- The Photoelectric Effect
Light falling on a metal plate can lead to emission of electrons, called the "photoelectric effect". Experiments show, for this to happen the frequency of the light needs to be above a threshold depending on the material. This finding was explained in 1905 by Albert Einstein who suggested that the light should be thought of as quanta whose energy is proportional to the frequency of the light, the constant of proportionality being Planck's constant. Einstein received the Nobel Prize in 1921 "for his services to Theoretical Physics, and especially for his discovery of the law of the photoelectric effect."
Recommended reading: Our post on the Photoelectric Effect and the Nobel Prize speech from 1921. - The Casimir Effect
This effect was first predicted by Hendrik Casimir who explained that as a consequence of quantum field theory, boundary conditions that can for example be set by conducting (uncharged!) plates, can result in measurable forces. This Casimir force is very weak and can be measured only at very small distances.
Recommended reading: Our post on the Casimir Effect and R. Jaffe's The Casimir Effect and the Quantum Vacuum. - The Doppler Effect
The Doppler effect, named after Christian Doppler, is the change in frequency of a wave when the source moves relative to the observer. The most common example is that of an approaching ambulance, where the pitch of the signal is higher when it moves towards you than when it moves away from you. This does not only happen for sound waves, but also for light and leads to red- or blueshifts respectively.
Recommended reading: The Physics Classroom Tutorial. - The Hall Effect
Electrons in a conducting plate that is brought into a magnetic field are subject to the Lorentz force. If the plate is oriented perpendicular to the magnetic field, a voltage can be measured between opposing ends of the plate which can be used to determine the strength of the magnetic field. First proposed by Edwin Hall, this voltage is called the Hall voltage, and the effect is called the Hall effect. If the plate is very thin, the temperature low, and the magnetic field very strong, a quantization of the conductivity can be measured, which is also known as the quantum Hall effect.
Recommended reading: Our post on The Quantum Hall Effect. - The Meissner-Ochsenfeld Effect
The Meissner-Ochsenfeld effect, discovered by Walther Meissner and his postdoc Robert Ochsenfeld in 1933, is the expulsion of a magnetic field from a superconductor. Most spectacularly, this can be used to let magnets levitate above superconductors since their field lines can not enter the superconductor. I assure you this has absolutely nothing to do with Yogic flying.
Recommended watching: Amazing Physics on YouTube. - The Butterfly Effect
"A legendary butterfly flapping its wings in Rio changes the weather in Chicago. (I have lived in Chicago and personally suspect that nothing can change the weather there.) It always appears to be the same butterfly whenever anyone tells of this example. One would think it possible to imagine, by a vast conceptual leap, some other example. Maybe a moth in Omaha, perhaps, or a starling in Sheboygan. Whatever winged creature is responsible, the point is that any small change in a chaotic system can, and typically does, have large and amplifying effects. Thus this sensitivity implies that the detailed initial conditions - how fast, at what angle, and precisely how the starling flapped its wings - would have to be known to infinite precision to predict the result."
- The Hawking Effect
Based on a semi-classical treatment of quantum fields in a black hole geometry, Stephen Hawking showed in 1975 that black holes emit thermal radiation with a temperature inverse to the black hole's mass. This emission process of the black hole is called the Hawking Effect. This result has lead to a great progress in understanding the physics of black holes, and is still subject of research.
Recommended reading: Black Hole Thermodynamics by David Harrison and P.K. Townsend's lecture notes on Black Holes. - The Zeeman Effect/Stark Effect
In the presence of a magnetic field, energy levels of electrons in atomic orbits that are usually degenerated (i.e. equal) can obtain different values, depending on their quantum number. As a consequence, spectral lines corresponding to transitions between these energy levels can split into several lines in the presence of a static magnetic field. This effect is named after the Dutch physicist Pieter Zeeman, who was awarded the 1902 physics Nobel prize for its discovery. The Zeeman effect is an important tool to measure magnetic fields in astronomy. For some historical reasons, the plain vanilla pattern of line splitting is called the Anomalous Zeeman effect.
A related effect, the splitting of spectral lines in strong electric fields, is called the Stark Effect, after Johannes Stark.
Recommended reading: HyperPhysics on the Zeeman effect and the Sodium doublet. - The Mikheyev-Smirnov-Wolfenstein Effect
The Mikheyev-Smirnov-Wolfenstein effect, commonly called MSW effect, is an in-medium modification of neutrino oscillation that can for example take place in the sun or the earth. It it a resonance effect that depends on the density of the medium and can significantly effect the conversion of one flavor into another. The effect is named after Stanislav Mikheyev, Alexei Smirnov and Lincoln Wolfenstein.
Recommended reading: The MSW effect and Solar Neutrinos. - The Sunyaev-Zel'dovich Effect
The Sunyaev-Zel'dovich effect, first described by Rashid Sunyaev and Yakov Zel'dovich, is the result of high energy electrons distorting the cosmic microwave background radiation through inverse Compton scattering, in which some of the energy of the electrons is transferred to the low energy CMB photons. Observed distortions of the cosmic microwave background spectrum are used to detect the density perturbations of the universe. Dense clusters of galaxies have been observed with use of this effect.
Recommended reading: Max Planck Society press release Crafoord Prize 2008 awarded to Rashid Sunyaev and The Sunyaev-Zel'dovich effect by Mark Birkinshaw. - Bonus: The Pauli Effect
Named after the Austrian theoretical physicist Wolfgang Pauli, the Pauli Effect is well known to every student of physics. It describes a spontaneous failure of technical equipment in the presence of theoretical physicists, who should therefore never be allowed on the vacuum pumps, lasers or oscilloscopes.
Recommended reading: Our post Happy Birthday Wolfgang Pauli.
TAGS: PHYSICS, EFFECTS
Regarding (1) and (2): Conductive surfaces have positive work functions into vacuum and therefore "behave." Diamond [100]-(2×1):H has a negative work function into vacuum. Boron p-doping affords large electrical conductivity,
ReplyDeleteWould diamond surfaces be naughty in (1) and (2)?
If I may...
ReplyDeleteThe Aharonov-Bohm effect (or Ehrenberg-Siday-Aharonov-Bohm effect): a charged particle is affected by EM fields in regions from which the particle is excluded.
