tag:blogger.com,1999:blog-1500584444083499721.post148308665305145584..comments2024-03-24T09:28:42.755+01:00Comments on CJ on Mathematics and Science: How to Fool Yourself with a PyrgeometerClaes Johnsonhttp://www.blogger.com/profile/07411413338950388898noreply@blogger.comBlogger29125tag:blogger.com,1999:blog-1500584444083499721.post-70817434044499013782015-05-24T10:26:08.283+02:002015-05-24T10:26:08.283+02:00"but not by warming the warmer object"
D..."but not by warming the warmer object"<br />Do you mean "warming" as the first object has to be gaining in temperature from the initial level? That will not happen as both object radiates energy as heat out of body and both will decrease temperature by time. Still, as the first object increase in cooling time it has got added some extra energy, hasn't it? If so, from where did it get the extra energy if not from the cooler second object?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-15375930830997309422015-05-23T21:28:48.991+02:002015-05-23T21:28:48.991+02:00Of course surrounding cooler objects will influenc...Of course surrounding cooler objects will influence how fast a warm objects cools off, but not by warming the warmer object, just slowing the cooling.Claes Johnsonhttps://www.blogger.com/profile/07411413338950388898noreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-37335419394929782692015-05-23T17:52:02.723+02:002015-05-23T17:52:02.723+02:00Experiment design: heat an object to an exact give...Experiment design: heat an object to an exact given temperature and measure how long time it takes until temperature is equal to surroundings. repeat the experiment, but now with a second object located close holding a lower temperature. Will the cooling time for the first object increase? If so the colder object have radiated energy to the hotter object.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-50665796689910894392014-11-28T11:07:36.660+01:002014-11-28T11:07:36.660+01:00The equation used in the calculations are absolute...The equation used in the calculations are absolutely wrong as Professor Johnson has written !!!<br /><br />They have mistaken the quantity E(net) for a radiation flux and of course it isn't !!<br /><br />E(net) is the difference between two discrete fluxes and is not amenable to comparison with sigmaT^4.<br /><br />You can easily verify this for yourself by plotting 2 temperature curves in a spread sheet using Planck's equation.<br /><br />E(net) is the area between the upper and lower curve BUT you can also generate a curve by subtracting every value for wavelengths of both curves.<br /><br />This generates a curve which superficially resembles a Planck curve but is not !<br /><br />You can prove this by rewriting Planck's equation for temperature then plot the 3 curves.<br /><br />Both of the real temperature curves plot as straight lines parallel to the x axis wavelength as a constant temperature - y axis value - no surprise here.<br /><br />The curve for E(net) plots as a varying temperature value instead of a constant PROVING beyond dispute the algebraic manipulation quoted is invalid !<br /><br />Remember sigmaT^4 also is the area under the Planck curve for the temperature value - it is the integral of the Planck equation !!<br /><br />You can easily demonstrate that many of the algebraic manipulations performed in climate science do not produce curves that are Planck curves hence they DO NOT equal sigmaT^4 !!<br /><br />If the result is not a Planck curve the integral of that curve is not a valid Stefan-Boltzmann equation relationship - UNLESS you happen to have developed some relationship which replaces Planck's equation.<br /><br />Professor Johnson is totally right !!Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-28516750481295917642014-05-05T01:13:47.484+02:002014-05-05T01:13:47.484+02:00(continued)
So, as explained above, the radiation...(continued)<br /><br />So, as explained above, the radiation that emanates from a blackbody has a frequency distribution, and Wien’s Displacement Law tells us that the peak frequency increases proportionally with the temperature. That peak frequency can be seen in the above Planck curves and we should note that the area under the curve represents the total radiative flux for the indicated temperature. But it is important to understand that not all of the electromagnetic energy in the emanating radiation actually came from thermal energy in the body itself. In the case of the Earth’s surface, much of the electromagnetic energy in the emanating radiation actually comes from the electromagnetic energy in radiation from the cooler atmosphere. This radiation is immediately re-emitted by the surface