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Comments (44)
Again, you're getting the fundamental and basic physics wrong. You've also evaded my question.
There is no such thing as a perfect whitebody. It is impossible. All those examples you mention are for narrow-band applications. Thermal radiation is wideband and occurs over the entire electromagnetic spectrum.
The piece in the wikipedia article links to papers such as http://arxiv.org/pdf/1109.5444.pdf in which thermal radiation (and absorption) are increased, not decreased!
Greenhouse warming of the Earth is an entirely different issue and I don't see how it's related. The Earth's surface is fairly cold in comparison to the Sun's.
One-way mirrors do not exist. http://web.archive.org/web/20050313084618/http://cu.imt.net/~jimloy/physics/mirror0.htm What are typically called 'one-way mirrors' are really just ordinary two-way partially-reflective mirrors connecting two rooms where one room is significantly dimmed compared to the other.
Well, firstly, you have to cool it down to below 2.7K in the first place. That most certainly requires 'active cooling'. Then you can either let it slowly equilibrate or keep it actively cold. But then you have to consider the carnot efficiency of the cooling system (which dictates energy consumption goes up as e/Tc, where Tc is the temperature of the computer and e is the energy dissipated by the computer). So you have to consider precisely how much energy the computer is going to use at a certain temperature and how much energy it will take to maintain it at that temperature.
EDIT: You've also mentioned in that thread you linked that "Assuming large scale quantum computing is possible, then the ultimate computer is thus a reversible massively entangled quantum device operating at absolute zero." Well, such a computer would not only be fragile, as you said, but it would also be impossible in the strong sense. It is impossible to reach absolute zero because doing so would require an infinite amount of energy: http://io9.com/5889074/why-cant-we-get-down-to-absolute-zero . For the exact same reason, it is impossible to construct a computer with full control over all the atoms. Every computer is going to have some level of noise and eventual decay.
Show instead of tell. I didn't yet answer your question about the initial energy cost of cooling the sphere because it's part of the initial construction cost and you haven't yet answered my questions yet about reflectivity vs emisison and how it relates to temperature.
Says what law - and more importantly - what is the exact limit then? Perfect super-conductivity may be impossible but there doesn't appear to be an intrinsic limit to how close one can get, and the same appears to apply for super-reflection. This whole discussion revolves around modeling technologies approaching said limits.
This helps my case - the incoming radiation is narrow-band microwave from the CMB. The outgoing radiation can be across the spectrum.
If the 'law' can be broken by materials which emit more than the law allows, this also suggests the 'law' can be broken in other ways as in super-reflectors.1
Ok.
If the earth's equilibrium temperature varies based on the surface albedo, this shows that reflectivity does matter and suggests a hypothetical super-reflector shielding for the CMB microwave could lead to lower than CMB temperatures. (because snow covering of the earth leads to lower equilibrium temperatures than a black-body at the same distance from the sun.)
Do you? I'm not clear on that - you haven't answered the earth counter example, which seems to show that even without active cooling, all it takes is albedo/reflectivity for an object's equilibrium temperature to be lower than that of a black body in the same radiation environment. Is there something special about low temps like 2.7k?
Apparently coherence in current quantum computers requires millikelvin temperatures, which is why I'm focusing on the limits approaching 0K. And from what I understand this is fundamental - as the limits of computing involve very long large coherent states only possible at temperatures approaching 0.
If we weren't considering quantum computing, then sure I don't see any point to active cooling below 2.7K. The the energy cost of bit erasures is ~CTc for some constant C, but the cooling cost goes as e/Tc. So this effectively cancels out - you don't get any net energy efficiency gain for cooling below the background temperature. (of course access to black holes much colder than the CMB changes that)
Yes - but again we are discussing limits analysis where said quantities approach zero, or infinity or whatever.
You can trivially prove this for yourself. High-energy gamma rays cannot be completely reflected by matter. All thermal radiation contains some high-energy gamma rays. Thus no material can perfectly reflect thermal radiation. QED.
No it's not. CMB radiation spans the entire EM spectrum. Thermal radiation is almost the exact opposite of narrow-band radiation.
It's not really broken though. It's just that radiation in these materials happens through mechanisms beyond conventional blackbody radiation. A common LED emits radiation far in excess of its thermal radiation. This doesn't mean that Stefan-Boltzmann is 'broken', it just means that an extra emission mechanism is working. A mechanism that requires free energy to run (unlike normal thermal radiation which requires no free energy). And sure enough, if you read that paper the extra mechanism requires extra free energy.
But you can't use an extra emission mechanism to reduce the emitted raditation.
