Built on Facts

Before we get to the heart of yesterday’s quiz, let me briefly define solid angle for those who may not be familiar with it. Regular angles are measured in degrees or radians, and solid angles are to angles as square meters are to meters. Solid angle is more or less angular area, and is measured in steradians (or sometimes square degrees). Here’s one steradian, from Wikipedia. There are 4π steradians in a sphere:

i-62dc2c15a322933efe52e24cb645b8a5-220px-Steradian.svg.png

Now let’s take a look at the sun. Make yourself a square frame made with 1-meter sides. Fly down and stand on the surface of the sun (this is a thought experiment, after all), and hold the frame so that the sunlight passes through the frame. The total power passing through the frame is just the power output of 1 square meter of the sun’s surface. The solid angle covered by the sun from the perspective of the frame is more or less that entire half of of the sky, so if we define a quantity called radiance which is watts per steradian per square meter then we have a particular value for the radiance at the frame. Now bring the frame back to earth and point it at the sun. The total power passing through the frame is now lower since it’s spread out by the usual 1/r^2 factor. But the angular diameter of the sun is also reduced by a factor of r, so the solid angle covered by the sun is also reduced by that 1/r^2 factor. If the power and the angular size have decreased by the same factor, then the total radiance is the same.

This glosses over some important issues involving the angle at which the rays pass through the frame, but with the understanding that those issues won’t end up affecting the result, it turns out to be a general principle of optics that the radiance is conserved. You can’t increase it with a passive optical system. When you burn paper with sunlight and a magnifying glass, you’ve increased the solid angle worth of sun seen by the paper. This means more power per area and thus more temperature, but the best you can possibly do is to increase the apparent size of the sun to just what is is when you were standing on it in the thought experiment. This will make the spot on the paper plenty hot, but no hotter than the surface of the sun.

But that geometric argument which I haven’t proved (The proof is not difficult, but it’s more trouble than it’s worth here. It’s in any good optics text.) might not be entirely convincing. What about, for instance, the case where we put the entire solar system in a huge elliptical reflector with the sun at one focus? What if we put a basketball-sized blackbody at the other focus? Won’t it be blasted away instantly with the entire power output of the sun?

No, it turns out. The reason is that the sun is not at the other focus. The sun is an extended body with only its center at the focus. The light emitted at the solar north pole is nearly half a million miles away from the focus, and except for the microscopic fraction of light leaving perfectly along the outward normal, the light will correspondingly miss the basketball by a comparable amount. Only a tiny, microscopic fraction of the sun’s output will actually be aligned precisely enough to hit the basketball. It will end up being heated to the temperature of the sun.

I’ll grant that it’s not obvious. It’s not terribly common to invoke the rules of geometric optics to look at thermodynamics. But nature doesn’t really worry about the boundaries between our textbook names.

Comments

  1. #1 Anonymous
    May 27, 2010

    What is the point of this article?

  2. #2 Paul Murray
    May 27, 2010

    Ok, at the surface of the sun the sun covers half the sky. Well, if we arrange reflectors so that the sun appears to cover the whole sky, can’t we get more power into our black-body? The temerature would be S * 2^.25 .

    Actually, the question kind of answers itself. If making the sun cover the whole sky could make an object hotter, then the sun itself just below the surface would have to be hotter than the sun, because all the rest of the sun appears to cover the whole sky from that POV.

    If indeed it were possible to heat up a body hotter than the surface of the sun with sunlight, this being a violation of the 3nd law – can anyone design a mechanism that would extract unlimited amounts of energy from the arrangement?

  3. #3 Carl Brannen
    May 27, 2010

    I once got hired by an engineer to use my metal lathe to make a small device out of polycarbonate that was intended to (I told him) violate thermodynamics in just this way. He wanted it to apply a higher intensity of light to a photo-resistor than existed in the ambient environment it was collecting them from. Of course it didn’t work as he expected. The whole thing was quite annoying (being paid to build something stupid), but I couldn’t resist getting some bucks out of my lathe.

