Electric universe theories here.

Ziggurat · Aug 2, 2009

Michael Mozina said:
If we run current through many ordinary materials like ordinary metals and neon like we find in a neon bulb, we get white light. The emissions of the neon bulb, in other words it's white light 'color' is not necessarily indicative of it's physical temperature.

Wow, are you clueless. First off, an incandescent bulb (ie, a bulb with a metal filament) glows because it gets HOT, Michael. The current is merely a way of making it hot. Secondly, neon lights don't give off white light (I'm still wondering what made you think this was ever the case), and they don't give off anything close to a blackbody spectrum.

You mean besides that xenon arc lamp we talked about earlier?

Xenon arc lamps give off a blackbody-like spectrum with a peak around the visible because they get bloody hot, Michael. Their color temperature IS their temperature. Fail again. But are you willing to concede you were completely wrong about them only emitting at their atomic spectral lines?

This is like insisting that because the air temperature above the ocean is 89 degrees that the temperature at the bottom of the ocean can't reach 33 degrees.

Are you seriously trying to argue that if the entire atmosphere was 89 degrees, the ocean would still stay 39 degrees at the bottom?

So how do you know that more dense layers under the photosphere aren't also "coupled to deep space again"?

I have no interest in debating with you about whether the blackbody spectrum we see comes from the photosphere or under it. But whatever the origin of it, it's opaque from the UV through the infrared. That's an absolute requirement, unless you want to violate the 2nd law of thermodynamics. Anything underneath cannot radiatively couple with deep space unless it's FAR hotter than that (in which case we'd see it, but we don't), but of course your solid layer is supposed to be colder. So it can't be. Quite simple, really.

You just *ASSUME* this to be the case.

No. I assume the 2nd law of thermodynamics is valid. Given that assumption, it is an unavoidable conclusion that anything underneath cannot be radiatively coupled to deep space.

You also "assume" that the color of that layer indicates it actual temperature rather than the average temperature of all the particles leaving the surface.

No, Michael. If that 6000 K blackbody spectrum is due to particles leaving the surface, than those particles leaving the surface are the layer that I'm talking about. I make no claims about whether or not that's the case, because it doesn't matter. There is something emitting a 6000K blackbody spectrum, and it's opaque from the UV on down.

Reality Check · Aug 2, 2009

Michael Mozina said:
As WHO expected?

As any one who knows anything about astronomy expected.

Michael Mozina said:
It is a bunch of coronal loops, and that's what caused the flare and blew material off the photosphere.

The WL movie has one emerging coronal loop that emerges over a few frames and vanishes. If you have an image of it flaring then please post it. Remember to point out where it flared from the area under the loop (the photosphere) rather than all of the other observed flared from the top of coronal loops.

There are other coronal loops but they are only visible in the 171A and 1600A movies. Those movies are of activity in the corona. The flare is shown in those movies. Those loops are what caused the flare and blew material off the corona.

Michael Mozina said:
You and Zig seem to be under the false impression that I've had all of this all figured out for many years. That is simply untrue. I wasn't even aware of this particular image "years ago".

Then when did you find this image of standard solar physics?
Why have you not published a paper on this astounding (to you) discovery?

Reality Check · Aug 2, 2009

Michael Mozina said:
How can we "analyze" anything if you can't find the image Tim?

Obviously Tim found the image.
The fact is that no one in this forum including you is scientifically analyzing the image. In order to do that we would have to know about the instrument that took the image and take some advanced astrnomy courses.
We are all interpreting the image. It is fairly obvious to me that this is a standard coronal loop.

What is your interpretation?

Maybe it is an electrical arc from your hypothetical, thermodynamically impossible iron surface/crust 4800 km below the photosphere that is of unknown temperature, composition and depth.
In that case look at the image closely. The coronal loop enters the photosphere almost vertically. This means that your electrical arc is not an "arc" - somehow it is shaped more like a croquet hoop with vertical sides and a circle on top.
This looks nothing like Birkeland's images. These have actual arcs.
Congratulations MM - you have just disproved what you call "Birkeland's solar model"!

Michael Mozina said:
...snip...

Are you expecting the photosphere to be totally opaque?

The single coronal loop in the frame is heating plasma above, on and (AFAIK) below the photosphere. The light from the heated plasma is visible in the frame because the photosphere is not totally opaque.

Reality Check · Aug 2, 2009

Michael Mozina said:
If we run current through many ordinary materials like ordinary metals and neon like we find in a neon bulb, we get white light. The emissions of the neon bulb, in other words it's white light 'color' is not necessarily indicative of it's physical temperature.

This seems so obviously wrong that not even MM would think it correct. Maybe he has a specific kind of "neon bulb" in mind.

What is a "neon bulb"?

If it is a neon lamp then you are obviously wrong. Neon in an electrical current emits a reddish-orange color.
If it is a neon tube then you are obviously wrong. Neon tubes have various colors.
If it is a fluorescent lamp (you have in the past cited the lights in your office as proof that neon emits white light!) then you are obviously wrong.

A fluorescent lamp or fluorescent tube is a gas-discharge lamp that uses electricity to excite mercury vapor. The excited mercury atoms produce short-wave ultraviolet light that then causes a phosphor to fluoresce, producing visible light.

brantc · Aug 2, 2009

Ziggurat said:
No, it wouldn't. You can't magically transform heat into other forms of energy, because that would violate the 2nd law of thermodynamics by decreasing entropy. And heat will flow from the hot visible layer of the sun to anything underneath it. Unless you can get rid of this heat, it's going to warm up until it reaches at least 6000 K. You cannot get rid of heat by simply dumping energy in some low-entropy form, as you propose. It doesn't work.

It would require violating the 2nd law of thermodynamics. Which is not a "slight shift".

No but you can make a plasma with electricity. It doesnt go from heat to electricity back to heat. A deuterium spectrum is similar to blackbody with a solid arc.

It is the electricity leaving the surface that heats the surface causing thermionic discharge leading to the ejection of matter over small areas.

It starts as electricity and ends as heat. Or really kinetic energy(the field) added to mass. And since blackbody is the property of solid matter and solid matter does not exist at 6000K I think that extending the idea of blackbody to 6000K is in error. Unless it is taken into consideration that the spectrum is really an solid arc/plasma lamp spectrum. .

From Aetherometry;
"A photon is a swing (a particle, a conjunction of waves, and an energy packet) of electromagnetic energy. Its particulate aspect relates to its linear momentum (its existence as a particle) and the pressure it exerts upon adjacent matter. Its quantization relates to its constant of angular momentum, and its quantized energy forms two distinct spectra - blackbody and ionizing. Photons do not travel through space, nor do they have a fibrous structure. Photons are globular, not fascicular, and they are created and destroyed on the spot - ie local productions. Rays are simply a probabilistic way of approximating the physical reality of the phase or excitation wave that transmits 'across space' the indirect stimulus for the production of light. In the case of blackbody photons, a mediating term must always intervene between the phase wave and the production of photons, or light; the mediating term is always a massbound charge.

Basic differences between the conventional and aetherometric conceptions of the photon

1. On the nature of photons
1.1. Currently, it is held that solar radiation consists of photons. Implied in this is the notion that photons travel through space, like fibers of light, with analogy to ballistic models for the projection of material particles - as if the photons were hurled across space.

It is the view of aetherometric theory that solar radiation does not consist of photons, but of the massfree electrical charges that compose the Ôsolar electrical field [http://aetherometry.com[/B]/abs-AS2v2B.html#abstractAS2-17A]. Moreover, it is also the view of aetherometric theory that photons are 'punctual' and local productions, that they do not travel through space but rather occupy a globular space where they are created and extinguished.

1.2. If photons do not travel through space, what is it that travels through space and is the cause of the transmission of the light stimulus, and ultimately of any local production of photons?

Aetherometry contends that what travels through space and transmits the light impulse is electrical radiation composed of massfree charges and their associated longitudinal waves (the true phase waves), not electromagnetic radiation composed of photons and their transverse waves. The wave transmission of all electromagnetic signals depends on the transmission of nonelectromagnetic energy, specifically the transmission of electric massfree charges (the propagation of Òthe fieldÓ)."

