The Cosmic Microwave Radiation (CMB) shows a blackbody spectrum of temperature 2.725 K
peaking at a wave length of about 0.2 cm beyond the far infrared spectrum. CMB is detected by radio-telescopes by resonance like radio antennas resonating with incoming radio waves thus generating a weak electrical signal which can be amplified into detection.
It would be difficult to detect CMB by thermal IR-imaging since the signal is very weak and thermal detection would require a detector at lower temperature than 2.725 K.
The concept of Downwelling Longwave Radiation DLR from the cold atmosphere to the warm Earth surface plays a key role in CO2 alarmism. CMB is here presented as an ultimate form of DLR with the argument that a picture of CMB shows that DLR is real. If even the cold dark space is contributing to global warming, then global warming must be real, right?
Let us now scrutinize this argument in the setting of mathematical model of blackbody radiation studied in Computational Blackbody Radiation, in the case of a radio-telescope as CMB-detector. The model takes the form of set oscillators with damping (see here for some more details)
- $U_{tt} - U_{xx} - \gamma U_{ttt} - h^2U_{xxt} = f$
where the subindices indicate differentiation with respect to space $x$ and time $t$, and
- $U_{tt} - U_{xx}$ represents the oscillators in a wave model
- $- \gamma U_{ttt}$ is a dissipative term modeling outgoing radiation
- $- h^2U_{xxt}$ is a dissipative modeling internal heating
- $f$ is incoming forcing/microwaves,
where $\gamma$ represents the constant in Planck's radiation law and $h$ represents a smallest mesh size, connected to dissipative losses as outgoing radiation and internal heating, respectively.
Microwaves are characterized by low frequency and long wave length (compared to visible and
infrared light) and in this case the dissipative loss of internal heating is small and is not detectable while the resonance can be detected after amplification just like a radio antenna is capable of detecting a weak radio wave by resonance followed by amplification.
Pictures of CMB are thus produced by an IR-camera in the form of a radio-telescope which works by resonance and not radiative heating. A CMB picture can therefore not be used as evidence that the weak glow of CMB acts as in a weak form of radiative heating named DLR or backradiation. This is because the CMB picture is not obtained from detection of radiative heating, but from resonance and amplification.
We conclude that a CMB picture is not any evidence of DLR, because no DLR is detected.
But there is no downward radiation from the atmosphere, because the atmosphere is colder than the Earth. The detector is irrelevant - zero is zero. Same with the CMB. No energy can travel from a colder body to a hotter one. So what people think is the CMB cannot be real. What is it?
SvaraRaderaLets keep this simple.
SvaraRaderaIs it in your opinion Claes, possible or not possible to always accurately (up until some justified limit) detect an IR-radiation spectra (only 3 answer alternatives available. Yes, no and don't know)?
Yes, but tjat does not mean it has a heating effect,
SvaraRaderapossibly as little as radiowaves are heating an antenna.
The natural follow up question would the be, in your opinion, is it possible to determine with an experimental setting, if there is a long wave radiation spectrum towards the earth (aka DLR)?
SvaraRaderaYour question is not precisely formulated: what is the meaning of " radiation
SvaraRaderaspectrum towards the Earth"? A spectrum is a curve on a piece of paper.
The same type of spectra as you answerd to in the previous question (IR-radiation spectra). If it is measured in the direction perpendicular to earths surface.
SvaraRaderaHelpful link when in question:
EM spectrum
Is your spectrum supposed to have a heating effect?
SvaraRaderaIf so why and how much?
Only the question of possibility to detect existence.
SvaraRaderaHow do you absorb radiation without there being a heating effect?
SvaraRaderaThe equation of a damping & radiating oscillator is
SvaraRaderaf = kx + dx’ + mx’’ + εx’’’ = (k + jdω - mω² - jω³)x
and its power is P =fx’ = (εω² - d)ω² + j(k - mω²)ω
If ω² = k/m then the reactive power vanishes (resonance). If ω² ≤ d/ε then the oscillator simply dissipates the forcing power. The oscillator is able to radiate only if ω² ≥ d/ε.
What could be the emitting threshold for the CMB?
Michele
A radio wave antenna absorbs energy by resonance but does not get heated because heat is high frequency decoherent waves and radio waves are low frequency coherent waves.
