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Sun & climate: moving in opposite directions

What the science says...

Select a level... Basic Intermediate Advanced

In the last 35 years of global warming, sun and climate have been going in opposite directions.

Climate Myth...

It's the sun
"Over the past few hundred years, there has been a steady increase in the numbers of sunspots, at the time when the Earth has been getting warmer. The data suggests solar activity is influencing the global climate causing the world to get warmer." (BBC)

Over the last 35 years the sun has shown a slight cooling trend. However global temperatures have been increasing. Since the sun and climate are going in opposite directions scientists conclude the sun cannot be the cause of recent global warming.

The only way to blame the sun for the current rise in temperatures is by cherry picking the data. This is done by showing only past periods when sun and climate move together and ignoring the last few decades when the two are moving in opposite directions. 


Figure 1: Annual global temperature change (thin light red) with 11 year moving average of temperature (thick dark red). Temperature from NASA GISS. Annual Total Solar Irradiance (thin light blue) with 11 year moving average of TSI (thick dark blue). TSI from 1880 to 1978 from Krivova et al 2007 (data). TSI from 1979 to 2009 from PMOD (see the PMOD index page for data updates).

Last updated on 22 February 2014 by LarryM. View Archives

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Comments 451 to 500 out of 1040:

  1. Gord -

    Let's get this straight -

    Considering the approx. 33 K difference between the approx. 255 K surface temperature with no greenhouse effect and the approx. 288 K surface temperature with the greenhouse effect about as much as it is (or recently has been):

    1.
    I correctly note that this 33 K warming is from the greenhouse effect, without any change in albedo, and I agree with you (as if I didn't know before) that albedo would change if the atmosphere in it's entirety is removed.

    2.
    NASA makes a mistake in their explanation, implying that removal of the atmosphere would leave the albedo unchanged, or that the 33 K warming comes from the greenhouse effect minus the cooling from the albedo increase, which is of course not true. I know that NO atmosphere means NO atmosphere, but I, assuming I am not more knowledgable than NASA, give them the benifit of the doubt, and choose to think that they did not actually intend what they mistakenly wrote. That should be quite clear to you, and you should have no reason to think I don't understand words.

    3.
    You seem to hold NASA's error against me or against climatology in general - as if it were up to me (or Kiehl and Trenberth (okay, maybe more likely), or...) what NASA says - and at the same time, tell me I have delusions of grandeur, when I said that I assumed that I was not more knowledgable on the subject than the people at NASA, and so chose to assume that they actually know that the 33 K warming is from the greenhouse effect with no change in albedo and that they were just sloppy in their writing. (I still do not think I am smarter or more knowledgable than the people at NASA in general - however I do think I much more knowledgable than you - you keep giving me evidence for that.)

    4.
    For NASA's mistake, you seem to suggest that I am wrong about the greenhouse effect. I have offered over the last several pages of comments and references therein a very good description of the greenhouse effect, but you look for erroneous descriptions from other sources to argue against me, to argue against things - things that I also agree are erroneous.

    5.
    Your entire point seems to be that there is no greenhouse effect. You were arguing against Kiehl and Trenberth's energy budget, and they do not make any mistake in assuming that the atmosphere contributes to albedo. I have made no mistake in that regard either. You also seem to be aware that there is some significant albedo. Yet you offer a calculation of temperature with zero albedo as support for your argument that there is no need to use the greenhouse effect to explain the temperature of the surface of the Earth.

    BOTTOM LINE:

    The albedo of the Earth, including atmospheric effects - is about 0.3.

    With that albedo, but in the absence of the greenhouse effect, the average temperature of the surface would be about 255 K, or possibly slightly colder.

    The average temperature of the surface is actually near 288 K.

    Meanwhile, the emission of radiation to space can be observed and it is less than the radiation that the surface would emit to space if it were completely exposed. The difference is especially associated with high cold cloud tops, and at wavelengths where absorption by atmospheric gases - H2O and CO2 in particular, is expected to be sizable based on known optical properties.
  2. Gord -

    Let's get this straight -

    Considering the approx. 33 K difference between the approx. 255 K surface temperature with no greenhouse effect and the approx. 288 K surface temperature with the greenhouse effect about as much as it is (or recently has been):

    1.
    I correctly note that this 33 K warming is from the greenhouse effect, without any change in albedo, and I agree with you (as if I didn't know before) that albedo would change if the atmosphere in it's entirety is removed.

    2.
    NASA makes a mistake in their explanation, implying that removal of the atmosphere would leave the albedo unchanged, or that the 33 K warming comes from the greenhouse effect minus the cooling from the albedo increase, which is of course not true. I know that NO atmosphere means NO atmosphere, but I, assuming I am not more knowledgable than NASA, give them the benifit of the doubt, and choose to think that they did not actually intend what they mistakenly wrote. That should be quite clear to you, and you should have no reason to think I don't understand words.

    3.
    You seem to hold NASA's error against me or against climatology in general - as if it were up to me (or Kiehl and Trenberth (okay, maybe more likely), or...) what NASA says - and at the same time, tell me I have delusions of grandeur, when I said that I assumed that I was not more knowledgable on the subject than the people at NASA, and so chose to assume that they actually know that the 33 K warming is from the greenhouse effect with no change in albedo and that they were just sloppy in their writing. (I still do not think I am smarter or more knowledgable than the people at NASA in general - however I do think I much more knowledgable than you - you keep giving me evidence for that.)

    4.
    For NASA's mistake, you seem to suggest that I am wrong about the greenhouse effect. I have offered over the last several pages of comments and references therein a very good description of the greenhouse effect, but you look for erroneous descriptions from other sources to argue against me, to argue against things - things that I also agree are erroneous.

    5.
    Your entire point seems to be that there is no greenhouse effect. You were arguing against Kiehl and Trenberth's energy budget, and they do not make any mistake in assuming that the atmosphere contributes to albedo. I have made no mistake in that regard either. You also seem to be aware that there is some significant albedo. Yet you offer a calculation of temperature with zero albedo as support for your argument that there is no need to use the greenhouse effect to explain the temperature of the surface of the Earth.

    BOTTOM LINE:

    The albedo of the Earth, including atmospheric effects - is about 0.3.

    With that albedo, but in the absence of the greenhouse effect, the average temperature of the surface would be about 255 K, or possibly slightly colder.

    The average temperature of the surface is actually near 288 K.

    Meanwhile, the emission of radiation to space can be observed and it is less than the radiation that the surface would emit to space if it were completely exposed. The difference is especially associated with high cold cloud tops, and at wavelengths where absorption by atmospheric gases - H2O and CO2 in particular, is expected to be sizable based on known optical properties.

    ----

    "It is very, very tedious explaining things to you and backing up my statements with links."

    Ha! I (perhaps quite unwisely) took the time to go over your mistaken ignorant babbling more or less point by point; I've used as pedantic an argument I ever would and then used another, I've explained the microscopic basis of macroscopic phenomena such as the second law of thermodynamics. I've explained here and offered references to other comments explaining just how the greenhouse effect and radiative energy transfer actually work (in a way that is easy to visualize, hence my occasional anthropomorphism, as in 'what the surface can see'), at times using the examples you offered. I've gone over the math of blackbody radiation. I might as well be your science tutor.

    And for the most part, all you ever do is repeat a bunch of ill-thought out, ill-informed assumptions that most serious physicists would laugh at.
  3. " I might as well be your science tutor."

    By the way, I QUIT.
  4. Patrick -

    Re: your posts #461 and #472:

    "2. Heat Radiation between hot and colder objects"

    "P = e*BC*A(T^4 - Tc^4)"

    When both the hot object and colder object have emissivities e and ec that could be less than 1, and there is a layer between with transmissivity T:

    P = (e*ec*T)*BC*A(T^4 - Tc^4)"

    ----------------------
    e*BC*A(T^4 - Tc^4) does not equal (e*ec*T)*BC*A(T^4 - Tc^4)

    PROOF:

    e does NOT equal e*ec*T and the units of P (watts) have now been changed...by Patrick !!!

    All justified by Patrick's delusional OPINION.
    ------------
    Mindless drivel that ANY algebra student could OBVIOUSLY see.

    Only a total "incompentent" and/or someone with "delusions of grandeur" would consider "modifying" the Stefan-Boltzmann Law and the units of Watts.

    Absolutely Hilarious!

    Adios, Patrick....you have "QUIT"...finally!
  5. Mizimi -

    RE: Your Post #467

    Your "simplified" model violated two fundamental Laws of Science and actual measurements!

    Enough said.
  6. Patrick 027 464

    Earth’s climate can be evaluated as a dynamic system for which the science of Control Theory applies. That means that all of the minutia of climate and weather, all of that stuff; forcings, Climate Science type feedbacks, GHG spectral absorption, ocean turnover, plant respiration, atmospheric carbon dioxide level, methane level, etc. etc. everything that Climatologists attend to (including all that you have described in excruciating detail) and even those things that haven’t been discovered yet; everything gets lumped together in a block in the Control Theory block diagram called ‘all internal factors that can alter average global temperature’. The only requirement is that none of them can add any energy to the block.

    The output from this block is average global temperature (agt). In Control Theory, feedback is a dimensionless number that is proportional to a change in output. In global climate it is a measure of the effect that change to agt has on how effective change to the energy entering the global climate system is at influencing agt. Feedback is a number that is added to 1.000 and the sum is multiplied by the input to the global climate system. Thus a feedback of 0.01 means that the input to the global climate system is multiplied by 1.01. If the feedback is -0.01 the input to the global climate system is multiplied by 0.99, etc. Note that feedback in Control Theory is dimensionless while feedbacks in Climate Science have units.

    Temperature data are readily available for the last and previous glacial periods (see, e.g. the few numerical data at 414 and graphs in my pdf linked from http://climaterealists.com/index.php?tid=145&linkbox=true . Since the agt trend changes the direction of slope from down to up and vice versa, Control Theory determines that the temperature can not be controlled by temperature feedback.

    A familiar example of a feedback system is the cruise control for a car. When working properly, it controls on speed and the speed of the vehicle is held fairly constant. Now if the speed changes substantially one must conclude that the unit is not controlling on speed. That is, there is no feedback from speed.

    In this analogy, agt is analogous to vehicle speed. Since earth’s temperature has changed as shown by the accepted temperature data from the last glacial period, it shows that there can be no significant positive (Control Theory) feedback from temperature in earth’s climate.

    Since there can be no significant Control Theory feedback, there can be no significant net Climate Science feedback either because that would influence temperature. Without Climate Science net feedback none of the twenty or so Global Climate Models that the IPCC uses predict significant global warming.
  7. Important correction to comment:

    http://www.skepticalscience.com/argument.php?p=19&t=474&&a=18#3338

    I used the same symbol for two different variables (transmissivity (unitless value) and temperature (units K or some other unit on an absolute scale), which caused confusion:

    P = (e*ec*T)*BC*A(T^4 - Tc^4)

    What variable shall we use for transmissivity? How about N:

    P = (e*ec*N)*BC*A(T^4 - Tc^4)

    Then let U = e*ec*N*BC*A

    The rest was correct:

    P = U * (T^4 - Tc^4)

    This can be rewritten as:

    P = U*T^4 - U*Tc^4

    Original point still valid. Units are correct.

    Inteligent people sometimes make simple slip-ups like that. This had the significance of a typo - I think I had meant to use a different variable and must have absent-mindedly used called transmissivity T. However, anyone reading that could have figured out what I was trying to communicate and made the correction for their self.

    And yes, I have been known to substitute 'their' for 'there', as might be noticed in prior comments.
  8. Patrick 027 468
    Thermalized means that the energy is mostly conducted to adjacent non-GHG molecules that do not significantly re-radiate. "When a radiant flux is absorbed, it is assumed to be mostly thermalized" Where then does all that radiation back to the surface come from?

    If the K&E graphic was not misleading there would be no need for all the clarification/explanation/rationalization.
  9. Dan -

    "A familiar example of a feedback system is the cruise control for a car. When working properly, it controls on speed and the speed of the vehicle is held fairly constant. Now if the speed changes substantially one must conclude that the unit is not controlling on speed. That is, there is no feedback from speed."

    That doesn't quite make sense - what if there was positive feedback - then the speed could certainly change.

    ----

    If the temperature goes up and down without forcing (which does happen - look at the all the wiggles about the longer-term trend of the last several decades), that doesn't mean that there are no positive feedbacks.

    (There are fluctuations in cloud cover patterns, humidity patterns, and evaporation and latent heating and temperature patterns associated with variability in circulation patterns in the atmospheres (which can also contribute to surface albedo fluctuations), so there are both thermal and mechanical (momentum) feedbacks through which circulation patterns cause changes in circulation patterns - but over time there tends to be a predictable statistical description of all that (which is the longer-term climate)... Heat can also be added to or removed from the surface environment by vertical transports in the ocean.)

    As I was explaining in the last comment to you, I think I saw where a major miscommunication may lie.

    In equilibrium (for the longer-term average over internal variability), the radiant energy out equals the radiant energy in. A solar TSI increase is a positive forcing, increasing the radiant energy input. An increase in CO2 is a positive forcing, decreasing the radiant energy output.

    In climatology, the equilibrium response without any feedbacks is generally described as the temperature increase required to change, directly through blackbody emission's dependence on temperature, the radiant energy output, so as to restore balance.

    However, I suspect that in Control theory, that is itself considered feeback. Remove that feedback, and what happens? The rate of net energy storage is constant - if the heat capacity doesn't change, temperature increases linearly over time to infinity. That is the climate response without any feedbacks, including the feedback of blackbody radiation as a function of temperature (let's call it the blackbody radiation feedback). The approx. 1 K increase due to a doubling of CO2 includes that feedback.

    And in the absence of forcings, this feedback is a negative feedback that must be stronger than the net positive feedback from all other feedbacks if the climate is to be stable.

    Consider the water vapor feedback again. Using the same values as before (and using a linear approximation to responses): in response to a positive radiant forcing, the temperature rises. Without any feedbacks including the blackbody radiation feedback, the temperature rises indefinitely. But with the blackbody radiation feedback, their is an equilibrium temperature increase of 1 K, and the difference between the equilibrium temperature and the temperature decays exponentially at a rate determined by the climate sensitivity and the heat capacity. If the water vapor feedback increases the equilibrium temperature change from 1 K to 2 K, then this implies that the water vapor response to a 2 K temperature change is sufficient to force a 1 K temperature increase - it would cause indefinite warming if the blackbody radiation feedback did not stop it.

    It might help to use some time-dependent formulas:

    Let T be the change in temperature from some reference value. Let R be the radiant forcing difference from some reference level. Let the blackbody radiant feedback be B, where B = b*T, and assume b is negative. Let F be the net of all other radiant feedbacks to the change in temperature, where F = f*T. Let the heat capacity (per unit area) be C. Time is t.