Tubbythin · Aug 2, 2009

brantc said:
No but you can make a plasma with electricity. It doesnt go from heat to electricity back to heat. A deuterium spectrum is similar to blackbody with a solid arc.

It is the electricity leaving the surface that heats the surface causing thermionic discharge leading to the ejection of matter over small areas.

It starts as electricity and ends as heat. Or really kinetic energy(the field) added to mass. And since blackbody is the property of solid matter and solid matter does not exist at 6000K I think that extending the idea of blackbody to 6000K is in error. Unless it is taken into consideration that the spectrum is really an solid arc/plasma lamp spectrum. .

So which solid body does the CMBR arise from?

From Aetherometry;
"A photon is a swing (a particle, a conjunction of waves, and an energy packet) of electromagnetic energy. Its particulate aspect relates to its linear momentum (its existence as a particle) and the pressure it exerts upon adjacent matter. Its quantization relates to its constant of angular momentum, and its quantized energy forms two distinct spectra - blackbody and ionizing. Photons do not travel through space, nor do they have a fibrous structure. Photons are globular, not fascicular, and they are created and destroyed on the spot - ie local productions. Rays are simply a probabilistic way of approximating the physical reality of the phase or excitation wave that transmits 'across space' the indirect stimulus for the production of light. In the case of blackbody photons, a mediating term must always intervene between the phase wave and the production of photons, or light; the mediating term is always a massbound charge.

Basic differences between the conventional and aetherometric conceptions of the photon

1. On the nature of photons
1.1. Currently, it is held that solar radiation consists of photons. Implied in this is the notion that photons travel through space, like fibers of light, with analogy to ballistic models for the projection of material particles - as if the photons were hurled across space.

It is the view of aetherometric theory that solar radiation does not consist of photons, but of the massfree electrical charges that compose the Ôsolar electrical field [http://aetherometry.com[/B]/abs-AS2v2B.html#abstractAS2-17A]. Moreover, it is also the view of aetherometric theory that photons are 'punctual' and local productions, that they do not travel through space but rather occupy a globular space where they are created and extinguished.

1.2. If photons do not travel through space, what is it that travels through space and is the cause of the transmission of the light stimulus, and ultimately of any local production of photons?

Aetherometry contends that what travels through space and transmits the light impulse is electrical radiation composed of massfree charges and their associated longitudinal waves (the true phase waves), not electromagnetic radiation composed of photons and their transverse waves. The wave transmission of all electromagnetic signals depends on the transmission of nonelectromagnetic energy, specifically the transmission of electric massfree charges (the propagation of Òthe fieldÓ)."

Yes. Let's ignore the entirety of 20th and 21st century physics. That'll solve the problem.

Dancing David · Aug 2, 2009

Michael Mozina said:
For the record, it's not only electrons that carry heat away from the surface it's protons too.

The only way to scientifically answer Zig's question would be to construct a series of concentric mass separated plasma layers and introduce the notion of "thermoclines" inside the various plasma layers. The plasma double layers would have to be arranged with the hottest and lightest plasma layers on top of cooler more dense plasma layers as we approach the surface.

The total energy output of the sun would have to be equal to the current solar estimates for energy output, however in this solar model, the majority of the light from plasma layers comes from the "current flow" through the double layer, not necessarily because it radiates at a specific temperature. The only requirement is that that total energy output is the same as standard theory but any other variations from standard theory are acceptable in this model.

You'll also note that Zig's fixation on the temperature of the photosphere fails to correctly predict a 20,000 degree chromosphere

Shift the goal posts, Your model still can't beat the radiated heat.

, or a million degree corona, so if we are going to be "fair" to all models, neither the standard solar model or the Birkeland solar model currently passes his "test" as it relates to the flow of energy and heat distribution. For some reason he gives standard theory a free pass. What's that double standard all about?

Nope, you haven't answered the question, and your mischaracterization Of Zig's position is funny.

You have not explained how the mechanism works, you have painted a very vauge picture, but no mechanisms , laboratory data or tests to show how it works.

Not a single lab result to show that the mechanism could even work.

You hoisted yourself on your own standard there.

Not one bit of lab evidence for your whole complex. (And please don't mention Birkeland or Alven they did not model your idea of the solid iron surface, now the epicycles of layers, cathode refrigeration and electron towing.)

Lab data for your refrigeration effect in plasma please.
Lab data on electron towing in plasma please.

Ziggurat · Aug 2, 2009

brantc said:
No but you can make a plasma with electricity.

Yes. What's your point? That doesn't rescue Michael's absurd cathode refrigeration model from thermodynamic impossibility.

A deuterium spectrum is similar to blackbody with a solid arc.

Does this mean you've changed your mind about the ability of plasmas to radiate at wavelengths other than their atomic emission lines? Or are you under the delusion that modern arc lamps radiate mostly from their electrodes? Because the arc part of an arc lamp sure as hell isn't solid.

And since blackbody is the property of solid matter and solid matter does not exist at 6000K I think that extending the idea of blackbody to 6000K is in error.

No, you are completely wrong. There is absolutely no requirement that a blackbody be a solid. Where on earth did you get that nonsensical notion?

It is the view of aetherometric theory that solar radiation does not consist of photons, but of the massfree electrical charges that compose the Ôsolar electrical field [http://aetherometry.com[/B]/abs-AS2v2B.html#abstractAS2-17A].

Don't stick tags in the middle of an url, it breaks the link. I looked at that page, and it's chock-full of crazy. The ideas it proposed make absolutely no sense, and its conception of standard physics is ridiculously wide of the mark. Why did you link to it, brantc? Did you just stumble upon it in a google search for "blackbody radiation"?

Michael Mozina · Aug 2, 2009

Dancing David said:
Shift the goal posts, Your model still can't beat the radiated heat.

I am not "shifting the goal posts", I'm simply noting that you're in no position to complain.

You can't explain that 20,000K chromosphere or that million degree corona with your model of heat distribution. That is because it is much more complicated than your "black body" model. Even if I can't mathematically justify the position at this time, that does not mean I never will, or it can never be done. That is not a justification for ignoring all the satellite images and the information contained in these images. It's taken me several years to even figure out enough specific details to even think about trying out a mathematical model. I've had to learn whole branches of science that I had never seen before like MHD theory. Nobody is shifting the goal posts, but you're in no position to brag or attempt to take the high ground over some "unanswered questions".

Nope, you haven't answered the question, and your mischaracterization Of Zig's position is funny.

Zig's position is irrational. He won't even fully cop to the fact that "magnetic reconnection" and "particle reconnection" and "circuit reconnection" identical physical processes in plasma. He'll "grant me" that Alfven's circuit theories and Bruce's discharge theories are "accurate for the sake of argument", only so he can ignore that piece of data *ENTIRELY* and ultimately never accept it. There is no talking to him. He won't look at any satellite images. He won't accept sol's math because he didn't want to hear it in the first place, just as I figured. To sol's credit at least, he is "true to science" above personalities, whereas zig is ignoring the science he doesn't want to deal with only so he can come up with a math formula to stump Michael Mozina. There is no theory in science that rises and falls on the personal math skills of Michael Mozina. His whole attitude is utterly irrational.

You have not explained how the mechanism works,

I have explained the basic physical process, thermoclines, mass separated plasma layers, heat carried in the form of charged particles, etc. No, I'm not prepared to offer you an oversimplified mathematical model at this time, but 'so what"? There are *MANY* things that standard solar theory cannot "explain" mathematically or even at a basic understanding of the physical process.

you have painted a very vauge picture, but no mechanisms , laboratory data or tests to show how it works.

I've shown you Birkeland's lab work. He was able to generate "hot loops" above a relatively "cooler" surface. He managed to move pieces of the cathode toward the chamber walls, not just "electrons". He managed to created high energy "jets", that again were "hotter" than the sphere itself. No, I can't explain every single detail of this model at this time, but like I said, so what? That is no excuse to ignore the satellite images altogether like this whole gang of EU antagonists have done.

Not a single lab result to show that the mechanism could even work.

That is simply untrue. Birkeland's sphere didn't melt when turned on, at least not "instantly". You seem to *INSIST* that the heat distribution of a Birkeland solar model be *EXACTLY* the same at the surface of the photosphere. That is a completely random and irrational expectation. They simply aren't the same model. The only thing that has to "match" in the final analysis is the total energy output and that is going to be mostly coming from those flying electrons and flying ions Birkeland was describing in his lab over 100 years ago.