The Sachs-Wolfe effect: gravitational redshifting of CMB photons
The Kapitza-Dirac effect: diffraction of a well-collimated particle beam by a standing wave of light.
(I think observing such a phenomenon as the latter should figure on my list of "to do or see physics experiments"! Life is so cool in the labs, for a theorist at least, seriously. I usually go to labs for a break from string theory or cosmology... it's totally refreshing and inspiring, and sometimes loads of damn good thrills!)
And wikipedia has its own list! Aaah.. the wikipedia effect! ;-)
What about "Special [lack of physics] effects"?
Ah, how could I forget about Aharonov Bohm! I considered SW, but it didn't score too highly.
ReplyDeleteI totally didn't know about the Wiki-list!
Either way, as in most cases there is way too much information on that Wiki page to be digestible. So I hope the above is an interesting selection.
Thanks,
B.
I just want to say that I really appreciate this kind of post, reminding me of basic physics stuff that is really quite interesting but that I'm not thinking about very often.
ReplyDeleteBy the way, do you have any feeling for the ratio of professional physicists among your readers? It could be fun to know how many really interested lay persons there are out there who follow blogs like this.
Hi Aka,
ReplyDeleteGlad you like it. I am afraid I can't quite answer your question about the ratio of professional physicists. For one, there are page views (on the average currently ~1,400 per day) and then there are visitors (~ 1,000 per day). A large fraction of these visitors comes by referral from one or the other search engine, and are one time visits. On the other hand, the search engine traffic also covers Stefan and me in many occasions, and probably others since we use the Google search box, see sidebar. I have no idea how large the number of returning frequent readers is. If I look at the IP address tracker, on a usual workday the fraction of addresses ending on .edu is around 5-10% or so. I first found that surprisingly high (at least 1/5th of the addresses is just 'unknown') but it probably just reflects how small the network around science blogs really is. One thing I can tell you for certain is that over the last years, I've again and again met people who tell me they read my blog. It's a bit weird sometimes actually. If I'm writing a post I don't think much about who is going to read it, then I realize at some point hundreds of people must have scratched their head about one of my more silly jokes. Either way, the visitor distribution doesn't reflect in the comments though. I am afraid, the blogosphere is still viewed with a certain suspicion by many people in the community. Which is sad, but not without a reason. It's still not clear to me where this is going to lead us.
Best,
B.
Re: the Pauli Effect. Wasn't the same effect attributed to Rutherford?
ReplyDeletei feel pity on idiots that havent heard of these.
ReplyDeletegood post, bee
A
Wasn't Rutherford a great experimentalist?
ReplyDeleteHow about the Quantum Zeno Effect.
ReplyDeleteGood choices, including the other ones like the weird Aharonov-Bohm effect. The Berry phase is interesting too, albeit not called an "effect" (but it is, anything that happens such and such way is, right?) I think that although the Unruh effect is related to Hawking radiation it is interesting to mention separately because it involves all acceleration (and thus all gravity fields, right, not just event horizons.)
ReplyDeleteWhat bugs me about relating Hawking radiation to Unruh radiation is this: The pop explanation I hear of HR is that virtual pairs are split in effect at the event horizon, and one goes on beyond the BH while the other falls back in. (Well, that's already inadequate I'm sure.) But if you are a Rindler observer, you define an equivalent to an "event horizon" at R = c^2/a. But that plane is normal space to the rest of us. It doesn't make sense for an objective event of virtual pair splitting to take place, and something really going off in one direction and another particle in the other direction.
The reality of it is explained somewhat in the Wikipedia article, so I'm not saying there's a real problem - maybe it just shows the inadequacy of the pop explanations anymore for the really offbeat stuff?
Hi Bee,
ReplyDeleteI must say I’ve been much affected by your effects:-) Many I’m familiar with while a few I will have to track down. Two of those of which I’m familiar are the Butterfly and Hawking effects. I find these interesting in seeing them together, since the first relates to chaos theory and the other as quantum theory in relation to black holes and entropy. That is the first although deterministic is however often too complex to predict and subject to wide variances and the other although nondeterministic (main stream view) is generally easier to predict within a more normally narrow range. I’ve always have found this interesting, since it demonstrates that while phenomena may be deterministic and causal, can render a larger practical uncertainty then that of quantum consequence. This then is not what many would consider as possible or as they understand.
Best,
Phil
Neil, It's been a long time since I read about this, but here's a couple of papers you might find interesting. Firstly, H. Dieter Zeh considering Hawking and Unruh radiation, and information loss:
ReplyDeletegr-qc/0507051
It says something like there is nothing special about black holes (your point). There is no information loss - it's just normal quantum decoherence.
Also Carlo Rovelli's paper on relational quantum mechanics:
quant-ph/9609002 which explains how *every* observer has a hidden region (inside a black hole, or otherwise) which might be normal space to another observer (your point again) so quantum behaviour can only be defined relative to the observer. "It doesn't make sense for an objective event of virtual pair splitting to take place" - as you say. This is all covered very well in Lee Smolin's book "Three Roads to Quantum Gravity".
Is the "Hawking effect" really an effect?
ReplyDeleteI always thought that effects are the (important or strange) results of experiments.
Is there any experimental/astronomical
proof for the radiation of black holes?
Regards
Georg
Hi Neil, Andrew, Georg,
ReplyDeleteI've thought for some while to maybe write a post about the Hawking effect, so I should probably do that at some point.
Georg, no there is no experimental confirmation that astrophysical black holes evaporate. I don't think though the word 'effect' has a precise definition for in which context it is allowed to be used. It is the same with many other words like e.g. 'principle', over the use of which one can discuss endlessly. It's just become to be known as the Hawking Effect, that is why it is on the list.
Neil': Yes, the pop explanation with pair production on the horizon is very misleading (but so are many other pop explanations). It is misleading e.g. because the typical wavelength of the emitted particles is the inverse temperature, and the temperature is the inverse of the radius (this holds in all dimensions. In 3+1 dimensions the radius is further proportional to the mass). This means the typical wavelength of the emitted 'particles' is about the horizon size, so they aren't very well localized, and it's more appropriate to talk about waves not particles to begin with. It is further misleading because one is immediately lead to ask where the energy for these particle pair 'comes from' (which ends in lengthy discussion about the energy definition of the grav. field itself), or resp. how it can be that the one particle falling in lowers the energy of the hole (which leads to a discussion about killing vectors), and so on. (In Hawking's calculation the backreaction (energy loss of the hole) was put in by hand later, i.e. the outgoing radiation doesn't actually carry any energy away). It is one of these cases where I find the pop explanation causes so many secondary problems that don't fit that I wonder whether one should use it at all.
