söndag 1 augusti 2010

BlackBody: Transformer of Radiation


The Earth as a blackbody transforms incoming high-frequency radiation from the Sun into outgoing low-frequency infrared radiation with total energy in = total energy out. The incoming spectrum peaks at 0.5 reflecting a Sun surface temperature of 5778 K, and the outgoing at 10 (mikrometer) corresponding to an Earth surface temperature of 288 K. As seen in the above figure, there is little overlap. 

The amplitude of the incoming radiation is down-scaled compared to the blackbody spectrum at the Sun surface given by Stefan-Boltzmann's Law, depending on the ratio between the Sun radius and  the distance from the Sun to the Earth, while the amplitude of the outgoing radiation from the Earth surface is directly given by Stefan-Boltzmann's Law. 

In other words, the Earth acts as a transformer of radiation, transforming both frequency and amplitude of the Sun surface radiation. High-frequency low-amplitude radiation is transformed into low-frequency high-amplitude radiation. 

The effect is in particular that the Earth emits more infrared radiation than what is coming in from the Sun. This effect does not come from any "backradiation" from the atmosphere, because the effect is there even without an atmosphere.

Does this mean that the atmosphere has no influence on the temperature of the Earth surface?
Of course  not! As we have seen in previous posts, the atmosphere acts as a retailer of outgoing radiation between the Earth surface and empty space, and as such has an influence on the temperature/price at the Earth surface (assuming the atmosphere is transparent to incoming radiation). But there is no "backradiation" with the retailer sending goods back to the producer (because a sufficient reason is lacking).

A mathematical analysis of blackbody radiation transformation is presented in Computational Blackbody Radiation.

This posts responds to a comment by ScienceofDoom to the previous post AGW Myth of Backradiation, where the above figure is presented. Recall that "backradiation" leads into a fiction of "radiative forcing from greenhouse gases" which has no physical counterpart. The forcing is the incoming radiation from the Sun and nothing else.

19 kommentarer:

  1. Solar radiation varies much more over the solar cycle (5 to 7 percent) in the ultraviolet range coupling chemically to the generation of ozone in the stratosphere and troposphere. The ozone is very effective at blocking UV, and since it is a greenhouse gas, it is be effective at transforming UV to infrared radiation. Presumably this would couple to greenhouse gasses at the top of the troposphere. Do you see a way to calculate the first order effects, would it slow the radiative cooling of the atmosphere or increase the cooling efficiency? Could it account for an increased sensitivity to solar variation beyond that expected from a purely radiative non-chemical coupling? Apologies if you have already discussed this, i just discovered your web site. Thank you.

    SvaraRadera
  2. This posts responds to a comment by ScienceofDoom to the previous post AGW Myth of Backradiation, where the above figure is presented. Recall that "backradiation" leads into a fiction of "radiative forcing from greenhouse gases" which has no physical counterpart.

    My comment in your earlier post referred to real measurements of this "fictitious" quantity.

    You don't have to do any maths to work it out. You can just measure it. I provided many measurements in The Amazing Case of “Back-Radiation”.

    So I'm confused by your comment. You don't believe it exists because you have calculated that it doesn't exist?

    You don't believe the measurements are real?

    SvaraRadera
  3. Well, the question what the measurements measure?

    Do you really insist that the Earth receives 168 W/m2 SW from the Sun plus 324 W/m2 LW as downwelling backradiation, a total of 492 W? And emits 390 W/m2 LW?

    To me this is just calculating backwards to get 390 because that is believed to fit with SB at +15 C.

    But in the version of SB I use the Earth is only emitting about 120 W/m2
    because this fits with the temperature drop from +15 to -18 C.

    Both theory and measurement must be interpreted, and correctly interpreted to make sense. According to Ockham, recirculating radiation
    needs to be motivated. What drives the recirculation?

    SvaraRadera
  4. "To me this is just calculating backwards to get 390 because that is believed to fit with SB at +15 C. "

    This isn't data fitting, this is measurement - with, for example, an Eppley pyrgeometer.

    And also the spectrum of this radiation with various FTIR. You can see some spectra in The Amazing Case of “Back Radiation” – Part Two

    Clearly, you have a conceptual problem with the idea, so rather than explaining why there is DLR (downward longwave radiation) it's best to focus on measurement.