  4. #4 Gerard Harbison
    May 27, 2010

    Regardless of what fraction of the sun’s area ia focussed on the basketball, they will come to the same temperature. It has nothing to do with the efficiency of the mirror.

    The whole point of thermo. is that the path can be completely ignored for a state variable like temperature.

  5. #5 oms
    May 28, 2010

    Question following your example: If you contain the entire solar system in an ellipsoidal reflector, and only a small fraction of the sun’s output falls upon the basketball, then where does the rest of the light which misses the basketball go?

  6. #6 Thomas
    May 28, 2010

    oms, the radiation that misses the basketball will hit the reflector again and be reflected onto the other focus, i.e. the sun, and the sun is large enough to absorb all of it. Fortunately this is a thought experiment so we can postulate that the mirror is removed before the entire sun boils away.

  7. #7 CS
    May 28, 2010

    A few points to make:

    1) For a parabolic mirror placed about 1 km from the collecting tower, the planes where the center and edge of the Sun form images are about 30 nm apart. For a 1 m mirror, the Sun would be focused down to an image size of ~ (30 nm)(1m/1000m) = 0.03 nm. This is purely a geometrical argument; finite wavelength effects would come in long before this.

    2) The energy density of the Sun’s radiation near Earth has a corresponding temperature of about 300 K (if it had the correct spectrum). If we had a 1 square kilometer array of mirrors focusing that radiation down to an area of 1 square meter (much larger than the geometric limit), the energy density of the radiation exceeds a temperature of 9000 K, hotter than the Sun. If you were able to focus the image down to 30 nm, the energy density would correspond to a temperature of 50 million K, exceeding the temperature at the center of the Sun. Focusing the Sun down to 0.03 nm yields a corresponding temperature of 150 million K. There is some looseness here in referring to the temperature of the radiation density in these cases as the photons are not in equilibrium with the surroundings, but the radiation would yield those temperatures if allowed to come to equilibrium.

    3) The formulation of the second law of thermodynamics as “heat cannot flow from lower temperature to higher temperature in an enclosed system” is a simplified form that applies only in circumstances where you can ignore things like changes in pressure, changes in volume, the chemical energy/potential (in the thermodynamic sense), etc. That last one involves changes in the number/types of particles and is very important here: the system heavily involves radiation and the number of photons is not conserved, so the chemical potential cannot be ignored. The formal definition of the second law for statistical mechanics states only that entropy cannot decrease in an enclosed system. That can still be satisfied even if the temperature of the tower is raised beyond that of the surface of the Sun, so the second law need not be violated.

    There is no fundamental difference between the case here and what happens with microwave ovens. The shape of the radiation spectrum in a microwave is closer to that of a blackbody around 20 K, but the energy density of the radiation itself is larger than would be expected from a blackbody cavity at 20 K, so the radiation density has a much higher corresponding temperature. The shape of the Sun’s spectrum corresponds to that of a blackbody around 6000 K, but the radiation density falls as the radiation moves away from the Sun until the density has a corresponding temperature of 300 K near Earth. Note it is the radiation’s energy density that is important, not the shape of the spectrum, which is why the Earth has a temperature closer to 300 K than 6000 K. By focusing the sunlight, you can increase the radiation’s energy density and produce higher temperatures, just like the microwave.

  8. #8 Gerard Harbison
    May 28, 2010


    3) The formulation of the second law of thermodynamics as “heat cannot flow from lower temperature to higher temperature in an enclosed system” is a simplified form that applies only in circumstances where you can ignore things like changes in pressure, changes in volume, the chemical energy/potential (in the thermodynamic sense), etc. That last one involves changes in the number/types of particles and is very important here: the system heavily involves radiation and the number of photons is not conserved, so the chemical potential cannot be ignored. The formal definition of the second law for statistical mechanics states only that entropy cannot decrease in an enclosed system. That can still be satisfied even if the temperature of the tower is raised beyond that of the surface of the Sun, so the second law need not be violated.