You keep making this same mistake. Thermal equilibrium temperature does not depend on surface reflectivity. https://www.researchgate.net/post/Is_it_possible_to_distinguish_thermal_bodies_in_equilibrium/1
This is a very basic physics error.
It makes no difference what type of computing you're considering. I suggest reading http://arxiv.org/pdf/quant-ph/9908043.pdf
Specifically, the limiting factor is not temperature at all but error rate of your computer hardware, quantum or not. The ultimate limit to efficiency is set by the error rate, not the temperature at which you can cool the system to.
For any system, even exotic? By what law? A simple google search seems to disagree - gamma rays are reflected today, in practice, (albeit with difficulty and inefficiently) by multilayer reflectors.
The vast majority of the energy peaks in microwave frequencies, but fine yes there is always some emission in higher frequencies - practical shielding would be complex and multilayer.
You keep bringing this up, but you can't explain how it applies to some basic examples such as the earth. How can you explain the fact that the temperature of planets such as earth, venus varies greatly and depends mostly on their albedo? Is it because the system is not in equilibrium? Then who cares about equilibrium? It almost never applies.
If the earth/sun system is not in equilibrium, then my hypothetical reflective object somewhere in deep space receiving radiation only from the CMB is certainly not in equilibrium either.
And finally the universe itself is expanding and is never in equilibrium - the CMB temperature is actually decaying to zero over time.
Until I see a good explanation of planetary albedo and temperature, I can't take your claim of "basic physics mistake" seriously.
Read that of course, and I'd recommend some of Mike Frank's stuff over it.1 Obviously the energy cost of bit erasure is the same for all types of computing. Quantum computing is different only in having much lower error/noise/temp tolerances due to decoherence issues.
These are directly linked.
Heat is just thermal noise. And noise and errors are fundamentally the same - uncertainty over states that can explode unless corrected. The error rate for the most advanced computers is absolutely limited by thermal noise (and quantum noise).
This is trivially obvious at extremes - ie the error rate of a computer at 10000K is 100% for most materials. The lowest error rates are only achievable by exotic matter configurations at very low temperatures.
The idealized perfect computer is one with zero entropy - ie ever quantum state stores meaningful information, and every transition at every time step is a planned computation.
Looking at it another way, using devices and transitions larger than the absolute physical limits is just an easy way to do error correction to handle thermal noise.
I still can't understand why you think the Earth system is representative here.... are you asking why the Earth isn't the same temperature as the Sun? Or the same temperature as the background of space? Because if you remove any one, it would equilibrate with the other. But you're proposing to put your system in deep space where there is only the background. If you did that to Earth, you'd find it would very rapidly equilibrate to close to 2.7 K, and the final temperature is irrespective of surface albedo.
Albedo doesn't have any relationship with final temperature. Only speed at which equilibrium is reached.
Again, I don't feel like I have to 'explain' anything here... perhaps you could explain, in clearer terms, why you think it bears any relationship to the system we are discussing?
It's great that you've read those, unfortunately it seems you haven't understood them at all.
Not in the way you probably think. Error rate depends on hardware design as well as temperature. You're confusing a set of concepts here. As errors are generated in the computation, the entropy (as measured internally) will increase, and thus the heat level will increase. If this is what you are saying, you are correct. But the rate of generation of these errors (bits/s) is not the same as the instantaneous system entropy (bits) - they're not even the same unit! You could have a quantum computer at infinitesimally low temperature and it would still probably generate errors and produce heat.
This is really just another way of saying that your computer is not 100% reversible (isentropic). This is because of inevitable uncertainties in construction (is the manufacturing process that created the computer itself a perfectly error-free computer? If so, how was the first perfectly error-free computer constructed?), uncertainties in the physics of operation, and inevitable interaction with the outside world. If you claim you can create a perfectly isentropic computer, then the burden of proof is on you to demonstrate such a system. You can't expect me to take it on faith that you can build a perfectly reversible computer!
I honestly can't understand how you can't understand it. :)
1. Take a spherical body and place it in a large completely empty universe. The body receives zero incoming radiation, but it emits thermal radiation until it cools to zero or something close to that - agreed? (quantum noise fluctuations or virtual particles perhaps impose some small nonzero temp, not sure) Albedo/reflectivity doesn't matter because there is no incoming radiation. Materials with higher emissivity will cool to zero faster.
2. Spherical body in an empty universe that contains a single directional light source that is very far away. The light source is not effected by the body in any significant way and does not prevent the body from emitting radiation. The source and the body will never reach equilibrium in the timescales we care about. The body absorbs radiation according to it's albedo and the incoming flux. The body emits radiation according to temperature and emissivity. It will evolve to a local equilibrium temperature that depends on these parameters.
