Black Body and our sun... [Archive] - Bad Astronomy and Universe Today Forum

Tim Thompson

26-August-2006, 10:23 PM

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. (http://grus.berkeley.edu/~jrg/ay202/node81.html)

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 (http://scienceworld.wolfram.com/physics/PlanckLaw.html) 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.

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.

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.

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 (http://scienceworld.wolfram.com/physics/Opacity.html), or optical depth (http://scienceworld.wolfram.com/physics/OpticalDepth.html) 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.

Van Rijn

26-August-2006, 11:45 PM

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)/guides/mtr/fw/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.

upriver

27-August-2006, 08:13 AM

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 (http://www.jet.efda.org/pages/content/fusion2.html) 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 (http://www.nrc-cnrc.gc.ca/highlights/2006/0602blackbody_e.html) 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 (http://trace.lmsal.com/POD/images/arcade_9_nov_2000.gif) of the sun at 171A through the photosphere.

Sun spot. (http://www.universetoday.com/am/uploads/2005-1006sun.jpg)

tusenfem

27-August-2006, 11:52 AM

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.

Tim Thompson

27-August-2006, 09:08 PM

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.

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.

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.

"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.

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.

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.

Again here is the blackbody surface (http://trace.lmsal.com/POD/images/arcade_9_nov_2000.gif) 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.

upriver

28-August-2006, 01:08 AM

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 (http://trace.lmsal.com/POD/images/arcade_9_nov_2000.gif) 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?

Van Rijn

28-August-2006, 04:29 AM

This is an image (http://trace.lmsal.com/POD/images/arcade_9_nov_2000.gif) 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.

tusenfem

28-August-2006, 10:49 AM

Okay, grabbing my copy of Rybicky and Lightman and looking up optical depth. On page 12 of the book you can read:

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.

upriver

29-August-2006, 04:04 AM

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 (http://trace.lmsal.com/POD/images/arcade_9_nov_2000.gif) 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; (http://en.wikipedia.org/wiki/Ultraviolet)

Visible (optical) spectrum: Violet | Blue | Green | Yellow | Orange | Red

So when I say optical, now you know.

Van Rijn

29-August-2006, 06:12 AM

This picture (http://trace.lmsal.com/POD/images/arcade_9_nov_2000.gif) 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.

tusenfem

30-August-2006, 06:53 AM

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).

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".

upriver

31-August-2006, 10:38 PM

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 (http://www.strangeeye.net/universe/pressurebroad.jpg), 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

papageno

31-August-2006, 11:54 PM

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 (http://nobelprize.org/nobel_prizes/physics/laureates/1918/planck-lecture.html):

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:

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.

korjik

01-September-2006, 12:25 AM

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.

Tim Thompson

01-September-2006, 01:06 AM

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.

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?

As you can see by this plamsa emission spectrum (http://www.strangeeye.net/universe/pressurebroad.jpg), 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?

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 (http://solarscience.msfc.nasa.gov/t_region.shtml), 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.

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.

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.

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.

tusenfem

01-September-2006, 09:36 AM

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.

upriver

02-September-2006, 09:47 AM

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,
Einstein’s 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 Stephan’s 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, Stephan’s 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/T171_010322_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.

papageno

02-September-2006, 03:57 PM

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:

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?

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"?

Nonetheless, physics has long since dismissed the importance of Kirchhoff’s work [9]. The basis for universality, no longer rests on the experimental proof [i.e. 9], but rather on Einstein’s 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, 1374–1383 (1969)
O. Theimer, "Derivation of the Blackbody Radiation Spectrum by Classical Statistical Mechanics", Phys. Rev. D 4, 1597–1601 (1971)
D. C. Cole, "Reinvestigation of the thermodynamics of blackbody radiation via classical physics", Phys. Rev. A 45, 8471–8489 (1992)

It has been held [i.e. 9] that with Einstein’s 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 Einstein’s derivation of the Planckian equations had moved the community beyond the limited confines of Kirchhoff’s enclosure. Einstein’s 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:

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.

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.

"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, Einstein’s 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:

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, Einstein’s 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, Einstein’s 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.

papageno

02-September-2006, 07:43 PM

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)

upriver

06-September-2006, 04:33 AM

tusenfem

06-September-2006, 08:13 AM

upriver

08-September-2006, 06:35 AM

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/PwrfromAEemissions.html#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."

tusenfem

08-September-2006, 08:08 AM

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 (http://climate.gsfc.nasa.gov/~cahalan/Radiation/Images/SolarIrrVblackbody.gif). 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.

upriver

10-September-2006, 05:04 AM

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.

Tim Thompson

10-September-2006, 06:17 AM

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.

korjik

10-September-2006, 06:22 AM

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

upriver

10-September-2006, 06:35 PM

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 (http://www.impulsedevices.com/media.html) that I shot.

I ran across this issue because I am specifying a spectrometer (http://www.jobinyvon.com/usadivisions/Mono/triax180190.htm) and a streak (http://www.sales.hamamatsu.com/en/products/system-division/ultra-fast/streak-systems/c5680.php)camera for our budget next year.

upriver

11-September-2006, 04:51 AM

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.

upriver

11-September-2006, 05:01 AM

The visible light from the sun is not generated in the photosphere, it only passes through the photosphere from below.

