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Comments 35301 to 35350:

35301. Tom Curtis at 11:57 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       JPostma @51 writes in response to my post @50:

       "   "sufficient to emit greater energy at source than is recieve         by incident radiation on the device"

       That violates the first law of thermodynamics, and though I've refrained from engaging in the type of ad-hominem attack continually thrown my way, a statement like this really does expose scientific incompetency and a clutching of straws. I do apologize for having to make that remark, but alas, it couldn't be passed over in kindness, this time. Those preeminent experimentalists did not indeed interpret that their apparatus was magically producing more energy than received, ostensibly finding an exception to the 1st Law of thermodynamics. They would have laughed at that. What they found is that they could get sunlight to induce its maximum temperature on a plane, a temperature which is well above +100C. It is not an unexpected result."

       (Initial quote italicized for clarity.)

       I will concede that my thought was poorly expressed so that it could be misunderstood.  Allow me to clarrify.

       The de Saussure hot box consisted of an insulated box painted black in the interior, above which were mounted two or more glass panels to allow in sunlight but trap heat (both convective or radiant):

       

       It should be noted that the addition of more than one glass panel in no way improves the traping of heat escape by convection.  One panel is sufficient to stop the mixing of the gasses from inside and outside the box, and hence to stop the convective transfer of heat from inside to outside the box.  It does, however, improve the traping of heat both by radiant transfer and by conduction through the glass.  In doing so, it brings the hotbox model closer to a model of the atmospheric greenhouse effect, in that no energy escapes to space by either conduction or convection.

       Using de Saussure hot boxes, de Saussure measured temperatures as high as 383 K, and John Herschel measured temperatures as high as 388.5 K, in the interior most comparment of the device.  The did so in regions where (due to latitude) peak surface insolation was almost certainly less than 1100 W/m^2, and probably less than 1000 W/m^2.  However, the black body radiation for 383 K radiates 1220 W/m^2, while at 388.5 K it radiates 1291.7 W/m^2.  That is, at the "source" (ie, the interior of the innermost compartment of the device), the energy radiated by black body radiation was at least 100 W/m^2 (9%) greater than the "incident radiation on the device" (ie, the solar energy falling on uppermost panel of glass).

       Postma does not like the result, so he disputes the written record (and in de Saussure's case, publicly demonstrated in London and Paris) of two of the foremost experimental scientists of their day as "anecdotal".  Consequently I refer to a recent test of a single panel hot box (solar cooker) in Jordan:

       

       The "black coated, fixed" cooker is the one most analogous to a de Saussure hot box.  As can be seen, the average peak temperature over three days measurements achieved in the water in the black kettle within the box was 341.4 K.  The equivalent black body radiation was 770.2 W/m^2.  In contrast, the peak insolation recorded on any day during the experiment was was 717.4 W/m^2, and the insolation at the 14th hour (ie, the time of peak temperature) was just 364.5 W/m^2, less than half of the black body energy of the water.  Without water to act as thermal ballast, the interior temperature of the hot box would have peaked earlier and time, and been greater than that recorded.

       The obvious conclusion is that the interior temperatures in de Saussure hot boxes can easily be high enough that the interior black body emission from the inner most compartment exceeds in energy that of the incident sunlight.  Postma says that this is impossible.  Indeed, it is essential to his claims that this is impossible.  It cannot be explained by the prevention of convection, and nor (given the high thermal conductivity of glass) can it be explained by insulation against conduction through the glass.  That means any explanation of the increased temperature must include a greenhouse effect.

       To illustrate this point, consider two hot box designs:

       

       The first hot box is sealed by a panel that is transparent to both visible and IR light, but impermiable to air.  Because it is impermiable to air, it prevents any mixing of external with internal air, and hence any escape of heat by convection.  Because it is transparent to IR light, it neither absorbs nor radiates IR light.  Therefore any IR radiation leaving the box must come from the floor of the box, as illustrated in (1) above.

       The second hot box is sealed by a panel that is transparent to visible light, but absorbs IR light perfectly (emissivity = 1 for IR).  Because it absorbs all IR radiation that falls on it, any IR radiaton from the floor of the box is absorbed by it.  Because absorptivity equals emissivity, that means that energy is then reradiated, with half of it going up, and half of it going down, back into the box, as illustrated in (2) above.

