Moderated Iron sun with Aether batteries...

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Got a number?

Talk to GeeMack if you want a number. I'm just pointing out the glaring conceptual error contained in the question, an error you have made before. But as expected, you have taken no responsibility to correct your conceptual error, or even acknowledge that you made it. You never learn. You refuse to learn.
 
Start with an infinite intensity light source at 171A and tell me at what "depth" does it become "opaque"?

As others have said, opacity is independent of light intensity. "Opaque" is ultimately a matter of degree; opacity is a measure of how opaque a thing is.

But on a more practical level, things can get very opaque very quickly. If we take GM's diagram, which showed 37% transmittance at 350 km and 4% transmittance at 400 km, then that 50 km of plasma blocked about 90% of the light. If the plasma didn't get denser with depth (it does) and kept the same level of ionization, we'd expect it to block another 90% of the light every 50 km.

So, starting at 400 km with 1/25 of the light getting through (4%)
depth . . . light coming through
400 km . . . . 1/25
450 km . . . . 1/250
500 km . . . . 1/2500
550 km . . . . 1/25,000
600 km . . . . 1/250,000
650 km . . . . 1/2,500,000
700 km . . . . 1/25,000,000
750 km . . . . 1/250,000,000
800 km . . . . 1/2.5E9
900 km . . . . 1/2.5E11
1000 km . . . . 1/2.5E13
1500 km . . . . 1/2.5E23
2000 km . . . . 1/2.5E33
3000 km . . . . 1/2.5E53

Thus, if we relied on photons simply passing through 3000 km of that plasma to provide us with light, the entire sun would emit an average of 1 photon every 4 years. That's pretty opaque.
 
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Photospheric Opacity and Composition

... and we should probably start with a 90/10 percent mixture of neon/(standard model elements) in terms of the plasma with a density that matches the standard model at the surface of the photosphere. ... The umbra however is not neon. It's silicon with roughly the same density as the photosphere, so that part would need to be calculated separately in a similar 90/10 mixture of mostly silicon.
It sounds loony, but OK. 90% neon, 10% standard solar model elements. Since the standard solar model says the photosphere is around 90% H, 9% He, plus trace - shall we just say 90%Ne, 10%H? Is that what you meant?
Yes, that's exactly what I meant by the way. That works for me. You'll eventually need the metals and other elements to explain the "white light', but I doubt they'd have much effect on the opacity in any relevant way, whatever simplified scenario works for you is fine by me. FYI, I appreciate what you're doing actually, and I'm looking forward to your results.
I still need to know whether you mean 90%Ne by mass or by number. In other words, you might mean that if I weigh some plasma, 90% of the weight comes from Ne and 10% from H. Or, you might mean that if I count atoms, 90% will be Ne and 10% will be H. Those are very different, because a typical Ne atom weighs 20x as much as an H atom (i.e. a proton).
I meant mass assuming other elements present so use mass. It might be a bit heavy on the hydrogen since we're not including other (heavier) elements, but that's fine IMO.
It will be interesting to see what comes out of all this, but I have my reservations.

We already know, as a matter of fact, that the mixture of plasma Mozina wants to investigate, both for the general photosphere and sunspot umbrae, exists nowhere in or on the sun. We can look at the sun and see what it is made of. We know its chemical composition (by number about 92% hydrogen, 8% helium and less than 1% everything else; see, e.g., Solar Astrophysics by Peter Foukal, 2nd edition 2004 section 5.6 and table 5-3; Asplund, et al., 2009). We have known that the sun is composed mostly of hydrogen since about 1930 (e.g., Russell, 1929; Stromgren, 1932; Eddington, 1932). The fact that the sun is made mostly of hydrogen is crucial, since we also know that, counterintuitively perhaps, the continuum opacity of stellar photospheres is dominated by the H- ion (e.g., Wildt, 1939; Massey & Bates, 1940; Chandrasekhar, 1945, a 5-part paper, all parts linked from this page; John, 1988; John, 1994; The Observation and Analysis of Stellar Photospheres by David Gray, 3rd edition 2005, pp. 154-157; Solar Astrophysics by Peter Foukal, 2nd edition 2004 section 5.3.2, pp. 149-150). Finally, there is quite good enough agreement between helioseismological observations, solar neutrino observations, and the standard astrophysical models of the sun, such that all of Mozina's alternate hypotheses are excluded with confidence (see, e.g., Bahcall & Ulrich, 1988; Bahcall, Pinsonneault, & Basu, 2001 and citations thereto for both papers).