What does that yellow stuff do on the top of the photosphere? Is it dark under the photosphere(sunspots)? Is it cooler or hotter under the photosphere?

Scientists Image the Three-dimensional Surface of the Sun
http://www.lmsal.com/Press/SPD2003.html

Here is the link to the The Institute for Solar Physics solar gallery.
http://www.solarphysics.kva.se/

captain swoop

11-September-2006, 12:05 PM

Is it dark under the photosphere(sunspots)?

Sunspots aren't dark.

papageno

11-September-2006, 01:07 PM

Since the Einstein coefficients give the probability for a transition, which can occur only if the frequency is nu = E/h, why does Dr. Robitaille think that the derivation requires a single atom to be a perfect absorber of radiation over the whole spectrum?

The derivation is for a theoretical blackbody continuous emission spectrum.

That's not the point.
Read again the relevant passage:

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, Einstein’s 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, Einstein’s perfectly absorbing atoms could be permitted.
(my emphasis)
Dr. Robitaille attributes to Einstein the assumption that an atom is a perfect absorber over the whole spectrum.
But this is a strawman, because the derivation does not require such an assumption.

As such it is applied to all cases without the understanding that emission for gas or liquid is different than condensed matter.
Part of the problem is that any spectrum hotter than 3500K is electrically driven is a discharge tube or plasma. We have no examples of thermally heated condensed matter above 3500, so its been assumed for a long time that this was correct, that bb emission extended to all matter.
In this way by default, a single atom is required to act the same as a brick of graphite.
I think that is the correct answer.

It does not even address my question: why does Dr. Robitaille say that Einstein's derivation requires that an atom is a perfect absorber over the whole spectrum?
This is about the theoretical derivation, which does not care about the practical limitations of experiments.

Do you remember how Planck used the Hertzian linear oscillator? He used them as a probe to determine the energy density in the cavity at a given frequency. To find the energy density at a different frequency, he used oscillators with different natural frequency.

Where does Dr. Robitaille actually show that the small piece of graphite dominates the radiation?
Where does he show that "inserting a small piece of graphite into the perfectly reflecting enclosure" equals "Kirchhoff lined the entire box with graphite"?

I dont know. I will email him. Also see below.

Poor old Kirchhoff! How long should he have waited?

I would expect thermal equilibrium could have been deduced by the rate of thermal change, and then wait a little longer. If the rate of change is too small, you probably could assume it(a highly polished metal enclosure) will never reach thermal equilibrium in a reasonable amount of time(5 yrs?).

I also said:

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?

Now if I were to examine this particular idea from an upriver ATM point of view, I would say it would never reach equilibrium because of one bit I ran across by John Bedini from his webpage on Nathan Stubblefield.

[snip!]

Since when is a web-page about free energy a reliable source?

What about my references?

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)

Of course its not peer reviewed, but its so strange, how could you make that up as an effect.

Your lack of imagination is not evidence.

If there was even the smallest effect related to this, the box would never be at equilibrium. And in the case of the graphite, I think it destroyed whatever effect was happening in the box(two polished parallel surfaces?).

There is a huge leap between "I think" and "It is". Can you bridge that gap with something substantial?

korjik

11-September-2006, 04:45 PM

If somebody can produce a BB emission(vs lines) on earth in a lab with a plasma like the photosphere I will retract my claim.

Just remembered what I wanted to say.

I think the that it is pretty much impossible to measure the blackbody radiation of photosphere conditions on the earth due to technical and/or financial considerations, not physics.

the technical considerations I can think of are that you would be trying to read the blackbody radiation of a transparent, hot gas without seeing the radiation of the background vaccum chamber. Just a quicky thought of mine on how to do it would be to inject the hot gas into a nitrogen cooled chamber. It would still be very hard to see even in an immense chamber.

The financial aspect would be that the above would be hideously expensive.

korjik

11-September-2006, 04:48 PM

Upriver, if the sun's surface is a solid, why does limb darkening fit a cooler temperature blackbody curve? Shouldnt it be a scattered constant temperature curve?

Tim Thompson

11-September-2006, 07:12 PM

The visible light from the sun is not generated in the photosphere, it only passes through the photosphere from below.
What does that yellow stuff do on the top of the photosphere?
It looks to me like that "yellow stuff" is the photosphere. The image is taken at different wavelengths that the images you were showing before, which reveal features well above this image, which are invisible here.

Is it dark under the photosphere (sunspots)?No. if you could see it revealed, "under the photosphere" would be considerably brighter than the photosphere, by virtue of being hotter. But the emission would tend toward ultraviolet & shorter wavelengths. "Under the photosphere", deep enough, would actually "look" relatively dark, because there would be few photons left, with wavelengths so long as visible light. Sunspots are relatively dark because they are relatively cool. They may extend below the photosphere, I am not sure about that. But I think they are for the most part entirely contained in the photosphere.

