Archive for October, 2010

This post covers some foundations which are often misunderstood.

Radiation emitted from a surface (or a gas) can go in all directions and also varies with wavelength, and so we start with a concept called spectral intensity.

This value has units of W/m².sr.μm, which in plainer language means Watts (energy per unit time) per square meter per solid angle per unit of wavelength. (“sr” in the units stands for “steradian“).

Most people are familiar with W/m² – and spectral intensity simply “narrows it down” further to the amount of energy in a direction and in a small bandwidth.

We’ll consider a planar opaque surface emitting radiation, as in the diagram below.


Hemispherical Radiation, Incropera and DeWitt (2007)

Hemispherical Radiation, Incropera and DeWitt (2007)


The total hemispherical emissive power, E, is the rate at which radiation is emitted per unit area at all possible wavelengths and in all possible directions. E has the more familiar units of W/m².

Most non-metals are “diffuse emitters” which means that the intensity doesn’t vary with the direction.

For a planar diffuse surface – if we integrate the spectral intensity over all directions we find that emissive power per μm is equal to π (pi) times the spectral intensity.

This result relies only on simple geometry, but doesn’t seem very useful until we can find out the value of spectral intensity. For that, we need Max Planck..


Most people have heard of Max Planck, Nobel prize winner in 1918. He derived the following equation (which looks a little daunting) for the spectral intensity of a blackbody:

Spectral Intensity, Max Planck

where T = absolute temperature (K); λ = wavelength; h = Planck’s constant = 6.626 x 10-34 J.s; k = Boltzmann’s constant = 1.381 x 10-23 J/K; c0 = the speed of light in a vacuum = 2.998 x 108 m/s.

What this means is that radiation emitted is a function only of the temperature of the body and varies with wavelength. For example:

Note the rapid increase in radiation as temperature increases.

What is a blackbody?

A blackbody:

  • absorbs all incident radiation, regardless of wavelength and direction
  • emits the maximum energy for any wavelength and temperature (i.e., a perfect emitter)
  • emits independently of direction

Think of the blackbody as simply “the reference point” with which other emitters/absorbers can be compared.


The Stefan-Boltzmann equation (for total emissive power) is “easily” derived by integrating the Planck equation across all wavelengths and using the geometrical relationship explained at the start (E=πI). The result is quite well known:

E = σT4

where σ=5.67 x 10-8 and T is absolute temperature of the body.

The result above is for a blackbody. The material properties of a given body can be measured to calculate its emissivity, which is a value between 0 and 1, where 1 is a blackbody.

So a real body emits radiation according to the following formula:

E = εσT4

where ε is the emissivity. (See later section on emissivity and note 1).

Note that so long as the Planck equation is true, the Stefan-Boltzmann relationship inevitably follows. It is simply a calculation of the total energy radiated, as implied by the Planck equation.

The Smallprint

The Planck law is true for radiant intensity into a vacuum and for a body in Local Thermodynamic Equilibrium (LTE).

So that means it can never be used in the real world

Or so many people who comment on blogs seem to think. Let’s take a closer look.

The Vacuum

The speed of light in a vacuum, c0 = 2.998 x 108 m/s. This value appears in the Planck equation and so we need to cater for it when the emission of radiation is into air. The speed of light in air, cair = c0/n, where n is the refractive index of air = 1.0008.

Here’s a comparison of the Planck curves at 300K into air and a vacuum:

Not easy to separate. If we expand one part of the graph:

We can see that at the peak intensity the difference is around 0.3%.

The total emissive power into air:

E = n²σT4, where n is the refractive index of air

So the total energy radiated from a blackbody into air = 1.0016 x the total energy into a vacuum.

This is why it’s a perfectly valid assumption not to bother with this adjustment for radiation into air. In glass it’s a different proposition..

Local Thermodynamic Equilibrium

The meaning, and requirement, of LTE (local thermodynamic equilibrium) is often misunderstood.

It does not mean that a body is at the same temperature as its surroundings. Or that a body is all at the same temperature (isothermal).

An explanation which might help illuminate the subject – from Thermal Radiation Heat Transfer, by Siegel & Howell, McGraw Hill (1981):

In a gas, the redistribution of absorbed energy occurs by various types of collisions between the atoms, molecules, electrons and ions that comprise the gas. Under most engineering conditions, this redistribution occurs quite rapidly, and the energy states of the gas will be populated in equilibrium distributions at any given locality. When this is true, the Planck spectral distribution correctly describes the emission from a blackbody..

Another definition, which might help some (and be obscure to others) is from Radiation and Climate, by Vardavas and Taylor, Oxford University Press (2007):

When collisions control the populations of the energy levels in a particular part of an atmosphere we have only local thermodynamic equilibrium, LTE, as the system is open to radiation loss. When collisions become infrequent then there is a decoupling between the radiation field and the thermodynamic state of the atmosphere and emission is determined by the radiation field itself, and we have no local thermodynamic equilibrium.

And an explanation about where LTE does not apply might help illuminate the subject, from Siegel & Howell:

Cases in which the LTE assumption breaks down are occasionally encountered.