Clearly, if we pick an unrealistic mix of elements, we get an unrealistic opacity as a result. 90% neon means a lot fewer H- ions and, perhaps, a lot less opacity. So if we find that the Mozina mixture is indeed much more translucent than we are claiming for the photosphere here, so what? Since the chosen mixture is very unphysical, so will the low opacity be representative only of the Mozina sun, as opposed to the real sun we look at. It will still remain to show that there is observational support for the Mozina mixture, and some objective reason not to believe the standard mixture, which has been built up over 80 years of careful observations of the sun. I suspect that Mozina will be as incapable of supporting his alternate hypothesis for the solar chemical abundances as he is incapable of just about everything else, but we will see.
 
Clearly, if we pick an unrealistic mix of elements, we get an unrealistic opacity as a result.

Ya, and that's my same beef with your claim by the way.

90% neon means a lot fewer H- ions and, perhaps, a lot less opacity. So if we find that the Mozina mixture is indeed much more translucent than we are claiming for the photosphere here, so what?

So what? Really?
 
Got a number?


Sure.

As others have said, opacity is independent of light intensity. "Opaque" is ultimately a matter of degree; opacity is a measure of how opaque a thing is.

But on a more practical level, things can get very opaque very quickly. If we take GM's diagram, which showed 37% transmittance at 350 km and 4% transmittance at 400 km, then that 50 km of plasma blocked about 90% of the light. If the plasma didn't get denser with depth (it does) and kept the same level of ionization, we'd expect it to block another 90% of the light every 50 km.

So, starting at 400 km with 1/25 of the light getting through (4%)
depth . . . light coming through
400 km . . . . 1/25
450 km . . . . 1/250
500 km . . . . 1/2500
550 km . . . . 1/25,000
600 km . . . . 1/250,000
650 km . . . . 1/2,500,000
700 km . . . . 1/25,000,000
750 km . . . . 1/250,000,000
800 km . . . . 1/2.5E9
900 km . . . . 1/2.5E11
1000 km . . . . 1/2.5E13
1500 km . . . . 1/2.5E23
2000 km . . . . 1/2.5E33
3000 km . . . . 1/2.5E53

Thus, if we relied on photons simply passing through 3000 km of that plasma to provide us with light, the entire sun would emit an average of 1 photon every 4 years. That's pretty opaque.


One photon every four years is escaping from your mythical thermodynamically impossible solid iron surface which you erroneously claim exists at .995R. Less than one photon every year if we look at your shallowest case scenario of 2100 kilometers. Oh, and that is giving you the most extreme benefit of the doubt by going with the physically impossible situation where the density of the plasma doesn't increase beyond that 400 kilometer depth.

There's your number, Michael. Now you can toss one quantitative figure into your made up sciency sounding prattle. You can say that you see a single photon every four years, so in the, what, eight years you've been constructing this work of fiction you've seen two, maybe three photons work their way up from your thermodynamically impossible solid iron surface.

Can you do a little math to support another way of looking at it? After all, you are working up some real numbers now that you know the Achilles heel of solar theory, the opacity issue. How's that coming for you?...

Now that I finally understand how to go about destroying mainstream theory, I'll start working on it. I think *THAT* little project might even motivate me to do a little math.
 
Ya, and that's my same beef with your claim by the way.



So what? Really?


So what? So since we know the composition of the Sun is quite different than the hypothetical you invented for the opacity discussion, the results of the calculations using that mixture have no bearing on reality. That's so what. :D
 
As others have said, opacity is independent of light intensity. "Opaque" is ultimately a matter of degree; opacity is a measure of how opaque a thing is.

Yes, I understand that, but with an infinite light source, it doesn't make any difference how far we go into the photosphere or what fraction of light that you arrive at. If the source is infinite, so is the light passing through.

"Practically speaking" you're absolutely correct of course, but GM definition of "opacity" is scientifically incorrect. No matter how many socks I add to mythical flashlight, if I can simply crank up the output of the flashlight by a factor of 10, the light through the "opaque layers" is unchanged. The intensity of light will make a difference when it comes to what can be "seen" through whatever plasma we're talking about. GM seems to confuse "percent" and "absolute" when it comes to being able to "see" through an "opaque" substance.
 
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At which depth does that infinite light source become "impossible to see" GM?


Whose face is that in that famous picture of the face on Mars? How tall is that bunny in the clouds, or those mountains you hallucinate in those thermal characteristic graphs?