Is it cooler or hotter under the photosphere?
Hotter. Where the photosphere temperature is about 5700 K, the core temperature is about 15,000,000 K. The temperature rises steadily from the photosphere down towards the core.

tusenfem

12-September-2006, 02:37 PM

From previous discussions on this topic, I still have not obtained an answer to some questions. The main question is: If the sun is a shell of solid iron (powered by some mysterious electric sun idea) how thick is this shell and do you think that this iron shell will be stable at a temperature of 5700 K?

upriver

13-September-2006, 05:24 AM

It does not even address my question: why does Dr. Robitaille say that Einstein's derivation requires that an atom is a perfect absorber over the whole spectrum?
This is about the theoretical derivation, which does not care about the practical limitations of experiments.

Do you remember how Planck used the Hertzian linear oscillator? He used them as a probe to determine the energy density in the cavity at a given frequency. To find the energy density at a different frequency, he used oscillators with different natural frequency.

I emailed Dr. Robitaille, as soon as I have a response, we will revisit this issue.

Since when is a web-page about free energy a reliable source?

Nathan Stubblefield by any accounts was a pioneer. It seems there was a whole field of "earth" energy that was being discovered.

http://en.wikipedia.org/wiki/Nathan_Stubblefield
http://smart90.com/nathanstubblefield

John Bedini. He invented the Audio Clarifier. He is also a "free energy" researcher. He hosted the Nathan Stubblefield page.
http://www.icehouse.net/john1/stubblefield.html
Forget the link on the top of the page and scrolldown.

"As a result of the disc manufacturing process the disc has inherent noise distortion which becomes more apparent as the disc is played over and over again; coupled with the polymer ability to hold electrostatic charges, it acutely masks the true dynamics of the digital media."

Bedini Electronics
http://www.bedini.com/

What about my references?

Quote:
Originally Posted by papageno
G. Machin and B Chu, "A transportable gallium melting point blackbody for radiation thermometry calibration", Meas. Sci. Technol. 9, 1653-1656 (1998)

E. Usadi, "Reflecting cavity blackbodies for radiometry", Metrologia 43, S1-S5 (2006)

"Reflecting cavity blackbodies for radiometry" is a return to the first case since it is not a 100% reflective apparatus. And its spherical!

"A transportable gallium melting point blackbody for radiation thermometry calibration" is based on a transportable source that maintains its accuracy.
It's interesting that gallium maintains it's melting point under "any" condition.

The issue is gas/plasma only emits lines. Condensed matter emits a BB curve that is distinguishable from a pressurized gas curve.

Can you bridge that gap with something substantial?

I guess I would have to mount two polished metal plates parallel and see what happens. From what I've read copper and steel.

Your lack of imagination is not evidence.

That's funny!!!

upriver

13-September-2006, 05:56 AM

Originally Posted by korjik
Upriver, if the sun's surface is a solid, why does limb darkening fit a cooler temperature blackbody curve?

What temperature are you talking about? Cooler than 6000K of the photosphere? I would expect it to be at the temperature of solid iron.

The angle dependance of thermal emission would naturally have that effect of darkening depending on the angle.
Siegel R., Howell J. Thermal radiation heat transfer. 4th ed.,
Taylor and Francis, New York, 2002,.

Shouldnt it be a scattered constant temperature curve?

Not sure what you mean.

upriver

13-September-2006, 08:00 AM

upriver

13-September-2006, 08:38 AM

Originally Posted by tusenfem
The main question is: If the sun is a shell of solid iron (powered by some mysterious electric sun idea) how thick is this shell

Since iron is more massive than the plasma that the MS sun is "composed' of, it would be the equivalent to the amount required to produce the 1.4 density figure that is based on the earths orbit. You did the math before.
It really doesnt matter since we know the surface is iron and it has not collapsed and it has a certain inferred density. There is some experimental evidence that arcs(z-pinch) produce hollow iron spheres. Supernova have iron cores. And a barrel shaped(z-pinch) remnants. Check out welding slag, etc.

I cant see inside the sun so anything I say is going to be a guess just like everyone else.

and do you think that this iron shell will be stable at a temperature of 5700 K?

The surface is not at 5700K That is the average temperature of the surface, not the spot temperature. The surface could be a glowing and still be solid(cooler) deeper. Only in the bright spots is that really true.

Forskern

13-September-2006, 08:42 AM

I would say that most of the suns visible light comes from the top layer of the photosphere. Clearly the center of the sunspot is either emitting UV or IR because its not emitting visible(396 to 630nm).

Actually the photosphere is defined loosely as "where most of the visible light comes from", so thats a safe assumption :P More soberly though, it is easy to compare the intensity (erg/Å/cm^2/sterad/s) measured in UV and visible light and confirm that the coronal emission is insignificant in the energy budget.

Also, the sunspots are giving off visible light just like the area around them, but at a lower BB temperature. That just makes them look black in photographs. While the active regions that are often associated with them give off more EUV than the rest of the solar disc, I'd still expect that most of the energy is still radiated in the visible, not in UV.

Now in this picture (http://trace.lmsal.com/POD/images/T171_010419_1330.jpg) taken at 17.1nm,there appear to be areas that are opaque to EUV in this layer that in the MS view is above the photosphere.

Where in that picture are those feature you believe are blocked out by opacity?

You mean to tell me that if I put a cool thing between two hot things it will stay cool?