       Now enter the laws of thermodynamics.  In particular, in this context the first law states that for any horizontal line drawn through a "box" above (horizontal plane for actual 3 D boxes), the energy going up equals the energy going down.  That is, 1U = 1D, 2u = 2D, 2u = 2d, and 2U = 2D + 2d = 2 x 2D.  (Note: 1U is energy flux U for box 1, etc.  It is not 1 x U.)  

       Further, the second law of thermodynamics states that for each such horizontal line, the entropy of the energy going up will not be less than the entropy of the energy going down.  Entropy, however, is the energy divided by the temperature.  The temperature of the light for black body radiation is just the temperature of the black body that emitted it.  Where it combines the light from two distinct black bodies, the entropy will be the energy weighted average of the entropies of the two black bodies.

       So, let's assume that 1D = 2D equals 1100 W/m^2.  Let us also assume the boxes are cubes with dimensions of 1 meter per side.  Then the temperature of the base of box 1 equals 373.2 K (~100 C), and the entropy of 1U = 1100/373.2 = 2.95 J/K.  In constrast, the black body emitting 1D was the Sun, with a surface temperature of approximately 5,750 K.  Consequently the entropy of 1D is 0.2 J/K, and as required the entropy of all downward energy at a given distance above the bottom of the box is less than the entropy of all upward energy at the same distance.  Indeed, the temperature of the bottom of the box would have to reach 5,750 K for that to not be the case - something it cannot do because of the first law of thermodynamics.

       In the second box, the temperature of the panel is also 373.2 K, and hence the entropy of 2u is 2.95 J/K.  The temperature of the base of box 2, however, rises to 443.8 K (~170 C).  The upward power from that base (2U) equals 2200 W/m^2.  The entropy of that energy is, therefore, 4.96 J/K.  That is comfortably greater than that of both 2D and 2d(=2u), and certainly greater than their combined entropy of 1.57 J/K.  Therefore the 2nd law of thermodynamics cannot forbid a situation such as illustrated in box 2, and the first law requires that the temperature of the floor of the box be 1.19 times greater than the temperature of the panel.

       (Postma, and others of similar belief, appear to confuse themselves by using imprecise statements of the 2nd law, to the effect that no body can be gain heat from a cooler body.  Heat, however, is net energy transfer.  In box 2, the floor of the box (443.8 K) gains heat from the Sun (5,750 K).  It then transfers heat to the panel (373.2 K).  There is energy flow from the panel to the floor, but the energy flow from the floor to the panel, so the net energy flow (heat flow) is from the floor to the panel.  This means that the floor is heated by the Sun, not the panel; but the floor is heated more by the Sun than it would be without the panel.  There are no entropy considerations preventing this unless the floor approaches temperatures near to that of the Sun's surface.)

       The important thing to note, however, is that mere prevention of convection cannot heat the floor more than sufficient to have a black body radiation equal in power to the incident radiation.  Neither can prevention of conduction where radiant heat can escape, as in the examples above, conduction is considered to be zero in both boxes.  Adding conduction can cool the floor temperature, but it cannot increase it in either case.  Therefore floor temperatures greater than the black body temperature for the power of incident solar radiation is proof that a greenhouse effect is in operation.  And just such temperatures have been observed historically by de Saussure and John Herschel, and more recently in testing of solar ovens.

35302. One Planet Only Forever at 11:21 AM on 17 July 2014
       El Niño in 2014: Still On the Way?
       

       The basics of how El Nino temporarily increases the global average and La Nina temporarily reduces the global average is the air circulation above the water surface is warmed above warmer water and cooled above cooler water:

       • when the ENSO is neutral, between the two, the global average could be said to not be influenced by the ENSO condition.
       • when the ENSO has shifterd toward El Nino there is a larger area of warmer surface waters which results in warmer air above it that winds can carry to other regions beyond the area of warmer water.
       • when the ENSO has shifterd toward La Nina there is a larger area of cooler surface waters which results in cooler air above it that winds can carry to other regions beyond the area of cooler water.
35303. Leto at 11:14 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       The posts by Postma and Devon raise the extraordinary image of a photon looking ahead along its path, measuring the temperature (or perhaps heat content) of the object it would strike, and then deciding retrospectively not to be emitted.

        

       I can see why Postma might not want to make his views too explicit.