You hoisted yourself on your own standard there.

No, not actually. I'm even willing to concede that standard solar theory is not based upon mythical entities with the exception of "magnetic reconnection", which is simply a mislabeling error related to "particle reconnection" or "circuit reconnection". Even still, there is nothing particularly "ad hoc" in terms of energy sources, basic energy transfer mechanisms etc. It's not stuffed with "dark energy" or "dark matter" pe se. The only bit of "pseudoscience' it contains at the moment is the notion of "magnetic reconnection" and Alfven explained all that 40 years ago.

Not one bit of lab evidence for your whole complex. (And please don't mention Birkeland

Birkeland built a working model DD. He did use the same physical model I'm proposing and he would certainly have understood the implications of finding those discharges patterns in the solar atmosphere.

or Alven they did not model your idea of the solid iron surface,

Alfven and Bruce applied the discharge theory to a standard solar model. EU theory works with *EITHER* solar model and it is dependent upon the accuracy of neither solar model.

now the epicycles of layers,

That is simply common sense and it is well supported in the lab. Gravitational fields and EM fields are *KNOWN* to create separation of elements. In fact many instruments we use rely upon these principles, particularly with respect to the EM fields influence on elements.

cathode refrigeration and electron towing.)

How did Birkeland explain parts of his cathode ending up on the sides of his chamber David? Please do not ask me about electron towing again until you answer that question.

Lab data for your refrigeration effect in plasma please.

Birkeland created many experiments where the temperature inside the loop was much higher than the sphere, and the sphere did not melt. You'll have to accept history here. No, I can't "scale the whole thing to size", at least not yet, but again, "so what"? That is still no excuse to ignore all of Bruce's work. It's still no excuse to ignore all of Alfven's work. It's certainly no excuse to ignore Birkeland's *empirical* lab work.

Michael Mozina · Aug 2, 2009

FYI....

It really is crunch time at work. I will be taking some time away from these boards for the next few weeks to wrap up some programming projects that I need to be ready for September. I may dabble here and there, but I will definitely be less active for awhile. Don't take it personally, and know that I will return when I'm done.

Ziggurat · Aug 2, 2009

Michael Mozina said:
I am not "shifting the goal posts", I'm simply noting that you're in no position to complain. You can't explain that 20,000K chromosphere or that million degree corona with your model of heat distribution. That is because it is much more complicated than your "black body" model.

You really don't get it. Nobody is claiming that blackbody radiation is all there is to it. Of course not. But simple thermodynamic considerations (including blackbody radiation) still constrain any possible model of how the sun operates. If a model cannot fit within those constraints, then it's clearly wrong. And your model cannot. Nothing about that says that blackbody radiation is the only thing going on.

Even if I can't mathematically justify the position at this time

It's more than that, Michael. Your position has been mathematically refuted.

That is not a justification for ignoring all the satellite images and the information contained in these images.

It's enough justification to discard your explanation as impossible.

It's taken me several years to even figure out enough specific details to even think about trying out a mathematical model.

That's sad. It took me all of 10 minutes.

Zig's position is irrational. He won't even fully cop to the fact that "magnetic reconnection" and "particle reconnection" and "circuit reconnection" identical physical processes in plasma.

I have zero interest in semantic debates, Michael.

He'll "grant me" that Alfven's circuit theories and Bruce's discharge theories are "accurate for the sake of argument", only so he can ignore that piece of data *ENTIRELY* and ultimately never accept it.

Nothing Alfven or Bruce wrote deals with your proposed cathode refrigeration, Michael.

There is no talking to him. He won't look at any satellite images.

Translation: I demand numbers, and am not satisfied with your impression of pretty pictures.

zig is ignoring the science he doesn't want to deal with only so he can come up with a math formula to stump Michael Mozina.

Because addition, subtraction, multiplication, and division are really so difficult.

That's all I needed to show how ridiculous your cathode refrigeration model is. If you're stumped by such simple math, that's not my fault.

There is no theory in science that rises and falls on the personal math skills of Michael Mozina.

Yes, I suppose that's true: your theory will never rise at all, because you can't (or won't) do the math to make even the most basic predictions with it.

I have explained the basic physical process, thermoclines, mass separated plasma layers, heat carried in the form of charged particles, etc.

And attached no numbers to anything, so that you cannot demonstrate that any of those things is sufficient to accomplish what you think it can accomplish. How can you evaluate a theory, Michael, when you can't even tell what it would actually predict? How can you tell what it would actually predict if you can't quantify that prediction? You can't.

I'm even willing to concede that standard solar theory is not based upon mythical entities with the exception of "magnetic reconnection", which is simply a mislabeling error related to "particle reconnection" or "circuit reconnection".

Wow. You've contradicted yourself in a single sentence. First magnetic reconnection is based upon mythical entities, but then it's based upon real entities which are just mislabeled. Your incoherence is as amusing as always.

Michael Mozina · Aug 2, 2009

Ziggurat said:
You really don't get it. Nobody is claiming that blackbody radiation is all there is to it. Of course not.

Of course you are. You are *INSISTING* that I "assume" that the whole photosphere is 6000K or more, and that all but the tiniest of energy radiates from anything *except* the surface of the photosphere. That's a bunch of baloney, right from the start. Birkeland's solar model radiated from the plasma around the sphere. It radiated heat from the surface of the sphere. It released heat in the form of moving charged particles. It didn't just radiate heat from a single plasma layer of the sphere.

But simple thermodynamic considerations

You mean *OVERSIMPLIED* considerations....

(including blackbody radiation) still constrain

Black body radiation constrains absolutely nothing. The only "requirements" in the final analysis is that the sun radiate a lot of different wavelengths in "roughly" (and I do mean roughly) 'black body" manner. One cursory glance at the SERTS data however and it's clear that not every photon originates at the surface of the photosphere at 6000K .

any possible model of how the sun operates.

Nope! Just *YOUR* "possible model". Not every solar model has to produce the heat exactly like your model in exactly the same way as your model. Your "black body", 6000K "surface radiation' model doesn't explain those iron ion or nickel ion wavelengths worth a damn.

If a model cannot fit within those constraints, then it's clearly wrong.

There you go, *INSISTING* that all solar models meet your personal expectations. Sorry but nature isn't required to operate by your personal standards.

And your model cannot. Nothing about that says that blackbody radiation is the only thing going on.

Then stop trying to *OVERSIMPLIFY* the process to a single surface as you keep trying to do.

It's more than that, Michael. Your position has been mathematically refuted.

You don't even seem to know what my position is in the first place. What temperature do I assume the silicon plasma layer radiates at near the bottom of that plasma layer?

You've simply whipped together an oversimplified mathematical model based upon *YOUR PERSONAL* expectations and beliefs about how our sun releases the energy that it releases on a daily basis.

It's enough justification to discard your explanation as impossible.

Um, even if we exclude my solar model, you will necessarily need to embrace EU theory of some sort to be congruent with Bruce and Alfven. So are you now an EU proponent based on a plasma solar model?

That's sad. It took me all of 10 minutes.

Ya, and I guess you figure 10 minutes worth of oversimplified math somehow means you can simply "ignore" those satellite images, to the point that you wouldn't even personally bother downloading them at all.

I have zero interest in semantic debates, Michael.

So you agree that they are all equally valid terms? Particle, circuit and magnetic reconnection are all the same physical process, correct?

Nothing Alfven or Bruce wrote deals with your proposed cathode refrigeration, Michael.

Pretty much everything Birkeland experimented with in his lab did deal with it. Did you even personally bother reading his work, or are you personally arguing from a place of ignorance on that issue?

Translation: I demand numbers, and am not satisfied with your impression of pretty pictures.

Likewise I demand physical explanations, and physical lab demonstrations and I am not satisfied with your 10 minute math oversimplification exercise.

Because addition, subtraction, multiplication, and division are really so difficult.

Er, no, it's the fact you expect me to use your personal expectations that are so difficult.

That's all I needed to show how ridiculous your cathode refrigeration model is. If you're stumped by such simple math, that's not my fault.

I'm not stumped by simple math, I'm stumped by complex processes of nature that can't be easily quantified in oversimplified ways.

