Guest Post by Willis Eschenbach
Dr. Judith Curry notes in a posting at her excellent blog Climate Etc. that there are folks out there that claim the poorly named planetary “greenhouse effect” doesn’t exist. And she is right, some folks do think that. I took a shot at explaining that the “greenhouse effect” is a real phenomenon, with my “Steel Greenhouse” post. I’d like to take another shot at clarifying how a planetary “greenhouse effect” works. This is another thought experiment.
Imagine a planet in space with no atmosphere. Surround it with a transparent shell a few kilometres above the surface, as shown in Figure 1.
Figure 1. An imaginary planet surrounded by a thin transparent shell a few kilometres above the surface (vertical scale exaggerated). The top of the transparent shell has been temporarily removed to clarify the physical layout. For our thought experiment, the transparent shell completely encloses the planet, with no holes. There is a vacuum both inside and outside the transparent shell.
To further the thought experiment, imagine that near the planet there is a sun, as bright and as distant from that planet as the Sun is from the Earth.
Next, we have a couple of simplifying assumptions. The first is that the surface areas of the planet and the shell (either the outside surface or the inside surface) are about equal. If the planet is the size of the earth and the transparent shell is say 1 kilometre above the surface, the difference in area is about a tenth of a percent. You can get the same answer by using the exact areas and watts rather than watts per square meter, but the difference is trivial. Assume that the shell is a meter above the surface, or a centimeter. The math is the same. So the simplification is warranted.
The second simplifying assumption is that the planet is a blackbody for longwave (infra-red or “greenhouse”) radiation. In fact the longwave emissivity/absorptivity of the Earth’s surface is generally over 0.95, so the assumption is fine for a first-order understanding. You can include the two factors yourselves if you wish, it makes little difference.
Let’s look at several possibilities using different kinds of shells. First, Fig. 2 shows a section through the planet with a perfectly transparent shell. This shell passes both long and shortwave radiation straight through without absorbing anything:
Figure 2. Section of a planet with a shell which is perfectly transparent to shortwave (solar) and longwave (“greenhouse”) radiation. Note that the distance from the shell to the planet is greatly exaggerated.
With the transparent shell, the planet is at -18°C. Since the shell is transparent and absorbs no energy at all, it is at the temperature of outer space (actually slightly above 0K, usually taken as 0K for ease of calculation). The planet absorbs 240 W/m2 and emits 240 W/m2. The shell emits and absorbs zero W/m2. Thus both the shell and the planet are in equilibrium, with the energy absorbed equal to the energy radiated.
Next, Figure 3 shows what happens when the shell is perfectly opaque to both short and longwave radiation. In this case all radiation is absorbed by the shell.
Figure 3. Planet with a shell which is perfectly opaque to shortwave (solar) and longwave (“greenhouse”) radiation.
The planet stays at the same temperature in Figs. 2 and 3. In Fig. 3, this is because the planet is heated by the radiation from the shell. With the opaque shell in Fig. 3, the shell takes up the same temperature as the planet. Again, energy balance is maintained, with both shell and planet showing 240 W/m2 in and out. The important thing to note here is that the shell radiates both outward and inward.
Finally, Fig. 4 shows the energy balance when the shell is transparent to shortwave (solar) and is opaque to longwave (“greenhouse”) radiation. This, of course, is what the Earth’s atmosphere does.
Here we see a curious thing. At equilibrium, the planetary temperature is much higher than before:
Figure 4. Planet with a shell that is transparent to shortwave (solar) radiation, but is opaque to longwave (“greenhouse”) radiation.
In the situation shown in Fig. 4, the sun directly warms the planet. In addition, the planet is warmed (just as in Fig. 3) by the radiation from the inner surface of the shell. As a result, the planetary surface ends up absorbing (and radiating) 480 W/m2. As a result the temperature of the surface of the planet is much higher than in the previous Figures.
Note that all parts of the system are still in equilibrium. The surface both receives and emits 480 W/m2. The shell receives and emits 240 W/m2. The entire planetary system also emits the amount that it receives. So the system is in balance.
