The Science of Doom

The surface of Venus is around 730K (457°C) – why is it so hot? Although there is still much to learn about Venus the basics have some simple explanations.

Energy Absorbed from the Sun

While earth is 150M km from the sun, Venus is only 108M km away, a ratio of 0.72.

The “solar constant” (as it has been historically called) at the earth is 1367 W/m², and, as energy intensity is proportional to the square of the distance away, or “r²”, the solar constant at Venus is 1367/(0.72²) = 2,636 W/m² – Venus is closer, so it receives more solar energy per m².

The earth has an average albedo of about 0.3, meaning that 30% of the solar radiation is reflected. This is by clouds, by aerosols in the atmosphere, and by the surface of the earth. (Watch out for “The Earth’s Energy Budget – Part Four – Albedo”). If Venus had the same albedo as the earth, the energy absorbed per m² of surface area would be, E = 2,636 * (1-0.3) / 4 = 461 W/m². [corrected – thanks to Bill Stoltzfus for pointing out my mistake]

For an explanation of why the value is divided by 4, see The Earth’s Energy Budget – Part One

This value equates to an “effective radiating temperature” of 300K (27°C). This is nothing like the surface of Venus. [corrected as well – thanks to Bill Stoltzfus for pointing out my mistake]

In any case, it turns out that Venus has a much higher albedo than the earth, with an albedo of 0.76 – meaning that 76% of the solar energy is reflected.

Redoing the calculation, E = 2,636 * (1-0.76) / 4 = 158 W/m² – which equates to an “effective radiating temperature” of 230K (-43°C). The same calculation for the earth gives 255K (-18°C) – see CO2 – An Insignificant Trace Gas? – Part One.

So in terms of a simple energy balance with the sun, Venus should be colder than the earth.

In the case of the earth, as laid out in the CO2 series, the reason the surface of the earth is so much warmer than predicted from simple energy balance is because various trace gases, including water vapor and CO2, absorb the upward radiation from the earth’s surface and reradiate it in all directions. As some of this is downward, the surface of the earth receives more energy than it would without these gases and so it is hotter. See CO2 – An Insignificant Trace Gas? Part Six – Visualization for some more insight into this.

So is Venus so much warmer at its surface because of the inappropriately-named “greenhouse” effect? Or is it for other reasons?

The Venusian Atmosphere

The atmosphere of Venus is quite unwelcoming for us earth-dwellers. The atmosphere is mostly CO2 (97%), with the balance made up mostly of nitrogen (N2), and trace amounts of water vapor and many other gases in minute quantities.

The mass of the Venusian atmosphere is around 100 times that of the earth, and consequently the pressure at the surface of Venus is much higher – at 92bar compared with 1bar for the earth.

Now some people say that the reason for the high temperature at the surface of Venus is because of the high atmospheric pressure and the depth of the atmosphere. For example, Steve Goddard on Wattsupwiththat and echoed by Lubos Motl. This explanation isn’t one that you can find in atmospheric physics text books.

A Quick Review of the Earth’s Surface

This is just to explain a few basics for some perspective.

From Trenberth and Kiehl (1997)

There’s much of interest in this diagram from Earth’s Annual Global Mean Energy Budget by Trenberth and Kiehl (1997) but we’ll focus on a few key elements for the purposes of this article.

The surface of the earth receives an average of about 170 W/m² from solar energy (with an additional 70W/m² of solar radiation absorbed by the atmosphere). The earth’s surface also receives an average of 324 W/m² of radiation from the atmosphere. So in total the earth’s surface receives about 490 W/m² (annual global average).

Now the average radiation from the surface of the earth is 396 W/m² (or 390 W/m² in the diagram above, which is close enough for our purposes). Convection and conduction remove the balance of around 100W/m². If you take a look at Tropospheric Basics you can see more about the temperature profile in the troposphere (lower atmosphere) and why convection is a more effective re-distributor of heat within the troposphere.

The principal point is that the warming of the air from the surface radiation, conduction and convection causes the air to expand. Air that expands is less dense, and so this air rises, moving heat by convection.

The temperature profile, or lapse rate, from convection can be easily calculated, both for dry air and moist air. Dry air is just under 10°C/km, while moist air depends on the amount of water vapor, but can be as low as 4°C/km. (And the environmental lapse rate, or what we find in practice, is around 6.5°C/km).

So in the case of the earth’s surface, it would be radiating out 490W/m², but for the fact that conduction and convection remove some of this energy from the surface, and then convection redistributes this energy up into the atmosphere.

The Surface of Venus

Energy radiated from a surface is proportional to the 4th power of absolute temperature. This is known as the Stefan-Boltzmann law but visualizing the 4th power of something isn’t that easy. However, calculators are readily available and so if you punch the numbers in you will see that for a surface of T=730K with an emissivity close to a blackbody:

E = 16,100 W/m²   – compare this with the surface of earth (288K, 15°C) of around 390 W/m²

This is over 40x the energy radiated from the surface of the earth – for a temperature only 2.5x greater. That’s the real world, very non-linear.. (And note that if the emissivity is not equal to 1, the energy radiated is simply the value above multiplied by the emissivity).

So if we think about the top of atmosphere of Venus, it is radiating round about 158 W/m². This balances the absorbed solar radiation. And yet the surface is radiating 16,100 W/m² – does the high pressure of the Venusian atmosphere explain it?

No

Think about it like this. For the surface of Venus to be radiating at 16,100 W/m² it has to be receiving this energy from somewhere. It receives a maximum of 158 W/m² from the sun (if all of the solar energy absorbed is absorbed in the surface and nothing in the atmosphere).

The explanation from others about a temperature gradient between the surface and the tropopause (top of the tropopause or lower atmosphere) only explains anything when the surface heats the atmosphere from below. In that case the atmosphere heats up, expands and rises – moving energy via bulk movements of air.

Can the atmosphere create heat from pressure and transmit this heat to the surface?

In the case of the earth’s surface, the extra radiation to the earth’s surface (caused by the “greenhouse” effect) heats the atmosphere from beneath and causes convection – with a lapse rate (or temperature profile) of between 4 – 10 °C/km. And convection moves some of this heat from the surface up into the atmosphere. In the case of Venus the argument that relegates the role of the “greenhouse” effect and promotes the role of atmospheric pressure doesn’t have a heat transfer mechanism.

Picture the starting condition where the surface is very cold. What heats it up?

There are three ways of moving heat – radiation, convection and conduction. Conduction in gases is extremely low and anyway the top of the atmosphere is around 230K – if the surface starts off colder what causes heat to flow to the surface to create such a huge emission of radiation?

Convection needs to work by warming a gas from below. Where is this mechanism if the surface is not already heated by the “greenhouse” effect?

And radiation has been ruled out (as the main mechanism) in these arguments from Steve Goddard and others.

How Can the Surface Get so Hot? An Over-Simplified Climate Model

Let’s take a look at the ignored radiation and the super “greenhouse” effect.

How can a surface get so hot from “back-radiation”? Isn’t that just as crazy an idea?

We will take a simple idea – as all models are at their start, just to demonstrate a point. There’s a little bit of maths, unfortunately, but possibly (if you haven’t seen this concept before), the concept might actually seem harder to grasp. Here’s a very simple (and not very realistic) model of a planetary surface and atmosphere (idea from Radiation and Climate from Vardavas and Taylor (2007)):

Simple climate model - atmosphere perfectly transparent to solar radiation, and totally opaque in the infra-red

The surface receives radiation from the sun, S. In the case of Venus this value would be (averaged across the surface), S = 158 W/m².

