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Thread: Does Lockheed Martin Understand Black Body Radiation?

  1. #181
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    The Infamous and Stupendous Faultline:
    I can't begin to imagine the scale of the features we see in the images. It all depends not only on magnification and distance of the viewer, but the level of enhancement done to make features stand out and be recognized for study. With none of that data around, I can't compare the structures to anything in my experience to get a handle on their size.



    MOZINA:
    Fine. We'll call the "big" for now. Why are those same patterns visible 6 hours later all shifted to the right?



    The Infamous and Stupendous Faultline:
    I don't imagine, though, that they compare in size to any mountain ranges on Earth. What if they are gaseous, and the size of the Earth? If they are that large, I can imagine a rolling cloud of gas that takes hours or days to move a noticable distance or change its appearance.




    MOZINA:
    Over a six hour window, you can typically see a great deal of change in cloud system, or a gaseous system. Over 8 days, you certainly would see changes.


    Let me reiterate that I am speaking of issues of scale. I only used Earth clouds as an analogy. If the clouds are the size of normal thunderheads, they'll change in minutes. If the clouds dwarf the size of Jupiter, you may not see changes for hours or days from millions of miles away.

    A possible explanation for the uniform rotation: The camera pivoted.

  2. #182
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    The 171 image, the 191 image

    A link to the 171 image.

    Let us pick five pixels in this image. Back onboard TRACE, we will assume the CCD recorded 10, 300, 1,000, 5,000, and 50,000 photons at the locations corresponding to these pixels. Let's assume that the transform from detected photons to pixel brightness is linear, with '10' in the gif image corresponding to 1,000 photons, and '200' to 5,000 photons (we'll also assume the gif has 256 brightness levels).

    On the 171 image, the first two points would be 'black' (below the threshold), the last as 'white' (above the threshold).

    Very important: 'black' in the image does NOT correspond to 'no photons detected!'

    We don't have the 191 image. However, let's assume the number of photons recorded at the same five locations (i.e. along the same five sightlines) is 5, 10,000, 1,000, 2,000, and 500 (respectively). On this image, produces as a gif the same way as the 171 image, the pixels would be black, white, grey, grey, and black (respectively).

    There's nothing 'magic' about any of these numbers; you could use any others (as long as they aren't outside the range of the CCD!).

    Next, estimating temperatures.

  3. #183
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    Quote Originally Posted by Baloo
    In fact they've explained on the their website what they've did with those images and is not "isolating a relative heat signature". Read it again.
    I've read it many times now and discussed it with them in some length actually.

    [quote]What is the peak wavelength emitted by a body having 1 million K? How much energy will put such a body in the specific Fe 171 A wavelength?[quote]

    The peak wavelength at that temperature is 2.898 X 10 ^9 meters. The energy is 1240/2.898 = 427.88 Ev/sec.

    No, is not about civility and politeness here: I've provided you a scientific explanation and you refuted it as useless whitout providing any reason. You didn't answered to my question: do you mainatin your statement that Planck's law is a "fifty cent equation"?
    No. In fact I maintain some of my comments hurt a lot of people's feelings and I'm trying to learn from my mistakes Baloo. If I hurt your feelings with that comment earlier (and clearly I did), I'm sorry.

    Nobody denies that is the Fe emitting the radiation. But is not blackbody radiation.
    It takes a particularly temperature and a lot of energy to ionize Fe ions and emit these photons. If the temperature of whole whole surface was a million degrees or more, we would "see" it. We do not see anything even in Yohkoh images. We therefore have to assume a lower temperature not a higher one.

    When we start comparing FE ions however, we are setting up a "spectrum" small though it might be. I think the method worked however, based on the results. I just think the got the colors backwards.

    And you really hope that you'll be able to SEE what is happening in the sun's core? That's not gonna happen.
    Well, never say never. We may "see" it through heliosciesmology before we see it any other way. I do think we'll find pretty powerful ways to see what is inside over time.

    "Demonstrate" includes, but is not limited at, quantitative estomations. You didn't provided any and by your statements regarding maths, you have no intention to do it.
    Actually, I do intend to provide some math, specifically the density RANGE calculations. I'm still playing with movement models, but it seems like everyone want a math formula and I can't think of a better place to give you one at the moment. Even still, Bruce put forth all sorts of math. So did Manuel. Not many folks listened to them. What makes you think a couple of math formulas from me are going to make a big difference here exactly?

    Yeap, that conclude the whole issue. And you claim that you have a scientific approach?
    Sure. I'm just one person. There are many things about this model I've never thought about before and that I have not focused on before. My specialty is satellite image recognition and interpretation. I never claimed to be an expert on everything.