And that’s it. That’s how the “greenhouse effect” works. It doesn’t require CO2. It doesn’t need an atmosphere. It works because a shell has two sides, and it radiates energy from both the inside and the outside.
The “greenhouse effect” does not violate any known laws of physics. Energy is neither created nor destroyed. All that happens is that a bit of the outgoing energy is returned to the surface of the planet. This leaves the surface warmer than it would be without that extra energy.
So yes, dear friends, the “greenhouse effect” is real, whether it is created by a transparent shell or an atmosphere.
And now, for those that have followed the story this far, a bonus question:
Why is the above diagram of a single-shell planetary “greenhouse” inadequate for explaining the climate system of the earth?
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So apparently the warmer body always reflects photons from a cooler one?
Imagine you have 3 objects at temperatures 10C, 11C and 12C. So the theory is that the object at 11C absorbs photons from that at 12C, but reflects photons from the object at 10C. In other words the emissivity of an object changes sharply according the temperatures of remote objects? This is Phantasy Physics not real physics at all.
“jimmi says:
November 30, 2010 at 11:19 pm
Do you think there is something which prevents a photon from a colder body being absorbed by a warmer one? If so what? How does the warmer body know where that photon came from?”
I don’t know. But if we follow your approach and manage to get more photons from a colder body to a warmer one, which should be possible, as the colder body can be much much larger, while a temperature difference may be rather small, then a warmer body should become even warmer according to your reasoning, and that would be against 2LOT.
Your thoughts?
No, even if you completely surround the warmer object you can never supply more inward photons than there are outgoing ones. Imagine a warm object radiating photons into the vacuum of space. The flux (power per unit area) would be given by the Stefan-Boltzmann equation.
flux = 5.67e-8 * T⁴
Now completely surround that object by another, but with a tiny gap so that energy has to radiate from one to the other. The flux from the second object will also be given by the Stefan-Boltzmann equation, but if the temperature is less, so will be the flux. You cannot make the second object any bigger, or closer, than one that completely surrounds the first, so you can never have the inward flux greater than the outward flux no matter what the geometry. Clearly though, if the outside object had a temperature only slightly greater than the inner one, the inner one would cool very slowly.
Um, that funny looking formula was suppose to be T-to-the-power-4, ie T superscript 4, but it has not come out properly
jimmi, do you think 2 glowing light bulbs produce about twice the number of photons or the same number as just one glowing light bulb?
Fractals will end the climate debate.
http://www.fractalnomics.com/2010/11/breaking-carbon-climate-spell-with.html
You must have some odd view of physics if you think that question reveals anything. Of course they have approximately twice the output of one bulb. If you think I was suggesting otherwise you have not understood anything. Anyway, I trust my previous comment shows that the idea of heating a small object with a very large colder one is invalid.
You have an omission in the description of Figure 3. It is correct only assuming that the shortwave albedo of the shell is the same as that of the earth.
Perhaps you should also emphasize more clearly that the LW emissivity of the shell is 1. The full opaqueness does not imply 100% emissivity as it allows for a lower albedo.
By these comments, I do not contradict your example.
@jimmi, I just wanted to verify this point. Here is a setup that I think will produce more photons from lower temperature (LT) objects to a higher temperature object (HT). Lets select an HT object with a temperature slightly higher than LT, so that it produces 10% more photons. Put an LT object in the focus point of a parabolic reflector, and then use a lense to focus reflected photons on HT object. This way we should be able to capture about 60% of all photons emitted by the LT object*. Put another LT2 object in a similar setup with output also directed to HT object. As a result we have total number of photons coming from LTs to HT at 120%, while HT is 110% of this number. Of course, if needed we can use more LT objects.
Comments?
Note: I used for reference this site for transmission/reflection losses:
http://www.telescope-optics.net/functions.htm
from NY says: December 2, 2010 at 6:36 am
… Put an LT object in the focus point of a parabolic reflector, and then use a lens to focus reflected photons on HT object….
It certainly is a clever idea, but ultimately it will fail for reasons that have been discussed before.