SvaraRaderaThe spectral analysis of only one oscillator is too simple to tell the truth. For a more complete spectral analysis of a set of oscillators in the form of the wave equation, telling more of the truth, see my article Computational Blackbody Radiation on my home page, from the Sky Dragon book, and the related article
SvaraRaderaNear Resonance with Small Damping.
What a truly terrible attempt to get yourself out of the hole you've dug. "heat is high frequency decoherent waves" is some of the purest nonsense you've yet come out with. That description of heat would shame a first year undergraduate.
SvaraRaderaIf downward radiation from the atmosphere doesn't exist, nor does the CMB, according to your own theory. So please explain what the apparent CMB is. Strangely enough you don't seem to want to answer this question.
Is it possible for you to write a concentrated presentation of your model? Maybe as a blog post. And preferably a section where you give your own perspective on the strengths and weaknesses of your model.
SvaraRaderaHave you done any test runs on well defined empirical situations to see if your model stands up to basic empirical data?
And come to think of it, it would be interesting if you could write a post where you summarize your science philosophical stance (concerning truth, reality and all that simple stuff ;-) ).
The guys who got the Nobel Prize for detecting CMB should be able to tell what they are detecting and thus what CMB is. But maybe they do not know.
SvaraRaderaI am very surprised that you can't answer a simple question. According to your theory of how radiation works, neither downward radiation from the atmosphere nor the CMB actually exist, because the surface of the Earth is warmer than both the atmosphere and the cosmic background. So, what is that image that you posted, purportedly of the CMB? It can't be the CMB because that is every bit as much a myth as the downward radiation from the atmosphere.
SvaraRaderaWhat I say it not so strange: For a blackbody to be heated by incoming blackbody radiation a certain condition must be satisfied: there must be frequencies of the incoming radiation above the cut-off of the absorbing blackbody, which effectively means that the emitter has higher temp than the absorber. CMB can thus only heat what has a temp lower than 2.725 K.
SvaraRaderaYet CMB can be recorded as resonance of antennas of radio-telescopes.
This is like reading/recording a text without getting impressed/heated, a common phenomenon.
What happens then with the energy that impinges earths surface? Does it just bounce?
SvaraRaderaYou have been claiming at length that downward radiation from the atmosphere does not exist. It is a myth. You seem now to be saying that it does exist, it just doesn't heat the Earth. These two positions are radically different. Please explain.
SvaraRaderaBy the way there if you're still talking about black bodies, there is no cut-off. And your analogy is woefully ridiculous. This is physics, not story time at playschool.
Comment from Anonym (accidentally deleted):
SvaraRaderaI think the worthless complexity is only very confusing.
It is important to point that not all the oscillators radiate. Even in resonance and so with the total forcing power used to create the useful effect, the oscillator completely absorbs the forcing power if d/ε ≤ k/m i.e., if the damping/emitting ratio is lesser than the resonant frequency.
There exists a range of frequencies where an oscillator behaves as a black hole, where we have only heat without radiation.
Of course, if the system is complex, the coefficients k – d – m – ε become arrays and we refer to their eigenvalues and eigenvectors.
Michele
The interaction of the oscillator with the damping is a phenomenon of near resonance, see http://www.csc.kth.se/~cgjoh/resonance.pdf
SvaraRaderaI describe very carefully in http://www.csc.kth.se/~cgjoh/blackbodyslayer.pdf
SvaraRaderahow electromagnetics waves with frequency below cut-off of the absorber
get re-emitted without heating effect, while frequencies above cut-off gets absorbed as heat. Read and reflect.
Your explanations are poor and self-contradictory. Please clarify whether you believe that the atmosphere does not radiate downwards at all (ie this "myth of backradiation" that you have been going on about) or whether it does indeed radiate downwards, but that that radiation does not heat the surface of the Earth.
SvaraRaderaYour statement "radiate downwards" has no clear meaning. Please give reference to this concept.
SvaraRaderaYou have been using the same terminology so I'm not sure why you don't understand now. Downwards = towards the surface of the Earth. Does the atmosphere radiate towards the surface of the Earth or not?
SvaraRaderaThe radiation is caused by the photonic pressure gradient, i.e., by the radiating density gradient, and, as far as I know, both the pressure gradient and the density gradient are only one way.