Regarding the question of horizon, if you want to insist on the analogy to Unruh (which I don't like and this is why the Unruh effect isn't on the list), then you'll find a horizon also for the Rindler observer: he is constantly accelerating, so 'runs away' from some signals that will never reach him. It's the null asymptotics that act as a horizon in this case.
Andrew: There is something special about black holes in classical GR: that is the singularity. It's the singularity that causes the inf. loss problem, not the horizon.
Best,
B.
Hello,
ReplyDeleteTo theoreticalminimum: to observe the Kapitza-Dirac effect - at least in its initial formulation, e.g., electrons waves being difracted by a grid of standing light waves - is quite hard and was accomplisehd only a few years ago:
http://www.nature.com/nature/journal/v413/n6852/full/413142a0.html
It is a good news for your list of "to see physics experiments", at least if you can find a brave experimental to reproduce this one for you :-)
Please, Bee, clarify: you do not like the Unruh effect or the way people use it to connect to Hawking radiation? If the first option, please, why?
Thanks,
C.M.
Bee said "... if you want to insist on the analogy to Unruh (which I don't like and this is why the Unruh effect isn't on the list),..."
ReplyDeleteWhat exactly is it you don't like? Do you not like the Unruh effect itself? (Do you think it's wrong? Do you think it's just wrongly explained?) Or is it just the analogy between Hawking and Unruh effects that you don't like? I thought the equivalence principle meant that the Hawking effect was subsumed into the Unruh effect. How wrong am I?
Aaah, what have I done by just stating my disliking? I also don't like Sushi, Brad Pitt's face or the pity state the global economy is in. Yet usually nobody cares. My disliking of the Unruh effect falls into the category of 'things I believe but can not prove', or in this case 'things I don't believe but can not prove'. I am not saying the Unruh effect is 'wrong' in the sense that the calculation is wrong. I've done it myself. It just rests on an observer-dependent definition of particles (the annihilation/creation operators), that I don't know why they should be physical, i.e. my disliking is one of interpretation. The Unruh effect takes place in flat space, there IS a preferred set of observers that can be used, these are the not-accelerated ones. With the Hawking effect, the situation is different - here the background is really curved. I grant that the Unruh effect serves as an analogy to the Hawking effect by replacing the acceleration with the surface gravity and babbling something about the equivalence principle. But this completely neglects the fact that the time-dependence of the background is a central ingredient in Hawking's calculation. Not sure how satisfactory this explanation is, but this is why I don't like it. Best,
ReplyDeleteB.
"Andrew: There is something special about black holes in classical GR: that is the singularity. It's the singularity that causes the inf. loss problem, not the horizon."
ReplyDeleteHi Bee, I'm not convinced about that. My understanding is that information loss happens with Unruh radiation (ie., it's just thermal) just as with Hawking radiation - because each effect depends on hidden regions. Like I say, Lee Smolin's book has four great chapters devoted to this - firstly on hidden regions: "All observers have their own hidden region. The hidden region of each observer consists of all those events that they will not be able to receive information from, no matter how long they wait. Each hidden region will include the interiors of all the black holes in the universe, but there may be other regions hidden as well." So nothing special about black holes or singularites. This is re-emphasized when discussing Unruh radiation: "Each photon detected by our accelerating observer's thermometer is correlated with one that is beyond her horizon. This means that part of the information she would need if she wanted to give a complete description of each photon she sees is inaccessible to her, because it resides in a photon that is in her hidden region. As a result, what she observes is intrinsically random ... Information about the exact states of the hot photons seen by the accelerating observer is missing because it is coded in the states of the photons in her hidden region. Because the randomness is a result of the presence of the hidden region, the entropy should incorporate some measure of how much of the world cannot be seen by the accelerating observer."
I feel very strongly that there is nothing so remarkable about black holes - as Neil points out in his comment and that Zeh paper I referenced - there are observer-independent hidden regions all over the universe. And information loss can occur elsewhere - not just in black holes. This is very clearly stated in that Zeh paper: "Even the global vacuum contains entanglement between both sides of a horizon, locally giving rise to thermal Hawking or Unruh radiation (that is, with or without a singularity) ... In principle, there is then
no difference to conventional decoherence (without black holes)." I think these kind of thoughts really downplay the importance of black holes as regards information loss and thermal radiation - hidden regions are everywhere.
Hi Andrew,
ReplyDeleteIt comes as no surprise for me that you are not convinced, as I have had this discussion numerous times with many people and is has been nothing but frustrating. So I am somewhat unhappy of being pulled into a remake of that fruitless exchange.
I think I read Lee's book but I actually can't recall what he said about Unruh or Hawking (neither can I recall where the book is), but to use the quotation you provide
"Each hidden region will include the interiors of all the black holes in the universe, but there may be other regions hidden as well."
So nothing special about black holes or singularites.
I am not sure you correctly identify the reason why people are talking about the information loss problem extensively. As long as the black hole is there, there is no problem. After all, the information could just be in the 'hidden region' or whatever. But Hawking tells us the black hole evaporates. Yet, there is no information in the Hawking evaporation (except for the temperature). It however leads the black hole to shrink and finally decay. Thus at some point the horizon has to be gone, and so the singularity, no more hidden regions. Where did the information go and how did it come out? That's the problem. (One can argue that in the late stages quantum gravity effects become relevant and spoil Hawking's semi-classical treatment, but then there isn't enough time left to emit all information, so that doesn't help either. Some people argue there could be a remnant left that keeps the information, but this is commonly considered a very unappealing scenario for it comes with other difficulties.)
I certainly find black holes remarkable, but I am not sure in how far this is relevant.
Best,
B.
Oh, criminiy, don't tell me about 'effects'. In engineering/ programming, one time honored way of taking over someone else's presentation is to make up some bullshit effect and throw it into the discussion.
