On Heat, the Laws of Thermodynamics and the Atmospheric Warming Effect

On average, Earth’s solar-heated global surface is warmer than the Moon’s by as much as 90 degrees Celsius! This is in spite of the fact that the mean solar flux – evened out globally and across the diurnal cycle – absorbed by the latter is almost 80% more intense than the one absorbed by the former.

The Earth’s global surface, absorbing on average 165 W/m2 from the Sun, has a mean temperature of ~288K (+15°C).

The Moon’s global surface, absorbing on average 295 W/m2 from the Sun, has a mean temperature of >200K (-75°C).

A pure solar radiative equilibrium for each of the two bodies (according to the Stefan-Boltzmann equation: Q = σT4, assuming emissivity (ε) = 1) would provide them with maximum steady-state mean global temps of 232K (-41°C) and 269K (-4°C) respectively.

As you can well gather from this, the Earth’s surface is 56 degrees warmer than its ideal solar radiative equilibrium temperature, while the lunar surface is at least 70 degrees colder than its ideal solar radiative equilibrium temperature. That’s a spread of no less than 126 degrees! On average …

Still, these two celestial bodies are at exactly the same distance from the Sun: 1AU.

So what could possibly account for this astounding difference between such close neighbours?

Very simple: The Earth has an atmosphere. The Moon doesn’t.

(Yes, the Earth also has an ocean, which of course makes all the difference in the world. But it’s still a secondary thing. The ocean wouldn’t be there in the first place if there were no atmosphere to keep it in place. The coupled ocean/atmosphere system makes our planet a very special place.)

The presence of our atmosphere resting on top of Earth’s solar-heated global surface clearly has a tremendous impact on planetary mean temperatures. It makes Earth a much, much warmer place on average than what it would’ve been without it. I say ‘on average’, because it actually also makes it a cooler place during the day, as compared to the Moon. In other words, it significantly cuts down on the diurnal temperature range, spreading the solar heat. For the same reason, it also strongly evens out the temperature gradient between the equator and the poles. The thicker the atmosphere, the more effective this evening-out process will be. On Mars, the temperature amplitudes are a fair bit larger than on Earth. On Venus they are all but nonexistent. On the Moon, where there is no real atmosphere to speak of, the amplitudes are very large indeed.

Thus, the cold lunar surface still maintaining a radiative balance with the incoming from the Sun is easily explained. Thanks to the natural exponential rise in radiative output from a surface with a linear temperature increase (T4), a surface with violently fluctuating temps could maintain a high average output even with a low average temperature without any problems, simply from the exceedingly high emission rates during those hot periods (and from the hot regions).

On Earth, the situation is very different. The radiative exchange doesn’t occur at one solid surface. The actual solid/liquid surface of our planet cannot maintain a purely radiative equilibrium with the Sun, simply because it’s in direct thermal contact with air, making other mechanisms for energy loss available. The atmosphere thus in effect makes any direct radiative exchange at the surface physically impossible. It’s ‘in the way’, lying between the solar-heated surface and the vacuum of space.

The atmosphere simply acts as an insulating layer.

So how does insulation really work?

First of all, insulation could never directly HEAT a warmer surface. A surface not already heated by some external/internal heat (power) source, could never become warmer in absolute terms by simply putting an insulating layer around it. If the warmer surface, as a result of being insulated, turns even warmer in absolute terms, meaning, its temperature rises above what it used to be before the insulating layer was put in place, then this could not be because of an extra energy input coming from the cooler layer of insulation. Because such an energy input making the receiver directly warmer would constitute ‘heat’ (or ‘work’). And ‘heat’ never moves spontaneously from cold to hot. Claiming (or merely implying) that it does would violate the Second Law of Thermodynamics.

Rather, insulation works in two ways:

  1. By reducing the temperature gradient away from a heated surface, consequently lowering the rate of energy loss per unit of time from the surface. The insulating layer accomplishes this by simply letting itself warm from the heat transferred to it from the warmer surface underneath.
  2. By obstructing (putting a limit to) convective and evaporative energy loss from the heated surface. The insulating layer can do this in several ways. Normally, like with clothes, blankets and walls/roofs, it does so by simply physically blocking the escape of lighter air by putting up a rigid barrier. In the atmosphere a slightly more subtle – and perhaps less intuitive – mechanism is at work. We’ll get back to this …

Before we move on from here, though, we need to have a couple of things clarified.

What is HEAT? And what does the 1st and 2nd Laws of Thermodynamics say?

We need to separate between two terms: ‘Net heat’ and ‘net energy’.

‘Net heat’ relates to the 1st Law of Thermodynamics. ‘Net energy’ relates to the 2nd. People tend to conflate the two terms. And by that creating confusion.