SvaraRaderaMichele
”The interaction of the oscillator with the damping is …”
SvaraRaderaClaes, you cannot omit the term of the oscillator equation containing x’’’ and so the emitting term.
Michele
You need to describe "radiate towards the surface of the Earth" in terms of Maxwell's equations for electromagnetic waves. Waves can go in any direction
SvaraRaderaand you need to specify in what sense the radiation carried by the waves is directed towards the earth surface. Equations are needed, not just vague words.
To Michele: why is the radiation directed to the earth surface?
SvaraRaderaHow remarkable, that you suddenly seem not to be able to understand simple concepts like "radiate towards the surface of the Earth". The words are not vague at all - they have very precise physical meanings. If you don't know the answer, you can just say that.
SvaraRaderaClaes, I think you realize that you are stuck in a corner and that you are talking nonsense when requiring Maxwell equations in explaining "radiating towards the surface of earth". Admit honestly that every black body emits radiation depending on its temperature. But that doesn´t mean that it directly can heat a warmer body because the warmer body can both absorb and reemit that radiation - absorptivity=emissivity.
SvaraRaderaI wonder how you explain multi-layer insulation on spacecraft. The effect is real - it reduces the rate of heat loss in eclipse - but how can a cold layer of opaque plastic affect the spacecraft surface in any way if its radiation cannot be absorbed by the warmer surface according to the 2nd Law? I have my own idea, based on the 1st and 2nd Laws, using a photon model, which explains why replacing the plastic with a shell of CO2 (not in contact) would not have this effect, but I would like to know how your wave model explains it.
SvaraRaderaI skimmed through part of your "Computational Blackbody Radiation" text (mostly the radiation model section).
SvaraRaderaIt looks like the model is purely classical. Have you done any thinking if there is some quantum corrections (classical quantum or quantum field/optic) that could be justified and change the result?
If you thought about it, then what is your conclusion?
Sorry about the previous post, I mixed up the Compton wavelengths here ( it's late) .
SvaraRaderaAnother thing that I wonder about is if you looked at what happens in a control volume (solid, water and air) when you have a setting where radiation impinges. Could there be emergent phenomena outside your model? What happens with the dissipated energy in a collective setting? Are there any recoil energies in the model (sorry if I doesn't see it, but as said it is late)?
Yes, the model is classical and that is its usefulness, since blackbody radiation is a macroscopic phenomeon. The model describes the interaction between (electromagnetic) waves, matter/osciilators and heat/internal energy of oscillators, and is useful as basis for an analysis and discussion.
SvaraRaderaWithout a mathematical model, the discussion becomes confused because the meaning of a mere phrase like "radiation towards the earth surface" is unclear and the phrase does not it become more precise it is shouted out. It is thus necessary to replace description in words by mathematical models. If the words have meaning this is possible, if we speak about physics. Without math the discussion can go around in circles for ever.
without intellectual honesty the discussion can go around for ever.
SvaraRaderaWho is dishonest?
SvaraRaderaNot all the thermal radiations are directed to the earth surface. The transfer is driven by grad(Pf) = grad (ρf) = grad(I/c) = (σ/c)grad(T^4). Then, there is the grad(T^4) which determines the radiating flux, not at all the T^4 alone and so the thermal radiation traveling toward the earth surface must come from a source having a temp higher than the earth surface one. Natural phenomena are always a consequence of the law of the strongest.
SvaraRaderaMichele
you've been claiming (or appeared to be claiming) all the time that there is no radiation from a colder body towards a warmer body.
SvaraRaderanow you've realised that there is.
you now say (or seem to say, since you keep it fuzzy) that there is radiation from a colder body towards a warmer body but that this radiation will not warm up the warmer body. fine!
why don't you discuss openly what are the consequences of your model and how energy and momentum conservation can still be true without absorption instead of trying to make the discussion impossible by asking stupid questions? why don't you critically inquire your own model instead of wasting your time in making analogies that are at the best not even funny and repeating on and on the very same thing?
why do you keep on using sloppy terminology (but then demand from people asking you questions to define everything) just increasesing confusion and creates discussions that are not at all interesting?
you're just pushing away everyone genuinely interested in your model and keeping only people /a priori/ with or against you.