    Conservation of energy:

    dT/dt = (R+F+B)/C

    dT/dt = [R + (f+b)*T]/C

    dT/dt = R/C + T*(f+b)/C

    -------------------------------

    For R = constant,

    Let T = Teq - A*exp(-Z*t)

    A*exp(-Z*t) = Teq - T

    then

    dT/dt = Z*[A*exp(-Z*t)]

    dT/dt = Z*(Teq - T)

    IF the climate is stable, Z is greater than 0.

    ---

    dT/dt = Z*Teq - Z*T

    and

    dT/dt = R/C + T*(f+b)/C

    ---

    therefore

    Z*Teq = R/C

    Z*T = -T*(f+b)/C

    ---

    Z = -(f+b)/C

    Z = R/(C*Teq)

    e-folding time of (Teq-T) = 1/Z = C*Teq/R = C * climate sensitivity

    -(f+b)/C = R/(C*Teq)

    -(f+b) = R/Teq

    climate sensitivity = Teq/R = -1/(f+b)

    Teq = -R/(f+b)

    -------------------------------

    IN SUMMARY (where R, T, and Teq are defined as relative to reference values):



    Whatever has happened up to time t, if R is set to a constant value, then:

    If the climate is stable, there is a constant equilibrium T = Teq, of the same sign as R, that T tends to approach over time through the exponential decay of (Teq - T).

    This exponential decay has a time scale (a time constant, or e-folding time) of 1/Z = C * Teq/R, which is the product of the heat capacity (per unit area) and the climate sensitivity.

    Teq = -R/(f+b)

    1/Z = C * Teq/R = -C/(f+b)

    ------

    (Teq-T)

    = exp[-Z*t]

    = exp[ -t / (C * Teq/R) ]

    = exp[ t*(f+b)/C ]

    ------

    The climate is stable if the climate sensitivity is either zero or positive and finite.

    The climate sensivity, defined as R/Teq, is equal to -1/(f+b), where B = b*T is the blackbody radiant feedback and b is negative, and F = f*T is the net effect of all other feedbacks.

    -------------------------------

    An increase in R from an initial value to a new value shifts Teq. If (Teq-T) was zero before the change in R, it will not be immediately zero after the change; (Teq-T) will then exponentially decay to zero. Notice that what happens to (Teq-T) is the same if, instead of changing the external radiative forcing R, there is an unforced change in T. An unforced change in T from Teq will tend to exponentially decay over time if the climate is stable.

    A stable climate is compatible with a positive value of f, provided that f is not greater than the absolute value of b.

    When climatologists say the net feedback is positive, they are refering to f; when they state the climate sensitivity Teq/R without feedbacks, they mean with f = 0; they still include the effect of a negative b.

    The climate sensitivity with f = 0 is -1/b

    ----------
    Revisiting an earlier example, considering water vapor as the only contributor to f:

    Let -1/b = 0.25 K / (W/m2) (which means that a 0.25 K increase in temperature increases the outgoing radiation by 1 W/m2 if f=0; b = -1/0.25 (W/m2)/K = -4 (W/m2)/K)

    Teq = 0 K when R = 0 W/m2 (these are both relative to a reference or baseline state).

    Then imposing a forcing R of 4 W/m2 increases the Teq by 1 K if f=0.

    If T = Teq before R was changed from 0 W/m2 to 4 W/m2, (Teq-T) will equal 1 K before T responds.

    As (Teq-T) decays to zero from 1 K, T rises from 0 to 1 K, at a rate depending on C and R/Teq for f=0.

    Before T has changed, C*T is increasing at a rate R = 4 W/m2. But for each 0.1 K increase in T, the radiative feedback -B increases 0.4 W/m2,But when T has increased 0.5 K, that rate has dropped to R+B = R+b*T = (4 - 2) W/m2 = 2 W/m2. For each halving of (Teq-T), the rate of energy gain is halved. Hence, if C is constant, (Teq-T) decays exponentially. The time for each halving of (Teq-T) is proportional to C and proportional to Teq/R.



    T (K), B (W/m2), R+B (W/m2), Teq-T (K),

    0.0, -0.0, 4.0, 1.0
    0.1, -0.4, 3.6, 0.9
    0.2, -0.8, 3.2, 0.8
    0.3, -1.2, 2.8, 0.7
    0.4, -1.6, 2.4, 0.6
    0.5, -2.0, 2.0, 0.5
    0.6, -2.4, 1.6, 0.4
    0.7, -2.8, 1.2, 0.3
    0.8, -3.2, 0.8, 0.2
    0.9, -3.6, 0.4, 0.1
    1.0, -4.0, 0.0, 0.0

    -----

    Let f = 0.5 * (-b) = 2 (W/m2)/K. Thus, as T increases 0.1 K, there is a net radiative feedback (except for B) of F = 0.2 W/m2. F+B for each 0.1 K increase is half of B alone, F+B = -0.2 W/m2.

    Because f+b has been halved, Teq will double to 2 K for R = 4 W/m2.

    Because F and B are both 0 when T = 0 K, the initial rate of T increase is the same regardless of what f and b are; C*T increases initially at a rate R = 4 W/m2. Since halving the sum f+b doubles Teq/R, T has to change twice as much, so it makes sense that the climate response time will now be twice as long (PS this makes it hard to determine climate sensitivity based on the earliest state of a climate response to a change in R).

    T (K), B (W/m2), B+F (W/m2), R+B+F (W/m2), Teq-T (K),

    0.0, -0.0, -0.0, 4.0, 2.0
    0.1, -0.4, -0.2, 3.8, 1.9
    0.2, -0.8, -0.4, 3.6, 1.8
    0.3, -1.2, -0.6, 3.4, 1.7
    0.4, -1.6, -0.8, 3.2, 1.6
    0.5, -2.0, -1.0, 3.0, 1.5
    0.6, -2.4, -1.2, 2.8, 1.4
    0.7, -2.8, -1.4, 2.6, 1.3
    0.8, -3.2, -1.6, 2.4, 1.2
    0.9, -3.6, -1.8, 2.2, 1.1
    1.0, -4.0, -2.0, 2.0, 1.0
    1.1, -4.4, -2.2, 1.8, 0.9
    1.2, -4.8, -2.4, 1.6, 0.8
    1.3, -5.2, -2.6, 1.4, 0.7
    1.4, -5.6, -2.8, 1.2, 0.6
    1.5, -6.0, -3.0, 1.0, 0.5
    1.6, -6.4, -3.2, 0.8, 0.4
    1.7, -6.8, -3.4, 0.6, 0.3
    1.8, -7.2, -3.6, 0.4, 0.2
    1.9, -7.6, -3.8, 0.2, 0.1
    2.0, -8.0, -4.0, 0.0, 0.0

    The rate of change remains proportional to Teq-T, but per unit Teq-T is half what it was when b+f had twice its value and Teq was only 1 K.

    _____________________________
  10. "Patrick 027 468
    Thermalized means that the energy is mostly conducted to adjacent non-GHG molecules that do not significantly re-radiate. "When a radiant flux is absorbed, it is assumed to be mostly thermalized" Where then does all that radiation back to the surface come from?"

    I explained that before when I explained what thermalization is.

    When a GHG molecule absorbs a photon, in most cases, it exchanges energy with other molecules before it would radiate a photon. This keeps the different gaseous substances in the same unit volume of air at about the same temperature in spite of different optical properties.

    But GHG molecules also radiate photons. They radiate photons, in each wavelength interval in each direction, at a temperature-dependent rate that statistically follows blackbody radiation as a function of temperature, multiplied by the emission cross section of each molecule.

    After a GHG molecule radiates a photon, it has less energy, but through molecular interactions, energy tends to be redistributed to keep all the gaseous substances near the same temperature in each unit volume. Hence, the air as a whole loses energy when the GHG molecules (and clouds, etc.) it contains emit photons. The net change in the enthalpy of the air (enthalpy is better to use with air because it expands as it gains heat, so that some of the energy actually does work, which increases the heat capacity (at constant pressure) relative to a constant volume value.) is caused by the absorbed radiant energy minus the emitted radiant energy, plus latent heat gain (if the latent heat present in water vapor is not counted toward the enthalpy of the air), and if within the first mm or so of the surface, conduction from the surface (and diffusion of water vapor from a wet surface, if water vapor latent heat is counted toward the enthalpy of the air).

    "If the K&E graphic was not misleading there would be no need for all the clarification/explanation/rationalization."

    Well, maybe it was intended for other scientists who would recognize what it means without much need to figure it out.
  11. "If the K&E graphic was not misleading there would be no need for all the clarification/explanation/rationalization."

    Actually, I think that's a bit like saying that this is misleading: that rain and snow, and other precipitation, comes from clouds which form when water vapor condenses (and sometimes freezes); the water vapor comes from evaporation from wet surfaces.

    Well, are dew and frost considered 'precipitation'? Sometimes clouds and precipitation evaporate before reaching the surface. Etc. Some plants get moisture directly from fog that is not actually precipitating.

    Aside from all that, though, there are a few missing links in the water vapor -> cloud -> precipiation chain:

    How does condensation take place without a surface (a small droplet has a high internal pressure because the surface tension of the spherical surface is squeezing it; this raises the equilibrium vapor pressure so that it tends to evaporate; this makes it very hard to form a droplet from pure gas, because it has to start out very small)? How do cloud droplets and ice crystals gather into particles large enough to fall out at significant rates (it is hard to grow large enough from continual condensation on particles from the vapor phase in the short time periods that are observed)?

    (Answers: generally aerosols. Some aerosols are more or less effective at nucleating cloud droplets. (If hydrophilic, they provide a surface so that the initial droplet can have less internal pressure from surface tension; perhaps much more important, if soluble, they provide a solute that can decrease the equilibrium vapor pressure, and more so at high concentrations - such as occurs when the smallest amount of water vapor condenses on a large enough aerosol. Some particles(such as salt) are considered hygroscopic because they pull moisture out of the air even when below 100% relative humidity. Moist particles go through a haze particle phase in which increasing size increases the relative humidity (relative to a flat surface of pure water) necessary to keep it in equilibrium (because the size increase dilutes the solute); once enough particles per unit volume reach some size where the necessary supersaturation for equilibrium decreases with increasing droplet size (due to the decrease in internal pressure from the surface tenstion), those particles continue to grow as cloud droplets, bringing the relative humidity back to about 100 %, and causing the evaporation of remaining haze particles. See also "Kohler curve".) Some are more or less effective at nucleating ice crystals. When the temperature gets down toward -40 deg C, homogeneous nucleation of ice within liquid water becomes more likely; the rate of homogeneous nucleation is a per unit volume rate, so larger droplets will freeze faster at higher temperatures than smaller droplets will. Larger droplets may also be more likely to contain an effect ice nucleus (?). When droplets hit ice particles, ice crystals may be nucleated. After some ice is nucleated, the surface of a droplet may freeze first, leaving liquid water within it, which can break the ice particle into smaller pieces when it freezes and expands. Different size particles have different terminal velocities and fall at different rates, so they may run into each other and sometimes combine. (Turbulence might have a local centrifuge effect that would amplify the differential motion of particles, though I'm not sure this is significant). Ice surfaces have a lower saturation vapor pressure at a give temperature than liquid water surfaces, so in the presence of a few ice particles, many small liquid droplets tend to evaporate as water vapor diffuses and is deposited to form a few larger ice particles that can then fall at a higher speed.)
  12. Correction:

    ... Larger droplets may also be more likely to contain an effectIVE ice nucleus (?)...
  13. Patrick -

    Re: 476

    Original Equation (Stefan-Boltzmann Law)

    P = e*BC*A(T^4 - Tc^4)

    which when expanded and giving ec as the emissivity of Tc gives:

    P = e*BC*A*T^4 - ec*BC*A*Tc^4

    If T = temp of the Earth and Tc = temp of the atmosphere, this is a clear subtraction of power produced by the Earth and the power produced by the atmosphere.

    Clearly, the emissivity of the atmosphere does not have any affect on the power produced by the Earth and vice-versa.

    ----------
    Here is the "Patrick's Law" equation pulled out of his imagination with no logical development shown:

    What variable shall we use for transmissivity? How about N:

    P = (e*ec*N)*BC*A(T^4 - Tc^4)

    Patrick said..."When both the hot object and colder object have emissivities e and ec that could be less than 1, and there is a layer between with transmissivity T:"
    (Patrick later changed T to N.)
    ----------
    Notice that "Patrick's Law" does not require a temperature for the layer between T and Tc!

    He just has N = transmissivity for this layer....THAT'S ALL!

    emissivity = 1 - Transmissivity - reflectivity
    http://www.optotherm.com/emiss-physics.htm

    Thus, the emissivity of this new layer of can be expressed as en = (1 - Tn - Rn)

    Substituting En into Patrick's Law gives:

    P = (e*ec*en)*BC*A(T^4 - Tc^4) and expanding gives:

    P = e*ec*en*BC*A*T^4 - e*ec*en*BC*A*Tc^4

    Now the power produced by the Earth is changed by the new atmospheric layer emissivity and the higher atmospheric layer emissivity!

    A clear violation of the Stefan-Boltzmann Law....and ABSOLUTELY HILARIOUS!

    Only a total "incompentent" and/or someone with "delusions of grandeur" would consider "modifying" the Stefan-Boltzmann Law.

    What utter drivel.
  14. Patrick 027 478
    "That doesn't quite make sense - what if there was positive feedback - then the speed could certainly change."

    Maybe it doesn't make sense because you have not grasped that feedback as used in Control Theory is different from feedback as used in Climate Science. See 475.
  15. Patrick 027 479

    So what fraction of the energy radiated from the surface and absorbed by the GHGs is thermalized?
  16. Dan -

    "Maybe it doesn't make sense because you have not grasped that feedback as used in Control Theory is different from feedback as used in Climate Science. See 475."

    Well, maybe, ... but what do you think of the feedback in climate science now? (Point being that you can't say that feedback by one definition is not positive because feedback by another definition is not positive - When climatologists say that the warming from a doubling of CO2 would be a bit over 1 K without feedbacks, they mean without feedbacks except for the increased thermal radiation to space as a function of temperature; adding positive feedback to make the warming 3 K may be entirely compatable with a negative feedback in Control theory.

    ----------

    "So what fraction of the energy radiated from the surface and absorbed by the GHGs is thermalized?"

    Essentially all of it.
  17. To whomever may wish to know:

    Regarding:
    -----------------------------
    P = (e*ec*N)*BC*A(T^4 - Tc^4)

    Then let U = e*ec*N*BC*A

    The rest was correct:

    P = U * (T^4 - Tc^4)

    This can be rewritten as:

    P = U*T^4 - U*Tc^4
    -----------------------------

    This refers to the net radiant power transfer between two layers, that is emitted and absorbed by those layers. It does not include emissions and absorptions in other layers. It assumes the layers are thin enough for each to be nearly isothermal within itself vertically (horizontal temperature variations are generally far to spread out compared to optical thickness to matter much, though there could be occasional exceptions on a small scale).


    Relative to perfect blackbody layers with entirely transparent space in between, the amoung of radiant power emitted by one layer and absorbed in another is proportional to the emissivity e1 of the emitting layer and the absorptivity of the absorbing layer, a2, and the transmissivity of the intervening space, N. The radiant power emitted by the second layer and absorbed in the first is proportional to e2 * a1 * N.

