Guest post by Willis Eschenbach
There is a lot of misinformation floating around the web about how the greenhouse effect works. It is variously described as a “blanket” that keeps the Earth warm, a “mirror” that reflects part of the heat back to Earth, or “a pane of glass” that somehow keeps energy from escaping. It is none of these things.
A planetary “greenhouse” is a curiosity, a trick of nature. It works solely because although a sphere only has one side, a shell has two sides. The trick has nothing to do with greenhouse gases. It does not require an atmosphere. In fact, a planetary greenhouse can be built entirely of steel. A thought experiment shows how a steel greenhouse would work.

Before we start, however, a digression regarding temperature. The radiation emitted by a blackbody varies with the fourth power of the temperature. As a result, for a blackbody, we can measure the temperature in units of radiation, which are watts per square meter (W/m2). For objects with temperatures found on the Earth, this radiation is in the range called “longwave” or “infrared” radiation. See the Appendix for the formula that relates temperature to radiation.
This means that we can denote the temperature of a blackbody using W/m2 as well as the traditional measures (Fahrenheit, Celsius, Kelvin). The advantage is that while temperature (degrees) is not conserved, energy (W/m2) is conserved. So we can check to see if energy lost is equal to energy gained, since energy is neither being created nor destroyed by the climate.
For our thought experiment, imagine a planet the size of the Earth, a perfect blackbody, heated from the interior at 235 watts per square meter of surface area. How is it heated from the interior? Doesn’t matter, we’ll say “radioactive elements”, that sounds scientific.
The planet is in interstellar space, with no atmosphere and no nearby stars. The equilibrium surface temperature of this planet is, of course, 235 W/m2. To maintain the equilibrium temperature, it constantly radiates this amount of energy out to space. Coincidentally, this is the amount of solar radiation that makes it past the clouds to warm the Earth. If we convert 235 W/m2 to one of our normal temperature scales, it is -19 Celsius (C), or -3° Fahrenheit (F), or 254 Kelvins (K). It’s a possible temperature that the Earth might have if there were no greenhouse effect. That is to say … cold.
Now imagine that the planet gets completely surrounded by a thin black steel shell, located a few thousand meters above the surface, as shown in a cutaway view in the picture above, and in Figure 1 below. What happens to the surface temperature of the planet? (To simplify calculations, we can assume the shell has the same outer surface area as the surface. For an earth-sized planet with a shell two kilometers above the surface everywhere, the difference in area is only six-hundredths of one percent. This assumption makes no difference to the argument presented.)
In order to maintain its thermal equilibrium, including the new shell, the whole system must still radiate 235 W/m2 out to space. To do this, the steel shell must warm until it is radiating at 235 watts per square meter. Of course, since a shell has an inside and an outside, it will also radiate 235 watts inward to the planet. The planet is now being heated by 235 W/m2 of energy from the interior, and 235 W/m2 from the shell. This will warm the planetary surface until it reaches a temperature where it radiates at 470 watts per square meter. In vacuum conditions as described, this would be a perfect greenhouse, with no losses of any kind. Figure 1 shows how it works.

Figure 1. Building a steel greenhouse. (A) Planet without greenhouse. Surface temperature is 235 W/m2 heated from the interior. (B) Planet surrounded by steel greenhouse shell. Shell radiates the same amount to space as the planet without the shell, 235 W/m2. It radiates the same inward, which warms the planet to 470 W/m2 (29 C, 83°F, 302 K). [Clarification added] Note that the distance from the shell to the planet is greatly exaggerated. In reality, it is close enough to the planet that the difference in the areas can be neglected in practice.
The trick can be repeated again, by surrounding the planet and the shell with a second outer shell. In this two shell case, the planetary surface temperature (in W/m2) will be three times the initial surface temperature.
In nature, planets have atmospheric shells, of course, rather than steel shells. The trick works the same way, however. Rather than being warmed from the inside, the Earth receives 235 W/m2 from the sun. Solar radiation passes through the atmosphere. Outgoing longwave radiation is absorbed by the atmosphere, just as it is absorbed by the steel shell shown in Fig. 1.
So that’s the trick of the greenhouse. It has nothing to do with blankets, mirrors, or greenhouse gases. It works even when it is built out of steel. It depends on the fact that a shell radiates in two directions, inwards and outwards. This radiation keeps the planet warmer than it would be without the shell.
