Moderated Iron sun with Aether batteries...

[Gaspode]

Discussion on the universe split to new thread.

Posted By: Gaspode

 

[Ziggurat]

Or it will detect 192/5A light from reflections off of solid material at a lower temperature. Like the mirror in TRACE'S telescope.
That mirror is no where near 1 million degrees. Same with the solar surface.

Basic physics.

So if the surface did exist then there is no problem with the reflection part.

We will get to the thermodynamic considerations.

It is your contention using your photosphere model that not enough energy escapes to allow a solid iron surface to exist under "your photosphere model".
And I'm sure your model requires complete coverage of the lower surface to achieve your goals(molten iron/plasma).

It is my contention that enough energy escapes from under the surface glow to allow a solid surface to exist in my surface glow model.

Your assumption is that the photosphere is the only surface that is emitting energy.
And there are various energy release mechanisms to prevent the iron surface from heating up..
Sun spots(holes), CME, coronal loops, IR from the solid surface below the glow, Solar wind, polar plumes, and not all of the visible comes from the photosphere. And neither does the UV which is the most energetic and variable part of the solar spectrum.

You really don't understand thermodynamics.

Absorptivity equals emissivity. There's no getting around that, it's a requirement for the 2nd law of thermodynamics. Violating it is equivalent to perpetual motion.

In order for your solid surface to emit IR radiation, it must also absorb IR radiation. That means that your solid surface will absorb more energy in the IR band than it emits, unless it's hotter than the photosphere. And sun spots don't help you out either, both because they're NOT black in the IR region (they're still quite bright, indicating temperatures well above any solids), and because they cover far too small a fraction of the sun's surface to be a significant factor in radiative loss. CME and coronal loops are also irrelevant, since they can't transport heat away from the surface. I've been through the numbers on solar wind, and it doesn't work. And UV? Sorry, but if your surface is cool, then it's not going to radiate any significant heat in the UV. And before you start blabbing about cathode discharge or other such nonsense, keep in mind that what the surface needs to do to keep from melting is get rid of heat, not just energy. The difference is critical for thermodynamics. The various cathode explanations are akin to claiming that you can cool off a can of gasoline by setting it on fire, since that releases energy.

There are no cooling mechanisms for your solid surface.

 

[brantc]

So we are on the same page at last: Any light can reflect off a reflective surface.
The problem is that the Sun is too hot for any such surface to exist.


I have no "contention".
The facts are that

1. There is overwhelming evidence that Sun is internally heated through fusion. The observed neutrinos types are a fingerprint of fusion. The observed neutrino flux matches enough fusion to produce the energy output of the Sun.
   This means that the temperature of the photosphere is the minimum temperature of the Sun before light escapes from it. This is simple thermodynamics - the closer you get to a source of heat the hotter you get.
2. The photosphere is at a temperature of ~6000 K. The temperature increases with depth to ~9400 K at a depth of a few hundred kilometers. This supports pont 1.

Thus the Sun is too hot for any solid surface to exist.

So that finishes the thermodynamic considerations.

The only "evidence" you have for fusion is supposedly neutrinos that change in certain ways(that you cant really see) on the way to the earth.

As far as your claim for a thermodynamically impossible(my model) model.
Your model has more problems than mine.

You have a cool layer that is heated from BOTH SIDES.

Mine is only heated from one side.

If you want to invoke some mechanism for heat transfer, then you can apply that same mechanism to my model.

 

[brantc]

Well, now, let's talk about basic physics and see if there really is "no problem".

First, what about "just like the mirror in TRACE'S telescope". The mirrors in TRACE, like any other astronomical mirror, are highly polished to enhance reflectivity. Is the "surface" of the sun (whatever it might be made of) highly polished? Unlikely, I think. And in the case of TRACE, the primary mirror is coated with special, normal incidence coatings that further enhance reflectivity at UV & EUV wavelengths, and to narrow the passband. Are there any reflectivity enhancing coatings on the surface of the sun? I think that too is unlikely.

Next, let's ask what this reflective surface is made of. In keeping with the primary consideration of this thread, let's just assume the surface is iron (or at least mostly iron). I don't know how iron reflects EUV wavelengths, but in a moment I will guess. Meanwhile, I do know that iron reflects visible light about 62% at the red end (7000Å) down to 52% at the blue end (4000Å). If we simply assume a linear trend to shorter wavelengths, we would get about 35% reflectivity around 192Å or 195Å. But that does assume a specular surface and we know that dispersion off of a non-specular surface will degrade the reflectivity, so 35% would be a best-case scenario guess.