    At local thermodynamic equilibrium, emmissivity = absorptivity (at each location and time, direction, and wavelength). It is generally a very safe assumption that each small unit volume (small enough to be nearly isothermal) is approximately in local thermodynamic equilibrium. (Fluorescence is an example of the type of radiation that can be emitted when a system is not in local thermodynamic equilibrium - the energy involved has not been thermalized.)

    Hence, a2 = e2 and a1 = e1.

    So, for the radiant power exchanged by two layers by emission and absorption, the radiant power in both directions, and thus, the net radiant power, are all proportional to e1*e2*N.

    In the above equations, e1 = e and e2 = ec, but more generally:

    Pjk = net P per unit area from layer j to layer k = ej*ek*Njk*BC* ( Tj^4 - Tk^4 )

    ------

    Of course, this formulation is for the most simple case where ej, ek, and Njk are not wavelength dependent, and also requires Njk to be the transmissivity for the total flux, integrated over direction (it also assumes the refractive index of each layer is close to 1).

    Still keeping the approximation of index of refraction = 1 (a very safe approximation for the atmosphere):

    For a single direction and wavelength, if the intervening space is made of layers l=j+1 to k-1 (assuming k is greater than j, otherwise ... you get the idea), and Nm is the transmissivity of layer m, then Njk = Nj+1 * Nj+2 * Nj+3 * ... * Nk-1 [where j+n,k+n are all subscripts], and in the absence of scattering or reflection, Nm = 1 - em.

    Identifying the direction by it's angle from verticle, q, and identifying the wavelength by L,

    and for each wavelength, measuring the layer thicknesses out so each layer has the same e and thus the same N,

    Ijk(L,q) = e'j*e'k*[(1 - e'j+1)*(1 - e'j+2)* ... *(1 - e'k-1)]*BC*[ Ibb(Tj) - Ibb(Tk) ]

    = e'^2 * (1-e')^(k-j-1) * BC * [ Ibb(Tj,L) - Ibb(Tk,L) ]

    Where Ijk(L,q) is the spectral intensity (monochromatic intensity per unit interval of the spectrum) of the radiation - the radiant power per unit solid angle per unit area normal to the direction of radiation, per unit interval of wavelength (it can also be formulated as per unit interval of frequency); Ibb(T,L) is the blackbody intensity at wavelength L at temperature T.

    **In this case, e' is the emissivity along a path at angle q through a layer, so it will be different for different q. What is e' if e is the emissivity along a vertical path through a single layer?

    Well, if N is the transmissity along a vertical path, and N' is the transmissivity in the direction q, then:

    N' = 1 - e'
    N = 1 - e

    and

    N' = N^sec(q),

    because sec(q) = 1/cos(q) = the path length through the layer in the direction q relative to a vertical path length,

    and the trasmissivity through a path is equal to the product of the transmissivities of subunits of the path.

    hence, e' = 1-N' = 1 - N^sec(q) = 1 - (1-e)^sec(q)

    ---

    Solid angle (symbol w used below) is measured in steradians (symbol sr), which is the projection (from the center of a sphere) of a solid angle onto a spherical surface area, divided by the square of the radius, and thus, like radians, steradians can be treated as unitless.

    The spectral power (monochromatic power per unit of spectrum) per unit horizontal area
    = Pjk(L)
    = (integral from q = 0 to q = pi/2)[ Ijk(L,q) * cos(q) * 2*pi*sin(q)*dq ].

    The integrand has a factor of cos(q) because Ijk(L,q) is a value per unit area normal to the direction q; that unit area is a fraction cos(q) of its projection onto a horizontal surface. The factor 2*pi*sin(q)*dq is the the solid angle dwencompassed by the interval of directions dq (it is a ring centered around q=0, of angular width dq from the center and angular length 2*pi (full circle) around the center.

    cos(q) * 2*pi*sin(q) = pi * sin(2*q)

    *** pi/2 is a measure in radians, in case anyone was confused by that; pi ~=3.1415927 and pi/2 is a 90 deg angle.

    Integrating from q = 0 to q = pi/2 encompasses all the contributions from a hemisphere of solid angle. It is not necessary to integrate over the whole sphere of directions because the quantity being integrated is a net value that is the sum of radiant intensities in opposing directions. The same result would be obtained by integrating the contributions to power per unit horizontal area from the intensity from each direction, over the whole sphere of solid angle.

    PS:

    For isotropic radiation I(q) = I, the radiant flux per unit area F =

    (integral from q = 0 to q = pi/2)[ I * cos(q) * 2*pi*sin(q)*dq ].

    = I * (integral from q = 0 to q = pi/2)[ pi*sin(2*q)*dq ]

    = I * pi * [ -cos(pi) + cos(0) ]/2

    = pi * I


    The solid angle in the hemisphere is:

    (integral from q = 0 to q = pi/2)[ 2*pi*sin(q)*dq ]

    = 2*pi*[-cos(pi/2) + cos(0) ]

    = 2*pi

    There are 2*pi steradians in a hemisphere.

    ---

    There is a less clunky way to figure this stuff out - we can let the layer thickness as measured by optical thickness become infinitesimally thin, and use calculus:

    still at a given wavelength:

    In a unit volume, there is a density of emission cross section, the emission cross section per unit volume = ecsv. This is equal to the absorption cross section per unit volume if at local thermodynamic equilibrium. ecsv from one substance is equal to the product of the emission cross section of that substance per unit mass times the mass of that substance per unit volume. The exv of different substances and additional amounts of substances add linearly to give the total.

    The exv is the effective blackbody surface area facing a given direction - each molecule contributes to ecsv with a cross-section. It is possible for ecsv to be direction-dependent (analogous to the molecules having elongated optically-defined shapes with some non-random alignment) but it is generally the case in gases that exv is isotropic (analogous to each molecule being spherical or to random alignments of molecules, so that on average, each molecule looks like a spherical blackbody that presents the same cross-sectional area in each direction).

    (Of course, the image of a molecule having some well-defined shape with a clear edge is just a mathematical equivalent to what it contributes, on average, to bulk optical properties.)

    (It is possible to imagine some macroscopic analogue to molecular cross-sections - imagine how radiation propagates through a room filled with blackbody spheres - except, there will be some diffraction around those spheres, which corresponds to a contribution to the scattering cross section on the molecular or small particle scale. Scattering on the molecular scale has a macroscopic analogue of mirrored and/or clear glass spheres or other shapes. The total (extinction) cross section (emission + scattering = extinction) can be different in different directions but must be the same in a pair of opposing directions; however, it would be possible to construct a macroscopic analogue in which the scattering and emission cross sections vary between pairs of opposing directions - for example, spheres that are mirrored on one side and perfect blackbodies on the other - but I think this might not have a molecular analogue, at least for radiation with wavelengths longer than the molecular scale (??).)

    A unit volume is equal to a unit path length with a given area. Thus, the escv, the emission cross section per unit volume is equal to the emission cross section (an area) per unit area per unit path length, and is thus a fraction of area per unit path length.

    We can consider an infinitesimally thin layer, of thickness dx, so that the fraction of area per unit path length, escv*dx, is nearly zero. Since the fraction of area of the whole layer dx is nearly zero, the fraction of area of any sublayer must be nearly zero, so the fraction of any sublayer cannot significantly block the fraction of another sublayer (the molecules that contribute to the emissivity are spread out enough so as to be very unlikely to align on top of each other and block each other from view in the direction x). Thus, essentially the entire emission cross section within such a thin layer dx is visible in the direction x. Thus, the emissivity of that layer in the direction x is equal to the fraction escv*dx.

    And the emissivity of the layer 2dx will be approximately 2*escv*dx, but actually just slightly less than that; the emissivity approaches 1 and cannot increase farther. The transmissivity (in the absence of scattering, and with the absorption cross section = emission cross section) is 1 - escv*dx for the layer dx in the x direction. The transmissivity through a layer n*dx is approximately 1 - n*escv*dx if n is not large, but it is exactly (1-escv*dx)^n;

    The transmissivity decreases by a fraction escv*dx for each dx.

    Where the transmissivity is N,

    dN/N = -escv*dx

    N = 1 when x = 0. Integrating:

    ln(N) - ln(1) = -escv *(x-0)

    ln(N) - 0 = -escv *x

    N = exp[-escv *x]

    For a radiant intensity at x, I(x), going in the x direction, I(x) = I(0)*N(x) = I(0) * exp[-escv*x], which I believe is refered to as Beer's Law - this applies to the portion of the radiation that was present at position 0 that reaches x, and does not include radiation emitted along the path from 0 to x.

    ------------

    For the spectral intensity Ixs emitted from a dx at position x that reaches position x = s (going in the positive x direction):

    Ixs = Ibb(T(x),L) * escv * dx * exp[-escv *(s-x)]

    Let Is is the intensity of radiation at s, in the x direction, emitted from the layer between x=0 and x=s:

    Is = (integral from x = 0 to x = s)[ Ibb(T(x),L) * escv * exp[-escv *(s-x)] * dx ]

    Which can be numerically integrated for any T(x) function, but if we assume T(x) is a constant T between 0 and s:

    Is = (integral from x = 0 to x = s)[ Ibb(T,L) * escv * exp[-escv *(s-x)] * dx ]

    Is = Ibb(T,L) * escv * exp[-escv *s] * (integral from x = 0 to x = s)[ exp[-escv *(-x)] * dx ]

    Is = Ibb(T,L) * escv * exp[-escv *s] * (1/escv) * [ exp[escv *s] - exp[escv *0] ]

    Is = Ibb(T,L) * exp[-escv *s] * [ exp[escv *s] - exp[escv *0] ]

    Is = Ibb(T,L) * [ exp[escv *(s-s)] - exp[escv *(0-s)] ]

    Is = Ibb(T,L) * [ exp[0] - exp[-escv *s] ]

    Is

    = Ibb(T,L) * [ 1 - exp[-escv *s] ]

    = Ibb(T,L) * [ 1 - Ns ]

    where Ns is the transmissivity of the layer from 0 to s in the x direction.

    Not surprisingly, this implies the emissivity of the layer, es, is 1 - Ns.

    Of course, the absorptivity of the layer is equal to the emissivity of the layer (local thermodynamic equilibrium)

    ------

    So the net radiative transfer in the x direction, from emissivity to absorptivity, between two layers g and h, seperated by layer s (where g, h, and s are the thicknesses of the layers in the x direction), defined as positive in the direction from g to h, is:

    Igsh(L) = eg * ah * Ns * Ibb(Tg,L) - eh * ag * Ns * Ibb(Th,L)

    where ej, aj, and Nj are the emissivity, absorptivity, and transmissivity of the layer j.

    Since ej = aj (local thermodynamic equilibrium):

    Igsh(L) = eg*eh*Ns * [ Ibb(Tg,L) - Ibb(Th,L) ]

    eg = 1 - exp[-ecsv *g]

    eh = 1 - exp[-ecsv *h]

    Ns = exp[-ecsv *s]

    ---

    What if we want to know the contribution of this radiant transfer (which is per unit area normal to the x direction) to the radiant transfer PGSH(L) per unit horizontal area, where G, S, and H are the layer thicknesses measured in the vertical direction?

    Let x be an angle q from vertical. Consider the solid angle dw.

    Note that h*cos(q) = H, so h = H*sec(q), and g = G*sec(q), and s = S*sec(q)


    d(PGSH(L))

    = Igsh(L) * cos(q) * dw

    = eg*eh*Ns * [ Ibb(Tg,L) - Ibb(Th,L) ] * cos(q) * dw

    = [ Ibb(Tg,L) - Ibb(Th,L) ] * (1 - exp[-ecsv *g]) * (1 - exp[-ecsv *h]) * exp[-ecsv *s] * cos(q) * dw

    = [ Ibb(Tg,L) - Ibb(Th,L) ] * (1 - exp[-ecsv *G*sec(q)]) * (1 - exp[-ecsv *H*sec(q)]) * exp[-ecsv *S*sec(q)] * cos(q) * dw

    Assuming layers G and H are isothermal over horizontal distance (generally a good approximation for the spatial scales involved in radiative energy transfers, except in some conditions), then the same value applies to all directions at an angle q from vertical, so dw can be replaced with the ring-shaped solid angle 2*pi*sin(q) * dq, and integration can be done over q from 0 to pi/2 to find PGSH(L).

    Of course, ecsv will vary over vertical distance (And anywhere the path runs into a cloud, or surface, etc.) Within the atmosphere, ecsv will tend to be proportional to air density at wavelengths dominated by well-mixed gases (in the absence of clouds), but at many wavelengths (in between absorption lines), decrease faster with height, while at some wavelengths (near line centers), decrease more slowly. Water vapor is concentrated downward, and there are clouds, etc.

    ---

    The mathmatics can be simplied by using optical depth or optical thickness-based coordinates. A unit 1 optical thickness is the distance x over which ecsv*x = 1, so that transmissivity is 1/exp(1).

    Various different vertical coordinates are used in atmospheric science, including geometric height z (as in x,y,z), pressure (x,y,p), potential temperature (x,y,theta), and sigma coordinates (x,y,sigma), where sigma is set to one value at the surface regardless of surface pressure, and one value at infinite height (p=0), and varies linearly with pressure in between.

    Any of those vertical coordinates can be mapped onto optical thickness coordinates and vice-versa based on the optical properties at each position. Different mappings will be necessary for different wavelengths, however, so integration over the spectrum to find total fluxes will require converting back to one of the other vertical coordinates. Pressure coordinates will tend to have nearly-constant heat capacity per unit vertical distance, which is convenient for converting an energy flux convergence (in the absence of horizontal net fluxes, -1 * the vertical derivative of the net upward energy flux (power) per unit area) to a rate of temperature increase.

    Redefining (to conserve on variables) G, H, and S to vertical layer thicknesses in optical thickness coordinates:

    eg = 1 - exp[-G*sec(q)]

    eh = 1 - exp[-H*sec(q)]

    Ns = exp[-S*sec(q)]

    d(PGSH(L))

    = eg*eh*Ns * [ Ibb(Tg,L) - Ibb(Th,L) ] * cos(q) * dw

    =

    [ Ibb(Tg,L) - Ibb(Th,L) ] * (1 - exp[-G*sec(q)]) * (1 - exp[-H*sec(q)]) * exp[-S*sec(q)] * cos(q) * dw


    ---

    For dw = 2*pi*sin(q) * dq,

    d(PGSH(L))

    =

    [ Ibb(Tg,L) - Ibb(Th,L) ] * (1 - exp[-G*sec(q)]) * (1 - exp[-H*sec(q)]) * exp[-S*sec(q)] * pi*sin(2*q) * dq

    Using optical thickness coordinates is convenient; the surface and space can be regarded as layers with infinite thicknesses, over which T is (for x,y, and time) constant.

