Moderated Iron sun with Aether batteries...

[Reality Check]

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?

 

[ben m]

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.

 

[Dancing David]

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.

 

[Dancing David]

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

 

[Dancing David]

It's not a matter of being 'unaware', I simply believe it's not a relevant factor. We'll just have to be patient and see.

Double Ruh roh.

 

[Dancing David]

Start with an infinite intensity light source at 171A and tell me at what "depth" does it become "opaque"?

The burrito so hot god can't eat it!

 

[Dancing David]

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

 

[tusenfem]

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.

 

[tusenfem]

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:

κ_ff = 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.

 

[tusenfem]

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.

 

[GlennB]

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)

 

[sol invictus]

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.

 

[Michael Mozina]

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.

 

[ben m]

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.)

 

[Tim Thompson]

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.

 

[GeeMack]

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!

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.

 

[Dancing David]

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.

 

[Michael Mozina]

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?

 

[ben m]

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."

 

[Michael Mozina]

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".