'06-09-11, 00:01 upriver	The visible light from the sun is not generated in the photosphere, it only passes through the photosphere from below. What does that yellow stuff do on the top of the photosphere? Is it dark under the photosphere(sunspots)? Is it cooler or hotter under the photosphere? Scientists Image the Three-dimensional Surface of the Sun http://www.lmsal.com/Press/SPD2003.html Here is the link to the The Institute for Solar Physics solar gallery. http://www.solarphysics.kva.se/
'06-09-11, 07:05 captain swoop	Is it dark under the photosphere(sunspots)? Sunspots aren't dark.
'06-09-11, 20:07 papageno	upriver wrote:   Originally Posted by papageno Since the Einstein coefficients give the probability for a transition, which can occur only if the frequency is nu = E/h, why does Dr. Robitaille think that the derivation requires a single atom to be a perfect absorber of radiation over the whole spectrum? The derivation is for a theoretical blackbody continuous emission spectrum. That's not the point. Read again the relevant passage: Dr. Robitaille wrote: In his derivation of the Planckian relation, Einstein has recourse to his well-known coefficients [10,11]. Thermal equilibrium and the quantized nature of light (E=hnu) are also used. All that is required appears to be 1) transitions within two states, 2) absorption, 3) spontaneous emission, and 4) stimulated emission. However, Einstein also requires that gaseous atoms act as perfect absorbers and emitters or radiation. In practice, of course, isolated atoms can never act in this manner. In all laboratories, isolated groups of atoms act to absorb and emit radiation in narrow bands and this only if they possess a dipole moment. This is well-established in the study of gaseous emissions [12]. As such, Einsteins requirement for a perfectly absorbing atom, knows no physical analogue on earth. In fact, the only perfectly absorbing materials known, exist in the condensed state. Nonetheless, for the sake of theoretical discussion, Einsteins perfectly absorbing atoms could be permitted. (my emphasis) Dr. Robitaille attributes to Einstein the assumption that an atom is a perfect absorber over the whole spectrum. But this is a strawman, because the derivation does not require such an assumption. upriver wrote: As such it is applied to all cases without the understanding that emission for gas or liquid is different than condensed matter. Part of the problem is that any spectrum hotter than 3500K is electrically driven is a discharge tube or plasma. We have no examples of thermally heated condensed matter above 3500, so its been assumed for a long time that this was correct, that bb emission extended to all matter. In this way by default, a single atom is required to act the same as a brick of graphite. I think that is the correct answer. It does not even address my question: why does Dr. Robitaille say that Einstein's derivation requires that an atom is a perfect absorber over the whole spectrum? This is about the theoretical derivation, which does not care about the practical limitations of experiments. Do you remember how Planck used the Hertzian linear oscillator? He used them as a probe to determine the energy density in the cavity at a given frequency. To find the energy density at a different frequency, he used oscillators with different natural frequency. upriver wrote:   Originally Posted by papageno Where does Dr. Robitaille actually show that the small piece of graphite dominates the radiation? Where does he show that "inserting a small piece of graphite into the perfectly reflecting enclosure" equals "Kirchhoff lined the entire box with graphite"? I dont know. I will email him. Also see below.   Originally Posted by papageno Poor old Kirchhoff! How long should he have waited? I would expect thermal equilibrium could have been deduced by the rate of thermal change, and then wait a little longer. If the rate of change is too small, you probably could assume it(a highly polished metal enclosure) will never reach thermal equilibrium in a reasonable amount of time(5 yrs?). I also said: papageno wrote: Where does Dr. Robitaille show that the time required to reach this equilibrium was short enough for Kirchhoff to wait and still have a meaningful experiment? upriver wrote: Now if I were to examine this particular idea from an upriver ATM point of view, I would say it would never reach equilibrium because of one bit I ran across by John Bedini from his webpage on Nathan Stubblefield. [snip!] Since when is a web-page about free energy a reliable source? What about my references? papageno wrote: G. Machin and B Chu, "A transportable gallium melting point blackbody for radiation thermometry calibration", Meas. Sci. Technol. 9, 1653-1656 (1998) E. Usadi, "Reflecting cavity blackbodies for radiometry", Metrologia 43, S1-S5 (2006) upriver wrote: Of course its not peer reviewed, but its so strange, how could you make that up as an effect. Your lack of imagination is not evidence. upriver wrote: If there was even the smallest effect related to this, the box would never be at equilibrium. And in the case of the graphite, I think it destroyed whatever effect was happening in the box(two polished parallel surfaces?). There is a huge leap between "I think" and "It is". Can you bridge that gap with something substantial?