    To figure out the total net radiant energy transfer emitted from one layer G and absorbed in all layers above Hj, one must integrate over the different Hj layers. If one wants to know the total net radiant energy absorbed from layer by all layers above and below, one must sum the net fluxes from the other layers. If one wants to know the net upward flux at a given position, one must integrate the net upward flux between each pair of layers across that position. Of course, one must also integrate over solid angle and over the spectrum (or the portion of the spectrum being considered).

    Fortunately, the qualitative effects are easy to visualize. One can see some fraction of perfect blackbody radiation radiation intensity from a layer according to that layer's optical thickness, the optical thickness of the space between you and that layer, and the angle; and putting yourself in the place of another layer at the viewing location, the fraction absorbed depends on your optical thickness in that direction - that, multiplied by the other factors, is how much 'you would see' of the other layer from that direction, and that other layer would see the same amount of 'you'. With radiation, 'what you see' (pretend you can see at any wavelength) is what you get.

    ---

    What about scattering?

    For the extinction cross section per unit volume xcsv
    and the scattering cross section per unit volume scsv,

    xcsv = ecsv + scsv.

    The transmissivity over distance s without scattering Ns = exp[-xcsv * s]

    There will, however, be a glow of diffuse scattered radiation, which will add to the radiant intensity in other directions; some scattered radiation still reaches some forward distance after scattering.

    ... it's complicated (but qualitatively easy to understand - scattering, and reflection in general, decrease transmission between layers going in any one direction but replace what they block from view with radiation from the same side of the scattering/reflection as the viewing position; for a mix of absorption and scattering - any blocking of radiation from behind will be replaced with a combination of radiation from the blocking layer and radiation from the same side of the blocking layer as the viewing position, but with varying intensities depending on the temperature of the emitting sources); to be continued if/when I get to it...
  18. Patrick 027 485

    There can be no significant NET Climate Science type positive feedback. So there could be at least as much positive as there is negative from the temperature increase. I expect that additional negative Climate Science type feedback will be discovered that is associated with clouds.

    I modeled the K&E graph except I included thermalization as an unknown. (first model April 29, updated today) I used average cloud temperature 258 K, cloud emissivity .5, cloud coverage 50% (see Zero-dimensional models at http://en.wikipedia.org/wiki/Climate_model) and varied the fraction of cloud radiation that reaches the ground from 50% to 90%. (I assumed cloud radiation to be graybody type from the water/ice particles) Thermalization varied from 16% for 50% cloud-to-ground to 24% for 90% cloud-to-ground. Considering all of the GHGs that the cloud radiation needs to penetrate to reach the ground, the fraction that makes it is probably closer to 50% than 90%. That which is not thermalized is needed as back radiation to the surface. I was surprised to see such a low number for thermalization but can't find anything wrong. Using their values for 'absorbed by atmosphere', thermals and evapotranspiration and correctly combining numbers I get acceptable agreement with what they got.
  19. Patrick -
    In your Post #461 you said...

    "Surface absorbs RSe from the sun and Rae from the atmosphere. Surface loses heat by Rea to the atmosphere, Res to space, and Cea (convection) to the atmosphere. The atmosphere absorbs RSa from the sun and Rea from the surface, and gains Cea from the surface, and loses energy by Rae to the surface and Ras to space.

    THESE EQUATIONS ARE BASED ON THE CONSERVATION OF ENERGY:

    Rate of energy gain by the surface = RSe + Rae - Res - Rea - Cea
    Rate of energy gain by the atmosphere = RSa + Rea + Cea - Ras

    In climatic equilibrium, the average of each rate of energy gain is zero.

    Where is energy being created or destroyed?"
    ----
    Answer:

    + Rae = energy absorbed by the Earth's surface. Rae came from the Atmosphere that got it's energy from the Earth's radiation.

    The Earth's radiation cannot cause it's own radiation to increase. It's called Energy Creation....a violation of Conservation of Energy.

    Further, Rae IS GREATER THAN Rse and Rse is the energy absorbed by Earth's surface from the SUN ...THE ONLY ENERGY SOURCE!..another Energy Creation and violation of Conservation of Energy.
    -------------------
    Your equations ARE BASED ON A VIOLATION OF CONSERVATION OF ENERGY.

    Like I have said...With Patrick, 1 + 1 = anything but 2.

    What great Comedy!
  20. CORRECTION TO http://www.skepticalscience.com/argument.php?p=19&t=488&&a=18#3338

    The Rae term was mistakenly missing in the atmospheric energy budget; here is the correction:

    THESE EQUATIONS ARE BASED ON THE CONSERVATION OF ENERGY:

    Rate of energy gain by the surface = RSe + Rae - Res - Rea - Cea

    Rate of energy gain by the atmosphere = RSa + Rea + Cea - Rae - Ras
  21. Re 467 Mizimi - yes, basically, GHGs reduce the rate of energy loss to space from the surface for a given temperature profile. Additional amounts can also reduce the energy loss to space from the lower atmosphere when opacity has become high enough.
  22. Following radiation from one point along a path, scattering removes the radiation from the direction of the original radiation.

    Considering the radiation I(0) in one particular direction x that is present at x = 0, the amount I(x) found at any other x (that has continued unabsorbed and unscattered; not including radiation that has been emitted or scattered into the path between 0 and x) again follows Beer's Law:

    I(x) = I(0) * exp(-xcsv * x)

    where xcsv is the extinction coefficient, equal to the extinction cross section per unit volume (a unit volume as measued in the coordinate system being used).

    xcsv = ecsv + scsv;

    extinction cross section = absorption cross section (assumed to be equal to emission cross section) + scattering cross section.

    optical thickness of path length x = xcsv * x

    The contributions of scattering and absorption/emission to optical thickness add linearly.

    The ratio of the scattering cross section to the extinction cross section is called the 'single-scatter albedo' (don't remember if that is supposed to be hyphenated or not).

    The loss to scattering and absorption per unit x can be found by taking the rate of change over x in the reverse direction:

    -dI/dx = xcsv * I(0) * exp(-xcsv * x)

    = xcsv * I

    = ecsv * I + scsv * I

    where the last two terms are the absorption per unit x and the scattering per unit x; the single scatter albedo is the ratio of the scattered I to the total loss of I without scattering and emission into the path.

    Scattered radiation of course continues in a different direction, and adds to I along some other path. If the new direction is not perpendicular to the original, then there has been either forward scattering (the angle between the directions is less than 90 deg) or back scattering. Note that when refering to solar radiation that is back scattered to space, the term may apply a bit differently, because as the sun gets farther from directly overhead at any one location, a greater portion of radiation that is forward scattered relative to the direct rays of the sun is still redirected upward.

    The directional distribution of scattered radiation varies; the scattering of sunlight by air molecules is dominated by Raleigh scattering, where the scattered radiation intensity is distributed in a symmetric dipole pattern aligned with the incident radiation, with equal amounts forward scattered and back scattered, and the lowest scattering intensities are found in the ring of directions in between. In at least some other types of scattering, the forward scattering dominates. For scattering by particles, Raleigh scattering occurs for particles much smaller than wavelength; Mie scattering occurs over a range, including when the particle size and wavelength are similar; geometric optics dominates for particles much larger than a wavelength (see p. 307, "Atmospheric Science - An Introductory Survey", Wallace and Hobbs, 1977 (but there is at least one other more recent edition, I believe)).
  23. Refinement to 487:

    Assume an average emissivity for earth’s surface at 0.98 then the energy radiated from the surface at 288°K average is 382 W/m2. K&E show 40 W/m2 or 10.5% getting all the way to space. Barrett at http://www.warwickhughes.com/papers/barrett_ee05.pdf , assuming 50% cloud cover, calculates 11.2% which corroborates this estimate. Since radiation from the clouds to the surface is not half-blocked by cloud cover, it can be said with reasonable confidence that the fraction of radiation that leaves the clouds going towards the surface that actually gets all the way from the average cloud cover to the surface is about 21%. Plugging this value in to the model results in the determination that only about 37 W/m2 or 11% of the radiation that leaves the surface but doesn’t go through the ‘window’ getting thermalized.
  24. Some interesting things about scattering:

    For geometric optics, aside from the effects of diffraction, the scattered light from a mirrored sphere(specular (as opposed to diffuse) reflection: angle of incidence = angle of reflection) is redirected equally in all directions; it is isotropic.

    If my understanding is correct on this point: while one might expect that, when in the realm of geometric optics, the extinction cross section of a sphere (either by scattering by specular or diffuse reflection, scattering by refraction, or absorption) would be equal to its actual cross sectional area. But it is actually twice that, because of diffraction around the edge. I would think that diffraction contributes essentially to just forward scattering, and for relatively large objects compared to wavelengths, the scattered rays are mostly not at large angles to the incident radiation (so it is hard to observe over relatively short distances where slight changes in direction have little effect on location).

    -----

    Depending on conditions, a photon may scatter many times (multiple scattering) between emission and absorption.

    When a given scattering cross section per unit volume produces mainly forward scattering within a narrow cone close to that of the incident rays, then multiple such scatterings might be approximated by using a wider scattering distribution with a smaller scattering cross section per unit volume (fewer scattering events with greater directional change per event).

    ----------

    In clear sky, outside the disk of the sun itself, the highest intensity radiation may be observed (careful!) in a narrow cone of solid angle just around the sun. I think this cone is too narrow and focussed to be explained by Raleigh scattering by air molecules - I presume it is by aerosols (dust, etc.). Aside from that, at least in the middle of the day, the brightest parts of the sky (the highest intensities) are generally nearer the horizon - this is because one looks through more air when looking near the horizon than straight up; the 'glow' of scattered radiation comes from a greater scattering cross section per unit area because of the longer optical path.

    (For the same reason, the incandescent glow of LW radiation from the sky will tend to appear brighter near the horizon - in that case, there is greater emission cross section per unit area, both from the total atmosphere (between the surface and space) and from the warmer, lower layers (at wavelengths with higher opacity, low level inversions could make the sky appear dimmer near the horizon). - However, with scattering, there is the added complexity of the distribution of scattering relative to incident rays; it may be necessary for significant multiple scatterings (including (generally) diffuse reflection from the surface) to occur to explain the visible brightness near the horizon if the sun is overhead (???))

    ----------

    A haze or thin cloud may appear significantly brighter when viewed toward the sun (careful!) than when viewed looking away from the sun. This indicates that forward scattering dominates over backscattering.

    What happens in thick clouds? For a thick cloud, each layer of cloud adds another round of scattering for forward scattered radiation. Also, even if an initially parallel set of rays is spread out over a small solid angle for an individual scattering, successive forward scatterings continue to spread the rays out more and more, and some 'net backscattering' can occur from several forward scatterings. Whatever has not left the cloud is still subject to the high density of scattering cross section. The initial scattering is concentrated within the cloud toward the source of the radiation, and while successive scattering allows further penetration, it also spreads out the directions of the rays, so if the cloud is thick enough, the rays may be going in opposite directions nearly equally before getting more than halfway through the cloud, and so more escapes out the side it came in than escapes the other side if the cloud is thick enough.
  25. ". Plugging this value in to the model results in the determination that only about 37 W/m2 or 11% of the radiation that leaves the surface but doesn’t go through the ‘window’ getting thermalized. "

    Setting aside scattering (which is a minor issue for LW radiation), the radiation emitted by the surface that does not directly reach space must either be absorbed by clouds or absorbed by something else (gases, or aerosols) - either way, it is essentially all thermalized.
  26. Near the surface, most is re-radiated by GHGs as required for the surface to radiate as it does. Only 37 W/m2 needs to get far from the surface as required to produce heat balance using the other K&T values.
  27. The other K&T updated values referred to are the 78 of incoming absorbed by the atmosphere, the 17 thermals and 80 latent. The 78 seems high. Reduction of this increases thermalization by the same amount and decreases back radiation by the same amount. Reducing the 78 to 10 increases thermalization by 68 to 105 and reduces back radiation from the atmosphere to 237. That which gets from average clouds to ground remains at 13.
  28. "Near the surface, most is re-radiated by GHGs as required for the surface"

    This can be a point of confusion: people throw around terms like 're-radiate' a bit too loosely.

    The rate of thermalization is sufficiently high in the great majority of the atmopshere by mass and by optical thickness that GHGs do not radiate by fluorescence - which is what some people seem to think (and they may get that idea from the term 're-radiate'); instead, GHGs and other greenhouse agents emit photons mainly from thermal energy, which they can get from the bulk air from molecular collisions.
  29. In some cases, it may be necessary to consider I not just along specific directions and at specific wavelengths, but also by polarization. At a given direction q and wavelength L, the intensity I(q,L) is the sum of contributions of I from each polarization P: I(q,L,P).

    Scattering and emission/absorption cross sections could be functions of polarizations in cases where the particles involved lack spherical symmetry and are not oriented at random. Within the atmosphere, ice crystals can have prefered orientations, and rain drops flatten slightly as they fall (though I wouldn't think the later has much significance overall).

    Reflection where there is a relatively sharp change in the index of refraction is also a function of polarization.

    -------
    Contributions to the intensity I(q,L,P) could also be seperated by phase - I don't know if there would be much point in this except in finding the brightness temperature of laser light that corresponds with the entropy; perfect blackbody radiation, as well as being isotropic across all q, is evenly distributed over polarizations P and phases (it is incoherent); so it would be of interest to know the temperature of a blackbody that can emit radiation with I(q,L,P) for a single phase equal to the intensity I(q,L,P) of some coherent radiation (defining brightness temperature with such specificity, any significant nonzero I that is either perfectly monochromatic (dL = 0), perfectly polarized (dP = 0), or perfectly coherent would have infinite brightness temperature and zero entropy - it would be all work and no heat).

    ---------

    Particles much smaller than a wavelength of radiation can change the overall index of refraction, but an even distribution of such particles would not cause a spatial variation in the index of refraction.

    When the index of refraction for the bulk medium (as opposed to just the scattering and absorption/emission agents) must be considered, then it will be more convenient to use frequency v instead of wavelength L to specify the part of spectrum considered. The monochromatic intensity is better given as the intensity per frequency interval dv because wavelength intervals dL can change. Alternatively, one could refer to L# as the wavelength the radiation of that frequency would have if in a vacuum (and then define the spectral intensity as the intensity per unit L#.

    Following I along a path when the real component of the index of refraction, n, changes, I in the absence of scattering, reflection, or absorption and emission, is not what is conserved along the path; it is I * n^2 that is conserved. This is because a unit of solid angle dw that envelopes a group of rays will be compressed or expanded by the square of the ratio of different n values following the refracted paths. This relates to total internal reflection: for a flat surface, there is a cone of accepantance through which some radiation can pass from high n to low n (100 % if there is a perfect antireflection coating (if and when such a thing exists); outside of that cone, there will be (in the absence of scattering and absorption) 100 % reflection. If the transition from high n to low n is gradual, the cone of acceptance applies, but instead of reflecting at a specific interface, ray paths outside the cone curve back gradually, as ray paths within the cone curve to spread out to fill the full hemisphere of solid angles (additional transition to lower n defines a narrower cone of accetance for the rays that reach over the larger variation of n without curving back or reflecting). Relative to wavelength, small-scale texterization can mimic a gradual variation of n and reduce reflection within the cone of acceptance in that way; larger-scale texterization can reduce reflection via successive intersections of a ray with the interface, so that the amount that transmits across the interface accumulates with each intersection; in that case, the cone of acceptance may not be so easily defined due to different angles, but ...
  30. I'm OK with radiated. I don't know which is less confusing to most people.
  31. (continued)...

