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Dave W.
Info Junkie
USA
26020 Posts |
 Posted - 03/30/2006 : 21:44:56 [Permalink]
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quote:
Originally posted by Michael Mozina
I have made up my mind. For purposes of this discussion, and based on the resolution and settings we're looking at in that composite image, I'd say the most if not all the light we can actually see in that particular image is directly related to an emission from an electrical arc.
Is it "most" or is it "all?" If it's not "all," then you'll still need to tell me how you distinguish what's a direct emission from what isn't.quote:
In most images, particularly large resolution images, we'd have to consider the other issues I mentioned. In this particular case, it really doesn't affect much since most of the light is a direct result of the light released in those arcs.
How many arcs? Before, you implied there was just one. Now you've switched to the plural.quote:
quote:
This contradicts your earlier contention that Lockheed is oversimplifying the problem.
Is this "take my comments out of context day" or what? They are oversimplying the process by not using *all* the images and adding them up.
But that method - adding all the images together - is vastly more simple than taking a ratio and somehow (more math, probably) determining temperature from it.quote:
They are also overcomplicating an otherwise simple matter of addition and trying to apply a complicated set of mathematics (which may not even apply) to what should be simple matter off adding up photons.
Adding up photons isn't meaningful. It's not even meaningful in a single TRACE image, since (for example) we can see from figure one that the emissions from the 1Mk Fe IX are more energetic (smaller wavelength) than emissions from 20 Mk Fe XX within TRACE's bandpass (thus further discrediting the notion that the coronal ions emit as black bodies, also).quote:
Worse yet, they oversimplified their overly complicated math presentation by not even bothering to consider the different absortion rates of the different filters.
And you haven't considered the exact same thing, Michael - at least not in any way that you can explain to me how to take the same thing into consideration in any image.quote:
What their math *really* seems to show us is that different wavelengths are absorbed at different rates.
Really? How can you tell that?quote:
Instead, they misrepresented the results of a overly complex math formula by oversimplifying the other aspects of this issue that must also be considered.
Except that, just like "the acceleration of the universe," you are completely incapable of actually taking those aspects into account yourself. You expect others to do so, however, and you know (somehow) that not doing so is somehow incorrect.quote:
All they had to do to isolate the heat and light signatures was to add *all* the images together.
Since brighter doesn't equal hotter, that methodology is invalid.quote:
By trying to compare ratios with complicated mathematical formulas, without regard to absortion rates of wavelengths...
Your methodology appears to disregard the "absortion rates of wavelengths" as well, Michael.quote:
...they turned what should have been a simple task into a complex excersize in futility that turns physics on its head and gives us invisible heat unicorns hiding in the shadows. What a bunch of boloney. This method doesn't work as advertized and even the author you cited hedged his bets and picked the loops to hold the highest temperature plasma anyway. Even he doesn't believe it works as advertized!
You don't even know what is "advertised," but you know it's baloney.
And I know that adding images together in the hopes of finding a "heat signature" is utterly bogus, as well. |
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Dave W.
Info Junkie
USA
26020 Posts |
Posted - 03/30/2006 : 22:05:31 [Permalink]
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quote:
Originally posted by Michael Mozina
This comment literally blew my mind Dave. If you have electrons being driven to extreme velocities by *anything* and it flows *anywhere* including along magnetic field lines that is the very *definition* of current flow. When electrons flow and move through material that *is* current flow. Holy Cow is that a major rationalization on your part. You have fast moving electrons following magnetic field lines but you refuse to see that as current flow. Hoy!
Yes, if one electron moves somewhere, it is "current flow." If two electrons move in opposite directions through a conductor, there is no net current flow. How tough is that to understand? If there were a gigantic net flow of electrons from one point to another along a loop, then one footprint of the loop should be dominated by free-free, free-bound and other electron/ion interaction radiation (as the electrons crash into the positively-charged "surface"), and the other footprint (the source of the electrons) should be relatively free of such a signature. Where is there evidence for that?
This brings up another interesting point. If there's a net flow of current Michael, from one footprint to another, why is it that the magnetograms don't show a black half and a white half to each individual "footprint" of an "arc?" Remember the right-hand rule for electric currents and the magnetic fields they generate? Of course you do. What the magnetograms show is that the magnetic fields of the loops are aligned along their lengths, whereas a direct-current "arc" wouldn't be aligned in such a manner.
Besides which, you've never addressed the issue of "electric current" not being a synonym for "electrical discharge." I've never actually disputed that there are electric currents within the corona (with such huge magnetic fields moving through a plasma, currents are a given). What I've disputed is the idea that such currents are necessarily the result of gigantic "sparks" from one point on the Sun to another.