without any of its electromagnetic energy being converted to thermal energy. The incident radiation from a cooler source does, however, slow the rate of radiative cooling of the warmer target because the target does not have to use as much of its own thermal energy in order to fulfil its “quota” of radiation as is determined by the area under its Planck curve. <br /><br />So, yes, the so-called “back radiation” does in fact slow down that portion of surface cooling which is itself due to radiation. However, because its energy does not go through the complicated process of being absorbed and converted to thermal energy, the back radiation can have no effect on the rate of non-radiative cooling of any planet’s surface. The only radiation that can increase the temperature of the surface must come from a hotter source, namely the Sun. Thus the Solar radiation getting through to a planet’s surface is the only radiation that plays a part in determining a planet’s surface temperature. Neither on Earth or Venus (or any other planet with a significant atmosphere) does that radiation account for the actual observed planetary surface temperatures, and this fact alone is sufficient to put to rest all the literature and “settled science” which blames global warming on back radiation from the radiating gases in the atmosphere. <br /><br />The reader might be thinking that, if back radiation slows surface cooling then it leads to warmer mean temperatures. In response we ask, “Cooling from what temperature?” The point is, the whole concept that back radiation is the cause of that “33 degrees of warming” is false, because back radiation does not add to the warming effect of the Sun.Doug Cottonhttps://www.blogger.com/profile/08564342660783793003noreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-47438277281306029552014-02-04T23:07:25.377+01:002014-02-04T23:07:25.377+01:00What is then your opinion J.R. as peer?What is then your opinion J.R. as peer?Claes Johnsonhttps://www.blogger.com/profile/07411413338950388898noreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-64185226298986651212014-02-04T22:01:14.650+01:002014-02-04T22:01:14.650+01:00I only got two words for ya .....
PEER REVIEW
Ot...I only got two words for ya .....<br /><br />PEER REVIEW<br /><br />Otherwise it's just another OPINION from a faceless blog any fool can writeLong Johnhttps://www.blogger.com/profile/10835162935515719974noreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-44745733657557738362013-07-22T23:12:10.310+02:002013-07-22T23:12:10.310+02:00Where do you get that zero Kelvin from?
"pyr...Where do you get that zero Kelvin from?<br /><br />"pyrgeometer measures a net transfer and then invents DLR by adding the net to outgoing radiation according to Stefan-Boltzmann for a blackbody emitting into a void at 0 K."<br /> <br />A blackbody does not know or care about the temperature of its surroundings.. As stated in Wikipedia and links there<br /><br />"The radiation has a specific spectrum and intensity that depends only on the temperature of the body."<br /><br />http://en.m.wikipedia.org/wiki/Black-body_radiation<br /><br />Furthermore, all objects - regardless of temperature - emit and absorb energy. <br /><br />Yes, a colder object sends energy (photons) to a warmer body which absorbs them.<br /><br />0^0https://www.blogger.com/profile/09590270703159793784noreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-12581120444830065162013-06-08T07:34:06.432+02:002013-06-08T07:34:06.432+02:00Your input is not constructive and of no interest....Your input is not constructive and of no interest. Please direct your energy to something else.Claes Johnsonhttps://www.blogger.com/profile/07411413338950388898noreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-68154979104665537992013-06-08T07:27:33.807+02:002013-06-08T07:27:33.807+02:00Read my book or web-site on black body radiation a...Read my book or web-site on black body radiation and you find my arguments and then reconsider your questions.Claes Johnsonhttps://www.blogger.com/profile/07411413338950388898noreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-88388150440569457392013-06-07T22:50:18.916+02:002013-06-07T22:50:18.916+02:00Again: Physics only concerns net transfer of heat ...<b>Again: Physics only concerns net transfer of heat energy from hot to cold, while two-way heat transfer is fiction, and a dangerous fiction because heat transfer from cold to hot violates the 2nd law.</b><br /><br />Another dangerous thing is to use sloppy terminology and ambiguous definitions.<br /><br />What exactly is heat energy? Heat is not energy. Do you understand the distinction?