    But just to satisfy your curiosity..

    The atmosphere with radiation-absorbing molecules like water vapor, CO2, ozone must also emit at the same wavelengths it absorbs - if the atmosphere is above absolute zero.

    However, if the measurements say that radiation from the atmosphere is actually zero then we can toss out the theory.

    I also don't understand your comment: "But in the version of SB I use the Earth is only emitting about 120 W/m2 because this fits with the temperature drop from +15 to -18 C.

    The value of upward flux from the surface that is measured always follows the Stefan-Boltzmann law of sigma x T^4. You can see results from the EBEX in Part Three. A surface with high emissivity (like most earth surface types) and 15'C actually emits radiation of 390 W/m^2. If 35'C it emits at 511 W/m^2, and so on.

    At the moment I'm still very curious. Maybe you don't know about the measurements of DLR. Maybe you are proposing a whole new theory of thermodynamics where Planck's law and Kirchhoff's law are thrown out.. Something will have to give.

    SvaraRadera
  5. I think the DLR meter is calibrated as if to have temperature 0 K. Then
    it may read 390 for Upwelling LR and why not 324 for DLR.

    But the atmosphere does not have temp 0 K, and the radiative heat transfer depends on the temperature difference. Right?

    I will write a post on this topic.

    SvaraRadera
  6. I can't see my earlier post, but it as if you have attempted to answer it.

    Many possible theories:
    1) Planck's law is flawed
    2) Stefan-Boltzmann law (just the integral of the Planck function) is flawed
    3) People who have been making pyrgeometers for half a century or more are still getting it wrong.

    But also there is still the spectrum of this radiation from various FTIR results from the atmosphere to explain. You can see some spectra in The Amazing Case of “Back Radiation” – Part Two

    With 1 and 2 you can make a name for yourself. You would probably surprise a few people with 3. And the FTIR results, I can't work out how to explain those away.

    SvaraRadera
  7. The confusion concerns SB: If A and B are two blackbodies in radiative contact of temperatures TA and TB with TA bigger than TB, then the heat radiated from A to B = sigma (TA^4 - TB^4). If A is the Earth at 15 C and B is the atmosphere at -15 C, then there is a radiative heat transfer according to SB of about 4 x 30 = 120 W/m2. Right?

    SvaraRadera
  8. Maybe the confusion comes from computing radiative fluxes by simply integrating radiation spectra?

    SvaraRadera
  9. Some of the DLR pyrometers use cryogenic cooling and some not; here is a comparison of 7 types:

    http://journals.ametsoc.org/doi/pdf/10.1175/1520-0426(2004)021%3C0268%3ATMIRCA%3E2.0.CO%3B2

    The thermister in these units increase resistance when pointed at a hotter body and decrease resistance when pointed at a colder body allowing measurement of colder bodies. But I wonder if this electronic compensation or the means of calibration are problematic?

    The reason why my IR sensitive camera can see an IR LED 50 feet away, but can't see any of the 342 Wm-2 IR "back-radiation" from a clear sky at night is because the camera sensor is hotter than the sky.

    SvaraRadera
  10. Claes,

    solar constant: 1370/4 = 342 W/m2 - 30% albedo =239.4 W/m2 average insolation during the day.

    0 W/m2 “average insolation” at night

    Average of day & “night” insolation = 120 W/m2

    Exactly what you calculated above using avg. temps over day & night

    SvaraRadera
  11. The day-night compensation is already included in the factor 4 dividing
    1370, so 120 does not come from 240/2. Instead it comes out from a balance between thermodynamics setting the lapse rate and radiation operating on the lapse rate.

    SvaraRadera
  12. "The confusion concerns SB: If A and B are two blackbodies in radiative contact of temperatures TA and TB with TA bigger than TB, then the heat radiated from A to B = sigma (TA^4 - TB^4). If A is the Earth at 15 C and B is the atmosphere at -15 C, then there is a radiative heat transfer according to SB of about 4 x 30 = 120 W/m2. Right?"

    The formula is correct but the maths seems slightly wrong.
    Blackbody radiation from 15'C (288K) source = 391W/m^2
    Blackbody radiation from -15'C (258K) source = 252W/m^2

    Difference = 139W/m^2.