First, the ‘net energy’ term. Net energy in a thermal energy exchange between two objects is simply the heat, the actual, detectable, working flow of energy moving from the hotter to the colder object (or more generally, from a hot to a colder place). ‘Heat’ in this sense is equivalent to bulk air moving from higher to lower pressure and to an electric current moving from higher to lower voltage. ‘Energy’ per se might move in both directions. ‘Heat’ (net energy) ALWAYS and ONLY flows spontaneously from hot to cold, high potential to low potential. This is the 2nd Law of Thermodynamics. Heat is defined in physics as energy in transit between two systems or regions at different temperatures, transferred solely as a result of the temperature difference. This definition holds for conductive, convective and radiative heat transfer. There are NO exceptions in nature. Heat, like work, is a method for changing the internal energy of a system (an object). The internal energy is reflected in the object’s temperature. The higher the internal energy, the higher the temperature. A hot object in thermal contact with a cooler one will lose internal energy, and thus cool, to the cooler object, by transferring energy to it as heat. Likewise, the cooler object with gain internal energy, thus warm, from the same transfer of heat from the hot object.

By the 1st Law of Thermodynamics, only a transfer of energy as ‘heat’ – Q – or ‘work’ – W – is able to change the internal energy – U – of a system (an object) and thus its temperature: ΔU = Q – W (ΔU: the change in system internal energy from one state to another; Q: heat transferred to the system (positive); W: work done by the system (also positive)).

This is where we arrive at the ‘net heat’ term. Because Q in the equation above is really the sum of two subterms: Qin and Qout.

Qin represents an energy gain for the system from heat transferred to it, while Qout represents an energy loss for the system from heat transferred from it. If these two subterms equal each other, the system’s ‘net heat’ (Q) is zero.

What’s important to know here, is that these two transfers of heat (the gain and the loss) do not make up one energy exchange. They don’t directly oppose one another. Heat cannot, as per its physical definition, flow both ways inside one exchange. So in this case, they involve three systems rather than two, and hence two separate heat transfers rather than one. But still in to and out of the one central system.

The easiest way to illustrate this is through Carnot’s heat engine.

A system can only gain heat from a hotter place, a heat source. In terms of a heat engine, this heat source would be the ‘hot reservoir’. The heat engine in turn (and as a result of the heat input) does work on some other nearby system, but cannot translate all the heat supplied from the ‘hot reservoir’ into mechanical energy, so gives up some residual part as heat to a colder place, the ‘cold reservoir’, normally its surroundings. Note, it never delivers any heat back to where it got it from in the first place, its hot reservoir. This is an important point. The difference between the heat input, Qin (QH), (from the hot reservoir) and the heat output, Qout (QC), (to the cold reservoir) equals the engine’s work output, W, And we have conservation of energy:

CarnotCarnot’s heat engine. Keep in mind that, for the heat engine, according to Clausius’ sign convention for the First Law (ΔU = Q – W), both W and QH are positive, while QC is negative. Looking at this diagram, this may seem counter-intuitive. All the energy available to the engine at any one time is the heat from its hot reservoir, QH. For conservation of energy, the total of the work and heat outputs must equal this input. The engine converts part of the supplied heat from its hot reservoir to work and simply discards the remaining as heat. The way Carnot saw it, the ‘engine’ is a working fluid expanding when supplied with heat, thereby pushing against a nearby object, like a piston, doing work on it. No device, however, can transform the heat supplied completely into such work. There is always some amount of ‘waste’ energy flowing into a lower temperature area (the cold reservoir) as heat. Dividing the work output with the heat input, W/QH, gives you the heat engine’s efficiency.

This setup provides us with a nice tool for explaining how our atmosphere forces the mean global temperature of Earth’s surface to equilibrate at a much higher level than what a pure solar radiative equilibrium would.

Earth’s relatively thin surface air layer, squeezed in between the actual solar-heated solid/liquid surface underneath and the overlying atmosphere at large weighing down on it from above, can very well be considered a proper heat engine. Its ‘hot reservoir’ is the surface, directly heating it through conduction and radiation, but also – upon surface solar heating – injecting light water molecules carrying so-called ‘latent heat of vaporisation’ into it through the process of evaporation. This highly dynamic layer of air basically acts as a ‘working fluid’, and I will here repeat some of what the caption above reads about the heat engine to make you see how this makes a pretty fitting analogy: “The way Carnot saw it, the ‘engine’ is a working fluid expanding when supplied with heat, thereby pushing against a nearby object, like a piston, doing work on it.”

In other words, the atmosphere at large is the ‘piston’. And the work being done is convection.

The convective process – the movement of bulk air – basically takes all the energy transferred from the surface to the surface air layer as conductive/radiative heat (and ‘latent heat’) and brings it into the atmosphere at large, up through the tropospheric column, towards the tropopause. This is how energy is moved in the lower (dense) part of the atmosphere. Heat (or introduce water vapour into) any volume of air in a gravity field and it will rise. It will move UP. Because it expands. Becomes less dense than the surrounding air.

