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″.
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
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).
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  and , 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 :
b = 2.5×10-7
And, from :
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:
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:
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.
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
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:
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.