35304. mrkt at 10:14 AM on 17 July 2014
       Rupert Murdoch doesn't understand climate change basics, and that's a problem
       

       Just a brief comment on the age question.  I was interested in who were the signers of the OISM petition in my state (Arkansas, USA) and did a brief search on a few.  Found ages for 50 (of 147 in the state).  The average age was 73.  Non-Scientist (@1) perception has some basis. On the other hand the average age at my Sierra club meeting last night was not much lower and these are not among the deniers.

35305. mancan18 at 09:50 AM on 17 July 2014
       Rupert Murdoch doesn't understand climate change basics, and that's a problem
       

       Australia is about to repeal the carbon tax and stop the introduction of an ETS and, instead, is preparing to introduce a process called "direct" action, a marketing exercise designed to placate voters concerned about climate change by pretending that you can act on climate change by burning more coal. Australia doing this makes it the first "developed" nation to undo the market mechanism that has been widely advised by economists as the most efficient method to transform an economy from a high emissions economy to a low emissions economy.

       In Australia, the main reason this has happened is the Murdoch press, which reaches about 83% of the reading public, taking a politically conservative editorial position and prominent anti-global warming stance. The other reason, I believe, is Australia's love of sport and its pub (going to a public bar in a hotel) culture. Many Australians, like everyone else, spend most of their time working and prefer to relax with their families, follow sport and socialise at a pub or club. They spend little of their relaxation time, which is valuable to them and rightly so, reading well researched articles about climate change, and get most of their climate change information from their most convenient source, a Murdoch controlled newspaper. Because many Australians read Murdoch newspapers for the sports reporting rather than the political commentary, the only political commentary many Australians get is from Murdoch newspapers. Because they feature three of Australia's most prominent conservative political commentators Andrew Bolt, Piers Ackerman and Miranda Devine, your average Australian gets a constant stream of climate change and global warming denial, which is reinforced by conservative shock-jocks on radio like Alan Jones and Ray Hadley. This narrative then drives the climate change debate in Australia's pub/club culture. This means the climate change narrative for most Australians is one of denial. Below the surface, there are many Australians who are concerned about climate change and do spend time researching the issue properly.
       Whether, this will be enough, to overcome the political hiatus, remains to be seen. It seems even events like record temperatures for record times over wide areas, disastrious bushfires in October and an unprecedented flood washing cars down the main street of a rural country town, doesn't seem to register with people as, perhaps, something to do with climate change. It is just seen as more of the historical "bushfires and flooding rains" nature of Australia.

35306. pschopf at 08:58 AM on 17 July 2014
       Atlas of Mortality and Economic Losses from Weather, Climate and Water Extremes
       

       Small point:  The WMO is the World Meteorological Organization - not Association.

       Moderator Response:

       [JH] My bad. Fixed. Thank you.

35307. DSL at 08:35 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       And again, Devon, the question Postma refused to answer: is the energy being radiated from Object A (at 10C) absorbed by Object B (at 50C)?

       As soon as we start talking about "heat," the discussion gets silly.  The term "heat" needs to be described in terms of energy exchange.  The term "net" needs to be incorporated.  The rhetorical ploy is to try to push the "argument" into terms that the general public understands: "cold things don't warm hot things."  The reality is that all things absorb/emit radiation.  That is, "all things exchange energy."  Thus, cold things do indeed supply energy to warm things, but the warm things get rid of more energy than they absorb.

35308. DSL at 08:29 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       Devon, do all things above absolute zero radiate?

       If so, then all such things are both emitting and absorbing radiated energy.  "Cold" things emit less energy than "hot" things, but all things are relative, and all things absorb energy. 

       A block of ice is radiating energy (else you wouldn't be able to see it).  A block of red hot iron is radiating.  Put the two in the same room.  Now put an identical block of red hot iron in an identical room but without the ice.  Which block of iron cools down more rapidly? 

       It's not that the room is heating the iron above its initial temperature.  Rather, it is decreasing the cooling efficiency of the iron relative to the situation with the block of ice. 

       
       In the same way, the atmosphere is absorbing energy radiated by the surface.  Some of that radiation is then radiated downward, back to the surface.  The surface is not a static energy source (well, the geothermal component is, but it's relatively small).  It is continually (except at night) warmed by the sun.  Energy once emitted by the surface returns to the surface and brings it to a temperature greater than solar energy alone would produce.  Keep in mind that downwelling radiation has been measured from the surface directly for about twenty years.  It's real.  The atmosphere is radiating toward the surface. 