Now the surface is at T_s and radiates to the atmosphere, which heats it up. The atmosphere is perfectly transparent to solar radiation, but totally opaque in the infra-red and all at one temperature, T_a. Therefore, the atmosphere radiates σT_a⁴ upwards and σT_a⁴ downwards.

Note that “totally opaque” means that no surface radiation makes it through this layer of the atmosphere. In this scenario we can calculate the surface radiation. If you are new to this kind of model, it is easiest to follow the small amount of maths against the graphic. First, using the simple energy balance at top of atmosphere, the outgoing radiation at the top of atmosphere equals the absorbed solar radiation averaged over the surface of the earth:

σT_a⁴ = S         [equation 1] – this is the Stefan-Boltzmann law, where σ = 5.67 x 10^-8

Second, the surface radiation balances the energy received at the surface – which is from the sun and the atmosphere:

S + σT_a⁴ = σT_s⁴ [equation 2]

Therefore, substituting [1] into [2], we get:

σT_a⁴ + σT_a⁴ = σT_s⁴

2σT_a⁴ = σT_s⁴, or 2S = σT_s⁴

and solving we find, T_s = (2S/σ)^1/4

In the case of S = 158 W/m², T_s = 273K

Now effective temperature at the top of atmosphere is 230K, so an opaque atmosphere has increased the surface temperature significantly – but not to 730K. (Barton Paul Levenson has a model like this, commented on in CO2 – An Insignificant Trace Gas? Part Eight – Saturation)

Now with a very optically thick atmosphere, we simply add more and more layers to our model. The equations get slightly harder to solve, but each time we add a new totally opaque layer the temperature rises yet more.

For example, with 3 totally opaque layers the solution to a similar set of equations (with 4 equations and 4 unknowns) is:

T_s = (4S/σ)^1/4, or T_s = 328K

It should be easy to see how the surface temperature gets extremely hot from radiation with many layers of opaque atmosphere (yet transparent to solar radiation).

So the Surface Temperature is Infinite, you Dummy!!

Well, if we can keep adding layers, and each one just increases the “back radiation” anyone can see that this can go on forever and the temperature will be infinite!

Obviously the model is wrong..

Not quite (well, if we could keep doing this, the model would be wrong). In the model above we have one totally opaque layer of atmosphere. But once we add multiple layers we are effectively dividing up the real atmosphere and saying that each layer is totally opaque. As we keep sub-dividing the atmosphere into more and more layers eventually they start to get optically thin and the radiation from the layer below will not be completely absorbed.

Radiation from CO2

All the above model does is demonstrate how the presence of significant radiatively absorbing gases can significantly increase the surface temperature. A 97% CO2 atmosphere is different from the model above for two reasons:

• CO2 doesn’t absorb at all terrestrial wavelengths so it isn’t a perfect absorber
• convection will moderate the surface temperature increase – once induced by radiation – as with the earth’s surface

So to calculate the effect of the CO2 atmosphere we have to solve the radiative transfer equations, which you can see in CO2 – An Insignificant Trace Gas? Part Three and Part Five (and the whole series).

These are fundamental equations of absorption and emission, but aren’t really solvable on a pocket calculator – despite so many people appearing to do just that in so many blogs.  Note as well that CO2 spectral lines broaden with pressure so that CO2 (and water vapor) become a much more effective absorber in the lower Venusian atmosphere than the earth’s.

And we have to consider that once we have very high temperatures at the surface, convection will begin to move heat more effectively. This essentially moderates the effect of radiation.

But for those who believe that high Venusian atmospheric pressure and the ideal gas laws cause the high 730K surface temperature – they have to explain how the heat is transferred to the surface so that it can radiate at 16,100 W/m².

Real Solutions

One early approach to using real atmospheric physics on this problem was by James Pollack (reference below) in 1969 who showed that that plausible amounts of water vapor and the very high levels of CO2 could explain the high temperatures – using the radiative transfer equations and a convective model.

Bullock and Grinspoon (reference below) did this more recent calculation of the temperature profile in the Venusian atmosphere:

And they note a few possible reasons for the divergence above 70km. The model produces this spectrum of outgoing radiation:

The Planck function for an “effective radiating temperature” of 232K is shown. Note that the much higher levels of flux (in comparison to the 232K curve) demonstrate that at lower wavelengths (higher wavenumbers) the atmosphere is less opaque. This tends to limit further temperature rises, as the presence of any “window” regions allows a higher surface temperature to radiate out efficiently to the atmosphere.

A Mental Model

One mental model for people new to the inappropriately-named “greenhouse” effect is to think about the sun as an internal heat source, and CO2 as some kind of insulator.

Picture an ambient temperature of 20°C and a surface which has a constant internal heat source. As you add more and more insulation around this surface the temperature will keep rising – as heat is less able to flow away from the surface. For a given insulation there will be an equilibrium temperature reached that we can calculate, and it will be a function of the properties of the insulation.

Even though the temperature might reach 100°C or 200°C doesn’t mean that energy is “created” in this model – and this is probably clear to everyone.

Whether or not mental models “work” doesn’t change the realities of physics, but of course everyone wants to understand a subject conceptually.

Conclusion

Venus follows the same physical laws as the earth, so explaining the high surface temperature should be possible, even though many details of the atmosphere of Venus are hidden from us.

Some people who have attempted to explain the high Venusian surface temperature have used the ideas about the relationships between pressure and temperature in ideal gases without the strong “greenhouse” effect of a 97% CO2 atmosphere.

However, these ideas seem to lack a heat transfer mechanism whereby the surface of Venus can radiate at 16,100 W/m². This is the missing element in ideas which eliminate or relegate the role of CO2. In contrast, high surface temperatures in very strongly absorbing atmospheres can be explained using the radiative effects. A simple model can demonstrate very high temperatures, but a thorough calculation does require solution of the radiative transfer equations.

Update – New articles – Convection, Venus, Thought Experiments and Tall Rooms Full of Gas – A Discussion

Venusian Mysteries – Part Two

Reference

A Nongray CO2-H2O Greenhouse Model of Venus, James Pollack, Icarus (1969)

The Recent Evolution of Climate on Venus, Mark Bullock and David Grinspoon, Icarus (2001)

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On Lunar Madness and Physics Basics, one commenter asked a very good question in response to a badly phrased answer from me.

He originally asked:

You agree that if earth had 100% nitrogen atmosphere (a non greenhouse gas), the “average” temperature of earth would be different (I’m not entirely sure it wouldn’t be lower) from the 255 K blackbody radiation would suggest.

I got the meaning wrong and said “right, it would be 255K” and wasn’t very specific with what I meant, which was a mistake as the article in question had just explained everything wrong with averages..

He responded with an interesting example of a fictional Latvia, where Latvia got all the solar energy (somehow) and the rest of the world none, and showed that the average temperature of the earth was 0.5K or almost -273°C (and Latvia was quite steamy).

And after I had written a comment as long as a post I thought it was probably a subject that would be worth making a post out of..

“Nice example, and I guess as I’ve been explaining everything that’s wrong with averages I should have been more careful in my explanation.

We will consider a real earth with an albedo..”

Absorbed Solar Radiation

The absorbed solar radiation has an average of 239 W/m². This is averaged over the surface of the earth, which is 5.1 x 10⁸ km² = 5.10 x 10¹⁴ m².