Examples are in very rarefied gases, where the rate and/or effectiveness of interparticle collisions in redistributing absorbed radiant energy is low; when rapid transients exist so that the populations of energy states of the particles cannot adjust to new conditions during the transient; where very sharp gradients occur so that local conditions depend on particles that arrive from adjacent localities at widely different conditions and may emit before reaching equilibrium and where extremely large radiative fluxes exists, so that absorption of energy and therefore populations of higher energy states occur so strongly that collisional processes cannot repopulate the lower states to an equilibrium density.

Now these LTE explanations are far removed from most people’s perceptions of what equilibrium means.

LTE is all about, in the vernacular:

Molecules banging into each other a lot so that normal energy states apply

And once this condition is met – which is almost always in the lower atmosphere – the Planck equation holds true. In the upper atmosphere this doesn’t hold true, because the density is so low. A subject for another time..

So much for Planck and Stefan-Boltzmann. But for real world surfaces (and gases) we need to know something about emissivity and absorptivity.

Emissivity, Absorptivity and Kirchhoff

There is an important relationship which is often derived. This relationship, Kirchhoff’s law, is that emissivity is equal to absorptivity, but comes with important provisos.

First, let’s explain what these two terms mean:

  • absorptivity is the proportion of incident radiation absorbed, and is a function of wavelength and direction; a blackbody has an absorptivity of 1 across all wavelengths and directions
  • emissivity is the proportion of radiation emitted compared with a blackbody, and is also a function of wavelength and direction

The provisos for Kirchhoff’s law are that the emissivity and absorptivity are equal only for a given wavelength and direction. Or in the case of diffuse surfaces, are true for wavelength only.

Now Kirchhoff’s law is easy to prove under very restrictive conditions. These conditions are:

  • thermodynamic equilibrium
  • isothermal enclosure

That is, the “thought experiment” which demonstrates the truth of Kirchhoff’s law is only true when there is a closed system with a body in equilibrium with its surroundings. Everything is at the same temperature and there is no heat exchanged with the outside world.

That’s quite a restrictive law! After all, it corresponds to no real world problem..

Here is how to think about Kirchhoff’s law.

The simple thought experiment demonstrates completely and absolutely that (under these restrictive conditions) emissivity = absorptivity (at a given wavelength and direction).

However, from experimental evidence we know that emissivity of a body is not affected by the incident radiation, or by any conditions of imbalance that occur between the body and its environment.

From experimental evidence we know that the absorptivity of a body is not affected by the amount of incident radiation, or by any imbalance between the body and its environment.

These results have been confirmed over 150 years.

As Siegel and Howell explain:

Thus the extension of Kirchhoff’s law to non-equilibrium systems is not a result of simple thermodynamic considerations. Rather it results from the physics of materials which allows them in most instances to maintain themselves in LTE and this have their properties not depend on the surrounding radiation field.

The important point is that thermodynamics considerations allow us to see that absorptivity = emissivity (both as a function of wavelength), and experimental considerations allow us to extend the results to non-equilibrium conditions.

This is why Kirchhoff’s law is accepted in thermodynamics.

Operatic Considerations

The hilarious paper by Gerlich and Tscheuschner poured fuel on the confused world of the blogosphere by pointing out just a few pieces of the puzzle (and not the rest) to the uninformed.

They explained some restrictive considerations for Planck’s law, the Stefan-Boltzmann equation, and for Kirchhoff’s law, and implied that as a result – well, who knows? Nothing is true? Not much is true?Nothing can be true? I had another look at the paper today but really can’t disentangle their various claims.

For example, they claim that because the Stefan-Boltzmann equation is the integral of the Planck equation over all wavelengths and directions:

Many pseudo-explanations in the context of global climatology are already falsified by these three fundamental observations of mathematical physics.

Except they don’t explain which ones. So no one can falsify their claim. And also, people without the necessary background who read their paper would easily reach the conclusion that the Stefan-Boltzmann equation had some serious flaws.

All part of their entertaining approach to physics.

I mention their papertainment because many claims in the blog world have probably arisen through uninformed people reading bits of their paper and reproducing them.


The fundamentals of radiation are well-known and backed up by a century and a half of experiments. There is nothing controversial about Planck’s law, Stefan-Boltzmann’s law or Kirchhoff’s law.

Everyone working in the field of atmospheric physics understands the applicability and limits of their use (e.g., the upper atmosphere).

This is not cutting edge stuff, instead it is the staple of every textbook in the field of radiation and radiant heat transfer.


Note 1 – Because emissivity is a function of wavelength, and because emission of radiation at any given wavelength varies with temperature, average emissivity is only valid for a given temperature.

For example, at 6000K most of the radiation from a blackbody has a wavelength of less than 4μm; while at 200K most of the radiation from a blackbody has a wavelength greater than 4μm.

Clearly the emissivity for 6000K will not be valid for the emissivity of the same material at a temperature of 200K.

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In Part One we looked at how solar radiation and DLR (or “back radiation”) were absorbed by the ocean. And we had a brief look at how little heat would move by conduction into the deeper ocean if the ocean was “still”.