ReplyDelete"Yes, but have you factored in the 'paper-doll effect'?"
"Uh, what's the paper-doll effect?"
"I'm glad you asked! If you consider the blah blah blah..."
There's a special circle in hell for people who do that.
That's funny, I haven't heard of that tactic before. What I have encountered is the tendency to give a problem a name, and then say yeah-yeah but that is just the well known soandso problem. As if that would make things better. Maybe I should make up some effects. The blog recurrence effect for example which occurs in blog threads where the comment section exceeds 100 contributions, upon which the discussion repeats because nobody reads the previous comments.
ReplyDeleteHi Bee,
ReplyDeleteMy intention was not to do a check of your opinions. I do like Sushi, but I have not a problem with your dislike of it.
I've asked you to clarify that statement just because the idea of someone not liking a physical effect is... unusual.
Anyway, concerning the Unruh effect, it is true that you may select a special class of observers in flat spacetime (the inertial ones). But it does not mean you should select their *point of view* as special, in the same way my opinion on Sushi does not select me as better than you, right?
So, let's take an event in spacetime, one that must be explained by whatever observer you want. I choose this event to be the decay of an accelerated proton. So let's suppose I remain seated in my (inertial) lab while I see a proton being brutally accelerated (and, apart from the proton, I see only vaccuum). I can very well explain the decay of this proton by my standard QFT-in-flat-spacetime methods. [A side comment: remember that *inertial* protons are stable, but accelerated ones may well decay]. But now... let's go to the frame of a bee who happened to be glued to the proton. This bee views the proton always stopped in her frame, and someone told her that protons should not decay while stopped! Very well then, but we know that the proton decay is a spacetime event that everyone must agree, just like a bomb explosion. So, how this bee will agree with me? There comes the Unruh effect: Bee, in her accelerated frame, is not just seeing the proton, but also a thermal radiation surrounding her... and it happens that eventually a particle from this thermal bath will hit this proton so hard that will break it.
When both observers (me and bee) compare our measurement of the proton lifetime (taking in account, of course, our different clock rates), we will finally agree. And only because of the Unruh effect, otherwise bee would never see that proton decaying.
My point is that you can not base your dislike of the Unruh effect in your dislike "of interpretation". This is a good physical effect, and as such is not based on personal preferences. Still, you know, who am I to say what you should like or not?
My best regards!
C.M.
P.S: sorry for the long comment. I tried to make it accessible to the lay reader.
Hi Clovis,
ReplyDeleteFirst, how do you accelerate the proton in flat space. No field there to do so. Second,
But now... let's go to the frame of a bee who happened to be glued to the proton. This bee views the proton always stopped in her frame, and someone told her that protons should not decay while stopped!
The bee will hopefully also know how to understand 'stopped', i.e. at rest with respect to an inertial frame. It is very clearly possible to distinguish with measurements in flat space an accelerated frame from a non-accelerated one. There is no reason why the bee should expect to see the same as the not-moving observer, it is not a (global) Lorentz transformation (under which the vev is invatiant). Best,
B.
it just occurred to me the central point of your argument is But it does not mean you should select their *point of view* as special, in the same way my opinion on Sushi does not select me as better than you, right?. what SR tells you (and Unruh is SR) is that all inertial observers are equally 'unspecial', not also the accelerated ones (this is also the cause of confusion with the twin paradox, you need to accelerate one of both to get them back together which breaks the symmetry, thus no paradox.) Best, B.
ReplyDeleteHi Bee,
ReplyDeleteGood point: how do we accelerate the proton? The picture I gave was somehow assuming that you had embarked it in an accelerating spaceship, just like it is usual in Gedankenexperiments in relativity. If you want to make life harder, you can use a electric field, since the proton has charge. Calculations get tougher, but eventually you will find the same result: you need to take the Unruh thermal bath in account in the accelerated frame, in order to have results consistent with the inertial observer.
And, yes, of course the bee will have means to notice she is no more inertial. So she knows that the proton can now, maybe, decay. But, in the spaceship-picture, how is she going to explain that? This way: she will also see an Unruh thermal bath that will interact with her proton.
Remember, please, that Unruh effect is not only about SR, it is quantum field theory. Special relativity teachs us how the concept of time and space are relative. What the Unruh effect teachs us is that somehow the concept of particle is also relative. For this very reason I would have included it in my 10-effecs-list. But there again, this is personal preference, which neither of us want to touch.
Clovis
Hi Bee,
ReplyDeleteYou said: "So I am somewhat unhappy of being pulled into a remake of that fruitless exchange."
That's not very friendly! I thought I was talking to Lubos Motl for a moment! (only joking!) But this is one of the most promising, unexplained, and important questions in physics today - I don't see how it can be a "fruitless exchange". It's possibly the most fruitful area.
You say: "As long as the black hole is there, there is no problem. After all, the information could just be in the 'hidden region' or whatever. But Hawking tells us the black hole evaporates. Yet, there is no information in the Hawking evaporation (except for the temperature)."
Well, this idea of quantum "information loss" is the dubious part. The idea is that in moving from a quantum pure state to a mixed state (explained so well here) we have some kind of information loss - essentially moving from a quantum superposition described by a wavefunction to a **classical** mixed state. But does this necessarily represent a loss of information? That's where Zeh's paper is so useful, because he realises that moving from a superposition state to a classical mixed state is just what happens in everyday quantum decoherence: "In principle, there is then no difference to conventional decoherence (without black holes)". So this whole idea of information loss is possibly over-hyped.
(I feel strongly about this because I can't shake the feeling that physicists are desperately trying to wring every last drop of theory out of black hole thermodynamics - because it's one area in which quantum gravity seems to be making some progress, some kind of promise. But in some cases - as in all this "information loss" hype - I think they go too far. There's nothing so remarkable about black holes - they're just another form of hidden region. And the apparent "information loss" is again nothing special - explained by everday decoherence.)
