Moderated Iron sun with Aether batteries...

[Dancing David]

That is another day.

All I'm saying is IF there was a solid surface nearby, the light at 171 would reflect off of it.

And it would travel 3.5 meters.
[/quote]

The flare is not on the objects of interest. It is external to the object of interest therefore it will reflect off of it.
A flare is a moving light source that is brighter than the surrounding light sources and is moving upward.
It may be below the corona but it is still moving independently from surrounding light sources therefore it can reflect off of other features(assuming they would reflect as opposed to scatter or absorb)..
[/quote]
3.5 meters.

It doesnt matter where the light comes from, it still reflects IF there was something that was solid nearby, the light emitted from the object @ 171 would reflect off of something solid just like light at 650nm, 700nm, 1600A or any other light.

And it will only travel 3.5 meters.

 

[Ziggurat]

Conversely you could also say that a blackbody spectrum is the property of solid matter.

You could say that, but you'd be wrong. Most solid matter is not black at all. And you don't need something to be solid in order to be black.

Blackbodies reflect light

Uh... no. No, they don't. That's why they're called black bodies: because they don't reflect light. If they reflected light, they'd be the color of whatever light they reflected. If they reflect red light only, they would be red bodies. If they reflected blue light only, they would be blue bodies. If they reflect all light, they are white bodies.

otherwise you couldnt see it, other than the sun which is emission mostly.

You're half wrong. Emission is indeed one of the two ways to see a black body. The other way to see something black is via its silhouette. Reflection, however, is NOT one of the ways to see a black body. Because, once again, if it's a black body, it doesn't reflect light! That's what black body means. Once again, you're wrong by definition, which is about as wrong as you can possibly be.

 

[Tubbythin]

Blackbodies reflect light, otherwise you couldnt see it, other than the sun which is emission mostly.

. No.

 

[DeiRenDopa]

The difference image begins with two images of *real things*. The fact that the sun is round means that the RD images will produce a "disk" rather than say a triangle or a square. It is the "real things" in the original images that provide the "shapes" we see in the RD images.

That's not correct.

If the Sun were a featureless disk, as we see it on the sky, without anything extending beyond the disk (e.g. no prominences), then RD images would be completely blank.

Further, if the Sun did have features, sunspots say, which remained the same between images, then RD images would be completely blank.

It's only because something - or some things - change between one image and the next that a RD images are not blank.

Oh, and what is a ""disk""? I mean, you use triangle and square with their normal meanings (presumably), what's wrong with the normal meaning of disk?

Also, what do you mean by "*real things*"? Are they different from ""real things""?

[...] how the mathematical RD process relates to "real things". [...] in reality the shapes in the image are directly related to real things

That's true, but an RD image will not show something that doesn't exist in the images from which it was derived.

And if you intend to derive the relative positions of features on the Sun (as I think you've said you want to), you need to use the original images, not the RD ones.

and real physical processes that occurred in the original images.

And that's why RD images are less reliable than the originals, in terms of determining the relative positions of features.

You do see why, don't you?

 

[GeeMack]

That's not correct.

If the Sun were a featureless disk, as we see it on the sky, without anything extending beyond the disk (e.g. no prominences), then RD images would be completely blank.

Further, if the Sun did have features, sunspots say, which remained the same between images, then RD images would be completely blank.

It's only because something - or some things - change between one image and the next that a RD images are not blank.

Oh, and what is a ""disk""? I mean, you use triangle and square with their normal meanings (presumably), what's wrong with the normal meaning of disk?

Also, what do you mean by "*real things*"? Are they different from ""real things""?


That's true, but an RD image will not show something that doesn't exist in the images from which it was derived.

And if you intend to derive the relative positions of features on the Sun (as I think you've said you want to), you need to use the original images, not the RD ones.


And that's why RD images are less reliable than the originals, in terms of determining the relative positions of features.

You do see why, don't you?