    You realize that if you're right about the crust then the helioseismology won't work and its results will be flawed? Then why do you use its results about the layers below sunspots?
    I totally and absolutely disagree. It only works BECAUSE of the transition layer. It's "observation" of the transition layer demontrates that the technology DOES work.

  4. #184
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    I'm going home for dinner now. I'm not sure whether I'll be offline or online tonight but I will locate Nereids reference about energy states and temperatures that SOHO is sensitive to. I think the link is on my portable at home anyway.

  5. #185

    Lightbulb

    Well, as usual, I can't keep up with the enormous volume of material. So let me just chip away at what I think matters. I may be repetitious, or fail to notice that something I say is already out there, buried in the pile.

    First ...
    Quote Originally Posted by Mozina
    My guess is the surface is approximately 2000K.
    This kind of thing really bothers me, and so does the fractured, many-topic nature of the whole discussion. At the risk of repeating myself ad nauseum, why bother fretting over the details of some model, any model, when it's just plain impossible from the get-go? It takes a mighty cavalier attitude towards physics to just "guess" that a surface, trapped under a 5700K plasma, is only 2000K. Physics simply does not allow such a fantastic thing to happen, no matter how you slice & dice the plasma or the surface.

    Although this appears as a simple sentence, already buried on on page 4, it's a huge thing to say, with enormous implications. It simply cannot go unchallenged. The high temperature of the photosphere renders a solid surface, any solid surface, of any alloy, to be simply impossible. It's just that easy. We call it "physics".

    Quote Originally Posted by Mozina
    I'm here because I want my 13 year old daughter to know that science works, that science is fair and science is about knowledge and open dialog and the open discussion of ideas. ... I want my daughter to know it's ok to be wrong once in a while
    Any 13 year old who can follow the science has my respect. Hello Daughter, whereever you are.

    Now let me make a general observation on the nature of things. "Blackbody" has nothing to do with the images linked on page 1. I say that because "blackbody" is a very specific term in physics, which refers to radiation with a very specific Planck Law spectrum. Emission from non-thermal processes, such as spectral line emission, is never blackbody radiation. Since we know that we are looking at images of line emission, we know that we are not looking at blackbody radiation at all. So let us leave that word behind.

    Now, Michael disputes (if I understand correctly), the explanation given for the various colors in the ratio image. I dispute the dispute, not necessarily meaning to be disputing disputatiously (that was not an easy sentence to think of, you know).

    It's really quite straight forward. The colors in the ratio image show the ratio of 191A to 171A line emission (so says the explanation, I think 191 is a typo for 195; the TRACE channels are at 171A for FeIX/X and 195A for FeXII). FeIX has lost 8 electrons, FeX has lost 9 electrons, and FeXII has lost 11 electrons. It takes higher temperature to knock off more electrons, so the ratio of line emission strength shows the ratio of temperatures as well (with corrections applied to account for the electron transition probability). I assume that the TRACE folks are not a bunch of bone-heads, and they probably got the ratio right, which means they also got the temperature right.

    But it's not all that surprising. One would expect the temperature at the foot of the magnetic loops to be lower than the upper loop temperature anyway, simply because the electrons get accelerated up the loop and gain energy. If the electrons in the loop were cooler, that would imply they had less energy than the electrons at the foot points, in which case one would have to wonder how they got up the loop in the first place. Besides, the foot points are commonly sunspots, which are cooler than their surroundings.

    Quote Originally Posted by Mozina
    What is the peak wavelength emitted by a body having 1 million K? How much energy will put such a body in the specific Fe 171 A wavelength?
    I think it is significant that on the one hand, you feel up to the task of claiming to be the world's leading expert on the sun and solar physics, yet on the other hand, cannot compute such a basic quantity for yourself. I get about 143 Angstroms, a bit shorter than the wavelengths we see here. But the question has no point. It is not the wavelength that matters, it's never the wavelength that matters, it's the spectral energy distribution that matters! You cannot meaningfully argue that radiation is "thermal" (i.e., due to temperature alone), unless you can show that it has a Planck Law spectrum. In this case, we already know that it does not, because we already know that it is line emission from ionized iron. And that's that, unless you are ready to argue that all of physics is wrong (you would not be the first).

    That wraps it up for this installment, I just don't have all the time you mega-posters seem to have.

  6. #186
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    temperatures

    First, this method produces only an estimate of the temperature.

    Second, there are a number of well-known limitations to this method, concerning such things as sightlines through more than one concentration ('blob') of gas, and the range of temperatures within which useful estimates are possible.

