*k_b).

Thanks,

Alex

Alex,

ReplyDeleteI have wondered the same before. I suspect it's because they can't deal with the dynamical collapse and almost all their arguments only hold in the static case. Then one needs to argue that the static case is in some sense the same as the dynamical case.

Peter,

ReplyDeleteThe problem isn't so much to write down the equations, but to actually calculate something. I don't know what saying that the problem is absent in some formulation helps, because we know that already. The question is how to deal with it practically.

In response to Alex V: I agree with your first paragraph but disagree with the last one.

ReplyDeleteA "constant radial observer" near a black hole is necessarily accelerating, since otherwise he would fall into the hole. In fact, the Hawking radiation seen by such an observer is almost entirely Unruh radiation. A freely falling observer near the the hole sees essentially no radiation. Far away from the hole the radially supported observer is almost inertial and sees "real" Hawking radiation unrelated to acceleration. These remarks are some of the main points made by Unruh in the paper discussed yesterday. And the difference between near and far zones with respect to "reality" (i.e., lack of relation to acceleration) of the Hawking radiation

strongly supports the main point that Sabine has been making.

Hi Stephen Fulling,

ReplyDeleteI'm not seeing any inconsistency between your comment, which was my prior understanding of things, and anything I wrote. I also don't see the inconsistency with what Sabine wrote...? I'm curious as to what specifically you disagree with though.

In the last two paragraphs I wrote above I was trying to express the fact that conceptually the Unruh effect and Hawking effect seem to be almost independent in my mind. The Unruh effect seems to occur in any spacetime for a particle detector with a large enough acceleration, with the clicks of the particle detector being produced from the energy it takes to accelerate the detector. While the Hawking effect occurs due to the non-static spacetime caused by an object collapsing to a black hole, with the energy to produce the particles coming from the mass of the black hole. I said the relationship between them is superficial because the calculations involved for either of them look similar and I guess are analogous in a way.

Alex,

ReplyDeleteIf you calculate what an observer sees who hovers near a black hole horizon at a fixed radius you reproduce the Unruh effect (leaving aside spherical harmonics) because to remain stationary there the observer needs to have a constant acceleration. Thus, the relation between the two effects is more than a mathematical coincidence - it's an expression of the equivalence principle.

Your statement

"an observer orbiting a black hole has zero acceleration and will not see particles according to the Unruh effect, where the temperature is proportional to acceleration. But this observer will see Hawking radiation."

is manifestly wrong because (as Stephen also wrote) to orbit the black hole at constant radius the observer needs to be accelerated. If you calculate what the observer sees, you get a result that you can either interpret as Unruh radiation (from the acceleration) or Hawking radiation (from the black hole). This has to be so, otherwise the observer could distinguish one from the other, in violation of the equivalence principle.

Sabine,

ReplyDeleteI was being sloppy in my descriptions. When I said "constant radial observer" I meant an observer following a constant r, theta, phi trajectory. I believe this would be a hovering observer in your terminology. Also, when I said an orbiting observer I meant an observer that is following an orbit which is a geodesic.

With that being said, a geodesic orbit around the black hole at a constant radius is possible. This means that there are orbiting observers that have zero acceleration. By acceleration I mean the magnitude of the acceleration 4-vector, which is a coordinate independent quantity.

When you say that an orbiting body must have an acceleration, I presume you mean the coordinate dependent 3d acceleration vector. So if you are saying that the Unruh effect depends on the magnitude of the 3d acceleration vector as opposed to the 4d acceleration vector, I think that is very problematic as one can set the 3d acceleration vector to zero by changing coordinates, i.e. this would violate the principle of general covariance. So I think it is incorrect to say that the Unruh effect depends on the magnitude of the 3d acceleration vector.

This has to be so, otherwise the observer could distinguish one from the other, in violation of the equivalence principle.It has been demonstrated that the response of particle detectors hovering over a black hole (constant r, theta, phi) is not equal to the response of a particle detector immersed in thermal radiation with temperature T = a/(2*pi), where a is the 4-acceleration magnitude it takes to hover at the given radius. http://arxiv.org/abs/1102.5564, and Birrell and Davies. In other words, one can distinguish between Hawking radiation and Unruh radiation if we know the acceleration of the particle detector. **I disagree with the overall conclusion of http://arxiv.org/abs/1102.5564 that this disagreement is a violation of the equivalence principle but the calculations seem to be correct**

If you calculate what an observer sees who hovers near a black hole horizon at a fixed radius you reproduce the Unruh effect (leaving aside spherical harmonics) because to remain stationary there the observer needs to have a constant acceleration. Thus, the relation between the two effects is more than a mathematical coincidence - it's an expression of the equivalence principle.In this statement you seem to be employing the 4-vector definition of acceleration. "near a black hole" here means as r->2M, where the response of the detector goes to infinity. anywhere away from r=2m the response are not equal. So I don't really see how this is an expression of the equivalence principle but maybe that's a matter of interpretation.