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-53901292983807587352013-06-07T21:30:34.884+02:002013-06-07T21:30:34.884+02:00Also, your logic doesn't make sense to me. Let...Also, your logic doesn't make sense to me. Let's take two stars, which are about the closest things we have to black bodies. Wolf 359 has a temperature of 2800 Kelvin, our Sun 5778. since they exist in the same universe, heat via radiation must be flowing between them. By your logic, since Wolf 359 is colder, it must not be emitting any heat (and therefore radiation) in our direction.thorhttps://www.blogger.com/profile/12658323151735653133noreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-62735376247139156062013-06-07T21:16:37.096+02:002013-06-07T21:16:37.096+02:00Obviously heat transfer from cold to hot does not ...Obviously heat transfer from cold to hot does not violate the second law because my refrigerator does it.<br /><br />The second law applies to a closed system in aggregate. Within that closed system, under some local partition, heat can indeed flow from cold to hot.<br /><br />But, if what you are saying is that not every algebraic expression one can form from the equations of physics has a physical manifestation, I agree with that. For a pyrgeometer, the physical interpretation is obvious. You don't like it, but it's obvious.thorhttps://www.blogger.com/profile/12658323151735653133noreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-50415141108423216392013-06-07T19:58:28.911+02:002013-06-07T19:58:28.911+02:00Again: Just because a certain algebraic operation ...Again: Just because a certain algebraic operation seems to be possible, that is to write <br />the heat transfer Q between two black bodies of temperature s T_h > T_c given as Q = sigma * (T_h - T_c) as the net transfer from hot to cold, formally as Q = sigma * T_h - sigma * T_c, does not mean that physically there is a two transfer of energy between hot and cold with in particular sigma *T_c the heat transfer from cold to hot. <br /><br />Again: Physics only concerns net transfer of heat energy from hot to cold, while two-way heat transfer is fiction, and a dangerous fiction because heat transfer from cold to hot violates the 2nd law. <br /><br />If you say that the heat transfer from cold to hot is conditional with always larger transfer from hot to cold, then you have to give the mechanism guaranteeing this condition to be satisfied, since it is not described in the literature. So whats is then your explanation of this mechanism?Claes Johnsonhttps://www.blogger.com/profile/07411413338950388898noreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-52686503897836777482013-06-07T19:27:14.868+02:002013-06-07T19:27:14.868+02:00I find the idea that physicists won't perform ...I find the idea that physicists won't perform an algebraic operation available to them quite hard to swallow. I find the idea that you think one can't perform an algebraic operation unless you already find it that way in the physics literature astounding.<br /><br />In any case, here's a link. Look at equations 1 and 2. You can see them go between the two forms with T4 in and T4 out. Note who the authors are. Turns out, Mahan wrote the book on Radiative Heat Ransfer.<br /><br />http://www.arm.gov/publications/proceedings/conf09/extended_abs/haeffelin_m.pdf<br /><br />http://www.amazon.com/Radiation-Heat-Transfer-Statistical-Approach/dp/0471212709/ref=sr_1_1?ie=UTF8&qid=1370625663&sr=8-1&keywords=J.+R.+Mahan+heat<br /><br />thorhttps://www.blogger.com/profile/12658323151735653133noreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-15019767773984176332013-02-24T21:51:33.237+01:002013-02-24T21:51:33.237+01:00You are confusing algebra with physics. An algebra...You are confusing algebra with physics. An algebraic manipulation does not necessarily have physical meaning. Your conception is too simplistic.Claes Johnsonhttps://www.blogger.com/profile/07411413338950388898noreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-5379555460079081922013-02-24T21:14:47.070+01:002013-02-24T21:14:47.070+01:00You want to be a "Professor of Applied Mathem...You want to be a "Professor of Applied Mathematics" and can't transform <br />E_net = sigma Ta^4 - sigma Te^4<br />to<br />E_net = sigma (Ta^4 - Te^4)?<br /><br />So you fail at mathematics for 8th grade pupils, you also fail to grasp simple physical concepts like conservation of energy and get lost in simple sums by making them unnecessarily complicated.