Yes, I suppose that's true: your theory will never rise at all, because you can't (or won't) do the math to make even the most basic predictions with it.

Not true. I'm going to start (in fact have started) predicting sunspots and CME's and such.

And attached no numbers to anything,

That's better than not attaching "physics" to anything. You seem to believe that a theory based in part on Maxwell's equations and in part on fluid dynamics cannot receive "inputs" at the level of physics. That is simply bizarre. Even an ocean might receive mechanical input from a fresh water stream and we may want to calculate the flow of fresh water at a certain distance from the input stream. In plasma, you can also have whole plasma filaments running through a plasma layer and nothing is particularly "homogeneous" inside a current carrying plasma double layer.

How can you evaluate a theory, Michael, when you can't even tell what it would actually predict?

How does you theory "predict" something like "solar wind", or those million degree coronal loops?

How can you tell what it would actually predict if you can't quantify that prediction? You can't.

So what? You can't even explain a sustained solar wind from the whole surface of the sphere, or something like a CME event. You're incapable of speaking actual "English" to an electrical engineer (magnetic reconnection?) or to someone involved in QM or particle physics. You can't even accept that there is an actual bridge between your personal "lingo" and their lingo (short circuit/particle reconnection).

Wow. You've contradicted yourself in a single sentence. First magnetic reconnection is based upon mythical entities,

It's based upon a mythical premise that magnetic lines "disconnect' or "reconnect". Only particles and circuits can do that, not magnetic lines. The magnetic field forms without beginning and without end. Induction is the "proper" scientific name for the transfer of kinetic energy to charged particles due to changes in magnetic fields. It would have been *infinitely* better to stick with Alfven's original terminology "circuits" and accept the fact the the term "magnetic reconnection" is in fact "pseudoscience" because it only "pseudo" conveys the physics. You can't even understand that there is no physical difference between what you are calling "magnetic reconnection" and what Aflven called a "short-circuit", and yes he personally did use that term.

but then it's based upon real entities which are just mislabeled. Your incoherence is as amusing as always.

The fact you can't even make the translation between your lingo and Alfven's lingo or Bruce's lingo demonstrates that there is more than an issue of "semantics'' at stake. You're hopelessly confused. Whereas I can make the translation to your personal "lingo", you cannot make the same translation to Alfven's "lingo" related to "circuits" and therefore you can't even begin to understand a single one of those countless circuit diagrams he drew.

Michael Mozina · Aug 2, 2009

Reality Check said:
[/URL]
If it is a fluorescent lamp (you have in the past cited the lights in your office as proof that neon emits white light!) then you are obviously wrong.

The only one that is even 'close' to the properties of the photosphere is the flourescent lamp, but the entire photon "spectrum" is composed of all the emissions from all the layers and all those coronal loops, etc. It's not a "perfect black body" spectrum, but when you add everything together, it's "in the ballpark".

Michael Mozina · Aug 2, 2009

Reality Check said:
As any one who knows anything about astronomy expected.

Really? I want be clear about what everyone thinks here, because it's not clear you are all of one mind on this topic. So you expected to see the loops come up through the photosphere and light up the photosphere in the areas around the loops? Are they evenly heated, if not where are they heated and how are they heated at that location?

The WL movie has one emerging coronal loop that emerges over a few frames and vanishes.

Actually, it's way more than a "single" coronal loop that emerges, and it only "vanishes" in white light. Other wavelengths show more of the physical processes that ultimately generated that one image.

If you have an image of it flaring then please post it. Remember to point out where it flared from the area under the loop (the photosphere) rather than all of the other observed flared from the top of coronal loops.

My DVD player won't let me copy and paste individual images from the DVD. Since you can do that with you player, please post the two images before that image and the image after it and I'll be happy to comment on all of them. There is ample evidence of what's going on in those four images.

There are other coronal loops but they are only visible in the 171A and 1600A movies.

Ok. There are lots of "loops" that are only visible in 171A. The loops traverse the surface of the photosphere. If the loops are heated uniformly, we should see them to *SOME* (if only 10 KM) distance below the surface of the photosphere. If not, you'll have to explain the heating mechanism that sustains them at high temperatures for hours on end and how you're sure that heat isn't coming from below the surface of the photosphere?

Those movies are of activity in the corona.

No. They are movies that show activity in the "transition region" (the surface), and they show us high into the solar atmosphere, including the chromosphere and corona. What by the way, what is that transitional region made of that allows those drifting ions to block 171A light at the limb of 171A images?

The flare is shown in those movies. Those loops are what caused the flare and blew material off the corona.

It blew material off of the photosphere too as I will demonstrate if you post the image before this one.

Then when did you find this image of standard solar physics?

Before you, before Tim, before tusenfem and before zig.

Why have you not published a paper on this astounding (to you) discovery?

I've been involved in 5 published papers already. Will a sixth *less than astounding* revelation that these are caused by electrical discharges in plasma somehow change your mind overnight? Did you read any of Bruce's materials?

Ziggurat · Aug 2, 2009

Michael Mozina said:
Of course you are. You are *INSISTING* that I "assume" that the whole photosphere is 6000K or more, and that all but the tiniest of energy radiates from anything *except* the surface of the photosphere.

No. I've stated this quite explicitly: it doesn't MATTER if the light we see is coming from the photosphere. But wherever it's coming from, EVEN IF it's coming from multiple layers, the source in total forms a roughly 6000 K blackbody.

You mean *OVERSIMPLIED* considerations....

No, Michael. Thermodynamics create hard restraints which every macroscopic system must obey. It's like energy conservation: the fact that it's simple doesn't means it's oversimplified.

Black body radiation constrains absolutely nothing.

Oh, but it most certainly does. If something emits as a blackbody, it must be opaque. That's a very hard restraint right there. If it's not a perfect blackbody, but only close, then we still know (again, this is a hard restraint) that it MUST be mostly opaque.

Not every solar model has to produce the heat exactly like your model in exactly the same way as your model. Your "black body", 6000K "surface radiation' model doesn't explain those iron ion or nickel ion wavelengths worth a damn.

Once again, you reveal your cluelessness. I'm not claiming that my blackbody analysis explains everything. I'm saying that things must operate within the constraints of those thermodynamic considerations. And the iron and nickel emissions you reference fall into that category without any problem.

There you go, *INSISTING* that all solar models meet your personal expectations.

It's not my personal expectations, Michael. It's basic thermodynamics. And yes, I do insist that all solar models be consistent with thermodynamics. You apparently do not.

Then stop trying to *OVERSIMPLIFY* the process to a single surface as you keep trying to do.

Call it a collection of surfaces if you want. Call it a layer. I don't care. But whatever it is, it's at approximately 6000 K, and it's opaque. Them's the undeniable, incontrovertible facts.

You don't even seem to know what my position is in the first place.

I know that you think that there's a solid surface underneath whatever it is that's at 6000 K, and you think that it's cooled by some form of cathode refrigeration.

Um, even if we exclude my solar model, you will necessarily need to embrace EU theory of some sort to be congruent with Bruce and Alfven. So are you now an EU proponent based on a plasma solar model?

That's a different subject, one which we can get to once we settle the issue of the thermodynamic impossibility of your solid surface model.

Ya, and I guess you figure 10 minutes worth of oversimplified math somehow means you can simply "ignore" those satellite images, to the point that you wouldn't even personally bother downloading them at all.

Well, yes. Just like if you told me that you had a perpetual motion machine, one second of consideration of the 1st and 2nd laws of thermodynamics would mean I could simply ignore pages of blueprints and a youtube video, too.

So you agree that they are all equally valid terms? Particle, circuit and magnetic reconnection are all the same physical process, correct?

Since "particle reconnection" is not a standard phrase in solar physics but "magnetic reconnection" is, the only thing I can conclude is that you mean the same thing by that phrase.

Oh, and I can also conclude that you have no objections other than semantic ones, because those are the only objections you ever raise.

Pretty much everything Birkeland experimented with in his lab did deal with it.

No, it did not.

I'm not stumped by simple math, I'm stumped by complex processes of nature that can't be easily quantified in oversimplified ways.