Thanks,

Alex

Alex:

ReplyDeleteI seem to have misunderstood what situation you had in mind. I don't know the paper you mention, I will have a look at this. As to the orbiting observer. In an orbit you can measure geodesic deviation on distances comparable to the radius, so I don't think you can use the equivalence principle in your argument. Also, if you have spherical symmetry, you have a preferred frame, so general coordinate covariance is not applicable. You can define acceleration unambiguously in radial coordinates. In any case, the acceleration I referred to is the absolute value of the four-acceleration. (It's the constant that appears in the argument of a time-dependent boost.) Having said that, I am not sure at this point what we actually disagree on... Best,

B.

Hi again Alex,

ReplyDeleteI vaguely seemed to recall that Unruh mentioned the orbiting observers in his paper, and turned out I recalled correctly. Have a look at page 886, right column, the paragraph that starts with "I would now like to return to the question..." which, I think, is exactly the question you asked. Best,

B.

Alex:

ReplyDelete" The Unruh effect seems to occur in any spacetime for a particle detector with a large enough acceleration, with the clicks of the particle detector being produced from the energy it takes to accelerate the detector. While the Hawking effect occurs due to the non-static spacetime caused by an object collapsing to a black hole, with the energy to produce the particles coming from the mass of the black hole. "As Sabine says, Einstein's principle of equivalence requires an observer undergoing generic "physical" acceleration (shown by the presence of gee-forces) to be locally indistinguishable from the situation involved in their being suspended above a gravity-source.

The energy required for this sort of "gee-forced-evidenced" acceleration can be nominally zero: the observer's mass can be arbitrarily small (until quantum limits come into play), or an observer can rest indefinitely on the inner rim of a large spinning centrifuge in deep space, without there being an obvious energy cost per second, or we can build a scaffolding framework around a black hole on which accelerated observers can stand.

If we stand on one of these frameworks and and lower a detector or a tv camera on a rope, it should see and report radiation in the area that we might otherwise insist was not there. Similarly, we can lower a simple lens on the end of a piece of fibre-optic cable, and when we look down the fibre-optics, see radiation in the region that we would not otherwise expect to be able to see: the stresses in the cable are associated with a deformation of the region of spacetime that helps the signals escape, so that part of the weight of the cable is due to the energy of the signals being transmitted up its length.

Sabine,

ReplyDeleteI'd tend to assume that the radiation seen by an orbiting observer is going to be equivalent to Zel'dovich radiation.

My comments concerned radially moving detectors (freely falling vs. hovering). For orbiting detectors I should defer to experts in general relativity. (I am a quantum theorist and mathematician.) But my understanding is that there are precious few closed geodesic orbits very near a black hole. An observer in a (say) circular orbit there (r < 3M) is not in free fall; there must be some outward nongravitational acceleration. But not as much as for a hovering (constant r) radial (constant angles) observer. I would expect, therefore, that the detected radiation will be intermediate between that seen by the hovering and the radially falling observer.

ReplyDeleteA hovering observer in the region r \approx 3M should encounter both Hawking and Unruh phenomena.

For the next half week I will be mired in grading finals against a deadline, so I'll probably be unable to continue this discussion.

Dear Sabine,

ReplyDeletethanks for your comments on this very interesting issue of where the radiation originates. I would like to ask you, when you mention that the stress tensor close to the horizon has been computed, can you give a reference? I would like to follow that computation closely.

Regarding non-locality, I am not convinced. It has been computed the Unruh vacuum for a Schwarzschild black hole in a rigorous way in http://arxiv.org/abs/0907.1034, within the Local QFT approach (or algebraic QFT). Anyway, this is just a comment in favor of locality.

Thanks again!

Alan

Unknown,

ReplyDeleteThis is a good reference

http://journals.aps.org/prd/abstract/10.1103/PhysRevD.21.2185

(and references therein) as well as this book

http://www.cambridge.org/be/academic/subjects/physics/theoretical-physics-and-mathematical-physics/quantum-fields-curved-space

Thanks Sabine, Candelas paper seems to give an explanation.

ReplyDeleteI had a question. We know that Ricci tensor around horizon is zero, then how come we get a positive energy-momentum tensor outside horizon?

ReplyDeleteSk,

ReplyDeleteThe stress-energy tensor of the Hawking flux is not a source to the background geometry. It's not a closed set of equations. The (missing) closure is known as 'backreaction'. Best,

B.