The incoming absorbed energy is therefore 1.2 x 10¹⁷ W.

Note: this average of 239 is a measured value of incoming – reflected solar radiation, divided by 4 for spatial geometry reasons, see The Earth’s Energy Budget – Part One. The total of 1.2 x 10¹⁷ W also equals the TSI of 1367 W/m² x “disc area” of the earth x (1-albedo).

This is also currently the measured average outgoing longwave radiation (OLR) at the top of atmosphere, within instrument error.

The Hoovering

We’ll come to the energy-grabbing Latvia later (sorry Latvian’s wasn’t my idea).

It’s a typical earth in all other respects but a cosmic being has just “hoovered up” the trace gases like CO2, CH4, NO2 and turned water vapor into a non-absorbing gas. (If water vapor was also hoovered up the oceans and lakes would be a ready source of water vapor and within a month or two the atmosphere would have the same water vapor as before).

As a result radiation >4μm just goes right through it. That is the whole point about radiation – if nothing absorbs it, it keeps on going.

The radiation emitted from the earth just after the Hoover incident is 396 W/m². Or, putting it another way, the total radiation emitted from the earth’s surface is 2.0 x 10¹⁷ W.

This total radiation is calculated by adding up the radiation from every square meter on the earth. And the average is simply the total divided by the area.

Note. In fact across a day and a year this value will change. Even year to year. So 2.0 x 10¹⁷ W is just the annual average across an appropriate time period. But it is appreciably higher than the absorbed solar 1.2 x 10¹⁷ W.

Energy Loss through Radiation and Climate Response

The net radiation loss from the planet starts at 0.8 x 10¹⁷ W = 2.0 x 10¹⁷ – 1.2 x 10¹⁷

Who knows what kind of climate response that would generate? And the time for the total response would also depend on how well-mixed the oceans were (because of their large heat capacity), but rather than trying to work out how long it will take, we can say that the earth will cool down over a period of time.

And no matter what happens to convection, lapse rates, and rainfall this cooling will continue. That’s because these aspects of the climate only distribute the heat. Nothing can stop the radiation loss from the surface because the atmosphere is no longer absorbing radiation. They might enhance or reduce the cooling by changing the surface temperature in some way – because radiation emitted by the surface is a function of temperature (proportional to T⁴). But while energy out > energy in, the climate system would be cooling.

Clouds and Ice Sheets

It’s possible (although unlikely) that all the clouds would disappear and in which case the net incoming – reflected radiation might increase, perhaps to 287 W/m². (This value is chosen by measuring the current climate’s solar reflection of clouds, see Clouds and Water Vapor).

If that happened, total absorbed energy = 1.5 x 10¹⁷ W.

However, as the earth cools the ice sheets will increase and there’s no doubt that the albedo of the earth’s surface will increase so who knows what exactly would happen.

But it seems like the maximum absorbed solar radiation would at most go from 1.2 x 10¹⁷ W to 1.5 x 10¹⁷ W and more likely it would reduce below 1.2 x 10¹⁷ W.

A New Equilibrium

Eventually the outgoing radiation would approximately match the incoming radiation and temperature would settle around a value – this would take centuries of course, maybe 1000s of years..

And depending on the ice sheet extent and whether any clouds still existed the value of outgoing radiation might be around 1.0 – 1.5 x 10¹⁷ W. This upper value would depend on the ice sheets not growing and all the clouds disappearing which seems impossible, but it’s just for illustration.

Remember that nothing in all this time can stop the emitted radiation from the surface making it to space. So the only changes in the energy balance can come from changes to the earth’s albedo (affecting absorbed solar radiation).

And given that when objects emit more energy than they absorb they cool down, the earth will certainly cool. The atmosphere cannot emit any radiation so any atmospheric changes will only change the distribution of energy around the climate system.

What would the temperature of the earth be?

I have no idea.

But let’s pick the 1.2 x 10¹⁷ W as our average incoming less reflected solar energy, and so when “equilibrium” is reached the earth’s total surface radiation will be at this value. (Strictly speaking equilibrium is never truly the case, but we are considering the case where measured over a few decades the average outgoing radiation is 1.2×10¹⁷ W).

Uniform Temperature

Suppose that all around the world the temperature of the surface was identical. It can’t happen, but just to put a stake in the ground, so to speak.

Average outgoing radiation / surface area = 235 W/m² – (that’s because I originally wrote down 1.2 x 10¹⁷ instead of 1.22 x 10¹⁷ and so the rounding error has caused a change from 239, but it doesn’t particularly matter, more a note for those in the “rounding police”).

And with a longwave emissivity close to 1 for most of the earth’s surface, this would be a temperature of 254K.

The Energy-Grabbing Latvians

As I mentioned before Latvians and Latvophiles, this wasn’t my idea. And anyway, it probably wasn’t your fault. I’ll adjust the commenter’s numbers slightly to account for the earth’s albedo.

In his example, mythical-Latvia has a surface area of 10,000 km² = 10¹⁰ m². And mythical-Latvia absorbs all of the energy with none left for the rest of the earth. The rest of the earth is at a temperature of 0K and Latvia is at a temperature of 5080K which means it radiates at 1.2 x 10⁷ W/m². Therefore the radiation from the whole earth = 1.2×10¹⁷ W and so the earth is in equilibrium (with the solar radiation absorbed), but the average temperature of the whole earth = 5080 * 1010 / 5.1×10¹⁴ = 0.1K.

Frosty.

And our commenter has nicely demonstrated the same point as in Lunar Madness and Physics Basics – you can have the same radiation from a surface with totally different average temperatures.

The Maximum “Average” Temperature

So we have had two approaches to calculating our equilibrium hoovered atmosphere, one with a temperature of 254K (-19°) and one with a temperature of 0.1K (-273°C) – both quite unrealistic situations it should be noted.

What’s the maximum “average” temperature? As already noted, at the wavelengths the earth radiates at (>4μ) it is quite close to a blackbody.

In this case, the maximum “average” temperature can’t be higher than what is known as the “effective blackbody temperature”, T_eff.

This value, T_eff, is a common convention to denote the effective radiating temperature of a body. This is just the temperature-radiation conversion from the Stefan-Boltzmann equation, T_eff = (E/σ)^1/4 – the rewritten version of the more familiar, E = σT⁴. It simply converts the energy radiated into a temperature. So 235 W/m² = 254K.

It doesn’t mean, as already explained, that T_eff is the “average” temperature. The “average” temperature (arithmetic mean) can be quite different from T_eff, demonstrating that average temperature is a troublesome value.

I created confusion using the concept of T_eff in my comment in the earlier article. By saying the earth would be at 255K without a radiatively absorbing atmosphere I really meant that in that situation the earth’s surface would be radiating (averaged globally annually) 239 W/m².

The Earth Without an Absorbing Atmosphere

With conventions out of the way – if the atmosphere didn’t absorb any terrestrial radiation the radiation from the surface would slowly fall from its current annual global average of 396 W/m² to around 240 W/m².

The climate would undergo dramatic changes of course and no one can say exactly what the equilibrium “effective blackbody radiating temperature” would be as we don’t know how much solar radiation would be reflected in this new climate. Clouds, ice – and aerosols – all play a part in reflecting the solar radiation from the atmosphere and the surface, and if these change the amount of energy absorbed changes.