Now, it is tempting to think that we could model the Earth in this manner, as a sphere surrounded by a single shell. This is called a “two-layer” model, with the two layers being the surface and the atmospheric shell. In fact, a number of simplified climate models have been built in this way. Unnoticed by their programmers, however, is that is not physically possible to model the Earth as a two-layer system. Figure 2 shows why. Note that in all cases, the system has to remain in thermal equilibrium. This means that the surface must radiate as much as it absorbs, and the shell must also radiate as much as it absorbs. In addition, radiation from the shell to space (upwelling longwave radiation or ULR) must equal radiation from the shell to the Earth (downwelling longwave radiation, or DLR)

Figure 2. Single-shell (“two-layer”) greenhouse system, including various losses. S is the sun, E is the Earth, and G is the atmospheric greenhouse shell around the Earth. The height of the shell is greatly exaggerated; in reality the shell is so close to the Earth that they have about the same area, and thus the small difference in area can be neglected. Fig. 2(a) shows a perfect greenhouse. W is the total watts/m2 available to the greenhouse system after albedo. Fig. 2(b) is the same as Fig. 2(a) plus radiation losses Lr which pass through the atmosphere. Fig. 2(c) is the same as Fig. 2(b), plus the effect of absorption losses La. Fig. 2(d) is the same as Fig. 2(c), plus the effect of thermal losses Lt.
Figure 2(a) shows the same situation as Figure 1(B), which is a perfect planetary greenhouse. In this case, however, it is heated by an amount of energy “W”, which is coming from the sun. The planet receives solar radiation in the amount of “W” from the sun, and longwave radiation “W” from the atmospheric shell. The surface temperature is thus 2W. All energy flows are in Watts/square metre (W/m2).
Figure 2(b) adds two losses. The first is the reflection from the Earth’s albedo (Wo – W). This is energy which never enters the system and is reflected back into space. We are still getting the same energy entering the system (W). The second loss Lr is from radiation which goes from the surface to space through the “atmospheric window”. Because of the second loss, the surface of the Earth does not get as warm as in a perfect system. In a perfect system, the temperature of the surface is 2W. But including the radiation loss Lr, the surface temperature drops to 2W – 2Lr.
Figure 2(c) adds another loss La, which is the solar radiation which is absorbed by the shell. This cuts the radiation hitting the surface down to W – La. Including this loss, the surface temperature is 2W – 2Lr – La.
Figure 2(d) adds the final loss. This is Lt, the thermal loss. It is the sensible energy (energy you can feel) and latent energy (evaporation) which is transported from the surface to the shell by convection. Including all of these losses, the surface temperature of the Earth is 2W – 2Lr – La – Lt.
Now, why can’t the Earth be modeled in this manner? A look at the size of the various losses shows why. Here is the canonical global energy budget, called the “Kiehl/Trenberth” budget, or the K/T budget.

Figure 3. The Kiehl/Trenberth Global Energy Budget. This is a “two layer” representation, with the surface and the atmosphere being the two layers. Lr, the radiation loss, is the 40 W/m2 labeled “Atmospheric Window”. La, the absorption loss, is the 67 W/m2 labelled “Absorbed by Atmosphere”. Lt, the thermal loss, is 102 W/m2. This is the sum of the 24 W/m2 labelled “Thermals” and the 78 W/m2 labelled “Evapo-transpiration”. W, the energy from the sun, is the incoming solar radiation of 342 W/m2 less the 107 W/m2 that is reflected by the surface and the clouds. This means that W is 235 W/m2. SOURCE
What’s wrong with this picture? Note that the temperature of the Earth is 390 W/m2, labeled as “Surface Radiation”. This is 15 C, or 59°F. But from Fig. 2(d), we know that the surface temperature of a greenhouse system with a single shell, as shown in the drawing, is 2W (470) – 2Lr (80) – La (67) – Lt (102) = 221 W/m2. This is far below the known surface temperature of the Earth. In other words, a single shell greenhouse system simply isn’t efficient enough to give a surface temperature which is warm enough to allow for the known losses.
So where is the problem with the K/T budget diagram? The hidden fault is that the upward radiation from the atmospheric layer does not equal the downward radiation. There is 195 W/m2 going to space from the atmospheric shell, and 324 W/m2 going down to the surface.