'06-09-11, 11:45 korjik	upriver wrote: If somebody can produce a BB emission(vs lines) on earth in a lab with a plasma like the photosphere I will retract my claim. Just remembered what I wanted to say. I think the that it is pretty much impossible to measure the blackbody radiation of photosphere conditions on the earth due to technical and/or financial considerations, not physics. the technical considerations I can think of are that you would be trying to read the blackbody radiation of a transparent, hot gas without seeing the radiation of the background vaccum chamber. Just a quicky thought of mine on how to do it would be to inject the hot gas into a nitrogen cooled chamber. It would still be very hard to see even in an immense chamber. The financial aspect would be that the above would be hideously expensive.
'06-09-11, 11:48 korjik	Upriver, if the sun's surface is a solid, why does limb darkening fit a cooler temperature blackbody curve? Shouldnt it be a scattered constant temperature curve?
'06-09-11, 14:12 Tim Thompson	15,000,000 Kelvins Tim Thompson wrote: The visible light from the sun is not generated in the photosphere, it only passes through the photosphere from below. upriver wrote: What does that yellow stuff do on the top of the photosphere? It looks to me like that "yellow stuff" is the photosphere. The image is taken at different wavelengths that the images you were showing before, which reveal features well above this image, which are invisible here. upriver wrote: Is it dark under the photosphere (sunspots)? No. if you could see it revealed, "under the photosphere" would be considerably brighter than the photosphere, by virtue of being hotter. But the emission would tend toward ultraviolet & shorter wavelengths. "Under the photosphere", deep enough, would actually "look" relatively dark, because there would be few photons left, with wavelengths so long as visible light. Sunspots are relatively dark because they are relatively cool. They may extend below the photosphere, I am not sure about that. But I think they are for the most part entirely contained in the photosphere. upriver wrote: Is it cooler or hotter under the photosphere? Hotter. Where the photosphere temperature is about 5700 K, the core temperature is about 15,000,000 K. The temperature rises steadily from the photosphere down towards the core.
'06-09-12, 09:37 tusenfem	From previous discussions on this topic, I still have not obtained an answer to some questions. The main question is: If the sun is a shell of solid iron (powered by some mysterious electric sun idea) how thick is this shell and do you think that this iron shell will be stable at a temperature of 5700 K?
'06-09-13, 00:24 upriver	It does not even address my question: why does Dr. Robitaille say that Einstein's derivation requires that an atom is a perfect absorber over the whole spectrum? This is about the theoretical derivation, which does not care about the practical limitations of experiments. Do you remember how Planck used the Hertzian linear oscillator? He used them as a probe to determine the energy density in the cavity at a given frequency. To find the energy density at a different frequency, he used oscillators with different natural frequency. I emailed Dr. Robitaille, as soon as I have a response, we will revisit this issue.   Since when is a web-page about free energy a reliable source? Nathan Stubblefield by any accounts was a pioneer. It seems there was a whole field of "earth" energy that was being discovered. http://en.wikipedia.org/wiki/Nathan_Stubblefield http://smart90.com/nathanstubblefield John Bedini. He invented the Audio Clarifier. He is also a "free energy" researcher. He hosted the Nathan Stubblefield page. http://www.icehouse.net/john1/stubblefield.html Forget the link on the top of the page and scrolldown. "As a result of the disc manufacturing process the disc has inherent noise distortion which becomes more apparent as the disc is played over and over again; coupled with the polymer ability to hold electrostatic charges, it acutely masks the true dynamics of the digital media." Bedini Electronics http://www.bedini.com/   What about my references?   Quote: Originally Posted by papageno G. Machin and B Chu, "A transportable gallium melting point blackbody for radiation thermometry calibration", Meas. Sci. Technol. 9, 1653-1656 (1998) E. Usadi, "Reflecting cavity blackbodies for radiometry", Metrologia 43, S1-S5 (2006) "Reflecting cavity blackbodies for radiometry" is a return to the first case since it is not a 100% reflective apparatus. And its spherical! "A transportable gallium melting point blackbody for radiation thermometry calibration" is based on a transportable source that maintains its accuracy. It's interesting that gallium maintains it's melting point under "any" condition. The issue is gas/plasma only emits lines. Condensed matter emits a BB curve that is distinguishable from a pressurized gas curve.   Can you bridge that gap with something substantial? I guess I would have to mount two polished metal plates parallel and see what happens. From what I've read copper and steel.   Your lack of imagination is not evidence. That's funny!!!
'06-09-13, 00:56 upriver	Originally Posted by korjik Upriver, if the sun's surface is a solid, why does limb darkening fit a cooler temperature blackbody curve? What temperature are you talking about? Cooler than 6000K of the photosphere? I would expect it to be at the temperature of solid iron. The angle dependance of thermal emission would naturally have that effect of darkening depending on the angle. Siegel R., Howell J. Thermal radiation heat transfer. 4th ed., Taylor and Francis, New York, 2002,.   Shouldnt it be a scattered constant temperature curve? Not sure what you mean.