How's that math coming along, Michael? Oh, not the math you just recently said you'd do, but the math that you confidently said you'd put together back in, what, 2005 on the BAUT forum? Maybe when you finally get around to that another photon will have popped it's way up from your mythical solid iron surface. :p
 
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Famous last words.....


Like when you "bet the farm" the STEREO satellites would prove your cockamamie claim back in, what was it, 2006? Oh, and we're still waiting for a single piece of quantitative support and your calculations you said you'd be providing real soon at the SFN forum? Yes, that was also in 2006 as I recall.

Famous last words, indeed. ;)
 
You didn't earlier.



Irrelevant, since there are no infinitely bright sources.

Of course. The point was to get GM to recognize that 90% is not 100%. What's the difference? Depending on the conditions, it can mean *everything*.
 
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Can you do a little math to support another way of looking at it? After all, you are working up some real numbers now that you know the Achilles heel of solar theory, the opacity issue. How's that coming for you?...


So Michael, since you aren't offering any mathematical objection, I take it you accept that the single photon every four years is a fair quantitative measure for the maximum possible light escaping form your mythical solid iron surface. You know, given the hypothetical but impossible situation of the density not increasing as you go deeper than 400 kilometers.
 
Of course. The point was to get GM to recognize that 90% is not 100%. What's the difference? Depending on the conditions, it can mean *everything*.


Well if I had ever said 90% is the same thing as 100% there might be a point to your continuing to babble about it, but since I haven't, it would make you a liar if you're suggesting I have. I'm sure you agree.

Oh, and how's that proof coming, Michael, you know, from those STEREO images that are going to finally show the world that there really is a solid surface on the Sun? How long has that program been going now? Maybe they're just a little slow on the draw over there at NASA. I haven't seen their press release announcing it to the world. But what could they possibly know about astrophysics anyway? :p
 
:popcorn1
Originally Posted by Michael Mozina
Now that I finally understand how to go about destroying mainstream theory, I'll start working on it. I think *THAT* little project might even motivate me to do a little math.
 
Argh! FYI, *I* am the one that cited the supercomputer simulation of a sunspot and explained how it ties in with the images and mass flow patterns. You folks ignored that math *ENTIRELY*. Now what? More math? What about the other math? What about all the images that show the mass flows coming *up and through* the photosphere at high velocity?
Argh! FYI *You* am the one that cited the supercomputer simulation of a sunspot and got it wrong.
The image you cited is an computer generated graph of the magnetic strength below two sunspots.
You did not explain how the magnetic strength graph mapped into any images or mass flow patterns.

See the comments before the question that you are ignoring in this post: How did you measure the curvature of penumbral filaments in the Hinode images?

P.S. Micheal Mozina's iron crust has been debunked!
 
I've been looking at plasma opacities using this program. It turns out, for reasons I don't understand well, that they are fairly sensitive to the abundances of heavier metals. That is, for plasmas that are roughly the Mozina mix of 90%Ne, 10%H by mass and with roughly the temperature and mass density of the photosphere, the opacity depends in a relatively strong way on the densities of metals, even when those densities are far below that of the Ne.

So Michael - can you tell me what you think the most important (by mass) other components are, and roughly how much of them there are compared to Ne? For example if Ne is 90%, how much Si is there? How much Fe? How much S? Etc.

Thanks.
 
How can we detect the 1 photon per year from your iron crust

Yet another question arises:
One photon every four years is escaping from your mythical thermodynamically impossible solid iron surface which you erroneously claim exists at .995R. Less than one photon every year if we look at your shallowest case scenario of 2100 kilometers. Oh, and that is giving you the most extreme benefit of the doubt by going with the physically impossible situation where the density of the plasma doesn't increase beyond that 400 kilometer depth.

We already know that you iron crust cannot exist (Micheal Mozina's iron crust has been debunked! ).

First asked 24 April 2010
Michael Mozina,
The opacity of the photoshere means that if all of the light that we see from the Sun was emitted from your impossible iron crust at 3000 km then we would see 1 photon every 4 years. The best that we can see from your crust would be 1 photon per year at 2100 km.

How can we detect the 1 photon per year from your iron crust against the background of the existing emission of the Sun?
 
I've been looking at plasma opacities using this program. It turns out, for reasons I don't understand well, that they are fairly sensitive to the abundances of heavier metals.

Insofar as the opacity comes from H- ions, heavy metals serve as the electron donors. But that's the sort of detail you need to know to get optical opacity. MM is trying to escape from a vacuum-ultraviolet opacity!