Why the hell would a spark not be uniformly heating its surroundings? In other words, why would a spark be responsible for the strange temperature profile of a loop (like the one you pointed out)? |
- Dave W. (Private Msg, EMail)
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Michael Mozina
SFN Regular
1647 Posts |
Posted - 03/31/2006 : 12:07:54 [Permalink]
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quote:
Originally posted by Dave W.
I'll assume you're frustrated because I'm not willing to bow down to your unscientific methodologies and agree that they are meaningful.
This only shows that we all hear what we wish to hear, and all see what we wish to see at times. No Dave, I'm frustrated because as it relates to this specific issue, you have been anything *but* scientific IMO. On most subjects, I've at least been able to relate to your position, even if I didn't agree with it. On this particular subject however, you seem to have *no* rational position at all. It seems more like a denial campaign of epic proportions from the outside looking in.
quote:
No, that's clearly not Lockheed's "logic," otherwise they wouldn't be using a ratio to determine temperature.
*My* method is simply an extension of what they take for granted. They take for granted (claim in fact) that each of these filters represents a specific temperature range. I'm simply adding up the photons from both images and noting that the combined imaged varies very little from either of the other two images. All three images pinpoint the photon release to the exact same location. There are small variations here and there, but typically the variations are simply light variations in the *same* coronal loops. This is simply a method to determine where the light (from both filters) originates. There is nearly complete agreement about the loops being brighter in all three images. These are simply line of sight images of what's going on in the atmosphere from a wider spectrum than one filter can see. We're simply adding up the line of sight photons from as large a spectrum as we can get out of the images we have to work with.
Lockheed's method has nothing to do with isolating photons in a line of sight manner. It's a completely different kind of methodology from line of sight concepts. Instead it is based on *comparing* two wavelenths in a new way. Because their method compares wavelenghts, there are other issues that must be considered and can *greatly* affect their outcome. Before we can begin to judge the usefulness of their method, we now must know something about absortion rates of both filters. We have to know for sure how photons are absorbed in that amosphere, otherwise all we will have shown with this method is that absortion is not uniform up and down the energy spectrum. Some wavelengths may be absorbed more readily than others. Some may scatter more than others. While this *could* affect line of sight methods as well, we can see from the results of our math that this wasn't really much of a factor. The loops overlay on one another, and line up with one another quite well. Whatever scattering took place, it seems to have been minimal, and whatever absorbtion took place, it really has little affect on combining line of sight photons. If anything, the loops may have been *brighter* than we actually see them, but either way, the light clearly comes from the loops in images A,B & C. The are all in agreement on the point, so the absortion issue isn't going to affect our line of sight pinpointing of the light sources.
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Unless you can show me where they've added images together (a meaningless exercise)
These are the kinds of comments that irk me, because you *know* that they are not true. By combining the wavelengths, we get a *greater range* of view, from a *wider spectrum* of photons. We see more of the photons, and we can see if they all originate from the same locations or from different locations by overlaying them on top of one another. There is certainly value and meaning in combining indivually small parts of the the spectrum into a combined view to look at a larger spectrum of energy. These kinds of comment are a clear and obvious indication that you are absolutely *not* approaching this issue scientifically.
quote:
and claimed that the material must be at least as hot as the highest temperature peak in a wavelength passband.
Here's another comment that isn't logical or scientific. Each of these pixels represents an *extremely* large area. There could be plasma from a *huge* assortment of temperature ranges in each and every pixel, since every pixel represents hundreds of kilometers.
If we look a the "least energetic" part of the spectrum, we know that plasma is *at least* that hot in *some* area of that pixel. Some plasma in that pixel area may be cooler, and some may be hotter. That all we could hope to deduce from a small, low energy range of the spectrum. Lockheed constantly associates these different filters with temperature. The give a range for each filter, and in paper after paper, the talk about temperatures in the millions of degrees. By *everyones* account, the brightest dots "tend" to represent plasma a temperature "X". Lockheed has see the temperature ranges. I'm simply using their ranges Dave. I don't even necessarily agree with them personally, but I have to start somewhere. For purposes of this discussion, it is irrelevant what temperatures we assign to the bright spots of the images, but the bright spots of the images are presumed to represent the "peaks" of the filters, which tend to be either 1-1.5 million degrees for FeIX, 4 million degrees if it's a Ca ion photon, or 10-20 Million Kelvin if these are Fe XX ions. All we know with absolute certainty, is that these areas are *much* hotter than the photosphere or chromosphere.
quote:
Okay, fine. All the temperature measurements are thus invalid for the purposes of this discussion, because neither one of us has "further corroboration" of any of the temperature readings done by anyone to support our respective points.