In order to get enough energy to allow for the known losses, the simplest model requires two atmospheric shells. A perfect greenhouse with two shells would give a surface temperature of 3W, or 705 W/m2. This is enough to allow for the known losses, and still give a surface temperature which matches that of the Earth. Figure 4 shows one such possible system.

Figure 4. Simplest greenhouse global energy budget capable of representing the Earth. Note that all major flows in the K/T energy budget have been maintained. There are two shells, which must be physically separate. These are the lower troposphere and the lowest stratosphere. They are separated by the tropopause.
This budget fulfills all of the requirements for thermal equilibrium. The same amount is radiated upwards and downwards by each layer. The amount absorbed by each part of the system equals the amount radiated. Further verification of this energy budget is the 147 W/m2 emission from just above the tropopause. This converts to a Celsius temperature of about -50 C, which is a typical temperature for the lowest part of the stratosphere.
I have written a simplified Excel radiation/evaporation/convection model which encapsulates the above system. It is available here. Click on the “File” menu on the webpage and select “Download”.
I invite people to play with the model. It is what I term a “Tinkertoy” model, which is what I call the simplest possible model that can represent a particular reality. One of the interesting results from the model is that there can be very little thermal loss between the inner and the outer atmospheric layers. If there is more than a trivial amount of leakage, the surface cools significantly. Among the other insights yielded by the model is that a change equivalent to a doubling of CO2 (an increase of 3.7 W/m2 downwelling radiation at the top of the atmosphere) can be canceled by a 1% increase in the upper and lower cloud reflections.
Experiment with the values, it is an interesting insight into the energy flows in the simplest possible climate model that can represent the Earth’s greenhouse system.
APPENDIX
The formula that relates the temperature to radiation is called the “Stefan-Bolzmann” equation:
R = sigma * epsilon * T^4
where r = radiation (W/m2)
sigma = the Stefan-Boltzmann constant, 5.67 * 10^-8
epsilon = the emissivity of the body, which for a blackbody = 1
T^4 = absolute temperature in Kelvins (which is Celsius plus 273.15) raised to the fourth power
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Joel Shore (07:21:07) :
That source really is too militant about semantics. For example,
“Does the atmosphere trap radiation? No, the atmosphere absorbs radiation emitted by the Earth.”
Is this really so necessary? “Trap” has no definitive physical meaning, but if a layman used ‘trap’ as a synonym for ‘absorb’, I would not get bent out of shape. Would you?
Willis: I should commend you on one point: that your picture goes beyond an isothermal slab atmosphere. Many websites don’t really mention that the IR emission that makes it out to space is from colder parts of the atmosphere.
julien (01:27:30) :
“Interesting model.
Is there a way to re-design this shell system using an external radiation flow ?”
Yes, replacing the shell with greenhouse gases.
I welcome this post. This is a physical model of the whole system. It not based on the chemistry of molecules such as CO2.
It is a model commonly used in climatology and is similar to that proposed by Professor Lindzen of MIT, as below.
http://www-eaps.mit.edu/faculty/lindzen/198_greenhouse.pdf
As many have said, it has problems.
1. Its basic assumptions are not close to reality.
2. Its logic does not follow it own fundamental assumption.
3. It gives a poor result and is then modified.
A thin metal shell is very different from a 30km layer of gas.
Energy is created from no-where.
85degreesF is way off.
A better shell model is described here.
http://freenet-homepage.de/klima/atmoseffect.htm
The increased temperature is due to the pressure of the air. This is called Charles Law, discovered over 200 years ago. The fact that more matter alone can increase temperature is not something we experience. So we look for other explanations.
Thats not a one way process either, as if you increase air pressure in a chamber, for eg, the temperature increases above ambient according to the pressure. THEN, when the optimum temperature is reached, temperature declines even if pressure increases
Those posting demands for removal of this posting need to spend some time in the library, or in a lab with an operating vacuum furnace or <10K cryostat. There is considerable peerreviewedlitchurcher (copyright ClimateAudit) to back up the veracity of Willis' analysis. For example, this free article popped up after 12 seconds of effort with google-
http://jpsj.ipap.jp/link?JPSJ/11/1184/
Y. Ishikawa, S. Sawada, "Effective use of radiation and thermal shields in a vacuum furnace," J. Phys. Soc. Japan, 11 (November 1956), pp 1184-1190.
They actually have experimental measurements of the efficacy of a stainless STEEL radiation shield in a vacuum furnace. Its nice to find measurements that verify theory.