35309. devon at 07:31 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       Could anyone give a real life example of cold heating hot, or energy being created from nothing as the K&T energy budgets suggest.
       
       So, if cold could in fact heat the already hot, then this would have been established as a theory and a universal fact of physics. I would invite someone here to give a real life example!

35310. KR at 05:20 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       Energy can't flow from a cold source to a warmer object? But most of the world around me is cooler than my body temperature, and that means Aaaahhh, I'm blind!

       Utter nonsense, Postma, utter nonsense.

35311. CBDunkerson at 05:04 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       Of course energy can't flow from a cold source to a warmer source... I mean, that's why when you open a freezer the light from the bulb in the back can't travel from the cold area at the back of the freezer to the warmer area of your eyes. :]

35312. Dikran Marsupial at 03:54 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       "It can be misinterpreted by inferring that a cold source can heat a warmer source. "

       This essentially explains why JPostma wouldn't give a direct answer to my question, but it isn't the one he thinks it is.  If some proposition X can be used to derive some conclusion Y, then if one accepts X then one must also accept Y, unless they can find an error in the chain of reasoning that links X to Y.  However, the possibility that there may be a specious chain of reasoning that links X to Y is not a good reason to refuse to explicitly state that you accept X if you actually do.  Refusal to do so is basically cowardice, not being confident that you will be able to demonstrate an error in the later chain of reasoning.

       Unfortunately I suspect JPostma knows at some level that if he maes a genuine attempt to engage in the thought experiment, he might not be able to spot a flaw in the chain of reasoning, but is unwilling to accept that his understanding of Y might be incorrect.

       This is why I often try to ask skeptics a series of carefully posed questions that are designed to define their position as clearly and unambiguously as possible, so that if they are right, we can all see if, and if they are wrong, then that will also be obvious.  However in my experience, this is generally met by evasion.  JPostma has provided a particularly clear example of this.

       Moderator Response:

       [TD] I believe Robert Murphy has asked the next question that Dikran would have asked.  JPostma, please give a simple yes or no answer to that question, and only then give any explanation that you'd like to give.

35313. John Hartz at 03:50 AM on 17 July 2014
       Rupert Murdoch doesn't understand climate change basics, and that's a problem
       

       Also see:

       Scientists Take Issue With Rupert Murdoch’s Remarks on Climate Change by Eric Roston, The Grid/Bloomberg News, July 16, 2014

35314. Robert Murphy at 03:47 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       Mr. Postma, when the photon from the cooler source hits the warmer source (which you acknowledge does happen), does the warmer source gain energy?  Does the photon have any effect on the warmer body? 

35315. Rob Honeycutt at 03:45 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       "It is unfortunate that so much time has to be wasted on BS like this."

       Unfortunate, but also illuminating to watch cognitive dissonance on steroids.

35316. Dikran Marsupial at 03:42 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       "It can be misinterpreted by inferring that a cold source can heat a warmer source. "

       O.K., give me an example of how that can be done.

       BTW, the net flow of heat energy is always from the warmer object to the cooler, so it is pretty irrelevant anyway. 

35317. JPostma at 03:39 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       "give me an example of how it could be misinterpreted"

       It can be misinterpreted by inferring that a cold source can heat a warmer source.  Of course, this is why the heat flow equation must be referenced with the answer, so that such misinferrence doesn't arise.

35318. KR at 03:39 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       Postma provides a great demonstration of denial here, in particular of the first law of thermodynamics, conservation of energy.

       At equilibrium, which may include averaged short term variations around that equilbrium, the energy leaving a mass (surface, atmosphere, red-faced denier) equals the energy entering that mass (photons, convection, junk food). Input = output, in other words. If there is an imbalance, as with the current climate and radiative imbalance, internal energy will change and with it temperature. 

       The lowest radiative input to a mass, the lowest contribution to that masses temperature, would be the mass surrounded by absolute zero, with the 3K universal background radiation of deep space a close second.

       Any object warmer than absolute zero, regardless of whether it is warmer or cooler than that mass, radiates photons according to its emissivity, and any of those photons absorbed by the mass in question increase it's energy. That added energy makes the mass warmer, until (via radiation, convection, conduction) the increased temperature causes the mass to lose energy at the same rate it gains it. 