    ... but for the flux per unit area (integrated over solid angle), the fraction that is trapped by total internal reflection remains the same.

    -------------

    CORRECTION: following ray paths without scattering, absorption and emission, or reflection, over variations in n:

    it is I / n^2 that is conserved.

    Not I * n^2 .

    -------------

    Considering the case where there is no reflection, scattering, or emission and absorption along a path over varying n, the conservation of I / n^2 implies that blackbody radiation Ibb(v,T) must be proportional to n^2, so that observed blackbody radiation at any given n, such as n=1 (vaccuum, approximately in air), has the same intensity regardless of the n at the location of emission.

    When following ray paths, the direction of the path obviously can bend as n varies, so the direction Q is a function of location (with location itself being a function of direction from another given location - outside of some simplified conditions, calculating location and direction requires integration of directional changes along the path from an initial location.) Q in general has two components, which can be the angles in spherical coordinates; the q used earlier is the angle from the top pole (the colatitude: latitude - 90 deg), which is all that is necessary if conditions do not vary in planes perpendicular to the axis of the coordinate system; more generally, the angle around the axis (analogous to longitude) could also be specified.

    n can depend on direction in materials, in which case n is a function of Q. I would guess it could also be a function of polarization P (is that what produces the double images through calcite?: note to self - look up birefringence). Many people are aware that in materials it is often a function of frequency v (why we have rainbows).

    Of course, n = c_vacuum / c_material, the ratio of the phase speed in a vacuum to the phase speed in the material (in the direction being considered) ...

    BUT when n is a function of frequency v, the group velocity - the direction and speed at which energy is carried by wave amplitude propagation - can be different from the phase direction and speed. Presumably, they will still be in the same direction (or directly opposite each other, as in the case of certain metamaterials with a negative index of refraction) when n does not vary with direction; I wonder what happens if n varies in direction; then perhaps group velocity might be at some angle to phase propagation - in which case, the ray path to follow when evaluating changes in I would be along the path defined by group velocity, ... and the n for which I / n^2 is conserved in the absence of reflection, scattering, and absorption and emission, is (??????).

    (PS it is actually common to deal with fluid mechanical waves in the atmosphere (and, I presume, the ocean) (gravity waves, inertio-gravity waves, Rossby waves, Rossby-gravity waves, Kelvin waves) for which the group velocity and phase propagation, relative to the air (as it moves or doesn't), are perpendicular to each other.)

    PS n is the real component of the index of refraction; the index of refraction can actually be a complex value, with an imaginary component that is related to the absorption cross section per unit volume (the absorption coefficient).

    ------------

    Fortunately, radiation transfer through the atmosphere is little affected by variations in the real component of the index of refraction of the bulk air (obviously it figures into the scattering properties of cloud droplets and aerosols). The most significant effects I am aware of is the twinkling of starlight and the shifting of the sunset and sunrise times.

    (Also, gravitational redshifting is not a big effect for radiation into and out of the Earth; I won't bother going into the relativistic effects on radiation here.)

    But it is interesting to consider what qualitative effect it would have, as well as the spherical geometry of the Earth (the layers of the atmosphere have slightly greater area than the surface, and so can emit the same total power at a slightly lower temperature than otherwise.)

    The effect of n slightly greater than 1 increases the length of the day as seen from the surface. This implies that, aside from atmospheric albedo contributions, the surface of the Earth intercepts a greater amount of sunlight than would actually pass through it's cross sectional area. Indeed, the implication of the index of refraction of the atmosphere being slightly greater than space is that the Earth would appear slightly bigger from space than it actually is - In general, any given spherical surface below the 'top' of the atmosphere will be magnified by the n greater than 1 of the air above it - but obviously, cannot appear any bigger than the spherical boundary of the layers of air that are magnifying it. This happens because, when looking at the edge of the Earth, the line of sight intersects surfaces of constant n within the atmosphere around the Earth at a glancing angle and is bent toward the Earth's surface; the visible edge of the Earth corresponds with the lines of sight that bend only enough to reach the surface nearly horizontally, and those lines of sight will reach the surface somewhat behind the front half of the Earth (as defined by viewing position). (This description applies to the view from an infinite distance; obviously, without refraction, much less than half the Earth would be visible from nearby, but refraction would extend the area visible by some amount.)

    If n were higher for solar radiation than terrestrial radiation, then it would increase the solar heating of the Earth and any given layer of atmosphere more than it would increase the LW radiative cooling to space from the same vertical levels. I don't know if that is the case, but it is a very small effect that can be ignored for most purposes anyway.

    Most of the radiation to space comes from within the troposphere, and a majority of direct solar heating occurs at the surface (or within the surface material, as in the ocean). For a very quick analysis to put some likely bound on the effect of this, suppose the radiation to space came entirely from a height of 16 km above the surface (that's within the stratosphere outside of the tropics, and close to the tropopause within the tropics), while all solar heating were at the surface. In that case, the effective surface area emitting to space would be about 2*(16/6371) ~= 0.5 % larger than the surface that is heated by the sun, which would allow the effective emitting temperature to be (0.5 % /4) = 0.125 % smaller than that required to radiate to space enough to balance solar heating with the same area as the surface. That would be about 2.55 / 8 ~= 0.32 K cooler. Not very much compared to a 33 K greenhouse effect overall.

    So both the increase in area with height and the refraction by the air can be ignored for climatologically-relavent radiative fluxes. Also note that the refraction would tend to counteract the effect of greater area with height, since it would magnify areas beneath greater amounts of atmosphere more than higher level areas.

    ------------
  32. ________________________

    MORE ABOUT REFRACTION: COMPLEX N:

    Let the complex index of refraction be N.

    Previously I used n to represent the real component of the index of refraction. It would be better to refer to that as nr.

    (From class notes):

    Imaginary component of the index of refraction ni:

    The absorption coefficient (equal to the absorption cross section per unit volume)

    =

    4 * pi * ni / L#

    where L# is the wavelength in a vacuum of radiation with the same frequency v.

    ----

    The real and imaginary components of the index of refraction,

    N = nr + i*ni ,

    do not vary independently of each other over v or L#. nr and ni are related by the Kramer-Kronig relationships.

    ----

    My understanding is that, When (magnetic) permeability is not different from a vacuum, the complex dielectric coefficient is equal to the square of the complex index of refraction.

    ------------------------
    **** IMPORTANT CLARIFICATION/CORRECTION ****

    The statement that I / nr^2 was conserved in the absence of absorption, reflection, or scattering is true at least in so far as the group velocity and phase propagation are in the same direction.

    However, my intent was that the change in I over a distance dx along a ray path would then be given by letting I# = I / nr^2, and then the differential formula would be:

    d(I#(Q,v,P)) = Ibb#(v,T) * ecsv - I#(Q,v,P) * (acsv + scsv) * dx + Iscat * dx

    where Ibb#(v,T) = Ibb(v,T) * nr^2 is the blackbody intensity for the index of refraction (given per unit frequency dv and per unit polarization dP) and:

    ecsv, acsv, and scsv are the emission, absorption, and scattering cross sections per unit volume,

    Iscat * dx is the radiation scattered into the path from other directions,

    and any reflection (removing and/or adding to I along the path) at an interface within dx is included in the scattering terms.

    And of course, ecsv = acsv if in local thermodynamic equilibrium.

    ---

    Such a relationship is constructed with I# = I/ nr^2 is based on Snell's law where nr*sin(q) = constant (not quite true, actually (???) - see below) is the relationship that determines q as a function of N, where N varies only in one direction z (planes of constant N are normal to the z direction) and q is the angle from the z direction.

    I# = I / nr^2 is derived from Snell's law, by determining that the solid angle dw that encompasses a group of rays expands or compresses with variation in the index of refraction, specifically so that dw is proportional to 1/nr^2.

    **** HOWEVER:

    When N is complex, Snell's law actually still uses the complex N, not just it's real component. I haven't entirely figured out what that means for q, though I have the impression that nr*sin(q) = constant should be at least approximately true.

    Snell's law itself is based on the requirement that the phase surfaces of incident and tranmitted waves line up at an interface, and that the phase speed is inversely proportional to N when N = nr. What is the phase speed when N has a nonzero imaginary component?

    And then there is also the complexity of what happens if group velocity is not in the same (or exact opposite) direction as the wave vector (the wave vector is normal to phase planes and thus is in the direction of phase propagation).

    --------

    So for the time being, let I# = I / n^2, where n is whatever N-related value that works in that relationship and also:

    d(I#(Q,v,P)) = Ibb#(v,T) * ecsv - I#(Q,v,P) * (acsv + scsv) * dx + Iscat * dx

    At least when N = nr and N is not a function of direction Q, n = N = nr.

    ________________________

    REFLECTION AND EVANESCENT WAVES:

    When the entirety of wave amplitude and wave energy is reflected by an interface (as in total internal reflection), some of the energy actually penetrates across the interface. The portion of the wave across the interface that is associated with the reflected portion of the wave is called an evanescent wave. The energy and amplitude of the evanescent wave decay exponentially away from the interface into the region where the reflected wave cannot propagate (in the time average, there is no group velocity component normal to the interface within the region of the evanescent wave). The evanescent wave is mathematically required for continuity of the electric and magnetic fields, and for the time-integrated divergence of the energy flux to be zero over a wave cycle when there is a constant incident energy flux (and no absorption or emission).

    If there is another interface, beyond which the wave could propagate, then waves energy can emanate from that interface to the extent that the evanescent wave penetrates to it. This flux of energy must pull energy from the evanescent wave, which results in reduced reflection from the first interface. This is how wave energy can 'leak' through a barrier. The same general concept applies to mechanical waves, including fluid mechanical waves in the atmosphere and ocean, and to quantum mechanical waves - electrons tunnel through barriers via evanescent waves.

    Absorption within the region of the evanescent wave allows some net energy flow across the interface and thus reduces the reflectivity, so that even when a nonzero 'transmissivity' is not allowed except for tunneling, the reflectivity can be less than 1.

    I'm not sure but I think absorption on the side of a transmitted wave might also reduce reflectivity even when trasmission is allowed. (??)

    ________________________

    I think I actually have the equations necessary to find some answers to questions just raised (if unable to find them elsewhere), but it will take time and it isn't pertinent to the matter here (I've already gone off on these tangents far enough).

    ________________________
    GROUP VELOCITY:

    Group Velocity in geometric space specifically is equal to the gradient of the angular frequency (omega = 2*pi*v) in the corresponding wave vector space (a wave vector is a vector with components of wave numer; wave number is equal to 2*pi / wavelength measured in the correpsonding dimension; the wavelength in the direction of phase propagation is equal to 2*pi divided by the magnitude of the wave vector).

    Where the wavevector = [k,l,m], where k,l,and m are the wave numbers in the x, y, and z directions

    Angular frequency = omega

    group velocity in x,y,z space = [ d(omega)/dk , d(omega)/dl , d(omega)/dm ]

    phase speeds cx, cy, cz in directions x,y,z

    cx = omega/k
    cy = omega/l
    cz = omega/m

    IN the direction of phase propagation, the phase speed c and wavelength L are related as:

    c = v*L

    --------

    IT IS POSSIBLE for some materials, at some values of v, to have a real component of the index of refraction less than 1. This (tends to or approximately??) corresponds to a phase speed that is greater than the speed of light in a vaccuum. This does not violate special relativity; the group velocity is the direction and speed of energy flux and information transport, and the group velocity will not be larger than the speed of light in a vaccuum.



    ________________________

    Relativity: Effects are gravitational lensing and gravitational redshift, and effects related to relevant motion (of the Earth and sun, for example.

    One effect of relative motion is that small dust particles orbiting the sun tend to fall in toward the sun over time (their semimajor axes shrink) because as they orbit, they recieve radiation more from their leading side because of their motion, so radiation pressure tends to slow them down. Of course, radiation pressure also pushes them out - on the other hand, any stable orbit would be such that the outward radiation pressure would only partly cancel the gravitational acceleration, thus causing a slower but stable orbit in the absence of the radiative pressure torque on the orbit from orbital velocity. Larger objects of a given density have more mass per unit surface area and are less affected by radiation pressure.


    ________________________

    Another way of looking at magnification by atmospheric refraction (considering mainly N = nr cases):

    For flat surfaces, leaving some vertical level moves upward across falling n values, total internal reflection keeps a portion of rays trapped; the other rays spread out to fill the full hemisphere solid angle, reducing the intensity and, if the radiation was initially isotropic, keeping the total upward flux per unit horizontal area proportional to n^2 at each level (so long as n only either decreases or remains constant with height).

    But for concentric spherical surfaces (with N decreasing outward to N = 1), the cone of acceptance defined for flat interfaces is narrower than the cone of rays that is able to escape upward to any given level, because as the rays curve over and downward, the interfaces - the locally defined horizontal surfaces - curve downward. Thus, the height to which any ray can reach is raised, and a greater solid angle of rays escapes all the way out to N = 1. This means a greater total upward flux per unit horizontal area reaches to any given height and to N = 1. But the intensity of the radiation still falls by the same amount as it passes to lower N (being proportional to N^2); so the greater flux requires a greater solid angle - hence, the underlying surfaces that are the source of the fluxes occupy a greater solid angle from any given viewing point - they appear larger - hence they are magnified.

    Some rays will still be trapped, so the magnification must be less than N^2.

    ----------

    From some class notes: N of the air, in the visible part of the spectrum, at STP (standard temperature and pressure) is about 1.0003. Near sea level conditions are broadly similar to STP.

    If N ~= 1.0003, then N^2 ~= 1.0006. In that case, the effect on I would be a 0.06 % increase relative to I in a vaccuum. The magnification of the surface as seen from space cannot be more than that (and is also limited by the vertical extent of different N values). The magnification of higher layers will tend to be less because N will tend to decrease with height toward 1.





    ------------------

    A particular important point about the climatic effects of increasing surface area and changes in N with height is that they would not be a source of significant positive or negative feedback (at least for Earthly conditions).

    As the greenhouse effect increases, the distribution of radiative cooling to space does shift upward. But a doubling of optical thickness per unit mass path would only shift the distribution within the atmosphere of transmissivity to space upward by about 5 km, give or take ~ 1 km (it is less at heights and locations where the temperature is colder, more where warmer). Doubling CO2 would have that effect only over the wavelengths in which it dominates (covering roughly 30 % of the total radiant power involved), and not quite even that, since at most wavelengths, emission cross sections are smaller with increasing height, at least for the troposphere and maybe lower stratosphere, and also, there would be some overlaps with clouds. Also, the tropopause level forcing would be due mostly to expansion in the wavelength interval in which CO2 significantly blocks radiation from the surface, water vapor, and clouds, given the shape of the CO2 absorption spectrum and that the central part of the CO2 band is saturated at the tropopause level with regards to further radiative forcing. The water vapor feedback is less than a doubling of water vapor, and water vapor density decreases much faster with height than air density within the troposphere. The height of the troposphere will shift upward to lower pressure as the temperature pattern shifts, so the highest cloud tops will be higher, but not by much in terms of affecting the emitting surface area.