I've made running difference graphs from current 4096 x 4096 171Å SDO images. I've made them from images at intervals from approximately 15 minutes to approximately 6 hours. They produce exactly the same kind of results as any other running difference graphs. The closer the time interval the less pronounced the optical illusion of lights and shadows. The longer the interval the more pronounced the difference. At a certain point where the difference is so great that a particular feature is separated from its previous position, rather than overlapping it somewhat, the illusion goes away just like with any other running difference graphs of thermal filtered data taken from the Sun's corona.

Oh, and the edge of the disk is of course in the same location in the output graphs as it is in the input source data.

And isn't it amazing that these long cadence running difference graphs, according to Michael's own affirmation spread across several pages of this thread, is the thing that will falsify his whole misunderstanding, yet it is I who have gathered the relevant data and created the running difference graphs and not him?

I've got some from Wednesday the 26th and from Thursday the 27th. Maybe I should post my results right after Michael posts his.

 

[Tim Thompson]

Magnetic Reconnection Redirected

On matters of magnetic reconnection raised in this thread, since they are arguably off-topic, I have responded with redirection to the magnetic reconnection thread: Magnetic Reconnection Redux XIII.

 

[brantc]

Even I know that the bit I bolded is cobblers.

Dave

A "blackbody" is purely an emission source, however if you were to shine a light on this emission source would you expect it to be invisible?

No matter that I know of is invisible.

 

[brantc]

You could say that, but you'd be wrong. Most solid matter is not black at all. And you don't need something to be solid in order to be black.

The only blackbody that is not solid matter is the "sun", supposedly.

A blackbody spectrum is the property of solid matter, or pressure broadened plasma, approaching solid matter, here on earth in a lab.

You cannot get a blackbody in a thin plasma here on earth. I know this from my work with cavitation.

Uh... no. No, they don't. That's why they're called black bodies: because they don't reflect light. If they reflected light, they'd be the color of whatever light they reflected. If they reflect red light only, they would be red bodies. If they reflected blue light only, they would be blue bodies. If they reflect all light, they are white bodies.

No solid matter is black enough to absorb 100% of the incident light.
There is always some small percentage of reflected light.

"On the other hand, researchers now say they have created the darkest material ever made by man. The material, a thin carbon coating, reflects less than 0.1 percent of incoming light. It absorbs the rest."
http://www.world-science.net/othernews/080223_black.htm

You're half wrong. Emission is indeed one of the two ways to see a black body. The other way to see something black is via its silhouette. Reflection, however, is NOT one of the ways to see a black body. Because, once again, if it's a black body, it doesn't reflect light! That's what black body means. Once again, you're wrong by definition, which is about as wrong as you can possibly be.

See above.

Once again reality and the lab trump theory and "definition". This perfect blackbody only exists in theory. The sun actually has a hump in the EUV which makes it a "not perfect blackbody".
http://folk.uio.no/paalb/sumer_atlas.html

http://climate.gsfc.nasa.gov/static/cahalan/Radiation/Images/SolarIrrVblackbody.gif


From the JET website.

"The Science of JET", by John Wesson".

"The initial idea was that of detecting the blackbody radiation from the thermal plasma ions. However, when the ICE spectra were measured they were not consistent with this expectation, having instead narrow equally-spaced emission lines, the spacing being proportional to the magnetic field, and intensities much larger than the blackbody level. The spectrum from a deuterium-tritium plasma is shown in Figure 13.4 (below). The observed frequencies depend on the magnitude of the magnetic field at the location of the emission and, surprisingly, it was found that in JET this meant that the emission comes from the edge of the plasma in the outer midplane."

Canada's Ultimate Light Ruler
"The core of the high-temperature blackbody is a hollow tube of a special form of graphite that can withstand intense heating. In order to produce the required UV radiation, the graphite core is heated to approximately 3230 ºC, a temperature at which almost all metals melt – hence the name an ultra high-temperature blackbody. The graphite is gradually heated over the course of several hours in the same way a stove element is heated – by running an electrical current through it. The graphite core is insulated with many concentric layers of carbon cloth which are water-cooled.

