'06-08-25, 23:35 upriver	Black Body and our sun... The Solar Spectrum A true BlackBody curve is a property of "condensed" matter. Gases emit line spectra. And blackbody only is emitted from a surface. A volume of gas will not emit a blackbody. Under pressure, it will emit a quasi-continuum. Argon looks like BB at approximately 100atm. Pressure only exists for a surface bounded volume of gas. In other words the 6000 degree blackbody from the sun really is a true blackbody that has absorbtion lines from the intervening plasma, telling us that the surface of the sun is truly "condensed" matter. From the JET website. "The Science of JET", by John Wesson". "The initial idea was that of detecting the blackbody radiation from the thermal plasma ions. However, when the ICE spectra were measured they were not consistent with this expectation, having instead narrow equally-spaced emission lines, the spacing being proportional to the magnetic field, and intensities much larger than the blackbody level. The spectrum from a deuterium-tritium plasma is shown in Figure 13.4 (below). The observed frequencies depend on the magnitude of the magnetic field at the location of the emission and, surprisingly, it was found that in JET this meant that the emission comes from the edge of the plasma in the outer midplane." So even in a fusion plasma you dont get blackbody, you get emission lines. It(BB) has to come from solid matter. No new physics.
'06-08-26, 20:50 papageno	upriver wrote: A true BlackBody curve is a property of "condensed" matter. Gases emit line spectra. No, not even condensed matter is a "true" blackbody. A perfect blackbody absorbs all EM incident radiation, and the best approximation we can make is a hole in the walls of a metallic cavity. That's why you find a lot of discussions about EM radiation in cavities when dealing with blackbody radiation. Here are a few links: Black Body Radiation Blackbody radiation at Hyperphysics A Heat Transfer Textbook
'06-08-26, 17:23 Tim Thompson	upriver wrote: A true BlackBody curve is a property of "condensed" matter. Gases emit line spectra. Already wrong. Anything, including gases, will emit blackbody radiation, if it is optically thick. Of course, as papageno points out, a real, true, honest to gosh blackbody will absorb 100% of incident electromagnetic radiation, and will emit an unmodified Planck Law spectrum. So the best we can hope for in the real universe is something that comes arbitrarily close to a blackbody, without actually being a 100% real live blackbody. upriver wrote: And blackbody only is emitted from a surface. A volume of gas will not emit a blackbody. Just a re-statement of the above, and equally false. A volume of gas can & will emit practical blackbody, if it is optically thick. upriver wrote: In other words the 6000 degree blackbody from the sun really is a true blackbody that has absorbtion lines from the intervening plasma, telling us that the surface of the sun is truly "condensed" matter. Wrong again. It tells us that the "surface" of the sun is optically thick, and neither more nor less than that. upriver wrote: So even in a fusion plasma you dont get blackbody, you get emission lines. It(BB) has to come from solid matter.. Based on a thoroughly false premise, the conclusion is false. It does not matter what the internal heat source is, fusion or otherwise is not relevant. Any internal heat source, embedded inside an optically thick medium will result in a blackbody (or very nearly blackbody) continuum. That's what happens in stars (all proven quite rigorously in Chandrasekhar's textbook, An Introduction to the Study of Stellar Structure, originally published in 1939, before stellar internal heat sources were known to be nuclear). "Optically thick" means that the opacity, or optical depth of the material is large enough to force a blackbody (or very nearly so) spectrum. This depends somewhat on local conditions, but a value greater than 1 should be sufficient.
'06-08-26, 18:45 Van Rijn	upriver wrote: Pressure only exists for a surface bounded volume of gas. We should all be breathing vacuum then. Gravity bounds gas and gas has weight (for any massive body, at least - a star certainly qualifies). It presses upon (applies pressure to) gas below it. See any reference on atmospheric pressure. For instance: http://ww2010.atmos.uiuc.edu/(Gh)/gu...w/prs/def.rxml A couple of thoughts, upriver. If you had posed these issues as questions instead of declarations, they could have made for a good discussion in Q&A. Also, do you really think such basic flaws in physics would have gone so unnoticed? If you think you've found a basic flaw, it is a good idea to research it carefully, because it is extremely likely you missed something.
'06-08-27, 03:13 upriver	Already wrong. Anything, including gases, will emit blackbody radiation, if it is optically thick. What does optically thick mean? Well most likely it will be dense. But just density will not do it because water is pretty clear. And so is our atmosphere. So it has to be ionized but just ionization will not do it because the plasma in the JET reactor is fusion hot and if it is optically opaque, its over longer distances than the reactor and it still emits lines. But if you do something like place ionized Argon, under 100atm of pressure, it begins to emit a quasi-continuium which is distinguishable from a true blackbody from say a blackbody graphite source, by the fact that the ends of the quasi-continuium/emission curve begin to rise. (Source Dr. Ott at NIST) If the Earth had no surface there would not? be 1atm(15psi) of pressure at this level. And you can see for miles even though the density is greater than the sun. In JET the density is approx. 1/1000 gram m-3. http://www.jet.efda.org/pages/content/fusion2.html So I wonder what the pressure of the plasma at the level of the photosphere is. "The average density of the photosphere is less than one-millionth of a gram per cubic centimeter." http://www.nasa.gov/worldbook/sun_worldbook.html So the blackbody curve(quasi-continuum) for gas is a combination of density/pressure and temperature. As is optical transparency. But over distance an optically thick, low pressure(photosphere) plasma emits lines. Again here is the blackbody surface of the sun at 171A through the photosphere. Sun spot.