The only remaining heat from the surface after the convective work has been done (well, it is of course a concurrent process), is the radiation going straight out to space (the ‘cold reservoir’) through the so-called atmospheric window. This can be considered the ‘residual heat’, even though it passes directly through the working fluid.

But – you would probably object at this point – the ‘piston’ that is the atmosphere is not a rigid barrier to the expanding surface air.

Yes, we have come to the point where we will explain how the atmosphere in fact insulates the solar-heated surface.

First of all, it has got nothing to do with thermal radiation. It has everything to do with the mass of the atmosphere. In two ways:

  1. The atmosphere having a mass means it’s got a ‘heat capacity’, meaning it can be heated. It can gain a temperature. Space can’t.
  2. The atmosphere having a mass and being held within a gravity field means it’s got a weight, also meaning it exerts a downward force on the surface/surface air layer, expressed by a specific pressure. Space doesn’t.

So, what would happen if we were to insert a massive atmosphere between a solar-heated planetary surface and the vacuum of space?

The surface would warm. A lot. Eventually equilibrating at a considerably higher steady-state temperature.

Yes. But how? And why?

Simply from the rate of energy shed by the surface (heat OUT, Qout) no longer being able to keep up with the rate of energy from the Sun absorbed by the surface (heat IN, Qin) … AT THE SAME TEMPERATURE (kinetic level) as before.

There will be an initial imbalance in the ‘net heat’ (Qin – Qout = Q) in the way that Qin stays largely the same, but Qout is reduced substantially, meaning that Q has grown larger than zero. Remember that the surface itself is only a ‘hot reservoir’, not a ‘heat engine’. It cannot and does not do work. So the ‘work’ term for the surface is always zero. Which means that the 1st Law of Thermodynamics applied to the surface in effect would look like this: ΔU = Q -> Qin – Qout. So if Q then is positive, ΔU will also be. Energy will accumulate. And the temperature of the surface will rise. Moving in this way from the old, imbalanced state to a new, equilibrated one at a higher mean temperature, Qout will once again equal Qin, and ΔU would have fallen back to 0.

But how come Qout was reduced in the first place?

Returning to the two points just above, it should be fairly easy to deduce what happened. At least the first step along the way.

Space does not have a mass. It therefore doesn’t have a ‘heat capacity’. In other words, it cannot have its internal energy increased by heat coming from the surface. For all intents and purposes, it doesn’t possess internal energy to begin with. No mass, no internal energy. So it cannot hold a temperature – in effect, a perfect cold reservoir at 0 K. The heat moving out from the surface and into the vacuum above will thus move along a steepest possible (maximum) temperature gradient. There will be no impedance to heat loss (except from the ‘heat capacity’ of the ground, but that’s a different story). And hence, as per the Stefan-Boltzmann equation, the radiation emitted by the surface (its ‘irradiance’) would directly equal its heat output: Q = εσT4.

An atmosphere on the other hand does have a mass and therefore a ‘heat capacity’. It will let itself be warmed by the surface. It will thus gain a temperature. And accordingly, there will be established a less than maximum temperature gradient away from the solar-heated surface.

A lowered temperature gradient means reduced heat loss. Simple as that.

So replacing space with an atmosphere will make less energy escape the surface, (still) constantly heated by the Sun, per unit of time. Leading to natural warming to regain balance between in and out.

But there’s more. Once the atmosphere is emplaced, energy is no longer primarily transported away from the surface through the propagation of electromagnetic waves – radiation. Other heat transfer mechanisms have become available, and actually become more important, mechanisms that can only operate in a massive medium like air, not in a vacuum. Three such mechanisms exist on Earth: conduction, evaporation and convection.

And what all of these mechanisms ultimately work in tandem to effectuate (also including the radiative surface>air heat transfer, BTW) is – as we’ve already seen – the upward movement of air: convection – the way energy is always brought away from a heated surface through a fluid mass like water or a volume of gas. To bring the energy up and away from the surface and into the atmosphere at large. Which means they are all, when it comes down to it, pretty sensitive to whatever might hamper such upward bulk movement. Because, the surface absorbs on average 165 J/s/m2 worth of solar heat, and that means, to balance this, the surface also needs to release on average 165 J/s/m2 back out. If this somehow doesn’t pan out, energy will pile up and the surface will necessarily warm.

Earth’s global convective engine simply needs to work at a sufficient pace to keep up with the constant solar input. And if there’s not enough accumulated surface energy to drive it, meaning, if the surface temperature isn’t high enough, then it won’t manage.

I’ll explain.