    The basic principle is straight-forward: an Fe atom can only emit at 171 if it has lost 8 electrons; at 191 if it has lost 11. If a gas (plasma) is in a state of local thermodynamic equilibrium (LTE), the proportions of atoms having lost 8 electrons to atoms having lost 11 depends only on the temperature (there are other factors; for such a thin plasma as the corona, they are unimportant) - the higher the temperature, the smaller the ratio (of Fe IX to Fe XII).

    Returning to our five sightlines. Dividing the number of 'Fe IX' photons detected be the number of 'Fe XII' photons detected, we get 20, 0.03, 1, 2.5, and 100 (respectively).

    How do we 'colour' these five pixels? Let's assume that, from quantum physics and spectroscopy, we know that a ratio of <0.9 corresponds to a temperature of >3 million K, and >10, <1 million K (the actual relationship will, of course, be different from these numbers). If we have only 3 colours (and 'black' - we'll come to that in a minute), let's say <0.9 is blue, >10 is red, and everything else is green.

    Our five pixels would thus be coloured red, blue, green, green, and red (respectively).

    Now, black.

    CCDs are not perfect; even with the door closed, they will still 'detect' some 'photons'. Further, the number detected always translates to 'pixel intensity' with some 'errors'. So, in practice, there is some minimum threshold, below which we can say 'number of photons detected indistinguishable from zero'. For a pixel in either image, where we detect 'zero' photons, to try to estimate of the temperature is a bad idea. We colour such pixels black. (The actual considerations that go into decisions about thresholds are, not surprisingly, more complex than this; however, you get the idea, I hope). Thus our first pixel (10, 5) will be black.

    Back to comparing the 171 image with the temperature data. Specifically, the second pixel (300, 10,000). In the 171 gif image, it would be black (in a corresponding 191 gif image, it would be white); at that location in the temperature array, it is ....

  7. #187
    Quote Originally Posted by Michael Mozina
    If you wish to teach me, take a single page from my website and explain the page a "better" way using the gas model.
    And here is the fundamental problem -- stated by many. The reason others aren't trying to explain the problem / images you see is because they don't believe it is a problem or that the images as they exist on your webpage are in a usable form. After multiple explanations of how such images are processed, how certain features are magnified, and how the images were taken in the first place, what we are left with is what many have said from the beginning: what you can SEE is not always what IS.

    If your model is correct, you should be able to 1) derive specific, testable hypotheses which must be 2) falsifiable. And you have to state what evidence would falsify them. That is the heart of the scientific method. [edited to add]I know your main arguments; from those arguments you need to provide testable implications and tell us how they could be falsified.

  8. #188
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    the other fundamental problem (other than solid iron?) is that it is not our job to explain the gas model to you, Michael. this, too, you have been told repeatedly. it isn't even our job to explain the holes in it. the gas model is, for a whole huge list of reasons that you're been told repeatedly, the standard model.

    note, please, that those reasons do not include any vast conspiracy to keep your model in the dark, so to speak. those reasons include the huge, huge number of predictions that the gas model can make that can then be used. no, I can't go into detail. repeat, English major. however, I do know enough about science to know that, if it didn't, a nuclear chemist in the latter third of the twentieth century wouldn't be the first to recognize that.

    okay. since the gas model does make all those successful predictions, it is assumed that, if you're going to overthrow it, you know enough about it to do so. ergo, it is not the job of mainstream science to educate you. it is your job to educate yourself. you say you've done a lot of study in this field. fine. so you should know how Newtonian physics predict a weight for the sun. you should know how (and this is a stab in the dark, guys, so if I'm wrong about this, let me know, huh?) that weight influences the very orbital mechanics that make those satellites studying the sun possible. you should know that there is only one kind of photon--I do, and I haven't studied the sun in any detail!

    pant, pant, pant.

    okay, I'm done ranting, but you need to be done shifting the burden of proof. it's your job to explain things, because it's your hypothesis that you're defending. and I still don't understand how the iron can stay solid.
    _____________________________________________
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  9. #189
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    Michael has many things to answer about, including fission problems that Faultline and I showed him earlier around page 6.

  10. #190
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    [/amused bystander]

    Is there a rule on this board about trollish behaviour?

    On the old BA board if a poster would not respond meaningfully to specific questions (A behaviour MM has exhibited throughout the many threads of this discussion) he/she would be judged a troll and banned.

    [amused bystander]

  11. #191
    Quote Originally Posted by Doe, John
    On the old BA board if a poster would not respond meaningfully to specific questions (A behaviour MM has exhibited throughout the many threads of this discussion) he/she would be judged a troll and banned.
    As per the FAQ (emphasis mine):

    13. Alternative Concepts

    If you have some idea which goes against commonly-held astronomical theory, then you are welcome to argue it here. Before you do, though READ THIS THREAD FIRST. This is very important. Then, if you still want to post your idea, you will do so politely, you will not call people names, and you will defend your arguments. Direct questions must be answered in a timely manner.