'06-08-27, 06:52 tusenfem	upriver wrote: What does optically thick mean? Well most likely it will be dense. But just density will not do it because water is pretty clear. And so is our atmosphere. Upriver, ever heard of looking into a textbook? Optically thick has nothing to do with density, albeit that more dense will more easily lead to optical thickness. You can compare it walking in rain, and you will see the streetlights, and walking in mist and you will only see a general broad glow from the direction of the streetlights. Please go to the library and take a look at "radiative processes in astrophysics" by Rybicky and Lightman, one of the best books on the subject. Also, the picture you showed of the sun is not the "blackbody" image of the sun at 171A, this is just a picture taken at that specific wavelength, nothing black body about it (though the pic does have a lot of black in it, so ... ) Now if you would like to get a temperature that belongs to this image, you have a problem, because you cannot do that from one wavelenght, you will have to take a pic at another wavelength, of which you know that the ratio of the strengths of these two lines is dependent on temperature, and then you can come up with a temperature.
'06-08-27, 16:08 Tim Thompson	Still wrong upriver wrote: What does optically thick mean? Well, scroll back 2 posts to my message, and look at all those cute, colorful, underlined words. Move the cursor over the words with your mouse. Click. If you have the wisdom to click on the suspiciously relevant looking words "optically thick", in my first sentence, you might (possibly) find out what "optically thick" actually means. upriver wrote: Well most likely it will be dense. But just density will not do it because water is pretty clear. And so is our atmosphere. Well, it is not exclusively dependent on density (as Dr. Tusenfem has pointed out). It has to do with absorption, and with that fact in mind, I will contradict you and point out that neither water, nor the atmosphere of Earth are at all "clear". They are both as opaque as brick walls; if I try to look up from Earth, at a wavelength of 100 microns, I will not be able to see out of the atmosphere at all. Likewise, in the wavelength range about 0.001 to 0.1 microns, the atmosphere is totally opaque. So, your statement is quite wrong, the atmosphere & water are not necessarily "pretty clear". They are "pretty clear", if you restrict your vision to eyeball ("visible") light wavelengths. But at other wavelengths, they are not "clear" at all. They are optically thick. That's why Earth emits essentially thermal (i.e., blackbody) radiation at long wavelengths, say in the range about 20 microns to about a millimeter. upriver wrote: If the Earth had no surface there would not be 1atm (15psi) of pressure at this level. And you can see for miles even though the density is greater than the sun. Of course, by careful selection of wavelength, you can arrange not to see beyond a few feet, so the fact that your eyes can see for miles means nothing at all. upriver wrote: "The average density of the photosphere is less than one-millionth of a gram per cubic centimeter." http://www.nasa.gov/worldbook/sun_worldbook.html True, but why is it relevant? In any mainstream theory of the sun, the photosphere is optically thin and does not produce blackbody radiation. Rather, it imposes an absorption spectrum on top of the nearly blackbody emission from the deeper, optically thick sun. upriver wrote: So the blackbody curve (quasi-continuum) for gas is a combination of density/pressure and temperature. As is optical transparency. Almost right. But your inclusion of the words "blackbody curve" makes it wrong. A "blackbody curve" is a Planck Law curve, and nothing else. It is the spectrum of the gas, which may well include both emission & absorption, as well as a continuum, which may or may not be a blackbody continuum, which is an overall function of density, pressure & temperature, as well as chemical composition. Likewise, the opacity and transparency are functions of the same variables. upriver wrote: But over distance an optically thick, low pressure (photosphere) plasma emits lines. Still wrong. The photosphere of the sun is certainly not optically thick. There is some line emission, but not much; the spectrum of the photosphere is dominated by absorption, since it is so much cooler than the lower layers of the sun. upriver wrote: Again here is the blackbody surface of the sun at 171A through the photosphere. Way wrong. As Dr. Tusenfem points out, this image has nothing to do with blackbody emission. The light you do see comes from ionized iron, and it is strictly line emission, highly non-thermal. The dark parts of the picture are simply places whre there is no iron line emission. And finally, no way are you seeing through to the photosphere anywhere in that image. It is entirely in the chromosphere, and well above the photosphere.
'06-08-28, 08:08 upriver	Still wrong. The photosphere of the sun is certainly not optically thick. For what distance? The whole 400km? The bottom, or the top where the tufting is?(50km) At what wavelengths? 171A(17.1nm)? 1000-1800nm? And how can you tell this without refering to the fusion model? In the work I'm doing right now, my problem is that the liquid makes it hard to see UV and x-rays or IR but we can take optical spectrum where as with ICF they can use any wavelength diagnostic. I think what your trying to tell me is that all the processes from the inside of the sun to the outside, produce a 6000K BB curve that is a better fit to condensed matter than to multiple processes (absorption, emission or other electronic process) by intervening plasma.   this image has nothing to do with blackbody emission This is an image of where the suns 6000K BB curve orginates from. It is the average temperature of the area of the surface(dark) and the area of the footprints(bright). SO the dark surface area may well be 2800K. At the base of the loops, in this picture, which is below the photosphere, is condensed matter. Beneath the loop at the lower center are features that are higher than the photosphere, chromosphere and corona put together. At a image temperature of 171A(Why iron plasma?). The photosphere, chromosphere and corona provide the intervening absorption lines. Coronal loops provide emission at their footprints and through out the loop. Optical emission that our eyes can see is provided by the photosphere. If its not a solid or a liquid, how does an unbounded gas sphere produce a rise in pressure with a rise in temperature?