There are two reasons why the planetary surface needs to be made even hotter than what the lowered idealised temperature gradient (the adiabatic lapse rate, the ALR) itself demands, and there are likewise two ways for it to become so:

  1. The air is simply heavy. And needs to be lifted against gravity. At a certain average speed. Setting a limit to mean convective uplift (buoyant acceleration) at a specific temperature gradient.* The heavier (denser: more mass per volume) a certain parcel of surface air, the hotter the surface needs to be to make the normal (environmental) lapse rate (ELR) on average match the ideal ALR. Otherwise the temp gradient would start ‘collapsing’, a highly unstable situation. We see the effect of this circumstance very clearly when for instance comparing Mars with Venus. (Lapse rates should be the topic for another post …)
  2. The atmosphere, through its sheer weight (and density), exerts a certain pressure on the surface, capping the mean evaporation rate from the ocean, covering 71% of Earth’s surface, at a specific temperature. The greater the surface pressure, the hotter the ocean surface needs to be.

*Newton’s Second Law: F = ma (F: the net force applied to the gas volume; m: the mass of the gas volume; a: the acceleration (the rate of change of velocity over time) of the gas volume).

F remains the same (the upward buoyant force from the expanding surface air layer – the ‘working fluid’ – minus the downward force from (the overlying weight of) the atmosphere at large – the ‘piston’), but m has increased, so a will reduce.

2 = 1*2

2 = 2*1


In the end it is next to impossible to quantify the contributions of these various atmospheric effects to our balmy mean global surface temperature of ~288K. This can only ever be a qualitative exposition of the mechanisms … But the fundamental physical principles behind them are all well-established and well-understood. Gas laws rather than radiative physics.

A brief summary

This is how the atmosphere makes the Earth’s surface warmer – much warmer – than the maximum pure solar radiative equilibrium temperature (because it sure does!):

  • It has a mass and therefore a ‘heat capacity’. This means it is able to warm. It does so by being directly convectively coupled with the solar-heated surface below it. Regardless of whether that atmosphere contains radiatively active gases (so-called ‘GHGs’) or not, it will warm – conductively > convectively; on our real Earth, like this: conductively/radiatively/evaporatively > convectively. The atmosphere is able to warm. Space isn’t. Therefore the atmosphere sets up a temperature gradient away from the solar-heated surface that has a finite (sub-max) steepness. Space doesn’t. The atmosphere thus INSULATES the surface. Energy is not able to escape the surface as fast as it’s coming in before it has warmed to a higher mean temperature than before the atmosphere was put in place.
  • It has a mass and therefore a weight (it’s in a gravity field, after all). Space doesn’t. This affects the surface energy escape rate in two ways: i) The expanding air lifting convectively from the surface air layer and into the atmosphere at large is heavy – it needs to be pushed upward against gravity. AT EQUAL TEMPERATURE, this circumstance makes it harder for energy to escape the surface convectively at the same rate with the atmosphere being denser (more mass per volume). ii) The atmosphere having a weight means it exerts a pressure on the solar-heated surface above 0. Unlike space. A higher atmospheric pressure/density makes it harder for energy to escape the surface than with a lower pressure AT EQUAL TEMPERATURE by suppressing the evaporation rate from the oceans. The weight of the atmosphere is not a rigid barrier. But it functions by the same principle – setting limits to convection/evaporation from a heated body.

There are indeed more things to say on this subject. Maybe in future posts …

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9 comments on “On Heat, the Laws of Thermodynamics and the Atmospheric Warming Effect

  1. JP Miller says:

    But the only way heat leaves our planet is via radiation. Ultimately, however heat gets moved from surface to atmosphere or from place to place in our oceans/ atmosphere, you have to address that. It may be possible that climate scientists have gotten the internal processes wrong and so have incorrect models for how that radiation propagates outward, but you have yet to demonstrate that it seems to me. And that’s what you need to show to convince me the “warmists” have it wrong.

    • okulaer says:

      Thanks for commenting.

      Well, I have several posts on this subject, each somehow expanding on the last and/or focusing on different aspects. I have also started a new series with the current top post. I don’t know how many parts there will be in the end, but I hope to shed some light on how the warmists get it wrong, yes.

    • okulaer says:

      But giving you a quick answer right here and now, you say: “But the only way heat leaves our planet is via radiation. Ultimately, however heat gets moved from surface to atmosphere or from place to place in our oceans/ atmosphere, you have to address that.”

      Well, you said it yourself already. The heat is eventually radiated to space. Courtesy of our IR-active gases. They don’t inhibit Earth’s radiative heat loss to space. They effectuate it. They make it possible.

  2. I came back here because “here” you say:

    “There will be no impedance to heat loss (except from the ‘heat capacity’ of the ground, but that’s a different story). And hence, as per the Stefan-Boltzmann equation, the radiation emitted by the surface (its ‘irradiance’) would directly equal its heat output: Q = εσT4.”