Oh, but the fundamental parameters of a model very frequently CAN be. For example, it takes very little math to figure out the approximate rate at which the sun is burning hydrogen through fusion. One could then calculate an approximate lifetime for the sun, by considering how long such a burn rate is sustainable. It would be similarly easy to calculate an approximate fission rate if you think the sun is powered by uranium. And so on. Simple calculations are often limited in showing the accuracy of a model, but they're damned useful in showing when a model is wildly inaccurate.

Not true. I'm going to start (in fact have started) predicting sunspots and CME's and such.

Talk is cheap. I'll believe it when I see it. And if you ever do any calculations, we can check if your numbers make any sense. Just like you COULD check the numbers I've already given to see if they make any sense, but you have not done so. You dismiss them without any indication of which numbers are wrong, why they are wrong, and what better numbers would be.

You seem to believe that a theory based in part on Maxwell's equations and in part on fluid dynamics cannot receive "inputs" at the level of physics.

No, Michael. Rather, I recognize that MHD doesn't include individual particle behavior. If you include such behavior, you're not using MHD anymore, you're using something else. It may resemble MHD, but it would not be MHD.

In plasma, you can also have whole plasma filaments running through a plasma layer and nothing is particularly "homogeneous" inside a current carrying plasma double layer.

Apparently you can't figure out the difference between homogeneous and continuum. A continuum can be inhomogeneous. You also seem unaware that double layers are an example of something that MHD doesn't handle correctly, precisely because it depends upon kinetic effects which are ignored by MHD. Oh, the irony.

So what? You can't even explain a sustained solar wind from the whole surface of the sphere, or something like a CME event.

I'm not trying to. I'm trying to check YOUR model for consistency with thermodynamics. Your model fails the consistency check. Present me with a model for the solar wind and I can check that model for similar consistency with thermodynamics.

It's based upon a mythical premise that magnetic lines "disconnect' or "reconnect". Only particles and circuits can do that, not magnetic lines.

It has been explained to you repeatedly what "reconnection" means in regards to a magnetic field. And what it means is most certainly possible. Whether or not you like that use of the term is irrelevant, and is nothing more than a semantic objection. You have not been able to challenge the actual physics involved.

And you have yet to quantify anything you say.

Reality Check · Aug 2, 2009

entire photon "spectrum" is composed of all the emissions from all the layers

Michael Mozina said:
The only one that is even 'close' to the properties of the photosphere is the flourescent lamp, but the entire photon "spectrum" is composed of all the emissions from all the layers and all those coronal loops, etc. It's not a "perfect black body" spectrum, but when you add everything together, it's "in the ballpark".

So now you have changed your mind - it is not a "mostly neon" photosphere emitting the nearly (not "perfect black body") black body spectrum that scientists measure. All your "neon tube" posts were mistakes or lies.

Now we have yet another unsupported assertion and that deserves a question:
First asked 3 August 2009

but the entire photon "spectrum" is composed of all the emissions from all the layers and all those coronal loops, etc. It's not a "perfect black body" spectrum, but when you add everything together, it's "in the ballpark".

Please present your calculation that all of your layers

Can produce radiation that escapes the Sun.
The radiation has the observed nearly black body spectrum.
That nearly black body spectrum has a temperature of ~6000 K (this means that the iron layer is at ~6000 K) and vapourizes).

Also:
First asked 3 August 2009
Your layers of "mostly X" will produce emission lines in the spectrum for element X. This means that emission lines in the Sun's spectrum will show the relative proportions of the elements.

Please give the citation that demonstrates that all of the elements in the Sun (Fe and below) are in roughly equal proportions or even more than hydrogen and helium.

Reality Check · Aug 2, 2009

Michael Mozina said:
Actually, it's way more than a "single" coronal loop that emerges, and it only "vanishes" in white light. Other wavelengths show more of the physical processes that ultimately generated that one image.

Actually it is one coronal loop at the photosphere in visible light movie.
The other movies are of coroanl loops in the coronal that are only related to the one coronal loop in white light be being part of the same activity.

Michael Mozina said:
My DVD player won't let me copy and paste individual images from the DVD. Since you can do that with you player, please post the two images before that image and the image after it and I'll be happy to comment on all of them. There is ample evidence of what's going on in those four images.

Print Screen key - press it.

Michael Mozina said:
Ok. There are lots of "loops" that are only visible in 171A. The loops traverse the surface of the photosphere. If the loops are heated uniformly, we should see them to *SOME* (if only 10 KM) distance below the surface of the photosphere. If not, you'll have to explain the heating mechanism that sustains them at high temperatures for hours on end and how you're sure that heat isn't coming from below the surface of the photosphere?

One more time for the stupid: 171A images are of activity in teh coroan.

Michael Mozina said:
No. They are movies that show activity in the "transition region" (the surface), and they show us high into the solar atmosphere, including the chromosphere and corona. What by the way, what is that transitional region made of that allows those drifting ions to block 171A light at the limb of 171A images?

No. They are movies that show activity in the transition region (1000's of km above the photosphere) and corona.
Theay are not movies that show activity at the photosphere.

Michael Mozina said:
It blew material off of the photosphere too as I will demonstrate if you post the image before this one.

You are the one with the claim - you post it.

Michael Mozina said:
I've been involved in 5 published papers already. Will a sixth *less than astounding* revelation that these are caused by electrical discharges in plasma somehow change your mind overnight? Did you read any of Bruce's materials?

You have been involved in 1 or 2 published papers (I forget which) and some preprints. If I am wrong then give citations to the published papers.

And yes - a published paper in a peer reviewed journal from you showing that solar flares are trhe result of electrical discharges in plasma (or from your iron crust 4800 km below the photosphere) would change my mind overnight (in fact instantly).
Of course according to plasma physics this is impossible - you need charge separation to have discharges and this does not happen for the required scale (100's of km) in the photosphere. So your first task is to rewrite plasma physics, e.g. start by disproving all of Hannes Alfvén's work.

I have read Bruce and he is fundamentally wrong. He states that there are dust particles in the photosphere to cause the electrical disharges. Even you (seem to) agree that the photosphere is at a temperature of ~6000 K.

ETA:
You did not answer my question: Then when did you find this image of standard solar physics?
Give a date. Even a year would be Ok.

GeeMack · Aug 2, 2009

Michael Mozina said:
Likewise I demand physical explanations, and physical lab demonstrations and I am not satisfied with your 10 minute math oversimplification exercise.

Yes, you demand physical explanations and lab demonstrations from others, but you are wholly lacking when it comes to providing them to support your own insane ideas. You've been asked dozens of times for the physical demonstration that shows how you can see 4000+ kilometers below the opaque photosphere by looking at a graph that was created using data obtained from several thousand kilometers above the photosphere. And to this day, after more than three years and dozens of requests to cough up the goods, you've offered no response to that concern. Well, except lying, bitching, moaning, evasion, crying, ignorance, and foot stomping, but you'll have to agree that those don't amount to a legitimate response.

When you're put to the task of living up to your own demands, you fail, 100%, every single time, Michael. Even you admit that you've pussied out when it comes to real physics with quantitative explanations and objective analyses. And you agree that your fantasy is based on nothing more than looks-like-a-bunny misunderstandings and true-because-I-say-so tantrums. In other words, it's all in your head.

So, since you can't meet your own standard of evidence, when are you going to remove that LMSAL running difference graph from your web site?

Reality Check · Aug 2, 2009

Michael Mozina said:
Really? I want be clear about what everyone thinks here, because it's not clear you are all of one mind on this topic. So you expected to see the loops come up through the photosphere and light up the photosphere in the areas around the loops? Are they evenly heated, if not where are they heated and how are they heated at that location?

Really? I will be clear about my opinion.
Coronal loops emerge from the photosphere (standard solar physics). This is what the white light image shows - a single coronal loop that has emerged from the photosphere. The magnetic field forming the loop will interact with magnetic fields of the sunspot and photosphere. Thus there may be visible effects on the surface of the photosphere, i.e. footprints on the sunspot.
This is what any astronomer would expect.

My opinion is that there could also be some heating of plasma below the photosphere and that this heated plasma could emit light that can be imaged.

BTW: You may want to look at Exploring Magnetism in Solar Flares: An Educator's Activity Guide for Grades 8-12.

Michael Mozina · Aug 3, 2009

Reality Check said:
Actually it is one coronal loop at the photosphere in visible light movie.