    I’m sorry, but that’s complete nonsense. If the two bodies are at different temperatures, we can always formulate a reversible path by which we transfer heat from the higher temperature body to the lower, with an increase in entropy. Therefore, the system is not at equilibrium. The Clausius formulation of the Second Law is rigorous.

    The entropy of substances must without exception increase with temperature. Therefore, if you have two bodies at different temperatures

  9. #9 Gerard Harbison
    May 28, 2010

    Sorry ; ignore the last two lines.

  10. #10 Gerard Harbison
    May 28, 2010

    There is no fundamental difference between the case here and what happens with microwave ovens. The shape of the radiation spectrum in a microwave is closer to that of a blackbody around 20 K, but the energy density of the radiation itself is larger than would be expected from a blackbody cavity at 20 K, so the radiation density has a much higher corresponding temperature. The shape of the Sun’s spectrum corresponds to that of a blackbody around 6000 K, but the radiation density falls as the radiation moves away from the Sun until the density has a corresponding temperature of 300 K near Earth. Note it is the radiation’s energy density that is important, not the shape of the spectrum, which is why the Earth has a temperature closer to 300 K than 6000 K. By focusing the sunlight, you can increase the radiation’s energy density and produce higher temperatures, just like the microwave.

    Just to add that this is also complete crap. If you took a black body at the top of the atmosphere, put it in radiative contact with the sun’s photosphere, and insulated it from the rest of the universe, it would also come to the temperature of the photosphere. It has zilch to do with the radiation density. The phrase ‘you could fry an egg on the sidewalk’ is not entirely without substance. Things can get remarkably hot on the Earth’s surface if you put them in efficient radiative contact with the sun. They cannot, however, get hotter than the sun.

    If you could figure out a way to put the object in radiative equilibrium with Sirius, and insulate it from everything else, it would get to the surface temperature of Sirius. I don’t care how few photons are coming in from Sirius, as long as there are enough to make statistical fluctuations insignificant. The radiative temperature is independent of the radiative flux.

    Fwiw, I teach stat. thermo. at the graduate level.

  11. #11 red pepper
    May 31, 2010

    There is no fundamental difference between the case here and what happens with microwave ovens. The shape of the radiation spectrum in a microwave is closer to that of a blackbody around 20 K, but the energy density of the radiation itself is larger than would be expected from a blackbody cavity at 20 K, so the radiation density has a much higher corresponding temperature. The shape of the Sun’s spectrum corresponds to that of a blackbody around 6000 K, but the radiation density falls as the radiation moves away from the Sun until the density has a corresponding temperature of 300 K near Earth. Note it is the radiation’s energy density that is important, not the shape of the spectrum, which is why the Earth has a temperature closer to 300 K than 6000 K. By focusing the sunlight, you can increase the radiation’s energy density and produce higher temperatures, just like the microwave.

  12. #12 CS
    June 1, 2010

    The Clausius formulation applies to classical thermodynamics, where various interacting bodies are in internal thermodynamic equilibrium. Temperature is a measure of something in internal thermodynamic equilibrium (local equilibrium is sufficient). Temperature is not defined and the Clausius formulation does not apply for non-equilibrium thermodynamics. The collection of photons coming from the Sun is not in internal equilibrium.

    Let’s do your blackbody calculation. Treating the Sun as a 5800 K blackbody, the radiation flux at the surface is 64,000,000 W/m^2 (Stefan-Boltzmann law). For energy to be conserved, the flux must fall by 1/r^2 as it expands spherically outward from the Sun. Using the ratio of the Earth’s orbital distance to the Sun’s radius (where the blackbody radiation is produced), the flux at Earth is (7.3 x10^7 W/m^2)(7 x10^5 km)^2/(1.5 x10^8 km)^2 = 1400 W/m^2. This is a well known (and measured) solar flux.