       Radiation from anything above absolute zero will warm an object compared to its absence - and the warmer the radiating object the warmer the receiving object will become, no matter what their relative temperatures are. The starting point is zero, and even a relatively cool radiator provides a non-zero addition to nearby masses energies. Cool objects do, indeed, cause nearby warm objects to rise in temperature. 

       ---

       Basic physics. Which Postma denies. His arguments are meaningless, Q.E.D. Not surprisingly, even many of the pseudo-skeptics disavow Postma and the 'Slayers', finding their imaginary physics an embarrassment...

35319. PhilippeChantreau at 03:37 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       It is unfortunate that so much time has to be wasted on BS like this.

35320. JPostma at 03:36 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       "give me an example of how it could be misinterpreted"

       It will be misterpreted precisely by avoiding the heat flow equation, and thus thinking that perhaps the cooler object heats the warmer object because the cooler emission "hit" it.  But of course, this is why it is important to reference the heat flow equation.

35321. JPostma at 03:34 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       More than avoid ambiguity, which is what would occur with a simple "yes or no" answer, I have provided the heat flow equation and some discussion of it to help clarify the fact that opposing emission is what determines equilibrium, and that the warmer object will not be heated, i.e., raise in temperature, due to the cooler object.  I am sorry that I can not provide ambiguous answers, and that my explanation of the answer seems like ambiguity to you.  But it is simple: cold does not heat hot, opposing emission determines equilibrium, etc.

35322. Dikran Marsupial at 03:31 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       JPostma writes "The problem is that answering only with a "yes or no" leaves the door open to misterpretation and sophist obfuscation of the physics"

       Sorry, but that is hilarious.  An answer of "yes" to the question "Does A emit any photons that strike B, "yes" or "no"?" leaves the door open to misunterpretation?  Give me a break!  Better still, give me an example of how it could be misinterpreted!

35323. JPostma at 03:29 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       "So, Mr Postma, for the last time, in the thought experiment described above, does object A emit any photon that strikes object B? Anything else than yes or no will be considered obfuscation by me."

       The problem is that answering only with a "yes or no" leaves the door open to misterpretation and sophist obfuscation of the physics, because without the heat flow equation the meaning of the answer can be interpreted arbitrarily.  Thus, I include the heat flow equation to help make it clear what the result of the physics is: heat not flowing from cold to hot, which means the warmer source does not increase in temperature.

35324. PhilippeChantreau at 03:26 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       I see the answer cam while I was writing my post. That yes has numerous consequences, which essentially negate your argument. "Following it up with Physics" is a fancy name for obfuscation. The condescending tone is rather amusing. I'm done here, that yes was all I needed to measure the depth of Mr Postma's need to receive Physics education, not give it

35325. Dikran Marsupial at 03:25 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       JPostma, I had repeatedly asked you just to reply "yes" or "no" to avoid all ambiguity, you chose not to, even when doing so would be a tacit admission that you were being deliberately obstructive.  I am always happy to discuss science, and be proven wrong if that is what I am, however I don't have endless enthusiasm for obfuscation, and I don't think it unreasonable for people to give a direct "yes" or "no" answer when explicitly asked for it.

       You have missed your chance to clearly explain your position.  My thought experiment was designed to very clearly distinguish between two possible explanations of your position, but you chose not to co-operate.

       Being willing to give direct answers to direct questions should be taken for granted in a scientific discussion, where the aim is to determine the truth.  The fact that you needed input from the moderator should tell you that your behaviour was unproductive.  Note the moderator did not delete your post, just gave advice.

35326. JPostma at 03:25 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       Q = sigma*(Th^4 - Tc^4) shows quite precisely that emission from a cold source does not heat up a warmer source.  The emission from the cold source is what balances with the emission from the warm source in order to determine equilibrium - the emissions cancel out, leaving equilibrium, whether or not we wish to discuss whether the photons from the cold source "strike" the hot source...there is a greater physics occuring than that, as found in the heat equation, which shows how emission actually balances or cancels out at equilbrium.

35327. PhilippeChantreau at 03:22 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       This is clear indeed, and rather boring. This is just a repeat of G&T nonsense and playing on words to spread confusion. We've talked at length about this years ago. Net heat transfer is different from energy transfer. Energy, under the form of photons, can be transferred from any object to any other object, regardless of temperature, as long as it is above absolute zero. Net heat, which in Thermodynamics is usually referred to as just "heat" can only go from hot to cold.