'06-09-13, 03:00 upriver	It looks to me like that "yellow stuff" is the photosphere. The image is taken at different wavelengths that the images you were showing before, which reveal features well above this image, which are invisible here. The images from the solar telescope are from 396nm to 630nm, which is visible light. The image I posted before are from 171 to 195A or 17.1nm to 19.5nm(EUV). With magnetogram pictures being 630nm. Interesting that they take pictures of the suns magnetic field with iron plasma(FeI) in visible light. And if I was to just look at those pictures, I would say that most of the suns visible light comes from the top layer of the photosphere. Clearly the center of the sunspot is either emitting UV or IR because its not emitting visible(396 to 630nm). Now in this picture taken at 17.1nm,there appear to be areas that are opaque to EUV in this layer that in the MS view is above the photosphere.     Quote: Originally Posted by upriver Is it dark under the photosphere (sunspots)? Originally Posted by Tim Thompson No. if you could see it revealed, "under the photosphere" would be considerably brighter than the photosphere, by virtue of being hotter. You mean to tell me that if I put a cool thing between two hot things it will stay cool? The quoted thickness for the photosphere is 400km. That requires new physics. Or a different model.   Originally Posted by Tim Thompson But the emission would tend toward ultraviolet & shorter wavelengths. Well the only UV comes from the coronal loops as you can see by the picture, because it certainly is not coming from inside the sun as the loops are providing the only illumination(reflected and direct). It is a filter at 17.3nm center with a bandwidth of .64nm or 6.4A. As a matter of fact if you look in the center of the picture you can see a feature blocking a flare. That feature is 20,000km high. And that feature is opaque to light from emission of 160,000 to 2 million degrees (FeIX).   "Under the photosphere", deep enough, would actually "look" relatively dark, because there would be few photons left, with wavelengths so long as visible light. So you saying that short wavelength radiation converts to visible right at the photosphere? So what makes the photosphere brighter? At wavelengths from 396nm to 630nm?   Sunspots are relatively dark because they are relatively cool. They may extend below the photosphere, I am not sure about that. But I think they are for the most part entirely contained in the photosphere " Sunspots are "dark" because they are cooler than their surroundings. A large sunspot might have a central temperature of 4,000 K (about 3,700 C or 6,700 F), much lower than the 5,800 K (about 5,500 C or 10,000 F) temperature of the adjacent photosphere." http://www.windows.ucar.edu/sun/atmo.../sunspots.html Not only that, a sunspot is deeper than the surrounding photosphere. The magnetic field extends through the sunspot into the surface of the sun. Both NASA and ESA indicate the the coronal loops originate below the photosphere.
'06-09-13, 03:38 upriver	Originally Posted by tusenfem The main question is: If the sun is a shell of solid iron (powered by some mysterious electric sun idea) how thick is this shell Since iron is more massive than the plasma that the MS sun is "composed' of, it would be the equivalent to the amount required to produce the 1.4 density figure that is based on the earths orbit. You did the math before. It really doesnt matter since we know the surface is iron and it has not collapsed and it has a certain inferred density. There is some experimental evidence that arcs(z-pinch) produce hollow iron spheres. Supernova have iron cores. And a barrel shaped(z-pinch) remnants. Check out welding slag, etc. I cant see inside the sun so anything I say is going to be a guess just like everyone else.   and do you think that this iron shell will be stable at a temperature of 5700 K? The surface is not at 5700K That is the average temperature of the surface, not the spot temperature. The surface could be a glowing and still be solid(cooler) deeper. Only in the bright spots is that really true.
'06-09-13, 03:42 Forskern	upriver wrote: I would say that most of the suns visible light comes from the top layer of the photosphere. Clearly the center of the sunspot is either emitting UV or IR because its not emitting visible(396 to 630nm). Actually the photosphere is defined loosely as "where most of the visible light comes from", so thats a safe assumption :P More soberly though, it is easy to compare the intensity (erg//cm^2/sterad/s) measured in UV and visible light and confirm that the coronal emission is insignificant in the energy budget. Also, the sunspots are giving off visible light just like the area around them, but at a lower BB temperature. That just makes them look black in photographs. While the active regions that are often associated with them give off more EUV than the rest of the solar disc, I'd still expect that most of the energy is still radiated in the visible, not in UV. upriver wrote: Now in this picture taken at 17.1nm,there appear to be areas that are opaque to EUV in this layer that in the MS view is above the photosphere. Where in that picture are those feature you believe are blocked out by opacity? upriver wrote: You mean to tell me that if I put a cool thing between two hot things it will stay cool? I think he means that when you turn of the heating, it will get colder. This is just what happens, the outer portion of the Sun (from about two thirds of the solar radius) is convectively heated. While the usual radiative heat transfer still works here, it is not strong enough to keep the spots at the same 6000 degrees as the convectively heated rest.