    An upward shift in the LW radiative cooling distribution of about 5 km would result in about a 0.1 K cooling; it seems that a vertical shift less than 5 km likely causes a warming of about 3 +/- 1 K, 20 to 40 times larger (PS the vertical shift includes all LW radiative feedbacks; if there were no positive LW feedbacks, the warming would be reduced, but so would the vertical shift. If SW feedbacks were excluded, then maybe the warming would be at the lower end of the range given, I think).

    What would truly be required to result in a 0.1 K cooling by this process is if all LW radiative cooling that occured within the troposphere and at the surface were confined to be below 5 km from the tropopause initially (or else, to have the tropopause level rise to accomodate the shift?), and then to have the whole distribution raised 5 km, so that there would then be no LW cooling below 5 km from the surface. This would actually cause warming of roughly 30 K, given a lapse rate of 6 K/km. Thus the cooling by area increasing with height would be roughly just 1/3 % of the warming.

    There is yet one other way to vertically shift the LW cooling distribution: Thermal expansion. The atmospheric mass in the troposphere would only expand about 1 % in response to a 3 K warming, which would only contribute a 0.002 K cooling for an initial 10 km effective level for LW cooling. (This mass expansion is a seperate issue from the increasing height of the tropopause mentioned in the previous paragraph - the later is a rise in the tropopause relative to the distribution of mass - essentially a transfer of mass from the stratosphere to the troposphere. This is actually an expected result of global warming in general, although nothing on the order of 5 km so far as I know).

    ---------

    Feedbacks involving changes in the index of refraction of the air, and forced changes in the index of refraction of the air, can also be expected to be insignificant.

    ________________________

    MORE ON THE 'I CAN SEE YOU AS MUCH AS YOU CAN SEE ME' PRINCIPLE (THAT IS REQUIRED if THE SECOND LAW OF THERMODYNAMICS IS TO APPLY):

    SPECULAR REFLECTION:

    For simplicity of illustration, consider an interface between materials, where on side A, N = NA, and on side B, N = NB, and in both cases, let the imaginary component be zero. Let N be invariant over direction within each material.

    Identify directions Q by two angles: Q = (q,h). q is the angle from the normal (perpendicular) to the interface (specifically, measured from the direction leaving the interface on the side that q is taken), and h is the angle around the normal to the surface; h going from 0 to 360 deg at constant q traces a circle in a plane parallel to the interface; rather than reversing the direction h is measured across the interface to keep the same overall coordinate system (right-handed or left-handed) for each side A and B, measure h on each side in the same sense (clockwise or counterclockwise) as viewed from just one side.

    Consider four ray paths that approach this interface. Two rays, 1 and 2, are incident from side B with directions Q1 and Q2, respectively. Two other rays, 3 and 4, are incident from side B with directions Q3 and Q4.

    Q1 = (qA,h0)
    Q2 = (qA,-h0)
    Q3 = (qB,-h0)
    Q4 = (qB,h0)

    So rays 1 and 2 have the same q = qA, rays 3 and 4 have the same q = qB, and 1 and 4 have the same h = h0, while 2 and 3 have the same h that is the opposite of the h of the other two rays.

    Let NA*sin(qA) = NB*sin(qB).

    A portion of each rays is reflected (denoted r) and a portion is transmitted (denoted t). For example, from the incident ray 1, the reflected ray is 1r and the transmitted ray is 1t.

    Notice:

    If all incident rays intersect the interface at the same point, then:

    ray 1t goes backwards along the path of ray 3
    ray 3t goes backwards along the path of ray 1

    ray 1r goes backwards along the path of ray 2
    ray 2r goes backwards along the path of ray 1

    ray 2t goes backwards along the path of ray 4
    ray 4t goes backwards along the path of ray 2

    ray 3r goes backwards along the path of ray 4
    ray 4r goes backwards along the path of ray 3

    etc.



    The incident rays have spectral (monochromatic) Intensities I# = I/N^2 of I1, I2, I3, I4.

    Along paths 1, 2, 3, and 4, the I/N^2 toward the interface are:


    I1
    I2
    I3
    I4

    IF the reflectivity from side A is RA and the reflectivity from side B is RB then:

    Along paths 1, 2, 3, and 4, the I/N^2 away from the interface are:

    RA*I2 + (1-RB)*I3
    RA*I1 + (1-RB)*I4
    RB*I4 + (1-RA)*I1
    RB*I3 + (1-RA)*I2

    Suppose each ray is emanating from a blackbody, and each blackbody has the same temperature. In that case, the net intensity (forwards - backwards) = 0 along each path so that there is no net heat transfer, assuming the second law of thermodynamics holds for the consequences of reflection and refraction.

    RA*I2 + (1-RB)*I3 - I1 = 0
    RA*I1 + (1-RB)*I4 - I2 = 0
    RB*I4 + (1-RA)*I1 - I3 = 0
    RB*I3 + (1-RA)*I2 - I4 = 0

    Also, I1 = I2 = I3 = I4. Then:

    RA + (1-RB) = 1
    RA + (1-RB) = 1
    RB + (1-RA) = 1
    RB + (1-RA) = 1

    Each relationship yields the same conclusion: RA = RB = R. Reflectivity is the same for any two rays approaching the same interface from opposite sides in which each of their transmitted rays goes backwards along the other incident ray.

    Reflectivity can vary with polarization P, so this only applies if either the incident rays are completely unpolarized or polarized specifically to fit some variation of N over P (as in perfect blackbody radiation), or if the intensity is evaluated for each polarization seperately.

    The formulas (one for parallel and one for perpendicular polarizations) for reflectivity (Fresnel relations) as a function of NA/NB and qA or qB (each q determines the other) give the same R for RA and RB for each polarization. The Fresnel relations are not determined by the second law of thermodynamics; they are determined (at least in part) with the constraint that the electric and magnetic fields only vary continuously in space; they each have only one value at any one location and time, including on the interface. I'm not sure if it is necessary for deriving the relations, but the Fresnel relations would also have to fit with the conservation of radiant energy (except for allowed sources and sinks such as emission and absorption).

    An interesting further point:

    Along ray 1, the I/N^2 coming back from the interface = R * I2 + (1-R) * I3

    What happens to the radiation going toward the interface along I1 ? R * I1 is reflected backward along ray path 2, and (1-R) * I1 is transmitted backward along ray path 3.

    In other words, the distribution of where the radiation going forward along a path reaches matches the distribution of the sources of the radiation that goes backwards along the same path (before weighting by the strength of each source).

    ------

    There should be a similar pattern of behavior for scattering. That is, if I is isotropic, then the I scattered out of a path must equal the I scattered into the path over the same unit of path dx. Otherwise, anisotropy in I# could spontaneously arise, which would break the second law of thermodynamics (clever use of such spontaneous anisotropy could run a perpetual motion machine).

    Of course, a scattering cross section per unit volume, scsv, can vary over direction, but it should (except for macroscopic scatterers where absorption cross section on one side can be matched with scattering cross section on the opposite side) be the same for a pair of opposing directions along a ray path.

    More generally:

    What I expect is that the distribution over directions of radiant intensity that is scatterd out of a ray going in the x direction over a distance dx should fit the distribution of the source directions of radiant intensity that is scattered into the ray in the same dx to go in the negative x direction, before weighting by anisotropy of the source direction intensities.

    When the same type of scattering has the same scattering cross section in all directions (and the scattering distribution has 0-fold rotational symmetry about the direction from which I is scattered), I think this can be shown to be true: Consider a fraction of radiation that is scattered by an angle q' into the solid angle dw in the direction Q2, from an initial direction Q1 and solid angle dw. The same angle of scattering applies to radiation that is scattered from the direction Q2 into the direction Q1. Thus the same fraction of radiation is scattered into Q2 from Q1 as is scattered from Q1 into Q2.

    ________

    CONCLUSION (in the expectation that more complex forms of scattering and reflection and dependence on complex N values that vary over direction, and where group velocity is the direction of propagation of intensity, while phase propagation may be in a different direction - that these situations still fit the general concepts illustrated above):

    At a given frequency v, and if necessary, polarization P (could be a function of position along paths):

    At a reference point O along a path in the direction Q (which can bend as refraction requires), with distance forward along the path given by x:

    Considering the forward intensity I#f in the solid angle dw, the distribution of where I#f goes is proportional to the derivative of the transmission with respect to x. It can be visualized by considered a distribution of distance x over dw, where each cross section per unit area normal to the path over the unit distance dx (including reflectivity at any interface within dx) occupies a fraction of dw equal to itself multiplied by the transmission over x. The solid angle dw is the sum of many fractions of dw that are each filled by cross section at a different distance. The fractions are the fractions of I#f that are intercepted at the corresponding distances. They are also the fractions of the total visible cross section from point O at the corresponding distances. This is essentially a distribution of dw over x, that describes the distribution over x of what is visible along x from point O and of how much visibility of I#f at O exists at x. d(dw)/dx can be integrated along x to give a visible cross section per unit area of 1. I#f(O) * d(dw)/dx is the interception at x per unit dx of whatever I#f passes through O. Integration of d(I#b(x)) * d(dw)/dx gives the total backward intensity I#b(O) that reaches O; it sums the proximate sources of I#b(O) over the fractions of dw that correspond to a distribution over x.

    This can be extended farther. Suppose we are interested in the absorption of I#f. For all the fractions of dw that are scattering or reflection cross sections per unit area, the fate of those fractions of I#f can be traced farther, through successive scatterings and reflections, until every last bit is absorbed. This will be a distribution of dw that may extend outside of the path and over other paths, perhaps over some volume. It is the distribution of the absorption of I#f(O). Assuming local thermodynamic equilibrium within each unit volume, It is also the distribution of the emission of I#b(O) - multiplying the distribution density by I#bb(T) and integrating over the distribution gives the I#b(O) value, where I#bb(T) is the blackbody intensity (normalized relative to refraction) as a function of T, however it varies over space.

    I#f(O) also has an emission distribution that can be found, tracing back in the opposite direction from point O.

    The net I# at point O in the forward direction is I#(O) = I#f(O) - I#b(O), and will be in some way proportional to the difference in temperature T (specifically, the difference in the weighted average of Ibb#(T) over the emission distribution) between the emission distribution of I#f(O) and the emission distribution of I#b(O).

    Assuming the forward direction is from higher to lower temperature within the emission distributions, if the two distributions extend over a larger range in T (at a given overall average T), then the net radiant transfer I#(O) will be larger; if the two distributions extend over a smaller range in T (at a given overall average T), then the net radiant transfer I#(O) will be smaller. I#(O) can be changed by either changing the emission distributions or the temperature distributions. Increasing overlap of the distributions will tend to reduce I#(O) (the distribution of emission for either I#b(O) and/or I#f(O) can wrap around the point O when there is sufficient scattering and/or reflection).

    EXAMPLES:

    -----------
    PURE EMISSION AND ABSORPTION (loss to space is 'absorption' by space for climatological purposes):

    The emission distributions of each of I#f(O) and I#b(O) are entirely along the ray path, with zero overlap. Each distribution density decays exponentially from point O in optical thickness coordinates; the distributions are more concentrated near point O in (x,y,z) space when there is greater cross section per unit (x,y,z) volume.

    If there is an emitting/absorbing surface (approximately infinite optical thickness per unit distance), that the ray path intersects, increasing the opacity of the intervening distance between the surface and O reduces the portion of the distribution that is on the surface, pulling it into the path between the surface and O while concentrating the distribution within the path toward point O.

    For a given temperature variation along the path, if the temperature is increasing in one direction, increasing the cross section per unit volume decreases the net I#(O). IF the temperature fluctuates sinusoidally about a constant average, then increasing the cross section per unit volume may have little effect on net I#(O), until the emission distribution becomes concentrated into a small number of wavelengths; if the temperature is antisymmetric about O, the maximum net I#(O) will occur when the emission distribution is mostly within 1 to 1/2 wavelength, so the nearest temperature maximum and minimum dominate the emission distribution; beyond that point, further concentration of the emission distribution reduces I#(O).

    -----------
    PURE SCATTERING within a volume between perfect blackbody surfaces at temperatures T1 and T2 (with 100 % emissivity and absorptivity):

    When the scattering cross section density is zero, net I#(O) is the difference between Ibb#(T1) and Ibb#(T2), only except for paths that are parallel to the surfaces, or with variations in N, never intersect both surfaces.

    The emission distributions will only be at the surfaces and not in the intervening space. The effect of scattering will be to redirect radiation so that a portion of the emission distribution for I#(O) from one side of O will come from the other side of O; this reduces the net I#(O) by mixing some of both Ibb#(T1) and Ibb#(T2) into both I#f(O) and I#b(O).

    Without multiple scattering, forward scattering makes no difference for direction Q that is everwhere normal to the surfaces (assuming constant N everywhere).

    Increasing either the scattering cross section density, the deflection angles of forward scattering, the portion of single scatter backscattering, or the angle from the normal perpendicular the surfaces, will increase the portion of the emission distribution for I#(O) from one side of O that is shifted to the other side of O.

    Interestingly, scattering could introduce some net I# into paths that do not intersect both surfaces, such as those that intersect the same surface twice due to total internal reflection. But increasing scattering will eventually reduce I# for even those cases.

    Partially reflecting surfaces between the emitting surfaces will have the same general effect as scattering; the emission distributions will not fill a volume with specular reflection alone, but they will spread out over a branching network of paths that penetrates space both back and forth.

    ****

    Having an emitting surface that have emissivity and absorptivity less than 100% would be analogous to placing either some partially reflective surfaces and/or a layer of high scattering cross section density immediately in front of the surface so that none of the points O considered would be between that layer and the emitting surface.

    -------------------------

    In the case of perfect transparency between surfaces with reflectivities R1 and R2, the I#f from surface 1 to surface 2 along any path that intersects both surfaces would be (PS this assumes that R1 and R2 are not direction dependent, or happen to be constant for all the directions that occur when tracing back and forth along a given path at a point O; at least in the case of no directional dependence, this applies to both specular and diffuse reflection):

    (1-R1) * Ibb#(T1) + R1*(1-R2)*Ibb#(T2) + R1*R2*(1-R1)*Ibb#(T1) + ...