I think he means that when you turn of the heating, it will get colder. This is just what happens, the outer portion of the Sun (from about two thirds of the solar radius) is convectively heated. While the usual radiative heat transfer still works here, it is not strong enough to keep the spots at the same 6000 degrees as the convectively heated rest.

captain swoop

13-September-2006, 10:06 AM

It really doesnt matter since we know the surface is iron and it has not collapsed ...

We don't know this at all.

tusenfem

13-September-2006, 12:24 PM

Since iron is more massive than the plasma that the MS sun is "composed' of, it would be the equivalent to the amount required to produce the 1.4 density figure that is based on the earths orbit. You did the math before.
It really doesnt matter since we know the surface is iron and it has not collapsed and it has a certain inferred density.

Uh, no, you keep on claiming that the sun has an iron surface, with no evidence to back up your claim.

There is some experimental evidence that arcs(z-pinch) produce hollow iron spheres. Supernova have iron cores. And a barrel shaped(z-pinch) remnants. Check out welding slag, etc.

I know of no such experimental evidence. Oh, I know you are obsessed by z-pinches, but having them create hollow iron spheres ... ? And I would hope that welding does not create hollow spheres, might not be real stable, I guess.

I cant see inside the sun so anything I say is going to be a guess just like everyone else.

So, just because you cannot see inside the sun, you can claim you are right? How about helioseismology, a well developed technique, which shows no evidence at all of a solid iron shell anywhere near the bottom of the photosphere.

The surface is not at 5700K That is the average temperature of the surface, not the spot temperature. The surface could be a glowing and still be solid(cooler) deeper. Only in the bright spots is that really true.
[/quote]

This does not make sense. So you want to have a BB spectrum of a body with a temperature of 5700 K, and since (in your view) a BB can only come from a solid, you need an iron shell of this temperature. Now you tell us that the shell may be cooler and only some hot spots hotter, but will the average of these give the nice BB at 5700 K?

And, just another comment, you do not see the top of the photosphere, you see the bottom, actually you see where tau = 2/3. That is where the photons come from, because that is the location from which they are able to escape.

And, one more comment, you cannot look at a line spectrum (your iron line showing the magnetic loops) and say that there are opague regions. It just does not work that way. In an emission line you see where the iron is, and it is dark where it is not.

papageno

13-September-2006, 02:28 PM

Since when is a web-page about free energy a reliable source?

Nathan Stubblefield by any accounts was a pioneer. It seems there was a whole field of "earth" energy that was being discovered.

[snip links!]

How about going to a library?

What about my references?

"Reflecting cavity blackbodies for radiometry" is a return to the first case since it is not a 100% reflective apparatus. And its spherical!

"A transportable gallium melting point blackbody for radiation thermometry calibration" is based on a transportable source that maintains its accuracy.
It's interesting that gallium maintains it's melting point under "any" condition.

The references were provided to disprove claims, based on Dr. Robitaille's paper, that graphite is the only available black-body:

Eventually, graphite’s behavior became the basis of the laws of Stefan [7], Wien [8] and Planck [3].

Obviously modern researchers do not agree with Dr. Robitaille's conclusion.

The issue is gas/plasma only emits lines. Condensed matter emits a BB curve that is distinguishable from a pressurized gas curve.

But black-body spectrum is not an emission spectrum: it is the spectrum of EM radiation in thermal equilibrium.
Just as Dr. Robitaille, you are confusing a metrological, technical, issue with a fundamental problem.

Can you bridge that gap with something substantial?

I guess I would have to mount two polished metal plates parallel and see what happens. From what I've read copper and steel.

Then come back when you have done the experiment and shown that graphite "destroyed whatever effect was happening in the box".

korjik

14-September-2006, 05:15 PM

What temperature are you talking about? Cooler than 6000K of the photosphere? I would expect it to be at the temperature of solid iron.

The angle dependance of thermal emission would naturally have that effect of darkening depending on the angle.
Siegel R., Howell J. Thermal radiation heat transfer. 4th ed.,
Taylor and Francis, New York, 2002,.

Not sure what you mean.

Limb darkening. Wiki has a pretty good article and a very good picture. Basically, the edges of the sun look cooler than the center. See the wiki article to see why, cause they explain it alot better.

This wouldnt work if we are seeing light from a solid surface. The temp of the surface sould be relatively constant from center to limb, so the center should be a 5700K blackbody and the limb should be a dimmed by scattering 5700K blackbody. What you get is a 5700K bb at the center and a cooler, and there for dimmer and more orange bb curve near the limb.

How does this fit into your theory?

Tim Thompson

14-September-2006, 07:10 PM

It looks to me like that "yellow stuff" is the photosphere. The image is taken at different wavelengths that the images you were showing before, which reveal features well above this image, which are invisible here.
The images from the solar telescope are from 396nm to 630nm, which is visible light. The image I posted before are from 171 to 195A or 17.1nm to 19.5nm (EUV). With magnetogram pictures being 630nm. Interesting that they take pictures of the suns magnetic field with iron plasma(FeI) in visible light.
So, we agree that the two images are at different wavelengths, and show different features. Hurrah.

In visible light, iron is seen strictly in absorption, never in emission. But an absorption spectrum reveals Zeeman splitting of the spectral lines, which allows a direct determination of the magnetic field strength. So iron can be used to measure & map the magnetic field strength. But the loops & filaments, as are visible in the extreme ultraviolet (EUV) images are not visible at these wavelengths.