       Postma's condescending tone does not match his unability to answer the very simple question posed by Dikran. Thers is no need to give additional information or "drown the fish" in a sea of words. The thought experiment described by Dikran is simple. The question is does A emit photons that strike B, not does A heat B or does A emit photons. The answer from Postma should be yes or no.

       So, Mr Postma, for the last time, in the thought experiment described above, does object A emit any photon that strikes object B? Anything else than yes or no will be considered obfuscation by me.

35328. JPostma at 03:20 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       Emission from 'A' in all directions equates precisely to 'yes' how is that not clear?  But it is of course important to follow that up with some physics, such as the Stefan-Boltzmann Law and heat flow equation, which indicates that emission from the cold object will not warm the warmer object.

       I find it surprising that there is an assumption that Dikran is the director of this conversation and that I must state answers only in the way he or she desires them - there is of course no point in conversation if my role here is only to say what Dikran requests of me.

       I am dissapointed both at Dikran and the moderator for thus prejudicial behaviour. I had thought to be free to discuss physics.

35329. Dikran Marsupial at 03:15 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       JPostma, it is a shame you take that attitude.  Had you just replied "yes", you would have doen yourself a big favour as it would have revealed where an apparent misunderstanding about your position has arisen.   Your loss entirely.

       "yes" is only three letters, you should ask yourself why you were unwilling to write them, rather than carry on with this obstructive behaviour.

35330. JPostma at 03:10 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       I have precisely stated the answer for you Dikran.  This is a silly game you play...as if you wish to move my mouth for me, or some dominance thing.  It is silly and not necessary.

       'A' emits photons in all directions. However, emission from 'A' does not heat a warmer object, as the heat flow equation shows that heating only occurs from hot to cold.

       I am sorry if you wish to attempt to find a different answer than this, but this is the answer.  You can not deny the Stefan-Boltzmann Law heat equation.  Are things clear for you yet, on the physics?  Do you wish to rather simply state your desired end-point, rather than directing me how to move my mouth for you?

       Moderator Response:

       [TD] JPostma, you are refusing to answer simple questions with simple answers.  You are merely repeating your statements regardless of the question, which constitutes sloganeering, which violates the rules of engagement for this site.

35331. Dikran Marsupial at 03:04 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       JPostma, I'm sorry, you are being deliberately obstructive.  I could not have taken more effort in posing the question for you in a way that made a simple, direct, unequivocal, unambiguous answer possible, but yet again you refuse to do so. 

       Here is your last chance to show that you are genuinely engaged in this discussion and are not just obfuscating:

       Does A emit any photons that strike B, "yes" or "no"?

       If your answer is anything other than "yes" or "no", it will be a tacit admission on your part that you are being deliberately obstructive.

35332. JPostma at 03:03 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       See the later paper here @PC.  A higher temperature than the maximum solar insolation temperature is not observed...which means that back-radiation/trapping does not contribute to surface heating.

35333. JPostma at 03:00 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       I am offering you as much information as I possibly can.  The answer is stated precisely by the heat flow equation which I have given you.

       Let me repeat:

       "As the example you presented was of a sphere in space, then the radiant emission power is given by P = A*sigma*T^4, and this goes out into space in all directions, as a uniform wavefront basically."

       and

       "We simply apply the heat flow equation using the Stefan-Boltzmann Law. The result is that heat flow is from hotter to cooler, of course. Thus, the cooler object heats up. The cooler one does not heat the warmer one, of course, and the equation for heat flow obviously does not imply this. Equilibrium is given when the respective emissions cancel eachother out, as in Q = sigma*(Tf^4 - Ti^4). Photons are of course emitted and the equation does not say that no photons are emitted from the cooler object, as you have attempted to conjecture. It is the balance of emission which determines equilibrium, not lack of emission."

       'A' emits photons in all directions.  However, emission from 'A' does not heat a warmer object, as the heat flow equation shows that heating only occurs from hot to cold.

       I do sincerely hope this is clear for you.

35334. PhilippeChantreau at 02:56 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       The link to J.Postma's paper in the OP appears broken. Considering how confident Mr Postma is of the need for everyone here to receive his teaching on Thermodynamics, I was curious to take a look at it and check the science publication where it appeared.