References therein go back even further into the annals of physics history.
A. W. Lawson, R. Fano, Rev. Sci. Inst. 18, (1947), pp 727.
A.E. De Barr, Rev. Sci. Inst. 19 (1948) pp 569.
J. B. Garrison, A. W. Lawson, Rev. Sci. Inst. 19 (1948) pp 574.
A. Braun und G. Busch, Helv. Phys. Act. 9 (1946), pp 33.
They are known as radiation shields, and are commonly used in high temperature furnaces and cryogenic containers. In both cases, radiation losses are very important.
Nice work, Willis.
Want to start this all over again ? ICE !!!!!!
P Wilson (08:02:49) :
The system reaches equilibrium when the shell emits 256 W/m2 to space, and then it also emits 256 to the surface, 512 in total, which must all come from the surface, the only source. What I tried to describe was the transition phase, aften placing the shell.
Ian Schumacher (06:40:44) : says
You may believe that there is no special filter that can be made for energetic particles of any kind, but your belief is wrong. As for a filter that lets in visible light but reflects infrared, these are quite common and ordinary. As several other posters have mentioned, they are called dichroic filters. They are also called dielectric mirrors, and sometime hot mirrors and cold mirrors. You will find them are some low voltage halogen bulbs. You may have noticed that most dentists have a lamp that is very bright, but without much heat. That’s because the lamp has filters built in that pass the visible light on to the patient work area, but allow the IR heat to escape out the backside of the lamp assembly.
The atmosphere works in a similar manner. It allows in visible light with relatively low visible absorption, but absorbs most infrared. For a thought experiment such as Willis’s article, it is very reasonable to postulate a shell that allows the sun to deliver an average of 235 W/m^2 on the earth’s surface, but which totally absorbs (and then re-radiates) the infrared emissions from the heated surface of the earth.
—————————————-
It is clear that you have seen diagrams that illustrate blackbodies as hollow spheres with a small hole in the shell. Real blackbodies used for scientific experiments and measurement are often built in this manner. The reason is actually pretty simple. A blackbody is defined as having an albedo (reflection) of zero. No real life surfaces have a truly zero reflection. But if you have a hollow sphere with just a tiny entrance hole, then any light (or infrared) that enters through the hole will bounce around many, many times before managing to find its way back out of that tiny hole. This makes the effective albedo looking into the hole much lower than the albedo of any one point on the surface of the sphere.
Since the albedo looking into the small hole is very close to zero, the radiation coming out of that hole will be very close to the theoretical predicted for a blackbody whose temperature is that of the inside of the hollow sphere.
It appears that you are somehow conflating this method of emulating a blackbody with Willis’s though experiment where sunlight passes through his steel sphere without being absorbed.
My recommendation to you it to, at least in the beginning, ignore the way that the sun gets to the surface. Just think about the even more simplified model where there is a isotope decay heat source at the center of the earth that generates sufficient energy to have a net outward radiative flux of 235 W/m^2 at the Earth’s surface.
I agree that its a good model: However, that piece of research Y. Ishikawa et al is about maintaining the furnace at 1300C, whereas the consensus here is that it will go way beyond 1300 because of these shields
lgl (08:34:09)
Ah thanks but… its transmitting 256 in two directions, though perhaps not 512 in total. If that surface is 10C it re-radiates 10c worth to the source, and 10C away. Not quite the same as 20C total, as the temp is fixed at 10C.
P Wilson wrote:
“Vincent
“However, some photons of energy from the cooler body will still strike the warmer body.”
I’ll put my eggs in the fridge, and see if they are cooked tomorrow morning””
Ask this question. In which scenario do the eggs cool faster – a) In a fridge or b) in space at absolute zero? That would be an analogy to Willis’s steel sphere. The presence of the sphere allows the planet to cool less quickly than if it was exposed directly to space at absolute zero.
You dodged the question I posed at the end of my previous post. If a cooler body emits a photon in the direction of a warmer body, can you explain why that photon cannot then strike that body and impart its energy? If you cannot, then your reply about eggs in the fridge is just an argument from ignorance. If you do manage to give a credible explanation that I can verify, then I will accept that I was wrong and you were right. Deal or no deal?
Here is another practical implementation of a radiation shield.