But here comes that basic physics. Photons that ionize iron don't reflect off of an iron surface, they ionize it and disappear in the process. Likewise, photons that carry enough energy to melt iron don't reflect off of an iron surface, they melt it and disappear in the process. The 192/5Å passband represents Fe XII (11-times ionized iron). So we would expect photons at that wavelength to preferentially not reflect off of an iron surface, but rather to ionize it. Furthermore, the 192/5Å passband represents the peak of thermal emission in the broad range of temperatures 500,000 - 2,000,000 Kelvins. Since the melting point of iron is 1811 Kelvins, and its boiling point is 3134 Kelvins, we would expect photons in the 192/5Å passband to melt & vaporize the iron surface and disappear in the process, and not to reflect off of the surface.

So, why does the mirror on TRACE not melt in the face of melting photons in the 192/5Å passband? It's all about intensity. The TRACE mirror has to deal with a few photons. One at a time, we don't necessarily expect the photons to ionize or melt anything (although the coatings are intended in part to prevent degradation of the primary mirror). But when photons gang up on a surface, they can and will provide enough thermal energy to melt & vaporize that surface, and then ionize the resulting vapor. Here at Earth, where we find TRACE, the incident solar EUV flux will be about 2 erg cm^-2 sec^-1, whereas at the sun we are looking at roughly 600,000 erg cm^-2 sec^-1 (which will be much higher near active regions). And do note this is EUV flux only, the bolometric flux at the photosphere is about 60,000,000,000 erg cm^-2 sec^-1. That's a lot of photons, and we cannot simply pretend that any solid surface is immune to the effect of that kind of photon bath.

So basic physics presents a problem after all. No, TRACE should not see any photons in the 192/5Å passband reflecting off of any iron surface. Rather, all of those photons should disappear in the process of melting, vaporizing and ionizing that surface. Furthermore, a little consideration shows that this should be the case no matter what the surface is made out of, since the temperature range 500,000 - 2,000,000 Kelvins vastly exceeds the boiling point of all known materials. So photons of the 192/5Å passband should not reflect off of any surface on the sun, regardless of its chemical/physical makeup.

EUV microscope explores nanoscale

A team of US, Russian and Ukrainian scientists is using a table-top extreme ultraviolet (EUV) illumination source to create an optical microscope that can image features as small as 100 nm. Operating in reflection mode and requiring little sample preparation, the EUV microscope can rapidly characterize the topography of microelectronic circuits, lithography masks and other material surfaces. (OPTICS EXPRESS 13 3983)
http://nanotechweb.org/cws/article/tech/22552

I never said the surface was immune to that flux of photons. But that doesnt mean the surface is not visible at those wavelengths. And its probably sputtering off great amounts of iron. But we know how much from the heavy atoms in the solar wind.

 

[Reality Check]

The only "evidence" you have for fusion is supposedly neutrinos that change in certain ways(that you cant really see) on the way to the earth.

The evidence for fusion is that fusion has a specific signature, i.e. gives off certain neutrinos. These neutrinos are detected in the types and amount that show that the sun is powered by fusion.
As for neutrino oscillation - the evidence for this has been established both from solar observations, atmospheric neutrinos, neutrinos from nuclear reactors and neutrino beams from particle accelerators.

The basic concepts behind the neutrino oscillation experiments are really simple, e.g.

1. Create a beam containing neutrinos of a known type.
2. Point the beam at a neutrino observatoriy some distance away.
3. Detect the neutrinos and determine what type they are.
4. If neutrino oscillation does not exist then you should only detect the original type of neutrino.
5. The experiments do not thus neutrino oscillation does exist.

The details though are quite complex.

As far as your claim for a thermodynamically impossible(my model) model.
Your model has more problems than mine.