    Now, place a 1 m^2 blackbody at the same distance. Now suppose you are right and this object reaches 5800 K when in radiative equilibrium with the Sun and isolated from everything else. The same calculation applies as above: at that temperature, a blackbody of that size radiates 7.3 x10^7 W of energy. However, it is absorbing only 1400 W of energy (assuming sunlight reaches all parts of the surface). Over 99.99% of the energy being radiated is appearing out of nowhere. This, I contend, violates the laws of thermodynamics — no free energy!

    Instead, a blackbody is in equilibrium when the power of the incoming radiation is equal to the power of the emitted radiation. By the very definition of a blackbody, the shape of the incoming spectrum is irrelevant, only the total power is relevant (it absorbs all light). A blackbody in equilibrium would emit 1 kW of power whether it was receiving 1 kW of sunlight or 1 kW of microwaves. As the intensity of the radiation near Earth is related to the energy density of the solar radiation (by a factor of c), my statement was not “complete crap” but is completely compatible with thermodynamics. The wikipedia article on blackbodies gives an explicit example, looking at Earth itself as a blackbody (which it is, to first order).

    Radiation is in internal equilibrium when it corresponds to the radiation state in a blackbody cavity of some temperature. Again, that involves the amplitude as well as the shape of the spectrum. If even one of the shape or amplitude does not match a blackbody cavity, the radiation is not in it’s highest entropy state and the radiation is not in equilibrium. The spectral shape of the Sun’s radiation does not change as it moves away from the star, but the amplitude surely does. In fact, useful energy can be extracted from this non-equilibrium radiation, which is the basis for photosynthesis. The number of higher energy photons (e.g. yellow and green light) is too high, while the number of lower energy photons is too low. Photosynthesis takes the high energy photons and extracts work; the waste product is eventually heat that produces lower energy photons in the end.

  13. #13 Martin
    June 1, 2010

    I totally agree with CS!

    Suppose that a certain amount of space with on-located in its material bodies surrounded by a closed adiabatic shell. On after a sufficiently long period of time between the material bodies in the cavity (closed shell), and radiation in the cavity is established thermodynamic balance.
    All the bodies will have the same temperature T and the radiation in the cavity – a spectral density of radiation, called equilibrium.

    This is a crucial point, equilibrium.
    There is no mention to the equilibrium in the problem as stated.

  14. #14 Robert P.
    June 1, 2010

    No, Gerard Harbison is correct. To the extent that the sun’s spectrum can be approximated by black body radiation (which for the purposes of this problem it can), solar radiation is in thermal equilibrium. Black body radiation is in thermal equilibrium, by definition. Its temperature is determined entirely by its spectrum, and has nothing to do with intensity.The temperature of the solar radiation reaching the earth is the same as that of the radiation leaving the surface of the sun.

  15. #15 CS
    June 1, 2010

    Sorry, Robert P., but you are incorrect. The intensity is an inherent part of the derivation of the the blackbody spectrum: is also a fundamental part of the Bose-Einstein statistics that go into the calculation. The Stefan-Boltzmann law is also a “law” with regards to blackbodies, not a suggestion.

    There is confusion here about what the shape of the radiation spectrum tells you: it indicates the temperature of the object that emitted it, not the temperature of the radiation itself. They are the same only within a blackbody cavity or at the surface of a blackbody that radiates externally. This is not the case for the solar system.

    If you still think I am wrong, you should point out where the blackbody example I gave above is incorrect.

  16. #16 Anonymous Coward
    June 1, 2010

    Re CS’s “If you still think I am wrong, you should point out where the blackbody example I gave above is incorrect.”

    Assuming you are referring to your example in #12, you misunderstand Harbison’s comment “If you took a black body at the top of the atmosphere, put it in radiative contact with the sun’s photosphere, and insulated it from the rest of the universe, it would also come to the temperature of the photosphere.” You missed the part of “insulated it from the rest of the universe”.