       I am very surprised by this statement: " the expected results of the artificial radiative greenhouse effect, using any argument or model for it, do not manifest, thus confirming their artificiality."

       The radiative transfer models do predict with remarkable accuracy not only the downwelling IR at the surface (both quantitatively and in wave length distribution) of which real time measurements are easily accesible for some locations, but at various other altitudes too. What data is there to support the statement that the expected results do not manifest?

35335. Dikran Marsupial at 02:54 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       JPostma, yet again you avoid giving a straightforward "yes" or "no" answer, so I shall rephrase it.

       
       Does A emit any photons that strike B, "yes" or "no"?

       There can only be two answers to this question, either it does or it doesn't, so any unwillingness to give a straight "yes" or "no" will be an indication of deliberate obstruction on your part, as far as I am concerned.

35336. JPostma at 02:52 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       Again I do apologize for answering with more information, but it is necessary to simply and kindly provide clarification of the physics, such that some erroneous interpretation that a cold source can heat a warm source is prevented from being assumed or inferred, which would of course be ridiculous in any case.

35337. JPostma at 02:49 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       We simply apply the heat flow equation using the Stefan-Boltzmann Law.  The result is that heat flow is from hotter to cooler, of course.  Thus, the cooler object heats up.  The cooler one does not heat the warmer one, of course, and the equation for heat flow obviously does not imply this.  Equilibrium is given when the respective emissions cancel eachother out, as in Q = sigma*(Tf^4 - Ti^4).  Photons are of course emitted and the equation does not say that no photons are emitted from the cooler object, as you have attempted to conjecture.  It is the balance of emission which determines equilibrium, not lack of emission.

35338. Dikran Marsupial at 02:43 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       JPostma, please give a simple "yes" or "no" answer to questions where this is possible, in order to avoid ambiguity.  I will take that answer as a "yes".

       Now consider a second black body object "B" which is introduced into the scenario.  B is a plane of infinite extent and infinite conductivity (so it is the same temperature everywhere).  B is maintained at a temperature fractionally higher than that of A.  B is placed to the right of A, and in close proximity, but not actually touching, so the only form of heat transfer possible will be radiative.

       Now, as B is warmer than A, is it your contention that A will not radiate any photons that will strike B (i.e. A will only emit photons to its left and none to its right)?

       This is a question where a "yes" or "no" answer is possible.  If "yes", then no further comment is necessary as we understand eachother; if "no", please clearly state that your answer is "no" and then provide an explanation.

35339. JPostma at 02:33 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       As the example you presented was of a sphere in space, then the radiant emission power is given by P = A*sigma*T^4, and this goes out into space in all directions, as a uniform wavefront basically.

35340. JPostma at 02:31 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       Of course, thermal radiant emission goes out in all directions not otherwise blocked to it.  I am glad that we can state such basic physics concepts.

35341. Dikran Marsupial at 02:26 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       JPostma wrote "even though I have statedd "yes", by supplying the actual Stefan-Boltzman equation, which of course you had referenced initially"

       supplying the Stefan Boltzmann equation does not equate to an unambiguous answer of "yes" as my question also involved the direction in which the photons were radiated, which is not covered by the Stefan-Botzmann law (only the power).  So despite the amount of words that have been written, you still have not answered my question, which was:

       Lets try an even more simple thought experiment. Consider a spherical blackbody object (A) at a temperature slightly above absolute zero, in a total vacuum with no other sources of radiation. Do you agree that the object will radiate photons in random directions according to the Stefan-Boltzmann law? A "yes" or "no" answer ought to be possible here, if "no" please state this explicitly and explain why.

       Note I have highlighted the part of the question you have not yet addressed to make sure you don't miss it this time.

       This is not a game (at least on my part).  The reason for wanting a direct answer was to make sure there could be no misundertanding of your position.  "yes" and "no" are utterly unequivocal, which is why I designed my question to be answerable with a simple "yes" or "no".

       
       So please, for the record, is your answer to my question (including the bit about the direction in which photons are radiated) "yes" or "no"?