As a slight aside, the 60% from the website you listed is 60% of the light entering the telescope, not 60% of the light leaving the object. You could certainly focus 60% of the light leaving an object onto some other object, but it would take a little different set of lenses/mirrors. It is easy to focus 60% of the sunlight entering the telescope into your eye; it is impossible to focus 60% of the light leaving the star into your eye!
Here is one slightly new argument. If you really wanted to focus light from the cooler object, you should put it at one focus of a perfectly reflecting ellipsoid and put the other (warmer) object at the other focus. Then you could get 100% of the light from the cooler object focused on the warm object — not just the 60% you suggested.
However:
1) this completely surrounds the warm object, so you can’t possibly add a second similar source. You could “cut a hole” in the ellipsoid so only 60% of the energy was focused on the warm object. But the size of the hole would limit you to another 40%, never an extra 60% if you added a second cool source.
2) By symmetry, all the light from the warm object is focused on the cool object. Since the warm object is emitting more energy, we will always have more energy from warm to cool than from cool to warm.
I’m afraid any more discussion would require being able to draw pictures and/or talk in person and/or perform integrals in spherical coordinates. Since none of these are feasible in this setting, I think it is time for me to drop out of the discussion.
“Tim Folkerts says:
December 2, 2010 at 11:06 am”
Tim, you constructed a straw man “ellipsoid” argument, bearing no relation to what I wrote. And wrong – 60% is for the light leaving the LT object. If you think this is too high – note, I wrote “Of course, if needed we can use more LT objects.”
If you have anything specific against what I described – please post.
NY,
I don’t think we will be able to resolve this issue in this venue – the tools are just not available. Maybe with webcams and mics we could get our points across, but short of that i don’t see how we can communicate our ideas clearly enough here.
I will just say that as I understand it (with degrees in math and physics) — the fields of optics, geometry, and thermodynamics all conclude that you cannot get more energy from the cool object to the hot object than from the hot object to to the cold object. I can sort of imagine the set-up of mirrors and lenses you want, but no matter how I picture it, I see physical and/or fundamental optics limitations that limit the results to no better than surrounding the hot object by a sphere of material at the low temperature.
With all due respect, since you are the one making an “extraordinary claim”, it is up to you to give extraordinary evidence — not just “this seems plausible to me”. I (and others) have already given standard responses.
(BTW, the elliptical mirror is not a strawman. It is actually a simplification of what you propose — it collects the light like your parabolic mirror AND focuses it like your lens. The parabolic mirror is a limiting case of an elliptical mirror where the second focus is at infinity.)
From NY,
I think I can deal with this without maths provided you get the idea of a limiting case, which is a standard way of dealing with physics problems. Suppose you were wanting to work out whether a given process was large enough to have a certain effect. You take the limit where you make that process as large as it possibly can be , and if it is still not large enough then there is no point in looking at other configurations.
So in the case I gave, you make the cooler object completely surround the warmer object with just a tiny gap between them. This means that absolutely all the energy leaving the cooler object must strike the warmer one, and you cannot do better than 100% of the energy. Even in this case the flux (power per unit area) from the cooler object towards the warmer one is not sufficient as it is less than the flux going the other way. i.e a cooler object surrounding a warmer one, with a small gap, will always result in the warmer one eventually cooling down – it just takes longer than if the warmer one were left to itself (congratulations you just invented the thermos flask)
Now I did not say what the cooler, surrounding object was made of, and that is because it does not matter – you can aggregate any number of smaller objects, mirrors etc in any configuration you like and you will never be able to come up with a case where more than 100% of the available energy hits the enclosed object. Using two or more heat sources makes no difference, provided they are all at the same temperature. Nor does it matter whether the objects are pure Black Bodies – again think of limiting cases e.g what if one object was a prefect mirror which only reflected radiation and neither absorbed nor emitted it – if the outer, surrounding object were a perfect mirror it would just reflect all radiation from the inner object back on itself, and nothing would change.