Since 171A is above the ionization threshhold for neon, we don't need to look at rare processes and species---the straightforward ionization cross section for neutral neon is something like 10^-18 cm^2. (see doi: 10.1098/rspa.1953.0172 Proc. R. Soc. Lond. A 22 October 1953 vol. 220 no. 1140 71-76) That gives you a channel for bound-free absorption which doesn't depend on plasma details the same way that longer wavelengths do. Given this cross section, a 171A photon can propagate through less than a 100 micrograms/cm^2 of neutral neon before being absorbed. Needless to say, there's a lot lot lot lot more than 100 ug/cm^2 of material between us and Mozina's supposed iron layer.

There's a reason they call this "vacuum ultraviolet"---it doesn't propagate through anything other than a vacuum. Come up to my lab and I'll show you, I've got lots of 1200A light propagating (or failing to propagate, depending on the purity) through argon.

So Michael - can you tell me what you think the most important (by mass) other components are, and roughly how much of them there are compared to Ne? For example if Ne is 90%, how much Si is there? How much Fe? How much S? Etc.

God forbid this discussion should have anything to do with the actual observed abundances of the photosphere. :)

Oh, and one more thing: Mozina's crazy model ("lots of photons are visible from a cooler, deeper layer in the Sun") predicts backwards limb darkening---the Sun would be brighter at the edge, when one optical depth on your line-of-sight is entirely in the hot photosphere, and dimmer in the middle where your line-of-sight supposedly sees down to the cold iron. Funny that the observations, which show the opposite, are entirely consistent with the mainstream temperature/opacity/etc vs depth.
 
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Argh! FYI, *I* am the one that cited the supercomputer simulation of a sunspot and explained how it ties in with the images and mass flow patterns. You folks ignored that math *ENTIRELY*. Now what? More math? What about the other math? What about all the images that show the mass flows coming *up and through* the photosphere at high velocity?



It doesn't. Thermodynamically you can't begin with the premise that the photosphere "Layer" is last atmospheric "layer" of the sun. Just as the chromosphere is hotter and less dense than the photosphere, the neon photosphere, is hotter than the silicon plasma. Just as the chromosphere's temperature is cooler at the bottom where it meets up with the photosphere, and hotter where it meets up with the corona, the photosphere is "hottest" at the top and cooler underneath (although not during sunspot activity). That cooler layer of silicon is what keeps the surface from melting and the movement of charged particle from the surface continuously moves heat away from the surface toward the heliosphere.

Posterity.
 
How many times must I explain this to you? I respect sol. I respect and *appreciate* his efforts right now. I'm very interested in the results and because he personally is doing them, I'm actually very excited. I'm excited because I know he will "do them right". He'll be "honest" (something you know nothing about) and he'll do it from a place of pure scientific curiosity, not from a place of ego or ignorance, or hostility towards me personally. I'm really looking forward to the numbers actually, and I intend to learn from his efforts to the best of my abilities.

Unlike you I don't profess to know the outcome for certain, but I know that whatever number he comes up with will not represent the point at which all light becomes magically invisible as you seem to believe.

I'm honestly very appreciative of his efforts and I hope to learn a great deal.

You personally have absolutely *NOTHING* to "teach" me.


Ruh roh
 
insofar as the opacity comes from h- ions, heavy metals serve as the electron donors. But that's the sort of detail you need to know to get optical opacity. Mm is trying to escape from a vacuum-ultraviolet opacity!

Since 171a is above the ionization threshhold for neon, we don't need to look at rare processes and species---the straightforward ionization cross section for neutral neon is something like 10^-18 cm^2. (see doi: 10.1098/rspa.1953.0172 proc. R. Soc. Lond. A 22 october 1953 vol. 220 no. 1140 71-76) that gives you a channel for bound-free absorption which doesn't depend on plasma details the same way that longer wavelengths do. Given this cross section, a 171a photon can propagate through less than a 100 micrograms/cm^2 of neutral neon before being absorbed. Needless to say, there's a lot lot lot lot more than 100 ug/cm^2 of material between us and mozina's supposed iron layer.

There's a reason they call this "vacuum ultraviolet"---it doesn't propagate through anything other than a vacuum. Come up to my lab and i'll show you, i've got lots of 1200a light propagating (or failing to propagate, depending on the purity) through argon.