There were some excellent comments in part one from Nick Stokes, Arthur Smith and Willis Eschenbach – probably others as well – take a look if you didn’t see them first time around.

We will shortly look at mixing and convection, but first we will consider some absolute basics.

The First Law of Thermodynamics

How does the ocean sustain its (high) temperature? Every second, every square meter of the ocean is radiating energy. The Stefan-Boltzmann relationship tells us the value:

j = εσT4, where ε is the emissivity of the ocean (0.99), σ = 5.67 x 10-8 and T is the temperature in K

For example:

  • if T = 20°C (293K), j = 415 W/m²
  • if T = 10°C (283K), j = 361 W/m²

Now, many people are confused about how temperatures change with heat imbalances. If, for some reason, more heat is absorbed by a system than is radiated/conducted/convected away – what happens?

More heat absorbed than lost = heat gained. Heat gained leads to an increase of temperature (see note 1). When the temperature of a body increases, it radiates, conducts and convects more heat away (see note 2). Eventually a new equilibrium is reached at a higher temperature. It is important to grasp this concept. Read it again if it isn’t quite clear. Ask a question for clarification..

Questions are welcome.

A Simple Model

Evaluating a very simple energy balance model might help to set the scene.

Here is the radiative input – solar radiation and “back radiation” from the atmosphere, with typical values for a tropical region:

The primary question – the raison d’êtra for this article –  is what happens if only the solar radiation heats the ocean? And compared with if the back-radiation also heats the ocean?

It’s easy to find the basic equilibrium point using the first law of thermodynamics. All you need to know is the energy in, and the equation which links energy radiated with temperature.

For radiation, this is the Stefan-Boltzmann law cited earlier. The starting temperature for the ocean surface in this example was set to 300K (27°C). Depending on whether solar and back-radiation or just solar is heating the surface, here is the surface temperature change:

Notice the difference in the temperature trends for the two cases.

Now the model doesn’t yet include convective heat transfer from the ocean to the atmosphere (or movement of heat from the tropics to the poles), which is why in the first graph the temperature gets so high. Convection will reduce this temperature to a more “real world” value.

The second graph has only solar radiation heating the ocean. Notice that the temperature drops to a very low value (-15°C) in just a few years. Clearly the climate would be very different if this was the case, and the people who advocate this model need to explain exactly how the ocean temperature manages to stay so much higher.

By the way, if we made the “well-mixed layer”, dmixed, of the ocean deeper it would increase the time for the temperature to change by any given amount. That’s because more ocean has more heat capacity. But it doesn’t change the fact of the energy imbalance, or the final equilibrium temperature.

The model is a very simplistic one. That’s all you need to demonstrate that DLR, or “back radiation” must be absorbed by the ocean and contributing to the ocean heat content.

Turbulence and the Mixed Layer

Let’s take a look at a slightly more complex model to demonstrate an important point. This simulation has four main elements:

  • radiation absorbed in the ocean at various depths, according to the results in Part One
  • conduction between layers in the ocean
  • convective heating from the ocean surface to the atmosphere, according to a simple model with a fixed air temperature

This model is not going to revolutionize climate models as it has many simplifications. The important factor - there is no convection between different ocean layers in this model.

Now conductivity in still water is very low (as explained in Part One).

The starting condition – the “boundary condition” – was for the temperature to start at 300K (27°C) for the first 100m, with the ocean depths below to be a constant 1°C.

The model is for illumination. Let’s see what happens:

The wide bars of blue and green are because the day/night variation is significant but squashed horizontally. If we expand one part of the graph to look at the first few days:

You can see that the day/night variation of the top 1mm and 10cm are significant.

Look back at the first graph which covers four years. Notice the purple line, 10m depth, the blue line, 3m depth; and the red line, 1m depth.

Why is the ocean 1-10m depth increasing to such a high temperature?

The reason is simple. This model is flawed- these results don’t occur in practice. (And yes, the ocean would boil from within..)

The equations that make up this model have used:

  • the radiation absorbed from the sun and the atmosphere (as described in part one)
  • the radiation emitted from the surface layer (the Stefan-Boltzmann equation)
  • conductivity transferring heat between layers

If these were the only mechanisms for transferring heat, the ocean 1m – 10m deep would be extremely hot in the tropics. This is because the ocean where the radiation is absorbed cannot radiate back out.

For a mental picture think of a large thick slab of PVC which is heated from electrical elements within the PVC. Because it is such a poor conductor of heat, the inner temperature will rise much higher than the surface temperature, so long as the heating continues..

The reason this doesn’t happen in practice in the ocean is due to convection.

If you heat a gas or liquid from below it heats up and expands. Because it is now less dense than the layer above it will rise. This is what happens in the atmosphere, and it also happens in the ocean. The ocean under the very surface layer heats up, expands and rises – overturning the top layer of the ocean. This is natural convection.

The other effect that takes place is forced convection as the wind speed “stirs” the top few meters of the ocean. Convection is the transfer of heat by bulk motion of a fluid. Essentially, the gas or liquid moves, taking heat with it.