GE released an improved incandescent light bulb in the late 1990’s (I knew the inorganic chemist that developed the coating process at GE R&D) that contained a small cylindrical bulb that enshrouded the filament. This assembly was then contained within a regular glass envelope. The small glass cylinder was coated with a multilayer dielectric stack (up to 30 layers!) to reflect near-IR radiation back onto the filament while allowing visible light through, resulting in additional heating of the filament and a concomitant reduction in electrical power (up to 30%) required to maintain the same filament temperature. It won a greenie award from the State of New York back when Cuomo was governor. I don’t know if they are still being sold.
Carrot eater,
I think that Kirchoff’s law says emissivity and absorptivity are equal at all wavelengths and angles at equilibrium. That is my interpretation.
Charlie,
I do believe there are filters allow through different wavelengths. I don’t believe there are filters that can let photons through in one direction and ‘trap’ them so that energy builds up to higher density then the incoming photon stream. The photons will interact with the environment and some of them will become high energy and escape and at equilibrium the energy density of the incoming stream and the ‘trapped’ photons will be the same.
Imagine I have a box with energetic particles of some kind bouncing around. If a one-way energy filter were to exist, then I put that filter in the middle of the box to split the box into two parts. Now our filter goes to work, letting high energy go one direction only. The box spontaneiously becomes warm on one-side and cold on the other.
We KNOW that particular filter doesn’t exist, but somehow we think we can get ‘around’ this problem if we are dealing with photons and not molecules (for instance).
What if we directed particles in through a small hole and had them bounce around. Wouldn’t that trap them? Only particles heading in very special directions could get out. Well yes, it does trap them, to a limit, but once the pressure builds up inside is equal to the pressure outside, we have reached equilibrium.
With photons it is the same. Yes we might use a filter that rejects low energy light, but this is no different then our small hole before. Just as particles in our previous example will keep bouncing around to find their way out, photon energy will redistribute to get out the frequency hole as well. So yes our device would trap photons, but to a limit. Once photon energy density is equal to incoming energy density then the limit has been reached. This is the blackbody limit.
Anyways I’m going to think about it some more. It really is a fascinating subject.
Ian Schumacher (09:16:30) :
“I think that Kirchoff’s law says emissivity and absorptivity are equal at all wavelengths and angles at equilibrium. That is my interpretation. ”
You are forgetting that emissivity and absorptivity themselves are not constant over all wavelengths. Please refer to a text that goes through the derivation, and explores the restrictions.
The emissivity at a given wavelength is generally equal to the absorptivity at that same wavelength. (I’m assuming for now that angle doesn’t matter). So, if Willis wants to use an emissivity of 1 at the long infrared, then he would have to use an absorptivity of 1 at the long infrared as well. And in fact, he does. The earth absorbs ALL of the infrared emitted by the steel shell: there is no reflection term there.
Willis does have less than perfect absorption by the Earth for visible light. This would imply an Earth emissivity of less than one in the visible range, but that doesn’t come up because the earth isn’t hot enough to visibly glow.
Vincent (09:04:39)
Yes a photon from a cool object can hit a warm object – just like a raindrop could be thrown on a glacier, although the tendency is for thermal equilibrium. With Longwave radiation the tendency is for the cooler object not to increase the temperature of the warm object, and if they are at the same air pressure, and ambient temperature -take 2 identical apples, one at 8C the other at 15C an inch apart, in a room at 20C – the likely outcome is that each apple reaches equilibrium with the ambient temperature of its own entropy, as if they were not in each other’s presence. the limiting factor is ambient temperature.
Eggs cool faster in space at absolute zero than in the fridge, but that doesn’t mean that cool fridges (eg at 7C) transfer heat that were put there at room temperature
Carrot eater,
My understanding was that e(f) = a(f) So yes e(f) is a function of frequency (or wavelength, whatever your preferred view), but Kirchoff’s law says at equilibrium then are equal at each frequency.
So while, at first there is a difference (i.e. when the system is not at equilibrium), but once equilibrium is reached then they are equal at all frequencies. Again that is my interpretation from what I have read.
To add to my energy density comment above. Energy density is a way to represent temperature. I like it better because temperature is very messy. Energy is conserved, etc, makes life a lot easier.
If we were to take a snapshot of a volume of space capturing the stream of photons from the sun hitting earth. We could calculate what the energy density is in that region of space. This will have an associated equivalent temperature (for an ideal gas, etc). Inside the earth, there is also an energy density (with an associated temperature).