God forbid this discussion should have anything to do with the actual observed abundances of the photosphere. :)

oh, and one more thing: Mozina's crazy model ("lots of photons are visible from a cooler, deeper layer in the sun") predicts backwards limb darkening---the sun would be brighter at the edge, when one optical depth on your line-of-sight is entirely in the hot photosphere, and dimmer in the middle where your line-of-sight supposedly sees down to the cold iron. Funny that the observations, which show the opposite, are entirely consistent with the mainstream temperature/opacity/etc vs depth.

jref
 
Excuse me, but you could have just as easily have used the term "circuit reconnection" or "particle reconnection" and not selected your own special "lingo" that Alfven himself referred to as "pseudoscience". Either terminology would have kept things congruent with electrical engineering, and/or particle physics theory and would have created no confusion at all. You only have yourself to blame for any "confusion" by clinging to "pseudoscience".

Yeah, Mikey, and as long as you don't show any real explanation about your "circuit/particle" reconnection, how e.g. the plasma is accelerated in the exhaust with velocities perpendicular to the magnetic field and ions and electrons in the same direction (so no double layer and certainly no double-double layer) then you can come back to the reconnection problem.

Alfvén was out of touch with reality when he made his pseudo claim and had he lived long enough and cared enough to keep on following MRx develepment he would have seen and accepted that it is real science. Unfortunately, he did neither.
 
Got a number?

Opacity is a number that goes into the equation of radiative transport, and it is given by:

I(x) = I0 e - κ x
in its most simplest form. This works for anything that light passes through, be it gas, plasma, liquid, solid. The only thing that needs to be determined is the value for κ and if you really want a number here:

κff = 0,64E23 ρ T-7/2 cm2/g

with the density ρ in grams per cubic centimeter and the temperature T in Kelvin. This is the opacity for free-free transitions, for which the electron is never actually attached to the ion. This is a special form of bremsstrahlung. Note, that in this definition for κ the exponent in the equation above gets a density ρ in it.

Then you can ask yourself, what happens when light of intensity I0 passes through that medium. Well, it will get weaker by a factor e - κ x, which it totally independent on I0 as all the physics is in κ, i.e. all the absorption, emission and scattering processes that can happen in the medium (including stimulated emission).

So, now you got your number.
 
I've been looking at plasma opacities using this program. It turns out, for reasons I don't understand well, that they are fairly sensitive to the abundances of heavier metals. That is, for plasmas that are roughly the Mozina mix of 90%Ne, 10%H by mass and with roughly the temperature and mass density of the photosphere, the opacity depends in a relatively strong way on the densities of metals, even when those densities are far below that of the Ne.

So Michael - can you tell me what you think the most important (by mass) other components are, and roughly how much of them there are compared to Ne? For example if Ne is 90%, how much Si is there? How much Fe? How much S? Etc.

Thanks.

Yeah, that should not come as a surprise, as the heavier the metals (anything above He) become, the more shells and thus the more transitions they will have. That all gets into the opacity, it is not a nice thing.
 
How can we detect the 1 photon per year from your iron crust against the background of the existing emission of the Sun?

I recall (maybe wrongly) that the hatching of fruit fly eggs can be triggered by a single photon. So, steer that photon over to the fruit fly farm and wait for one to hatch. (bit of a waste of the photon I suppose. I mean you cudda worked on your suntan with that photon) :)
 
Insofar as the opacity comes from H- ions, heavy metals serve as the electron donors. But that's the sort of detail you need to know to get optical opacity. MM is trying to escape from a vacuum-ultraviolet opacity!

Since 171A is above the ionization threshhold for neon, we don't need to look at rare processes and species---the straightforward ionization cross section for neutral neon is something like 10^-18 cm^2. (see doi: 10.1098/rspa.1953.0172 Proc. R. Soc. Lond. A 22 October 1953 vol. 220 no. 1140 71-76) That gives you a channel for bound-free absorption which doesn't depend on plasma details the same way that longer wavelengths do. Given this cross section, a 171A photon can propagate through less than a 100 micrograms/cm^2 of neutral neon before being absorbed. Needless to say, there's a lot lot lot lot more than 100 ug/cm^2 of material between us and Mozina's supposed iron layer.

Thanks Ben - that's extremely helpful.

  1. Neon ionization: The temperature we're taking for this plasma is of order 6000K, which corresponds to about .5eV. Because neon is a noble gas, its ionization energy is high - around 20eV - and its first excited state is a large fraction of that. That means very little of the neon in the Mozina plasma will be ionized, because there is a huge Boltzmann suppression e-20eV/.5eV~10-17.5. So nearly all the neon will be neutral and in its ground state.