Hi Andrew,
ReplyDeleteIf you can imagine and ASCII equivalent of vomiting on the keyboard, please visualize you just read it. It is far worse being compared to Lubos than being insulted by him, and it's not funny even if you add it's a joke. And since you brought it up, has it maybe occurred to you (or other readers) that I am increasingly reluctant to write about or even mention anything but the majority opinion on physics questions (like e.g. the Hawking and Unruh effect) because I risk that Lubos picks it up, and he and his followers giddily call me names without even trying to find out what I was saying and why (or whether it was actually my opinion I was talking about or maybe somebody elses etc)?
But to come back to the topic, I think you are disagreeing with yourself. You start by saying "this is one of the most promising, unexplained, and important questions in physics today" and end by explaining "in all this "information loss" hype - I think they go too far. There's nothing so remarkable about black holes". Not that it's actually relevant, but it would help me to understand your motivation if you could maybe clarify your point of view.
I apologize, but I currently don't have the time to read John's text, or the paper you mention. I will do so at some point and come back to it. Yes, one can also approach the problem by asking for the meaning of 'information' to begin with, in case that was what you were saying.
Either way, if I read your last paragraph it seems that we share the same sense of this issue has been blown up above proportion. It has certainly generated a vast amount of very creative approaches to tackle the problem. Best,
B.
Hi Clovis,
ReplyDeleteRemember, please, that Unruh effect is not only about SR, it is quantum field theory. Special relativity teachs us how the concept of time and space are relative. What the Unruh effect teachs us is that somehow the concept of particle is also relative.
What I was trying to say in my first comment is that the concept of a particle for the Unruh observer is a matter of interpretation, and it's not a question you can settle with thought experiments. In QFT we make a mode expansion into plane waves, the coefficients are the annihilation and creation operators that define the particle content. For the Unruh effect you make a different expansion, corresponding to the coordinate system of the accelerated observer (the Rindler coordinates). This way you have different annihilation/creation operators, thus a different particle definition, and if you do the maths, you'll find that the resting observer's vacuum is a thermal heat bath if you decompose it in the particle picture you'd get from using an expansion in the Rindler modes. That's what the Unruh effect tells you. It does not tell you though whether these operators actually do correspond to physical particles, and what I've tried to tell you above is that you can't argue SR 'teaches' you that, since they aren't both inertial observers and you can't claim an equivalence between them. Since nobody has ever measured this, it is so far a matter of interpretation: does this expansion correspond to physically 'real' particles or not? I happen to believe if you accelerate your spaceship fast enough and measure your vacuum there wouldn't be any Unruh effect. Unfortunately, everybody I've ever talked to disagrees with me. If you have an argument that I haven't yet heard, I would be grateful to be cured from my constant frowning of the foreheard, and happily join the enlightened.
I've asked you to clarify that statement just because the idea of someone not liking a physical effect is... unusual.
I am not disliking a 'physical effect', so far it's a merely 'mathematical effect'.
If you want to make life harder, you can use a electric field, since the proton has charge.
As I tried to indicate above an electric field carries energy and thus spacetime would no longer be flat. But besides this, if you have a field to accelerate the proton, QFT tells you you have a background of a photon-soup from which you can extract energy and create particles etc. It's a completely different question to ask what a charged particle will do when moving through an electric field than asking what an accelerated observer would see in 'vacuum'. I want to point out once again that the situation for the Hawking effect is different because there is a time-dependent and not time-inversal-invariant background.
Best,
B.
Hi Bee,
ReplyDelete"What I was trying to say in my first comment is that the concept of a particle for the Unruh observer is a matter of interpretation,"
- Yes, I see, and to convince you of the contrary was my initial motivation. But do not worry, I'll soon give up.
- You have good points concerning the way we accelerate the proton. The reason I do not want to enter in all the details is that they take too much time to give us the same answer. There are a lot of thing to worry about once you have a strong electric field. After you deal with them, the part of the cross section that is due to the acceleration radiation will be very tiny. Nevertheless, you will only get consistent results between the accelerated and inertial observers when you consider the Unruh thermal bath (even though the difference may be tiny due to the so much bigger influence of other effects).
But the lesson remains: without Unruh effect, the accelerated observer can not agree with the inertial one. So, if you do not believe in the Unruh effect (as a real physical effect), you certainly have a problem of lack of consistency in plain and standard quantum field theory.
- But, if you can still sleep well with that, of course we will agree that people have not measured it - at least not in definite ways. And, if not measured, no one gets the prize, right?
There are somehow indirect effects which people claim to have measured, e.g., the spin-flip of electrons in accelerator rings, but the Unruh thermal bath by itself is still waiting.
- I would very much like to continue discussing it. It is a fascinating topic. But the space and method here is too slow. So, unless we meet in some conference, where 15min talking would be worth more than 3 hours of blog-typing, my best shot is to point you to this paper, to appear in RMP:
http://arxiv.org/abs/0710.5373
My best regards,
Clovis
Bee,
ReplyDeleteBy the way, as you very probably know, Unruh himself is in the same country as you. Email or call or visit him. It is almost a lost of time to be discuss his effect in a blog when you have him so near :-)
Clovis
Hi Clovis,
ReplyDeleteThanks for the reference, which looks interesting. I am genuinely sorry, but currently don't have the time to read 53 pages. In case you are referring to a specific calculation, it would greatly help if you could be more specific. Regarding the charged particle in the electric field: I always understood the Unruh effect to be a vacuum effect. If there is an electric field, it is not a vacuum, so I don't actually know what we are arguing about. You say you will only get consistent results between the accelerated and inertial observers when you consider the Unruh thermal bath , which is an argument I can't follow without a derivation. To derive the result for the 'accelerated observer' what was the field decomposition used there? I find it hard to believe one can argue the problem of a particle definition away by considering different observables like e.g. a cross-section instead of a number count.
as you very probably know, Unruh himself is in the same country as you. Email or call or visit him. It is almost a lost of time to be discuss his effect in a blog when you have him so near
As you very probably know, Canada is a very large country, and Vancouver is as near as 4,000 km. Besides this, I believe I've met Unruh several times (he'd probably not recall it, but it's hard to ignore him). The Unruh effect isn't exactly my main interest area, and the time I am willing to spend in arguing about it is rather limited.
Best,
B.
" I happen to believe if you accelerate your spaceship fast enough and measure your vacuum there wouldn't be any Unruh effect. Unfortunately, everybody I've ever talked to disagrees with me."