The only problem is with your model of the sun is that it is physically impossible as has been pointed out many times before. That makes any assertions you make about it moot. You should read the posts:

• Is your solid iron surface thermodynamically possible? First asked 8 July 2009. See this post for a fuller explanation of the thermodynamic problems with MM's solid iron surface.
• Why this iron crust is thermodynamically impossible
• Iron Sun Surface Thermodynamically Impossible V

The current solar model does describe the sun well in general. It has many problems like all scientific theories, e.g. the several theories to explain the coronal heating problem do not quite work (yet).
N.B. there is no problem with fusion powering the sun in the current solar model.

You have a cool layer that is heated from BOTH SIDES.

Mine is only heated from one side.

If you want to invoke some mechanism for heat transfer, then you can apply that same mechanism to my model.

Yours is also heated from both sides enough to melt

• Fusion from below: ~13,000,000 K at the core.
• Plasma above whatever depth you have arbitrarily placed your surface:
  • ~5700 K at the top of the photosphere.
  • ~9400 K a few hundred kilometers below the top of the photosphere.

The minimum temperature at both sides of your layer is ~5700 K. And since we do not see a solid surface at the top of the photosphere your layer is heated by much more.

You need to specify the mechanism that stops your model's layer from exploding!

 

[GeeMack]

I never said the surface was immune to that flux of photons. But that doesnt mean the surface is not visible at those wavelengths. And its probably sputtering off great amounts of iron. But we know how much from the heavy atoms in the solar wind.

The surface is not visible because the surface does not exist. That fact has been explained dozens, maybe hundreds of times in this thread alone, and is irrefutably supported by several very well developed branches of astrophysical and related sciences including thermodynamics, general relativity, and helioseismology. Your continued arguments from incredulity and ignorance do not support your claim.

 

[Ziggurat]

As far as your claim for a thermodynamically impossible(my model) model.
Your model has more problems than mine.

You have a cool layer that is heated from BOTH SIDES.

No, it isn't. The corona is a higher temperature, but it's also transparent, which means that it doesn't stop outgoing radiation and provides very little ingoing radiation. The photosphere radiatively couples to deep space, not the corona. But anything under the photosphere cannot radiatively couple to deep space, but only to the photosphere. Again, basic thermodynamics fail.

Mine is only heated from one side.

No, brantc. Your model is heated from ALL sides.

If you want to invoke some mechanism for heat transfer, then you can apply that same mechanism to my model.

Obviously not, since the corona is transparent but the 5700K photosphere is not. Thanks for demonstrating that you haven't learned a thing.

 

[brantc]

The surface is not visible because the surface does not exist. That fact has been explained dozens, maybe hundreds of times in this thread alone, and is irrefutably supported by several very well developed branches of astrophysical and related sciences including thermodynamics, general relativity, and helioseismology. Your continued arguments from incredulity and ignorance do not support your claim.

No. I dont think so.

For instance lets do a little media surfing.

"Science News"
By Alexandra Witze
July 31st, 2010; Vol.178 #3 (p. 18)
Physicists do know a lot about the sun and how it works: Hydrogen atoms fuse in its core, forging helium and heavier elements and spewing out energy in the process. But over the past several years, scientists have dramatically overhauled estimates of the sun’s chemical makeup. In particular, they say there may be far less of key elements such as oxygen, carbon and nitrogen than previously thought. These changes are major enough to throw into question other basic assumptions about the sun, such as ideas about how sound waves travel through its interior, ringing it like a gong."

More like a bounded sphere as opposed to a decreasing density plasma sphere. They had to change the model from a plasma sphere to a gong sphere.

"These recalculations seemed to improve things, by tightening estimates for the abundances of elements like iron and silicon. “It was only when we started applying it to more important elements like oxygen and carbon, the most abundant metals in the sun, that we quite quickly realized our results were going to be very different,” says Martin Asplund, an astronomer at the Max Planck Institute for Astrophysics in Garching, Germany.

Asplund’s team published estimates of solar photospheric oxygen abundance that were 30 to 40 percent lower than commonly accepted values, with similar changes for carbon, nitrogen and neon. The researchers also analyzed the other elements in the sun, most of which required only minor revisions. In all, the team found a solar metal content not of 2 percent, but of 1.4 percent. The researchers summarized their latest work last year in the Annual Review of Astronomy and Astrophysics."