But without an atmosphere that absorbs longwave radiation there is no way that the radiation from the surface can be greater than the radiation from the top of the atmosphere. And that means that eventually the emission of radiation from the surface would be approximately equal to the absorbed solar radiation.

Therefore, the value of global annual average radiation might be 290W/m² (unlikely), or it might be less than 239W/m² (more likely).

The world would be much colder.

With an annual surface radiation of 239W/m², the “average temperature” would be  -18°C or colder.

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On a couple of unrelated articles on this blog, people have been asking questions about the moon. This is because a lot of people have read an article called A Greenhouse Effect on the Moon from www.ilovemycarbondioxide.com that makes some confused claims.

The article starts:

We’ve been told that the earth’s surface is quite a bit warmer than calculations predict. Theory has it that heat-trapping “greenhouse gases” account for a 33° Celsius disparity. But it turns out that our airless moon is also quite a bit warmer than predicted.

And finishes with:

The Earth is not “unusually” warm. It is the application of the predictive equation that is faulty. The ability of common substances to store heat makes a mockery of blackbody estimates. The belief that radiating trace gases explain why earth’s surface temperature deviates from a simple mathematical formula is based on deeply erroneous assumptions about theoretical vs. real bodies.

A long time ago a friend told me that the way the Bank of England trains people to spot counterfeit notes is to give them real notes to spend time getting used to the feel, texture, weight and so on. They don’t give them lots of counterfeits because it’s not as effective.

I have no idea if the story is true but I always thought that it was a useful concept for approaching any subject. Best to spend the time helping people understand the real theory – as all scientific “facts” are called – rather than spend 5% of the time on the real theory and introduce them to 19 flawed theories.

Therefore, most of this article will focus on building understanding of the basics rather than pointing out the many flaws in the article. We will look at the temperature of a moon-like body by way of very simple models.

These models are in Excel because it’s quick and easy.

The Model

The concept is very simple. This is an idealized moon-like surface for illustration.

For my moon-like body, we will consider one square meter of surface. This is because lateral heat flow within the surface will be extremely low and so we don’t want or need to build a GCM to solve this problem.

Solar radiation is absorbed by this surface and heats up. The surface has a definite heat capacity which we vary in the model to see how the results change.

The sun moves slowly through the sky so the amount of solar radiation incident on the surface varies over the course of the lunar “day”. The surface has an “absorptivity” for solar radiation – the proportion of solar radiation absorbed vs the proportion reflected.

When the sun is directly overhead the solar radiation incident is 1367 W/m² and when the sun is on the horizon the solar radiation is zero – then for the whole “night” the radiation stays zero. Therefore, I’m considering the “equator”.

For reasons of laziness I set the lunar day to be 28 days, but the exact value doesn’t matter.

And the absorptivity was set to 0.9 (which means 90% of incident solar radiation is absorbed and 10% is reflected). Also the emissivity was set to the same value, but in this example it could be different. With different values similar results would occur but with different equilibrium temperatures. See Note 1.

The simple maths for the model is at the end of the post as many people don’t like seeing equations.

The Results

Now, if the surface had no heat capacity (or as mathematicians might say, “as the heat capacity tends to zero”) then the surface would instantaneously heat up until the radiation emitted matched the absorbed radiation.

So in that unrealistic case, the temperature would follow this curve:

Moon-like surface, zero heat capacity

So during the moon-like night, the surface drops immediately to absolute zero, and during the “day” the emission of radiation exactly matches the absorption. (For mathematically inclined readers this follows a cos θ relationship – see maths section at end).

Note that this isn’t like the earth or any real body. It’s just a useful thought experiment to show what would happen if the surface had no heat capacity.

Under this condition:

• absorption of solar radiation = 391.7 W/m² (averaged over many cycles)
• emission of lunar radiation = 391.7 W/m² (averaged over many cycles)
• mean temperature = 169.3K
• min temperature = 0K
• max temperature = 394K

Energy in = energy out – so no surprises there.

Let’s start increasing the heat capacity and see what happens – per m², 10,000J/K heat capacity:

Moon-like surface with 10,000 J/K heat capacity per m^2

• absorption of solar radiation = 391.7 W/m² (averaged over many cycles)
• emission of lunar radiation = 391.7 W/m² (averaged over many cycles)
• mean temperature = 195.3K
• min temperature = 38K
• max temperature = 397K

Per m², 50,000J/K heat capacity:

Moon-like surface with 50,000 J/K heat capacity per m^2

• absorption of solar radiation = 391.7 W/m² (averaged over many cycles)
• emission of lunar radiation = 391.5 W/m² (averaged over many cycles)
• mean temperature = 211.3K
• min temperature = 64K
• max temperature = 394K

Per m²,  500,000J/K heat capacity:

Moon-like surface with 500,000 J/K heat capacity per m^2

• absorption of solar radiation = 391.7 W/m² (averaged over many cycles)
• emission of lunar radiation = 390.0 W/m² (averaged over many cycles)
• mean temperature = 247.7K
• min temperature = 133K
• max temperature = 393K

Per m²,  5,000,000J/K heat capacity:

Moon-like surface, 5,000,000 J/K per m^2

• absorption of solar radiation = 391.7 W/m² (averaged over many cycles)
• emission of lunar radiation = 391.7 W/m² (averaged over many cycles)
• mean temperature = 290.9K
• min temperature = 247K
• max temperature = 342K

Hopefully, for most people, the fact that the temperature range is reducing as heat capacity increases is reasonably intuitive. If you want to heat up a cupful of water it takes less time than heating a swimming pool. If you want to cool down both through the same surface area it will take longer for the swimming pool to cool down.

Summary of Results

Notice that in each case the average value of absorption = emission – to within 1%.

The 1% is just a result of imperfect starting conditions. If the chosen simulation starting temperature was exactly right, or there were enough “spin up” cycles to get into the steady state before the averaging was done then the absorption = emission exactly.

It’s probably not surprising to anyone that absorption = emission over a set number of cycles because otherwise the overall trend in temperature would be increasing or decreasing.

Next a plot of mean, min and max temperature as the heat capacity increases, note the log axis for heat capacity:

The reason for plotting the heat capacity on a “log” or logarithmic axis was because the heat capacity is increased by a an order of magnitude each time. Linear plots make the results of this kind of simulation less clear.

The mean temperature is simply the arithmetic average of temperature over every single time step. (All the numbers added up and divided by the number of results).

So the mean temperature does increase when the surface has an increased heat capacity!

It looks like the ilovemyco2 writers were correct and the whole greenhouse effect was just a result of heat capacity of the oceans and land.

Time for me to pack my bags and head off into the sunset..

But wait, hold on a minute..

There’s something very strange going on. The temperature is increasing, but the average emission of radiation has stayed exactly the same:

How can temperature increase without the radiation increasing? Radiation is emitted in proportion to the 4th power of temperature – for a blackbody (ε=1), E = σ . T⁴, where σ = 5.67×10^-8

If the temperature goes up, radiation must go up as well. Is there something wrong with the model?

No. And for those who’ve read Why Global Mean Surface Temperature Should be Relegated, Or Mostly Ignored this example won’t be surprising.

Take 3 “temperatures”:  1, 10, 100.