What would happen if the cooler object did not completely surround the warmer one? Well there is no problem there either, because in that case you have indirectly introduced a third object – call it “space” or “the vacuum” i.e if there are holes in the outer object radiation can leak away into the vacuum. The effect? Well now both objects are cooling – some of the energy will radiate directly into the vacuum, some will come from one object (does not matter which), and take an indirect route out as a result of being reflected and/or absorbed and re-radiated from the other object – either way the energy eventually leaks away. So in this case both objects are cooling and there is no problem w.r.t LOT2.
The model presented at the top of the thread cannot be a quantitative description of the GHE, in the sense that it is too simple to give a value for the temperature increment, but it illustrates one property that can give rise to a greenhouse effect, and that is that the incoming radiation is unhindered, whereas the outgoing radiation has difficulty leaving. The model has no problems with LOT2, and neither does the actual GHE, which is a real and entirely naturally occurring effect (which may or may not have been altered by human activities – but that is a separate discussion)
@jimmi, December 2, 2010 at 7:28 pm …
Energy radiation (Stefan–Boltzmann law) is proportional to 4th power of temperature per unit of time per unit of surface area. In the case you describe when colder object surrounds warmer object with a tiny gap, the radiating surfaces are about the same, thus colder object radiates less than the warmer one.
This is why I asked you in a previous post if you think 2 glowing light bulb radiate approximately twice as one, apparently it was not clear why I asked the question, but you agreed that twice the output is approximately right. So, unlike your example, which essentially is one colder bulb and one warmer bulb, I constructed 2 colder bulbs (objects) with the output directed with some losses against one warmer bulb (object). I hope it is clear now; your example is invalid.
From NY
Sorry, but my example is as many light bulbs as you want so as to surround the warmer object – not just 2 – as many as you want as long as each is cooler than the central object and as long as the central object is completely surrounded so that every square metre of the central object is irradiated. I actually said that previously, “using two or more heat sources makes no difference “. And if you do not surround the warmer object it does not work due to losses to the vacuum.
@NY
The crux here is the difference between
a) “radiating more from it’s own surface” and
b) “radiating more to the other surface“.
No one is refuting (a). A large cool object can certainly radiate more total energy than a small warm object. If the cool object is 10% cooler but twice as big, then it will indeed radiate more energy (about 30% more) than the slightly warmer, smaller object.
But (b) is believed to be impossible (based on very fundamental optics and thermodynamics). In my example above, the cool object was radiating 30% more total energy. BUT some of that energy is not going to hit the warm object, since there must be a gap between the two. Some of the photons will “miss” the warm object and get reabsorbed by the cool object. On the other hand, all of the photons from the warm object will get absorbed by the cool object, since the cool object completely surrounds the warm object.
You would get something like
* Warm object emits 100 J per second — all of which gets absorbed by the cool object
* The cool object emits 130 J/s — 65 hits the warm object and 65 hits the cold object object thru the “gaps”
No matter how you arrange it, the “gaps” will let more energy “miss” than you gain by making the surrounding cooler object bigger. And no amount of mirrors or lenses can change that, either.
Steve says:
“So the air in a typical room has absolutely no IR absorption in your universe? And apparently it is 100% O2 and N2! Why don’t you check out this graph. Note that the IR band begins just before the 1, and “thermal IR” is commonly considered to be the range of 3 to 15 micrometers.”
Absorption: YES
Reemission: NO
Up to 10km height, animated molecules give away the extra energy they absorbed before by collision
@Tim Folkerts says:
“If the surrounding surface gets twice as big in radius, the area increase by 4x. But the intensity of light from any part of the surface decreases by 1/4. The net effect is that the surrounding surface provides the same intensity of light (= the same number of photons = the same W/m^2) to an object inside no mater how far away the surrounding surface is.
Put another way, the surrounding surface emits more and more light as it gets bigger (more total photons), but fewer and fewer of them hit the object inside (ie more and more miss the object and simply return to the surround surface at some other location). The two effects exactly cancel out.
If you still don’t get it, then I give up.”