'06-08-27, 23:29 Van Rijn	upriver wrote: This is an image of where the suns 6000K BB curve orginates from. It is the average temperature of the area of the surface(dark) and the area of the footprints(bright). SO the dark surface area may well be 2800K. No. That is for one single wavelength. It is an image of hot iron ion emissions (about a million degrees) from above the photosphere. The only thing you can tell from that image is where there are emissions at 171A. It's like taking an image with a blue filter. If you use a blue filter, you can tell where things are blue and where things aren't. That's it. You can't tell what's red or green, infrared or ultraviolet. Compare that to a blackbody spectrum, a continuous spectrum that is a mix of wavelengths. These are different things. See here, for example: http://www.astronomynotes.com/light/s4.htm Please read this. You're making assumptions that there is information in the image that simply isn't there. By the way, this was all covered before with Michael Mozina. You might want to look it up.   If its not a solid or a liquid, how does an unbounded gas sphere produce a rise in pressure with a rise in temperature? It is bounded. It has mass and is self bounded by gravity. In fact, it is the counterbalancing forces between gravity and the energy from fusion that keeps it at the size it is. You really need to study some physics and astronomy.
'06-08-28, 05:49 tusenfem	Okay, grabbing my copy of Rybicky and Lightman and looking up optical depth. On page 12 of the book you can read: Rybicky and Lightman wrote: optical depth defined as: d taunu = alphanu ds or taunu(s) = int alpha[sub]nu[/alpha](s') ds' The optical depth defined above is measured along the path of a traveling ray; occasionally, tau is measured backward along the ray and a minus sign appears in the equation. (skip) A medium is said to be optically thick or opaque when tau, integrated along a typical path through the medium, satisfies taunu > 1. So, now we have the definition for optically thick. The alpha in the equation has not been defined yet, it is the absorption coefficient of the medium (see e.g. Einstein coefficients). So at a certain frequency nu you will not be able to see any deeper into an object (e.g. the sun) then where tau = 1. Now for different frequencies this layer can be at different physical depths. Now the "bottom layer" of the photosphere, the region where the photons of the sun come from (hence the name), has to be at tau = 1. However, for reasons that I will not go into, it is usually taken that the photons come from a layer at tau = 0.7. Now with your question Upriver:   For what distance? The whole 400km? The bottom, or the top where the tufting is?(50km) At what wavelengths? 171A(17.1nm)? 1000-1800nm? And how can you tell this without refering to the fusion model? Just do the calculations, and you will find that yes over the whole 400 km it is optically thin, or transparant. And this has nothing to do with the production of the energy somewhere deep in the sun. It is a property of the plasma of the sun, dependent on density, species, ionization etc. etc.
'06-08-28, 23:04 upriver	So let see. If I get a plasma glowing inside of my little reactor, with the right gas density(5 torr) and voltage(1000v) it glows so much you can barely see through it but on the spectrograph it shows lines. So what kind of optical thickness shows a blackbody curve? And if your saying that the core of the sun is responsible for the suns BB emission then it would have to be attenuated by just the right amount. At the right frequencies. Unlikely.   Just do the calculations, and you will find that yes over the whole 400 km it is optically thin, or transparant. And this has nothing to do with the production of the energy somewhere deep in the sun. It is a property of the plasma of the sun, dependent on density, species, ionization etc. etc. Well its been argued before on this board that you cannot see below the photosphere and that the image that I posted is from above the photosphere. This picture is below the photosphere. And I'm saying because the BB emission curve is such a good fit to a condensed matter curve, it cannot be a pressurized gas emission, it has to be condensed matter producing it. Hence proof for the solar surface being condensed matter(iron). Again; If you do something like place ionized Argon, under 100atm of pressure, it begins to emit a quasi-continuium which is distinguishable from a true blackbody, from a blackbody graphite source, by the fact that the ends of the quasi-continuium/emission curve begin to rise. From EIT Wavelength selection: EIT He II 30.4 nm: chromosphere, erupting prominences Fe IX 17.1 nm: high contrast in coronal loops Fe XII 19.5 nm: typical quiet corona Fe XV 28.4 nm: hotter, 2.5 MK corona These are the wavelengths that the photosphere is supposed to be optically "thick" to. Now if this is optically thick does it exhibit BB behaviour? (As Tim says) And even so, I dont think it is optically thick at that wavelength because of the 17.1(171A) pictures. All of the 17.1nm etc. pictures show the surface of the sun. Which is the source of the BB emission curve. Not the million degree core. Again; If you do something like place ionized Argon, under 100atm of pressure, it begins to emit a quasi-continuium which is distinguishable from a true blackbody, from a blackbody graphite source, by the fact that the ends of the quasi-continuium/emission curve begin to rise. I dont think anybody really understands what this means. Just think about it for awhile. It means gas(plamsa) only emits lines unless under pressure and only condensed matter will have a BB emission that is close to "theoretically perfect". Oh, just so you know from Wiki; Visible (optical) spectrum: Violet | Blue | Green | Yellow | Orange | Red So when I say optical, now you know.