Price & Weller (1986) commented:

Under summer heating conditions with vanishing wind, the trapping depth of the thermal response is only about 1m (mean depth value), and the surface amplitude is as large as 2ºC or 3ºC. But, more commonly, when light or moderate winds are present, solar heating is wind mixed vertically to a considerably greater depth than is reached directly by radiation: the trapping depth is typically 10m, and the surface amplitude is reduced in inverse proportion to typically 0.2ºC. Given that the surface heating and wind stress are known, then the key to understanding and forecasting the diurnal cycle of the ocean is to learn how the trapping depth is set by the competing effects of a stabilizing surface heat flux and a destabilizing surface stress.

Here are the results from a model with another slight improvement. This includes natural convection. The mechanism is very rudimentary at this stage. It simply analyzes the temperature profile at each time step and if the temperature is inverted from normal buoyancy a much higher value of thermal conductivity is used to simulate convection.

The “bumpiness” you see in the temperature profile is because the model has multiple “slabs”, each with an average temperature. This could be reduced by a finer vertical grid.

During the early afternoon with peak solar radiation, the ocean becomes stratified. Why?

Because lots of heat is being absorbed in the first few meters with some then transported upwards to the surface via convection – but while the solar radiation value is high this heat keeps “pouring in” lower down. However, once the sun sets the surface will cool via radiation to the atmosphere and so become less buoyant. With no solar radiation now being absorbed lower down, the top few meters completely mix – from natural convection.

I did have a paper with a perfect set of measurements to illustrate these points. It showed day/night and seasonal variation. Sadly I put it down somewhere. Many hours of hunting for the physical paper and for the file on my PC but it is still lost..

Note that the large variation of surface temperature (4-5°C) is just a result of the convective mixing element in the model being too simplistic and moving heat much faster than happens in reality.

Kondo and Sasano (1979) said:

In the upper part of the ocean, a mixed layer with homogeneous density (or nearly homogeneous temperature) distribution is formed during the night due to free convection associated with heat loss from the sea surface and to forced convection by wind mixing.

During the daytime, the absorption of solar radiation which occurs mostly near the sea surface causes the temperature to rise, and a stable layer is formed there; as a consequence, turbulent transport is reduced.

Daily mean depth of the mixed layer increases with the wind speed. When the wind speed is lower than about 7-8 m/s, the mixed layer disappears about noon but it develops again in the later afternoon. A mixed layer can be sustained all day under high wind speeds..


The subject of convection and oceans is a fascinating one and I hope to cover much more. However, convection is a complex subject, the most complex mechanism of heat transfer “by a mile”.

There are also some complexities with the skin layer of the ocean which are worth taking a closer look at in a future article.

This article uses some very simple models to demonstrate that energy radiated from the atmosphere is being absorbed in the ocean surface and affecting its temperature. If it wasn’t the ocean surface would freeze. Therefore, if atmospheric radiation increases (for example, from an increase in “greenhouse” gases), then, all other things being equal, this will increase the ocean temperature.

The models also demonstrate that conduction of heat on its own cannot explain the temperature profiles we see in the ocean. Natural convection and wind speed both create convection, which is a much more effective heat transport mechanism in gases and liquids than conduction.

Updates: Does Back Radiation “Heat” the Ocean? – Part Three

Does Back Radiation “Heat” the Ocean? – Part Four


Diurnal Cycling: Observations and Models of the Upper Ocean Response to Diurnal Heating, Cooling and Wind Mixing, James Price & Robert Weller, Journal of Geophysical Research (1986)

On Wind Driven Current and Temperature Profiles with Diurnal Period in the Oceanic Planetary Boundary Layer, Kondo and Sasano, Journal of Physical Oceanography (1979)


Note 1 – For the purists, heat retained can go into chemical energy, it can go into mechanisms like melting ice, or evaporating water which don’t immediately increase temperature.

Note 2 – For the purists, the actual heat transfer mechanism depends on the physical circumstances. For example, in a vacuum, only radiation can transfer heat.

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I have done a partial update of the Roadmap section – creating a few sub-pages and listed the relevant articles under the sub-pages.

It is a work in progress, the idea is to make it possible for new visitors to find useful articles. Most blogs have a high bias towards the last few articles.

I have split off:

CO2 – an 8-part series on CO2 as well as a few other related articles

Science Roads Less Traveled – science basics and alternative theories explained

“Back Radiation” – the often misunderstood subject of radiation emitted by the atmosphere

Just a note as well for new visitors. There are many articles explaining some climate science basics. Many people assume from this – and from other simplistic coverage on the internet – that climate science is full of over-simplistic models.

I don’t want to encompass all in a sweeping generalization.. but.. almost all comments I see on this subject are attacking simplistic models aimed at educating rather than models actually used in climate science.

For example, models aiming to give simple education on the radiative effect of CO2 range from:

  • ultra-simplistic/misleading – CO2 works like a “greenhouse”
  • simplistic – CO2 is an “insulator” trapping heat
  • basic radiative model – blackbody radiator of the surface, atmosphere & solar combination

But in a real climate model, there are equations from fundamental physics like:

And in atmospheric radiation textbooks:


From Vardavas & Taylor (2007)

From Vardavas & Taylor (2007)


Providing a set of equations doesn’t prove anything is right.