Now, the visible images are "multispectral", which means they are broadband, and cover a wide range of spectral lines in one image. On the other hand, the EUV images are narrow band images, restricted to a single spectral line of highly ionized iron. So, they reveal where the iron is, and they reveal the loops & filaments. But they do not reveal, for instance, how much iron there is, unless that emission is compared to line emission from other species. If you interpret the images to mean that there is only iron present, or perhaps that there is mostly iron present, then you are wrong. The images do not contain enough information to do that.

And if I was to just look at those pictures, I would say that most of the suns visible light comes from the top layer of the photosphere.
And you would be right. but there is a lot of difference between saying that, and saying that the visible light is physically generated, or created, in the photosphere.

When you look at the photosphere, you are looking at the surface of last scattering, where the mean free path of the photon has at last become long enough that it can escape altogether, and make the trip from the sun to us. Below the photosphere, the photons are scattered, with increasing strength (shorter mean free path), the farther down you go. As the photons migrate upward from the deep interior, they loose energy, and become collectively thermalized. The loss of energy means they present a lower temperature when thermalized by scattering. In the photosphere, the atmosphere of the sun is sufficiently sparse that line absorption can take place. So, the thermal spectrum is deformed, and we get what we see, a thermal spectrum with line absorption superimposed on it. The trip from the center of the sun, to the photosphere, takes a typical photon about 1,000,000 years.

Clearly the center of the sunspot is either emitting UV or IR because its not emitting visible (396 to 630nm).
Well, actually it emits a lot of visible light, but an exposure set to match the sorrounding, much brighter photosphere, makes it look relatively dark. And since it is much cooler, then it should emit more IR than the surrounding atmosphere.

Now in this picture (http://trace.lmsal.com/POD/images/T171_010419_1330.jpg) taken at 17.1nm, there appear to be areas that are opaque to EUV in this layer that in the MS view is above the photosphere.
Well, they could be opaque. Or maybe there just isn't enough 17.1 nm emission to register, in an exposure designed to reveal the much brighter loops. In fact, that is the likely explanation, as the plasma in the magnetic field loops should be very much hotter than the plasma below.

You mean to tell me that if I put a cool thing between two hot things it will stay cool? The quoted thickness for the photosphere is 400km. That requires new physics. Or a different model.
Yes, that is what I mean to tell you. No, it does not require any new physics, and it does not require a new model. It only requires a knowledge of both physics and solar models that you sadly lack.

There is a thing in my kitchen, a modern convenience called a "refrigerator". The inside stays very cold, despite being immersed in a sometimes very hot kitchen. It works because it is cooled by non-spontaneous processes. It is cooled by a heat pump, that pumps heat "uphill", away from the cold and towards that hot, in exactly the opposite direction that one would expect heat to naturally flow. I have yet to see anyone suggest that new physics is required to understand the refrigerator.

In the case of the photosphere of the sun, it is just that easy. The photosphere remains cooler than the underlying, hotter interior, because of purely spontaneous, ordinary thermodynamics. Because the sun is in thermal equilibrium (or very nearly so), each successive spherical shell of arbitrary (but equal) radial extent must hold that same total thermal energy as the shell below (or above). But the shells are successively larger in volume, so the thermal energy density must be lower, in the successively outward shells. So the temperature must be lower, hence the sun gets cooler from the center to the photosphere.

But starting about the bottom of the convective zone, perhaps 1/4 of the way down from the photosphere, into the sun, the magnetic field is increasingy less confined by the progressively less dense plasma. At and above the photosphere, the strongly time varying magnetic field pumps the tenuous plasma in the transition zone, chromosphere & corona into much higher temperatures than the photosphere. It is in essence a heat pump, though very different from the refrigerator. In this case, the pump is non-thermal, so that time varying magnetic fields (which necessarily generate equally time varying electric fields) accelerate the plasma, and hence vastly increases the temperature of the higher layers of the atmosphere; temperature is proportional to kinetic energy, so anything that increases kinetic energy, also increases temperature. The pump is radial, so in this case, the temperature has a preferred, outward radial direction.

Well the only UV comes from the coronal loops as you can see by the picture (http://trace.lmsal.com/POD/images/T171_000527_144107.gif), because it certainly is not coming from inside the sun as the loops are providing the only illumination (reflected and direct). It is a filter at 17.3nm center with a bandwidth of .64nm or 6.4A. As a matter of fact if you look in the center of the picture you can see a feature blocking a flare. That feature is 20,000km high. And that feature is opaque to light from emission of 160,000 to 2 million degrees (FeIX).
I fail to see why this is a relevant point. Who said the UV was not generated in the loops, except perhaps yourself?

So you saying that short wavelength radiation converts to visible right at the photosphere?
Yes.

So what makes the photosphere brighter? At wavelengths from 396 nm to 630 nm?
Its temperature. Peak thermal emission, at the temperature of the photosphere, should be at about 550 nm. No UV from below will get out, and the photosphere should generate very little of either UV or IR because of its temperature.

"Sunspots are "dark" because they are cooler than their surroundings. A large sunspot might have a central temperature of 4,000 K (about 3,700° C or 6,700° F), much lower than the 5,800 K (about 5,500° C or 10,000° F) temperature of the adjacent photosphere."
http://www.windows.ucar.edu/sun/atmosphere/sunspots.html
I do believe I said essentially the same thing.