35342. JPostma at 02:20 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       Dikran, I am afraid that the equivocation is on your part, as you refuse to acknowledge the direct answer, and even the simple "yes" as you have requested it, which I have supplied to you.  This seems to be some game you are playing now, and have been for some time in fact.  I am sorry that I do not wish to play such a simple thing with you.  I have in fact more fully answered the question than stating "yes", even though I have statedd "yes", by supplying the actual Stefan-Boltzman equation, which of course you had referenced initially.  I am sorry if these satisfactions still do not fulfill you.

35343. Dikran Marsupial at 02:16 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       JPostma, why not simply reply "yes", rather than equivocate?  You are giving the impression of trying to avoid discussing the thought experiment by being unwilling to give simple direct answers to simple direct questions.  There is no need for equations for such a simple question where "yes" fully answers the question.

       So, is it a "yes" or a "no"?

35344. JPostma at 02:14 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       I have used the Stefan-Botlzmann law to give you the answer, as you referenced it, so I am sorry if it was missed within the body of text.

35345. JPostma at 02:13 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       I did answer that precisely, in fact, to quote:

       "An object radiates power P = A*sigma*T^4 where A is the surface area and T is the temperature. If you have two objects in a simplified geometry then you get the heat flow as the difference between their emissions, with heat flowing only from the warmer to the cooler."

       Perhaps I am giving too much information, and it distracts you from being able to infer the answer.  The answer, is of course, and I am sorry if you do not see the answer in the physics and in the math, that yes, and object radiates power as function of its surface temperature.  I do hope this helps, and I hope that you can begin to recognize the greater answer in the math and physics.

35346. Dikran Marsupial at 02:09 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       "All of the questions have been answered. "

       no, you have not answered mine, posed at 71, here it is again for your convenience

       Lets try an even more simple thought experiment. Consider a spherical blackbody object (A) at a temperature slightly above absolute zero, in a total vacuum with no other sources of radiation. Do you agree that the object will radiate photons in random directions according to the Stefan-Boltzmann law? A "yes" or "no" answer ought to be possible here, if "no" please state this explicitly and explain why.

       This ought to be uncontentious, so you should just be able to reply "yes", in which case we can move on to the next step.

35347. JPostma at 02:05 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       All of the questions have been answered.  I am honestly trying to help you guys out, because it seems to be clear that a few basic thermodynamics and physics concepts might be unknown to you.  I truly do not mean to sound patronizing or anything like that in making that statement, that is not my intention at all.  So I do apologize for that.  I have simply been assuming a certain level of ability to understand physics and thermodynamic concepts, such as the basic equation for heat flow under radiant emission, equilibrium conditions, etc.

       The lump of iron is heated by the furnace.  The iron does not heat the furnace.  The heat flow equation shows exactly how and when the iron is radiating, as would the most basic conceptual understanding of the physics stated in words.

       I hope this helped, and again, apologies for assuming certain things to be understood.

35348. Dikran Marsupial at 01:59 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       JPostma, I will not rise to your ineffectual attempts to be patronising ("Please let me know if you still require my help with this"), as I said the question is very easily answered under the modern statistical intepretation of thermodynamics, the fact that you can't (or won't) give a direct answer to MA Roger's question says it all.

       Now, perhaps you would like to answer my question (it is the start of a thought experiment, but I thought it would be best to establish something we both ought to be able to agree on as a solid foundation)?

35349. JPostma at 01:55 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       I have just stated the answer, directly.  I am sorry if you missed it again.

       Please familiarize yourself with the heat flow equation Q = sigma*(Tf^4 - Ti^4).  Is the answer still not clear?  It also answers your latest question.  Please let me know if you still require my help with this, and I will try to make it clear.  An object radiates power P = A*sigma*T^4 where A is the surface area and T is the temperature.  If you have two objects in a simplified geometry then you get the heat flow as the difference between their emissions, with heat flowing only from the warmer to the cooler.  I hope this helps.

35350. Dikran Marsupial at 01:47 AM on 17 July 2014
       Joseph E. Postma and the Greenhouse Effect
       

       JPostma, repeating "The iron will radiate what is required of it" doesn't make it any less evasive. 

       Lets try an even more simple thought experiment.  Consider a spherical blackbody object (A) at a temperature slightly above absolute zero, in a total vacuum with no other sources of radiation.  Do you agree that the object will radiate photons in random directions according to the Stefan-Boltzmann law?  A "yes" or "no" answer ought to be possible here, if "no" please state this explicitly and explain why.

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