But my intent is to highlight that simple models are for illumination. It is easy to prove that simple models are simplistic.

The science of atmospheres and climate is much more sophisticated than these models designed for illumination.

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Probably many, most or all of my readers wonder why I continue with this theme when it’s so completely obvious..

Well, most people haven’t studied thermodynamics and so an erroneous idea can easily be accepted as true.

All I want to present here is the simple proof that thermodynamics textbooks don’t teach the false ideas circulating the internet about the second law of thermodynamics.

So for those prepared to think and question – it should be reasonably easy, even if discomforting, to realize that an idea they have accepted is just not true. For those committed to their cause, well, even if Clausius were to rise from the dead and explain it..

On another blog someone said:

Provide your reference that he said heat can spontaneously flow from cold to hot. And not from a climate ‘science’ text.

I had cited the diagram from Fundamentals of Heat and Mass Transfer by Incropera and DeWitt (2007). It’s not a climate science book as the title indicates.

However, despite my pressing (you can read the long painful exchange that follows) I didn’t find out what the blog owner actually thought that the writers of this book were saying. Perhaps the blog owner never grasped the key element of the difference between the real law and the imaginary one.

So I should explain again the difference between the real and imaginary second law of thermodynamics once again. I’m relying on the various proponents of the imaginary law because I can’t find it in any textbooks. Feel free to correct me if you understand this law in detail.

The Real Second Law of Thermodynamics

1a. Net heat flows from the hotter to the colder

1b. Entropy of a closed system can never reduce

1c. In a radiative exchange, both hotter and colder bodies emit radiation

1d. In a radiative exchange, the colder body absorbs the energy from the hotter body

1e. In a radiative exchange, the hotter body absorbs the energy from the colder body

1f. This energy from the colder body increases the temperature compared with the case where the energy was not absorbed

1g. Due to the higher energy radiated from a hotter body, the consequence is that net heat flows from the hotter to the colder (see note 1)

The Imaginary Second Law of Thermodynamics

2a. – as 1a

2b.  - as 1b

2c.  - as 1c

2d.  - as 1d

2e. In a radiative exchange, the hotter body does not absorb the energy from the colder body as this would be a violation of the second law of thermodynamics

Hopefully everyone can clearly see the difference between the two “points of view”. Everyone agrees that net heat flows from hotter to colder. There is no dispute about that.

What the Equations Look Like for Both Cases

Now, let’s take a look at the radiative exchange that would take place under the two cases and compare them with a textbook. Even if you find maths a little difficult to follow, the concept will be as simple as “two oranges minus one orange” vs “two oranges” so stay with me..

Here is the example we will consider:


Radiant heat transfer

Radiant heat transfer


We will keep it very simple for those not so familiar with maths. In typical examples, we have to consider the view factor – this is a result of geometry – the ratio of energy radiated from body 1 that reaches body 2, and the reverse. In our example, we can ignore that by considering two very long plates close together.

E1 is the energy radiated from body 1 (per unit area) and we consider the case when all of it reaches body 2, E2 is the energy radiated from body 2 (per unit area) and we consider that all of it reaches body 1.

We define Enet1 as the change in energy experienced by body 1 (per unit area). And Enet2 as the change in energy experienced by body 2 (per unit area).

Radiation Exchange under The Real Second Law

E1 = εσT14; E2= εσT24 (Stefan-Boltzmann law)

Enet1 = E2 – E1 = εσT24 – εσT14

Enet2 = E1 – E2 = εσT14 – εσT24

Therefore, Enet1 = -Enet2

Under The Imaginary Second Law

Enet1 = – E1 = -εσT14

Enet2 = E1 – E2 = εσT144 – εσT24

Therefore, Enet1 ≠-Enet2 ; note that ≠ means “not equal to”

This should be uncontroversial. All I have done is written down mathematically what the two sides are saying. If we took into account view factors and areas then the formulae would like slightly more cluttered with terms like A1F12.

In the case of the real second law, the net energy absorbed by body 2 is the net energy lost by body 1.

In the case of the imaginary second law, there is some energy floating around. No advocates have so far explained what happens to it. Probably it floats off into space where it can eventually be absorbed by a colder body.

Alert readers will be able to see the tiny problem with this scenario..

What the Textbooks Say

First of all, what they don’t say is:

When energy is transferred by radiation from a colder body to a hotter body, it is important to understand that this incident radiation cannot be absorbed – otherwise it would be a clear violation of the second law of thermodynamics

I could leave it there really. Why don’t the books say this?

Engineering Calculations in Radiative Heat Transfer, by Gray and Müller (1974)

Note that if the imaginary second law advocates were correct, then the text would have to restrict the conditions under which equation 2.1 and 2.2 were correct – i.e., that they were only correct for the energy gain for the colder body and NOT correct for the energy loss of the hotter body.

Heat and Mass Transfer, by Eckert and Drake (1959)

Note the highlighted area.