'06-08-29, 01:12 Van Rijn	upriver wrote: This picture is below the photosphere. This again. upriver, what is your evidence that this is below the photosphere? Please be specific. I really don't think you're paying attention to the information and references that have been provided now and many times in the past.
'06-08-30, 01:53 tusenfem	Taken from the book Astrophysics I by Bowers & Deeming. This is one of those books that should be on every astrophysicist's bookshelve. (with its companion II). Bowers & Deeming wrote: Page 74, Section 5.5 Black Body Radiation A distribution of photons contained within an enclosure whose walls are maintained at constant termperture T will eventually reach a state of thermodynamic equilibrium at the same temperature. This equilibrium is referred to as black-body radiation. — skip equations -- Now, in the deep interiors of a star, the temperatures are verrz high, and thus the radiation densitz is high, but the net outward flux is small compared to the general level of radiation. Thus we might feel justified in assuming that, in the deep interior, the radiation field is close to isotropic and close to the Planck black-body radiation spectrum in form, corresponding to the local temperature of the gas. — skip equations etc. -- Thus we requirie that a temperature be definable throughout a region that is large enough that the thermodynamic enclosure approximation can be used, but small enough that the properties of the region are uniform. This is one version of the general hypothesis called local thermal equilibrium (LTE). — skip -- Here our LTE assumption is that the radiation field is locally Planckian, with temperature T Okay, this short quote shows what the general idea is, why a start will emit an Planckian spectrum. And it also shows why Upriver does not measure one of these in his Argon plasma in his plasma device. Even though it may radiate like mad, and you cannot look through it, it can hardly be considered to be in a thermal equilibrium in a large enough volume. Therefore, you just see the line spectrum of Argon. Then about the picture, I wonder why you think it is under the photosphere, when clearly you see protuberances in the picture, which are big loops of magnetic field outside of the sun's "surface".
'06-08-31, 17:38 upriver	Okay, this short quote shows what the general idea is, why a start will emit an Planckian spectrum. And it also shows why Upriver does not measure one of these in his Argon plasma in his plasma device. Even though it may radiate like mad, and you cannot look through it, it can hardly be considered to be in a thermal equilibrium in a large enough volume. Therefore, you just see the line spectrum of Argon. Are you saying that because the sun is in thermal equilibrium in some form, that causes gas(plasma) to emit a BB curve? BB emission is a property of condensed matter. As you can see by this plamsa emission spectrum, as the drive pressure increases, the emission becomes a quasi-continuium.   Then about the picture, I wonder why you think it is under the photosphere, when clearly you see protuberances in the picture, which are big loops of magnetic field outside of the sun's "surface". Because both ESA and NASA say that the loops originate below the photosphere. And the bright features at the loop footprints in the picture, clearly have all the characteristics of an arc origination point. The photosphere does not have the charge density to support the loops. Spend some time looking at that picture, without fusion model glasses on. The loops are 75 to 100,000km high. Underneath are features that are 10 to 30,000km high. And they are opaque to UV. The loops are iron plasma. In this movie, it shows a loop forming and as that loop forms you can see it poking through the photosphere. http://trace.lmsal.com/POD/movies/BastilleSlinky.mov http://trace.lmsal.com/POD/TRACEpodarchive3.html
'06-08-31, 18:54 papageno	upriver wrote: BB emission is a property of condensed matter. WRONG! The black-body spectrum is the spectrum of heat radiation. You just need good ol' thermodynamics to realize it: From Planck's Nobel Lecture: Planck wrote: Since Gustav Kirchhoff has shown that the state of the heat radiation which takes place in a cavity bounded by any emitting and absorbing substances of uniform temperature is entirely independent upon the nature of the substances, a universal function was demonstrated which was dependent only upon temperature and wavelength, but in no way upon the properties of any substance. (my emphasis) It does not matter what the substance is with which the radiation is in equilibrium: the universal function — the black-body spectrum — does not depend on it. That's why Planck used simplified models of physical systems to derive his formula: Planck wrote: Heinrich Hertz's linear oscillator, whose laws of emission, for a given frequency, Hertz had just previously completely developed, seemed to me to be a particularly suitable device for this purpose. If a number of such Hertzian oscillators are set up within a cavity surrounded by a sphere of reflecting walls, then by analogy with audio oscillators and resonators, energy will be exchanged between them by the output and absorption of electromagnetic waves, and finally stationary radiation corresponding to Kirchhoff's Law, the so-called black-body radiation, should be set up within the cavity. [...] The noteworthy result was found that this connection [between the energy of a resonator of specific natural period of vibration and the energy radiation of the corresponding spectral region in the surrounding field under conditions of stationary energy exchange] was in no way dependent upon the nature of the resonator, particularly its attenuation constants - a circumstance which I welcomed happily since the whole problem thus became simpler, for instead of the energy of radiation, the energy of the resonator could be taken and, thereby, a complex system, composed of many degrees of freedom, could be replaced by a simple system of one degree of freedom.