  2. Photon energy: The wavelength of interest is 171A, or 72eV of energy per photon. That's far above the ionization energy for neon (20eV). Therefore, as Ben says, the photons in this band can scatter by ionizing neon atoms. The cross section for that in the 171A range is a bit larger than 10^-18 cm^2. Since I'm trying to systematically underestimate the opacity, I'll use 10^-18.

  3. Bound-free opacity: We now have the information necessary to compute one of the contributions to opacity, the one coming from ionizing neutral neon. The opacity is κ=nσ/ρ, where n is the number density (of neutral neon in this case), σ is the cross section for scattering with the photon, and ρ is the mass density. The number density is the mass density 10^-7 g/cm^3 divided by the mass per neon atom (3.5E-23g), which gives n=3*10^15 neon atoms/cm^3. Multiplying both sides by ρ, we have κρ=1/(3.5*10^2 cm).

  4. Attenuation of 171A radiation: The intensity after passing through some thickness of material is I(x)=I0e-xρκ, where I0 is the intensity at the source or when the radiation first enters the plasma and x is the thickness. So what this tells us is that for every 350cm=3.5m of plasma our 171A radiation passes through, its intensity is attenuated by a factor of e=2.71. To give some sense of what that means, after propagating through 1km of plasma, the intensity will be reduced by a factor of about e^(-1000/3.5)=10^-124.

  5. Necessary intensity of source for visibility: Therefore, for one photon to make it through a 1km thickness of Mozina plasma, we'd need about 10^124 photons to be emitted by the source. Each photon carries 10^-17J of energy. So that's 10^107J of energy emitted by the source, which is vastly more energy than there is in the entire observable universe. In other words it is impossible for even one photon of 171A radiation to propagate through 1km of the Mozina plasma, no matter what the source.

I may well have made an error somewhere in there, so I invite critiques. Thanks to all - I learned from this.
 
I've been looking at plasma opacities using this program. It turns out, for reasons I don't understand well, that they are fairly sensitive to the abundances of heavier metals. That is, for plasmas that are roughly the Mozina mix of 90%Ne, 10%H by mass and with roughly the temperature and mass density of the photosphere, the opacity depends in a relatively strong way on the densities of metals, even when those densities are far below that of the Ne.

So Michael - can you tell me what you think the most important (by mass) other components are, and roughly how much of them there are compared to Ne? For example if Ne is 90%, how much Si is there? How much Fe? How much S? Etc.

Thanks.

That sensitivity is likely to be related to the things Ben mentioned and the valence shell configurations of the elements in question. It goes back to that conversation I had yesterday with Zig. Neither neon nor hydrogen has a valence shell that is capable of absorbing that particular wavelength of light.

Since you have been so gracious to use this solar model to do these opacity calculations, my "best" suggestion would be to go to the solar wind data, and use the ratios you find there. Any elements present in that data, must come up and through that neon layer at pretty much those ratios. Silicon might be in the neon in greater quantities, but that should be a separate calculation anyway so I wouldn't bother trying to adjust that number. IMO the most "scientifically valid" way I can think of to answer that question would be to suggest you methodically pick elements from the solar wind data in whatever ratios you find.
 
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That sensitivity is likely to be related to the things Ben mentioned and the valence shell configurations of the elements in question. It goes back to that conversation I had yesterday with Zig. Neither neon nor hydrogen has a valence shell that is capable of absorbing that particular wavelength of light.

Wrong. Atoms can absorb a photon by transitioning from bound to free. This is a continuum process and absorbs a continuum of wavelengths, not just discrete ones.

(ETA: and imagine for a moment that you don't understand atomic theory well enough to know why neon atoms absorb continuum VUV photons. Instead, you could turn to the data to find out whether neon atoms absorb continuum VUV photons. They do. Go look it up. That's where I got 10^-18 number from. If your mental picture of atoms tells you that they don't, then your mental picture is obviously wrong.)
 
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Standard Volcanic Surface

Necessary intensity of source for visibility: Therefore, for one photon to make it through a 1km thickness of Mozina plasma, we'd need about 10124 photons to be emitted by the source. Each photon carries 10-17 J of energy. So that's 10107 J of energy emitted by the source, which is vastly more energy than there is in the entire observable universe. In other words it is impossible for even one photon of 171A radiation to propagate through 1km of the Mozina plasma, no matter what the source.