    Your 1 m^2 blackbody is radiating into (if I follow your calculations correctly and assume you’re only accounting for radiation out the side facing the sun) 2 pi of solid angle. That 2 pi of solid angle is much greater than the angle occupied by the sun, so you have failed to “insulate it from the rest of the universe”.

    Your blackbody is in “equilibrium” at 1kW of radiated power in the sense that it does not change in time. It is not in thermal equilibrium with either the sun or the cosmic microwave background. Rather, it is at some intermediate temperature related to its relative coupling strength to those fields.

    As for defining radiation fields being in “thermal equilibrium”, I’m uncomfortable with that language for this problem: the distribution of photon occupation #’s are completely different for different k-vectors (ferinstance, those pointing at and away from the sun in your example). In such a circumstance, I can’t see how “thermal equilibrium” makes any sense, since one direction of propagation is not in equilibrium with the other. This is a general phenomena for radiative fields in free space. One notable exception to this – i.e. where the EM field is in thermal equilibrium – is the pedagogical example of the field inside a closed blackbody cavity.

  17. #17 CS
    June 2, 2010

    In my example, I assumed radiation was free to flow off to infinity; there was just no incoming external radiation. Just the Sun and a blackbody some distance away in an empty, open universe. This is what I had assumed to be meant by “isolated.”

    If instead, you take the Sun and blackbody to be a completely isolated system where energy cannot enter or leave, then I agree the surface of the Sun and blackbody would reach the same temperature, but it would also be substantially hotter than the current temperature. The only reason the Sun’s surface is 6000 K is because it is not isolated — it continually sheds energy off into space. The energy produced by the fusion reactions in the Sun has to go somewhere; the end result is that the temperature of the entire system rises until the structure of the Sun becomes so distorted that fusion ceases.

    If you want to say the surface of the Sun has a fixed temperature of 6000 K that is completely independent of the rest of the system, the solar system is a completely enclosed system (surrounded by mirrors or a blackbody cavity), then a blackbody anywhere else in that system will reach a temperature of 6000 K as well. However, this rather artificial setup has absolutely nothing to do with the issue of the temperature of the solar tower in the OP and sheds no light on the issue. My setup is much more appropriate (which is why I assumed that was what was meant).

    The last paragraph of yours is the point I was trying to get across. This Clausius formulation applies to objects in internal thermal equilibrium. As you point out, the radiation here is not in equilibrium, so the Clausius formulation is being inappropriately used to get incorrect results. The statistical version of the 2nd law (in terms of entropy) should be used in this case and this version does not preclude a solar tower from reaching temperatures hotter than the Sun.

    I can give another example. Take an array of photovoltaic cells (solar arrays) to generate a voltage. By putting the cells in series, you can create an arbitrarily large voltage. By giving the cells a larger surface area (or using more of them), an arbitrarily large current (or power) can be made available at that voltage. Now attach that voltage to an arc welder, which is driven only by an electric current. These arc welders create plasmas that can have temperatures exceeding 20,000 K. The only input here is the solar radiation, the same as the focused mirrors case. Yet that radiation is generating temperatures hotter than the body that created that radiation. This is not a violation of the 2nd law of thermodynamics as suggested by previous postings. If that is the case, then arc welders attached to solar power sources are essentially perpetual motion machines.

  18. #18 Anonymous Coward
    June 2, 2010

    Re: CS at #17

    I guess I don’t understand your point.

    Your arc welder example is different than the optics example because the power conversion step is, at some level, a heat engine. This makes it irreversible, of “efficiency” less than one, and entropy increasing. This is different than light propagation through optics, which is reversible and does not increase entropy.

    Certainly your arc welder example in #17 is possible, and doesn’t violate any law of thermodynamics, but I don’t see the relevance here.