I'm curious how many of your fellow EU antagonists share that viewpoint?

GeeMack · Aug 4, 2009

Michael Mozina said:
I'm curious how many of your fellow EU antagonists share that viewpoint?

I'm curious how many of your fellow science and math antagonists share the viewpoint that you can see stuff over 4000 kilometers below the Sun's photosphere by looking at a graphical representation of a series of mathematical computations that was created using data acquired several thousand kilometers above the photosphere. My bet would be that, even though being in your league they'd misunderstand science and be scared of math, there aren't any!

When do you figure to take down that LMSAL running difference graph from your web site, Michael? We've already determined, those of us who are sane, intelligent, and not delusional have anyway, that it doesn't show the nonsense that you once claimed it showed.

You're a riot, kid. I'm still waiting for the day you admit you're just a sick troll, and you've been keeping up this ridiculous stupid scientist-wannabe act for so many years just to have a good joke on everyone.

Michael Mozina · Aug 4, 2009

GeeMack said:
When do you figure to take down that LMSAL running difference graph from your web site, Michael?

I don't have any running difference "graphs" on my website, only running difference "images". In RD "images" we can observe real things like real stars, and real comets, and real planets. When are you going to figure out that it's not a graph and you can see real objects in these images? Flying stuff? Ya, flying stuff too!

And for the record, was that one loop or many?

sol invictus · Aug 4, 2009

Ziggurat said:
Only thermal radiation, unless you've got some other light source.

Thermal, in a vacuum? At what temperature?

No, you'd only get thermal radiation if you did something like accelerate your detector at a constant rate. If it's inertial you'll just get quantum fluctuations with a spectrum determined by the characteristics of your detector and interactions of whatever you're detecting.

I don't see how it could. What kind of photodetector would you use?

Well, let's see. A static background B field conserves energy but not momentum (of charges passing through it), classically. So you'd be best off with something that measures the lateral acceleration of a charge, like (say) the scattering angle as an electron passes through the field. In a very weak field, you'd see that the electrons (all else being equal) change direction in discrete jumps as they pass through the field - or more precisely, the Fourier transform of a histogram of the angles they scatter through would be related to the FT of the field, but with the effects of discrete photon numbers visible (because the particles absorb integral numbers of photons from the field). For a strong field that discreteness is of course very hard to see, but it must be there nonetheless.

Michael Mozina · Aug 4, 2009

brantc said:
It starts as electricity and ends as heat. Or really kinetic energy(the field) added to mass. And since blackbody is the property of solid matter and solid matter does not exist at 6000K I think that extending the idea of blackbody to 6000K is in error. Unless it is taken into consideration that the spectrum is really an solid arc/plasma lamp spectrum.

You hit the nail on the head with that comment IMO. The smaller discharges (under the photosphere) tend to serve the role of the solid plasma lamp as these discharge filaments ionize heavier elements in loops all along the surface. Some of those ionized (and dusty) heavier elements make their way into the atmosphere and photosphere and that is what creates a more "white light' spectrum in the mostly neon photosphere.

The overall energy release of a Birkeland model would need to jive with standard theory, but in almost every other respect, it would work "very differently" than standard theory. There will be a "flow pattern" of particles moving away from the surface, and through the various plasma layers in the atmosphere. I would expect mass separation and thermoclines to form between and in plasma double layers, but the bulk of the heat is carried away, and contained in the flowing electrons and ions, not simply in the surface features.

I would also add here that nobody is claiming that the surface is "solid iron", it simply has a "crust". The only temperature that is 'critical' is the melting point of carbon which is closer to 4000K than 6000K. Anything below 4000K and solids could and would begin to form. The other debate over "optical depth" is really moot IMO since were simply talking about "how far" below the photosphere the surface might be located and that video of the flare demonstrates that it *MUST* originate far below the photosphere.

Ziggurat · Aug 4, 2009

sol invictus said:
Well, let's see. A static background B field conserves energy but not momentum (of charges passing through it), classically. So you'd be best off with something that measures the lateral acceleration of a charge, like (say) the scattering angle as an electron passes through the field. In a very weak field, you'd see that the electrons (all else being equal) change direction in discrete jumps as they pass through the field - or more precisely, the Fourier transform of a histogram of the angles they scatter through would be related to the FT of the field, but with the effects of discrete photon numbers visible (because the particles absorb integral numbers of photons from the field). For a strong field that discreteness is of course very hard to see, but it must be there nonetheless.

I don't think so. An electron moving in a magnetic field will emit synchrotron radiation, whose frequency (and hence momentum) is characterized by the field strength and not by the field size (which characterizes the photons forming the field). This is what changes the momentum of the electron, not absorption of photons in the field, whose frequency (and momentum) are continuous anyways in the case of an infinite volume. And even in a finite volume, the emitted radiation doesn't need to form standing waves, so the momentum transfer need not be discrete either.

Ziggurat · Aug 4, 2009

Michael Mozina said:
You hit the nail on the head with that comment IMO.

He summarized his misconceptions quite nicely, I'll give him that much. But he's still under the delusion (and so, apparently, are you) that only solids can act as blackbodies. This has already been falsified, Michael.

I would also add here that nobody is claiming that the surface is "solid iron", it simply has a "crust". The only temperature that is 'critical' is the melting point of carbon which is closer to 4000K than 6000K. Anything below 4000K and solids could and would begin to form.

And yet, 4000 K is still thermodynamically impossible. Your cathode refrigeration model doesn't work. Which is why you can't put any numbers on it.

sol invictus · Aug 4, 2009

Ziggurat said:
I don't think so. An electron moving in a magnetic field will emit synchrotron radiation, whose frequency (and hence momentum) is characterized by the field strength and not by the field size (which characterizes the photons forming the field).

But I'm not talking about the radiation the charge emits - I'm talking about the radiation it absorbs from the field.

This is what changes the momentum of the electron, not absorption of photons in the field, whose frequency (and momentum) are continuous anyways in the case of an infinite volume.

It's synchrotron radiation which causes the acceleration of the electron, which caused the synchrotron radiation? No - it's the absorption of photons from the B field which causes the acceleration, which in turn can lead to synchrotron radiation. And the effect of the field is entirely accounted for by individual processes in which a field photon interacts with the electron - and the spectrum of those photons is determined by the field profile (and their quantity by its intensity).

One way to understand this is by conservation of momentum. A static B field with some spatial profile violates conservation of momentum (ignoring the backreaction of the particle on the field), because it violates translation invariance. How much momentum can it transfer because of that? One way to find out is to take its FT (come to think of it the FT of the vector potential is better, because it's more simply related to the momentum transfer).

And even in a finite volume, the emitted radiation doesn't need to form standing waves, so the momentum transfer need not be discrete either.

You seem to be mixing two things here - the momentum transfer due to radiation with the momentum transfer due to absorption. They're not the same. And even the absorbed radiation doesn't have a discrete FT - it's not a plane wave, it's a localized lump.

Ziggurat · Aug 4, 2009

sol invictus said:
You seem to be mixing two things here - the momentum transfer due to radiation with the momentum transfer due to absorption. They're not the same. And even the absorbed radiation doesn't have a discrete FT - it's not a plane wave, it's a localized lump.

If it's not a plane wave, then it doesn't have a well quantized momentum. Which means each such photon doesn't transfer discrete momentum either. So you can't measure the number of photons it absorbs by counting discrete changes in the electron's momentum, because they won't be there.

sol invictus · Aug 4, 2009

Ziggurat said:
If it's not a plane wave, then it doesn't have a well quantized momentum.

Each photon carries a certain momentum determined by its frequency. The probability of absorbing a photon of a particular momentum is related to the FT of the classical field, which is a smooth function. By doing the experiment many times, you'll be able to reconstruct that function, and see the effects of the quantization (for example, the FT will fall off like a power high momentum, but quantization will mean all occupation numbers will be zero or exponentially small).

Which means each such photon doesn't transfer discrete momentum either.

Oh, but it does. It's the probability p(k) of absorbing a photon with momentum k that's continuous, not the k of any given photon.

So you can't measure the number of photons it absorbs by counting discrete changes in the electron's momentum, because they won't be there.