'06-08-31, 19:25 korjik	Upriver, you do realize that if you take a picture of Fe IX ions, you are generally going to see Fe IX ions? Yes, the picture is Iron plasma, but that is because that is what they are taking a picture of, not because it is the only thing there. Filters are used to bring out details, because the otherwise the huge amount of light would overwhelm whatever you are using to make the pic. Also, brightness does not imply opaqueness (is that a word?). All that brightness does is saturate the instrument, making the contrast undetectable. If you were to shine a flashlight thru your argon plasma, and then filter the argon emission lines, you would see the flashlight just fine. A truly opaque object would block or scatter the light, and you wouldnt be able to see the flashlight no matter what you do. the photosphere dosent need any charge to support a loop. the loops are magnetic field poking up from below. You really should learn some more physics before you try to interpret these things. I am not looking 'with fusion glasses on' because the fusion is six hundred thousand kilometers away from the surface. All I see is plasma, gas and magnetic fields.
'06-09-01, 08:06 Tim Thompson	Careless upriver wrote: Are you saying that because the sun is in thermal equilibrium in some form, that causes gas(plasma) to emit a BB curve? I suspect he is. But in any case, I am willing to say that. upriver wrote: BB emission is a property of condensed matter. You keep saying that. And everybody keeps telling you that you are wrong. But you keep saying it anyway. Do you not care about being right? Or maybe you think you are the only one who knows the secret? Once again: No, BB emission is not a property of condensed matter. It is certainly possible for "condensed matter" to emit BB radiation. But it is also possible for "condensed matter" to emit non-BB radiation. And it is thoroughly possible for non-"condensed matter" to emit BB radiation. But, let us humor you and pretend for a moment that you are correct. What makes you think that the sun is not "condensed matter"? What is the minimum density required to be "condensed matter"? At about 1/2 a solar radius, the density of the sun is 1 gm/cm3, which is the density of liquid water. Is that "condensed" enough? At about 3/4 of a solar radius, the density is about 0.1 gm/cm3. Is that "condensed" enough? upriver wrote: As you can see by this plamsa emission spectrum, as the drive pressure increases, the emission becomes a quasi-continuium. Totally irrelevant. Or, maybe you really believe that an entire generation of scientists is just too stupid to tell the difference between real BB and some "quasi continuum"? Are you the only one who can see the difference? upriver wrote: Because both ESA and NASA say that the loops originate below the photosphere. And the bright features at the loop footprints in the picture, clearly have all the characteristics of an arc origination point. The photosphere does not have the charge density to support the loops. So, since the bright points in the picture look to you like "arc origination points", we need think no further? They are not the origination points, even if they do look that way to you. The image you show us is an image of the transition region, which sits above the chromosphere. The bottom of your image is about 1500 km above the top of the photosphere, assuming that it is the bottom of the transition zone. It may be higher than that, since it is only dark because the iron is not hot enough to be ionized enough to emit at 171 Angstroms. upriver wrote: Spend some time looking at that picture, without fusion model glasses on. Well, you could "spend some time" trying to get at least one thing right. Besides, as I said before, the "fusion model" is not relevant. It does not matter what the heat source is, all that matters is the thermal equilbrium of the sun. upriver wrote: The loops are 75 to 100,000km high. Underneath are features that are 10 to 30,000km high. And they are opaque to UV. The tops of the loops range from 75,000 to 100,000 km above the base of the photosphere, which is well below the bottom of the image. The loops are not opaque to UV, as your own image clearly shows. If the loops were "opaque", we would not be able to see loops behind loops, as we clearly do. upriver wrote: The loops are iron plasma. No they are not. The loops are mostly hydrogen, but hydrogen does not emit at 171 Angstroms, so we don't see it. Iron has a lot of electrons, so it can be highly ionized. That makes it a good tracer for the magnetic field lines that make up the loop. That's why images are made in a single line of the iron spectrum. But the loops are mostly not iron. You are far too careless, and will learn nothing until you actually bother to care about the difference between being right & being wrong.
'06-09-01, 04:36 tusenfem	upriver wrote: 1. Are you saying that because the sun is in thermal equilibrium in some form, that causes gas(plasma) to emit a BB curve? 2. The photosphere does not have the charge density to support the loops. 3. And they are opaque to UV. 1. Yes, I am saying that and if you would take a book on stellar structure, or just the Bowers and Deeming book from which I quoted, you could read up on it and edumacate yourself, albeit that LTE does not make for light reading. 2. And what should be the charge density that can support the loops, can you give us an estimate? Or do you mean, as so many ATM posters that a neutral plasma does not have charges? Or what? 3. The loops are not opaque to UV, what you are seeing is the CCDs being overexposed and therefore you can no longer look through the loop, as you could before. Looking at pretty pictures is a nice hobby, but with lacking physical knowlegde it does not lead to any insight.