Basic Heat Transfer, M. Necati Özisik (1977)

Note the circled equations – matching the equations for the “real second law” and not matching the equations for the “imaginary second law”. Note the highlighted area.

Heat Transfer, by Max Jakob (1957)

Note the highlighted section, same comment as for the first book.

Principles of Heat Transfer, Kreith (1965)

Note the highlighted sections. The second highlight once again confirms the equation shown at the start, that under “the real second law” conditions, Enet1 = – Enet2. Under the “imaginary second law” conditions this equation doesn’t hold.

Fundamentals of Heat and Mass Transfer, Incropera and DeWitt (2007)

Note the circled section. This is false, according to the advocates of the imaginary second law of thermodynamics.

And the very familiar diagram shown many times before:


From "Fundamentals of Heat and Mass Transfer, 6th edition", Incropera and DeWitt (2007)

From "Fundamentals of Heat and Mass Transfer, 6th edition", Incropera and DeWitt (2007)



There are some obvious explanations:

1. Professors in the field of heat transfer write rubbish that is easily refuted by checking the second law – heat cannot flow from a colder to a hotter body.

2. Climate science advocates have crept into libraries around the world, and undiscovered until now, have doctored all of the heat transfer text books.

3. (My personal favorite) Science of Doom is refuted because these writers all agree that net heat flows from the hotter to the colder.

4. Look, a raven.

Relevant articles - The Real Second Law of Thermodynamics


Note 1 – Strictly speaking a hotter body might radiate less than a colder body – in the case where the emissivity of the hotter body was much lower than the emissivity of the colder body. But under those conditions, the hotter body would also absorb much less of the irradiation from the colder body (because absorptivity = emissivity). And so net heat flow would still be from the hotter to the colder.

To keep explanations to a minimum in the body of the article in 1e and 1f I also didn’t state that the proportion of energy absorbed by each would depend on the absorptivity of each body.

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In the three part series on DLR (also known as “back radiation”, also known as atmospheric radiation), Part One looked at the network of stations that measured DLR and some of the measurements, Part Two reviewed the spectra of this radiation, and Part Three asked whether this radiation changed the temperature of the surface.

Very recently, on another blog, someone asked whether I thought “back radiation” heated the ocean. I know from a prominent blog that a very popular idea in blog-land is that the atmospheric radiation doesn’t heat the ocean. I have never seen any evidence for the idea. That doesn’t mean there isn’t any..

See note 1 on “heat”.

The Basic Idea

From what I’ve seen people write about this idea, including the link above, the rough argument goes like this:

  • solar radiation penetrates tens of meters into the ocean
  • atmospheric radiation – much longer wavelengths – penetrates only 1μm into the ocean

Therefore, solar radiation heats the ocean, but atmospheric radiation only heats the top few molecules. So DLR is unable to transfer any heat into the bulk of the ocean, instead the energy goes into evaporating the top layer into water vapor. This water vapor then goes to make clouds which act as a negative feedback. And so, more back-radiation from more CO2 can only have a cooling effect.

There are a few assumptions in there. Perhaps someone has some evidence of the assumptions, but at least, I can see why it is popular.

Solar Radiation

As regular readers of this blog know, plus anyone else with a passing knowledge of atmospheric physics, solar radiation is centered around a wavelength of 0.5μm. The energy in wavelengths greater than 4μm is less than 1% of the total solar energy and conventionally, we call solar radiation shortwave.

99% of the energy in atmospheric radiation has longer wavelengths than 4μm and along with terrestrial radiation we call this longwave.

Most surfaces, liquids and gases have a strong wavelength dependence for the absorption or reflection of radiation.

Here is the best one I could find for the ocean. It’s from Wikipedia, not necessarily a reliable source, but I checked the graph against a few papers and it matched up. The papers didn’t provide such a nice graph..

Absorption coefficient for the ocean - Wikipedia

Absorption coefficient for the ocean - Wikipedia

Figure 1

Note the logarithmic axes.

The first obvious point is that absorption varies hugely with the wavelength of incident radiation.

I’ll explain a few basics here, but if the maths is confusing, don’t worry, the graphs and explanation will attempt to put it all together. The basic equation of transmission relies on the Beer-Lambert law:

I = I0.exp(-kd)

where I is the radiation transmitted, I0 is the incident radiation at that wavelength, d is the depth, and k is the property of the ocean at this wavelength

It’s not easy to visualize if you haven’t seen this kind of equation before. So imagine 100 units of radiation incident at the surface at one wavelength where the absorption coefficient, k = 1:

Figure 2

So at 1m, 37% of the original radiation is transmitted (and therefore 63% is absorbed).

At 2m, 14% of the radiation is transmitted.

At 3m, 5% is transmitted

At 10m, 0.005% is transmitted, so 99.995% has been absorbed.

(Note for the detail-oriented people, I have used the case where k=1/m).

Hopefully, this concept is reasonably easy to grasp. Now let’s look at the results of the whole picture using the absorption coefficient vs wavelength from earlier.

Figure 3

The top graph shows the amount of radiation making it to various depths, vs wavelength. As you can see, the longer (and UV) wavelengths drop off very quickly. Wavelengths around 500nm make it the furthest into the ocean depths.