"The work of Asplund’s team throws a wrench in that picture. If models of the solar interior are adjusted to fit the 30 to 40 percent lower oxygen abundance in the photosphere, then they no longer match up with helioseismologists’ observations. For instance, the models now calculate incorrect values for the speed of sound and the density within the sun, as compared with those actually measured."
http://www.sciencenews.org/view/feature/id/61127/title/Beneath_that_blazing_facade

So what that means is that with less heavier elements the sound speed does not match calculations. That means a model that has a heavier "core" fits better.

In addition to that from Nature;

The proton shrinks in size

Tiny change in radius has huge implications.
http://www.nature.com/news/2010/100707/full/news.2010.337.html


Particle physics: 'Honey, I shrunk the proton'
"Even if the deviation is negligible on a day-to-day scale, it possibly has significant consequences. Researchers are unable to say precisely what these may be, however. What is certain is that this changes the Rydberg constant. Quantum physicists use this constant to calculate which energy packets atoms and molecules absorb and emit when they change their states. These energy packets correspond to the spectral lines of the elements. The calculations for the spectral lines now shift noticeably and no longer match the experimental findings."
http://www.physorg.com/news197727820.html

This means that spectral calculations may also be off.

That means that calculated opacities will be off. That means photospheric opacities may be wrong. Or that my model is more likely to be right.

So what I am seeing is that as the measurements get more accurate the models, accepted models("very well developed branches"), break.

So I wonder how much heat energy is actually put out by my surface glow layer if it is a layer . How much heat/light is put out by the surface?
10-7 g/cm3 is the density of the photosphere. How much heat do you think a vacuum puts out?

So if I do an experiment where I levitate a iron slug with a magnetic field in a 6K plasma, do you think the slug will eventually melt?
Or do you think it will re-radiate enough energy in IR to prevent it from melting....

The Levitated Dipole eXperiment
http://www.psfc.mit.edu/ldx/

The surface glow may put out a spectrum that says 5.8K but a large percentage of white light, UV, IR comes from the surface in the form of flares.


From a paper that I dont necessarily agree with the conclusions but he has some interesting things to say.

"Seismology:
The Sun is a laboratory of seismology [17]. Yet, on Earth, seismology is a science of the condensed state. It is interesting to highlight how the gaseous models of the Sun fail to properly fit seismological data. In the work by Bahcall et. al. [18] for instance, experimental and theoretical siesmological findings are compared as a function of Solar radius. Precise fits are obtained for most of the solar sphere. In fact, it is surprising how the interior of the Sun can be so accurately fitted, given that all the data is being acquired from the solar surface. At the same time, this work is unable to fit the data
in the exterior 5% of the Sun [18]. Yet, this is precisely the point from which all the data is being collected. The reason that this region cannot be fitted is that the gaseous models are claiming that the photosphere has a density on the order of 10-7 g/cm3. This is lower than practical vacuums on Earth.
Thus, the gaseous models are trying to conduct seismology in a vacuum by insisting on a photospheric density unable to sustain seismic activity."
http://thermalphysics.org/Sun.evidence.1.pdf

 

[Reality Check]

No. I dont think so.

For instance lets do a little media surfing
...various news reports...

No. I don't think that any of these news reports address the fact that your idea is physically impossible according to quite basic physics as pointed out to you in many posts.

All you have is a list of normal science using the normal scientific process, i.e.. as more evidence the models break and better models that fit the evidence are created.


The composition of the photosphere does not matter.

• It's temperature is measured to be ~5700 K at the top and increase with depth (~9400 K at a few 100 kilometers).
• There is overwhelming evidence that the Sun is has a core undergoing fusion and and the models predict a temperature of ~13,000,000 K.

This evidence rules out any solid surface. Thus your idea is not only broken because of the existing evidence - it is totally vaporized!

That means that calculated opacities will be off. That means photospheric opacities may be wrong. Or that my model is more likely to be right.

You do not have a "model" - that suggests that you can produce numeric predictions like the scientific solar model.
What you have is an idea that is easily seen to be physically impossible.

 

[Reality Check]

From a paper that I dont necessarily agree with the conclusions but he has some interesting things to say.
...
http://thermalphysics.org/Sun.evidence.1.pdf

The author is distinguished but has no formal background in astronomy: "Pierre-Marie L. Robitaille, Ph.D. is currently a Professor of Radiology at The Ohio State University in Columbus, Ohio.".