'06-09-02, 04:47 upriver	I'm sure this paper has popped up before. It took me awhile for it to sink in but after talking to Dr. Ott at NIST and looking at spectrum for sonoluminescence, I began to understand that blackbody calculations for gas were base on flawed interpetation and were strictly theoretical(this is not my finding). That gas or plasma exhibits a BB curve under pressure. There is no experimental evidence for BB due physical distance. Gas and plasma emit lines only. And if you go looking for data(not calculations), it will reflect that finding. Please do not use stars as proof. An Analysis of Universality in Blackbody Radiation Pierre-Marie Robitaille, Ph.D.* http://arxiv.org/ftp/physics/papers/0507/0507007.pdf "In fact, by adding a perfect absorber to his perfectly reflecting box, it was as if Kirchhoff lined the entire box with graphite. He had unknowingly returned to the first case. Consequently, universality remains without any experimental basis." "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."   What is the minimum density required to be "condensed matter"? I cant find any references where somebody has taken the spectrum of a gas as it gone from as gas to a liquid to a solid. As I posted before, Dr. Ott at NIST indicated that Argon begins to show a quasi-continuum at 100Atm. Liquid will show a lower temperature than expected because of convection. " Since the frequency and amount of photons released by an object is related only to the amount of energy in the vibrational degrees of freedom Evib ," The Little Heat Engine: Heat Transfer in Solids, Liquids and Gases http://www.thermalphysics.org/heat/heatenginepict.html "Consequently, we can see that Stephans law does not hold for gases.7 In fact, thermal emission for the diatomic gas (like CO and NO) occurs in discrete bands of the electromagnetic spectrum and in a manner not simply related to temperature (see Figure 14 , Figure 15 and Figure 16.)8,16 The situation becomes even more interesting if the gas is not molecular, but rather monatomic in nature (like Ar or He for instance). In that case, when moving from the liquid to the gas phase, G1 looses both its rotational, and more importantly, its vibrational, degrees of freedom, Ebond= Evib = Erot=0. Neglecting electronic emission, a monatomic gas cannot emit significant radiation. Indeed, for such a gas, Stephans law no longer has any real meaning."   The loops are mostly hydrogen, Coronal rain. http://trace.lmsal.com/POD/TRACEpodarchive3.html http://trace.lmsal.com/POD/images/TLya_990529_14.gif "we see material falling down at a temperature of approximately 10,000 degrees, that is a factor of 100 cooler than the hot corona. The matter falls in clumps, showing up along only an occasional loop." Clumps of hydrogen falling? Mountains in the background. This is good because you can see how most of the surface is glowing, where as under the loops you can see the build up of coronal rain(cooled iron plasma). I would have a hard time believing that those features are only opaque in UV(if this were the T region). I expect features(20,000km) that are larger than the transition region, photosphere and chromosphere would show up in some other wavelength above the photosphere. http://trace.lmsal.com/POD/images/T171_010419_1330.jpg My new favorite picture. This picture is great. http://trace.lmsal.com/POD/images/T1...322_043253.jpg A closeup of the suns surface. If you can suspend disbelief for just a moment and carefully study the picture, you can see where loops emerge. Notice that there are continuous "trenches". The dark areas are higher than the light areas. And its(surface) is opaque to (E)UV. That really is the condensed matter surface of the sun.
'06-09-02, 10:57 papageno	upriver wrote: An Analysis of Universality in Blackbody Radiation Pierre-Marie Robitaille, Ph.D.* http://arxiv.org/ftp/physics/papers/0507/0507007.pdf "In fact, by adding a perfect absorber to his perfectly reflecting box, it was as if Kirchhoff lined the entire box with graphite. He had unknowingly returned to the first case. Consequently, universality remains without any experimental basis." Let me quote the relevant passage: Dr. Robitaille wrote: Kirchhoff then sought to extend his findings [1,2,5]. He constructed a second box from metal, but this time the enclosure had perfectly reflecting walls (e =0, k =0). Under this second scenario, Kirchhoff was never able to reproduce the results he had obtained with the graphite box. No matter how long he waited, the emitted spectrum was always dominated by the object enclosed in the metallic box. The second condition was unable to produce the desired spectrum. Poor old Kirchhoff! How long should he have waited? The condition for Kirchhoff's law is that the radiation in the cavity is in thermal equilibrium with the wall of the cavity. This requires that the radiation exchanges energy with the material of the cavity. But if the walls are reflecting, there is not much energy exchanged, so it can take a long time before an equilibrium is reached. 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? Dr. Robitaille wrote: As a result, Kirchhoff resorted to inserting a small piece of graphite into the perfectly reflecting enclosure [5]. Once the graphite particle was added, the spectrum changed to that of the classic blackbody. Kirchhoff believed he had achieved universality. Both he, and later, Planck, viewed the piece of graphite as a "catalyst" which acted only to increase the speed at which equilibrium was achieved [5]. If only time was being compressed, it would be mathematically appropriate to remove the graphite particle and to assume that the perfect reflector was indeed a valid condition for the generation of blackbody radiation. However, given the nature of graphite, it is clear that the graphite particle was in fact acting as a perfect absorber. Universality was based on the validity of the experiment with the perfect reflector, yet, in retrospect, and given a modern day understanding of catalysis and of the speed of light, the position that the graphite particle acted as a catalyst is untenable. In fact, by adding a perfect absorber to his perfectly reflecting box, it was as if Kirchhoff lined the entire box with graphite. He had unknowingly returned to the first case. Consequently, universality remains without any experimental basis. 