Have patience, Tim 😉
Imagine two coins opposite to each other. one small and warm, one 10 times bigger and slightly colder. Then put a lens between an direct the photons of the slightly colder coin exactly onto the surface of the smaller coin. You would transfer roughtly 10 times the energy the warm coin radiates onto the warm coin…
No, this is one of the cases I was discussing with FromNY – you actually have three objects here – the two coins and “space” – both coins are cooling because they are both radiating to space – the fact that some of the energy from the larger, cooler object goes via the smaller one is irrelevant.
Also, regarding FromNY’s light bulbs – note that we have accidentally changed the example from the initial one, which was just considering warmer and cooler objects – if we now start considering powered objects, then we also need to consider what is happening to the ultimate power supply – in other words FromNY needs to think about what happens when he puts all these light bulbs in place, waits from them to warm up, and then switches off the power supply
bessokeks says:
“Imagine two coins opposite to each other. one small and warm, one 10 times bigger and slightly colder. Then put a lens between an direct the photons of the slightly colder coin exactly onto the surface of the smaller coin. You would transfer roughtly 10 times the energy the warm coin radiates onto the warm coin…”
Last I checked, lenses work in both directions. If the lens captures light from he cool object and focuses it on the warm one, then it will simultaneously focus light from the warm object onto the cool one. If you focus 10x more light from cold to warm, you will be focusing 10x more light from warm to cold.
Or to clarify what you were trying to say before …”You would transfer roughly 10 times the energy AS YOU DID WITHOUT THE LENS FROM THE COOL COIN onto the warm coil. You would also transfer roughly 10 times the energy AS YOU DID WITHOUT THE LENS from the warm coin onto the cool coin.” If the Warm coin had been “winning” before, it will still be “winning” if all the numbers are multiplied by 10.
Following up on jimmi’s comment, the lens reduces the energy that is going to “to space” and increases the energy going to “the other coin”. BOTH coins get more energy than before from the other. The coins will stay warmer longer, but they will not violate the 2nd Law.
Since you are the one trying to violate the 2nd Law of Thermodynamics, I leave it to you to propose a specific lens arrangement and coin sizes, and then do the calculations that show us that your system truly focus light in such a one-sided way as to send more photon energy from cool to warm than from warm to cool. Give us the hard numbers of energy flux from cold to hot and from hot to cold.
If you can do that, you will have yourself a Nobel Prize. If you can’t do that, then you might as well stop now trying to come up with “this seems like it should work” scenarios.
Tim Folkerts says:
December 4, 2010 at 4:36 pm
“this seems like it should work” scenarios.
Here’s a very simple one. A massive self-gravitating planet is surrounded at a distance by an unbroken shell. The outer shell is warmed by the sun, the planet in turn is warmed by the shell. Photons emitted from the shell gain energy falling down the gravitational well onto the planet. Photons emitted from the planet lose energy climbing back out. Then, in the steady state, the planet is hotter than the shell (as measured by thermometers on each). The temperature ratio is equal to the gravitational blue shift. The time rate is also higher on the planet, in the same proportion. In all, the brightness of the shell is boosted by the fourth power of the blue shift, the rest coming from a compression in the apparent angular size of the shell overhead (aberration of light). This does not violate the second law, even though the hotter object (the planet) has been heated to that temperature by a cooler object (the shell).
Tim Folkerts — December 4, 2010 at 6:28 am
Tim, (a) follows from law, and (b) is demonstrated, for example, in example by bessokeks:
“Imagine two coins opposite to each other. one small and warm, one 10 times bigger and slightly colder. Then put a lens between an direct the photons of the slightly colder coin exactly onto the surface of the smaller coin. You would transfer roughtly 10 times the energy the warm coin radiates onto the warm coin…”
Tim says, “Last I checked, lenses work in both directions. If the lens captures light from he cool object and focuses it on the warm one, then it will simultaneously focus light from the warm object onto the cool one. If you focus 10x more light from cold to warm, you will be focusing 10x more light from warm to cold.”
It is interesting to follow your argument, for now you directly dispute jimmi’s claim:
“jimmi says:
November 29, 2010 at 2:19 pm
There is absolutely nothing to stop a photon from a colder object being absorbed by a warmer object, it is just that there are more going in the opposite direction.”