    Regarding your comment at #7, when you talk about using a 1 km array of mirrors and focussing the image down to 30 nm, is that something that you think is possible using optics?

  19. #19 CS
    June 2, 2010

    No, that is not likely to be feasible. As I noted in that post, though, that was a geometrical argument only and finite wavelength effects should be taken into account on those scales. As I also stated, even focusing it down to 1 square meter can yield temperatures hotter than the surface of the Sun, and that is certainly possible with optics.

    With the optics example, the heating occurs when the population of photons changes it state to higher entropy, which is not a reversible process either. The path through the optics may be reversible, but the interaction with the solar tower is not.

  20. #20 Anonymous Coward
    June 3, 2010

    Re: CS at #19

    Regarding your last example (with a 1 square km array of mirrors focussing down the sunlight to a 1 square meter image area), is that something you think is possible with optics?

    Keep in mind that the sun is – as pointed out in the original article – not a point source of light, but a finite size (an angular size of about a half degree as viewed from the earth, if memory serves me correctly).

  21. #21 CS
    June 4, 2010

    I am not sure if you are suggesting it is impossible or even difficult to do that, neither of which is the case. Many of the current solar towers use flat mirrors, not even bothering to try to get further reduction in the area of impact by the reflections. At one kilometer, the Sun’s reflection off of a 1 m^2 flat mirror would be about 10 m wide. Do you not think a parabolic (or even just spherical) mirror can reduce that width by a factor of 10? A cheap 10 cm diameter magnifying glass has no difficulty focusing the Sun down to 1 mm or so, a much greater reduction than needed with the mirrors here. That magnifying glass is also subject to various aberrations that a parabolic mirror is not subject to. The optics in most cameras and telescopes are more complex and of better tolerances than needed here.

    I want to go back to the issue of the focused reflections versus photovoltaic arrays. The argument going on here was that a blackbody cannot be heated greater than the Sun by the solar radiation because it would violate the 2nd law of thermodynamics. This necessarily implies that the incoming radiation would have a higher entropy before interacting with the body than the blackbody + photon population after the interaction (that is what a violation of the 2nd law means). The mechanism of interaction was not discussed, there was simply a statement that heating a blackbody above 6000 K violates the 2nd law regardless of mechanism because “heat cannot flow from cold to hot.” The photovoltaic array/arc welder gave an explicit mechanism for heating something above 6000 K from the same radiation (the mechanism does not add any additional energy to the system other than the energy of the radiation). This means that either the photovoltaic array violates the 2nd law or there is a flaw in the reasoning behind this statement.

    The flaw I see in the reasoning is using the temperature of the Sun for the cold case, whereas I suggest the “temperature” of the radiation is not the same and is what should be used for any “heat cannot flow from cold to hot” discussions. But, as you pointed out, the radiation is not in any sort of local equilibrium so temperature is not really defined; without a temperature, there is no way to describe it in a quantitative way as “cold” or “hot” in making the comparison with the blackbody.

  22. #22 Anonymous Coward
    June 4, 2010

    Re: CS at #21

    > At one kilometer, the Sun’s reflection off of a 1 m^2 flat mirror would be about 10 m wide.

    I disagree. It would be wider, due to the finite angular size of the sun when viewed from the earth.

    Re: using optics to reduce the spot size

    It is certainly true that a 10 cm diameter magnifying glass (of an appropriate focal length) can focus the image of the sun to a very small size under certain conditions (small focal length, small lens-image distance).

    I do not believe it is true that a magnifying glass (or parabolic mirror) can reduce the image of the sun to a meter spot size if the lens (or parabolic mirror) is a km away from the image plane.

    In this case it doesn’t have anything to do with abberations or diffraction or anything tricky like that. It’s just geometric optics.

    I would recommend reviewing image formation by optics to see how image size is related to object size (and relative distances from the optic), and plugging in some numbers. I expect that will be quite illuminating, no pun intended.