ReplyDeleteI agree with you. :-) Unfortunately I have nothing clever to say about this, because the reasons for my scepticism seem to be exactly the same as yours....
Hi Bee,
ReplyDelete(I know you get terrible grief from Lubos, please keep just ignoring it all).
You said: "But to come back to the topic, I think you are disagreeing with yourself. You start by saying 'this is one of the most promising, unexplained, and important questions in physics today' and end by explaining 'in all this information loss hype - I think they go too far'. There's nothing so remarkable about black holes'. Not that it's actually relevant, but it would help me to understand your motivation if you could maybe clarify your point of view."
I'd love to explain myself! I think this topic is so important because once we start analyzing black hole information (entropy) we get results such as the holographic principle (which I find fascinating) and how that leads to adS/CFT - very fruitful indeed! But when you consider someone like Stephen Hawking who has spent virtually his whole career analyzing black hole thermodynamics, and now we're getting new results out such as the holographic principle, I just wonder how much theory is left to be pulled out of analyzing black holes. When are these guys going to hold up their hands and say "OK, boys (and girls), we've analyzed black holes to death, now let's move on to something else." I can see the attraction of continuing to analyze black holes because it seems to be one area in which quantum gravity is making progress, but I fear we're getting to the stage of over-analysis, and all this fuss about apparent "information loss" seems to be a classic case of over-analysis - imagining a problem when there is none.
So moving on to consider apparent information loss, does this sound right...
Let's imagine we have an entangled pair of photons at the black hole event horizon. One gets sucked into the hole, while the other gets emitted as Hawking radiation. The general view is that entanglement between these particles is lost, hence the resultant radiation is uncorrelated (thermal). Now let's imagine an observer that travels with the particle that enters into the black hole. From his point of view, he can still receive information about the other photon outside the black hole! There is nothing stopping that information from reaching him (this shows how "hidden regions" must be defined relative to observers). From his point of view, he can see that the particles are still entangled - there is no "information loss" for him. But for an observer outside the black hole, the radiation appears thermal, though in reality that emitted particle is actually correlated with the particle inside the black hole about which we can obtain no information. So "for all practical purposes" - as Zeh says - the radiation is random. But it's not truly random - it actually remains correlated with something about which we can never obtain information. So there is probably no "information loss" - it's all a fuss about nothing (or am I totally wrong in my analysis?)
As just stated, for the observer which enters the black hole, he sees no "hidden region". Hidden regions are defined relative to the observer. This idea of observers inside and outside black holes having different views of reality is described in this absolutely fascinating New Scientist article. (At the end of that article it stresses that particles inside and outside the black hole could remain entangled.)
Hi Bee,
ReplyDelete"Thanks for the reference, which looks interesting. I am genuinely sorry, but currently don't have the time to read 53 pages. In case you are referring to a specific calculation, it would greatly help if you could be more specific."
- The way you state it, you do not have a problem about any specific calculation, but you doubt the whole concept of the particle content of the theory be dependent of the observer. So I can only urge you to take a look at that review, the authors make an effort to convince the reader of that and to clarify some controversies that make people confused.
Concerning the accelerated proton example, I think you may convince yourself by asking: let me be in the same spaceship the proton is. How would I describe the phenomena?
You should be able to describe this proton decaying in whatever frame you choose, so describe it in the proton frame. Since you claim not to have time nor interest, you can either go the papers (the review gives the references for the proton or other examples), try by yourself, or just be coherent enough for not asking for a derivation and, at the same time, to declare not to be interested.
Good luck,
C.
Hi Clovis,
ReplyDeleteI think we're just running in circles, so I think we better leave it at this. I have already explained you that if you're sitting in the spaceship you clearly are an accelerated observer. You can measure this. You do not expect an equivalence. If you have a particle in rest with the spacecraft, you would not expect it to behave as one in rest in an intertial frame. You can take all your observables that you have and make the trafo into Rindler coordinates using the usual tensor transformation law, yet that doesn't change the particle content. For this you need to do more, namely changing the basis in which you decompose your quantum field, make the Bogoliubov trafo etc - we've had that before and it seems you know the details of the calculation, so I think it is unnecessary to be more specific. What I am trying to tell you is not that the calculation is wrong, but that it rests on an additional assumption for which there is no experimental evidence whatsoever. I will give the paper a read at some point. Thanks,
B.
Hi Dr Who,
ReplyDeleteWow, a first :-) What stuns me about this admittedly is how little disagreement there is on the question. It makes me feel like I must be completely stupid, but whenever I look at the calculation its based on the same assumption that people just seem to take for granted? And if I start arguing with somebody they will come with all kinds of elaborated thought experiments and detector constructions and so on, yet, the underlying assumption is still the same, no matter how many layers you shovel over it. Either way, I've had this discussion too many times I think. Best,
B.
Well, Bee, this is the part where people start mumbling about the principle of equivalence. My teacher defined the principle of equivalence as "the belief that it is possible to make the curvature tensor equal to zero by means of a coordinate transformation", and he praised it as being very clear and very falsifiable. That is, clearly false.....
ReplyDeleteHi Dr Who,
ReplyDeleteWell. I can't recall how many times I've come across one or the other abuse of the equivalence principle but this one is news to me. I find it very unfortunate if textbooks motivate the Hawking effect with the Unruh effect, the Unruh paper was even published after Hawking's calculation. The situation for the Hawking effect is completely different: there are two asymptotically flat regions, both have inertial observers that have some vacuum. Yet both vacua aren't identical, so if it was 'empty' initially the endstate will be a thermal heat bath. If you motivate the Hawking effect with the Unruh effect it gives a completely wrong picture of the cause for the effect. Best,
B.
Hi Andrew,
ReplyDeletebut I fear we're getting to the stage of over-analysis, and all this fuss about apparent "information loss" seems to be a classic case of over-analysis - imagining a problem when there is none.
Yeah indeed, I share this fear. The problem with this problem is that there have been so many different attempts to explain it that 95% of the publications on the matter have to be wrong. Unfortunately, there hasn't been any natural selection taking place since nobody has actually seen any black hole evaporating (and I doubt this will be the case in the soon future).