Now we average them -> average = 111/3 = 37K

And calculate the energy radiated, E = 37 ⁴ x 5.67×10^-8 = 1,874,161 x 5.67×10^-8 = 0.11 W/m²

Alright, let’s do it the other way. Let’s calculate the energy radiated for each temperature:

• 1⁴ x 5.67×10^-8 = 1 x 5.67×10^-8 = 5.67×10^-8
• 10⁴ x 5.67×10^-8 = 10,000 x 5.67×10^-8 = 5.67×10^-4
• 100⁴ x 5.67×10^-8 =100,000,000 x 5.67×10^-8 = 5.67

And now average the energy radiated -> average = (5.6705670567/3) = 1.89 W/m²

One method gives 18x the other method – how can this be and which one is right?

Just for the many people would prefer to see the calculation without the Stefan-Boltzmann constant of 5.67×10^8 everywhere – in that case we compare 37⁴ = 1,874,161 with the alternative method of (1⁴ + 10⁴ + 100⁴)/3 = 100,010,001/3= 33,336,667

Also (of course) a factor of 18 between the two methods of calculating the “average”.

There’s nothing surprising about this – average a series of numbers and raising the average to the 4th power will almost always give a different answer to first calculating the 4th power for each of a series of numbers and averaging the results.

Now the moon has some extreme temperature ranges in the examples shown and, therefore, the “mean” temperature changes significantly.

The earth by contrast, with less extreme temperatures has this result –

• the “average” temperature = 15°C, and converting that to the “average” radiation = 390 W/m²
• calculated the correct and painful way, the individually calculated values of radiation from each and every surface temperature around the globe every few hours over a year.. then averaged = 396 W/m²

Conclusion

So the reason that the moon – with a surface with a real heat capacity – appears to have a warmer climate “than predicted” is just a mathematical error. A trap for the unwary.

The right way to calculate a planet’s average radiation is to calculate it for each and every location and average the results. The wrong way is to calculate the average temperature and then convert that to a radiation. In the case of the earth’s surface, it’s not such a noticeable problem.

In the case of the moon, because of the wide variation in temperature, the incorrect method produces a large error.

So there’s no “lunar explanation” for the inappropriately-named “greenhouse” effect.

In the case of the earth there is anyway a huge difference from the moon. The solar radiation absorbed at the top of the earth’s atmosphere – about 240W/m² is approximately balanced by the outgoing longwave radiation of the same amount. But the radiation from the surface of the earth of 396W/m² is much larger than this top of atmosphere value of 240W/m².

That’s the greenhouse effect.

But ilovemyco2 – hats off to you for enthralling and exciting so many people with a simple mathematical puzzle.

Maths in the Model

E_in = S . cosθ . α   – for -90° < θ < 90°

E_in = 0   otherwise

where E_in = energy absorbed by the surface in J/s, S = the solar irradiance in W/m², θ = angle of the sun from the zenith, α = absorptivity of the surface at the solar radiation wavelengths.

E_out = ε . σ . T⁴

where E_out = energy radiated by the surface in J/s, ε = emissivity of the surface at the wavelengths it is radiating at, σ = 5.67 x 10^-8, and T is the temperature in K (absolute temperature). This is the Stefan-Boltzmann equation.

and for each time step, Δt:

ΔT = (E_in – E_out)/C

where C = heat capacity of a 1m² surface in J/K and ΔT is the change in temperature.

For people who like even more detail:

The assumption is that the conductivity of heat into the surface is very high with some kind of insulating layer below the “heat capacity” layer. This makes the calculation slightly easier to understand than using thermal diffusivity.

And the conductivity of heat laterally is very low to avoid considering thermal equalization between adjacent surfaces.

Neither of these assumptions has any significant effect on the “experiment”, or on the principles that it demonstrates.

Note 1

Emissivity and absorptivity are inherent properties of the material in question and are wavelength dependent. In the case of a surface like the earth, the surface receives solar radiation centered around 0.5μm and radiates out with wavelengths centered on 10μm. See, for example, The Sun and Max Planck Agree. So there is no reason to expect that absorptivity = emissivity (because we are considering the properties at different wavelengths).

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In Radiation Basics and the Imaginary Second Law of Thermodynamics I covered a fair amount of ground because it started in answer to another question/point from a commenter.

We all agree that the net effect of radiation between hot and colder bodies is that heat flows from the hotter to the colder body, but many people have become convinced that this means radiation from a colder body has no effect on a hotter body.

After explaining a few basics about emission and absorption, I concluded:

Therefore, there is no room in this theory for the crazy idea that colder bodies have no effect on hotter bodies. To demonstrate the opposite, the interested student would have to find a flaw in one of the two basic elements of thermodynamics described above.

And just a note, there’s no point reciting a mantra (e.g., “The second law says this doesn’t happen”) upon reading this.

Instead, be constructive. Explain what happens to the emitting body and the absorbing body with reference to these elementary thermodynamics theories.

One of our regular commenters has finally explained what happens.

This was a great day of joy because in three other articles on this blog (The Imaginary Second Law of Thermodynamics, How Much Work Can One Molecule Do? and On the Miseducation of the Uninformed by Gerlich and Tscheuschner (2009)) and one on another blog the subject has been much discussed.

I have asked many many times:

What happens to radiation from the colder body when it “reaches” the hotter body?

But I had never been given an answer – until now. (Note: the general consensus from the imaginary second law advocates – as much as I can determine – is that the colder body does emit radiation so at least there is agreement on the first step).

Finally, the answer is revealed:

Back to thermodynamics and electromagnetic radiation.

Scienceofdoom and others think that the hot surface has no option but to absorb a photon from the cold surface.

I think a lot of this radiation is in fact scattered from the hot surface and is not absorbed.

“A Lot” is Not Absorbed?

Before we dive into the fascinating topic of absorptivity and absorption, I hope people don’t think I am being pedantic for drawing attention to the fact that one of our most prominent advocates of the theory (the Imaginary Second Law) has actually failed to support it.

Science is about detail.

If no radiation from the colder body is absorbed by the hotter body then the imaginary second law stands. That is, if any radiation emitted by a colder body “reaches” the hotter body and is absorbed by the hotter body then the colder body has transferred energy to the hotter body.

The colder body has had “an effect”.

It’s hard to be certain about the imaginary second law of thermodynamics because I can’t find it in a text book. In fact, on another note, it has been a day of double joy, because another imaginary second law advocate stepped up to the plate when presented with this from a thermodynamics textbook:

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

And said:

It is absolutely in error.

Which was wonderful to hear because up until now everyone else had simply ignored the question as to whether Incropera and DeWitt didn’t understand the basics of radiation (for reasons that are all too easy to imagine).

But I digress.

I can’t be 100% certain what the imaginary second law of thermodynamics teaches but it has appeared up until now – from the comments by many advocates – that colder bodies have no effect on hotter bodies.

How can it be that the first time someone explains what happens to the incident radiation they agree, at least in principle, with the rest of the world? We (the rest of the world) all think that cold bodies have an effect on hotter bodies.

Perhaps more advocates can comment and vote on this idea..

[Stop press – after writing this a newer entrant to the field has promoted a new idea but this one can wait until another day]

Absorptivity and Absorption

There are a few basic concepts in traditional thermodynamics that are worth explaining, even though they might appear a little tedious.

I was already thinking about writing an article on this after I had earlier explained:

According to Kirchoff’s law, emissivity = absorptivity for a given substance at the same wavelength (and for some surfaces direction needs to be defined also).

And the same commenter responded:

You have a particularly naive view of heat transfer.

When a green plant absorbs some em radiation to make starch does it simultaneously emit the same radiation?

