tag:blogger.com,1999:blog-1500584444083499721.comments2018-02-18T21:51:07.358+01:00Claes Johnson on Mathematics and ScienceClaes Johnsonhttp://www.blogger.com/profile/07411413338950388898noreply@blogger.comBlogger4350125tag:blogger.com,1999:blog-1500584444083499721.post-52000305049645403362018-01-27T16:17:44.156+01:002018-01-27T16:17:44.156+01:00Åsa W fick smäll på fingrarna för begreppet "...Åsa W fick smäll på fingrarna för begreppet "faktaresistens", hennes meddebattör liknde det vid antibiotikaresistens, alltså att den "faktaresistente" indirekt jämställs med en sjukdomsframkallande bakterie. Intressant, ungefär som begreppet "förnekare". <br /><br /> Bengt Abelssonhttps://www.blogger.com/profile/03138267313188917270noreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-21953767443521010182018-01-22T15:50:34.180+01:002018-01-22T15:50:34.180+01:00Ungefär som att be en Imam att byta religion.........Ungefär som att be en Imam att byta religion......Jan Lindströmhttps://www.blogger.com/profile/14242998903871902230noreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-76435839515137329872018-01-20T07:30:04.010+01:002018-01-20T07:30:04.010+01:00prof premraj pushpakaran writes -- 2018 marks the ...prof premraj pushpakaran writes -- 2018 marks the 100th birth year of Julian Seymour Schwinger!!!prof prem raj pushpakaranhttps://www.blogger.com/profile/14561237920972677898noreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-62015628759585754302018-01-18T16:04:04.482+01:002018-01-18T16:04:04.482+01:00Well, what I menar to ”endorse” was the gravito-th...Well, what I menar to ”endorse” was the gravito-thermal effect and not the specific idea of a mean value temperature on the Moon of 198 K suggesting an effect of the presence of the atmosphere as high as 90 C. I still think that a value 15-20 C suggested by a direct application of Stefan-Boltzmann could be more relevant, but the whole idea of comparing with a hypothetical Earth without atmosphere is questionable. In the same sprit one could ask about an Earth temperature without a Sun, for which an answer could be given but it would say very little...Claes Johnsonhttps://www.blogger.com/profile/07411413338950388898noreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-87108546461238169362018-01-18T09:30:00.034+01:002018-01-18T09:30:00.034+01:00This post
http://claesjohnson.blogspot.no/2017/07/...This post<br />http://claesjohnson.blogspot.no/2017/07/inflated-greenhouse-effect.html Petter Tuvneshttps://www.blogger.com/profile/01317185038067784642noreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-13375997829754085322018-01-17T22:22:06.356+01:002018-01-17T22:22:06.356+01:00Which post?Which post?Claes Johnsonhttps://www.blogger.com/profile/07411413338950388898noreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-7402359522427951232018-01-15T16:48:04.769+01:002018-01-15T16:48:04.769+01:00In an earlier post you endorsed the Nikolov & ...In an earlier post you endorsed the Nikolov & Zeller paper with the 90 K moon comparison.<br />Have you changed your mind completely now?Petter Tuvneshttps://www.blogger.com/profile/01317185038067784642noreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-62073641685719547162017-12-28T08:49:46.582+01:002017-12-28T08:49:46.582+01:00Get this Non-Contact #infraredthermometer that del...Get this Non-Contact #infraredthermometer that delivers quick, easy, and accurate readings. Just point and shoot this Smart Sensor device for temperature readings.<br /><a href="https://www.amazon.com/Professional-Clinical-Non-contact-Infrared-Thermometer/dp/B006JTNXDW" rel="nofollow"> 2 in 1 Professional Infrared Thermometer </a>harish devhttps://www.blogger.com/profile/01453582554007504213noreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-76683074101492545812017-12-28T08:49:38.818+01:002017-12-28T08:49:38.818+01:00Get this Non-Contact #infraredthermometer that del...Get this Non-Contact #infraredthermometer that delivers quick, easy, and accurate readings. Just point and shoot this Smart Sensor device for temperature readings.<br /><a href="https://www.amazon.com/Professional-Clinical-Non-contact-Infrared-Thermometer/dp/B006JTNXDW" rel="nofollow"> 2 in 1 Professional Infrared Thermometer </a>harish devhttps://www.blogger.com/profile/01453582554007504213noreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-43303664122720818992017-12-21T10:27:07.394+01:002017-12-21T10:27:07.394+01:00You are making a fundamental mistake.