    Two details that may help later:
    Zero Kelvin cannot be used for either temperature in the S-B equation Zero Kelvin is an asymptote not a position such cannot be raised to a power higher than 1 without arithmetically creating Not A Number. Sounds trivial but the standards folk are now creating a method to make 1 Kelvin the standard for just this reason. That will make all temperatures logarithmic with Zero Kelvin just as far away from 1 Kelvin as is infinite Kelvin. The effects on “abstract hard sciences” will likely. be profound. No need to change anything yet. 0.0001 Kelvin is still OK for temperature.

    For thermal EMR the space impedance between any two thermalizations is the impedance of space 377 ohms reactive. Through the atmosphere it is somewhat less and a combination of reactive and resistive. The resistive and atmospheric thermal mass give rise to a time constant for any change in flux. This time constant is why Climate Clowns think the atmosphere absorbs thermal radiative flux, it does not! It truly attenuates any modulation on that flux produced by modulation of the radiance at either end. Yes, the modulation can be the opposite vector from the flux, also externally going at (c).

    If you wish to chat about the distinction between power and force or the result, energy and work, give me a shout anywhere!

  3. RealOldOne2 (the real one) says:

    Based on your excellent ‘ ‘To heat a planetary surface’ for dummies; Part 4′ article, showing why “thermodynamic ‘energy transfer’ is always unidirectional”, shouldn’t you update/revise the part of this article about ‘net energy’ to reflect that?

    Specifically your one sentence: “‘Energy’ per se might move in both directions.”
    As you pointed out in the above referenced ‘Part 4’ article, when two objects are radiatively interacting, energy flow is always unidirectional, being the resultant single EM wave interaction/combination between the separate EM waves of both objects.

    I suspect that you wrote this article before the other one, and having considered this further, your later article reflects your current understanding that ‘energy transfer’/flow is always only unidirectional, since a a real thermal energy flow/transfer/flux from a colder object to a warmer object would violate the 2nd Law.

    I have had individuals claim that you are contradicting yourself, in this article admitting bidirectional energy flow exists, and in your ‘Part 4’ article saying that there is no bidirectional energy flow.

    Thanks,
    The ‘real’ RealOldOne2

    • “when two objects are radiatively interacting, energy flow is always unidirectional, being the resultant single EM wave interaction/combination between the separate EM waves of both objects.”

      It appears that you have contradicted yourself. You describe “separate EM waves of both objects” which is, and can only be, a bidirectional energy flow.

      “the resultant single EM wave”, this is terribly confused. You seem to be suggesting that the electromagnetic waves from which the resultant is formed no longer exist?

      For two equal wavelength opposing energy waves you claim to have created a single resultant standing wave, light that is now stationary and not travelling at the speed of, well, light. That’s rather impressive.

      Your “single resultant” wave would also be of greater energy than the two comprising electromagnetic waves, therefore it’s wavelength would change. That would make many aspects of astronomy impossible, and indeed the measurement of atmospheric microwave emissions on the sunlit side of the planet would not be possible. Evidence shows otherwise.

      Dr Roy Spencer

      • RealOldOne2 says:

        First I’d like to thank you for the excellent work you, Dr. Christy and your team do on the UAH satellite temperature work. Second, I’d like to thank you for taking the time to post your blog. I’ve learned much from reading it.

        “It appears that you have contradicted yourself. You describe “separate EM waves of both objects” which is, and can only be, a bidirectional energy flow”

        While it may appear that way to you, the fact that EM waves interfere/superimpose says otherwise. Each EM wave by itself has the potential to be a real energy flow/transfer of thermal energy, but there is only a real thermal energy transfer if it is in thermal contact with another object which is at a lower temperature. Kristian(Okulaer) explains why:

        The energy exchange in a radiative heat transfer is continuous, simultaneous and instantaneous, the radiation field through which the heat is transferred completely integrated and indivisible. Which means there is no way you could ever detect any surface effect of separate emittances, separate waves of radiation (or ‘photons’ if you will) moving around the field. ONLY THE HEAT, the spontaneously occurring vector (net) sum of them all, moving through the field in one direction – from hot to cold – is a real transfer of energy, directly detectable and sensible. It is equivalent to a waterfall, wind or to an electric current, all moving spontaneously from high to low potential. SoD and the climate establishment need to stop pretending that there are two separate fluxes of energy operating distinct from the other one within one and the same radiation field, one and the same thermal exchange. As if they were both ‘heats’ in their own right.

        https://okulaer.wordpress.com/2014/08/11/why-atmospheric-radiative-gh-warming-is-a-chimaera/

        So it appears to me that you have succumbed to climate establishment’s groupthink, what Okulaer calls:

        … an appallingly naive, simplistic, and quite frankly absurd view of how things work in the real world. But it still has effectively managed to infect the minds of practically every person alive today…

        ibid.

        As Okulaer explained in the heat transfer example in that article, there is no way around the fact that your misguided belief in real bidirectional thermal energy flow/flux/transfer means that the increase in internal energy and temperature of the hotter object is a result of a transfer of thermal energy/heat from a colder object to a hotter object, which is a direct violation of the 2nd Law of Thermodynamics. I challenge you to please quote the portion of what Kristian said in that article and show why it is wrong.

