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  #151 (permalink)  
Old 25-September-2005, 10:19 PM
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Wolverine Wolverine is offline
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Originally Posted by Michael Mozina
You might explain that to Van as well. I do hear you and I will back off but Van needs to respond in kind.
My warning extends to all participants here. Van Rijn has not personally attacked you in this thread that I can see.
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  #152 (permalink)  
Old 25-September-2005, 10:50 PM
seohtu seohtu is offline
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In the interest of science, I will break my vow of silence and accept whatever wrath this incurs

Mr. Mozina, when I asked about additional images under different filter sets, I was inquiring if such existed of the arcs/filaments (can I just call 'em filaments? It just comes naturally to me that way) in question, as a means of comparing the iron content to that of other elements. Obviously, when filtered specifically for a given element, that element will prove to be the most abundant visible. If we could filter for a spectral signature for lead, it would be the most abundant element in those images also. But, with proper calibration, a true estimation of the proportions of the various elements could be acquired from such a set of images.

The reason this seems significant to me is that if it came out that there was only evidence of say, < 1% of the material in the "filaments" being iron, through analysis of properly calibrated images, that would, by definition, affect your proposed model. Images of that nature do not appear on your website, and I am asking if you have searched for such, and, if so, what you have found.

As a neophyte, I am intensely curious also about what the curent model's explanation for the arcs might be, and do hereby beg anybody who knows to present as persuasive an argument as possible about their nature under the current model. The description of the shifted magnetic fields causing a difference in electrical potential, resulting in massive electrical equatorial arcs *does* seem almost intuitive to me if approached from the (vaguely defined) model that has been described.

Note to all involved: it is no flattery to convince me one way or the other - I lack the foundation most of you so eloquently display. But you could consider this a kind of "public outreach" to the uninitiated, if you chose.

Many thanks. Sincerely,
Derrick Baumer
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  #153 (permalink)  
Old 25-September-2005, 10:52 PM
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Quote:
Originally Posted by Michael Mozina
Let me explain how you are playing both sides of the street here, so maybe you will see where I'm coming from. On one hand you accept that we "see" various photons from a distance *AND* that we can identify them by their very unique wavelength. This is something I agree with actually. Each photon has a unique signature that is related to the valence shells and energy states and quantum discharges of from various atoms. We both agree that it is possible to identify the element by the wavelength.
I'm not playing both sides, what you're missing (altough it has been pointed to you many times) is that light is emitted in more than just one way:

1. Light emitted by atom/molecule transition from one energy level to another; since those levels are quantized the emitted spectrum is discreet and unique for each chemical element, therefore this kind of emission is used in spectroscopy.

2. Blackbody radiation, emitted in a continuous spectrum and depending only on the temperature. (I must point out that both those mechanisms have the same origin: the energy is emitted in quanta, so you accept or reject them both.)

3. Synchrotron/cyclotron radiation: produced by an accelerated charged particle. It has a continuous spectrum.

There are other mechanisms that produce light, but already you may see that our sun give a spectra that is a superposition from (at least) three other spectra, two continuous and one discreet.


Quote:
Originally Posted by Michael Mozina
You then turn right around and hand me a fifty cent math formula that presumably renders the first argument mute. Do you see the problem here Baloo? We can't say that photon wavelength is related to the element and heat and then turn right around and suggest it does not matter.
It does matter as long as you're aware of all the means by which the said photon could be generated. You're not so you think that all the photons are generated in the same way, which is false.

Why don't you ask dr. Manuel or dr. Bruce what do they think about this "50 cents" formula? You'll be surprised.

Quote:
Originally Posted by Michael Mozina
You are trying to suggest that ANY element can emit 171A, 195A, and 284A wavelengths is raised to the right temperature.
A body having about 102000 °K will give a continuous spectrum having a peak intensity at exactly 284 A. You can do your own calculations for what spectra will have a body with a specific temperature using the formula at the bottom of this page: http://scienceworld.wolfram.com/phys...cementLaw.html

Of course a highly ionised Fe ion will give also a photon having exactly 284 A; fortunately for us the sun has only 6000 °K at the surface, therefore it's thermal emission at 284 A is very weak and is not masking the Fe ions emission.