At 3230 ºC any oxygen would react instantly with the graphite, causing a fire. So, during operation the entire core is flushed with argon, a non-reactive gas. At operating temperature, the high-temperature blackbody produces an intense beam of light that's emitted from a tiny eight millimetre hole – a beam of light that will soon be Canada's ultimate light ruler."
http://www.nrc-cnrc.gc.ca/eng/news/nrc/2006/02/06/black-body.html

 

[Dave_46]

A "blackbody" is purely an emission source, however if you were to shine a light on this emission source would you expect it to be invisible?

No matter that I know of is invisible.

I would expect NO light to be reflected from a black body, I would expect it to appear black. A black body absorbs 100% of incident light.

I have worked with instruments (heat flux meters) which have an emissivity of about 0.95 - 0.97. Not quite full black body, and pretty well impossible to examine with the naked eye, or using a magnifying glass to try to examine the surface. If there was detail visible then it was time for a re-paint, as it was too far from being a black body

Unless it is at a temperature where it is emitting visible light then I would expect a black body to appear, wait for it, BLACK.

Dave

 

[sol invictus]

The only blackbody that is not solid matter is the "sun", supposedly.

Nonsense.

A blackbody spectrum is the property of solid matter, or pressure broadened plasma, approaching solid matter, here on earth in a lab.

Again, nonsense.

You cannot get a blackbody in a thin plasma here on earth. I know this from my work with cavitation.

I don't know if you've noticed - the sun isn't very "thin", brantc. It's more than a million km across.

A "blackbody" is purely an emission source, however if you were to shine a light on this emission source would you expect it to be invisible?

No matter that I know of is invisible.

Why in the world would you think a perfectly black object would be "invisible"? How often do you see military vehicles painted black? You aren't just getting basic thermodynamics completely wrong, you're failing the common sense test.

 

[Toke]

I have the impression that a blackbody is a theoretical object only.

When considering the heat radiation emited from my forearm I ignore a lot of stuff to focus on the surface temperature and the emission it causes.

 

[Ziggurat]

The only blackbody that is not solid matter is the "sun", supposedly.

Nope. Ever seen pictures from the deep ocean?

A blackbody spectrum is the property of solid matter

No, it isn't. It's a property of the electromagnetic spectrum. Crack open a thermodynamics textbook sometime. You will find that the derivation of the black body spectrum has nothing to do with any properties of solids. I've even got several texbooks sitting on my shelf right now that I can consult. Which would you rather hear from, Kittel & Kroemer, Reif, or Huang?

You're out of your depth, brantc.

or pressure broadened plasma, approaching solid matter

A "pressure broadened plasma" (presumably you mean a plasma with spectral lines broadened by interactions) doesn't approach solid matter, it approaches denser plasma.

You cannot get a blackbody in a thin plasma here on earth. I know this from my work with cavitation.

What counts as thin, brantc? Well, thinner than the optical depth, obviously. The optical depth can vary quite a lot. But it's smaller than the size of the arc in many high-intensity discharge lamps. In sea water, it's a lot longer than a foot, but much shorter than the depth of the ocean. How long is the optical depth for the photosphere? How far down do you think your solid surface is?

No solid matter is black enough to absorb 100% of the incident light.
There is always some small percentage of reflected light.

So what? So most stuff isn't a perfect black body. The absorptivity and emissivity is generally less than one. If it's close to one, then the difference generally doesn't matter. And it sure as hell doesn't make a difference in the case of the surface of the sun, and the impossibility of that surface being solid, or of seeing light reflected off a solid surface underneath the photosphere. And it doesn't matter what the absorptivity of your proposed solid surface is, what matters is the absorptivity of what we can see. Which is close enough to one to rule out your proposal.

Once again reality and the lab trump theory and "definition". This perfect blackbody only exists in theory.