        “…“the resultant single EM wave”, this is terribly confused. You seem to be suggesting that the electromagnetic waves from which the resultant is formed no longer exist?”

        Actually it is not confused at all if you understand the thermodynamics of radiant heat transfer and EM wave superposition/interference. This has been well accepted for a long time:

        The distinguishing characteristic of radiant heat is, that it travels in rays like light, whence the name radiant. These rays have all the physical properties of rays of light, and are capable of reflexion, refraction, interference, and polarization.

        Maxwell, ‘The Theory of Heat’, 1902

        Kristian shows this single resultant EM wave, which represents the thermal energy transfer/flow in this graphic:
        y
        in the article that I referenced, ‘To Heat a Planetary Surface’ for dummies; Part 4:
        https://okulaer.wordpress.com/2015/02/19/to-heat-a-planetary-surface-for-dummies-part-4/

        In this article Kristian also stated:

        The oppositely moving green and red waves portrayed here would not be observed in the real world. They basically do not exist as physical entities. They exist separately and independently only up to the point where they meet. After this point, their individual physical existence is lost. Their potentials carry on, though. But only the ‘net result’ of the two, the blue standing wave pattern, is to be observed. Only that is realised.

        When I read your comment, I went to that article to see if you had commented and attempted to refute anything Okulaer’s article, and I see that you hadn’t. Why not, since the substance of my comment depended on what Okulaer said there? Again, I challenge you to please quote the portion of what Kristian said in that article and show why it is wrong.

        “For two equal wavelength opposing energy waves you claim to have created a single resultant standing wave, light that is now stationary and not travelling at the speed of, well, light. That’s rather impressive.”

        My comment said nothing about two equal wavelength opposing energy waves. Where did you get this? This leads me to suspect that this comment may not have come from the real Dr. Roy Spencer, but from an impersonator. Also, in his own blog, you reply as “Roy Spencer”, whereas you reply as “Dr Roy Spencer”, another anomaly. Regardless of that, the stationary standing wave represents the energy being transferred. No temperature difference, no energy transfer:

        Consider two objects at different temperatures that are brought together. Energy is transferred from the hotter one to the cooler one, until both objects reach thermal equilibrium (ie., both become the same temperature). … The transfer of energy is caused by the temperature difference, and ceases once the temperatures are equal.

        https://www.boundless.com/physics/textbooks/boundless-physics-textbook/heat-and-heat-transfer-13/introduction-110/heat-as-energy-transfer-390-10942/

        When two things are in thermal contact but no thermal energy is exchanged between them, they’re in thermal equilibrium. If two things are in thermal equilibrium, they have the same temperature.

        http://www.dummies.com/how-to/content/transferring-heat-through-radiation.html

        Dr. Spencer, your claim that two objects which are at exactly the same non-zero temperature are continuously making a 100% efficient, lossless transfer of thermal energy via radiation back and forth endlessly, ie., a isothermal, reversible thermal energy transfer, would be a violation of the 2nd Law because it would be a transfer of thermal energy/heat in which the entropy of the universe would not increase, thus violating the Entropy law:

        Entropy and the Second Law of Thermodynamics: … How the Universe Works … In trying to synthesize the ideas of Kelvin, Joule, and Carnot – that is, that energy is conserved in thermodynamic processes and that heat always “flows downhill” in temperature – Rudolf Clausius invented the idea of entropy in such a way that the change in entropy is the ratio of the heat exchanged in any process and the absolute temperature at which that heat is exchanged. That is, he defined the change in entropy DS of an object which either absorbs or gives off heat Q at some temperature T as simply the ratio Q/T. With this new concept, he was able to put the idea that heat will always flow from the higher to the lower temperature into a mathematical framework. If a quantity of heat Q flows naturally from a higher temperature object to a lower temperature object – something that we ALWAYS observe, the entropy gained by the cooler object during the transfer is greater than the entropy lost by the warmer one since Q/Tc > Q/Th. So he could state that the principle that drives all natural thermodynamic processes is that the effect of any heat transfer is a net increase in the combined entropy of the two objects. And that new principle establishes the direction that natural processes proceed. All natural processes occur in such a way that the total entropy of the universe increases. The only heat transfer that could occur and leave the entropy of the universe unchanged is one that occurs between two objects which are at the same temperature – but that is not possible, since no heat would transfer. So a reversible isothermal heat transfer that would leave the entropy of the universe constant is just an idealization – and hence could not occur. All other processes – meaning, all real processes – have the effect of increasing the entropy of the universe. That is the second law of thermodynamics.