I may well have made an error somewhere in there, so I invite critiques. Thanks to all - I learned from this.

On the opacity front, it goes not well for X-rays through Mozina's photosphere. While that adventure plays itself out, let us ask about another, little explored aspect of the Mozina Sun.

How is "a standard volcanic surface" ill-defined?
Easy: There is no such thing as a "standard volcanic surface" recognized in science. Each volcanic surface is peculiar to its specific environment. The terrestrial crust is dominated by silicates, whereas you propose some kind of surface dominated by iron. Even on Earth, volcanic terrain depends heavily of the nature of the extruded magma, dominated by silicates; cinder cones or shield volcanos develop very different kinds of terrain, very different kinds of surface; which is supposed to be the "standard" volcanic surface? You can't just throw out some vague reference to "standards" that are not "standard" except for you. Be specific. What are the chemical compositions of the surface and magma? What are the temperatures of the surface and magma?

Let us explore a bit further just what we are supposed to think that a "standard volcanic surface" is supposed to mean, since Mozina has provided us with no additional information on the subject. I assume, in the absence of any indication to the contrary, that "standard" must mean "Earth-like" or "terrestrial". Mozina uses the word iron so much in reference to his surface that I have come to believe that he actually means iron. However, neither of the two standard crusts on Earth (oceanic and continental) have much in the way of iron in them. Both are dominated by Silicon & Oxygen. See the Abundances of Elements in Earth's Crust page from Hyperphysics, which simply averages the two together, and we see 46.6% Oxygen, 27.7% Silicon (by weight), and everything else in the descending single digits, with iron down around 5%. The Wikipedia Earth page references a reliable geology text book and gives molecular abundances, for both oceanic & continental crusts (SiO2 and AlO3 dominate both, with FeO and Fe2O3 combining to around 6.3% continental and 8.5% oceanic (probably by weight, though the page is not explicit on that). These are in line with abundances I see in various reference sources of my own (e.g., Allen's Astrophysical Quantities, AIP Press, 4th edition 2000; Mantle Convection in the Earth and Planets, Cambridge University Press, 2001).

There are physical & chemical differences between oceanic & continental crust, so as to render "standard" an ambiguous concept. But we can simply average the two together, as the differences are not all that huge for our purpose here, and just pretend that the combination of the two will represent our "standard volcanic surface", as Mozina likes to call it. First & foremost, we see that our standard surface has no more than 10% iron, and is anywhere from 30% to 50% SiO2, with AlO3 in second place. Already we see that the elemental composition of what appears to be the intended standard is vastly different from the only elemental composition we have been given for the "crust" of the sun, namely "iron". Hence, by what appears to be Mozina's own choice for "standard", the solar "volcanic surface" is quite non-standard. But we also notice that, with nice comfy temperatures of a few hundred Kelvins to roughly 1000 Kelvins, the standard crust is completely dominated by amorphous & crystalline molecular minerals, whereas on the sun, sporting a healthy 6000 Kelvins, the surface can have no molecules at all (the only regions of the sun cool enough for molecules to form are sunspot umbrae, around 3000 Kelvins).

Now, of course I am simply trying to guess what I think Mozina intends to mean by "standard", since he does not say. And considering his penchant for using Earth as a standard for gamma rays and magnetic reconnection, I simply assume that he means likewise for "standard volcanic surface". Since this raises some serious problems, we need the expert on "standard volcanic surfaces" himself, Mozina, to tell us what he thinks the volcanic surface of the sun looks like, in enough detail for us to explore this aspect of the "iron sun" as well.
 
  1. Necessary intensity of source for visibility: Therefore, for one photon to make it through a 1km thickness of Mozina plasma, we'd need about 10^124 photons to be emitted by the source. Each photon carries 10^-17J of energy. So that's 10^107J of energy emitted by the source, which is vastly more energy than there is in the entire observable universe. In other words it is impossible for even one photon of 171A radiation to propagate through 1km of the Mozina plasma, no matter what the source.


Okay, in order for that one photon to make its way up from Michael's mythical solid iron surface, we find it would take, how'd sol put it, "vastly more energy than there is in the entire observable universe".

Oh, wait, I said that wrong. That fictional iron surface is 3000+ kilometers down. It takes "vastly more energy than there is in the entire observable universe" to get that single photon through one kilometer of Michael's plasma.

