Monday, June 7, 2010

Hertzberg, Schreuder, and Siddons

Our correspondent Peter Newnam points us to a new paper that's being talked about on the Internet, Greenhouse Effect on the Moon?, by Hertzberg et al. The gist of the paper is that no planet acts like an ideal black body, therefore any calculation that is based upon the assumption that a planet is a black body is unreliable.

We use ideal objects like black bodies, the frictionless planes, and thin lenses all the time in physics and engineering. We use these ideal objects to estimate the behavior of the real world, and as we do so, we try to keep track of the error we are likely to make as a result of our ideal assumptions. Our hope is that the error is small compared to the accuracy we require of our calculation.

When Hertzberg et al. say that planets are not ideal black bodies, they are of course correct. But in the case of the moon, we see that the black-body model they present is pretty good. The actual temperature change on the moon is 80 K to 370 K, while the black-body model they present predicts a change of 35 K to 385 K. But Hertzberg et al. point out that the average temperature of the moon is roughly 40 K higher than predicted by a black-body model, and they claim that this difference invalidates the black-body analysis of planetary temperatures.

If we want a better understanding of the moon's surface temperature, we must add the heat capacity and conductivity of the surface rocks, as well as the absorption spectrum of these rocks for short-wave and long-wave radiation. Hertzberg et al. present a graph of the moon's surface temperature versus time during the day-night cycle. We see that the surface reaches its hottest temperature after the sun has passed the zenith, which indicates that the surface rock is taking time to heat up, by virtue of its heat capacity. The surface does not have a chance to reach its equilibrium temperature in the presence of the strongest sunlight. So we are not surprised that the moon fails to reach its black-body equilibrium temperature during the day. During the night, we see the temperature is still dropping at night when the sun rises again.

The albedo of the moon for short-wave radiation is around 0.1. It absorbs 90% of incident short-wave radiation from the sun. The emissivity of the moon's surface for long-wave radiation appears to be around 0.9. It emits 90% of the radiation that a black body would radiate. If the incident radiation from the sun is a uniform 350 W/m2, we expect the moon's surface temperature to be around 280 K. But the incident heat is not uniform. The moon spends half its time very cold, at around 80 K, in which state it radiates only 0.6% of the heat it would radiate at 280 K. For less than a third of the time, the temperature is above 300 K, reaching a maximum of 370 K. At 370 K, it radiates only 300% of the heat it would radiate at 280 K. We see that the excess of heat radiated during the day cannot compensate for the deficiency of heat radiated in the cool of the night. The moon does not radiate its own heat as efficiently as it absorbs the sun's heat, and so it is warmer than a constant-temperature black-body calculation predicts.

Hertzberg et al. conclude, "The ability of common substances to store heat makes a mockery of blackbody estimates." Quite the opposite is the case. If the moon conducted and stored heat well, its surface temperature would vary less, and the black-body estimate would be more accurate.


  1. As a fellow critic who also assessed this new paper I can only give your review a C-, sadly. What lets down your welcome assessment is that the writer has missed the whole point. He estimates, for instance, that the moon's emissivity is 0.9, but apparently doesn't wonder why modelers give the earth (and all planets in fact) an emissivity of 1. And his conclusion is completely ass-backwards. The more stable the temperature as insolation changes, the less like a blackbody an object will be. Because such an object is constantly out of equilibrium with the heat source.

  2. Thank you for your comment. But I'm sorry to say that I can't understand what you are talking about. Could you illustrate your case with some quantities and equations? For example, I have no idea what you mean when you say "constantly out of equilibrium with the heat source."

  3. Kevan,
    The point being made by Siddons et al. is that there are NO SUCH equations applicable to this set of circumstances.

    Startling to most people as this may seem, but no physics textbook, NOT ONE, will stipulate the conductivity of an imaginary blackbody.

    This is because such a property is foreign to it. A blackbody can have NO conductivity-please try to grasp the logic of this this self-evident statement.

    Heat storage would compromise its perfect temperature response. If some amount of heat is going INTO a body, this means that less heat is escaping as radiation — yet a blackbody is DEFINED as a body that emits 100% of the radiation that falls on it. Heat storage and blackbodies are strangers to each other.

    It took me some time to understand this concept, too. But understand this: if it takes X hours of irradiation to make a real body emit 0.99% of what a blackbody would emit (for 100% has never been achieved), then DURING that time lag the real body cannot be regarded as a blackbody, by definition.

    It is by such strict definition that a "blackbody" emits 100% of the thermal radiation that falls on it!

    Thus, what the 'Moon Paper' is getting at is that by it's very nature, every part of a blackbody is in equilibrium with the radiant energy it is exposed to.

    If 1000 W/m² of sunlight is beaming onto one part and 0 W/m² on another, those two parts will emit 1000 and 0 respectively, corresponding to 364 and 0 Kelvin.