<br /><br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-65427807742320408102012-02-07T03:13:57.069+01:002012-02-07T03:13:57.069+01:00For calibration of pyrgeometers using other instru...For calibration of pyrgeometers using other instruments see http://www.arm.gov/publications/proceedings/conf16/extended_abs/stoffel_t.pdfAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-68658477370244201772012-01-24T12:27:19.219+01:002012-01-24T12:27:19.219+01:00Neither a pyrgeometer nor the Earth's surface ...Neither a pyrgeometer nor the Earth's surface are anything like a perfect blackbody. A blackbody is assumed to be one surrounded by a vacuum, eg the whole Earth+atmosphere system surrounded by space. There can be no means of energy transfer other than radiation. In other words, the blackbody is perfectly insulated. <br /><br />But the Earth's surface (including oceans) is nothing like perfectly insulated. It can gain or lose thermal energy both from the atmosphere and the sub-surface. Indeed it does this every day when SW radiation is converted to thermal energy and then conducted into deeper layers of the surface, only to exit again that night. Neither the surface nor the instrument needs to emit all the thermal energy by radiation. Hence any application of SBL is doomed to be significantly inaccurate. What a joke that assumed -18 deg.C surface temperature is. And what a joke are the "measurements" of backradiation.Doug Cottonhttp://climate-change-theory.comnoreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-858315655404145522011-09-04T14:09:53.049+02:002011-09-04T14:09:53.049+02:00Yes, we agree.Yes, we agree.Claes Johnsonhttps://www.blogger.com/profile/07411413338950388898noreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-44525678929821215392011-09-04T13:14:25.504+02:002011-09-04T13:14:25.504+02:00I think that the improper use of the pyrgeometer a...I think that the improper use of the pyrgeometer and/or any other room temperature antenna unquestionably shows the total and contextual misunderstanding of the metrology and of the thermal radiation.<br />The aforesaid tools are simple dynamic calorimeter, i.e., devices that measure instantaneously the output electric property induced by the difference between the sensor temperature and the room temperature and hence the heat flux that the sensor exchanges with the optically sighted region. <br />Until now, all that is correct for the physics and the metrology.<br />But here starts the fairy. The heat flux is assumed as NET FLUX due to the difference between the incoming and outgoing long wave radiation flux and that allows to compute the incoming flux. <br />This is neither a physic measure nor a proof. I think it is only a banal trick! <br />MicheleAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-30033226080869835112011-08-14T01:12:47.987+02:002011-08-14T01:12:47.987+02:00When reality does not correspond to the map (model...When reality does not correspond to the map (models), of course reality must be wrong... [irony]<br /><br />To be honest, I'm a little freaked over contributing to this professors salary (I'm a Swedish citizen).Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-61741054071974214722011-08-12T13:43:32.904+02:002011-08-12T13:43:32.904+02:00Huh? By summing up DLR(lambda) over the wavelength...Huh? By summing up DLR(lambda) over the wavelengths measured you get a total incoming energy E_in which matches what the pyrgeometer sees. That same constant value that's independent of the T of the pyrgeometer. You can call it a fiction if you like, but an invariant quantity observed many different ways independent of the measuring instrument is normally regarded as something with real physical reality.Arthurhttps://www.blogger.com/profile/06249922708053689717noreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-38187755987410069782011-08-12T13:40:14.357+02:002011-08-12T13:40:14.357+02:00I don't deny that you can design an instrument...I don't deny that you can design an instrument which is sensitive to IR light. It is the connection to transfer of heat energy which I doubt based on evidence I present.Claes Johnsonhttps://www.blogger.com/profile/07411413338950388898noreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-90308671663370631712011-08-12T13:31:46.862+02:002011-08-12T13:31:46.862+02:00Huh? I earlier pointed you to a page at ScienceOfD...Huh? I earlier pointed you to a page at ScienceOfDoom that has half a dozen different infrared spectroscopy measurements of DLR, and you say it's a fiction because Wikipedia doesn't link to it? Here's the ScienceOfDoom link again.<br /><br />http://scienceofdoom.com/2010/07/24/the-amazing-case-of-back-radiation-part-two/<br /><br />Those measurements were made by pointing an infrared spectrometer up at the sky and seeing what the spectrum of DLR was. Do you deny these measurements exist?Arthurhttps://www.blogger.com/profile/06249922708053689717noreply@blogger.com