    People will attack your arguments with glee and fervor here; that's what science and scientists do. If you cannot handle that sort of attack, then maybe you need to rethink your theory, too. Remember: you came here. It's our job to attack new theories. Those that are strong will survive, and may become part of mainstream science.

    Additionally, keep promotion of your theories and ideas to only those Against the Mainstream threads which discuss them. Hijacking other discussions to draw attention to your ideas will not be allowed.

    If it appears that you are using circular reasoning, depending on long-debunked arguments, or breaking any of these other rules, you will receive one warning, and if that warning goes unheeded, you will be banned.

  12. #192
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    Quote Originally Posted by Michael Mozina
    Quote Originally Posted by Baloo
    What is the peak wavelength emitted by a body having 1 million K? How much energy will put such a body in the specific Fe 171 A wavelength?
    The peak wavelength at that temperature is 2.898 X 10 ^9 meters. The energy is 1240/2.898 = 427.88 Ev/sec.
    I guess you meant 2.898x10^-9 meters. That means about 29 A, far from 171 A at which the images where taken. Why don't you compare the amount of energy released at 171 A by a blackbody having 1million K with the actual energy recorded by those satelites? Oh, sorry, that involves numbers and math...

    Since it seems that you've start using that "fifty cents" Planck law I have to ask you:
    1. What is the peak wavelength at which the photosphere (6000 K) emitts?
    2. Do you still maintain your statement that a sunspot (having 4000 K) is NOT giving off visible light?

    Quote Originally Posted by Michael Mozina
    Quote Originally Posted by Baloo
    No, is not about civility and politeness here: I've provided you a scientific explanation and you refuted it as useless whitout providing any reason. You didn't answered to my question: do you mainatin your statement that Planck's law is a "fifty cent equation"?
    No. In fact I maintain some of my comments hurt a lot of people's feelings and I'm trying to learn from my mistakes Baloo. If I hurt your feelings with that comment earlier (and clearly I did), I'm sorry.
    How about acknoweldging some of your mistakes? It took about 4 months and hundreds of explanations of the same fact from a couple of dozen of this board's members until you started to admit that blackbody radiation plays a role in the sun's emission, but you never stated that you've changed your opinion. Is not only about people feelings here; is about your way of analize data provided to you: you have dissmised explanations whitout bothering to understand them and this is not science and is not in the spirit of this board. Now it appears that you've started to use some of those explanations (like Planck's law), but whitout acknoweledging the fact that you've changed your mind and that your initial rejection was plain wrong.
    I don't know about others, but I have no desire to take another 4 months just to make you understand another basic concept involved in data analysis. This is not science and is not good faith from you to behave that way and I'm not willing to waste my time for someone that wants to defend his pet theory more than learning something new.
    Quote Originally Posted by Michael Mozina
    When we start comparing FE ions however, we are setting up a "spectrum" small though it might be.
    It's an emissionspectrum, not a blackbody one! Like it has been already explained to you in those images the temperature is not derived from the spectrum but from the fact that a specific Fe ion exists only in a well defined temperature range: therefore when such a ion is observed (by its characteristic discrete emission/absorption spectrum we know that the temperature in that area is in that specific range.
    Quote Originally Posted by Michael Mozina
    Even still, Bruce put forth all sorts of math. So did Manuel. Not many folks listened to them. What makes you think a couple of math formulas from me are going to make a big difference here exactly?
    They'll make a huge difference if they predict the right numbers. This is the final prove for theory: checking it against the observed and measured phenomena.

    Quote Originally Posted by Michael Mozina
    Quote Originally Posted by Baloo
    You realize that if you're right about the crust then the helioseismology won't work and its results will be flawed? Then why do you use its results about the layers below sunspots?
    I totally and absolutely disagree. It only works BECAUSE of the transition layer. It's "observation" of the transition layer demontrates that the technology DOES work.
    Do you think that they've directly measured the speed of sound inside the sun? Do you realize that the speed of sound will be different in a neon layer than in a hidrogen one? How can you trust their results if you claim that they have no ideea in which kind of material the sound is propagating? Is like using sonar technology conceveid for water to detect a submerged object in a sea of oil: the results will be completely wrong because the parameters that your system assumes to exists (as density, temperature range, etc) and upon it has been modeled, conceived and built are wrong.
    The fact that the technology works proves that they've been properly modeled the environment; and guess what? They've modeled taking into account a hidrogen layer, not a neon one.