    But maths aside, as you explain perfectly how 2 bodies at different temperatures exchange radiation I'm confused why radiation from the atmosphere seems like a fiction to you..

    Back to the measurement techniques. The pyrgeometer receives radiation from a source that it is pointed towards - could be the sky, could be the surface. And then a calibration factor for the temperature of the pyrogeometer itself needs to be taken into account.

    So if you think that the Stefan-Boltzmann law is a fiction then the readings from pyrgeometers can be waved away. I'm sure you don't think this.

    Which just adds to my confusion as to your true question.

    Well, maybe you can satisfy my curiosity and confirm that you do believe the Stefan-Boltzmann law is accurate?

    And that radiatively-absorbing gases also emit radiation (at the same wavelengths at which they absorb) if they are above absolute zero?

    SvaraRadera
  13. So we agree on SB. Fine. What you do not understand is that A and B are
    in electromagnetic radiative contact and what A and B radiate is relative to
    the other. There is nothing like A separately spitting out photons independent of B. This is not physics, only fantasy.

    SvaraRadera
  14. "What you do not understand is that A and B are in electromagnetic radiative contact and what A and B radiate is relative to the other. There is nothing like A separately spitting out photons independent of B. This is not physics, only fantasy."

    Can you explain this further. What I learnt and still find in the physics books is "the fantasy" unless I misunderstand some specific condition that is applied. Or what you mean by "EM radiative contact".

    For example:
    -the sun and the earth are in EM radiative contact, but the sun still "spits out" photons as if it was a blackbody at 5800K and the earth as if it was a blackbody at temperatures ranging from 220-320K.

    I hope you can clarify what you mean.

    SvaraRadera
  15. Light is electromagnetic waves. The Sun emitting light generates electromagnetic waves covering space of which the Earth occupies a part
    and thus is in electromagnetic contact with the Sun. In this contact the Earth is a receiver and not emitter, and in particular does not emit any photon particles reaching the Sun. But the Earth is an emitter (electromagnetic waves) vs space minus the Sun (and other hotter objects).

    SvaraRadera
  16. That hasn't clarified anything for me.

    Body A emits radiation according to emissivity(A) x 5.67x10^-8 x T(A)^4

    Body B emits radiation according to emissivity(B) x 5.67x10^-8 x T(B)^4

    Under what specific conditions are the above equations false?
    And how are they modified?

    SvaraRadera
  17. Radiation requires both an emitter/source and absorber/receiver. It is like
    a lively discussion around a dinner table. Everybody speaks and listens
    and the important aspect is the transfer of information/energy from
    persons who to more to persons knowing less.

    Your statements "Body A emits...." are vague/incomplete because you only specify the emitter and not the absorber.

    SvaraRadera
  18. Radiation only requires an emitter. There is no requirement for an absorber.

    So the statement I made was not vague or incomplete at all. The Stefan-Boltzmann equation is just the integral of the Planck function which is a function of the emitting body only.

    I think Planck was correct and was a physicist not a fantasist.

    If you want to know how much energy is transferred in a system then you need to know many other conditions, including view factors, absorptivity as a function of wavelength and direction. You would also need to know the energy sources (if any) of each body and their temperatures. And if you want to know transient response and not just final equilibrium you would need to know many other properties of each body.

    But I don't think you have been clear.

    While you claim that my statements are "fantasy", "vague", "incomplete" you have yet to clarify or define what you actually think.

    Let's try and be clear for others who are reading.

    A) I believe that a body will emit thermal radiation at a rate = emissivity x 5.67x10-8 x T^4 in W/m^2, where T is absolute temperature.

    B) I believe this is just the integral of the Planck function over all wavelengths and all directions.

    Is A complete?
    Is B accurate?

    What precisely do you believe about the physics of emission of radiation?

    SvaraRadera
  19. A is incomplete because the receiver is not specified. Yes, B is the integral
    of the Planck function. Radiative heat energy transfer between two blackbodies of temp TA and TB with TA bigger than TB, is prop to TA^4 -TB^4. One cannot speak about only emission, at least I can't.

    SvaraRadera