The change for any given electron is obviously discrete. A histogram of the changes for many electrons will be continuous in the limit of an infinite number of repetitions - is that what you mean? - but that's just as expected.

But I'm not sure what we're disagreeing over. Surely you agree that QED (plus the rest of the standard model) is the correct theory for electromagnetic interactions. Surely you agree that all QED processes arise from the cumulative effects of the only interaction that exists in the theory, which is the absorption or emission of a single photon by a charged particle.

But then the effects of a magnetic field on a particle can be understood that way - by the absorption of the photons making up that field. And the spectrum of those photons must be closely related to the FT of the field (that would be easy enough to check - just write down the Feynman rules in momentum space, and think about how the background field enters).

Ziggurat · Aug 4, 2009

sol invictus said:
Oh, but it does. It's the probability p(k) of absorbing a photon with momentum k that's continuous, not the k of any given photon.

But if you have a continuous distribution of momenta that can be absorbed, then the momentum change in the electron is continuous. There's no way to tell if you've absorbed four photons of one frequency, five of another, or some mix. All you will be probing is the magnetic field strength. Which is not enough to determine how many photons are in the field.

But I'm not sure what we're disagreeing over.

I'm saying that the only way to determine how many photons make up the field is to measure the field strength AND size, and then take the Fourier transform, because there's no way to directly measure those photons.

But then the effects of a magnetic field on a particle can be understood that way - by the absorption of the photons making up that field. And the spectrum of those photons must be closely related to the FT of the field (that would be easy enough to check - just write down the Feynman rules in momentum space, and think about how the background field enters).

The FT is a non-local property of the field. But the measurement you propose is essentially a local measurement. So how can it be sensitive to the differences? I don't think it can. I don't think it matters that one can do the calculations in terms of individual photons: I don't think you can extract the number of photons in the field from the measurement, because multiple scenarios will produce the same measured result.

Michael Mozina · Aug 4, 2009

sol invictus said:
But I'm not sure what we're disagreeing over. Surely you agree that QED (plus the rest of the standard model) is the correct theory for electromagnetic interactions. Surely you agree that all QED processes arise from the cumulative effects of the only interaction that exists in the theory, which is the absorption or emission of a single photon by a charged particle.

The problem is that I brought up the idea, not you. If you had said that photons were the carrier particle of the EM field, this conversation would probably be over already, and probably never questioned in the first place. I for one however have benefited from the discussion, so by all means, do continue.

Ziggurat · Aug 4, 2009

Michael Mozina said:
The problem is that I brought up the idea, not you. If you had said that photons were the carrier particle of the EM field, this conversation would probably be over already, and probably never questioned in the first place.

Don't flatter yourself. And the exchange we're having isn't about whether or not photons are the carrier particle for EM fields.

sol invictus · Aug 4, 2009

Ziggurat said:
But if you have a continuous distribution of momenta that can be absorbed, then the momentum change in the electron is continuous.

You mean, the possible momentum changes? Then yes.

There's no way to tell if you've absorbed four photons of one frequency, five of another, or some mix.

I don't think that's the case. For one thing, you could use something more sophisticated than an electron - something with some structure, that's more sensitive to some frequencies than others. For another you can vary the field profile and strength, and do many experiments. And in reality you have to allow some input from theory to interpret the results, and the theory predicts a certain spectrum of momentum changes due to photon absorption, and you can go out and check the predictions of the theory.

All you will be probing is the magnetic field strength. Which is not enough to determine how many photons are in the field.

Not just the strength, because even classically the profile of the field (say the volume where it's non-zero) affects the momentum transferred. But yes, the strength is part of it as well.

I'm saying that the only way to determine how many photons make up the field is to measure the field strength AND size, and then take the Fourier transform, because there's no way to directly measure those photons.

So you agree (?) I have a theory in which I can assign a finite number of photons to the field configuration, and you agree that every single prediction of that theory - calculated using discrete interactions with finite numbers of photons - is correct, but you don't agree there's a way to measure that number or be sure it's accurate. If that's your position, I can't argue with it (nor am I very interested in trying).

The FT is a non-local property of the field. But the measurement you propose is essentially a local measurement.

Why? I can shoot electrons through from any angle, through the whole thing, through part of it, etc. How is that local?

I don't think you can extract the number of photons in the field from the measurement, because multiple scenarios will produce the same measured result.

Well, there is a significant subtlety here regarding infrared divergences. Any time you have massless particles like photons, the number of "soft" photons - photons with very low momentum and energy - gets hard to define. But that's a problem that's solved the minute you consider a detector with finite energy resolution, it goes away if the universe were huge but finite (which tells you right away it can't matter), and it's there even when the field is precisely zero. So I don't think it prevents one from answering this question, although I'll agree my argument is somewhat vulnerable on that point.

As an example more topical to this thread - if I asked you how many photons there are in the sun, or how many it emits per second, would you agree that's a sensible question with a finite answer?

Ziggurat · Aug 4, 2009

sol invictus said:
I don't think that's the case. For one thing, you could use something more sophisticated than an electron - something with some structure, that's more sensitive to some frequencies than others.

In other words, try to use multiple electrons. I don't see how that will change anything fundamentally. Nor is it clear what such a device might be like, nor why frequency selectivity will matter when the field is, in effect, static.

For another you can vary the field profile and strength, and do many experiments.

And if it's a local measurement, I don't think field profile differences will have any effect on your measurement.

And in reality you have to allow some input from theory to interpret the results, and the theory predicts a certain spectrum of momentum changes due to photon absorption, and you can go out and check the predictions of the theory.

That's not enough. If the theory will give you the same predicted experimental results for multiple field profiles (and I think it will), then you cannot determine the number of photons from the measurement.

So you agree (?) I have a theory in which I can assign a finite number of photons to the field configuration, and you agree that every single prediction of that theory - calculated using discrete interactions with finite numbers of photons - is correct, but you don't agree there's a way to measure that number or be sure it's accurate. If that's your position, I can't argue with it (nor am I very interested in trying).

You're missing a step, Sol. It is not enough to predict the experimental results, because what's in question isn't the theory. What's in question is the number of photons. And in order for the theory to calculate this, the theory needs the field profiles as an input. If I'm right that multiple field profiles will produce the same experimental outcome, then local experiments cannot determine the number of photons, because different field profiles will have different numbers of photons.

Why? I can shoot electrons through from any angle, through the whole thing, through part of it, etc. How is that local?

You can do a local measurement in multiple places, but each measurement is still local. What happens to the momentum of the electron doesn't depend upon the field in places the electron never goes. That's the sense in which it's local. But the Fourier transform, and hence the number of photons in the field, does depend upon the entire field profile. And if you want to do multiple measurements in multiple places to try to reconstruct all this stuff, well, all you're really doing is measuring the field itself, along with its profile, which you still need to take the Fourier transform of in order to get an answer for the number of photons. But the detector itself isn't counting photons. A single local measurement will not tell you the number of photons, even in a probabilistic sense.

As an example more topical to this thread - if I asked you how many photons there are in the sun, or how many it emits per second, would you agree that's a sensible question with a finite answer?

Yes.

GeeMack · Aug 4, 2009

Michael Mozina said:
I don't have any running difference "graphs" on my website, only running difference "images". In RD "images" we can observe real things like real stars, and real comets, and real planets. When are you going to figure out that it's not a graph and you can see real objects in these images? Flying stuff? Ya, flying stuff too!

First, a running difference image is a graph. That has been shown time and again in this thread and on other forums. Each pixel graphically represents the difference in the brightness values between corresponding pixels in a pair of source images. And that's all. Matching corresponding bright pixels in the source images will create the exact same output value as matching corresponding dark pixels. There are no actual objects in a running difference image. The method of constructing the output makes that impossible. That is a fact and has not been refuted. Your unsubstantiated assertion to the contrary does not constitute a refutation in the real world of sane people and legitimate science.

Second, the LMSAL running difference image you tout as evidence for your insane fantasy was created using data obtained several thousand kilometers away from where you claim your physically impossible solid surface exists. That has also been clearly shown in this thread and in discussions on other forums. So not only is it physically impossible for that surface to exist, it is impossible that your interpretation of the graphic is correct. That is also a fact that has not been refuted. And again, your simple assertion to the contrary is not a valid refutation in the world of legitimate science and sane, intelligent people.