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"? Dr. Robitaille wrote: Nonetheless, physics has long since dismissed the importance of Kirchhoffs work [9]. The basis for universality, no longer rests on the experimental proof [i.e. 9], but rather on Einsteins theoretical formulation of the Planckian relation [10, 11]. From a quick search on PROLA: T. H. Boyer, "Derivation of the Blackbody Radiation Spectrum without Quantum Assumptions", Phys. Rev. 182, 13741383 (1969) O. Theimer, "Derivation of the Blackbody Radiation Spectrum by Classical Statistical Mechanics", Phys. Rev. D 4, 15971601 (1971) D. C. Cole, "Reinvestigation of the thermodynamics of blackbody radiation via classical physics", Phys. Rev. A 45, 84718489 (1992) Dr. Robitaille wrote: It has been held [i.e. 9] that with Einsteins derivation, universality was established beyond doubt based strictly on a theoretical platform. Consequently, there appears to no longer be any use for the experimental proof formulated by Kirchhoff [1,2,5]. Physics has argued [9] that Einsteins derivation of the Planckian equations had moved the community beyond the limited confines of Kirchhoffs enclosure. Einsteins derivation, at least on the surface, appeared totally independent of the nature of the emitting compound. Blackbody radiation was finally free of the constraints of enclosure. Ref. 9 in his bibliography: Schirrmacher wrote: Returning to radiation theory, it is also due to Einstein that the disputes on the proof of Kirchhoff's law evaporated. His 1916 paper on "Radiation emission and absorption according to quantum theory" finally let the 19th century proto quantum problem of theproof of the relation between absorption and emission disappear. On the basis of Planck's oscillator model, Einstein defined emission and absorption coefficients A(n,m) and B(n,m) as the probabilities that in the course of a transition from energy level m to n radiation with the energy hnu = E(m) - E(n) is emitted and the absorption of such an energy quantum gives rise to the change of state, resp.154 In terms of these coefficients, however, no simple equivalent exists to Kirchhoff's law. A universal function cannot be found, as the theory of thermal equilibrium cannot fully be recovered in the quantum description.155 In this sense Einstein threw away the ladder Planck had climbed. Schirrmacher wrote: During this life cycle of Kirchhoff's law, however, neither the statement of the law nor its foundational roots remained constant. Rather, a number of different interpretations and foundations were found, that, though often coexisting at the same time, still exhibit a distinct development of new understandings and new identifications of the issues physicists and mathematicians felt obliged to prove. upriver, quoting Dr. Robitaille wrote: "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." Let me quote 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. 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? In conclusion, I think that Dr. Robitaille is confusing a problem of metrology with a problem of fundaments.
'06-09-02, 14:43 papageno	Addendum: 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)
'06-09-05, 23:33 upriver	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. 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.   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.   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?). 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. "They all noticed that the interior of the cabin was "toasty warm", as if heated by a strong fire. Moved to locate the source of this heat, town officials found " two highly polished metal mirrors which faced each other, radiating a very great heat in rippling waves" This is a great discovery. It fulfills what Nathan reported in his last testimony." http://www.icehouse.net/john1/stubblefield.html Halfway down the page. Of course its not peer reviewed, but its so strange, how could you make that up as an effect. 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?). It also ties in with ground radio(read up on ground radio), and the work of the Correas. Ground radio. http://www.icehouse.net/john1/groundradio.html http://www.borderlands.com/newstuff/...FelixRadio.htm
'06-09-06, 03:13 tusenfem	How interesting that these "free energy sources" are sooooo abundant on the web, but none of them get build, not even by the proponents of these devices. Oh, I know, they cannot build them, because then they will be liquidated by the oil industry and gubbermint.
'06-09-08, 01:35 upriver	How interesting that these "free energy sources" are sooooo abundant on the web, but none of them get build, not even by the proponents of these devices. Oh, I know, they cannot build them, because then they will be liquidated by the oil industry and gubbermint. What free energy device? I never said anything about free energy. Ground radio works. Try it. As much as I would love for there to be free energy, based on the aether or magnets, 99% of it is false. There are a couple of effects that are real that could be used if they were taken advantage of. One of the most fascinating is the PAGD. "Our point of departure was a serendipitous observation - made while studying sustained X-ray production - of quasi-regular discontinuities in glow discharges having a minimal positive column at very high vacua (10E-5 to 10E-7 Torr) and at low to medium voltages (10-50 kV DC). These events, which were associated with X-ray bursts, spontaneously originated localized cathode discharge jets that triggered the plasma glow in a fashion quite distinct from the flashing of a photocathode or from an externally pulsed plasma glow. It would soon become apparent that these discontinuities were elicited by spontaneous electronic emissions from the cathode under conditions of current saturation of the plasma glow, and could be triggered with much lower applied DC field strengths. The discharge was distinct from the VAD regime in that the plasma channel was self-starting, self-extinguishing, and the regime was pulsatory (79)" http://www.aetherometry.com/PAGD/Pwr...l#anchor135950 Could be a mechanism for T-Tauri, Harbig-Haro object oscillations. Or solar flares. You could go here and debunk a new magnet device. I signed up to critique their technology. 1 day left. http://www.steorn.net/frontpage/default.aspx?p=1 "Steorn is making three claims for its technology: 1. The technology has a coefficient of performance greater than 100%. 2. The operation of the technology (i.e. the creation of energy) is not derived from the degradation of its component parts. 3. There is no identifiable environmental source of the energy (as might be witnessed by a cooling of ambient air temperature). The sum of these claims is that our technology creates free energy. This represents a significant challenge to our current understanding of the universe and clearly such claims require independent validation from credible third parties. During 2005 Steorn embarked on a process of independent validation and approached a wide selection of academic institutions. The vast majority of these institutions refused to even look at the technology, however several did. Those who were prepared to complete testing have all confirmed our claims; however none will publicly go on record. In early 2006 Steorn decided to seek validation from the scientific community in a more public forum, and as a result have published the challenge in The Economist. The company is seeking a jury of twelve qualified experimental physicists to define the tests required, the test centres to be used, monitor the analysis and then publish the results."