    =

    SUM(j=0 to infinity)[ Ibb#(T1) * (1-R1)*(R1*R2)^j + Ibb#(T2) * R1*(1-R2)*(R1*R2)^j ]

    ---

    And the I#b would be:

    SUM(j=0 to infinity)[ Ibb#(T2) * (1-R2)*(R1*R2)^j + Ibb#(T1) * R2*(1-R1)*(R1*R2)^j ]

    ---

    So the net I# would be:

    SUM(j=0 to infinity)[ Ibb#(T1) * (1-R1)*(R1*R2)^j + Ibb#(T2) * R1*(1-R2)*(R1*R2)^j ]
    -
    SUM(j=0 to infinity)[ Ibb#(T1) * R2*(1-R1)*(R1*R2)^j + Ibb#(T2) * (1-R2)*(R1*R2)^j ]

    =

    SUM(j=0 to infinity)[ Ibb#(T1) * (1-R1)*(1-R2)*(R1*R2)^j - Ibb#(T2) * (1-R1)*(1-R2)*(R1*R2)^j ]

    =

    [ Ibb#(T1) - Ibb#(T2) ] * SUM(j=0 to infinity)[(1-R1)*(1-R2)*(R1*R2)^j ]


    =

    [ Ibb#(T1) - Ibb#(T2) ] * (1-R1)*(1-R2) / [1-(R1*R2)]

    -------

    When R2 = 0,

    I# =

    [ Ibb#(T1) - Ibb#(T2) ] * (1-R1)

    ---

    When R2 = R1 = R,

    I# =

    [ Ibb#(T1) - Ibb#(T2) ] * (1-R)*(1-R) / [(1-R)*(1+R)]

    =

    [ Ibb#(T1) - Ibb#(T2) ] * (1-R)/(1+R)

    -------

    Increasing either R1 or R2 decreases I#.

    -------------------------

    MIX OF SCATTERING/REFLECTION AND ABSORPTION/EMISSION (between two opaque surfaces with nonzero emissivities and absorptivities):

    -----------
    Weak scattering and reflection relative to absorption/emission:

    Adding scattering and reflection concentrate the emission distributions closer to O, pulling them off of and away from any opaque surfaces, and spread them out from the ray path into a volume of space, and can cause some overlap of the two distributions. Specular reflection will not cause the emission distributions to fill a volume but will have other effects broadly similar to scattering. For the same scattering cross section density, single scatter dominated by forward scattering with small deflections has the weakest effect. Most of the photons may only scatter or reflect once if the emission/absorption cross section density is enough relative to scattering/reflection cross section density.

    -----------
    Weak emission/absorption relative to scattering/reflection:

    Adding emission/absorption cross section density pulls the emission distributions off of the opaque surfaces; that portion which is lifted off the surfaces is concentrated near the point O. Adding more emission/absorption cross section pulls more of the emission distributions off the opaque surfaces and increases the concentration near O.

    With sufficient scattering relative to emission, multiple scattering will tend to diffuse an intensely anisotropic I# into nearly isotropic I#; this will make the emission distributions for both I#f(O) and I#b(O) into nearly spherical regions that are both centered near O, so that there will be great overlap of the two distributions, reducing the net I#(O). With less emission/absorption cross section density, the spheres expand; with zero emission/absorption within space between opaque surfaces, the distributions are left on the surfaces (both distributions nearly evenly divided among surfaces if scattering and/or reflection is sufficient).

    -----------
    Varying proportions of emission/absorption cross section and scattering/reflection cross section:

    All cross sections tend to restrict the emission distributions, but within the larger distribution, pockets of higher densities of emission cross section will have a greater density of the emission distribution. Pockets of sufficiently high densities of scattering and reflection cross sections may reflect and deflect the emission distribution around themselves.

    ------------------------
  33. "The forces of American colonialism began to drop containers that produce a sound explosion, a very huge sound. I remind you that they said that their strategy is based on shock and awe. Those failed ones manufactured a type of container that has an explosive substance, which they drop. They cause a very huge explosion in terms of sound, as if the universe was shaken. After a while, you go out and you don't find anything. You find some nails, screws, pieces of metal, but the important thing here is the sound. Those failed
    ones think that through the huge sound explosion, people would be shocked and consequently would collapse and be defeated. What happened? The contrary."

    ....a "Baghdad Ali" quote.

    Sounds a lot like Patrick?
  34. In so far as the differential equation following a path locally in the direction x over distance dx, for I# in the direction x:

    At specific v and P, per unit spectrum and polarization dv and dP (P could have more than one dimension, actually):

    d(I#)
    =
    IL#(G) * Lcsv * dx
    + Ibb#(T) * ecsv * dx
    - I# * (acsv+scsv) * dx
    + Is# * scsv * dx
    - I# * R
    + Ir# * R

    The terms on the right hand side (if this were written out in one line)

    1.
    non-thermal emission into the path (such as fluorescence), where Lcsv is a cross section density for that process and IL#(G) is a function of the the energy available for such a process and the nature of such a process.

    2.
    thermal emission into the path, where ecsv is the emission cross section density and Ibb#(T) is the blackbody intensity.

    3.
    absorption and scattering out of the path, where acsv and scsv are the absorption and scattering cross section densities, which sum to give the extinction cross section density. acsv = ecsv at local thermodynamic equilibrium (it could be possible to define them as equal even when not at local thermodynamic equilibrium since a non-thermal emission term is also included).

    4. scattering into the path, where scsv is the same scattering cross section density, but Is# depends on I# in all directions at that location (including backwards along the same path) and the type of scattering.

    5. specular reflection out of the path, where R is the reflectivity of any interface encountered within dx.

    6. specular reflection into the path, where Ir# depends on the I# going backward along the path taken by reflection out of the path.

    Inclusion of reflection at a discontinous interface (relative to the scale of the wavelength) in the differential equation works so long as dx is small enough that the other terms are very small.
  35. A condition where the phase of the wave could matter (?) is when there is significant nonlinearity - In nonlinear optics, passage of very high amplitude electromagnetic waves through a medium can alter the optical properties of the medium, allowing photon-photon interaction.

    When evaluting I# at specific values of P, the P at one location is not necessarily the same P everywhere, and when finding the emission distribution, there could be multiple P values at the same location corresponding to different paths that photons took between locations and the different changes to P along the way. P is obviously relative to the orientation of processes that depend on P.

    ----------

    For radiative energy transfer in the atmosphere, there are some useful approximations (for at least Earthly conditions):

    1. Assume local thermodynamic equilibrium, at least below some height level (that is at least above the tropopause). Ignore non-thermal emission. Assume emission cross section = absorption cross section.

    2. For wavelengths shorter than about 4 microns (SW radiation), assume the only emission is from the sun.

    3. For wavelengths longer than about 4 microns (LW radiation), assume there is no solar contribution (a bit less accurate than assumption 2, but still a good approximation).

    4. Aside from absorption and emission, assume N = 1 within the atmosphere on a macroscopic scale (larger than scattering mechanism scales), and ignore gravitational lensing, so that radiation propagates in straight lines within the atmosphere and space, except for scattering and reflection.

    5. Also ignore gravitational redshift, and the blueshifting and redshifting of solar energy at sunrise and sunset, etc.

    6. Below some level that is above the vast majority of atmospheric emission and absorption at most wavelengths, ignore the curvature of the Earth and atmosphere, so that radiation propagating in straight lines is assumed to have a constant angle relative to the local vertical, and the total horizontal area around the globe at any height can be assumed constant over vertical distance.

    7. Except at the land and ocean surface, assume zero scattering and zero reflection for LW radiation within the atmsophere. (Note that even when there is some significant single-scatter albedo, the multiple scatterings required (when that is the case) for radiation to scatter back from, for example, a cloud, can result in very low albedo due to absorption between each scattering.)

    8. For some purposes, scattering and reflection at the surface can also be neglected for LW radiation.
  36. Additional notes and observations of scattering of solar radiation:

    When there is a hole through which direct solar rays pass, scattering along the path of the beam allows the beam to glow with scattered radiation - it can be seen from outside itself. This can be observed in a dusty room with sunlight coming through a window. A shadow cast through the air can also be seen from outside itself by the same mechanism. Variations in direct solar ray intensity can be seen to the extent that they have optical thickness along the line of sight and there is not too much optical thickness along the line of sight between the viewer and variation being observed. These variations are called crepescular rays and can be seen when the direct sun is blocked by a layer of clouds with holes, or there are patches of clouds casting shadows, or when the sun is behind a cloud with an irregular edge.

    One particularly interesting case is the shadow cast be a long thin straight contrail (the cloud left by a jet when conditions allow). Such a contrail casts a shadow that is a thin planar slice through the air; along lines of sight nearly parallel to this shadow, a dark streak can be seen through the sky; it will be darkest to viewers within the shadow. But the shadow will not be observed along lines of sight in most other directions because the shadow is thin in those directions.

    Clear air atmospheric scattering is generally stronger for shorter wavelengths. This is of course why the midday clear sky is generally blue (it can be white near the horizon - possible contributors to that: less blue light reaches to air near the surface, while there is nonzero scattering of other wavelengths, and lines of sight near the horizon pass through a greater thickness of lower-level air as well as the total atmosphere; some aerosols with different scattering properties may also be abundant in the lowest level air).

    White objects near the horizon may appear slightly red because blue light is scattered out of the line of sight. However, some blue light is scattered into the line of sight from an overhead sun and/or overhead thin clouds. Thick clouds overhead can reduce this effect, so that distant clouds lit by the sun will appear more red.

    Of course, the extinction of blue light and the nonzero scattering of other wavelengths explain the colors of sunset and sunrise. The overhead clear sky still appears blue, and the sky near the horizon can appear blue after sunset and before sunrise - this is because of the curvature of the Earth; the solid/liquid Earth casts a shadow on it's own atmosphere, but the edge of the shadow is at some finite height within the atmosphere at dawn and dusk, and when reaching the Earth nearly horizontally, sunlight travels through a smaller mass of air when it goes by higher above the ground, so there is plenty of blue light to scatter.

    ---------

    Specular reflection (as in a mirror) can be observed for some smooth surfaces, such as calm water. Wavy water still has locally specular reflection but images will break up and be distorted.

    Diffuse reflection takes an incident beam of light and reflects it over a range of directions. It is a form of scattering.

    Lambertian reflection is diffuse reflection in which the reflected radiation is isotropic.

    Reflected radiation can be a mix of Lambertian and specular or nearly specular (images would appear fuzzy), or more complex.

    I have actually noticed in lawn grass in sunny conditions that the grass appears brighter just around the shadow of my head - this means that there is a concentration of reflected radiation going back near the direction from which it came - similar to the reflecting surfaces used for traffic signs. I noticed a similar phenomenon in gravel.

    Have you ever noticed crepescular rays when looking down into murky water?

    When the surface has a high albedo, the sky can appear brighter than otherwise because it can scatter back to the surface some of the radiation reflected from the surface. I have seen a picture of a blue light on the base of clouds over a patch of shallower water (which appears bright blue because there is less absorption of sunlight through the thinner water layer and there is reflection from the underlying surface) - perhaps an atoll.

    (Clear) water is generally blue because, while nearly transparent (to visible wavelengths) in thin layers, it does absorb light, and more strongly at longer visible wavelengths than for blue light. Of course, looking near the horizon, one sees less into the water and more the reflection of the sky off the water. When there are waves, one can see into the water best on the side of the wave closest to normal to the line of sight, while mainly a reflection of the sky will be seen from the other side of the wave or where the line of sight just grazes the water surface.

    PS for LW radiation:

    Then nonzero LW albedo of the surface:

    If this is specular reflection (As might be expected for relatively calm water), then, except in an inversion with sufficient opacity, the reflected radiation will be absorbed over a shorter distance in the air because a greater portion of it will come from angles near the horizon where the LW glow fo the air will (except for a low level inversion with sufficient opacity) generally appear brightest near the horizon.
  37. "I have actually noticed in lawn grass in sunny conditions that the grass appears brighter just around the shadow of my head - this means that there is a concentration of reflected radiation going back near the direction from which it came - similar to the reflecting surfaces used for traffic signs. I noticed a similar phenomenon in gravel."

    But the process that produces this effect is different for the traffic signs (and the effect itself will probably be a little different). I think that type of reflecting surface has inverted square pyramids, where each flat triangular surface is at a 45 deg angle to the plane of the larger surface, and each flat triangular surface has specular reflection.

    In contrast, what may be happening with the grass and the gravel is that the individual surfaces have nearly Lambertian reflection - they will look at bright from any direction - but the surfaces themselves are angled differently and thus have more or less sunlight per unit area to reflect. When looking toward the shadow of one's own head, one would see more of the surfaces that are nearly normal to the direct solar rays.

    ----

    Sometimes the brightest and deepest colors can be seen when looking at the diffuse transmission through leaves and petals. Highly recommended viewing.
  38. LW radiation:
    "If this is specular reflection (As might be expected for relatively calm water), then, except in an inversion with sufficient opacity, the reflected radiation will be absorbed over a shorter distance in the air"

    ... as opposed to Lambertian reflection.
  39. "I think that type of reflecting surface has inverted square pyramids" ... Oh, maybe that's just bicycle reflectors - but you get the idea.

    --------------

    And now the moment very few people have been waiting for: What is the spectral (monochromatic) Ibb?

    For N = 1:

    Where the blackbody flux per unit area
    = π*Ibb = sigma * T^4
    where sigma (also known as BC above)
    =2 * π^5 * k^4 /(15 * c^2 * h^3)

    and

    Where c, h, and k, and sigma are:

    (Where there are two values, the second value is from a physics textbook (or calculated from physics textbook values), and the first value is from Wikipedia ( http://en.wikipedia.org/wiki/Boltzmann's_constant , http://en.wikipedia.org/wiki/Avogadro_constant , http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_constant ):

    c = 2.99792458 E+8 m/s

    h = 6.62606896 E-34 J*s
    h = 6.626075 E-34 J*s

    k = 1.3806504 E-23 J/K
    k = 1.380658 E-23 J/K

    sigma = 5.670400 E-8 W/(m2 K4)
    sigma = 5.670511 E-8 W/(m2 K4)



    Ibb(T,L)
    = |d[Ibb(T)]/dL|
    = ( 2*h*c^2 / L^5 ) / ( exp[h*c/(L*k*T)] - 1 )


    Ibb(T,v)
    = |d[Ibb(T)]/dv|
    = ( 2*h*v^3 / c^2 ) / ( exp[h*v/(k*T)] - 1 )

    where:

    c = L * v

    v = c/L

    |dv| = c / L^2 * |dL|

    L is the wavelength in a vacuum, v is the frequency.

    The L of maximum Ibb(T,L) is equal to
    (2897 microns/K )/ T
    (Wien's displacement law, from class notes but can be found elsewhere.)

    The L of maximum Ibb(T,L) is 10 microns at T = 289.7 K.
  40. "I think that type of reflecting surface has inverted square pyramids" ... Oh, maybe that's just bicycle reflectors - but you get the idea.

    --------------

    And now the moment very few people have been waiting for: What is the spectral (monochromatic) Ibb?