OK, I consider a mathematical model and I say that any meaningful discussion must start from a mathematical model. Words can do anything
SvaraRaderabut that is not of interest in physics. I have given my equations, now give yours and we can discuss. My math model is not fuzzy, my math analysis is not fuzzy, words explaining equations can be illuminating, words without equations are fuzzy and lead to neverending meaningless fuzzy plays with fuzzy words, as in this discussion.
Claes wrote:
SvaraRaderaMy math model is not fuzzy, my math analysis is not fuzzy, words explaining equations can be illuminating, words without equations are fuzzy and lead to neverending meaningless fuzzy plays with fuzzy words, as in this discussion.
But one has to remember that a mathematical model, is just a model. If there is some empirical fact that contradicts the model, then guess what... It's the old question about the map and the terrain.
Having said that, all in all it's a tough situation. A mathematical model is for this reason limiting, it could be partially wrong in that there are phenomena not included.
So we are at least to consider using heuristics in understanding, in a way I guess that this is to a large extent what engineering is all about. But using heuristics is also limiting, and can be clouded by bias, so all in all it's a loose-loose situation.
I must agree with commentator lorenzo that the reasoning is fuzzy. Maybe not in a mathematical sense, but in the motivation of why; 1) the components in the model are motivated and 2) that the components are exhaustive concerned with all ongoing phenomena.
( 1) not so much as 2) )
Call me naive but I think that there must be tons of published data that could settle if there is validity in this model. If it's so experiments developing space shuttle insulation or small scale exploration of green house gases in a laboratory setting.
If you Claes want people to take your model seriously I guess the best method (maybe the only) is to start with for you to show that it can be used to explain these simpler situations. I personally don't see any other way to bring validity to your model.
You are making the claim that the model has physical relevance, so the burden of proof is upon you.
Sincerely,
Dol
The model is similar to that used by Planck in his proof.
SvaraRaderaAnyone accepting Planck's model would be able to accept the variant
I consider. Anyone not understanding Planck's would have difficulties with mine. The choice is yours.
The models are similar but the analysis is different. Planck uses statistics of quanta, I use finite precision computation.
Claes wrote:
SvaraRaderaI use finite precision computation.
Two questions
1) Is there an easy argument to justify that a finite precision computation gives a true understanding of what is going on? This is for me not at all clear. Numerical stability issues should be a model problem, not an ontological one (what we can and can not compute should have no impact on reality, as it is).
Which brings me to the second question.
2) Have you done any validity checks of your model and related it to empirical situations? This is connected to what I earlier wrote.
Dol wrote:
Call me naive but I think that there must be tons of published data that could settle if there is validity in this model. If it's so experiments developing space shuttle insulation or small scale exploration of green house gases in a laboratory setting.
Sincerely,
Dol
Since the model gives a Planck law and Planck's law has been experimentally verified, my analysis is supported by observation.
SvaraRaderaThe model is as simple as possible with all essential elements. Yes, the finite precision computation is an assumption, and there is not yet any direct experimental verification of the assumption. But its is in principle possible to verify or falsify in direct experiment, while Planck's assumption about statistics of quanta is impossible to test directly (which may explain its popularity).
Claes wrote:
SvaraRaderaSince the model gives a Planck law and Planck's law has been experimentally verified, my analysis is supported by observation.
Well, as the saying goes, correlation doesn't imply causation.
That your model (or Planck's original for that matter) hits the right equation could be a mere fluke. It does not imply that the underlying reasoning has anything to do with reality. It's unfortunate that Planck's assumption isn't testable, but then maybe in the same way fortunate that yours is.
Further one must consider the possibility that the Planck assumption is correct, and that "mother nature" is so cruel that the inner nature is epistemologically unreachable. And believe me, that is not the case I'm hoping for.
So the question remains, have you taken any initiative in getting your model and your assumptions tested against true empirical situations?
I have one another question to, has the model been peer-reviewed yet?
Further I agree with you that maybe one jumps the gun in assuming a catastrophic global warming effect based on current knowledge. But truly that judgement call is undecidable for me at the moment since I have no deeper knowledge in the reasoning that is made on the research-level of the relevant fields (There is of course another issue and that is if the global availability of fossil fuel is feasible accounted for in the current modelling. But this isn't an issue in the current debate according to your model). And what's more important, this kind of warming effect could be correct albeit the underlying reasoning could be faulty.