'06-09-08, 03:08 tusenfem	Okay. All very entertaining, but let's get back to the black body radation from the sun. Naturally there are small differences from BB as can be seen here. Especially in the X-ray range there is too much, because X-rays are also generated by solar flares and such on the (oh my goodness do I dare to write this word) surface of the sun. But the underlying driver for this thread is of course that when you can make doubtful that plasmas emit BB radiation, and that that only can happen by solid state, then you would be able to claim that the sun has an iron shell as a surface which emits the BB.
'06-09-10, 00:04 upriver	But the underlying driver for this thread is of course that when you can make doubtful that plasmas emit BB radiation, and that that only can happen by solid state, then you would be able to claim that the sun has an iron shell as a surface which emits the BB. Exactly. I'm saying that the photosphere will only produce a line spectrum if it is truly a plasma of its stated conditions. The one that you show is a matter BB emission with absorbtion lines. Here is my page of spectrums. http://www.strangeeye.net/universe/blackbody.htm You said build a model of the ES. The STEREO mission will be the clincher, though. The other MS claim is that the BB spectrum is somehow produced by the core of the sun and as a result of all the intervening process becomes 6000 degrees. 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. I looked at a lot of papers looking for BB spectrum for gas for work(6months). I couldnt find anything. And that's when I began to realize what Dr. Robitaille was talking about. My phone call to NIST confirmed some of the properties of gas/plasma. The only way you can produce BB from gas or plasma is when its under pressure.
'06-09-10, 01:17 Tim Thompson	Voila upriver wrote: The only way you can produce BB from gas or plasma is when its under pressure. That's the point. The visible light from the sun is not generated in the photosphere, it only passes through the photosphere from below. Nobody says that the BB spectrum of the sun actually comes from the photosphere. The photosphere is evident by absorption, and not emission, in the solar spectrum. The BB emission from the highly pressurized plasma in the solar interior is deformed by its passage through the absorbing medium of the photosphere, which imposes absorption features on it.
'06-09-10, 01:22 korjik	upriver wrote: Exactly. I'm saying that the photosphere will only produce a line spectrum if it is truly a plasma of its stated conditions. The one that you show is a matter BB emission with absorbtion lines. Here is my page of spectrums. http://www.strangeeye.net/universe/blackbody.htm You said build a model of the ES. The STEREO mission will be the clincher, though. The other MS claim is that the BB spectrum is somehow produced by the core of the sun and as a result of all the intervening process becomes 6000 degrees. 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. I looked at a lot of papers looking for BB spectrum for gas for work(6months). I couldnt find anything. And that's when I began to realize what Dr. Robitaille was talking about. My phone call to NIST confirmed some of the properties of gas/plasma. The only way you can produce BB from gas or plasma is when its under pressure. You do realize that the third image on the page you made shows a blackbody radiation curve of an argon plasma under pressure? As a matter of fact, it shows a pretty good fit until you get close to the argon emmission, which dosent look anything like a blackbody curve at any pressure. (please note the fancy dotted line showing a figured blackbody curve) The apparent lines get smeared out by collisional processes when you get to such high temp and pressure (6.6 atmospheres at the top). The second graph shouldnt even be there. Trying to find a blackbody curve at the megahertz range in a magnetically confined plasma wont work. I am suprised they even tried. I dont have the bandwith to download the PDFs, but if I were to hazard a guess, they are seeing the cyclotron waves you would expect to see in a magnetized plasma. Lastly   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. If you look on your own page, you will see the graph of a blackbody emitting Ar plasma
'06-09-10, 13:35 upriver	If you look on your own page, you will see the graph of a blackbody emitting Ar plasma But that is exactly the point. That plasma is in a sonoluminesence bubble. Its a plasma under pressure. You will notice that as you increase the drive pressure, the emission becomes more BB like. And as I said, only a gas or plasma under pressure will produce a BB emission. Its hard to find ionized pressurized gas in a lab. (mercury vapor, high pressure sodium). At this point I think I have looked at every sono and laser induced bubble dynamics paper on the web and in our company library(alot). Sonoluminesence is acoustically induced cavitation in a liquid, that is driven in such a way to collase the bubble a produce a pulse of light. Here are some movies that I shot. I ran across this issue because I am specifying a spectrometer and a streak camera for our budget next year.
'06-09-10, 23:51 upriver	The second graph shouldnt even be there. Trying to find a blackbody curve at the megahertz range in a magnetically confined plasma wont work. I am suprised they even tried. I dont have the bandwith to download the PDFs, but if I were to hazard a guess, they are seeing the cyclotron waves you would expect to see in a magnetized plasma. Are you saying that you would expect to see a blackbody in an electrostatic plasma at that frequency? I dont know much about this potion of the show but will catch on.