Quote:
Originally Posted by Michael Mozina
Other than iron, show me one element that is capable of releasing all three of these wavelenghts using real nuclear chemical data and show me the materials used.
Any element heated at 100000-160000 °K will produce a spectra having a peak intensity at a wavelength between 171-284 A.
And there is more; actually any body that surround you is emitting at 171 A: your computer screen, your chair, yourself. By using that "50 cents" formula you can compute how much energy you're releasing at this wavelength. (you'll need to know your temperature: it should be around 300 °K )

Quote:
Originally Posted by Michael Mozina
I am not "pretending" anything. I could (wouldnt recommend it) look at the sun with my own two eyes and "see the light" any time I wish. I do see neon, and I see it in the penumbral filaments and not in the umbra.
Let see what you've stated so far:
Quote:
Originally Posted by Michael Mozina
Quote:
Originally Posted by Baloo
What is, in your opinion, the temperature of the sun's photosphere?
5800 - 6000K
Quote:
Originally Posted by Michael Mozina
Quote:
Originally Posted by Baloo
what wavelength have the photons emitted by a surface having a 6000° Kelvin temperature?
In this case I think most of the photons being released at 6000 degrees come from neon plasma
...
Seems like a great question and it warrants study.
So you've said that you don't know how a neon surface having 6000°K look like. Then how do you know that is a neon surface having 6000°K in those pictures?


Quote:
Originally Posted by Michael Mozina
Quote:
Originally Posted by Baloo
Neon gives off visible light? Could you tell me in which conditions the neon produce visible light?
Electricity is typically required. The next time you're in a pub, take a look at the neon signs. You can get them in all colors, including white.
Mister Mozina, do I need to explain you the difference between an incadescent and a fluorescent light bulb?
The neon inside a fluorescent bulb don't give off VISIBLE light; that's why the fluorescent bulbs are coated with phosphor. The phosphor is the one actually shining in a neon lamp... since the sun give visible light would you concede that is actually phosphor on it's surface and not neon?

Quote:
Originally Posted by Michael Mozina
Exactly Baloo. They have to be iron ions flowing through those arcs. You just explained it yourself perfectly. There can be no other light source here other than iron ion photons. That *IS* the light source and HAS TO BE the light source.
Mister Mozina, nobody disagreed that there is iron in those arcs. But due to the used filters in those pictures all the other elements (that are emitting on other wavelengths) are not visible. Is like using sunglasses at night: you'll see the white cars, but the blue, brown, black, red cars will be invisible to you.

Last edited by Baloo; 25-September-2005 at 10:57 PM. Reason: Remove double quote
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  #154 (permalink)  
Old 25-September-2005, 11:08 PM
seohtu seohtu is offline
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I'm curious - do the blackbody radiation formulae apply in the case of the filaments? The argument seems to be that the amount of radiation with that particular spectral signature, *if the source is ionized iron* (correct me if that isn't the proper term), cannot be produced without the kind of heat generated by massive electrical arching through them. That argument is being countered by the possibility that the filaments are emitting at that frequency through a basic blackbody radiation mechanism? If so, the calculations would be based on the area of the filament in question, right? Not the area of the sun? Do the numbers still hold up then?

As usual, I won't be offended if I'm ignored.

Derrick Baumer
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  #155 (permalink)  
Old 25-September-2005, 11:26 PM
seohtu seohtu is offline
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Quote:
Originally Posted by Baloo
Mister Mozina, do I need to explain you the difference between an incadescent and a fluorescent light bulb?
The neon inside a fluorescent bulb don't give off VISIBLE light; that's why the fluorescent bulbs are coated with phosphor. The phosphor is the one actually shining in a neon lamp... since the sun give visible light would you concede that is actually phosphor on it's surface and not neon?
I may be mistaken, but I don't think so, in my understanding that neon lights are not the same as fluorescent lights as used commonly in commercial installations, and that neon, when electrically excited, does, indeed emit at visible wavelengths, but not in multiple colors as suggested by Mr. Mozina. Different colors are achieved by using different elements. For instance, the famous green "neon" outline of that skyscraper in Dallas (or is it Houston?) is actually argon, not neon. Some effects can be achieved by using colored glass...