But at the same time, maybe you jumps the gun dismissing their supposed argument based on your model whose validity isn't truly settled as I understands it.
I do consider myself as an hardcore empirical skeptic. In my opinion we must rely all assumptions on empiric knowledge, but unfortunate for us, it isn't certain that empirical knowledge is attainable.
Sincerely,
Dol
It is fine to be a skeptic, but it is even better when combined with something constructive. I do my best, but there are lots of responsible scientists and physicists who do their best to suppress and cover up, and that is destructive to science. Ask some prominent physicist about CMB and DLR and global warming, record the answer and put it up on the web. That would be a constructive step to take for a serious skeptic. I have 100 articles peer reviewed and I know how little that matters. If you have something of real interest to say, eventually you will have impact, but the resistance is directly proportional to the novelty and real interest of your results, since in academia the game is to prevent anything from changing unless it is under your own control.
SvaraRaderaClaes wrote:
SvaraRaderaAsk some prominent physicist about CMB and DLR and global warming, record the answer and put it up on the web.
Of course I could do that. But that's not the current issue and for the current discussion not even a relevant one. If they are wrong doesn't mean that your model is correct.
The question about peer-reviewing was not either a specially relevant one, just curiosity on my behalf.
The relevant question still is, have you taken any initiative in getting your model and your assumptions tested against true empirical situations?
Sincerely,
Dol
Thanks Dol for your good attemps to receive clarity in this blog. Please Claes, answer Dol's question:
SvaraRadera"So the question remains, have you taken any initiative in getting your model and your assumptions tested against true empirical situations?
"
I repeat that the model gives a Planck law and Planck's law has been empirically tested. The finite precision computation shows up as a small dissipative effect
SvaraRaderaof standard form with a coefficient which can be matched to experiment since it
directly connects to the cut-off in Planck's law. But please make the experiment to ask some more or less prominent physicist about DLR and ponder the answer.
Yes of course I have taken steps, and will report when the results are clear.
SvaraRaderaWhat steps has Anonym taken?
It seems to me that the disproof of 'radiative forcing' stands on correctly writing the energy balance equation (1st Law), based on the prohibition of the 2nd Law. These laws have extensive proof in experiment.
SvaraRaderaIt is interesting to propose mathematical models of physical mechanisms by which these Laws are obeyed, followed by experimental verification so we know that these Laws hold and are correctly applied to the problem. However, there is no reason to believe these Laws don't hold, and in this case, their application to the problem is incredibly straightforward. The burden of experimental disproof is clearly on the warmists.
The CMB disproof has fallen flat, by this article. What else do you have?
Claes wrote:
SvaraRaderaI repeat that the model gives a Planck law and Planck's law has been empirically tested.
As I wrote earlier, correlation with Planck's law doesn't mean that your model gives the right cause. Something you must be all familiar with. Or do you have any complaints against this view?
Also Planck's own statistical model gives Planck's law...
Claes wrote:
But please make the experiment to ask some more or less prominent physicist about DLR and ponder the answer.
I will most likely do this next year when I will follow a course on heat transfer.
But that's besides the point. Right now I'm asking you, about your model.
Claes wrote:
Yes of course I have taken steps, and will report when the results are clear.
That's very interesting to hear. As a starter do you care to share what kind of empirical situations you are testing?
Wait and see. Go ahead with your studies.
SvaraRaderaClaes wrote:
SvaraRaderaGo ahead with your studies.
Are you implying that I'm not fit to discuss the matter?
/Dol
No, but I don't find the discussion constructive.
SvaraRaderaProper experiment must have isolation of variables to show a single cause has the measured effect. The CMB experiment does not isolate the experimentally known phenomenon of resonance. That it even CAN influence the measurement significantly is enough to invalidate the experiment. Try again.
SvaraRaderaThe nagging concern seems to be "what happens to the photons if they cannot be absorbed?" Failing to account for them would not invalidate the fact the surface cannot have heat transferred from a colder source. However, if I understand this model correctly, it does answer this by separating energy from information using a wave model. I believe it can also be accounted for with a photon model, starting with the experimental observation that whatever light is not absorbed is either transmitted or reflected.
There ought to be an experiment that could decide between the two but regardless, radiative forcing still cannot stand up to the 2nd Law.