The energy density in space from the sun can not be lower than the energy density in the earth atmosphere, otherwise energy is being transferred from a low energy density to a higher one. Or in other words, heat is being transferred from a low temperature region to a higher one. This can’t be done, no matter what kind of filters we put in place.
Can it? Can energy density heating an object be less then the energy density of the object itself? At this time, I don’t believe so.
Okay, I can admit when I’m wrong. And I am.
The radiative process described in the first model work as stated… with one caveat… W/m2 is not temperature, but is rather a consequence of temperature.
I know its probably just a slip of the tounge, but posts have repeatedly stated that the model results in a doubling of the temperature (from 235 to 470). That is not the case, it results in a doubling of the energy flux. Temperature is the 4th root of this and so this results in a temperature increase of only about 18.9%. I’m sure this is well understood by the author (as he did correctly state this at one point).
Also, although additional shells would increase the energy flux by 235 per shot (in this model), they increase the temperature by every decreasing amounts (as compared to the energy flux increase). For example, a second shell would increase the temp by an additional 10.7% (the energy flux increases by 50%), a third shell by a further 7.5% etc.
So, looking at this from the point of view of actual temperature, it starts to look a lot more reasonable. The so-called “doubling” of the temp was throwing me off.
Of course, the increases stated only work for an idealized blackbody, which doesn’t exist, and other losses would further mute the effects, but you’re right… the basic pricipal is sound.
Sorry to have been so harsh in my (incorrect) rebutal. Please accept my apologies.
P Wilson
“Eggs cool faster in space at absolute zero than in the fridge, but that doesn’t mean that cool fridges (eg at 7C) transfer heat that were put there at room temperature”
I think we agree then. I thought for a moment you were saying that Willis’ model was impossible because it violated the second law of thermodynamics.
Greenhouse effect = picture-book science.
Tom Vonk,
Consider two parallel planes of infinite extent with different temperatures, Ta and Tb. Are you saying that the energy transfer per unit area between the planes cannot be described by the Stefan-Boltzmann equation: E=sigma*(Ta**4-Tb**4) because they are not hemispheres? Each point on the surface of either plane emits over 2pi steradians and also sees radiation coming from 2pi steradians. For a steel shell close to sphere with the radius of the Earth, a parallel plane approximation is pretty good.
P Wilson- “However, that piece of research Y. Ishikawa et al is about maintaining the furnace at 1300C, whereas the consensus here is that it will go way beyond 1300 because of these shields”
That’s irrelevant when discussing the physics of the situation. The paper derives the general solution for n layers, then explores the solutions for the cases of 1 layer and 2 layers, including effects due to variations in spacing between layers and due to variations in emissivity. I suspect the general solution in the paper will also predict much higher source temperatures as n increases.
P Wilson (08:57:43) :
Yes, 256+256=512, always. “re-radiates 10c worth” has no meaning, stick to energy, and see Anton Eagle (10:12:57) :
B t w, thanks to Anton. We are all making errors all the time, it’s the cost of creativity.
Ian Schumacher (10:12:35) :
“My understanding was that e(f) = a(f)”
Yes (assuming angle doesn’t matter), and that’s all you need. Willis obeys this throughout. For the earth, e(IR range) = a(IR range) = 1. a(visible) < 1 so e(visible) would also be < 1, but it never comes up.
I'm not sure where you're going with energy density. You're best off going back to the temperature of the two bodies. The sun is much hotter than the earth; thus NET radiation exchange will be from sun to earth, by a huge margin. No need to complicate things beyond that.
P Wilson (04:36:11) :
Keep going with the analogy of making aluminum shields for lightbulbs.
Observe that to come as close to possible to being isothermal, you would have to wrap the bulb with foil, (unfortunately you would then have to deal with substantial conduction, but of course we can ignore such rigorous treatment for now. I doubt anyone will notice).
Now for the second half of the experiment, make that shield a foot radius from the bulb. (Note that the top will rapidly get very hot, which will be from convection, but again don’t worry that anyone will notice such a minor detail. If need be, we can draw a vacuum down on the room to prevent that.)
While waiting for this new shield to come to thermal equilibrium with the bulb you might want to call out for a pizza, this could take a while, or, if you are really hungry, check the fridge to see if your eggs are done.
When finished with the experiment, you can make party hats for everyone else from the excess foil. It seems they already have sufficient Kool-Aid.
Cheers.
Hmm looks like there is no scientific consensus even in the basic mechanism of “global warming” 😮