The point being that in the case of neon lights, as opposed to fluorescent lights, it actually is the neon that is emitting the light under electrical current, or argon, or whatever other esoteric element they came up with.

Sincerely,
Derrick Baumer
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Last edited by seohtu; 25-September-2005 at 11:29 PM. Reason: Typo alert!
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  #156 (permalink)  
Old 25-September-2005, 11:55 PM
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Quote:
Originally Posted by seohtu
I'm curious - do the blackbody radiation formulae apply in the case of the filaments? The argument seems to be that the amount of radiation with that particular spectral signature, *if the source is ionized iron* (correct me if that isn't the proper term), cannot be produced without the kind of heat generated by massive electrical arching through them. That argument is being countered by the possibility that the filaments are emitting at that frequency through a basic blackbody radiation mechanism? If so, the calculations would be based on the area of the filament in question, right? Not the area of the sun? Do the numbers still hold up then?
Take a look here: http://online.cctt.org/physicslab/co...andspectra.asp (scroll down a little bit) and here: http://coolcosmos.ipac.caltech.edu/c...rial/spec.html for a short tutorial concerning the blackbody radiation vs. emission/absorption spectra.
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  #157 (permalink)  
Old 26-September-2005, 12:22 AM
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Thank you for those links, Baloo. Now I don't have to learn anything else new until tomorrow. Phew!

Am I correct in my understanding, then, that one side is arguing that we are looking at continuous spectra in the provided images, which would be visible equally regardless of the elemental composition or source of heat in the filaments, at various wavelengths, and the other side is arguing that we are looking at emission spectra from ionized iron, which could only occur under extreme electrical current or other source of extreme heat, and should only be highly visible at specific frequencies?

The easies validation on this specific point, then, would be to locate a spectral analysis of such an arc projecting from the sun, without the light from the sun interfering with the reading. Do we have the technology to do this? Has it been done? Can we acquire that data? Should cartoonists be required to be drawn cartoon characters fully dressed? Where would one look for this kind of data? As the theory being presented is Mr. Mozina's, do I understand correctly that it would be his responsibility to acquire such data (if it is, indeed, relevant) and explain the results of its analysis in terms of his theory?
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  #158 (permalink)  
Old 26-September-2005, 12:24 AM
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Quote:
Originally Posted by seohtu
I may be mistaken, but I don't think so, in my understanding that neon lights are not the same as fluorescent lights as used commonly in commercial installations, and that neon, when electrically excited, does, indeed emit at visible wavelengths, but not in multiple colors as suggested by Mr. Mozina. Different colors are achieved by using different elements. For instance, the famous green "neon" outline of that skyscraper in Dallas (or is it Houston?) is actually argon, not neon. Some effects can be achieved by using colored glass...

The point being that in the case of neon lights, as opposed to fluorescent lights, it actually is the neon that is emitting the light under electrical current, or argon, or whatever other esoteric element they came up with.

Sincerely,
Derrick Baumer
I stand corrected.
What I've said is indeed true only for argon/mercury fluorescent lamps.

Concerning Neon emission: "The neon spectrum is dominated by red lines.
That's why neon signs are red."

(from here)

Doesn't look like sun light to me...
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  #159 (permalink)  
Old 26-September-2005, 12:34 AM
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Quote:
Originally Posted by Baloo
I'm not playing both sides, what you're missing (altough it has been pointed to you many times) is that light is emitted in more than just one way:

1. Light emitted by atom/molecule transition from one energy level to another; since those levels are quantized the emitted spectrum is discreet and unique for each chemical element, therefore this kind of emission is used in spectroscopy.
Ok, I'm with you so far Baloo. That is what interested me in the SERTS data in fact. It listed three specific iron ions, the same three that SOHO and TRACE can image. By definition, the images we see in these satellites are from iron ions.

Quote:
2. Blackbody radiation, emitted in a continuous spectrum and depending only on the temperature. (I must point out that both those mechanisms have the same origin: the energy is emitted in quanta, so you accept or reject them both.)
I absolutely accept they are emitted in quanta. My whole point is based on that premise in fact. I'll see if I cant find where we disagree, because to this point in time we're totally and completely on the page except for any notion that ALL elements emit ALL quantas.