'06-09-13, 05:06 captain swoop	upriver wrote: It really doesnt matter since we know the surface is iron and it has not collapsed ... We don't know this at all.
'06-09-13, 07:24 tusenfem	upriver wrote: Since iron is more massive than the plasma that the MS sun is "composed' of, it would be the equivalent to the amount required to produce the 1.4 density figure that is based on the earths orbit. You did the math before. It really doesnt matter since we know the surface is iron and it has not collapsed and it has a certain inferred density. Uh, no, you keep on claiming that the sun has an iron surface, with no evidence to back up your claim. upriver wrote: There is some experimental evidence that arcs(z-pinch) produce hollow iron spheres. Supernova have iron cores. And a barrel shaped(z-pinch) remnants. Check out welding slag, etc. I know of no such experimental evidence. Oh, I know you are obsessed by z-pinches, but having them create hollow iron spheres ... ? And I would hope that welding does not create hollow spheres, might not be real stable, I guess. upriver wrote: I cant see inside the sun so anything I say is going to be a guess just like everyone else. So, just because you cannot see inside the sun, you can claim you are right? How about helioseismology, a well developed technique, which shows no evidence at all of a solid iron shell anywhere near the bottom of the photosphere. upriver wrote: The surface is not at 5700K That is the average temperature of the surface, not the spot temperature. The surface could be a glowing and still be solid(cooler) deeper. Only in the bright spots is that really true. [/quote] This does not make sense. So you want to have a BB spectrum of a body with a temperature of 5700 K, and since (in your view) a BB can only come from a solid, you need an iron shell of this temperature. Now you tell us that the shell may be cooler and only some hot spots hotter, but will the average of these give the nice BB at 5700 K? And, just another comment, you do not see the top of the photosphere, you see the bottom, actually you see where tau = 2/3. That is where the photons come from, because that is the location from which they are able to escape. And, one more comment, you cannot look at a line spectrum (your iron line showing the magnetic loops) and say that there are opague regions. It just does not work that way. In an emission line you see where the iron is, and it is dark where it is not.
'06-09-13, 09:28 papageno	upriver wrote:   Originally Posted by papageno Since when is a web-page about free energy a reliable source? Nathan Stubblefield by any accounts was a pioneer. It seems there was a whole field of "earth" energy that was being discovered. [snip links!] How about going to a library? upriver wrote:   Originally Posted by papageno What about my references? "Reflecting cavity blackbodies for radiometry" is a return to the first case since it is not a 100% reflective apparatus. And its spherical! "A transportable gallium melting point blackbody for radiation thermometry calibration" is based on a transportable source that maintains its accuracy. It's interesting that gallium maintains it's melting point under "any" condition. The references were provided to disprove claims, based on Dr. Robitaille's paper, that graphite is the only available black-body: Dr. Robitaille wrote: Eventually, graphites behavior became the basis of the laws of Stefan [7], Wien [8] and Planck [3]. Obviously modern researchers do not agree with Dr. Robitaille's conclusion. upriver wrote: The issue is gas/plasma only emits lines. Condensed matter emits a BB curve that is distinguishable from a pressurized gas curve. But black-body spectrum is not an emission spectrum: it is the spectrum of EM radiation in thermal equilibrium. Just as Dr. Robitaille, you are confusing a metrological, technical, issue with a fundamental problem. upriver wrote:   Originally Posted by papageno Can you bridge that gap with something substantial? I guess I would have to mount two polished metal plates parallel and see what happens. From what I've read copper and steel. Then come back when you have done the experiment and shown that graphite "destroyed whatever effect was happening in the box".
'06-09-14, 12:15 korjik	upriver wrote: What temperature are you talking about? Cooler than 6000K of the photosphere? I would expect it to be at the temperature of solid iron. The angle dependance of thermal emission would naturally have that effect of darkening depending on the angle. Siegel R., Howell J. Thermal radiation heat transfer. 4th ed., Taylor and Francis, New York, 2002,. Not sure what you mean. Limb darkening. Wiki has a pretty good article and a very good picture. Basically, the edges of the sun look cooler than the center. See the wiki article to see why, cause they explain it alot better. This wouldnt work if we are seeing light from a solid surface. The temp of the surface sould be relatively constant from center to limb, so the center should be a 5700K blackbody and the limb should be a dimmed by scattering 5700K blackbody. What you get is a 5700K bb at the center and a cooler, and there for dimmer and more orange bb curve near the limb. How does this fit into your theory?