3. The earth sun system - it is effectively equivalent to 2. The sun is not infinitely far away, but as far as the earth's temp is concerned the sun is just a photon source - the earth has no effect on the sun's temp and the objects are not in equilibrium. This situation is equivalent to 2.
We have hard data for situation 3 which shows that the balance between incoming radiation absorbed vs outgoing radiation emitted can differ for a complex composite object based on alebdo/reflectivity vs emissitivity. The end result is the object's local equilibrium temperature depends on these material parameters and can differ significantly from that of the black body temperature for the same input irradiance conditions.
4. Object in deep space. It receives incoming radiation from the CMB - which is just an infinite omnidirectional light source like 3. The directionality shouldn't change anything, the energy spectrum shouldn't change anything, so it's equivalent to 3 and 2. The object's resting temperature can be lower than the CMB blackbody 'temperature' (which after all isn't the temp of an actual object, it's tautologically just the temperature of a simple blackbody absorbing/emitting the CMB).
So what am I missing here? - seriously - still waiting to see how #3 could possibly differ.
Whatever principle it is that allows the earth's resting temp to vary based on surface albedo can be exploited to passively cool the earth, and thus can be exploited to passively cool other objects.
Google is now good enough that it gets some useful hits for "temperature lower than the CMB". In particular on this thread from researchgate I found some useful info. Most of the discussion is preoccupied with negative temps, but one or two of the replies agree with my interpretation and they are unchallenged:
Rüdiger Mitdank · Humboldt-Universität zu Berlin:
The cosmic Background Radiation is in a very good approximation a black Body Radiation. Every Body which is in a thermal Equilibrium with this Radiation source has this temperature. If you have another Radiation sources, usually hotter bodies like suns, the temperature increases. If due to the surface reflectivity the Absorption is low, the Body temperature approximates to a lower value, that Emission and Absorption are equal. Therefore it might be possible, that bodies consisting of ice or snow and having a clean surface, have a temperature below cosmic Background temperature. This occurs only far away from any other Radiation source. I would look for comets out of our sun system.
The final temp for the earth in the (earth, sun, background) 'equilibrium' does depend on the surface albedo.
I recommended Frank's work because it has the most clear unifying explanations of computational entropy/information. A deterministic computer is just an approximation - real systems are probabilistic (and quantum) and eventually we will move to those models of computation. The total entropy is always conserved, with some of entropy budget being the usable computational bits(qbits) and some being unknown/error bits such as thermal noise (but this generalization can also cover quantum noise). The 'erasure' of a bit really is just intentional randomization.
The idea of a hard error comes from the deterministic approximation, which assumes that the state of every bit is exactly known. In a prob circuit, we have instead a distribution over bit states, and circuit ops transform these distributions.
Uncertainty in the construction can be modeled as a learning/inference problem. Instead of simple deterministic circuits, think of learning probabilistic circuits (there are no 'errors' so to speak, just distributions and various types of uncertainty). As inference/learning reduces uncertainty over variables of interest, reversible learning must generate an equivalent amount of final excess noise/garbage bits. Noise bits in excess of the internal desired noise bit reserve would need to be expelled - this is the more sophisticated form of cooling.
Each device has an IO stream that is exactly bit conserved and thus reversible from it's perspective. Same principle that applies to each local circuit element applies to each device.
The final limitation is incoming entropy - noise from the outside world. This inflow must be balanced by a matching bit outflow from the internal noise bit reserves. This minimal noise flow (temperature) places ultimate limits on the computational capability of the system in terms of SNR and thus (analog/probabilistic) bit ops.
For the moment I'm just going to ignore everything else in this debate (I have other time/energy committments...) and just focus on this particular question, since it's one of the most fundamental questions we disagree on.
You are wrong, plain and simple. Rüdiger Mitdank is also wrong, despite his qualifications (I have equivalent qualifications, for that matter). Either that or he has failed to clearly express what he means.
If it were true that you could maintain an object colder than the background without consuming energy (just by altering surface absorption!), then you could have a free energy device. Just construct a heat engine with one end touching the object and the other end being a large black radiator.
Yea - several examples from wikipedia for the temperature of a planet indicate that albedo and emissivity can differ (it's implied on this page, directly stated on this next page).
Here under effective temperature they have a model for a planet's surface temperature where the emissivity is 1 but the albedo can be greater than 0.
Notice that if you plug in an albedo of 1 into that equation, you get a surface temperature of 0K!