Not only that, a sunspot is deeper than the surrounding photosphere. The magnetic field extends through the sunspot into the surface of the sun. Both NASA and ESA indicate the the coronal loops originate below the photosphere.
Indeed so. A sunspot is essentially a "bubble" of magnetic field, which extends far below, and far above the photosphere. The loops in your trace images may be sunspot magnetic fields.

Now, as far as I can see, there is nothing about sunspots, or loops & filaments, or about the temperature and/or structure of the photosphere, that is in any way a problem for the mainstream model of the sun. And, of course, we are now in the "shotgun" approach, with so many topics afoot that it is no longer possible to understand what the point of the discussion is supposed to be.

Maybe you can help us out by concentrating on one, or at most two points, which you consider to be most important. The Big Deals, the points where it should be "obvious" that the mainstream view must be wrong. Do you have any?

upriver

15-September-2006, 08:16 AM

Helioseismology.

Because the sun is denser in the middle in the gas model, sound intensity should drop off from the center of the sun. But that is not the case. There is clearly a condensed matter surface to propagate the sound waves.

Helioseismology shouldn't work in a vacuum better than the best vacuums on earth.

"The particle motion associated with f-modes and p-modes is essentially confined to a region outside the solar core"
http://www.stat.berkeley.edu/~stark/Seminars/Aaas/helio.htm#what

The iron shell. If you look at HelioSeismology(HS), the shell thickness is approxamitely half the radius.

Flare radio brightness temperatures exceed 109 K and the peak in the radio spectrum is as high as 35 GHz: both these two features and the hard X–ray data require very high densities of nonthermal electrons, possibly as high as 1010 cm 3 above 20 keV at the peak of the flare.
http://www.astro.umd.edu/%7Ewhite/papers/03_norh_020723.pdf

How does the photosphere etc. support these kind of electron densities?

But black-body spectrum is not an emission spectrum: it is the spectrum of EM radiation in thermal equilibrium.

What?????????

"The Planck law gives the intensity radiated by a blackbody as a function of frequency (or wavelength)."
http://scienceworld.wolfram.com/physics/PlanckLaw.html

The sun is heated. By what ever process. Its temperature is not changing if thats what you mean by "in thermal equilibrium." In the MS model the photosphere radiates as much as it receives from underneath. Is the sun in thermal equilibrium with space? No. How about with its interior? No. It could be considered in LTE if you look at a small enough area.

"A material's emission spectrum is the amount of electromagnetic radiation of each frequency it emits when it is heated (or more generally when it is excited)."
http://en.wikipedia.org/wiki/Emission_spectrum

"In physics, a black body is an object that absorbs all electromagnetic radiation that falls onto it. Despite the name, black bodies are not actually black as they radiate energy as well."
http://en.wikipedia.org/wiki/Black_body

More to come.

captain swoop

15-September-2006, 09:55 AM

Iron Sun? Solid shell needed for Helioseismology to work?

hhm! I think we have been here before!

tusenfem

15-September-2006, 12:32 PM

But black-body spectrum is not an emission spectrum: it is the spectrum of EM radiation in thermal equilibrium.

What?????????

I guess it all depends on definitions. An emission spectrum is in astrophysics usually meant to be spectrum made up of emission lines. As the link to wiki you put in, you see examples of emission spectra, and see, is it just lines.

The iron shell. If you look at HelioSeismology(HS), the shell thickness is approxamitely half the radius.

So now the iron shell is what..., half the thickness of the sun?
Then how can you claim to see the iron shell in your iron lines if there is half the sun above it?
And I guess you want to say here that the shell radius is half the sun's radius. If the thickness is half the sun's radius then you will have at least a factor of 10 more mass in the sun.

Helioseismology shouldn't work in a vacuum better than the best vacuums on earth.

I would hardly call the sun starting from the photosphere and going inward a vacuum.

It could be considered in LTE if you look at a small enough area.

LTE works for volumes not for areas. Overall the sun is in LTE, untill you get to the location where tau=2/3, from which the photons can escape. And because the photons take so long to get from inside to outside they will also be in thermal equilibrium (it takes much longer for a photon to reach the photosphere, than you wanted Kirchhoff to wait until equilibrium).

How does the photosphere etc. support these kind of electron densities?

Although the paper states that the density is pretty high, you seem to forget that they are talking about solar flares. These plasmas need not be supported by the photosphere, but arise from "below the surfaced of the sun" and grow big and explode. The high density is sustainde by a relatively strong magnetic field.

papageno

15-September-2006, 11:24 PM

I have nothing to add to what tusenfem said regarding upriver's misconceptions about black-body radiation.

upriver

19-September-2006, 10:51 PM

I will be working on a CD recording project for the next couple of weeks so my time is limited. I will respond when I can. Maybe on Sundays.

Upriver

papageno

19-September-2006, 11:16 PM

No answers from Dr. Robitaille?

upriver

25-September-2006, 08:08 AM

No word from Dr. Robitaille. I will be calling him.