Really? The cosmic microwave background looks like a perfect black body. But in any case, it doesn't matter. All we need is close enough to a black body to rule out your proposal, and that's what we observe.

The sun actually has a hump in the EUV which makes it a "not perfect blackbody".

Again, so what? The deviations from black body are not significant enough to accommodate your proposal. And if you really understood the significance of that hump, you'd realize that it indicates that the sun gets HOTTER, not colder, as you go deeper. Which, once again, rules out a solid surface under the photosphere.

 

[Ziggurat]

I have the impression that a blackbody is a theoretical object only.

First off, the "black body" radiation spectrum is a property of the electromagnetic field itself. If you have, for example, a closed box at thermal equilibrium, the radiation inside the box will have a "black body" spectrum, regardless of whether or not the walls of the box are black. You can think of it as a photon gas, which has a specific equilibrium spectrum depending on its temperature. And you can derive this spectrum without reference to the walls of your box.

OK, so if it's a property of the electromagnetic field itself, why do we call it "black body" radiation? Well, suppose we have a surface in thermal equilibrium with our photon gas. The number of photons hitting that surface must be the same as the number of photons coming from that surface (both reflected and emitted), or else it's not equilibrium. If that surface is black (meaning it absorbs all photons which hit it), then it MUST emit the same spectrum that is incident upon it. That's why it's called "black body". Furthermore, even if we take away the incident photons, the surface will keep emitting photons according to that spectrum.

If the surface is not perfectly black but slightly reflective, then at equilibrium we still need the same total number of photons coming from the surface, but now some of them come from reflection, so the number coming from emission must drop. If the surface is, say, 60% reflective, then it will be 40% absorptive, and it will emit 40% of the black body spectrum. So even if we're interested in surfaces other than perfect black bodies, the black body spectrum is STILL the baseline we use. And we know how far off that baseline we need to go by how reflective the surface is.

In other words, the fact that a perfect black body doesn't exist (except for maybe the CMB) doesn't matter. Black body radiation is still what we use to describe thermal emissions even for non-ideal sources.

 

[brantc]

I would expect NO light to be reflected from a black body, I would expect it to appear black. A black body absorbs 100% of incident light.
Dave

If you can show me something (matter) with 0% reflectivity, then I will agree with you.

That is what brought this about. I was told that blackbodies dont reflect any light (100% absorb). The closest I can find is 99.9% absorption.

I'm still waiting for that experiment that shows 100% absorption of light.

 

[Ziggurat]

I was told that blackbodies dont reflect any light (100% absorb). The closest I can find is 99.9% absorption.

Oh, it's actually easy to get higher absorptivity than that. But you STILL don't get it, brantc. What you're arguing is that nothing is a perfect black body. But so the hell what? That's got nothing to do with what black body means (ie, you were still wrong by definition), it's got nothing to do with the validity of thermodynamics (which doesn't require perfect black bodies), and in the particular case under discussion, the sun is observably close enough to a black body that there's no chance in hell you can make the observation you claimed of light reflecting from a solid surface underneath the photosphere.

 

[brantc]

Nope. Ever seen pictures from the deep ocean?

Yep. Water is closer to a solid than a plasma or gas?

No, it isn't. It's a property of the electromagnetic spectrum. Crack open a thermodynamics textbook sometime. You will find that the derivation of the black body spectrum has nothing to do with any properties of solids. I've even got several texbooks sitting on my shelf right now that I can consult. Which would you rather hear from, Kittel & Kroemer, Reif, or Huang?

You're out of your depth, brantc.

Since using the sun would be circular logic, can you show me anything else that is a plasma or gas emitting a blackbody spectrum characteristic of its temperature?
I showed you an example of where it was expected and was not the case. Theory failed.

A "pressure broadened plasma" (presumably you mean a plasma with spectral lines broadened by interactions) doesn't approach solid matter, it approaches denser plasma.