The bottom graph shows the total energy making it through to each depth. You can see that even at 1mm (10-3m) around 13% has been absorbed and by 1m more than 50% has been absorbed. By 10m, 80% of solar radiation has been absorbed.

The graph was constructed using an idealized scenario – solar radiation less reflection at the top of atmosphere (average around 30% reflected), no absorption in the atmosphere and the sun directly overhead. The reason for using “no atmospheric absorption” is just to make it possible to construct a simple model, it doesn’t have much effect on any of the main results.

If we considered the sun at say 45° from the zenith, it would make some difference because the sun’s rays would now be coming into the ocean at an angle. So a depth of 1m would correspond to the solar radiation travelling through 1.4m of water (1 / cos(45°)).

For comparison here is more accurate data:

From "Light Absorption in Sea Water", Wozniak (2007)

From "Light Absorption in Sea Water", Wozniak (2007)

Figure 4

On the left the “surface” line represents the real solar spectrum at the surface – after absorption of the solar radiation by various trace gases (water vapor, CO2, methane, etc). On the right, the amount of energy measured at various depths in one location. Note the log scale on the vertical axis for the right hand graph. (Note as well that the irradiance in these graphs is in W/m².nm, whereas the calculated graphs earlier are in W/m².μm).

From "Light Absorption in Sea Water", Wozniak (2007)

From "Light Absorption in Sea Water", Wozniak (2007)

Figure 5

And two more locations measured. Note that the Black Sea is much more absorbing – solar absorption varies with sediment as well as other ocean properties.

DLR or “Back radiation”

The radiation from the atmosphere doesn’t look too much like a “Planck curve”. Different heights in the atmosphere are responsible for radiating at different wavelengths – dependent on the concentration of water vapor, CO2, methane, and other trace gases.

Here is a typical DLR spectrum (note that the horizontal axis needs to be mentally reversed to match other graphs):

Pacific, Lubin (1995)

Pacific, Lubin (1995)

Figure 6

You can see more of these in The Amazing Case of Back Radiation – Part Two.

But for interest I took the case of an ideal blackbody at 0°C radiating to the surface and used the absorption coefficients from figure 1 to see how much radiation was transmitted through to different depths:

Figure 7

As you can see, most of the “back radiation” is absorbed in the first 10μm, and 20% is absorbed even in the first 1μm.

I could produce a more accurate calculation by using a spectrum like the Pacific spectrum in fig 6 and running that through the same calculations, but it wouldn’t change the results in any significant way.

So we can see that while around half the solar radiation is absorbed in the first meter and 80% in the first 10 meters, 90% of the DLR is absorbed in the first 10μm.

So now we need to ask what kind of result this implies.

Heating Surfaces and Conduction

When you heat the surface of a body that has a colder bulk temperature (or a colder temperature on the “other side” of the body) then heat flows through the body.

Conduction is driven by temperature differences. Once you establish a temperature difference you inevitably get heat transfer by conduction – for example, see Heat Transfer Basics – Part Zero.

The equation for heat transfer by conduction:

q = kA . ΔT/Δx

where k is the material property called conductivity, ΔT is the temperature difference, Δx is the distance between the two temperatures, and q is the heat transferred.

However, conduction is a very inefficient heat transfer mechanism through still water.

For still water, k ≈ 0.6 W/m.K (the ≈ symbol means “is approximately equal to”).

So, as a rough guide, if you had a temperature difference of 20°C across 50m, you would get heat conduction of 0.24 W/m². And with 20°C across 10m of water, you would only get heat conduction of 1.2 W/m².

However, the ocean surface is also turbulent for a variety of reasons, and in Part Two we will look at how that affects heat transfer via some simulations and a few papers. We will also look at the very important first law of thermodynamics and see what that implies for absorption of back radiation.

Update - Does Back-Radiation “Heat” the Ocean? – Part Two


Light Absorption in Sea Water, Wozniak & Dera, Atmospheric and Oceanographic Sciences Library (2007)


Note 1 - To avoid upsetting the purists, when we say “does back-radiation heat the ocean?” what we mean is, “does back-radiation affect the temperature of the ocean?”

Some people get upset if we use the term heat, and object that heat is the net of the two way process of energy exchange. It’s not too important for most of us. I only mention it to make it clear that if the colder atmosphere transfers energy to the ocean then more energy goes in the reverse direction.

It is a dull point.

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Following discussions about absorption of radiation I thought some examples might help illustrate one simple, but often misunderstood, aspect of the subject.

Many people believe that radiation from a colder atmosphere cannot be absorbed by a warmer surface. Usually they are at a loss to explain exactly why – for good reason.

However, some have the vague idea that radiation from a colder atmosphere has different wavelengths compared with radiation from a warmer atmosphere. And, therefore, that’s probably it. End of story. Unfortunately for people with this idea, it’s not actually solved the problem at all..

The specific question I posed to one commenter some time ago was very specific:

If 10μm photons from a 10°C atmosphere are 80% absorbed by a 0°C surface, what is the ratio of 10μm photons from a -10°C atmosphere absorbed by that same surface?