Ok. I started this thread talking about how gas/plasma only emits lines and not blackbody. Blackbody emission(curve) is the domain of condensed matter.
Even compressed plasma has a quasi-continium emission(lines/curve) that are/is distinguishable from a true blackbody emission(curve).

Now your telling me how backbody radiation should not be an called emission.
Emission is to be used only for lines.

Lines pertain to gas, blackbody pertains to condensed matter.
Blackbody emission means condensed matter by default, whereas line emission means gas or plasma.

An emission is an emission. What label you put on it to define what type of emission it is depends on the field of interest. Just because I dont use it in a "astrophysics" way you think its a misconception.
You guys are giving yourselves more credit than you deserve to think that is a misconception on my part.

And what is it with you and your libraries, papageno? You think libraries are an infallible source of knowledge and wisdom, and that some how will cure me of my genius. Libraries are a preset knowledge filter, I think I am old enough to not have my knowledge chewed for me.
Let me repeat my PM to you. I was reading encylopedias(all of them) in kindergarten. I grew up in libraries.
I dont like math just because people mistake it for reality, when all it is supposed to be is a descriptor for existing reality and a predictor for engineering.
Blackholes are the biggest boonedoggle ever because they got people to accept any kind of theoretical monkey business that came around.

A magnetar with a quantum magnetic gas bubble that collapses produces GRB's. (MS)
The barrel shaped supernova remnants are the result of a Bennett pinch that produced a GRB at Rmin.(EU)
Can you guys see the difference between those 2 statements?

This universe is composed of normal matter, any intelligent person can see that if they ignore the beautiful calculations.

So give me a little credit and really look at what I'm saying as there might be something to it. Once we have a logical model with workable mechanisms then we can twiddle knobs.

Like we have observed "magnetic slinkys, but the mechanism by which they exist differs depending on the model. EU say that they are current flows. MS people say they are frozen-in fields left over from the who knows when. MS have calculations to back it up. I dont. You have to either have a bunch of money or work at a university or LLNL or some place like that to have access to the good 2D plasma code to run simulatons. To get 3D Bennett pinch code. Dream on. To write is beyond my expertise right now, but I am modifing a simulator for work, for I think there is a mechanism that exisits that is responsible for the electric current flow of the slinky and for the current flow to the sun.

And I guess you want to say here that the shell radius is half the sun's radius. If the thickness is half the sun's radius then you will have at least a factor of 10 more mass in the sun.

No. The shell is thick enough to provide the observed density for the sun.
The half radius figure was a guess from looking at the helioseismology computer interpetations.

Quote:
Originally Posted by upriver
Helioseismology shouldn't work in a vacuum better than the best vacuums on earth.

I would hardly call the sun starting from the photosphere and going inward a vacuum.

Not dense enough to support sound propagaton.

Iron Sun? Solid shell needed for Helioseismology to work?

hhm! I think we have been here before!

Reflections are the result of an impedance mismatch. If you know what that means then you know why an iron sun fits. In addition to that you cannot have a "cavity like resonator" that is a ball of gas. Think about it. The "sound" would just decay in amplitude. Even EM waves.

you are looking at the surface of last scattering

No, you are looking at the surface of thermalization.

korjik

25-September-2006, 05:08 PM

Please explain limb darkening and how it would occur on a solid surface. If you think this is out of the topic of the thread (in which case you threadjacked yourself) then please explain why the third graph on the web page you linked to in post 26 shows argon emitting a blackbody radiation curve.

korjik

25-September-2006, 05:12 PM

Blackholes are the biggest boonedoggle ever because they got people to accept any kind of theoretical monkey business that came around.

A magnetar with a quantum magnetic gas bubble that collapses produces GRB's. (MS)
The barrel shaped supernova remnants are the result of a Bennett pinch that produced a GRB at Rmin.(EU)
Can you guys see the difference between those 2 statements?

This universe is composed of normal matter, any intelligent person can see that if they ignore the beautiful calculations.

So give me a little credit and really look at what I'm saying as there might be something to it. Once we have a logical model with workable mechanisms then we can twiddle knobs.

Like we have observed "magnetic slinkys, but the mechanism by which they exist differs depending on the model. EU say that they are current flows. MS people say they are frozen-in fields left over from the who knows when. MS have calculations to back it up. I dont. You have to either have a bunch of money or work at a university or LLNL or some place like that to have access to the good 2D plasma code to run simulatons. To get 3D Bennett pinch code. Dream on. To write is beyond my expertise right now, but I am modifing a simulator for work, for I think there is a mechanism that exisits that is responsible for the electric current flow of the slinky and for the current flow to the sun.

In the interest of keeping this thread alive, could you please keep to the topic of this thread. This thread is about blackbody radiation and the surface of the sun. Bringing in other topics just clouds the arguement here.

korjik

25-September-2006, 05:16 PM

This universe is composed of normal matter, any intelligent person can see that if they ignore the beautiful calculations.

So give me a little credit and really look at what I'm saying as there might be something to it. Once we have a logical model with workable mechanisms then we can twiddle knobs.

we have a logical mechanism. it fits very nicely and the knobs are being twiddled. you are the one wanting to toss it for no reason.

Not dense enough to support sound propagaton.