        California Polytechnic State University, http://www.calpoly.edu/~rbrown/entropy.html

        Radiation exerts a force/pressure:

        Radiation pressure is defined as the force per unit area exerted by electromagnetic radiation

        http://scienceworld.wolfram.com/physics/RadiationPressure.html

        With two exactly equal temperature objects, the radiation force/pressure is exactly equal but opposing, so there is no transfer of energy. It’s just like if you have a linear force pushing on a block on a frictionless surface the block will move in the direction of the applied force. (analogous to a single EM wave) The instant you apply an equal but opposite force, all motion ceases because you now have an equal but opposite force (analogous to the two opposing EM waves resulting in a single resultant EM standing wave for which there is no energy motion/transfer).

        Here’s another explanation:

        The temperature is the driving force for heat transfer, just as the voltage difference is the driving force for electric current flow and pressure difference is the driving force for fluid flow. The rate of heat transfer in a certain direction depends on the magnitude of the temperature gradient (the temperature difference per unit length or the rate of change of temperature) in that direction. The larger the temperature gradient, the higher the rate of heat transfer.

        http://web.archive.org/web/20150528053416/
        http://www.cdeep.iitb.ac.in/nptel/Mechanical/Heat%20and%20Mass%20Transfer/Conduction/Module%201/main/1.3.1.html
        India Institute of Technology

        These two water tanks show the IIT pressure difference/fluid flow example:

        When the water levels/pressures are equal, there is no water flow in the pipe connecting the two tanks. The claim of bidirectional thermal energy transfer between two objects in radiative contact is the same as claiming that there is continuous bidirectional water flow in the pipe connecting two water tanks with exactly the same water level, at the rate as if there was no opposing pressure in each other tank. Separately, each tank would have a water flow out of the pipe at the bottom, but once the two tanks are simultaneously interacting, all water transfer/flow out of both tanks ceases, just an energy flow ceases when two objects are at the same temperature.

        “Your “single resultant” wave would also be of greater energy than the two comprising electromagnetic waves, therefore it’s wavelength would change.”

        No it wouldn’t, because it neither travels nor carries any energy:

        If you have a source emitting an EM wave through a certain space and another source emitting another EM wave through the same space at the same frequency, but in the opposite direction, you get a standing wave. A standing wave does not travel or carry energy, so its average Poynting vector is zero. You can think of it as the Poynting vector of the one wave canceling the other.

        PhysicsForums,
        https://www.physicsforums.com/threads/cancellation-of-poynting-vector-components.507450/

        And as Okulaer’s previously referenced graphic shows that this standing wave is exactly the same frequency as the EM waves of the two equal temperature object. That’s how wave superposition/interference works.

        “That would make many aspects of astronomy impossible”

        Actually no it wouldn’t. Okulaer explains why in a comment on your own blog:
        http://www.drroyspencer.com/2017/05/uah-global-temperature-update-for-april-2017-0-27-deg-c/#comment-247505

        Your blind spot seems to be missing the fundamental fact that there are not two separate, real, independent thermal energy/heat transfers, as that would be a violation of the 2nd Law.

        Here is another heat transfer example similar to Okulaer’s example, which explains why your incorrect view that there is real bidirectional thermal energy transfer is a violation of both the 1st and 2nd Laws.

        Givens:
        – We have an object with a surface area of 1m².
        – The object is in a vacuum with the surroundings being space with a temperature of 0K.
        – the object and the radiation shield are blackbodies so ϵ=1.
        – The S-B constant, σ=5.67×10⁻⁸.
        – Temps rounded to whole numbers.
        – The shield is very close to but not touching the object so the surface area of each side of the very thin radiation shield is also 1m².
        – The only Energy-in to the object is an internal heat/energy source of 240W/m² within the object.
        – The object is initially radiating to 0K outer space.
        – The only heat transfer mechanism is radiative heat transfer, so there is no conductive, convective or latent heat of vaporization.

        So with those givens, at initial thermal equilibrium the Energy-out must equal the Energy-in = 240W/m². The S-B equation, q=ϵσ(Th⁴-Tc⁴) , thus tells us that the initial temperature of the object is 255K(-18C). There is no external Energy-in to the object because the temperature of space surrounding it is absolute zero.

        Again, the ONLY energy source to the object that exists in our system is 240W/m², period. Since one watt is defined as 1 Joule/sec, that means that the only energy source and the only energy being transferred through any point in our system is 240Joules/sec.

        At initial thermal equilibrium the 240Joules/sec of energy which is internally generated is transferred away from the surface of the object to the 0K surroundings.

        We now surround the object with a radiation shield which has an initial temperature of 0K, so no new energy is added to our system. The purpose of a radiation shield is to reduce heat loss from an object.

        The shield now initially receives the 240W/m²(240 Joules/sec) of energy/heat which is being transferred away from the object so the internal energy of the shield begins to increase which causes the temperature of the shield to increase. As the shield temperature increases, the heat transfer from the object is reduced, since the cold temperature (Tc) in the S-B equation is no longer zero, q=ϵσ(Th⁴-Tc⁴) . The reduction of heat transferred away from the object means that less than 240W/m² of energy is being transferred away which causes an accumulation of internal energy within the object. This accumulation of internal energy causes an increase in temperature of the object.