    When you grasp this aspect of what "blackbody" actually means you see the absurdity of incorporating this notion into any equations related to the Earth's energy budget. The Earth’s accepted ‘base’ temperature should never be referenced to a blackbody model to begin with.

    I trust that better explains what I mean when I say "constantly out of equilibrium with the heat source."

  4. You say a black body is defined as one that "a body that emits 100% of the radiation that falls on it." Where did you get that idea? Not only is your definition absolutely wrong, but it defies the First Law of thermodynamics. Your definition could not possibly be correct even if you were in a position to invent a definition of your own for a term that has been well-established in physics for a hundred years.

    If "black body" was defined as you suggest, we could use a metal pipe to conduct heat away from your "black body" while it sits under a heat lamp, and still have it radiate all the heat it absorbs from the heat lamp. We would get more heat out of the object than we put in. Either that, or the object would cool to absolute zero and then it would not be able to radiate any heat, contradicting your own definition.

    In physics, a "black body" is one that absorbs all radiation that is incident upon it. That's all there is to it. Now, the laws of quantum physics dictate that this same black body must itself radiate power according to Stefan's Law (see Black Bodies). Furthermore, the distribution of this radiated power with frequency must follow a particular curve, with a peak at a frequency that is proportional to absolute temperature (see see Black Bodies again).

    A black body can be warming up, cooling down, conducting, non-conducting, whatever you like in that regard.

    The only reliable way to make a perfect black body is to drill a small hole in a well-insulated oven with conducting walls. The inside of the oven is at one temperature. Any radiation entering the hole will be absorbed. The spectrum of power emitted by the hole is the black-body spectrum.

    The moon absorbs roughly 90% of long-wave radiation, and so it also emits 90% as much long-wave radiation as a black body of the same temperature.

    I think you will find that my conclusion is correct: the more conducting a body and the greater its heat capacity, the more accurate a single-temperature black-body approximation becomes in the face of variable incident radiation.

  5. I'm sorry to disagree with you but you continue to miss the point. You estimate the moon's emissivity as 0.9, but apparently don't wonder why modelers give the earth (and all planets in fact) an emissivity of 1.

    Think harder on this point: The more stable the temperature as insolation changes, the less like a blackbody an object will be. Because such an object is constantly out of equilibrium with the heat source.

    Try thinking thermodynamically like this: imagine trying to heat a fire with the very radiation that it causes. What your implying is that back-radiation makes a fire cool down more slowly, whatever that means. To remove other complicating factors, think of a fire as a self-luminous blackbody shining in a vacuum. Say it’s at 1000°C, thus emitting about 149,000 W/m². Now encase it in a perfectly reflecting sphere. What will its temperature reach? The same: 1000°C.

    As a light-trapping laboratory blackbody demonstrates, confined radiation only lifts a real body's emission closer to a theoretical blackbody emission. To my knowledge, confined radiation doesn't raise a body's actual temperature, only its emission at that temperature. An important distinction.

    That reflection or reradiation induces no heating effect is proved by its very absence. Ask any greenhouse proponent to stipulate the temperature an object will reach if its radiation is confined and he’ll change the subject. Because, even though it is the central pillar of greenhouse physics, no equation exists for this scenario. Because the phenomenon doesn’t exist. Thus it can't be quantified.

    Also, think of hot coffee in a thermos flask-it offers a prosaic yet vital clue. The radiative confinement that a thermos's reflective lining provides only helps to sustain a given temperature; it doesn't increase that temperature. If you want to make the coffee hotter, you'll have to add more energy. As for the earth, however, is there any evidence of radiative confinement in the first place? No. The earth emits to space the same amount of radiant energy it gets from the sun. (At least according to official sources.)

  6. No, I'm not worried about what "the modelers" say, whoever they are. I never said anything about "back radiation". I am not an expert on what greenhouse proponents believe or say.

    Let me address your fire example, because you appear to have performed a calculation and obtained a the correct results: 149 kW/m2 at 1000°C.

    Fire is a chemical reaction, not a substance in itself. Most of the light emitted by fire is not black-body radiation (see fire). The light emitted by heated coals in the fire is, however, black-body radiation.

    When you surround a fire with a reflecting sphere, all matter in the sphere that is not transparent gets hotter. The matter in the flame will be hotter also, because the stuff that's burning is hotter before it burns. Your idea of "the fire" as a thing that is a constant body of matter is incorrect. The fire is a place through which matter is passing.

    You say, "To my knowledge, confined radiation doesn't raise a body's actual temperature, only its emission at that temperature." Radiation that strikes a black body is absorbed. When it is absorbed, it agitates an atom, thus heating the atom up. The radiation that is absorbed can be of any wavelength: FM radio waves, microwaves, infra-red, or green light from a laser. They all are absorbed by a black body and turn into heat.

    Black-body radiation is the radiation that results from the movement of electrons around agitated atoms. A body will radiate according to Stefan's law even if it receives no external radiation at all.