Dont you think that someone in the last 40 or so years may have checked that? maybe using the math that you think is so worthless?

papageno

25-September-2006, 08:17 PM

Now your telling me how backbody radiation should not be an called emission.
Emission is to be used only for lines.

Actually, I have always told you that the black-body spectrum is the spectrum of EM radiation in thermal equilibrium.

Lines pertain to gas, blackbody pertains to condensed matter.
Blackbody emission means condensed matter by default, whereas line emission means gas or plasma.

You persist in mistaking a technical issue with a fundamental problem.

And what is it with you and your libraries, papageno? You think libraries are an infallible source of knowledge and wisdom, and that some how will cure me of my genius. Libraries are a preset knowledge filter, I think I am old enough to not have my knowledge chewed for me.

Considering how you fall for the Iron Sun idea and the EU/PC ideas, I would say you are mistaken in your self-assesment.

Where are the results of your experiment with a reflective cavity?

captain swoop

26-September-2006, 12:48 AM

Celestial Mechanic

26-September-2006, 05:33 AM

And what is it with you and your libraries, papageno? You think libraries are an infallible source of knowledge and wisdom, and that some how will cure me of my genius. Libraries are a preset knowledge filter, I think I am old enough to not have my knowledge chewed for me.This statement says more to me than all of your other posts together.Sadly, I must agree with you, captain swoop.

No, upriver, libraries are not "infallible" sources of "knowledge and wisdom" anymore than the Internet is. I've seen some pretty goofy books in university libraries, such as a 19th century text by Karl Theodore von Heisel that asserted that pi was equal to 256/81. (I hope I remembered the author's name correctly.) But a critical reader can usually separate the gems from the dross, and any decent library is going to have more gems than dross. If a library acts as a "preset knowledge filter", it is generally filtering out things that have been tried or theorized and found not to work. It can be a great time-saver. Imagine having to wade through scholarly dissertations on phlogiston along with the real chemistry books!

Tim Thompson

26-September-2006, 03:58 PM

Blackbody emission(curve) is the domain of condensed matter.
Since when is the sun not "condensed matter"? I see no conflict between this "fact" and the observed radiative property of the sun. Do you? What is the conflict you see? And do remember that the mainstream position is that the BB emission comes from below the photosphere, and not from the photosphere itself.

Even compressed plasma has a quasi-continium emission (lines/curve) that are/is distinguishable from a true blackbody emission (curve).
That is a false statement. The "quasi continuum" is easily distinguished from a true BB curve, if the data are good enough. Always.

Blackholes are the biggest boonedoggle ever because they got people to accept any kind of theoretical monkey business that came around.
This is irrelevant to your own topic. It is also a fairy tale.

A magnetar with a quantum magnetic gas bubble that collapses produces GRB's. (MS)
This is irrelevant to your own topic. It is also not "MS" in any way, shape, manner or form, so this too must fall into the "fairly tale" category.

The barrel shaped supernova remnants are the result of a Bennett pinch that produced a GRB at Rmin.(EU)
This is irrelevant to your own topic. If you actually knew anything at all about Bennett pinches, you could immediately make a model of one with pencil & paper (no money or national lab required) and see if it made sense. You don't, so you can't, and this is just some blind assertion of no value.

This universe is composed of normal matter, any intelligent person can see that if they ignore the beautiful calculations.
This is irrelevant to your own topic. it is also a fairy tale. Any intelligent person can clearly & easily see that the universe is not made up entirely of normal matter, the simple observations are compelling & obvious. The data are inescapable. But it does take knowledge to understand this.

So, just out of curiosity, do you have anything relevant to add? I mean, something associated with the idea of "Black Body and our sun ..."?

upriver

03-October-2006, 01:36 AM

So, just out of curiosity, do you have anything relevant to add? I mean, something associated with the idea of "Black Body and our sun ..."?

NO. The BB emission curve is better fit by solid matter not convecting plasma.

I will just continue on my way. If I'm going to be required to show math to support my idea against the "standard solar model" which is completely math(aside from neutrinos and helioseismology) I have not the math skills that are required. I am going to learn math but not for this.
I spent the last week reading on the Standard Model and then I went to the TRACE website and looked at pictures.

Sorry, the pictures win. The sun has a solid surface. If math allows you to ignore what your eyes see then so be it.
The standard model is just speculation no matter how good the math is. Helioseismology and neutrinos can be explained by other models.

I am going to take the advice of everyone and concentrate on learning the math to simulate a GRB using a z-pinch. It may take some time but what else is there. Now if you can simulate a GRB with a pencil, I need your help.

If a library acts as a "preset knowledge filter", it is generally filtering out things that have been tried or theorized and found not to work. It can be a great time-saver.

IF you are looking for that type of knowledge. The status quo. l doubt you would find anything that was revolutionary in a library. Its been pre approved.

The freshest work in my field is pretty much only available online or in subscribed journals.
I know looking in a library would have not gotten me my patent.

"[0002] The present invention relates generally to sonoluminescence and, more particularly, to an acoustic driver assembly for use with a sonoluminescence cavitation chamber."
http://www.freshpatents.com/Brant-James-Callahan-NevadaCity-invdirc.php

And they certainly do not talk about PEAR research. http://www.princeton.edu/~pear/

When I think I have a even a little grasp on what is required for my project, I will post for your approval.