" still want to punch the damned book. Not because I didn’t understand it, but because it convinced so many other people they did understand it."

ReplyDeleteI think that this is true for almost all popular physics books. I mean, these books almost always are without math and only use metaphors because that's the only way to explain complex ideas that are often mathematically abstract. I mean, String Theory, as far as I understand, are not literally vibrating strings. It is a metaphor for a mathematical construct not meant to be taken literally. But to many science writers do no makes this more point well.

ReplyDelete"I mean, String Theory, as far as I understand, are not literally vibrating strings. It is a metaphor for a mathematical construct not meant to be taken literally."I don't know, not being a stringy person. Maybe someone who actually works in the field can comment. My impression is that they are indeed one-dimensional objects which vibrate. A similar question goes for branes. It seems to me that people really do take them literally, not as some sort of metaphor. I recall reading about "the tension of the brane", which seems pretty literal for me (not to mention the tension in my brain while reading this stuff).

One sometimes hears that spin is not really angular momentum, but only mathematically similar. However, the Einstein-de Haas effect shows that spin really is angular momentum. (Maybe, on this point, there is some confusion with isospin, which I'm sure Sabine remembers from Greiner.)

By the way, Sabine, that reminds me: I recently heard of a very mathematical book by Greiner (and perhaps a co-author) involving, IIRC, GR (or some theory encompassing GR). Heard of it? Interested?

Just found this. My question is a bit related to the one of Steve further up, but not the same.

ReplyDeleteIf the Hawking radiation is "smeared out" over a large vicinity around the black hole, would it then be possible for a negative-energy particle to interact with other particles (matter or light) before it enters the black hole, and if so, could it "steal" energy from them (as it seems to be able to steal energy from the black hole)?

If so, could this be a process delaying the evaporation of primordial black holes? (Within a galaxy, there's lot of light and matter around. Their cousins in intergalactic space or even a void would have it easier in that case. By the astronomical timescales after which stellar-mass black holes would "produce" more Hawking radiation than they gain by infalling particles, the vicinity would be nearly empty, so it would matter less for them)

PS I originally wanted to ask this question concerning an energy-lacking background, but you said that background was fixed.

Ambi Valent,

ReplyDeleteThe answer to your question is 'in principle, yes', but the interaction with other matter is normally not taken into account. The way that these calculations are done, one merely deals with one field that doesn't self-interact. In other words, nobody ever considers the possibility, so not much is known about it.

I disagree. If the gravitational collapse stops some \(10^{-100000}\) Planck lengths away from the Schwarzschild radius, Hawking radiation stops after a very short time. So, if there will be Hawking radiation after this short time, it has to be emitted from this thin region around the BH, and not at all from somewhere far away.

ReplyDeleteSee Paranjape, Padmanabhan, Radiation from collapsing shells, semiclassical backreaction and black hole formation, Phys.Rev.D 80:044011 (2009), arxiv:0906.1768v2, but this is also simply the point that stable stars do not Hawking-radiate, and this is quite rigid, because for a stable star the vacuum state is stable too.

Thus, if there is Hawking radiation from a black hole, the origin is the further collapse after that radius has been reached, because this is the only remaining difference.

Ilja,

ReplyDeleteIt is correct that the cause of Hawking radiation is the time-dependence of the background geometry. It does not follow from this however that Hawking radiation must be created at the horizon, or a thin region around the black hole. This is simply wrong, as you can convince yourself easily. Go and look up, for example, the stress-energy-tensor of the Hawking radiation. You will find that it is *not* peaked around the horizon. Best,

B.

I don't understand this. What does the stress-energy tensor of the Hawking radiation tell us about its origin? All I need for creating regions with high local energy of radiation not originated there is a lense. And a BH is an extremely powerful lense. So, from the radiation created at the collapsing surface a lot may be catched for a longer time around the photon orbit, before flying away or falling back, not?

ReplyDeleteI have anyway no trust in any computation of such an energy-momentum tensor. Sure that there is some computation not endangered by the trans-Planckian problem?

Sabine,

ReplyDeletethat's a great post!

One question I couldn't find an answer to:

Does an infalling observer notice any hawking radiation at all? Or does it effect only external observers?

baronm1100,

DeleteIt can be vice versa: external observers see no Hawking radiation but infalling bodies really see full of it that could be the extra cause with the degeneration pressure so that no event horizon ever formed. I see the value of Mersini-Houghton's work.

Sabine, thank you so much for this post.

ReplyDeleteI love how you follow Einstein's idea of "things should be made as simple as possible, but not any simpler".