  13. #193
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    Moz, everything you are suggesting is speculative. Pardon me for calling your hand, but, show the math. It's OK to speculate and contest on logical grounds, but, your arguments have no substance [i.e., no math]. So show the math. Is that too much too ask? Without that, it appears to be just a bunch of 'handwaving' and off topic deflections.

  14. #194
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    Quote Originally Posted by Faultline
    The Infamous and Stupendous Faultline...
    It's not allowed to call names on this board. Please attack the idea, not the person.

  15. #195
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    May I add a request? I have yet to see a substantive argument in this thread that offers a VALID objection to any mainstream theories. What I do see are twisted versions of 'logic' without substance. I elect to ignore any 'theory' without 'the math'. Frankly, I'm tired of hearing this 'Alternative Cosmology' nonsense.

    I have repeatedly, and patiently, given reliable sources. And all I have got is 'handwaving' objections. Show the math, for a change. [Hint: try not ignoring the solid observational evidence that happens to blow your boat out of the water.]

  16. #196
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    Quote Originally Posted by Michael Mozina
    It takes a bit more time and effort to spend 15 years of so studying satellite images to and be able to explain them all, and show how they all tie together. I've done just that. I didn't do it over a weekend, or over a year, but I've studied the sun for decades of my life, and particularly through satellite imagery. There are no math formulas that are going to precisely explain these images. At best they will be approximations of a theoretical method to explain the phenomenon.
    For your information: a theory is useless if it cannot make predictions. What do you think a prediction is? Pointing to a image and saying "that's what it should look like" is NOT a prediction; instead providing quantitative estimations that could be checked against observed facts is an acceptable prediction.
    Quote Originally Posted by Michael Mozina
    There is a difference however between a mathematical theory and practical reality. I'm more interested in the later than the former. I will take the former if it applies, but first you need to show me how it applies to real images from the real sun.
    I've provided you a link to the Planck's law. You've dissmised it as useless whitout providing any reason. I cannot prove you that it applies to the real world, is up to you to check the assumption upon which the equation has been derived and to show them wrong.
    The fact that the physical world is fairly described by our math is proved by everything that surrounds you: your computer works because the flow of current in it is correctly described by our math, your TV set display an image because the electromagnetic wave propagation is properly described by Maxwell' equations, your stereo play music because we know how to mathematically describe the propagation of sound waves.

    Quote Originally Posted by Michael Mozina
    Not all knowledge is mathematical in nature Baloo.
    You cannot build a theory in physics whitout math. It is a basic tool.

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    I affirm Baloo's objection. No math, no credibility.

  18. #198
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    Although this question has nothing to do with black body radiation, it is part of the iron sun model. If there is an iron shell around the sun, then why do we see upwelling and downwelling hydrogen on the surface of the sun, which is supposed to be of iron?

    @Tim very nice discussion of line ratios. I was planning on something similar (having used it in the Io plasma torus) but you got there before me
    All comments made in red are moderator comments. Please, read the rules of the forum here and read the additional rules for ATM, and for conspiracy theories. If you think a post is inappropriate, don't comment on it in thread but report it using the /!\ button in the lower left corner of each message. But most of all, have fun!

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  19. #199
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    Lockheed is no longer funding that project. Does that qualify as a signal of their faith in the future of that endeavor? But by all means, don't let that stop you from proving them wrong by inventing an anti-grav drive in your basement. I will gladly help you file the patents.

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    Quote Originally Posted by Baloo
    It's not allowed to call names on this board. Please attack the idea, not the person.
    Can I not even attack myself?

    LOL

  21. #201
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    Quote Originally Posted by Faultline
    Can I not even attack myself?
    [ATM arguing mode]
    I've seen whit my own eyes how one member of this board has attacked another whitout provocation. Clearly this shows that ad hominems are tolerated here.
    [/ATM arguing mode]
    Hey, its quite easy if I choose to ignore some 'unimportant' facts (such as the ofender and the ofended being one and the same person).

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    something specific, and quantitative

    I'd like to put this suggestion to Michael:

    Please pick an area within your 'Sun has a solid iron/ferrite surface' idea. Present a quantitative case that supports your idea, in this area.

    The case could be an OOM (order of magnitude) calculation; it could be some (relevant) equations or math; it could simply be a 'numbers consistency' case.

    Should the area you choose be a Manuel, Birkeland, or Bruce one, please point to the specific quantitative aspects that support your idea.

    In the meantime, please don't post anything else, on your idea, in this thread.

    Once you have presented your case, others will attack it with glee and fervour, and you will answer questions that are asked.

    For all others who wish to post to this thread, I ask only that we wait until Michael has had a chance to read and reply to this suggestion.