    For N = 1:

    Where the blackbody flux per unit area
    = π*Ibb = sigma * T^4
    where sigma (also known as BC above)
    =2 * π^5 * k^4 /(15 * c^2 * h^3)

    and

    Where c, h, and k, and sigma are:

    (Where there are two values, the second value is from a physics textbook (or calculated from physics textbook values), and the first value is from Wikipedia ( http://en.wikipedia.org/wiki/Boltzmann's_constant , http://en.wikipedia.org/wiki/Avogadro_constant , http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_constant ):

    c = 2.99792458 E+8 m/s

    h = 6.62606896 E-34 J*s
    h = 6.626075 E-34 J*s

    k = 1.3806504 E-23 J/K
    k = 1.380658 E-23 J/K

    sigma = 5.670400 E-8 W/(m2 K4)
    sigma = 5.670511 E-8 W/(m2 K4)



    Ibb(T,L)
    = |d[Ibb(T)]/dL|
    = ( 2*h*c^2 / L^5 ) / ( exp[h*c/(L*k*T)] - 1 )


    Ibb(T,v)
    = |d[Ibb(T)]/dv|
    = ( 2*h*v^3 / c^2 ) / ( exp[h*v/(k*T)] - 1 )

    where:

    c = L * v

    v = c/L

    |dv| = c / L^2 * |dL|

    L is the wavelength in a vacuum, v is the frequency.

    The L of maximum Ibb(T,L) is equal to
    (2897 microns/K )/ T
    (Wien's displacement law, from class notes but can be found elsewhere.)

    The L of maximum Ibb(T,L) is 10 microns at T = 289.7 K.
  41. The STUPIDITY of AGW.
    ----
    Trenberth's Energy Budget

    Incoming Solar Radiation = 342 w/m^2
    Solar Radiation Absorbed by atmosphere = 67 w/m^2
    -------------------
    (342 - 67) Leaves 275 w/m^2 available.

    Reflected by Clouds etc. = 77 w/m^2
    Reflected by Surface = 30 w/m^2
    Total due to reflection = 107 w/m^2

    The percentage of reflected energy is 107/275 = 0.389 or 38.9%.

    Leaves 168 w/m^2 absorbed by the Surface of the Earth.

    168 w/m^2 and an emissivity of 1, gives a temperature of 233.31K or -39.69 deg C.
    --------------------
    Now what happens if the reflected energy was decreased by 1% to 37.9%?

    0.379 X 275 = 104.23 w/m^2 so an additional (107 - 104.23 = 2.77 w/m^2) is available to heat the Earth.

    168 + 2.77 = 170.77 w/m^2 is now absorbed by the Surface of the Earth.

    170.77 w/m^2 and an emissivity of 1, gives a temperature of 234.26K or -38.74 deg C.
    --------------

    The Earth just warmed by (39.69 - 38.74) 0.95 deg C !!

    That's just due to a ONE PERCENT change in reflected energy!!

    -----------------
    Why the Hell is anybody talking about CO2, positive feed-back loops, carbon taxes etc. to explain something so easily explained?

    Especially since the AGW'ers admit that their "computer models" can't and don't handle CLOUDS well and the SUN is the ONLY ENERGY SOURCE!
    -----------------
    AGW is UTTER STUPIDITY no matter how you look at it!
  42. sure, Gord, and why is anybody talking about the graviational pull of the moon and sun when the tides can be so easily explained by waves and currents? Your posts are UTTERLY STUPID. But I hope most people reading this do not need me to point it out.
  43. Patrick -

    Re: your Post #511

    You remind me of "Baghdad Ali".

    Although, your logic may be just a tad inferior to his.

    Don't understand?...It's OK....everybody else will.
  44. Gord
    Re: "Don't understand?...It's OK....everybody else will."
    Sorry. I don't understand. I don't what is "Baghdad Ali".
  45. I am adding my first comment to this blog so please treat me gently. I found a reference to a Physics Today paper which is relevant. I cannot reach the paper - at least not without paying. The abstract says the paper claims that up to 69% of global warming may be attributed to solar variations. The reference is

    Scafetta, N., and B. J. West, Is climate sensitive to solar variability?, Physics Today, March 2008, 50-51, 2008.

    Any comments on that paper?
  46. expat: try this site :
    http://www.fel.duke.edu/~scafetta/pdf/opinion0308.pdf
  47. Mizimi - thanks, I was able to read the paper. I wont pretend to understand the details of the math, but it seems to provide an alternate explanation (to AGW) for much of the warming which has been observed. What am I missing here?
    Response: There are a number of problems with Scafetta's paper which are outlined in Solar variability does not explain late-20th-century warming (Duffy 2009). To summarise them briefly:
    1. If climate was so sensitive to solar variations, there would be a much strong 11 year cycle in the temperature record.
    2. If climate was so sensitive to solar variations, past climate change should've been much greater such as during the Maunder Minimum when solar levels were much lower.
    3. Solar activity has shown no long term trend since 1979.
    4. Even if you use the ACRIM solar data which shows a slight warming trend, the increase in solar energy is much less than the actual build-up of energy in the world's oceans. The only explanation of the build-up of ocean heat is that the energy radiating back out to space is less which is confirmed by satellite measurements of outgoing radiation.
    5. If solar variations were causing the warming, it fails to explain why the large build-up of greenhouse gases have had such little effect.
    6. Solar warming would mean the stratosphere should show a long term warming trend. In fact, the stratosphere has shown a long term cooling trend which is what has been observed by radiosondes and satellites.
    I will add that Duffy's paper doesn't mention that independent reconstructions of solar activity show greater agreement with the PMOD data (which shows slight solar cooling over the past 50 years) than with the ACRIM data (which shows slight warming). Eg - the sun has shown a slight cooling trend while global temperatures have been rising.
  48. 1. The paper you refer to is not a scientific paper, expat. It's an "Opinion" piece that Physics Today occasionally publish.

    2. This opinion shouldn't be read without reading the responses published in the October 2008 issue of Physics Today, and the Opinion published in the January 2009 issue of Physics Today which John Cook has provided a link to in his response to your post just above:

    Philip B. Duffy, Benjamin D. Santer, and Tom M. L. Wigley (2009)
    "Solar variability does not explain late-20th-century warming" Physics Today, January 2009, page 48

    3. The latter Opinion highlights some of the flaws of the Scafetta/West hypothesis, a major one being that the solar parameters have been flat, tending in a cooling direction since the early 1980's. Changes in solar output cannot have made significant contributions to the very marked warming of the last 30-odd years.

    4. It's worth remarking on the nature of Scafetta/West methodology. They use "phenomenological" approaches to scientific questions [*] in which the physics of the phenomenon is set aside, and the data are treated as a "black box" to which various elements of numerology are applied. More specifically, in the case of their attribution of solar contributions to surface temperature, they make the assumption that the only contribution to pre-industrial surface temperature variation is solar (i.e. they ignore volcanic, greenhouse gas, ocean circulation etc. contributions) and thus derive a solar climate sensitivity by curve fitting that assumes all variation is solar-induced, and then extrapolate this forward into the 20th century.

    5. This practice has the result of making the earth's surface temperature extremely sensitive to solar irradiation variation, and has the curious implication that the earth's climate sensitivity to a radiative forcing equivalent to a doubling of atmosphere CO2 is very high; 4.5 – 5.5 oC. Scafetta/West don't address the latter since their analysis is essentially non-physical/non-mechanistic – it's a curve fitting/extrapolation exercise based on a what is almost certainly a false premise (i.e. that the sole contribution to pre-industrial surface temperture variation is solar).

    6. Several solar scientists have made detailed physical, empirical and theoretical analyses of solar irradiance contributions to earth's surface temperature. These pretty uniformly indicate that the solar contribution to 20th century warming has been small:

    e.g.
    J. L. Lean and D. H. Rind (2008) "How natural and anthropogenic influences alter global and regional surface temperatures: 1889 to 2006", Geophys. Res. Lett.35, L18701., who conclude their analysis with:

    "For the ninety years from 1906 to 1996, the average slope of the anthropogenic–related temperature change in Figure 3d is 0.045 K per decade whereas Allen et al. [2006] concluded that the rate is 0.03–0.05 K per decade for this same period. Solar-induced warming is almost an order of magnitude smaller. It contributes 10%, not 65% [Scafetta and West, 2006, 2008], of surface warming in the past 100 years and, if anything, a very slight overall cooling in the past 25 years (Table 1), not 20–30% of the warming.


    6. In addition to the other Physics Today letters/opinion articles noted in #2 above, anyone considering the relevance of Scafetta/West's phenomenological numerology should read a paper published last month that (amongst other things!) addresses the methodologies of Scafetta/West and shows that these are flawed in relation to a realistic analysis of solar irradiance contributions to earth surface temperature variations. The methods (a) don't result in meaningful reproduction of the 20th century temperature variation, and (b) grossly overestimate the potential solar contributions.

    Benestad, R. E., and G. A. Schmidt (2009), Solar trends and global warming, J. Geophys. Res., 114, D14101, doi:10.1029/2008JD011639.

    Abstract: We use a suite of global climate model simulations for the 20th century to assess the contribution of solar forcing to the past trends in the global mean temperature. In particular, we examine how robust different published methodologies are at detecting and attributing solar-related climate change in the presence of intrinsic climate variability and multiple forcings. We demonstrate that naive application of linear analytical methods such as regression gives nonrobust results. We also demonstrate that the methodologies used by Scafetta and West (2005, 2006a, 2006b, 2007, 2008) are not robust to these same factors and that their error bars are significantly larger than reported. Our analysis shows that the most likely contribution from solar forcing a global warming is 7 ± 1% for the 20th century and is negligible for warming since 1980.

    -------------------------------------------------------[*] To understand where Scafetta/West are coming from it's worth pointing out that they have no background in climate science, nor any empirical science as such…they are numerologists that make phenomenological/statistical analyses of data sets covering subjects from teenage birth statistics, to fatigue crack growth, to human gait, to cardiac rhythms, to wealth distribution…. That's perfectly fine of course! The problem (to my mind) is their attempt to use premise-sensitive numerological analyses to ascribe attributions (solar contributions to surface warming, in this case), under circumstances where this simply isn't justified.

    Here's some of the other stuff they apply these methods to (cited to indicate the general nature of their work):

    Scafetta, N; Marchi, D; West, BJ (2009)
    Understanding the complexity of human gait dynamics
    Chaos, 19: Art. No. 026108

    Froehlich, KF; Graham, MR; Buchman, TG; et al. (2008)
    Physiological noise versus white noise to drive a variable ventilator in a porcine model of lung injury
    Canadian Journal Of Anaesthesia, 55: 577-586

    Scafetta, N; Moon, RE; West, BJ (2007)
    Fractal response of physiological signals to stress conditions, environmental changes, and neurodegenerative diseases
    Complexity, 12: 12-17

    Scafetta, N; Ray, A; West, BJ (2006)
    Correlation regimes in fluctuations of fatigue crack growth
    Physica A-Statistical Mechanics And Its Applications, 359: 1-23

    Scafetta, N; Restrepo, E; West, BJ (2003)
    Seasonality of birth and conception to teenagers in Texas
    Social Biology, 50: 1-22

    West, BJ; Scafetta, N; Cooke, WH; et al. (2004)
    Influence of progressive central hypovolemia on Holder exponent distributions of cardiac interbeat intervals
    Annals Of Biomedical Engineering, 32: 1077-1087

    Scafetta, N; Picozzi, S; West, BJ (2004)
    An out-of-equilibrium model of the distributions of wealth
    Quantitative Finance, 4: 353-364
  49. Thanks to you, Chris and Mizimi. I wondered why I had seen scant reference to that paper.

    I do have another question - it relates to the tropical tropospheric hot spot or not. Here is a reference. http://sciencespeak.com/MissingSignature.pdf

    Maybe you can help me do due diligence on this one also. I had put it on the "Satellites show little to no warming of the troposphere" topic but had no response so far.
  50. More on the role of the SUN and the Greenhouse Effect
    -----------------------------------------------------
    First, what the AGW'ers say:

    Greenhouse Effect
    "In the absence of the greenhouse effect and an atmosphere, the Earth's average surface temperature of 14 deg C (57 deg F) could be as low as -18 deg C (-0.4 deg F), the black body
    temperature of the Earth."
    http://en.wikipedia.org/wiki/Greenhouse_effect
    NOTE: THE ABOVE USES THE TERM "BLACK BODY".

    This calculation uses an albedo of 0.3.
    A "black body" actually has an albedo = 0, not 0.3 !
    This calculation uses a Sun temp of 5505 deg C or 5778 deg K.

    TE = TS ( ( (1-a)^0.5 * Rs)/(2*D) ) )^0.5)
    Where TE is blackbody temp of the Earth in K
    TS is the surface temp of the SUN in K (5778 K)
    Rs is radius of the Sun (6.96 X 10^8 meters)
    D is distance between the Sun and Earth in meters (1.5 X 10^11)
    a is albedo of the Earth and is 0.3 for a NON-black body

    Result:
    TE = 254.90 Kelvin
    TE = -18.25 Celsius
    --------------------------------------------------------
    Sun temp
    "The Sun's outer visible layer is called the photosphere and has a temperature of 6,000°C (11,000°F)."
    (6000 deg C = 6273 deg K)
    http://www.solarviews.com/eng/sun.htm

    TE = TS ( ( (1-a)^0.5 * Rs)/(2*D) ) )^0.5)
    Where TE is blackbody temp of the Earth in K
    TS is the surface temp of the SUN in K (6273 K)
    Rs is radius of the Sun (6.96 X 10^8 meters)
    D is distance between the Sun and Earth in meters (1.5 X 10^11)
    a is albedo of the Earth and is zero for a black body

    Result:
    TE = 302.55 Kelvin
    TE = 29.40 Celsius
    --------------------------------------------------------

    Temperature on the Surface of the Sun
    There are five sources for the surface temp of the Sun (6000,5500,5700,6000 and 5600 deg C).
    The average is 5800 deg C or 6073 K.
    http://hypertextbook.com/facts/1997/GlyniseFinney.shtml

    TE = TS ( ( (1-a)^0.5 * Rs)/(2*D) ) )^0.5)
    Where TE is blackbody temp of the Earth in K
    TS is the surface temp of the SUN in K (6073 K)
    Rs is radius of the Sun (6.96 X 10^8 meters)
    D is distance between the Sun and Earth in meters (1.5 X 10^11)
    a is albedo of the Earth and is zero for a black body

    Result:
    TE = 292.91 Kelvin
    TE = 19.76 Celsius
    --------------------------------------------------------
    The calculations using a max Sun temp of 6273K and average Sun temp of 6073K, and correctly using an albedo = 0 for a Black Body completely falsifies the statement:

    "In the absence of the greenhouse effect and an atmosphere, the Earth's average surface temperature of 14 deg C (57 deg F) could be as low as -18 deg C (-0.4 deg F), the black body temperature of the Earth"

    In fact, the addition of an atmosphere actually LOWERED the "black body" Earth temp (29.4 deg C (max) or 19.76 deg C (average)) to +14 deg C.
    --------------------------------------------------------
    Never, ever forget that the SUN is the ONLY ENERGY SOURCE.

    The Earth and the Atmosphere ARE NOT ENERGY SOURCES!

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