'06-09-14, 14:10 Tim Thompson	Mainstream Wins Again Tim Thompson wrote: It looks to me like that "yellow stuff" is the photosphere. The image is taken at different wavelengths that the images you were showing before, which reveal features well above this image, which are invisible here. upriver wrote: The images from the solar telescope are from 396nm to 630nm, which is visible light. The image I posted before are from 171 to 195A or 17.1nm to 19.5nm (EUV). With magnetogram pictures being 630nm. Interesting that they take pictures of the suns magnetic field with iron plasma(FeI) in visible light. So, we agree that the two images are at different wavelengths, and show different features. Hurrah. In visible light, iron is seen strictly in absorption, never in emission. But an absorption spectrum reveals Zeeman splitting of the spectral lines, which allows a direct determination of the magnetic field strength. So iron can be used to measure & map the magnetic field strength. But the loops & filaments, as are visible in the extreme ultraviolet (EUV) images are not visible at these wavelengths. Now, the visible images are "multispectral", which means they are broadband, and cover a wide range of spectral lines in one image. On the other hand, the EUV images are narrow band images, restricted to a single spectral line of highly ionized iron. So, they reveal where the iron is, and they reveal the loops & filaments. But they do not reveal, for instance, how much iron there is, unless that emission is compared to line emission from other species. If you interpret the images to mean that there is only iron present, or perhaps that there is mostly iron present, then you are wrong. The images do not contain enough information to do that. upriver wrote: And if I was to just look at those pictures, I would say that most of the suns visible light comes from the top layer of the photosphere. And you would be right. but there is a lot of difference between saying that, and saying that the visible light is physically generated, or created, in the photosphere. When you look at the photosphere, you are looking at the surface of last scattering, where the mean free path of the photon has at last become long enough that it can escape altogether, and make the trip from the sun to us. Below the photosphere, the photons are scattered, with increasing strength (shorter mean free path), the farther down you go. As the photons migrate upward from the deep interior, they loose energy, and become collectively thermalized. The loss of energy means they present a lower temperature when thermalized by scattering. In the photosphere, the atmosphere of the sun is sufficiently sparse that line absorption can take place. So, the thermal spectrum is deformed, and we get what we see, a thermal spectrum with line absorption superimposed on it. The trip from the center of the sun, to the photosphere, takes a typical photon about 1,000,000 years. upriver wrote: Clearly the center of the sunspot is either emitting UV or IR because its not emitting visible (396 to 630nm). Well, actually it emits a lot of visible light, but an exposure set to match the sorrounding, much brighter photosphere, makes it look relatively dark. And since it is much cooler, then it should emit more IR than the surrounding atmosphere. upriver wrote: Now in this picture taken at 17.1nm, there appear to be areas that are opaque to EUV in this layer that in the MS view is above the photosphere. Well, they could be opaque. Or maybe there just isn't enough 17.1 nm emission to register, in an exposure designed to reveal the much brighter loops. In fact, that is the likely explanation, as the plasma in the magnetic field loops should be very much hotter than the plasma below. upriver wrote: You mean to tell me that if I put a cool thing between two hot things it will stay cool? The quoted thickness for the photosphere is 400km. That requires new physics. Or a different model. Yes, that is what I mean to tell you. No, it does not require any new physics, and it does not require a new model. It only requires a knowledge of both physics and solar models that you sadly lack. There is a thing in my kitchen, a modern convenience called a "refrigerator". The inside stays very cold, despite being immersed in a sometimes very hot kitchen. It works because it is cooled by non-spontaneous processes. It is cooled by a heat pump, that pumps heat "uphill", away from the cold and towards that hot, in exactly the opposite direction that one would expect heat to naturally flow. I have yet to see anyone suggest that new physics is required to understand the refrigerator. In the case of the photosphere of the sun, it is just that easy. The photosphere remains cooler than the underlying, hotter interior, because of purely spontaneous, ordinary thermodynamics. Because the sun is in thermal equilibrium (or very nearly so), each successive spherical shell of arbitrary (but equal) radial extent must hold that same total thermal energy as the shell below (or above). But the shells are successively larger in volume, so the thermal energy density must be lower, in the successively outward shells. So the temperature must be lower, hence the sun gets cooler from the center to the photosphere. But starting about the bottom of the convective zone, perhaps 1/4 of the way down from the photosphere, into the sun, the magnetic field is increasingy less confined by the progressively