You have not grasped the fact that if the absorption of 4um radiation was exactly balanced by the emission of 4um radiation the net effect would be zero.

Which would shoot a massive hole in the case for AGW.

I prefer to think of my understanding of physics as traditional rather than naive. So, a few basics need explaining so that readers can judge for themselves.

Just a comment, as well, for new people looking into this subject. The basics often seem a little dull, but hopefully I can persuade a few to dig deep and work hard to get the very basics clear. I can promise unrelenting joy down the track as you realize that you understand more than most other people who had jumped ahead and are now writing confused comments.. (assuming this will give you unrelenting joy).

Let’s look at snow in sunlight and use it to “shed light” on a dull subject.

First of all, there are two important factors involved in the absorption of radiation:

• the radiation incident on the snow (i.e., the radiation value as a function of wavelength)
• the absorptivity – the property of the snow (as a function of wavelength) which determines how much radiation is absorbed and how much is reflected

We will focus on the snow being heated by sunlight and ignore terrestrial radiation – even though the material will also be heated by terrestrial radiation. This is because we aren’t trying to work out a complete energy balance for this particular material, just illustrating the important points.

Solar radiation has a spectrum which looks something like this:

Solar radiation at top of atmosphere and the earth's surface, Taylor (2005)

This shows how the radiation incident on the snow varies with wavelength. The actual amplitude is dependent on the time of day, the latitude, the amount of clouds, and so on.

The important point is that radiation from the sun peaks at a wavelength of around 0.5μm and above 2μm is much reduced (91% of solar radiation is below 2μm). Therefore, to find out how much radiation the material actually absorbs we need to know the absorptivity at these wavelengths.

Here are some examples of reflectivity and absorptivity of various materials across quite a wide spectral range:

Reflectivity vs wavelength for various surfaces, Incropera (2007)

The scale on the left is reflectivity and on the right (harder to see) is absorptivity from 1.0 at the bottom up to 0.0 at the top.  Absorptivity = 1-reflectivity.

Absorptivity is a function of wavelength and is the proportion of incident radiation at that wavelength which is absorbed.

Absorptivity is an inherent property of that material

Let’s take snow as an example. Sunlight on snow will be mostly reflected and not absorbed. That’s because the incident sunlight is mostly between 0.2μm to 2μm – and if you check the reflectivity/absorptivity graph above you will see that the absorptivity is quite low (and reflectivity quite high).

Snow has a high albedo for sunlight – around 60% – 80% is reflected, meaning only 20%-40% is absorbed.

Now let’s consider the snow at a temperature of 0°C (273K). How much thermal radiation does it emit?

If it was a blackbody, it would emit radiation as the Planck function at 273K:

Blackbody radiation at 273K (0'C)

The first thing you notice is that the snow is radiating at completely different wavelengths to the solar radiation. The solar radiation is mostly between 0.2μm to 2.0μm, while the snow is radiating between 5μm and 50μm.

So we need to know the emissivity between these wavelengths to work out the actual emission of radiation from the snow. Emissivity is a value between 0 and 1 which says how close to a blackbody the material is at that wavelength. Emissivity is equal to absorptivity – see the next section – so we can just look up the absorptivity instead.

In the graph from Incropera the absorptivity of snow at higher wavelengths is not shown. But it’s clear that it has changed a lot (and in fact absorptivity is very high – and reflectivity very low – at these higher wavelengths, which climate scientists call “longwave”).

Kirchhoff’s Law

Kirchoff’s law says that emissivity = absorptivity as a function of wavelength – and sometimes direction. That is, these two intrinsic properties of any material have the same value at any wavelength. (The derivation of this formula isn’t something that will be discussed here).

Kirchoff didn’t say that emission = absorption

That’s because emissivity is not the same as emission. And absorptivity is not the same as absorption.

Of course, if a body is only gaining and losing energy by radiation (i.e., no conduction or convection), and the body is not heating up or cooling down then absorption will equal emission. This is due to the first law of thermodynamics or conservation of energy.

But if absorption increases, the body will heat up until the new value of emission balances the increase in absorption. However, the absorption might be in one wavelength range and the emission in a totally different one.

It’s not so difficult to understand, but it does require that you grasp hold of the basics.

So (digressing) back to our commenter, it’s clear that his “reasons” for ditching Kirchhoff’s law weren’t because Kirchhoff was wrong..

And Kirchhoff’s law is very strong. It would need a monumental effort to overturn this part of thermodynamics basics. Reasonable people might expect that if over-turning Kirchhoff’s law is necessary to support the imaginary second law of thermodynamics, then this might imply that the imaginary law is, well, imaginary..

However, even stranger concepts are necessary to support the imaginary second law.

Would Sir like to Absorb this Radiation? No? Very Good Sir, I’ll take it Back.

Now that we have covered a few basics, perhaps some points might start to make sense.

And perhaps this comment might seem a little flawed:

Back to thermodynamics and electromagnetic radiation.

Scienceofdoom and others think that the hot surface has no option but to absorb a photon from the cold surface.

I think a lot of this radiation is in fact scattered from the hot surface and is not absorbed.

Let’s examine the idea that radiation from the cold surface is “not absorbed”, and see it in its comedic glory.

Consider a surface of 0°C (273K).

And now consider one body at 10°C radiating towards this 0°C surface. According to the imaginary second law advocates the radiation from the 10°C body is accepted.

Now consider a similar scenario but the 10°C body has been replaced with a
-10°C body. According to the imaginary second law advocates the radiation from the colder -10°C body is not accepted. And according to its strongest advocate, “it is scattered and not absorbed“.

Here is the comparison spectrum for the two radiating bodies:

Blackbody radiation curves for -10'C (263K) and +10'C (283K)

Notice the very similar radiation curves.

And remember that absorptivity is simply a function of the material receiving the radiation.

And ask yourself, how can the 0°C surface reflect the 10μm radiation from the colder -10°C body, and yet absorb 10μm radiation from the hotter +10°C body?

If the absorptivity at 10μm is 0.9 then it will accept 90% of the radiation at this wavelength from a +10°C body and 90% of the radiation at this wavelength from a -10°C body. It can’t check the menu and “send it back”.

Likewise for each wavelength in question.

[By the way, the fact that the body at -10°C emits radiation that is absorbed by the 0°C surface doesn’t mean that the real second law of thermodynamics is violated. Simple, the 0°C surface is also radiating, and at a higher intensity than the -10°C surface. The net is from the hotter to the colder.]

Conclusion

It took many many requests to finally hear the explanation as to why radiation from a colder body has no effect on a hotter body.

It’s not an explanation that will stand the test of time – except for the wrong reasons. It requires the advocate to believe amazing things about materials. Perhaps that’s why it took so long to get the answer.

Yet more ridiculous ideas have recently been proposed. All in the cause of supporting the imaginary second law of thermodynamics. (These need considering in another post).

Just a digression on the perpetual motion machine (because I don’t want to write a whole post on it). For some reason, perhaps the Gerlich and Tscheuschner miseducation, many confused people think that the absorption and re-emission of longwave radiation by the atmosphere constitutes a perpetual motion machine – and therefore this proves the inappropriately-named “greenhouse” effect can’t exist! Well, we all agree that there is no perpetual motion machine.

But why would the atmosphere radiating towards the earth constitute “perpetual motion”?

Think for a minute before answering, if you claim this.