    Why am I making this suggestion, I hear you ask? Because I feel that unless we quickly get to something specific, and quantitative, this thread will continue to go nowhere. By asking Michael to address an area in his idea of his choosing, he has the opportunity to put the strongest foot forward. By asking that it be a quantitative case, we all have the opportunity to see the strength of the bones upon which the pretty raiments have been draped.

  23. #203
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    Quote Originally Posted by Nereid
    I'd like to put this suggestion to Michael:

    Please pick an area within your 'Sun has a solid iron/ferrite surface' idea. Present a quantitative case that supports your idea, in this area.

    The case could be an OOM (order of magnitude) calculation; it could be some (relevant) equations or math; it could simply be a 'numbers consistency' case.

    Should the area you choose be a Manuel, Birkeland, or Bruce one, please point to the specific quantitative aspects that support your idea.

    In the meantime, please don't post anything else, on your idea, in this thread.

    Once you have presented your case, others will attack it with glee and fervour, and you will answer questions that are asked.

    For all others who wish to post to this thread, I ask only that we wait until Michael has had a chance to read and reply to this suggestion.

    Why am I making this suggestion, I hear you ask? Because I feel that unless we quickly get to something specific, and quantitative, this thread will continue to go nowhere. By asking Michael to address an area in his idea of his choosing, he has the opportunity to put the strongest foot forward. By asking that it be a quantitative case, we all have the opportunity to see the strength of the bones upon which the pretty raiments have been draped.
    I actually agree with you 100% Nereid. It's time to try a completely new approach here.

    This whole conversation has seemed a bit surreal to me to be honest. I've seen just about everything in this thread. I've even seen Van come to my defense. Now how could I ever be frustrated at Van again?

    R.A.F., you are a true gentleman!

    I would also like to compliment you personally Nereid for taking the time to clearly explain the processes involved in creating the colorized image. You did in fact provide a far more detailed explanation of the processes inolved in creating that image than Lockheed. I appreciated your efforts. If we could get our hands on the original FITS files and/or the original 195A image, I'll bet you and I could sit down and apply your methods and demonstrate that Lockheed got the color scheme backwards. The brightest areas of the original image are in fact colored differently than the rest of the loops, so I'm reasonably confident the technology works as described, I simply think they got it backwards.

    From my perspective, we haven't even gotten to first base on the satellite imagery analysis since we cannot even agree to the light source in these images or agree to a heating mechanism for the outer plasma layers. We have not agreed on anything meaningful that would allow us to do any sort of satellite imagery analysis. If we cannot even agree that light and heat are concentrated in the coronal loops then there is simply no way to begin to analyze these satellite images IMO. The light we see in many of these filters comes from the coronal loops, and they are quite hot compared to the surrounding plasmas. Even that seems to be in dispute around here.

    I will most likely focus the OOM calculations on the colonal loops, using some of Bruce's materials. The coronal loops provide the light source in these satellite images and these upwelling plasma flows carry much of the "heat" to the outer plasma layers. Until we can all agree on the light source for these images and agree that coronal loops provide a significant heating mechanism for the out plasma layers, we will never get anywhere IMO.

    I'm busy today at work, but I will put together something over the weekend. It will be short and sweet but to the point and include a few images.

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    I think we can all agree that there are aspects of the sun we would like to study more closely. Here is to some big probes made of borazon doing a suicide dive into the sun.

  25. #205
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    Quote Originally Posted by Baloo
    It's an emissionspectrum, not a blackbody one! Like it has been already explained to you in those images the temperature is not derived from the spectrum but from the fact that a specific Fe ion exists only in a well defined temperature range: therefore when such a ion is observed (by its characteristic discrete emission/absorption spectrum we know that the temperature in that area is in that specific range.
    But that is the whole point Baloo. We *do* know the temperature range in the lit up areas. We do not know the temperature range in the dark areas, but it is likely to be significantly less than 1 million degrees since the photosphere is 6000K and the surface is darker than the coronal loops. If the dark areas were in fact "hotter" than the coronal loops, then Trace would show them glowing brightly and Yohkoh would not see the x-ray emissions concentrated in the same exact coronal loops.

    Your explanation is illogical IMO. Somehow you have this mystery "black body" that shows NO high temperature photons coming from anywhere from around the coronal loop in either Trace or Yohkoh images, but somehow the million degree coronal loops are cooler than a part of the surface that is dark to both Yohkoh and Trace. It seems to me you folks wish to have your cake and eat it too. Where's the heat in these satellite images? Why does Yohkoh show all the x-ray concentrations *in the some coronal loop* rather than around the loop? Why isn't the around around the loop lit up in their Yohkoh or Trace images if these areas are hot?