Well there's your honest scientific answer, Michael, from the guy who sticks to the facts and whose ability you trust to calculate this stuff. Darn the luck, eh? We find out after running the numbers that it's impossible for you to be seeing anything from your crackpot iron surface. Looks like you're wrong about seeing surface terrain in those running difference images, or in any other images for that matter. Seems I've heard that somewhere before! :p

And the results of that sort of coincide with what we already know about thermodynamics and how it's impossible for that surface to even exist in the first place. I know you don't like numbers, so let's make this easy for you. We have impossible plus impossible, or is that impossible times impossible? Not looking good for your insane solid surface conjecture, Michael. But we aren't forgetting that you did say...

Now that I finally understand how to go about destroying mainstream theory, I'll start working on it. I think *THAT* little project might even motivate me to do a little math.


... so I'm sure if for any reason you don't agree with the numbers that we have so far, you'll be working up a little math of your own. :D
 
Thanks Ben - that's extremely helpful.

  1. Neon ionization: The temperature we're taking for this plasma is of order 6000K, which corresponds to about .5eV. Because neon is a noble gas, its ionization energy is high - around 20eV - and its first excited state is a large fraction of that. That means very little of the neon in the Mozina plasma will be ionized, because there is a huge Boltzmann suppression e-20eV/.5eV~10-17.5. So nearly all the neon will be neutral and in its ground state.

  2. Photon energy: The wavelength of interest is 171A, or 72eV of energy per photon. That's far above the ionization energy for neon (20eV). Therefore, as Ben says, the photons in this band can scatter by ionizing neon atoms. The cross section for that in the 171A range is a bit larger than 10^-18 cm^2. Since I'm trying to systematically underestimate the opacity, I'll use 10^-18.

  3. Bound-free opacity: We now have the information necessary to compute one of the contributions to opacity, the one coming from ionizing neutral neon. The opacity is κ=nσ/ρ, where n is the number density (of neutral neon in this case), σ is the cross section for scattering with the photon, and ρ is the mass density. The number density is the mass density 10^-7 g/cm^3 divided by the mass per neon atom (3.5E-23g), which gives n=3*10^15 neon atoms/cm^3. Multiplying both sides by ρ, we have κρ=1/(3.5*10^2 cm).

  4. Attenuation of 171A radiation: The intensity after passing through some thickness of material is I(x)=I0e-xρκ, where I0 is the intensity at the source or when the radiation first enters the plasma and x is the thickness. So what this tells us is that for every 350cm=3.5m of plasma our 171A radiation passes through, its intensity is attenuated by a factor of e=2.71. To give some sense of what that means, after propagating through 1km of plasma, the intensity will be reduced by a factor of about e^(-1000/3.5)=10^-124.

  5. Necessary intensity of source for visibility: Therefore, for one photon to make it through a 1km thickness of Mozina plasma, we'd need about 10^124 photons to be emitted by the source. Each photon carries 10^-17J of energy. So that's 10^107J of energy emitted by the source, which is vastly more energy than there is in the entire observable universe. In other words it is impossible for even one photon of 171A radiation to propagate through 1km of the Mozina plasma, no matter what the source.

I may well have made an error somewhere in there, so I invite critiques. Thanks to all - I learned from this.

Wowza, wowza, wowza.
 
Thanks Ben - that's extremely helpful.

  1. Neon ionization: The temperature we're taking for this plasma is of order 6000K, which corresponds to about .5eV. Because neon is a noble gas, its ionization energy is high - around 20eV - and its first excited state is a large fraction of that. That means very little of the neon in the Mozina plasma will be ionized, because there is a huge Boltzmann suppression e-20eV/.5eV~10-17.5. So nearly all the neon will be neutral and in its ground state.


  1. FYI, I missed this post earlier. That is highly *unlikely* IMO, especially since the surface is discharging to the heliosphere *THROUGH* the neon. What makes you think *any* of the neon will be in the ground state?
 
Well, that was interesting. I will point out, as a final point of fact, that MM needn't hope that the *neon* is specifically the problem, and that he can plug in some different photosphere composition and hope for a different answer. Helium should have a very similar cross section to neon; everything else (and I do mean everything) will be higher.

Michael, think carefully about your past few year's work. Go back and look again at every 171A image you've ever "analyzed" and say: "This emission is actually from the top of the photosphere, not from deep below; whatever I did to convince myself otherwise, I got it wrong."
 
Sorry Ben, but you simply can't treat an "electric sun" the way you might treat your solar model. There's almost no "non ionized" neon in that layer since all the layers are "current carrying plasma".
 
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