That is false. I gave you a external corroboration of the temperatures ranges with that TRACE/Yohkoh composite. Yohkoh represents further corroboration that high energy photons are indeed associated with the coronal loops, not with the *entire* corona. I handed you corroborating evidence, and you have not dealt with it *scientifically*.
quote:
The intrinsic brightness of a pixel doesn't directly relate to the temperature of the material found there.
False. The photosphere is less than 6000 Kelvin. The chromosophere is measured in the tens of thousands of degrees Kelvin. Based on the laws of physics, the brightly lit parts of this image are not in these temperature ranges. The most brightly lit regions of these images must be a minimum of 160,000 Kelvin, and could potentially be 20 Million Kelvin, but based on the laws of physics, the light is related to temperurates, and the |
| Edited by - Michael Mozina on 03/31/2006 13:51:39 |
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Michael Mozina
SFN Regular
1647 Posts |
Posted - 03/31/2006 : 12:32:57 [Permalink]
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quote:
Originally posted by Dave W.
quote:
Originally posted by Michael Mozina
Where in any of the three images (four if you include the Yohkoh/Trace overlay) is there *any* observational evidence to support Lockheed's methods actually works as advertized? How will you provide *any* evidence at all to demonstrate that this method Lockheed is using is actually capable of showing heat signatures instead of just demonstrating differing photon absortion rates of two different wavelengths?
Nice try at shifting the burden of proof, but I'm only interested in your claim that "brighter equals hotter," and showing it to be the crap that it is.
You are attempting to prove I'm wrong by introducing a new method into the discussion Dave. Lockhed and I both agree that for light to shine brightly in these images, it has to be greater than 160,000 degrees. If you now wish to introduce evidence that dark areas can be greater than these temperatures, you then must demonstrate that some other method can be used to determine temperature and that this method works accurately. You have not done so. You handwaved in a statement that Nitta made that was true. The light *could* be related to a density increase. I agree with that point that he made. I also agreed with him that the high temperature plasma was likely to be found in the coronal loop, and that composite image I showed you demonstrates that we were both right about this point as well. You also *alledge* that we can compare wavelengths in a new way, and thereby extract temperature information about the dark regions. Before I agree that this method works, I A) need the see the math, B) need to know how the differentiated between photons from FeIX vs. Calcium ions, vs. FeXX or some other ion in these images, and C) accounted for potentially different absortion rates for different wavelengths. If you can demonstrate the method works as adverstized and can answer some of the logical questions one would ask about such new methods, I'll be happy to discuss it with you. If however you *don't* know what math they used, and you *don't* know if they accounted for absorbtion issues, and you *don't* know how they differentiated varies photons from one another, I have no reason to take you seriously or to put any faith in the method to begin with. Based upon the authors stance on the location of high temperature plasma, it isn't clear that the author himself puts a lot of faith in the method he described. He even stated rather bluntly that there was not a consensus that the method does work, even inside of Lockheed.
quote:
You'll note that I am not claiming that Lockheed's method is valid. Neither you nor I actually know what Lockheed's method is, so neither of us can verify or disprove that it "works as advertised."
But Dave, you are claiming that some dark areas of this image are hotter than the brightest regions of this image. You've given me no logical reason to believe you or that your invisible heat exists. The burden of proof is on you, since I cannot disprove an invisible heat source. |
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Michael Mozina
SFN Regular
1647 Posts |
Posted - 03/31/2006 : 12:45:38 [Permalink]
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quote:
Originally posted by Dave W.
Is it "most" or is it "all?" If it's not "all," then you'll still need to tell me how you distinguish what's a direct emission from what isn't.
I'll tell you what Dave, since neither of us know for sure exactly what the percentage might be, let's just go with a 20% figure for absortion/reflection/emission from the atmosphere. No pixel with less than 50 photons counts as a "hit". Will that work for you?
quote:
How many arcs? Before, you implied there was just one. Now you've switched to the plural.
Oh man, this is painful. You're looking for *any* excuse at this point to avoid the actual issues. It's a given in my model that "little" arcs occur all along the surface all the time, adn that large *groups* of coronal loop ocassionally shot high above the surface. Every image I've ever posted here shows *numberous* loops traversing the atmosphere Dave. Get real.
I did the math for you Dave. I added up the photons for you and we can all see for ourselves that they line up pretty well. There are some diffrences of course, but the shots were taken several seconds apart, and no one expected them to be exactly alike in the first place.