You gave no evidence of the PDF ever being published. However it was published in Progress in Physics which has a dubious reputation. The paper by itself makes the journal quite dubious.

Pierre-Marie L. Robitaille makes basic mistakes, e.g.

• "The fact that this spectrum is continuous in nature leads to difficulties for the gaseous models [1]. This is because gases are known to emit radiation only in discrete bands [12]".
  Plasma is known to emit continuous spectrum due to free electrons.
• "The atmosphere of the Earth does not collapse due to the relatively rigid oceanic and continental surfaces. Within the gaseous models of the stars however, there is no mechanism to introduce the rigid surface required to maintain gas pressure."
  There is no solid surface needed to maintain pressure in any body of gas (or plasma). All you need is a force to keep the body of gas together - like gravity.

 

[Dancing David]

Hi Brantc!
http://thermalphysics.org/Sun.evidence.1.pdf

Really? he claims to have disporoved the universality of black body radiation, in a paper that references his papre that references another paper, one of which was published.

Um sure whatever.

 

[brantc]

Hi Brantc!
http://thermalphysics.org/Sun.evidence.1.pdf

Really? he claims to have disporoved the universality of black body radiation, in a paper that references his papre that references another paper, one of which was published.

Um sure whatever.

A true blackbody spectrum comes only from a solid body. That box in kirchoffs experiment will never be thermally equilibrated.

Plasma only emits lines. Water is not a blackbody as the emission changes with the nadir angle.


Thats his basic argument.

And its true.

Thats why the solar spectrum is the result of a solid body + arc overlayed with plasma.

 

[brantc]

No, it isn't. The corona is a higher temperature, but it's also transparent, which means that it doesn't stop outgoing radiation and provides very little ingoing radiation. The photosphere radiatively couples to deep space, not the corona. But anything under the photosphere cannot radiatively couple to deep space, but only to the photosphere. Again, basic thermodynamics fail.
No, brantc. Your model is heated from ALL sides.

Obviously not, since the corona is transparent but the 5700K photosphere is not. Thanks for demonstrating that you haven't learned a thing.

Sure. With your model of the photosphere. But we are not talking about that are we.

Since the loop foot prints are visible in white light then you can see through the photosphere which is just a layer of plasma about 200 to 400km thick.

Not your surface of last scattering after photons have traveled ten bizillion years randomly to the surface. And then the have emerged with exactly the right spectrum to simulate a perfect blackbody.
LOL.... Its a decreasing density plasma ball with a "convection" and some other kind of layer. And like thats not going to show up in the spectrum....


As far as the photosphere goes its so thin that its thermodynamically impossible for that layer to trap 100% of all out going radiation.

Its so thin that its total heat emission is lower than the total heat emission from the solid surface below.

Here let me say that again.


Its so thin that its total heat emission is lower than the total heat emission from the solid surface below.


RC...

The thermosphere is 1500K- 2500K how come the earth doesnt overheat???????


"The thermosphere is the biggest of all the layers of the earth's atmosphere directly above the mesosphere and directly below the exosphere. Within this layer, ultraviolet radiation causes ionization.

The highly diluted gas in this layer can reach 2,500 °C (4,530 °F) during the day. Even though the temperature is so high, one would not feel warm in the thermosphere, because it is so near vacuum that there is not enough contact with the few atoms of gas to transfer much heat. A normal thermometer would read significantly below 0 °C (32 °F), due to the energy lost by thermal radiation overtaking the energy acquired from the atmospheric gas by direct contact."
http://en.wikipedia.org/wiki/Thermosphere


Dont you get this yet?????? The photosphere on the sun is of lower density than the thermosphere......

 

[Tim Thompson]

Solar Photosphere vs Terrestrial Thermosphere

Don't you get this yet?????? The photosphere on the sun is of lower density than the thermosphere ......

Not really. The thermosphere mass density, in the high temperature regions you are talking about, ranges between 10^-8 and 10^-9 gm/cm³, whereas the mass density of the solar photosphere, at optical depth 1 and temperature 6520 Kelvins, is about 3x10^-7 gm/cm³, 30 to 300 times the mass density of the thermosphere. The solar photosphere does not get that tenuous until about 400 km above the optical depth 1 layer, where its mass density reaches about 10^-8 gm/cm³ and its temperature falls to about 4600 Kelvins (it falls to a local minimum of about 4400 Kelvins at 500 km and then increases again to over 6000 Kelvins at the top, where the mass density is a meager 10^-11 gm/cm³). So you are off by at least an order of magnitude there.