    I will provide the OOM calculations that Bruce talked about with these coronal loops and then we will try this discussion again.

  26. #206
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    I will continue the rest of these discussions after I have posted some calculations as requested.

  27. #207
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    Quote Originally Posted by Michael Mozina
    I will continue the rest of these discussions after I have posted some calculations as requested.
    Good. When you're ready, PM me (or an admin or another moderator), and I'll unlock this thread.

  28. #208
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    I have received a PM from Michael, stating that he's ready to present his quantitative material. He says it will likely be in two posts, so please wait for the second post before replying.

  29. #209
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    Oom

    Quote Originally Posted by Baloo
    I guess you meant 2.898x10^-9 meters. That means about 29 A, far from 171 A at which the images where taken. Why don't you compare the amount of energy released at 171 A by a blackbody having 1million &#176;K with the actual energy recorded by those satelites? Oh, sorry, that involves numbers and math...

    (On the subject of the value of me personally providing mathematical support for my own ideas)

    …..They'll make a huge difference if they predict the right numbers. This is the final prove for theory: checking it against the observed and measured phenomena.
    You are absolutely correct Baloo. It’s rather ironic IMO that I left off the (–) sign on my first OOM calculation when posting my answer to this forum. The universe has quite a humbling quality to it. Like I said to Van earlier, we’re all human.

    Actually this is exactly the point I have been trying to explain with the Trace/Yohkoh composite image. Yes, I did in fact mean 29A, and this is a highly significant mathematical number and wavelength as it relates to the Yohkoh SXT imaging equipment. Evidently I am the only one that recognizes the significance of that wavelength as it relates to the Yohkoh/Trace composite image, and I finally understand that I am completely to blame for this confusion for not providing you with the necessary information in a mathematically precise way. I failed to provide the necessary mathematical data or explain the mathematical correlation between the satellite images and the wavelengths in question. Hopefully this post will rectify that oversight on my part.

    Yohkoh is/was an incredible piece of engineering IMO. In addition to an assortment of gamma ray filters, Yohkoh carried a broad x-ray spectrum SXT instrument composed of five filters that could see x-ray photons between 2.4A and 46A wavelengths.

    http://www.lmsal.com/SXT/html2/yag/r...00000000000000
    http://www.solarviews.com/cap/sun/moss8.htm
    The same image as "moss7" but with the co-temporal Yohkoh Soft X-ray Telescope image overlaid. Note that the patches of moss seen in the previous images (See Trace Spacecraft Discovers Moss on the Sun) occur only beneath areas of high intensity soft x-ray emission as imaged by the SXT instrument.
    AR 8227 30-May-1998 14:41:01 UT,
    N20 E04 TRACE 171 Angstrom filter. 27.6 sec exposure.
    SXT Al/Mg filter image taken at 14:40:34 UT.
    These wavelengths of photons in the x-ray spectrum can and are observed by Yohkoh’s SXT broadband imaging equipment. The filter used in this specific composite image in question is the Al/Mg filter which has a visual range of 2.4-32A. This means that any photons in these peak wavelengths and heat ranges will be observed in this composite image in yellow:
    .
    http://www.solarviews.com/browse/sun/moss8.jpg

    The SXT filter used in this composite image “sees” that all photons in the 2.4-32A range are concentrated inside the coronal loops, just as the Trace and SOHO satellites observe photons from 171A, 195A and 284A are concentrated within the coronal loops. However, the surface is quite a bit darker to Yohkoh’s SXT instrument as well as to Trace and SOHO at 171A, 195A and 284A. If that surface was peaking in the 29A wavelength range as you suggest, Yohkoh’s SXT broadband equipment would see that background glowing brightly. It would not be dark in Yohkoh’s images in the same areas that it is dark to Trace. The first OOM calculation we seem to agree on, combined with the Yohkoh/Trace composite image I provided you, demonstrate that the dark surface is not hotter than the coronal loops, but is quite a bit cooler than the coronal loops.


    http://www.gsfc.nasa.gov/gsfc/spaces...g%20progre.mov
    http://www.lmsal.com/YPOP/sxt_movie.html

    If a million degree surface was emitting photons, the background in that image would be glowing white, rather than being darker than the materials inside coronal loops.