        The shield temperature and object temperature continue to increase until a new thermal equilibrium is reached. The temperature of the shield will then be 255K(-18C).

        At the new thermal equilibrium, the Energy-out from the object to the shield must equal the Energy-out from the shield to space which must equal the 240W/m²(240 Joules/sec) of internally generated energy/heat.

        The new equilibrium temperature of the object as calculated by the S-B equation is 303K(30C):
        q = ϵ σ (Th⁴ – Tc⁴) → 240 = (1) (5.67 x 10⁻⁸) (Th⁴ – 255⁴) →
        Th⁴ = (240 / 5.67 x 10⁻⁸) + 255⁴ → Th = 303K

        That is my correct description of what happens from the thermodynamic/heat transfer perspective. In my correct understanding:

        1) The increase in temperature from 255K to 303K is solely due to the accumulation of internal energy from the internal heat source of 240W/m²(240 Joules/sec), which is the only energy source to the object which exists in our system.

        2) The internal heat source remains Energy-in of 240W/m²(240 Joules/sec) , the heat/energy transferred away from the surface (to the shield) remains 240W/m² (240 Joules/sec) and the heat/energy transferred away from the shield to space is 240W/m² (240 Joules/sec).

        3) Before, during and after the process of adding the shield and reaching the new equilibrium temperature, energy flow is always and only uni-directional flowing away from the higher temperature/energy object(s) to the lower temperature/energy surroundings/shield and never flowing from the colder surroundings/shield to the warmer objects. This satisfies the 2nd Law, just as the Thermodynamics textbook says:

        The second law involves the fact that processes proceed in a certain direction but not in the opposite direction. A hot cup of coffee cools by virtue of heat transfer to the surroundings, but heat will not flow from the cooler surroundings to the hotter cup of coffee.

        Fundamentals of Classical Thermodynamics, VanWylen and Sonntag, Chap.6 ‘The Second Law of Thermodynamics’, p.155.

        4) Before, during and after the process of adding the radiation shield, energy is neither created nor destroyed. It is conserved, satisfying the 1st Law.

        My correct understanding of this problem is supported by Heat Transfer textbooks. Section “8-8 Radiation Shields” of J.P.Holman, ‘Heat Transfer’ says:

        These shields do not deliver or remove any heat from the overall system; they only place another resistance in the heat-flow path so that the overall heat transfer is retarded.

        J.P.Holman, ‘Heat Transfer’, McGraw Hill, 2nd ed., textbook

        Your wrong understanding of bidirectional thermal energy transfer is that at the new thermal equilibrium, the temperature increase of the always warmer object is solely due to a new energy/heat flow of 240W/m² (Joules/sec) being transferred from the always colder shield to the always warmer object. So your wrong understanding at final equilibrium now has two Energy-ins to the object, the original 240W/m² (240 Joules/sec) from the internal heat source plus a new Energy-in of 240W/m² (240 Joules/sec) being transferred from the colder radiation shield.

        Thus your wrong understanding has created 240W/m² (240 Joules/sec) of energy out of thin air.

        And your wrong view has 480 Joules/sec of energy transferring away from the 303K object, which is twice as much energy/heat as is coming from the only energy/heat source of our system.

        Your wrong view is not consistent with a radiation shield, whose purpose is to reduce heat transfer from an object, because your wrong view has the heat transfer away from the object actually increasing.

        In summary:
        Your wrong view created a new energy-in source to the object of 240W/m² out of thin air. Thus your wrong understanding violates the 1st Law, because it wrongly has the object receiving 480W/m^2 of energy where the only energy source is 240W/m^2, so conservation of energy, the 1st Law has been violated.

        Your wrong understanding violates the 2nd Law as it has the cause of the increase in temperature of the always warmer object coming from the transfer of heat from an always colder object (shield) to an always warmer object.

        This shows that my understanding is correct and the wrong understanding of bidirectional thermal energy/heat transfer is wrong.

        Please quote the portion of my heat transfer example that you think is incorrect and show the science of why it is incorrect.

        Okulaer & my view that there is no real bidirectional energy flow is supported by Ozawa(2003) ‘The Second Law of Thermodynamics and the Global Climate System’, a peer reviewed paper written expressly from the perspective of the SLoT: https://a.disquscdn.com/uploads/mediaembed/images/3163/2606/original.jpg

        Figure 5(a) shows the “Global-mean (surface-area mean) energy flux components (i.e., shortwave radiation, longwave radiation, vertical turbulent heat transport) of the energy budget for the Earth. It shows no energy transfer component from the colder atmosphere to the warmer surface of the Earth, as your bidirectional energy flow would require.

        Cheers.

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