As far as the photosphere goes its so thin that its thermodynamically impossible for that layer to trap 100% of all out going radiation.

That does not really make any physical sense. Anything that is in thermal equilibrium, whether thin or thick, will not trap any of the out going radiation, not even 1%, let alone 100%. It will absorb energy on one side, and emit the same amount of energy on the other side, minus the energy needed to maintain its kinetic temperature. Now, I suspect what you think you are saying (in which case you should actually say it instead of something else), is that the photosphere is transparent at some wavelength. That's an easy assertion to make, but not to prove. If you think it is, all you have to do is specify the constituent gases, their density & temperature, and compute the opacity/transparency at a given wavelength. The trick is that if your model photosphere is not consistent with observations, then nobody will believe you, and for good reason.

Since the loop foot prints are visible in white light then you can see through the photosphere which is just a layer of plasma about 200 to 400km thick.

Non-sequitor. What does being visible in white light, or any other light, have to do with "through the photosphere"? how do you know that that the bright patches do not originate where they appear to originate, namely above the photosphere, where they appear in the images?

The thermosphere is 1500K- 2500K how come the earth doesnt overheat???????

First, get the temperatures right. Zero Celsius is +273.5 Kelvin (let's call it 273 for simplicity), so 1500 - 2500 Celsius from the Wikipedia page is 1227 - 2227 Kelvins, a difference of 11% at the high end, which is significant.

Second, you have to realize that there are two different kinds of temperature here. The quoted temperature for the thermosphere is a kinetic temperature, derived from the kinetic energy of the molecules in the atmosphere. A thermometer in that dilute gas does not register the high temperature because only a few molecules hit it per second. So the actual energy transfer from the gas to any object in the gas is miniscule, despite the high temperature. However, the quoted temperature for the solar photosphere is not a kinetic temperature, it is a radiative temperature, which means it represents the energy of the photons, not the kinetic energy of the particles. Radiative temperature & kinetic temperature are not at all the same thing, but you are treating them as such, and that is wrong.

If you hold a thermometer in the thermosphere gas, it gains energy from infrequent molecular collisions, and the photons of sunlight at the Earth (about 1365 Watts/meter², the solar constant). However, if you hold a thermometer in the gas of the solar photosphere, it gains energy from more frequent atomic collisions (at least 10 times the mass density) and the photons of sunlight at the photosphere (about 63,000,000 Watts/meter²). That's why the Earth does not over heat, the power in the photon bath at the solar photosphere is 46,154 times more intense per unit area than here on Earth. Put Earth in the photosphere, and you can bet it will overheat.

 

[Dancing David]

A true blackbody spectrum comes only from a solid body. That box in kirchoffs experiment will never be thermally equilibrated.

Plasma only emits lines. Water is not a blackbody as the emission changes with the nadir angle.


Thats his basic argument.

And its true.

Thats why the solar spectrum is the result of a solid body + arc overlayed with plasma.

And if you ignore what has been said before about the emission of free electrons, then it makes you look like you don't really know much about plasma or physics.

Seriously Brantc, you got served on this issue before.

 

[GeeMack]

Sure. With your model of the photosphere. But we are not talking about that are we.

Apparently you are talking about a model which isn't based in physical reality.

Since the loop foot prints are visible in white light then you can see through the photosphere which is just a layer of plasma about 200 to 400km thick.

You can't see through the photosphere.

Not your surface of last scattering after photons have traveled ten bizillion years randomly to the surface. And then the have emerged with exactly the right spectrum to simulate a perfect blackbody.
LOL.... Its a decreasing density plasma ball with a "convection" and some other kind of layer. And like thats not going to show up in the spectrum....


As far as the photosphere goes its so thin that its thermodynamically impossible for that layer to trap 100% of all out going radiation.

Its so thin that its total heat emission is lower than the total heat emission from the solid surface below.

The photosphere is, by definition, the region in the Sun's atmosphere where the plasma goes from being transparent to being opaque. You can't see through the photosphere.

Here let me say that again.


Its so thin that its total heat emission is lower than the total heat emission from the solid surface below.