Perhaps Tim would now address jimmi’s questions — November 30, 2010 at 11:38 am
“Where does the energy go if it is not absorbed? Do photons carry a little sign saying “please don’t eat me?” How does the absorbing body know the photons are from a colder object?”
In another formulation,
“jimmi says:
November 30, 2010 at 11:19 pm
Do you think there is something which prevents a photon from a colder body being absorbed by a warmer one? If so what? How does the warmer body know where that photon came from?”
To those who remain unconvinced of the greenhouse effect, focus on Willis’ claim: “It doesn’t require CO2. It doesn’t need an atmosphere.” His statements zero in on the operational feature which troubles many; how can a cold object (atmosphere) warm a warmer object (earth)?
Consider an airless planet orbiting an isolated star, one side always facing the star.
The night-side of that planet transfers less heat to space than is received by the day-side (the bulk difference is reflected or radiated back to the sun). Therefore, for that sector of space, the planet is a restriction in the rate of heat transfer from the star. With less heat leaving the star’s surface and presumably no decrease in heat arriving from the interior, the star’s surface temperature will be higher.
Add to that planet an atmosphere with Tyndall gases (‘greenhouse’ gases) and the planet will be warmer for the same reason the star was warmer – restriction in outgoing heat transfer.
All the explanations for the “Greenhouse Theory” I have ever seen or read about rely on the atmosphere being able to stop – or restrict convection. This “Steel Greenhouse” is no different. Well no, – it is different as it has got no atmosphere but a vacuum, so in stead a presence of air, the shell (or replacement atmosphere) this time is opaque to outgoing long wave radiation. Stopping outgoing long wave radiation should work just as well, after all this model is not warming an atmosphere just the planetary surface
This, by the way is not the first time I have seen this kind of planetary contraption and the first time I saw it, yes – it had many shells (all imaginary ones) but that just made the whole thing even more unrealistic and I did wonder at the time; “Why do these people make things so complicated?”
– Well if you have a vacuum then of course you do not have to explain away convection. – Could that be the reason?
In this “Steel Greenhouse” there are lots of numbers, some are relating to energy in W/m² and some are relating to temperatures in °K and some in °C. – I would like to know how they worked any one of those numbers out. And by the way as there seems to be 240 W/m² hitting the planetary surface from the Sun and a further 240 W/m² from the inside of the shield, why not carry this “thought experiment” a bit further? What is there to stop the now 480 W/m² from bouncing down to the planetary surface again and again ad infinitum doubling it’s square wattage with each bounce?
And lastly, can anybody work out how 240 W/m ² can possibly leave the outside of the shield?
How did that wattage get through the shield?
Paul Birch says: December 5, 2010 at 3:22 am
Here’s a very simple one. A massive self-gravitating planet is surrounded at a distance by an unbroken shell. The outer shell is warmed by the sun, the planet in turn is warmed by the shell. Photons emitted from the shell gain energy falling down the gravitational well onto the planet. Photons emitted from the planet lose energy climbing back out. Then, in the steady state, the planet is hotter than the shell (as measured by thermometers on each)….
You are one of the few people I have ever heard of speaking general relativity as “simple”. 🙂
We are getting far afield, but I don’t think your example invalidates anything I said.
In relativity all the measurements are — well — relative. Suppose the the system has reached equilibrium. The temperature of the two objects as measured in any given reference would be the same.”
* Mr P. on the planet would measure the same spectrum (and total energy flux) for the upgoing radiation as the downgoing radiation –> he concludes the planet and the shell are the same temperature.
* Mr S. on the shell would measure the same spectrum (and total energy flux) for the upgoing radiation as the downgoing radiation –> he concludes the planet and the shell are the same temperature (although this temperature is a little less than Mr P would measure for the two)
These two numbers for the temperatures (as measured by Mr P vs Mr S) are different because they are measuring the temperature in different frames. In relativity it is perfectly OK to get different numbers in different frames. In any specific reference frame, however, the temperatures will indeed be equal if the radiation in and out is balanced.