less dense plasma. At and above the photosphere, the strongly time varying magnetic field pumps the tenuous plasma in the transition zone, chromosphere & corona into much higher temperatures than the photosphere. It is in essence a heat pump, though very different from the refrigerator. In this case, the pump is non-thermal, so that time varying magnetic fields (which necessarily generate equally time varying electric fields) accelerate the plasma, and hence vastly increases the temperature of the higher layers of the atmosphere; temperature is proportional to kinetic energy, so anything that increases kinetic energy, also increases temperature. The pump is radial, so in this case, the temperature has a preferred, outward radial direction. upriver wrote: Well the only UV comes from the coronal loops as you can see by the picture, because it certainly is not coming from inside the sun as the loops are providing the only illumination (reflected and direct). It is a filter at 17.3nm center with a bandwidth of .64nm or 6.4A. As a matter of fact if you look in the center of the picture you can see a feature blocking a flare. That feature is 20,000km high. And that feature is opaque to light from emission of 160,000 to 2 million degrees (FeIX). I fail to see why this is a relevant point. Who said the UV was not generated in the loops, except perhaps yourself? upriver wrote: So you saying that short wavelength radiation converts to visible right at the photosphere? Yes. upriver wrote: So what makes the photosphere brighter? At wavelengths from 396 nm to 630 nm? Its temperature. Peak thermal emission, at the temperature of the photosphere, should be at about 550 nm. No UV from below will get out, and the photosphere should generate very little of either UV or IR because of its temperature. upriver wrote: "Sunspots are "dark" because they are cooler than their surroundings. A large sunspot might have a central temperature of 4,000 K (about 3,700 C or 6,700 F), much lower than the 5,800 K (about 5,500 C or 10,000 F) temperature of the adjacent photosphere." http://www.windows.ucar.edu/sun/atmo.../sunspots.html I do believe I said essentially the same thing. upriver wrote: Not only that, a sunspot is deeper than the surrounding photosphere. The magnetic field extends through the sunspot into the surface of the sun. Both NASA and ESA indicate the the coronal loops originate below the photosphere. Indeed so. A sunspot is essentially a "bubble" of magnetic field, which extends far below, and far above the photosphere. The loops in your trace images may be sunspot magnetic fields. Now, as far as I can see, there is nothing about sunspots, or loops & filaments, or about the temperature and/or structure of the photosphere, that is in any way a problem for the mainstream model of the sun. And, of course, we are now in the "shotgun" approach, with so many topics afoot that it is no longer possible to understand what the point of the discussion is supposed to be. Maybe you can help us out by concentrating on one, or at most two points, which you consider to be most important. The Big Deals, the points where it should be "obvious" that the mainstream view must be wrong. Do you have any?
'06-09-15, 03:16 upriver	Helioseismology. Because the sun is denser in the middle in the gas model, sound intensity should drop off from the center of the sun. But that is not the case. There is clearly a condensed matter surface to propagate the sound waves. Helioseismology shouldn't work in a vacuum better than the best vacuums on earth. "The particle motion associated with f-modes and p-modes is essentially confined to a region outside the solar core" http://www.stat.berkeley.edu/~stark/...helio.htm#what The iron shell. If you look at HelioSeismology(HS), the shell thickness is approxamitely half the radius. Flare radio brightness temperatures exceed 109 K and the peak in the radio spectrum is as high as 35 GHz: both these two features and the hard Xray data require very high densities of nonthermal electrons, possibly as high as 1010 cm 3 above 20 keV at the peak of the flare. http://www.astro.umd.edu/%7Ewhite/pa...orh_020723.pdf How does the photosphere etc. support these kind of electron densities?   But black-body spectrum is not an emission spectrum: it is the spectrum of EM radiation in thermal equilibrium. What????????? "The Planck law gives the intensity radiated by a blackbody as a function of frequency (or wavelength)." http://scienceworld.wolfram.com/physics/PlanckLaw.html The sun is heated. By what ever process. Its temperature is not changing if thats what you mean by "in thermal equilibrium." In the MS model the photosphere radiates as much as it receives from underneath. Is the sun in thermal equilibrium with space? No. How about with its interior? No. It could be considered in LTE if you look at a small enough area. "A material's emission spectrum is the amount of electromagnetic radiation of each frequency it emits when it is heated (or more generally when it is excited)." http://en.wikipedia.org/wiki/Emission_spectrum "In physics, a black body is an object that absorbs all electromagnetic radiation that falls onto it. Despite the name, black bodies are not actually black as they radiate energy as well." http://en.wikipedia.org/wiki/Black_body More to come.
'06-09-15, 04:55 captain swoop	Iron Sun? Solid shell needed for Helioseismology to work? hhm! I think we have been here before!