The generalized stefan boltzmann law is thus the local equilibrium where irradiance/power absorbed equals irradiance/power emitted:
Ja = Je
Ja = Jin*(1 - a)
J_e = eoT^4
T = (J_in * (1-a) / (e*o)) ^-4
J_in is the incoming irradiance from the light source, a is the material albedo, e is the material emissivity, o is SB const, T is temp.
This math comes directly from the wikpedia page, I've just converted from power units to irradiance. replacing the star's irradiance term of L/(16 * PI * D^2) with a constant for an omni light source (CMB).
On retrospect, one way I could see this being wrong is if the albedo and emissivity are always required to be the same for a particular wavelength. In the earth example the albedo of relevance is for high energy photons from the sun whereas the relevant emissivity is lower energy infrared. Is that your explanation?
Hmm perhaps, but I don't see how that's a 'free' energy device.
The 'background' is a virtual/hypothetical object anyway - the CMB actually is just a flux of photons. The concept of temperature for photons and the CMB is contrived - defined tautologically based on an ideal black body emitter. The actual 'background temperature' for a complex greybody in the CMB depends on albedo vs emissivity - as shown by the math from wikipedia.
One can construct a heat engine to extract solar energy using a reflective high albedo (low temp resevoir) object and a low albedo black object. Clearly this energy is not free, it comes from the sun. There is no fundamental difference between photons from the sun and photons from the CMB, correct?
So in theory the same principle should apply, unless there is some QM limitation at low temps like 2.7K. Another way you could be correct is if the low CMB temp is somehow 'special' in a QM sense. I suggested that earlier but you didn't bite. For example, if the CMB represents some minimal lower barrier for emittable photon energy, then the math model I quoted from wikipedia then breaks down at these low temps.
But barring some QM exception like that, the CMB is just like the sun - a source of photons.
I've never seen someone so confused about the basic physics.
Let's untangle these concepts.
Effective temperature is not actual temperature. It's merely the temperature of a blackbody with the same emitted radiation power. As such, it depends on two assumptions:
Of course if these assumptions aren't true then the temperature estimate is going to be wrong. Going back to my LED example, a glowing LED might have an 'effective temperature' of thousands of degrees K. This doesn't mean anything at all.
The source of your confusion could be that emitted and received radiation sometimes have different spectra. This is indeed true. It's true of the Earth, for instance. But at equilibrium, absorption and emission are exactly equal at all wavelengths. Please read this: https://en.wikipedia.org/wiki/Kirchhoff%27s_law_of_thermal_radiation
Irrelevant, as I said. (The concept of albedo isn't very useful for studying thermal equilibrium, I suggest you ignore it)
Yes, this is the definition of being at the same temperature, if you didn't know. (Assuming, of course, that the radiation is thermal in origin and radiation is the only heat transfer process at work, which it is in our example). If you disagree with this then you are simply wrong by definition and there is nothing more to say.
You seem to think that temperature is some concept that exists outside of thermal equilibrium. This is a very common mistake. Temperature is only defined for a system at thermal equilibrium, and when two objects are in thermal equilibrium with one another, they are by definition at the same temperature. It does not matter at all how fast their atoms are moving or what they are made of.
No it's not. It's based on analysis of the spectrum, which is almost perfectly the spectrum of an ideal black body.
The physics would be exactly the same if it were an actual sheet of black material at 2.7 K covering the universe.
That was indeed a source of initial confusion as I stated above, and I read Kirchnoff's Law. I said:
However this still doesn't explain how passive temps lower than 2.7K are impossible. Passive albedo cooling works for the earth because snow/ice is highly reflective (inefficient absorber/emitter) at the higher frequencies where most of the sun's energy is concentrated, and yet it is still an efficient absorber/emitter at the lower infrared frequencies. - correct?
Now - what prevents the same principle for operating at lower temps? If ice can reflect efficiently at 500 nm and emit efficiently at 10um, why can't some hypothetical object reflect efficiently at ~cm range CMB microwave and emit efficiently at even lower frequencies?
You said: "But at equilibrium, absorption and emission are exactly equal at all wavelengths."
But clearly, this isn't the case for the sun earth system - and the law according to wikipedia is wavelength dependent. So I don't really understand your sentence.
You are probably going to say ... 2nd law of thermodynamics, but sorry even assuming that said empirical law is actually axiomatically fundamental, I don't see how it automatically rules out these scenarios.
This isn't an explanation, you still haven't explained what is different in my examples.
Sure - but notice that it's infinitely far away, so the concept of equilibrium goes out the window.
Also - aren't black holes an exception? An object using a black hole as a heat sink could presumably achieve temps lower than 2.7K.