Quote:

http://www.thefreedictionary.com/synchrotron
Quote:
3. Synchrotron/cyclotron radiation: produced by an accelerated charged particle. It has a continuous spectrum.
syn·chro·tron Pronunciation (sngkr-trn, sn-)
n.
An accelerator in which charged particles are accelerated around a fixed circular path by an electric field and held to the path by an increasing magnetic field.
I want to be sure we're talking about the same mechanism here. You are suggesting the sun is producing strong electromagnetic fields "without" the flow of electricity? How?

Quote:
There are other mechanisms that produce light, but already you may see that our sun give a spectra that is a superposition from (at least) three other spectra, two continuous and one discreet.
Well Baloo, first I need to understand how you intend to do this without electrical flow and how your idea is superior to mine based on direct observation. I will keep an open mind, but as you might suspect, I will also ask some questions along the way. I first need to see you apply this method of yours to images and real world things. I will accept it may apply, but you'll have to demonstrate how it is a "better" explanation than the one I offered. I particularly am interested in what continues create these ever increasing magnetic fields all the way into the tops of these arcs if it isn't electricity. I think that is where you will run into your biggest obsticle.

Quote:
It does matter as long as you're aware of all the means by which the said photon could be generated. You're not so you think that all the photons are generated in the same way, which is false.
No, I actually concede your point here Baloo. I will concede that it is "possible" that these quanta are produced by ever increasing magnetic fields, but what causes that increase in the field in the first place?

If you would like to show me how your admittedly viable option applies to these images, I will consider it. I will concede that it "could" apply, but not that it "does" apply.

Quote:
Why don't you ask dr. Manuel or dr. Bruce what do they think about this "50 cents" formula? You'll be surprised.
As it relates to your example, I'm sure they are rather pleased with your presentation. Point noted. Now if you will be so kind as to show me how it applies to our sun......

Quote:
A body having about 102000 °K will give a continuous spectrum having a peak intensity at exactly 284 A.
Woah. What "kind" of body? Let's establish the makup of this body and some of it's properies. Before we go any further, let's do a reality check here. I will willingly concede there are viable alternatives, though you have yet to explain the increasing magnetic field without the flow of electricity. Even still, we should compare what we see what these two options and see which fits the observation best, correct?

Before I go any further, I would like to start there. Agreed?

What accelerates these emissions and concentrates them in the arcs that is not electrical in nature?
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  #160 (permalink)  
Old 26-September-2005, 12:39 AM
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Quote:
Originally Posted by seohtu
Am I correct in my understanding, then, that one side is arguing that we are looking at continuous spectra in the provided images, which would be visible equally regardless of the elemental composition or source of heat in the filaments, at various wavelengths, and the other side is arguing that we are looking at emission spectra from ionized iron, which could only occur under extreme electrical current or other source of extreme heat, and should only be highly visible at specific frequencies?
Not quite...is actually a long story. You may take a look here and here and here to see what is all about...you could jump to the last pages in each thread since they're pretty long.
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Old 26-September-2005, 12:46 AM
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Quote:
Originally Posted by Baloo
Take a look here: http://online.cctt.org/physicslab/co...andspectra.asp (scroll down a little bit) and here: http://coolcosmos.ipac.caltech.edu/c...rial/spec.html for a short tutorial concerning the blackbody radiation vs. emission/absorption spectra.
http://acs.omnibooksonline.com/papers/2001_P2.16.pdf

When you are ready to discuss the g_band frequency and how it relates to the photosophere, I believe I am ready. That frequency just so happens to form with neon and helium, which in my model would define the boundry zone between the neon and helium layers. If ever there was a strong argument for my model this is it. Thanks for dropping that bit of info in my lap. I should have made that link already and your question piqued my curiousity. As you see this particular frequency is linked to helium and neon and the border areas of my model, specifically the heat conducting tops of the phenumbral filament layer, where neon meets helium.
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Old 26-September-2005, 12:47 AM
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Am I illustrating my ignorance by asking what is the evidence that these projections are accompanied by increasing magnetic fields? I don't doubt that such evidence exists - I am simply unaware of it.