It was eventually conceded that there would be no difference – 10μm photons from a -10°C will also be 80% absorbed. This material property of a surface is called absorptivity and is the proportion of radiation absorbed vs reflected at each wavelength.

Basic physics tells us that the energy of a 10μm photon is always that same, no matter what temperature source it has come from – see note 1.

Here’s an example of the reflectivity/absorptivity of many different materials just for interest:


Reflectivity vs wavelength for various surfaces, Incropera (2007)

Reflectivity vs wavelength for various surfaces, Incropera & DeWitt (2007)


Clearly materials have very different abilities to absorb /reflect different wavelength photons. Is this the explanation?


The important point to understand is that even though radiation emitted from different temperature sources have different peak wavelengths, there is a large spread of wavelengths:

The peak wavelength of +10°C radiation is 10.2μm, while that of the -10°C radiation is 11.0μm – but, as you can see, both sources emit photons over a very similar range of wavelengths.


Let’s now take a look at the proportion of radiation absorbed from both of these sources.

First, with the case where the surface absorptivity is higher at shorter wavelengths – this should favor absorbing more energy from a hotter source and less from a colder source:

The top graph shows the absorptivity as a function of wavelength, and the bottom graph shows the consequent absorption of energy for the two cases.

Because absorptivity is higher at shorter wavelengths, there is a slight bias towards absorbing energy from the hotter +10°C source – but the effect is almost unnoticeable.

The actual numbers:

  • 43% of the -10°C radiation is absorbed
  • 46% of the +10°C radiation is absorbed

So let’s try something more ‘brutal’, with all of the energy from wavelengths shorter than 10.5μm absorbed and none from wavelengths longer than 10.5um absorbed (all reflected).

As you can see, the proportion absorbed of the energy from the hotter source vs colder source appears very similar. It is simply a result of the fact that +10°C and -10°C radiation have almost identical proportions of energy between any given wavelengths – the main difference is that radiation from +10°C has a higher total energy.

The actual numbers:

  • 22% of the -10°C is absorbed
  • 27% of the +10°C is absorbed

So – as is very obvious to most people already – there is no possible surface which can absorb a significant proportion of 10°C radiation and yet reflect all of the -10°C radiation.

And If There Was Such a Surface

Suppose that we could somehow construct a surface which absorbed a significant proportion of radiation from a +10°C source, and yet reflect almost all radiation from a -10°C source.

Well, that would just create a new problem. Because now, when our surface heats up to 11°C the radiation from the 10°C source would still be absorbed. And yet, the radiation is now from a colder source than the surface. Red alert for all the people who say this can’t happen.


The claim that radiation from a colder source is not absorbed by a warmer surface has no physical basis. People who claim it don’t understand one or all of these facts of basic physics:

a) Radiation incident on a surface has to be absorbed, reflected or transmitted through the surface. This last (transmitted) is not possible with a surface like the earth (it is relevant for something like a thin piece of glass or a body of gas), therefore radiation is either absorbed or reflected.

b) The material property of a surface which determines the proportion of radiation absorbed or reflected is called the absorptivity, and it is a function of wavelength of the incident photons. (See note 2)

c) The energy of any given photon is only dependent on its wavelength, not on the temperature of the source that emitted it.

d) Radiation emitted by the atmosphere has a spectrum of wavelengths and the difference between a -10°C emitter and a +10°C emitter (for example) is not very significant (total energy varies significantly, but not the proportion of energy between any two wavelengths). See note 3.

The only way that radiation from a colder source could not be absorbed by a warmer surface is for one of these basic principles to be wrong.

These have all been established for at least 100 years. But no one has really checked them out that thoroughly. Remember, it’s highly unlikely that you have just misunderstood the Second Law of Thermodynamics.

See also: The Real Second Law of Thermodynamics

Intelligent Materials and the Imaginary Second Law of Thermodynamics

The First Law of Thermodynamics Meets the Imaginary Second Law

The Amazing Case of “Back Radiation” – Part Three

and Amazing Things we Find in Textbooks – The Real Second Law of Thermodynamics

Note 1 – Already explained in a little more detail in The Amazing Case of “Back Radiation” – Part Three – the energy of a photon is only dependent on the wavelength of that photon:

Energy = hc/λ

where h = Planck’s constant = 6.6×10-34 J.s, c = the speed of light = 3×108 m/s and λ = wavelength.

Note 2 – Absorptivity/reflectivity is also a function of the direction of the incident radiation with some surfaces.

Note 3 – For those fascinated by actual numbers – the energy from a blackbody source at -10°C = 272 W/m² compared with that from a +10°C source = 364 W/m² – the colder source providing only 75% of the total energy of the warmer source. But take a look at the proportion of total energy in various wavelength ranges:

  • Between 8-10 μm  10.7% (-10°C)   12.2% (10°C)
  • Between 10-12 μm  11.9% (-10°C)   12.7% (10°C)
  • Between 12-14 μm  11.2% (-10°C)   12.4% (10°C)
  • Between 14-16 μm   9.8% (-10°C)     9.5% (10°C)

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