    Because of the failure on my part to provide the necessary mathematical specifications on the satellites in question and because of my failure to provide a mathematical connection between satellite images and the wavelengths in dispute, few folks have understood the ideas I have been trying to convey. My apologies to you all and to Nereid especially, you were right all along. I can clearly see now that there is a strong need for me to demonstrate the mathematical data necessary to continue this discussion. In the future I will attempt to introduce more mathematical representations of data into key areas of our discussion, but I will likely pick instances that I feel are crucial to our discussions and that are quantifiable with actual data or direct observation rather than jump at every theoretical request tossed my way. I would however like to eventually take a shot at the movement and force calculation ranges.

    http://articles.adsabs.harvard.edu//...00932.000.html
    http://solar.nro.nao.ac.jp/meeting/n...palswamy_2.pdf

    The fact that Yohkoh and other instruments observe that the higher energy photons are concentrated in the coronal loops, suggest that the coronal loops begin as intense, hot filaments near the transition layer. As these filaments grow in intensity they form larger coronal loops. These loops are in fact the source of the highest energy photons emitted from the sun, and are the light source in the three filters on SOHO and TRACE that relate to iron ions. The coronal loops are the primary light source in Yohkoh SXT images. The highest concentration of these superheated photons in Yohkoh images originate within the coronal loops, and less intensely from outside of the loops. The brightest concentration of high energy, high heat photons are located within the coronal loops. We have a visual, 3D confirmation of the heat and energy concentration patterns that can be seen observationally in Yohkoh images. Yohkoh is and must be the final arbitrator between different theoretical mathematical models of heat concentration issues related to the solar atmosphere.

    Since I left off the minus sign of the first OOM calculation I provided (serves me right for trying to rush off to dinner), and I didn’t even read your second question properly, I believe that I still owe you some additional mathematical support for my model and for my point.

    I thought I’d take a stab at the total output of a 1 million degree black body the size of the sun to see how well that jives with general OOM calculations related to the sun’s total energy output.

    Total Energy Output of a Black Body the Diameter of the Sun at 1 Million K.

    http://eesc.columbia.edu/courses/ees...ctures/energy/
    Intensity of radiation from a black body is dependent on T.
    The total energy (flux) released by a black body is proportional to the fourth power of its absolute temperature. This is because the internal energy of the body is related to its temperature, i.e. the more energy the atoms and molecules of the substance have (vibrational energy for example), the higher the temperature of the body. This is known as the Stephan-Boltzman law named after the men who formulated it. If we know the temperature of a black body we can calculate the energy it will radiate by the following equation.
    I = σT4
    Where I = the intensity of the radiation
    T is temperature in degrees Kelvin
    σ is a constant (Stephan-Boltzman) and is 5.67x10-8 Watts/m2.
    So if the transition region is 1,000,000 K
    I = (5.67x10^-8) (1,000,000)^4 = 5.67 x 10^16 Watts/ m^2
    For simplicity, we’ll just assume the photosphere is close to the transitional layer. To calculate the number of square meters on the sun:
    Surface area of a sphere = 4πr2
    4π (6.96x108)2 = 6.087 x 10^18 square meters
    Output in watts: = (6.087x10^18) (5.67x10^16) =

    3.45 X 10^35 Watts or Joules/Sec
    Now let us compare this number to Wikipedia’s OOM calculation of the solar output:

    http://en.wikipedia.org/wiki/Orders_...nitude_(energy)
    3.827 &#215; 10^26 J — energy output of the Sun in one second

    That is a difference of about nine orders of magnitude. Coincidently when we plug in the actual surface temperature of the photosphere into these same calculations, we come up with the correct number. I would therefore have to conclude mathematically that Lockheed simply does not comprehend black body radiation concepts and the implication of these concepts as it relates to satellite imagery. Most of the high energy photons are located inside the coronal loops, or the points where the coronal loops meet the transitional layer. This transition layer that is relatively cool in comparison to the coronal loops themselves. These images, particularly the broad spectrum x-ray images from STX suggest that Lockheed grossly misrepresents the heat concentration patterns in these images. All of the images from the SXT broadband instrument show that the heat and the emissions of these high energy photons are concentrated inside of the coronal loops, and fewer emissions are found in the surrounding atmosphere around the loops. Far few emissions are found on the transitional surface of the sun than are seen in the coronal loops. IMO, Lockheed simply does not begin to understand the basics of black body radiation and how these principles relate to satellite imagery. If they did fully understand how these concepts related to SXT and Trace images, they would have corrected their problem immediately when I first pointed it out to them, especially after I sent them their own composite image as proof. Instead they defended their explanation, and defended it quite poorly I might add.
    So why does this matter? Well, for one thing it demonstrates that the temperature of the transition region must be at or near or even below the temperature of the photosphere. It suggests the transitional region is below, not above the photosphere, which just so happens to jive with recent heliosiesmology findings of a transitional layer at 4800km beneath the visible photosphere.
    Last edited by Michael Mozina; 2005-Oct-10 at 02:23 AM.

  30. #210
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    It seemed to fit in one.
    Last edited by Michael Mozina; 2005-Oct-10 at 04:04 AM.

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