Yes, that true. If you follow the broadening curve it leads to a blackbody spectrum when the density approaches solid matter.

What counts as thin, brantc? Well, thinner than the optical depth, obviously. The optical depth can vary quite a lot. But it's smaller than the size of the arc in many high-intensity discharge lamps. In sea water, it's a lot longer than a foot, but much shorter than the depth of the ocean. How long is the optical depth for the photosphere? How far down do you think your solid surface is?

I dont care, pick a density.

The surface is about 1000km.

So what? So most stuff isn't a perfect black body. The absorptivity and emissivity is generally less than one. If it's close to one, then the difference generally doesn't matter. And it sure as hell doesn't make a difference in the case of the surface of the sun, and the impossibility of that surface being solid, or of seeing light reflected off a solid surface underneath the photosphere. And it doesn't matter what the absorptivity of your proposed solid surface is, what matters is the absorptivity of what we can see. Which is close enough to one to rule out your proposal.

Really? The cosmic microwave background looks like a perfect black body. But in any case, it doesn't matter. All we need is close enough to a black body to rule out your proposal, and that's what we observe.

Again, so what? The deviations from black body are not significant enough to accommodate your proposal. And if you really understood the significance of that hump, you'd realize that it indicates that the sun gets HOTTER, not colder, as you go deeper. Which, once again, rules out a solid surface under the photosphere.

Yes. The sun gets to an observable 10000 just below the photosphere.

The hump is an UV/EUV portion of the spectrum that can be explained by thermionic emission(arc activity) at the loop foot prints at the surface.

It cannot be explained by the "slow photon to the surface train" process which would produce an "exceedingly smooth" blackbody spectrum.

What was my proposal again???

 

[Toke]

In other words, the fact that a perfect black body doesn't exist (except for maybe the CMB) doesn't matter. Black body radiation is still what we use to describe thermal emissions even for non-ideal sources.

I knew that a blackbody is a model, thank you for the elaboration on reflectivity.

 

[Ziggurat]

Yep. Water is closer to a solid than a plasma or gas?

Way to move the goalpost, brantc. I don't care what you think it's closer to, water is not a solid.

Since using the sun would be circular logic, can you show me anything else that is a plasma or gas emitting a blackbody spectrum characteristic of its temperature?

We've been over this before, brantc. Almost a year ago. Have you learned nothing?

Yes, that true. If you follow the broadening curve it leads to a blackbody spectrum when the density approaches solid matter.

No, brantc. You get a blackbody spectrum when the physical depth exceeds the optical depth. The optical depth is shorter for denser plasma, but pick any density and you can make the plasma thick enough to produce a black body spectra.

I dont care, pick a density.

I didn't ask for a density, I asked for an optical depth.

The surface is about 1000km.

And how does that compare to the optical depth?

The hump is an UV/EUV portion of the spectrum that can be explained by thermionic emission(arc activity) at the loop foot prints at the surface.

No, it cannot.

It cannot be explained by the "slow photon to the surface train" process which would produce an "exceedingly smooth" blackbody spectrum.

Congratulations: you've just discovered that the optically thin and very hot corona is also a source of EUV light.

What was my proposal again???

You don't remember?

 

[Toke]

Since using the sun would be circular logic, can you show me anything else that is a plasma or gas emitting a blackbody spectrum characteristic of its temperature?

Last week on my refresher smoke diver course I saw a video of a candle left in a couch.
Within 5 minutes the smoke layer were down to waist height and there were bits of ceiling dropping down through the smoke.
It is a reasonable good example of a gas emiting blackbody heat and light radiation consistent with it's temperature.

Another would be a large vat of burning kerosene, lots of blackbody radiation, ranging from heat to light.

 

[Reality Check]

Yes. The sun gets to an observable 10000 just below the photosphere.

No. The Sun gets to an observable 9400 K just below that 5700 K photosphere.
This rules out any solid surface (black or not ) to provide any of the reflections that you are talking about.