At the mom...You are making a fundamental mistake. <br /><br />At the moment I don't have time to go into detail, but for your conclusions about energy transport by heat, following conclusions from Fourier's law, you are assuming a general validity of Fourier's law. And more specific, you are begging the question about the direction of the flow, since the assumption of one-directional flow is an pre-assumption for the validity of the Fourier law.<br /><br />But Fourier's law is not true in general, it is an approximation, as been known for long. It has since way back been noted that Fourier's law is incomplete due to lack of inertial response to thermal perturbations. More specific, this makes the law violate microscopic reversibility, and based on how you model yourself, I don't think that is something you would promote violation of. When it comes to radiative exchange, between spatially separated bodies, thing would get even worse... Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-82154501217179521412017-12-19T08:01:43.768+01:002017-12-19T08:01:43.768+01:00My analysis of radiative heat transfer is presente...My analysis of radiative heat transfer is presented in detail at https://computationalblackbody.wordpress.com . And yes, heat transport by conduction and radiation are similar as energy transport scaling with temperature difference warm to cold.Claes Johnsonhttps://www.blogger.com/profile/07411413338950388898noreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-35744638618669307842017-12-19T00:36:55.300+01:002017-12-19T00:36:55.300+01:00You do understand that Fourier's law is a phen...You do understand that Fourier's law is a phenomenological law, right?<br /><br />The thermal conductivity \(\kappa\) in a material is a fundamentally different property than the \(\beta\) (as introduced by Kevin above) in radiative energy transport.<br /><br />Would be interesting to see if you actually can explain the difference between thermal conduction and thermal radiation.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-84378171868332511722017-12-18T21:32:41.898+01:002017-12-18T21:32:41.898+01:00What does ”radiate onto” mean? Transfer of heat en...What does ”radiate onto” mean? Transfer of heat energy? From cold to warm?Claes Johnsonhttps://www.blogger.com/profile/07411413338950388898noreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-47836541908165834622017-12-18T20:49:43.247+01:002017-12-18T20:49:43.247+01:00Lets say I have two light bulbs, one stronger (hot...Lets say I have two light bulbs, one stronger (hotter) than the other, and the strong one is turned off, while the weak one is turned on. Then the weaker one radiates onto the stronger one. Then I turn on the stronger one. Then you say that the weaker one stops radiating onto the stronger one; is that correct?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-65275034847652641092017-12-18T11:05:50.489+01:002017-12-18T11:05:50.489+01:00You speak about "energy density" of some...You speak about "energy density" of something. I consider transfer of heat energy by radiation between bodies, which is similar to transfer by conduction in a dependence on temperature difference: The one-way transfer of heat energy from the inside of your house to a colder exterior by conduction depends on the temperature difference between inside and outside, and the same holds for radiation. Claes Johnsonhttps://www.blogger.com/profile/07411413338950388898noreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-58801721085163508732017-12-17T12:40:56.939+01:002017-12-17T12:40:56.939+01:00I wonder if you read the comment by Kevin above. I...I wonder if you read the comment by Kevin above. It clearly shows how you misunderstand the situation.<br /><br />The energy density for the radiation being in thermal equilibrium is of course only dependent on the temperature of the body it is interacting with. The ambient temperature can not be involved here, otherwise there is no thermal equilibrium. This is very reasonable since the time it takes for the radiation to reach equilibrium is much quicker than the time it takes for the body to change its temperature. The heat capacity of the radiation is many, many, many orders of magnitudes smaller then that for solids, liquids and ordinary gases (if I remember correctly there is a ratio of about 10^-12 to liquids for constant volume capacities). <br /><br />The energy density for the radiation in equilibrium can easily be shown to follow a T^4 dependence, where T is the temperature of the body. When the radiation that is in thermal equilibrium leaves, it has the same energy density at that instance. So the assumption that this is a formula for radiation to a 0K background is very flawed. It must be independent of the temperature of the surroundings. How could it not??Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-17757137866533147072017-12-16T20:42:50.125+01:002017-12-16T20:42:50.125+01:00It is stated that the outgoing radiation is postul...It is stated that the outgoing radiation is postulated to be sigma T^4 with T the pyrgeometer/Earth temperature from which the incoming radiation is computed. But the outgoing sigma T^4 is the radiation to an environment at 0 K which is not the temperature of the atmosphere (which is what is effectively being measured). So you have not properly understood the setting.Claes Johnsonhttps://www.blogger.com/profile/07411413338950388898noreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-41871250725977696892017-12-16T17:58:38.865+01:002017-12-16T17:58:38.865+01:00Ah, forgot the point :)
The point is that the Fun...Ah, forgot the point :)<br /><br />The point is that the Fundamental law of calculus has completely nothing to do with the device.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-28227916172809009722017-12-16T17:56:53.750+01:002017-12-16T17:56:53.750+01:00I think you misunderstand the pyrgeometer.