'06-09-15, 07:32 tusenfem	But black-body spectrum is not an emission spectrum: it is the spectrum of EM radiation in thermal equilibrium. What????????? I guess it all depends on definitions. An emission spectrum is in astrophysics usually meant to be spectrum made up of emission lines. As the link to wiki you put in, you see examples of emission spectra, and see, is it just lines.   The iron shell. If you look at HelioSeismology(HS), the shell thickness is approxamitely half the radius. So now the iron shell is what..., half the thickness of the sun? Then how can you claim to see the iron shell in your iron lines if there is half the sun above it? And I guess you want to say here that the shell radius is half the sun's radius. If the thickness is half the sun's radius then you will have at least a factor of 10 more mass in the sun.   Helioseismology shouldn't work in a vacuum better than the best vacuums on earth. I would hardly call the sun starting from the photosphere and going inward a vacuum.   It could be considered in LTE if you look at a small enough area. LTE works for volumes not for areas. Overall the sun is in LTE, untill you get to the location where tau=2/3, from which the photons can escape. And because the photons take so long to get from inside to outside they will also be in thermal equilibrium (it takes much longer for a photon to reach the photosphere, than you wanted Kirchhoff to wait until equilibrium).   How does the photosphere etc. support these kind of electron densities? Although the paper states that the density is pretty high, you seem to forget that they are talking about solar flares. These plasmas need not be supported by the photosphere, but arise from "below the surfaced of the sun" and grow big and explode. The high density is sustainde by a relatively strong magnetic field.
'06-09-15, 18:24 papageno	I have nothing to add to what tusenfem said regarding upriver's misconceptions about black-body radiation.
'06-09-19, 17:51 upriver	I will be working on a CD recording project for the next couple of weeks so my time is limited. I will respond when I can. Maybe on Sundays. Upriver
'06-09-19, 18:16 papageno	No answers from Dr. Robitaille?
'06-09-25, 03:08 upriver	No word from Dr. Robitaille. I will be calling him. Ok. I started this thread talking about how gas/plasma only emits lines and not blackbody. Blackbody emission(curve) is the domain of condensed matter. Even compressed plasma has a quasi-continium emission(lines/curve) that are/is distinguishable from a true blackbody emission(curve). Now your telling me how backbody radiation should not be an called emission. Emission is to be used only for lines. Lines pertain to gas, blackbody pertains to condensed matter. Blackbody emission means condensed matter by default, whereas line emission means gas or plasma. An emission is an emission. What label you put on it to define what type of emission it is depends on the field of interest. Just because I dont use it in a "astrophysics" way you think its a misconception. You guys are giving yourselves more credit than you deserve to think that is a misconception on my part. And what is it with you and your libraries, papageno? You think libraries are an infallible source of knowledge and wisdom, and that some how will cure me of my genius. Libraries are a preset knowledge filter, I think I am old enough to not have my knowledge chewed for me. Let me repeat my PM to you. I was reading encylopedias(all of them) in kindergarten. I grew up in libraries. I dont like math just because people mistake it for reality, when all it is supposed to be is a descriptor for existing reality and a predictor for engineering. Blackholes are the biggest boonedoggle ever because they got people to accept any kind of theoretical monkey business that came around. A magnetar with a quantum magnetic gas bubble that collapses produces GRB's. (MS) The barrel shaped supernova remnants are the result of a Bennett pinch that produced a GRB at Rmin.(EU) Can you guys see the difference between those 2 statements? This universe is composed of normal matter, any intelligent person can see that if they ignore the beautiful calculations. So give me a little credit and really look at what I'm saying as there might be something to it. Once we have a logical model with workable mechanisms then we can twiddle knobs. Like we have observed "magnetic slinkys, but the mechanism by which they exist differs depending on the model. EU say that they are current flows. MS people say they are frozen-in fields left over from the who knows when. MS have calculations to back it up. I dont. You have to either have a bunch of money or work at a university or LLNL or some place like that to have access to the good 2D plasma code to run simulatons. To get 3D Bennett pinch code. Dream on. To write is beyond my expertise right now, but I am modifing a simulator for work, for I think there is a mechanism that exisits that is responsible for the electric current flow of the slinky and for the current flow to the sun.   And I guess you want to say here that the shell radius is half the sun's radius. If the thickness is half the sun's radius then you will have at least a factor of 10 more mass in the sun. No. The shell is thick enough to provide the observed density for the sun. The half radius figure was a guess from looking at the helioseismology computer interpetations.     Quote: Originally Posted by upriver Helioseismology shouldn't work in a vacuum better than the best vacuums on earth. I would hardly call the sun starting from the photosphere and going inward a vacuum. Not dense enough to support sound propagaton.   Iron Sun? Solid shell needed for Helioseismology to work? hhm! I think we have been here before! Reflections are the result of an impedance mismatch. If you know what that means then you know why an iron sun fits. In addition to that you cannot have a "cavity like resonator" that is a ball of gas. Think about it. The "sound" would just decay in amplitude. Even EM waves.   you are looking at the surface of last scattering No, you are looking at the surface of thermalization.