Either way, I am afraid you do indeed get the problem wrong. As I said above, as long as there is a hidden region, there is no problem. If you say (or you say somebody else says or whatever) the radiation is random for all practical purposes but correlated with something we can't see, then please note that the 'something we can't see' ends up in the singularity and finito. The 'hidden' part never comes out, the black hole evaporates, and it's gone - that is the problem. Besides this, forget about quantum mechanics for a moment, it makes matters unnecessarily complicated. Let Alice have a kilogram of lead and a kilogram of iron, and tell her to throw one of both into the black hole. Let Bob measure the radiation that comes out of the black hole, all the way until it's completely evaporated. How can he find out whether it was lead or iron (I'd call that information)?
Best,
B.
So, Dr Who, so refrigeration equipment that works while cruising inertially in Minkowski spacetime will encounter a Unruh radiation thermal bath when accelerated, and not be able to achieve the same low temperature?
ReplyDeleteHi Bee,
ReplyDeleteThanks for your response - I'll make this my last comment! I'm a bit sorry other people haven't commented on this.
You seem to belong to the school that says information is lost in black holes? However, I still don't accept it (and I see Stephen Hawking changed his mind on this subject now believing that information is not lost after all, famously losing his bet - see that New Scientist article).
As you say, if there is no way of telling if it's iron or lead then information would certainly have been lost. I'm not convinced, though, that saying - as you say - that information "ends up in the singularity and finito" is a conclusive explanation as to whether or not information is lost. Quantum gravity theorists don't seem to deal with singularities - saying "gravity becomes repulsive at small distances", avoiding any sort of breakdown of determinism. I also don't think you can ignore the quantum. I'll end by quoting from that New Scientist article I referenced: Since no information loss can occur in a swarm of ordinary quantum particles, there can be no mysterious information loss in a black hole either. "The boundary theory respects the rules of quantum mechanics," says Maldacena. "It keeps track of all the information." Thanks again.
Hi Bee,
ReplyDeleteStill I will comment on two things we do not seem to be speaking the same language.
- "I have already explained you that if you're sitting in the spaceship you clearly are an accelerated observer. You can measure this. You do not expect an equivalence."
When I say that both observers (inertial and accelerated) must agree, I am not saying they should describe the process equally - instead I've been arquing the contrary. Both will describe it very differently. Still, they must agree if the proton have decayed or not. Right?
To do an analogy, since you invoked the twin "paradox" earlier, we know that both twins will, in the end, know who is aging more, and exactly by how many years. So they have consistency, they agree. It is in just the same sense I am talking about consistency in the proton case.
- "What I am trying to tell you is not that the calculation is wrong, but that it rests on an additional assumption for which there is no experimental evidence whatsoever."
Tracing back, I believe your problem is about acceptance that different time-like killing vectors yealds to different particle content. Even though you do not like to compare Unruh and Hawking effects, you look to miss the point that both effects depend on the acceptance of this. That's why people usually compare them.
But enough of "mumbling" about thought experiments here. Sometimes people need to carefully think by themselves and external input will not help anymore.
Best regards,
C.
Hi Andrew,
ReplyDeleteYou continue to misunderstand me. I don't 'belong to the school that says information is lost in black holes' (not sure I belong to any school). On the contrary, I am convinced information is not lost in black holes. I was merely trying to explain why your solution to the information loss problem doesn't even address the problem to begin with. Neither did I say I want to ignore quantum, I brought up the example to clarify that the problem occurs because the information can't come out, rather than it being 'hidden'. You finally tough the point I was trying to make in the very beginning, the problem is the singularity. QG should erase the singularity. Alas, no information loss. Best,
B.
This comment has been removed by the author.
ReplyDeleteI tried looking for the 10 greatest experiments in concert with this blog entry. I have it somewhere on my site.
ReplyDeleteI have enjoyed Andrews contributions to your conversation on the Blackhole Paradox.
Sometimes when people do not see in the same way as another person there is a "potential there" for bringing perspective to an "ole situation."
Sometimes one has to give themself permission and freedoms to explore the "creative aspect of this situation," does not in any way disregard the scientific knowledge of that process.
If you had seen the whole context within a "bulk perspective" how would this have changed what constitutes the "nature of the graviton" as a expression of that bulk? Within context of the nature of the Blackhole?
I am not saying "it is this way," yet, providing a "new framework" is necessary sometimes.
Hi Clovis,
ReplyDeleteWell, I do agree that sometimes people need to carefully think by themselves. To repeat once again: you argue that for the accelerated observer to agree with the non-accelerated one (e.g. on the decay of the proton) you need the Unruh effect, because otherwise it would not be consistent. Your argument is based on saying the accelerated observer should be able to "describe it in the proton frame". This is of course true. But for this to be possible you don't need the Unruh effect. All you need is SR and the tensor transformation. Now tell me exactly which quantity you transform within standard SR into the accelerated frame that does not agree with experiment. The mistake you make is (once again) to assume that the accelerated observer would be puzzled to see the proton decay in his restframe, as a solution you claim the Unruh effect is necessary. I have tried to tell you this is not the case, because the observer hopefully is smart enough to realize he is not in an inertial frame.
Tracing back, I believe your problem is about acceptance that different time-like killing vectors yealds to different particle content.
Yes.
Even though you do not like to compare Unruh and Hawking effects, you look to miss the point that both effects depend on the acceptance of this. That's why people usually compare them.
And that's what I usually dislike. If you check Hawking's calculation you will see that it's for a non-static background. And it's so for a good reason.
Best,
B.
Information is lost on this blog, though:
ReplyDelete"Comment deleted
This post has been removed by the author.
11:47 AM, March 27, 2008"
:)
Anyway, I was wondering about the following detailed balance argument about black holes in thermal equilibrium with thermal radiation. The black holes mass should remain stationary when it is in a box filled with thermal radiation at a temperature equal to the Hawking Temperature.
Now, the Hawking temperature is such that typical photons have a wavelength of the order of the Schwarzschild radius. But this means that the photons have a significant probability of being scattered by the black hole.
So, it seems that black holes should radiate less energy than a black body at the Hawking temperature.
In regards to Andrews link to the article by Amanda Geftner.
ReplyDeleteI was curious as well to how one may actually describe the process of the horizon and work toward that horizon in terms of it's geometry?