There are no IR photons. My model satisfies a 2nd law.
SvaraRaderaClaes wrote
SvaraRaderaNo, but I don't find the discussion constructive.
I'm sorry that you feel that way.
Never the less, this got me inspired to make the discussion more constructive for you.
I think I got something really interesting for you here.
One of the big questions seem be under what circumstances a surface is possible to emit heat rays (light rays, or what one like to call them).
I started to look into what Planck himself is saying on the matter and did quite quickly find his book called The Theory of Heat Radiation. This looked interesting so of course I had to check it out. Already in the first chapter called Fundamental fact and definitions we find something very interesting and highly relevant for the current discussion.
On page 6 in the edition I'm looking into Planck writes. [...] But the empirical principal law that the emission of any volume-element depends entirely on what takes places inside of this element holds true in all cases (Prevost's principle). A body A at 100^{\circ} C emits toward a body B at 0^{\circ} C exactly the same amount of radiation as toward an equally large and similarly situated body B' at 1000^{\circ} C.
This empirical law seems to stand in contradiction with what you claim must be true on base of your own model.
Sincerely,
Dol
Yes, this is of interest and I will comment in a new post.
SvaraRaderaMay be, Planck is wrong. A physical phenomenon must be studied "in itinere" instant by instant and not "a posteriori" as what is balanced and consolidated, e.g. in one year.
SvaraRaderaSimplifying.
In the space between the two plain plates exists, because their temp, a EM field perturbed and therefore in every point of said space acts a plain EM wave defined in the space-time as
δ²f/ δx² = (1/c²) δ²f/ δt²
whit the general solution
f(x,t) = Asin(kx – ωt) + Bsin(kx + ωt)
that is a linear combination of two waves (according to superposition principle because the linearity of the problem) that are propagating in opposite directions both perpendicular to the two plates (x axes). That’s, the total effect is found instant by instant in the same place taking into account the boundary effects.
1) If there is the left plate alone, is f(x,t) = Asin(kx – ωt) i.e. a wave propagating rightward.
2) If there is the right plate alone, is f(x,t) = Bsin(kx + ωt) i.e. a wave propagating leftward.
3) If there are both the left and the right plate, and B = A (in the our case the two plates have the same temperature) is f(x,t) = Asin(kx – ωt) + Asin(kx + ωt) = 2Asin(kx)cos(ωt). So there is no longer a travelling wave because the position and time dependence have been separated. Therefore, the sum of two counter - propagating waves (of equal amplitude A and frequency) creates a standing wave whit amplitude 2A and there isn’t propagation of energy at any point. It’s wrong to say that the two plates emit and absorb the same flux of heat. Both A and B have earlier emitted until the saturation of the space between each other. After, neither A and B emit because between them doesn’t exist any wave that carries energy, all the waves are in “surplace”, the whole space is filled of standing energy.
4) If A < B, then f(x,t) = Asin(kx - ωt) + Asin(kx + ωt)+ (B – A)sin(kx + ωt) = 2Asin(kx)cos(ωt) + (B – A)sin(kx + ωt), i.e., there are been created both a standing wave of amplitude 2A and a traveling leftward wave (B – A)sin(kx + ωt). Also in this case it’s wrong to say that the two plates emit and absorb. A doesn’t emit and B emit only what the traveling wave carries from B to A.
That’s, the traveling effect of counter wave Asin(kx-ωt) (e.g. the imaginary and unreal back radiation) vanishes on the way and there exists a reduced traveling effect of stronger wave B. Of course, the energy that doesn’t travel isn’t destroyed but simply stored into the space as potential energy of the stationary wave and it can be suddenly supplied as soon as an imbalance arises.
Michele
Yes, what you say makes sense.
SvaraRaderaMichele, standing waves only appear when there is no adaption at the receiving end and for blackbodies there is completely adaption (by definition), so your arguments are not valid in this case
SvaraRaderaSlabadang!
SvaraRaderaHej Claes!
Med anledning av din kommunikation med R Spencer har du snart samlat ihop till en egen tråd på TCS. Vi kanske tror att vi börjat fatta vad det är du hävdar. Vi frågar efter dig!!
http://www.theclimatescam.se/2012/02/23/ryssvarmen-under-nadens-ar-2010/comment-page-1/#comment-270680