'06-09-25, 12:08 korjik	Please explain limb darkening and how it would occur on a solid surface. If you think this is out of the topic of the thread (in which case you threadjacked yourself) then please explain why the third graph on the web page you linked to in post 26 shows argon emitting a blackbody radiation curve.
'06-09-25, 12:12 korjik	upriver wrote: Blackholes are the biggest boonedoggle ever because they got people to accept any kind of theoretical monkey business that came around. A magnetar with a quantum magnetic gas bubble that collapses produces GRB's. (MS) The barrel shaped supernova remnants are the result of a Bennett pinch that produced a GRB at Rmin.(EU) Can you guys see the difference between those 2 statements? This universe is composed of normal matter, any intelligent person can see that if they ignore the beautiful calculations. So give me a little credit and really look at what I'm saying as there might be something to it. Once we have a logical model with workable mechanisms then we can twiddle knobs. Like we have observed "magnetic slinkys, but the mechanism by which they exist differs depending on the model. EU say that they are current flows. MS people say they are frozen-in fields left over from the who knows when. MS have calculations to back it up. I dont. You have to either have a bunch of money or work at a university or LLNL or some place like that to have access to the good 2D plasma code to run simulatons. To get 3D Bennett pinch code. Dream on. To write is beyond my expertise right now, but I am modifing a simulator for work, for I think there is a mechanism that exisits that is responsible for the electric current flow of the slinky and for the current flow to the sun. In the interest of keeping this thread alive, could you please keep to the topic of this thread. This thread is about blackbody radiation and the surface of the sun. Bringing in other topics just clouds the arguement here.
'06-09-25, 12:16 korjik	upriver wrote: This universe is composed of normal matter, any intelligent person can see that if they ignore the beautiful calculations. So give me a little credit and really look at what I'm saying as there might be something to it. Once we have a logical model with workable mechanisms then we can twiddle knobs. we have a logical mechanism. it fits very nicely and the knobs are being twiddled. you are the one wanting to toss it for no reason.   Not dense enough to support sound propagaton. Dont you think that someone in the last 40 or so years may have checked that? maybe using the math that you think is so worthless?
'06-09-25, 15:17 papageno	upriver wrote: Now your telling me how backbody radiation should not be an called emission. Emission is to be used only for lines. Actually, I have always told you that the black-body spectrum is the spectrum of EM radiation in thermal equilibrium. upriver wrote: Lines pertain to gas, blackbody pertains to condensed matter. Blackbody emission means condensed matter by default, whereas line emission means gas or plasma. You persist in mistaking a technical issue with a fundamental problem. upriver wrote: And what is it with you and your libraries, papageno? You think libraries are an infallible source of knowledge and wisdom, and that some how will cure me of my genius. Libraries are a preset knowledge filter, I think I am old enough to not have my knowledge chewed for me. Considering how you fall for the Iron Sun idea and the EU/PC ideas, I would say you are mistaken in your self-assesment. Where are the results of your experiment with a reflective cavity?
'06-09-25, 19:48 captain swoop	And what is it with you and your libraries, papageno? You think libraries are an infallible source of knowledge and wisdom, and that some how will cure me of my genius. Libraries are a preset knowledge filter, I think I am old enough to not have my knowledge chewed for me. this statement says more to me than all of your other posts together.
'06-09-26, 00:33 Celestial Mechanic	captain swoop wrote:   Originally Posted by upriver And what is it with you and your libraries, papageno? You think libraries are an infallible source of knowledge and wisdom, and that some how will cure me of my genius. Libraries are a preset knowledge filter, I think I am old enough to not have my knowledge chewed for me. This statement says more to me than all of your other posts together. Sadly, I must agree with you, captain swoop. No, upriver, libraries are not "infallible" sources of "knowledge and wisdom" anymore than the Internet is. I've seen some pretty goofy books in university libraries, such as a 19th century text by Karl Theodore von Heisel that asserted that pi was equal to 256/81. (I hope I remembered the author's name correctly.) But a critical reader can usually separate the gems from the dross, and any decent library is going to have more gems than dross. If a library acts as a "preset knowledge filter", it is generally filtering out things that have been tried or theorized and found not to work. It can be a great time-saver. Imagine having to wade through scholarly dissertations on phlogiston along with the real chemistry books!