Plato:Previously, I left a comment in relation to Susskind's thought experiment about the elephant and Bob on the back of the elephant B moving toward the horizon of the blackhole. My thoughts were about the "entanglement process" and how Alice on the back of elephant A would reveal aspects of the nature of the blackhole as elephant B move closer to that horizon.
Susskind gave it an amazing twist in my view when he added the elephant. I gave the poem the Elephant and Six men for this purpose. As well as, to the nature of your SciBar gathering earlier.
Always open for corrections.
island,
ReplyDeleteAs a blog developer, you have the option to delete your own comments too:)
I saved the author of this site the option to delete a double post. My bad:)
Bee,
ReplyDelete- "But for this to be possible you don't need the Unruh effect. All you need is SR and the tensor transformation."
Wrong. You know, in the Unruh effect the energy-momentum tensor is zero for the inertial guy. It is zero also for the accelerated one, since a zero tensor is zero in whatever frame you transform it. Still, the accelerated guy will see a bath of particles. QFT in curved spacetimes is not only about tranforming tensors. Your transform-the-tensor argument would never give you the Unruh effect. And of course it wouldn't: that tensor (calculated in the inertial frame) is based on the concept of radiance (hence particles) that this inertial observer uses. No new information from that.
- "The mistake you make is (once again) to assume that the accelerated observer would be puzzled to see the proton decay in his restframe, as a solution you claim the Unruh effect is necessary."
Again and again, I've not claimed the accelerated observer will be puzzled. By heavens, he will swear that he saw a particle from the bath to hit the proton, how could he be puzzled? What I say is that he will explain the decay due to interaction with the thermal bath. And I ask: if you do not believe so, it is your job to show how he will describe the decay. Since you are not interested in doing this job, nor in trusting my word, we can not go far here, I'm sorry.
C.
Hi Clovis,
ReplyDeleteYour argumentation is an excellent example for a circular argument, I am beginning to like it.
Wrong. You know, in the Unruh effect the energy-momentum tensor is zero for the inertial guy. It is zero also for the accelerated one, since a zero tensor is zero in whatever frame you transform it. Still, the accelerated guy will see a bath of particles. QFT in curved spacetimes is not only about tranforming tensors. Your transform-the-tensor argument would never give you the Unruh effect.
Yes, that's what I am trying to tell you all the time: The tensor transformation of zero is zero. What is wrong with that? There is nothing inconsistent about it, and no need for an additional effect. If I summarize your above argument which was meant to explain the necessity of the Unruh effect it says: one needs the Unruh effect, because without the Unruh effect one wouldn't get an Unruh effect.
Best,
B.
Hi Count:
ReplyDeleteNot that it is of any actual relevance, but almost all of the 'posts deleted by the author' you see are corrections of typos or likewise. Best,
B.
Bee,
ReplyDelete- "Yes, that's what I am trying to tell you all the time: The tensor transformation of zero is zero. What is wrong with that? There is nothing inconsistent about it, and no need for an additional effect. If I summarize your above argument which was meant to explain the necessity of the Unruh effect it says: one needs the Unruh effect, because without the Unruh effect one wouldn't get an Unruh effect."
Nice. If that is all you have understood of my arguments, I genuinely give up.
Bye,
Clovis
Hi Clovis,
ReplyDeleteWell, now you know why I find this discussion frustrating. Nevertheless, thanks for dropping in anyhow. Best,
B.
Bee,
ReplyDeleteThere are some things you may think you understand, when in fact you have got only a superficial knowledge of them. To recognize so, when it happens, is a good step in order to learn better. It happens to us all, I have came across such a situation so many times. This time I believe it is hapenning to you.
The frustration of this discussion is not related to lack of objective and verifiable statements. It is just lack of a more humble position.
My best wishes,
Clovis
OK here is my interpretation of the Bee-Clovis Unruh effect discussion. Clovis claims the photon decay needs a thermal bath explanation while Bee claims it just needs the math of a thermal bath explanation. It does sound like one of those quantum computer debates where someone like Deutsch is claiming real many-worlds paths for the computations and others say it's just the math of many-worlds-like paths. I like the Deutsch many-worlds interpretation so does that mean I'm on the Clovis side of the discussion?
ReplyDeleteA thermal photon bath suggests a mixed, decoherent whatever state, one point of disagreement 'tween the inertial and accelerated observer. Another point of disagreement seems to be what a thermometer would read. How about entropy?
ReplyDeleteBeen away for a few days, come back to find an interesting post that mentions Hawking effect without Unruh effect, think must comment, find almost none of the other 10 effects mentioned, LOTS of discussion of why Unruh is/not same as Hawking. Less than 100 posts, so I read them all -- but quickly.
ReplyDeleteTo me, the Unruh effect says a thermal state gives the same measurement results as the vacuum state observed by an accelerating observer. In this approach there are no worries about particle interpretations of QFT. There is definition of a thermal state in terms of particles. Instead, I find congenial a discussion in which "Quantum fluctuations" are transformed into "quantum fluctuations plus thermal fluctuations" by moving to a non-inertial reference frame. On this way of thinking, the equivalence principle says that quantum fluctuations and thermal fluctuations should therefore be presented in a manifestly covariant way -- QFT does not, Planck's constant and Boltzmann's constant have rather different roles in the algebra, therefore something may be wrong with the presentation.
To me, this is significant: it is precisely the dissonance between the Hawking and Unruh effects in some interpretations that requires them to be mentioned together. None of the other mentioned effects seem to me to be close to being as much a challenge to existing Physics as Hawking/Unruh.
I think myself that the difference between the Unruh effect and the Hawking effect because of the flat and curved space-time backgrounds, respectively, is not too significant. Simple-mindedly, the mathematics is just that we have to construct a "Fourier" transform in each of the two space-times, in different coordinate systems, which is somewhat harder to do in the Hawking case, and not conceptually identical to the Unruh case, but not, I think, conceptually too distant to think the effects unconnected. Wigner's definition of particles in terms of Hilbert spaces that are invariant under Poincare group transformations does not survive general covariance too well.
So, Bee, a full post on the Hawking/Unruh effect would be interesting, if y-you d-dare.
That should be: There is no definition here of a thermal state in terms of particles.
ReplyDelete