I think (God help us) that Baloo is trying to establish what is apparently an accepted scientific fact that

"Continuous spectra (also called a thermal or blackbody spectra) are emitted by any object that radiates heat (has a temperature). The light is spread out into a continuous band with every wavelength having some amount of radiation. For example, when sunlight is passed through a prism, it's light is spread out into it's colors."
--- http://coolcosmos.ipac.caltech.edu/c...rial/spec.html

and that this means the specific wavelength you are referring to can, indeed, be emitted by any element. It would just be stronger in relation if the source were electrically-charged iron. I may be projecting my own question onto his by asking if a thorough spectral analysis of these projections would be able to differentiate between the continuous spectra emitted by *any* element in the arcs and the specific spectra produced by ionized iron. It appears that simply detecting energy at those wavelengths does not necessarily prove a strong presence of iron.
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  #163 (permalink)  
Old 26-September-2005, 12:48 AM
Michael Mozina Michael Mozina is offline
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Quote:
Originally Posted by Baloo
I stand corrected.
What I've said is indeed true only for argon/mercury fluorescent lamps.

Concerning Neon emission: "The neon spectrum is dominated by red lines.
That's why neon signs are red."

(from here)

Doesn't look like sun light to me...
Yes, but when you mix it with other elements, neon can shine in nearly any color and the greater amount of electrical current, the greater neon intesifies toward "white" rather than red.
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  #164 (permalink)  
Old 26-September-2005, 12:51 AM
Michael Mozina Michael Mozina is offline
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Quote:
Originally Posted by Michael Mozina
http://acs.omnibooksonline.com/papers/2001_P2.16.pdf

When you are ready to discuss the g_band frequency and how it relates to the photosophere, I believe I am ready. That frequency just so happens to form with neon and helium, which in my model would define the boundry zone between the neon and helium layers. If ever there was a strong argument for my model this is it. Thanks for dropping that bit of info in my lap. I should have made that link already and your question piqued my curiousity. As you see this particular frequency is linked to helium and neon and the border areas of my model, specifically the heat conducting tops of the phenumbral filament layer, where neon meets helium.
Well, I jumped the gun. I misread something in that article. I'm not ready after all. I'll keep working on it however.
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  #165 (permalink)  
Old 26-September-2005, 01:00 AM
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Quote:
Originally Posted by Michael Mozina
If you have a scientific explanation for the light source, be a man about it and put it on the table. If not, go away.
I'm guessing you are talking about the running difference movie. Ok, since this one doesn't take much interpretation, I'll answer it. It is quite simple: When you watch the movie ... you're looking at it. The surface of the sun is the light source. Which is, through many lines of evidence, a hot gas of mostly hydrogen and helium.

Incidently, a "running difference" image is made to show and emphasize differences between frames. It is not the same as a conventional photograph. It is a perfect example of why you cannot just "look at the picture." Unless you have the full context and understand what it means, your assumptions will likely be incorrect.

Now, I have a hunch you won't agree with my answer, but will you "be a man" and accept that I answered it and put it to rest? Or will you deny it and ask endless questions based on your interpretations?

Quote:
All your comments are loaded with stupid put downs. That last line about "I suggest you read...." is a perfect example.
I didn't know that was a put down.

If somebody points out an error in my statements I take it as constructive criticism and study the subject and drop out of the discussion until I understand where I made my error. If somebody suggests I study a subject, and if I don't know why, I'll ask for details. Then, where appropriate, I'll study the subject to correct my misunderstandings.

You have said a number of things that lead me to suspect you are unfamiliar with certain subjects. For instance:

Quote:
There is no denying that 2D concepts of Newtonian gravity work well relative to the movement plane of the solar system.
Considering that "Newtonian gravity" works quite well in three dimensions, this suggests you are unfamiliar with the subject. If you have another concept that you wish to present, that's fine, but first you need to understand the material you are attempting to overthrow.

It comes down to this: I'm far from the only one who has noticed apparent misunderstandings in your statements. Given that you are the one that is trying to overthrow a good chunk of well established science, don't you think it would help to research these subjects in more detail?
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  #166 (permalink)  
Old 26-September-2005, 01:38 AM