Here, let me say that again. The photosphere is defined as the region where the solar atmosphere transitions from being transparent to being opaque. You cannot see through it. Nobody can. Is there some particular reason you insist on trying to change the perfectly good, valid, legitimately scientific definition of the word?

You're continuing to try to describe a fantasy Sun, one whose existence has been demonstrated time and again to be impossible. There are several very highly developed branches of physics which show, to the satisfaction of every professional solar physicist on Earth, that your claim is false, your incessant arguments from incredulity and ignorance notwithstanding.

 

[Reality Check]

RC...

The thermosphere is 1500K- 2500K how come the earth doesnt overheat???????

brantc...
So obvous that I am going to leave it as anexercise for you (hints: inverse square law & 93 milloion miles)

Dont you get this yet?????? The photosphere on the sun is of lower density than the thermosphere......

The photosphere is also of lower density than:

• The surface of the Earth.
• The core of the Earth.
• The computer that I am writing this post on.
• Me !

Your point is??????

There is the Wikipedia quote without your strange highlighting
Thermosphere

The thermosphere is the biggest of all the layers of the earth's atmosphere directly above the mesosphere and directly below the exosphere. Within this layer, ultraviolet radiation causes ionization.
...
The highly diluted gas in this layer can reach 2,500 °C (4,530 °F) during the day. Even though the temperature is so high, one would not feel warm in the thermosphere, because it is so near vacuum that there is not enough contact with the few atoms of gas to transfer much heat. A normal thermometer would read significantly below 0 °C (32 °F), due to the energy lost by thermal radiation overtaking the energy acquired from the atmospheric gas by direct contact.

Fairly standard physics:

• The thermosphere is a rarefied gas with particles moving at a high speed. This corresponds with a temperature of 2,500 °C during the day.
• A normal thermometer would lose a lot of heat through radiation and gain a little heat through contact and so will have a low temperature reading.

 

[brantc]

And if you ignore what has been said before about the emission of free electrons, then it makes you look like you don't really know much about plasma or physics.

Seriously Brantc, you got served on this issue before.

Ok. Refresh me. Serve me again.

 

[brantc]

Apparently you are talking about a model which isn't based in physical reality.
You can't see through the photosphere.

The photosphere is, by definition, the region in the Sun's atmosphere where the plasma goes from being transparent to being opaque. You can't see through the photosphere.

I know what your definition is. And just because you say it doesn't make it so.

My model is different than your definition.

Here, let me say that again. The photosphere is defined as the region where the solar atmosphere transitions from being transparent to being opaque. You cannot see through it. Nobody can. Is there some particular reason you insist on trying to change the perfectly good, valid, legitimately scientific definition of the word?

I am not changing your definition of the word.
My definition of the luminous region that covers the surface of the sun is surface glow. You can use the word photosphere however you want.

You're continuing to try to describe a fantasy Sun, one whose existence has been demonstrated time and again to be impossible. There are several very highly developed branches of physics which show, to the satisfaction of every professional solar physicist on Earth, that your claim is false, your incessant arguments from incredulity and ignorance notwithstanding.

The internal fusion model is dead.

You have not demonstrated that my model is impossible. Far from it.

As a matter of fact the latest data I have showed indicates that your model is not heavy enough. Metals do not match helioseismology data.
According to the latest measurements of the proton there are problems with opacity calculations as well.

You know nothing about your model except from what you see on the surface(maybe neutrinos)...

 

[brantc]

You can't see through the photosphere.

It has already been agreed that the footprint of the loops are below the photosphere.

TRACE observes white light flares above the limb at the base of certain loops. According to Tim this Ghost limb, where the flare footprints take place is an artifact of the telescope.

So if this is an artifact and then if you correct for the artifact (2 degree wedge) then that places the white light flares under the photosphere at the loops footprints. I have already said that the loop footprints should produce white light flares.


Here again are pictures of the loop footprints which emit from IR to EUV.
Notice the solar moss also emits at EUV.
http://trace.lmsal.com/POD/images/T171_001017_033928.jpg
http://trace.lmsal.com/POD/images/arcade_9_nov_2000.gif

The loops are 50,000 to 75,000km high. That means the structure under them is higher than the transition region, photosphere and chromosphere.

So we have loop footprints, solar moss and structures under the loops in these pictures.