  23. #23 CS
    June 4, 2010

    If a flat mirror is small enough to ignore its width and if the incoming light from an object has a 1/2 degree angular size, the reflected light leaves with an angular spread of the same 1/2 degree. Over a distance of 1 km, that reflected light spreads to a width of just under 9 m. If the mirror is not small, the width of the reflected light is expanded by the width of the mirror. So, for a 1 m wide mirror, the width of the reflected image of the Sun about 1 km away is ~ 9+1 m = 10 m, like I said.

  24. #24 Anonymous Coward
    June 4, 2010

    Re: CS at #23

    I stand corrected on the spot-size-of-a-flat-mirror-at-1km calculation.

    But with regards to the focussing issue:

    > Do you not think a parabolic (or even just spherical) mirror can reduce that width by a factor of 10?

    Not if it’s on the order 1 km away from the focal plane. Plug in the numbers for image formation yourself to see.

    The answer will – in this case – also illustrate why it makes sense to build a solar collector with an array of flat mirrors! (As long as those mirrors are small compared to 10 meters, for the “1 km away” configuration.) Which makes it a pretty interesting calculation.

  25. #25 CS
    June 7, 2010

    But if I do the calculation for a parabolic mirror with a focal distance of 1 km, I find the image is… 9 meters. What’s this on my face? Is that… egg?

    OK, so I really should have done the calculation before going off on the focused light. It probably doesn’t help that I referred to the “image size” in post #7, when I meant the depth of the image due to the center of the Sun being closer than the edges (so their images form at different distances). I had not tried calculating the height of the image, which would have saved me some embarrassment….

    Now that I look back on this thread, I think I have been arguing different issues with Gerard Harbison and you with regards to the 2nd Law. I agree that, if the whole system was completely thermally isolated, a blackbody would come to the same temperature of the Sun. I had taken the “cold cannot heat hot” as arguing that work cannot be extracted by the radiation in the current context (Sun and blackbody isolated in space, but not an isolated thermodynamic system). The Earth or space are essentially reservoirs at 300 K or 3 K and the radiation is in a state that is not in equilibrium with those temperatures. By the photovoltaic cell example, it is clear that you can get a blackbody to higher temperature than the Sun without violating the second law (again, this is what I thought some were arguing against). This is made possible by the lower temperature reservoirs — this would not work if the Earth was also at 6000 K, as the photovoltaic cells would generate more radiation than they absorb from the Sun here and could not supply a voltage.

    Post #14 was still in error, though. For a blackbody, the intensity of the incoming radiation not only cannot be ignored in determining the equilibrium temperature, it is the only thing that matters. Without focusing the Sun’s rays, a blackbody isolated from Earth would come to an equilibrium temperature of 280 K or so at this distance from the Sun due to the drop in solar radiation intensity. The blackbody cannot reach 6000 K without focusing the radiation or extracting work from it.

  26. #26 Glen
    June 8, 2010

    This discussion sounds like we are arguing over the details of a perpetual motion machine – details are not necessary, since they are impossible, but it gives space for the mistaken to lose the principles in a fog of non-sequiteurs. (Such as the variable uses of the term ‘isolated’ and the irrelevant discussion of an electrically powered arc-welder: yes, powered by the sun, but not in thermal radiative contact with it.)

    CS – you mentioned the Stefan-Boltzman law. This says that the net radiation loss rate is proportional to the difference in body temperature raised to the fourth power, so the net energy transfer is zero (i.e. equilibrium) when the temperatures are equal.

    The only reason why the Earth has an equilibrium temp of apprx 280 K is that the Earth is largely in thermal contact with the cold depths of interstellar space as well as with the Sun, so it is not a simple two body problem as you suggest.

  27. #27 CS
    June 11, 2010

    Glen:

    That was actually my point: the Sun and Earth are in a universe where radiation is free to stream away, not in some enclosed system.

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