It cle...I think you misunderstand the pyrgeometer.<br /><br />It clearly says in the text in the picture that it is the outgoing energy (red arrows) that is assumed to be calculated with the Stefan-Boltzmann law.<br /><br />Hence the incoming is not radiation following a Stefan-Boltzmann law, so there is not a relation of the type Q = sT2^4 - sT1^4.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-81693729525719452782017-12-15T01:20:45.719+01:002017-12-15T01:20:45.719+01:00Yes, that looks like non-interpretable voodoo (for...Yes, that looks like non-interpretable voodoo (for instance, the integral in (*) should have the limits 0 to \(\infty\), right?). <br />Isn't it better that we look at the physical origin of the blackbody radiation energy density and its relation to \(Q\) instead?<br /><br />We only need to use classical thermodynamics for this.<br /><br />First, we note that for radiation to be in thermal equilibrium with matter, the matter only needs to be sufficiently optically dense, so that enough interaction takes place for the radiation to equilibrate. This means that both solids, liquids, gases and even a hollow cavity can function to equilibrate the radiation, given sufficient optical depth. This is essentially the physical meaning of a blackbody.<br /><br />To make the discussion meaningful, we make the ordinary assumption that the matter functions as a thermal bath for the radiation. That is that the temperature of the matter changes on a much larger timescale, so that we can treat it as having a constant temperature.<br /><br />We now note that the most important property of blackbody radiation is that it is radiation in <b>thermal equilibrium</b> with matter at temperature \(T\). Using only <b>classical thermodynamics</b>, we can then quite easily show that the energy density of such radiation is only dependent on the temperature of the matter, according to the formula<br /><br />\(u(T) = \beta T^4\),<br /><br />where \(\beta\) is some constant. We also note that we don't need the Planck spectrum for this result, the result relies only on the facts that electromagnetic radiation carries energy and momentum which gives us a relation between the radiation pressure and the energy density.<br /><br />When the radiation in thermal equilibrium escapes the matter, the energy density of the radiation maintains the same as when in equilibrium with the matter, since it no longer interacts. Looking at the enclosing area this escaping energy density can be interpreted as a radiation flux.<br /><br />For incoming radiation, the situation we have is reversed. The incoming radiation is not interacting until it starts to equilibrate with the thermal bath. Because of our assumption of sufficient optical thickness the incoming radiation will eventually equilibrate.<br /><br />We now get the quantity that you call \(Q\), comes naturally from ordinary energy conservation over the enclosing surface.<br /><br />We note that this physical picture is a much more general one, since the incoming radiation not necessarily needs to be thermal radiation. This better reflects a real situation when two bodies at different temperatures exchange energy by radiation, black or otherwise.Kevinnoreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-7674003275292384182017-12-10T18:24:36.757+01:002017-12-10T18:24:36.757+01:00ELECTROMAGNETIC WAVES, (photons) colliding against...ELECTROMAGNETIC WAVES, (photons) colliding against charged particles, photoelectrical effect, of the plasma-jets HH which emerge from the black holes central galactic and new young stars in formation, protostars, forming in the jet "knots" more shining each a certain distance... GRAVITATIONAL WAVES (gravitophotons), unknown yet, come from the deep Universe since the beginning of the times. Is not a relativistic affaire "curvature" of that called "space-time".tonyonhttps://www.blogger.com/profile/08253501266473243514noreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-6607088574789454552017-12-09T21:54:50.053+01:002017-12-09T21:54:50.053+01:00A certain temperature has a maximum frequency, thu...A certain temperature has a maximum frequency, thus temperature and frequency are not completely independent. check the Planck curves.Petter Tuvneshttps://www.blogger.com/profile/01317185038067784642noreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-81229103165867061952017-12-07T00:21:36.127+01:002017-12-07T00:21:36.127+01:00Petter Tuvnes, you wrote
"Temperature and fr...Petter Tuvnes, you wrote<br /><br />"Temperature and frequency are related."<br /><br />How is temperature and frequency related? <br /><br />Are they not both independent variables in Planck's radiation law?Mikenoreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-53369940276298769702017-12-06T07:47:33.367+01:002017-12-06T07:47:33.367+01:00Anonym's assertion will imply that low frequen...Anonym's assertion will imply that low frequency radiation from a low temperature object will make a hotter "blackbody" hotter still. Not credible.Petter Tuvneshttps://www.blogger.com/profile/01317185038067784642noreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-21347851526107452432017-12-05T12:24:03.126+01:002017-12-05T12:24:03.126+01:00You are missing that a black-body thermalize the r...You are missing that a black-body thermalize the radiation it absorbs. Internal conversion processes during this thermalization (e.g. phonon collision) can convert two lower energy states to a higher energy state, that is later relaxed at a higher frequency.Anonymousnoreply@blogger.com