Right now the earth is around the same temperature it was 100 years ago and also 1000 years ago. Is that a perpetual motion machine – a machine that can’t exist? No. The sun warms the earth. And the sun is powered by internal reactions.

Ok. So if the sun turns off what happens? The earth cools down.

For people who think that the earth’s surface is radiating towards the colder atmosphere, and the colder atmosphere is radiating less energy back towards the earth, and the earth is absorbing this radiation.. we expect the same thing to happen when the sun turns off. The earth will cool down. Just a little slower.

No perpetual motion machine.

Well, the second law of thermodynamics is quite a basic one but misunderstood by many who think they are supporting it.

For newcomers to this debate the approach I have taken is to take a specific example and ask the advocates of their theory how specific well-understood physical properties can possibly support their argument. Mostly I get no response.

I have finally had two answers. One is considered here and it’s hard to understand how anyone can believe it.

See the followup article – The First Law of Thermodynamics Meets the Imaginary Second Law

And the later article – The Real Second Law of Thermodynamics

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This post will suffer from the unfortunate effect of too much maths – something I try to avoid in most posts and certainly did in The Imaginary Second Law of Thermodynamics. It’s especially unfortunate as the blog has recent new found interest thanks to the very kind and unexpected words of Steve McIntyre of Climate Audit.

However, a little maths seems essential. Why?

Some of the questions and triumphant points some commentators have made can only be properly answered by a real example, with real equations.

It’s something I commented on in American Thinker Smoking Gun – Gary Thompson’s comments examined, where I explained that a particular theory is not usually a generalized statement about effects but rather a theory is usually a set of mathematical equations to be applied under certain well-defined circumstances.

In The Imaginary Second Law of Thermodynamics the example was of a sun radiating into the nothingness of space, when a new star was brought into the picture. And the new star was hotter. So the question was, would the radiation from the colder star actually have an effect on the hotter star.

Some Gerlich and Tscheuschner apostles thoughtfully spent some time trying to enforce some discipline about terminology of heat vs energy and whether radiation was a vector – but forgot to answer the actual question.

Recently a commentator suggested that the real answer lay in considering two stars of equal temperature which were brought into proximity.

As for the sun and the other star, only the sun at T10,000 will go up in T if you insert another T11,000 star, never mind how hard one thinks about it. Maybe this becomes obvious once you make the sun and the new star equally hot at T10,000 . They don’t become both hotter, which is what is predicted in the thought experiment. If the sun heats up the newly inserted star it should not matter really if that new star is at T11,000 or T10,000 does it?

And I said:

Unless all of the radiation is reflected it will increase the surface temperature. It might be 0.1K, 1K, 0.0001K – it all depends on the W/m^2 at that point – and the absorptivity at the wavelengths of the incident radiation.

If that isn’t the case, then you have a situation where incident radiation is absorbed but has zero effect.

And our commentator responded:

Congratulations on sticking with it! I think you just discovered the endless source of energy we are all so desperately looking for. When you expect two equally hot bodies to keep heating each other, where is the limit? and could we not syphon off some off that excess heat.

Which bring us to here. Many people gets confused around these basic points, which is why we need a post with some maths. The maths can prove the point, unlike “talk”.

Conceptual understanding is what everyone seeks. I hope that this article brings some conceptual understanding even though it has a core maths section.

Some Unfortunate but Necessary Maths

Let’s first of all consider one “star” out in the nothingness of space.

The star has a radius of 1000m (1km) and a temperature of 1000k (727°C). This temperature is identical all over the surface and is powered from internal stellar processes. This internal heat generation is constant and not dependent on any changes in surface temperature. We will also assume – only necessary for the second part of the experiment – that the thermal conductivity of this star is extremely high. This means that any radiation absorbed on one part of the surface will conduct rapidly around the surface of the star – to avoid any localized heating.

We also assume that its emissivity is 1 – it is a blackbody across all wavelengths.

A few derived facts about this star, which we will give, in true mathematical style, the exotic name of “1”.

Surface area:

A₁ = 4πr² = 1.256×10⁷m²

Flux from the surface of star 1, from the Stefan-Boltzmann equation:

F₁ = εσT⁴ = 1 x 5.67×10^-8 x 1000⁴ = 56,700 W/m²

How the radiation emission varies with wavelength:

Total thermal energy radiated:

E₁ = A₁. F₁ = 7.12 x 10¹¹W

If this is the thermal energy radiated, and star 1 is at equilibrium, then the heat generated within the star must also be this value. After all, if the heat generated was higher then the star’s surface temperature would keep increasing until steady state was reached.

For example, if the internal energy source increased its output (for some reason) to 8 x 10¹¹W then the output of the star would eventually reach this value. So F₁(new) = 8 x 10¹¹/A₁ = 6.37 x 10⁴ W/m²

And from the Stefan-Boltzmann equation, T = 1,030K. Just an example, for illustration.

And now, two stars brought into some proximity

So what happens when two identical stars are brought into some proximity? According to our commentator, nothing happens. After all, if “something” happens, it can only be thermal runaway.

The only way we can find out is to use the maths of basic thermodynamics. For people who go into “fight or flight” response when presented with some maths, the conclusion – to relieve your stress – is that the system doesn’t go into thermal runaway, but both stars end up at a slightly higher temperature. Deep breaths. See a later section for “conceptual” understanding.

We define E₁ = the energy from star 1 before star 2 (an identical star) appears on the scene.
And E₁‘ = the energy from star 1 after star 2 appears on the scene.

The distance between the two stars = d

The radius of each star (the same) = r = 1000m

Consider star 2, radiating thermal energy. Some proportion of star 2’s thermal radiation is incident on star 1, which has an absorptivity (= emissivity) of 1.

To calculate how much of star 2’s thermal radiation is incident on star 2, we use the very simple but accurate idea of a large sphere at radius d from star 2. This large sphere has a surface area of 4πd².

On this large sphere we have a small 2-d disk of area πr², which is the area projected by the other star on this very large sphere. And so the proportion of radiation from star 2 which is incident on star 1:

b = πr² / 4πd² = r² / 4d² [equation 1]

This value, b, will be a constant for given values of r and d.

So, our big question, when star 2 and star 1 are “wheeled in” closer to each other, at a distance d from each other – what happens?

Well, some of star 2’s radiation is incident on star 1. And some of star 1’s radiation is incident on star 2.

Will this – according to the crazy theories I have been promoting – lead to thermal runaway? Star 1 heats up star 2, which heats up star 1, which heats up star 2.. thermal runaway! The end of all things?

Thermal Runaway? Or a Slight Temperature Increase of Both Stars?

To work out the answer, it’s all about the maths. Not that the subject can’t be understood conceptually. It can be. But for those who are convinced this is wrong, “conceptual” just leads to “talk”. Whereas maths has to be disputed by specifics.

When our two stars were an infinite distance from each other in the vastness of space, E₁ = E₂ – with the values calculated above.

Now that the two stars are only a distance, d, from each other, a new source of thermal energy is added.

Consider star 1. If this star absorbs thermal radiation from elsewhere, it must emit more radiation or its temperature will rise. If its temperature rises then it will emit more radiation. (See note 1 at end).

So:

E₁‘ = E1 + E₂‘ b   [equation 2]

E₂‘ = E2 + E₁‘ b   [equation 3]

This is simply showing mathematically what I have already expressed in words.