Third, even if we disregard the fact that a running difference graph can't possibly show any real things, and the fact that it is impossible to see anything thousands of kilometers from where the source data was obtained, there are a few thousand kilometers of opaque plasma between that location and where you claim to see that physically impossible surface. That has also been clearly shown in this thread and in many others on other forums. You can't see through that opaque layer. That also makes it impossible for your interpretation of the graphic to be correct. There is another fact which has not been refuted. Your repeated assertion that things are otherwise does not amount to a refutation in the eyes of legitimate science and sane, rational people.

Now, we know the above to be facts. Yet rather than address those facts, you choose to quibble with my using the term "graph" to describe a graph?

sol invictus · Aug 4, 2009

Ziggurat said:
And if it's a local measurement, I don't think field profile differences will have any effect on your measurement.

I'm confused. If I shoot an electron through a B field, the momentum transferred will depend on a line integral of the field dotted into the electron's velocity along its path. That obviously depends on the field profile as well as its overall magnitude, and it's not very local. And by doing many such experiments, shooting from different angles etc., I can certainly reconstruct the full profile. And I can also reconstruct the number of photons that must be in there - it's like reconstructing the number of marbles in a jar by sending in some probe that bounces around for a while and comes out, and then doing it over and over again. Sure, a single such measurement isn't enough, and it might be hard to distinguish one bounce from two smaller bounces. But do it enough times, with enough different probes, and be clever enough, and you can count them all (that's more or less how vision works, actually).

That's not enough. If the theory will give you the same predicted experimental results for multiple field profiles (and I think it will), then you cannot determine the number of photons from the measurement.

Can you give an example of two field profiles that would have different numbers of photons (according to me) that will give the same results for all such experiments? Hell, even forgetting about having different numbers of photons? If the profiles are different in any way, even my simple electron scattering experiments are certainly going to detect it.

You can do a local measurement in multiple places, but each measurement is still local. What happens to the momentum of the electron doesn't depend upon the field in places the electron never goes.

You'll find that all works out fine when you do the calculation. For example if the field is very extended along some axis and very thin along another, and the electron flies through in the thin direction, the low momentum part of the FT coming from the extended direction won't affect it much, that's true. But that works just as well in momentum space as it does in position space, as I'm sure you know.

And if you want to do multiple measurements in multiple places to try to reconstruct all this stuff, well, all you're really doing is measuring the field itself, along with its profile, which you still need to take the Fourier transform of in order to get an answer for the number of photons. But the detector itself isn't counting photons. A single local measurement will not tell you the number of photons, even in a probabilistic sense.

Who said anything about a single local measurement? You'd need lots of these experiments, and they're really not very local in any sense I can think of.

Yes.

Then maybe we should discuss that instead. What is it about the sun, as opposed to a static magnetic field, that makes you think the number of photons in it is well-defined?

Ziggurat · Aug 4, 2009

sol invictus said:
I'm confused. If I shoot an electron through a B field, the momentum transferred will depend on a line integral of the field dotted into the electron's velocity along its path. That obviously depends on the field profile as well as its overall magnitude, and it's not very local.

The momentum transfer only depends upon the field along the path. But the Fourier transform depends on more than the path, it depends upon the field everywhere.

And by doing many such experiments, shooting from different angles etc., I can certainly reconstruct the full profile.

Yes. You are thereby measuring the field strength and profile, and you can use that to take a Fourier transform and deduce the number of photons that the field is made of. But at no point can you take one of your measurements and say, "Ah, this means that there are X photons in the field". It doesn't work that way. There's no way to get the number of photons except by taking the Fourier transform of the field.

Sure, a single such measurement isn't enough, and it might be hard to distinguish one bounce from two smaller bounces.

Hard? I think it's impossible.

Can you give an example of two field profiles that would have different numbers of photons (according to me) that will give the same results for all such experiments?

All I need is for the fields to match along the path of the electron. If they differ in other locations, the experiment will produce the same outcome, but the Fourier transforms will not be the same, and the number of photons in the field will not be the same.

If the profiles are different in any way, even my simple electron scattering experiments are certainly going to detect it.

Yes, if you move your scattering experiment to different locations. In which case, you're doing exactly what I said before: you're measuring the field strength and profile, and reconstructing a Fourier transform from this. Your individual measurements are never counting photons.

Then maybe we should discuss that instead. What is it about the sun, as opposed to a static magnetic field, that makes you think the number of photons in it is well-defined?

Well, for starters the sun has a reasonably well-defined size. A 1 Tesla field does not have any defined size - that information was not given.

Furthermore, while it is possible to treat static macroscopic fields as a collection of photons, it's a bloody stupid way to treat the problem because it's needlessly complicated. But treating solar radiation that way doesn't have the same problem. The radiative aspects of it are the characteristics of primary interest to begin with.

sol invictus · Aug 4, 2009

Ziggurat said:
Yes, if you move your scattering experiment to different locations. In which case, you're doing exactly what I said before: you're measuring the field strength and profile, and reconstructing a Fourier transform from this. Your individual measurements are never counting photons.

Well, as far as I can tell your position is identical to the one I said I wasn't interested in debating: sure. you have a theory that water is made of molecules, and your theory correctly predicts every aspect of water we've ever measured, but you can't really count the number of molecules in a drop even approximately.

Well, for starters the sun has a reasonably well-defined size. A 1 Tesla field does not have any defined size - that information was not given.

That's why I added that information in my very first post on this topic.

Furthermore, while it is possible to treat static macroscopic fields as a collection of photons, it's a bloody stupid way to treat the problem because it's needlessly complicated.

For most purposes, yes - but not for all. For example if you shine a laser beam (or a single photon for that matter) through a magnetic field, each photon in that beam will have a chance of scattering off the field due to the Euler-Heisenberg interaction. The easiest way to treat that is to draw Feynman diagrams in which the effect of the field arises from field photons interacting with the laser photons.

Ziggurat · Aug 4, 2009

sol invictus said:
For most purposes, yes - but not for all. For example if you shine a laser beam through a magnetic field, each photon in that beam will have a chance of scattering due to the Euler-Heisenberg interaction. The easiest way to treat that is to draw Feynman diagrams in which the effect of the field arises from field photons interacting with the laser photons.

That may be the easiest way to conceptualize the problem, but is that really the easiest way to calculate it? We're venturing into territory I'm not familiar with, but my intuition suggests that the final answer for the probability of a given laser photon scattering should just depend on some sort of path integral of the magnetic field along the beam's path. It would seem quite strange to me that you'd be better off taking the Fourier transform of the whole field in order to actually figure out how many photons you've got in the field.

sol invictus · Aug 4, 2009

Ziggurat said:
That may be the easiest way to conceptualize the problem, but is that really the easiest way to calculate it?

It's pretty easy, at least if the field has a simple profile. If not it's hard, but so is any other method.

We're venturing into territory I'm not familiar with, but my intuition suggests that the final answer for the probability of a given laser photon scattering should just depend on some sort of path integral of the magnetic field along the beam's path.

Your intuition is correct to the extent that you could write an n-loop effective action, which will include the most important piece of the interaction that matters here, and then treat the B field classically as a source in that effective action. So long as the field is sufficiently classical that's a good approximation, but it remains an approximation.

It would seem quite strange to me that you'd be better off taking the Fourier transform of the whole field in order to actually figure out how many photons you've got in the field.

You don't need to figure out that number at any step - but if you work in momentum space (which is usually easier, because propagators are simple in momentum space) you do need to take the FT of the field.

Electric universe theories here.

Penultimate Amazing

Penultimate Amazing

Penultimate Amazing

Penultimate Amazing

Muse

Illuminator

Penultimate Amazing

Penultimate Amazing

Banned

Banned

Penultimate Amazing

Banned

Banned

Banned

Penultimate Amazing

Penultimate Amazing

Penultimate Amazing

Banned

Penultimate Amazing

Banned

Banned

Banned

Philosopher

Banned

Penultimate Amazing

Penultimate Amazing

Philosopher

Penultimate Amazing

Philosopher

Penultimate Amazing

Banned

Penultimate Amazing

Philosopher

Penultimate Amazing

Banned

Philosopher

Penultimate Amazing

Philosopher

Penultimate Amazing

Philosopher