And because the stars are identical:

E₁ = E₂ [equation 4],  and

E₁‘ = E₂‘   [equation 5]

So, from [2] and [4],  E₁‘ = E₁ + E₁‘ b, or (rearranging):

E₁ = E₁‘(1-b), so E₁‘ = E₁ / (1-b)   [equation 6]

So once new equilibrium is reached, we can calculate the new radiation value, and from the Stefen-Boltzmann equation, we can calculate the new temperature.

These equations don’t tell us how long it takes to reach equilibrium, as we don’t know the heat capacity of the stars.

Let’s put some numbers in and see what the results are:

Let d=1000km = 1,000,000m or 1×10⁶m

Therefore, from [1]:

b =1000²/4x(1×10⁶)²

b = 2.5×10^-7

And, from [6]:

E₁‘ = E₁/0.99999975 = 1.00000025 E₁

Do we even want to work out the change in temperature required to increase the radiation from the star by this tiny amount? Just for interest, the new surface temperature = 1000.00006 K

But this is the new equilibrium for both stars.

Note that there is no thermal runaway.

The approach can now be subject to criticism. (So far no one has checked my maths, so it’s quite likely to have a mistake which changes the numerical result). I can’t see how there can be a mistake which would change the main result that no thermal runaway occurs. Or that would change the result so that no change in temperature occurs.

For more interest, suppose the stars were only 10km or 10,000m apart. Strictly speaking, while the distance between the stars is “much greater” than the radius of the stars we can use my equations above. The mathematical expression for this “much greater” is, d>>r. Once the stars are close enough together the maths gets super-complicated. This is because the distance from one point on one star to one point on another star is no longer “d”. For example, as a minimum it will be d-2r (the two closest points)

No one wants to see this kind of “integral” (as the required maths is called). Least of all, me, I might add.

Well, we’ll ignore the complexities and how it might change the result, just to get a sense of roughly what the values are.

If d=10,000, b=0.0025 and so E₁‘ = E₁ / 0.9975 =  1.0025 E₁

Consequently the change in surface temperature to increase the temperature by this amount, T=1000.6K

Not very exciting, and still no thermal runaway.

Conceptual Understanding and Some Radiation Theory

Understanding this conceptually for most people won’t be too difficult. If you add energy to a body it will warm up. And it will emit more radiation. There will be a new equilibrium.

Two bodies doing this to each other will also just reach a new equilibrium – they can’t go into thermal runaway. Of course, no one believes that thermal runaway will result, least of all the person who made the original comment – that was their whole point. They just didn’t realize that a new equilibrium could exist. The only way I can prove it is mathematically.

Conceptual thinking is very valuable. Maths is very tedious. But because Gerlich and Tscheuschner have made such a huge contribution to the misunderstanding of basic thermodynamics it needs some extended explanation, including some maths.

Many people have got confused about the subject because

Heat flows from the hotter body to the colder body

We all agree.

Many people have taken the statement about heat flow and imagined that thermal radiation from a colder body cannot have any effect on a hotter body. This is where they go wrong.

A body with a temperature above absolute zero will radiate according to its emissivity (and according to the 4th power of temperature). This property is dependent on wavelength and sometimes on direction. The emissivity of a body is also equal to the absorptivity at these same wavelengths and directions.

The wavelength dependence of emissivity and absorptivity is very striking:

Reflectivity vs wavelength for various surfaces, Incropera (2007)

Absorptivity is the scale on the right from 1 at the bottom to zero at the top and is 1-reflectivity. (See note 2).

Here you can see that snow is highly reflective at solar wavelengths (shortwave) and absorbs little radiation, whereas it has a high absorptivity at longer wavelengths (and therefore does not reflect much longwave radiation).

The same goes for white paint. It reflects sunlight but absorbs terrestrial radiation.

The equation for how much radiation is emitted by a body – εσT⁴ – does not include any terms for where the radiation might end up. So whether this radiation will be incident on a colder or hotter body, it has no effect on the radiation from the source. (See note 3).

Similarly, when radiation is incident on a body the only factor which affects how much radiation is absorbed and how much radiation is reflected is the absorptivity of the body at that direction and wavelength. The body cannot put out traffic cones because the incident radiation has been emitted by a colder body.

This is elementary thermodynamics. Emissivity and temperature determine the radiation from a body. Absorptivity determines how much incident radiation is absorbed.

Therefore, elementary thermodynamics shows that a cold body can radiate onto the surface of a hotter body. And the hotter body will absorb the radiation – assuming it has absorptivity at that wavelength and direction.

And once thermal radiation is absorbed it must heat the body, or slow down a loss of heat which is taking place. It cannot have no effect. This would be contrary to the first law of thermodynamics.

Two bodies at different temperatures in proximity both radiate towards each other. Heat flow is determined by the net effect. As standard textbooks indicate:

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

Why the Original Misconception?

I think that the original comment about two bodies with the same temperature being unable to heat each other is an easy misconception for two reasons:

First, the most likely mental image immediately conjured up is of two pots of water at say 50ºC. When these two pots of water are mixed together the temperature is obviously still at 50ºC.

Second, the two stars are probably pictured as already in equilibrium at the original temperature. Well, if that’s the case then nothing will change. The change only occurs when they are brought closer together and so the mutual radiation from each has a slight increase on the temperature of the other.

It’s just my guess. But what actually happens in the thought experiment probably isn’t intuitively obvious.

Conclusion

When two bodies have an energy source which has created a constant surface temperature and they are subsequently brought into proximity with each other, there will be an increase in each other’s temperature. But no thermal runaway takes place, they just reach a new equilibrium.

Basic thermodynamics explains that bodies emit thermal radiation according to temperature (to the fourth power) and according to emissivity. Not according to the temperature of a different body that might happen to absorb this radiation.

And basic thermodynamics also explains that bodies absorb thermal radiation according to their absorptivity at the wavelengths (and directions) of the incident radiation. Not according to the temperature (or any other properties) of the originating body.

Therefore, there is no room in this theory for the crazy idea that colder bodies have no effect on hotter bodies. To demonstrate the opposite, the interested student would have to find a flaw in one of the two basic elements of thermodynamics described above. And just a note, there’s no point reciting a mantra (e.g., “The second law says this doesn’t happen”) upon reading this. Instead, be constructive. Explain what happens to the emitting body and the absorbing body with reference to these elementary thermodynamics theories.

Update – now that one advocate has given some explanation, a new article: Intelligent Materials and the Imaginary Second Law of Thermodynamics

Notes

1. I said earlier: “If this star absorbs thermal radiation from elsewhere, it must emit more radiation or its temperature will rise. If its temperature rises then it will radiate more energy.” Strictly speaking when radiation is absorbed it might go into other forms of energy. For example, if ice receives incident radiation it may melt, and all of the heat is absorbed into changing the state of the ice to water, not to increasing the temperature.

2. Incident radiation can also be transmitted, e.g. through a thin layer of glass, or through a given concentration of CO2, but this won’t be the case with radiation into a body like a star. The total of reflected energy plus absorbed energy plus transmitted energy has to equal the value of the incident radiation.

3. The Stefan-Boltzmann equation is the integral of the Planck function across all wavelengths and directions:

Spectral Intensity, Planck

Where h, c₀ and k are constants, T is temperature and λ is the wavelength.

The Planck function describes how spectral intensity changes with wavelength (or frequency) for a blackbody. If the emissivity as a function of wavelength is known it can be used in conjunction with the Planck function to determine the actual flux.

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