If you think there is some hidden heat, or that the dark regions are hotter than the brightest regions, then you need to demonstrate this in some mathematically precise way that includes explanations of all the variables. By all the variables, I'm mostly refering to how you will explain which photons are represent which *exact* temperatures of *exact* ions, and how you account for absorbtion rates. If you can't do that, I'm not buying your invisible heat source story. |
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Dave W.
Info Junkie
USA
26020 Posts |
Posted - 03/31/2006 : 14:35:39 [Permalink]
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quote:
Originally posted by Michael Mozina
You are attempting to prove I'm wrong by introducing a new method into the discussion Dave.
No, I'm asking you to show that your assertion that brighter always equals hotter is correct. Given the huge temperature range of TRACE's filters, and our (yours and mine) inability to do even a single brightness-to-temperature calculation, I don't see how you could possibly prove that claim.quote:
Lockhed and I both agree that for light to shine brightly in these images, it has to be greater than 160,000 degrees.
Once again, you are putting words in Lockheed's collective mouth. Lockheed's data clearly shows that for light to shine at all in these images, it must be at least 160,000 kelvin. If it shines "brightly," it just means more photons collected, not necessarily a higher temperature, than dim areas. A pixel with a value of 1 can mean an average 20 MK, while a pixel with a value of 255 can mean an average temperature of 1 MK. You've got no way to determine which is which, Michael, since even the dimmest non-zero pixel in a Yohkoh image will relate to some higher temperature than most TRACE pixels.quote:
If you now wish to introduce evidence that dark areas can be greater than these temperatures, you then must demonstrate that some other method can be used to determine temperature and that this method works accurately.
No, I want you to demonstrate that the darker areas cannot be a higher temperature than the bright areas, because that's what you claimed. You need to show that Schryver is wrong, and do so without claiming that the corona can be modeled as a black body, which is a false premise.quote:
You have not done so. You handwaved in a statement that Nitta made that was true.
That wasn't in defense of any claim of mine, it was for no other reason than to show that your "brighter equals hotter" claim is false.quote:
The light *could* be related to a density increase. I agree with that point that he made.
Then why is it that you find it totally impossible to say, "I was wrong about brighter pixels always meaning hotter temperatures"?quote:
I also agreed with him that the high temperature plasma was likely to be found in the coronal loop, and that composite image I showed you demonstrates that we were both right about this point as well.
I don't care.quote:
You also *alledge* that we can compare wavelengths in a new way, and thereby extract temperature information about the dark regions.
I've never alleged any such thing: Nitta alleges it, as do other folks at Lockheed - you're the one claiming they're all wrong.quote:
Before I agree that this method works, I A) need the see the math...
Ask Nitta.quote:
...B) need to know how the differentiated between photons from FeIX vs. Calcium ions, vs. FeXX or some other ion in these images...
Since there's no way to do so, I don't know why you'd demand such a thing.quote:
...and C) accounted for potentially different absortion rates for different wavelengths.
Something which you refuse to do with your own model.quote:
If you can demonstrate the method works as adverstized and can answer some of the logical questions one would ask about such new methods, I'll be happy to discuss it with you. If however you *don't* know what math they used, and you *don't* know if they accounted for absorbtion issues, and you *don't* know how they differentiated varies photons from one another, I have no reason to take you seriously or to put any faith in the method to begin with.
And this is just a red herring, since I never asked you to do any such thing. You're just refusing to understand my point.quote:
Based upon the authors stance on the location of high temperature plasma, it isn't clear that the author himself puts a lot of faith in the method he described. He even stated rather bluntly that there was not a consensus that the method does work, even inside of Lockheed.
All of which is irrelevant.quote:
But Dave, you are claiming that some dark areas of this image are hotter than the brightest regions of this image.
No, I'm not - I'm asking you to prove your claim that the brighter areas must be hotter.quote:
You've given me no logical reason to believe you or that your invisible heat exists. The burden of proof is on you, since I cannot disprove an invisible heat source.
No, you're supposed to prove your own claims, Michael, and stop trying to shift the burden of proof over to me in a pathetic attempt to distract people from your hastily-made generalizations and unscientific pronouncements. |
- Dave W. (Private Msg, EMail)
Evidently, I rock!
Why not question something for a change?
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Dave W.
Info Junkie
USA
26020 Posts |
Posted - 03/31/2006 : 14:49:43 [Permalink]
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quote:
Originally posted by Michael Mozina
I'll tell you what Dave, since neither of us know for sure exactly what the percentage might be, let's just go with a 20% figure for absortion/reflection/emission from the atmosphere. No pixel with less than 50 photons counts as a "hit". Will that work for you?
You really don't get it, do you? In order to properly analyze the image, I need to know which pixels are reflections/re-emissions, and which are direct emissions.quote:
quote:
How many arcs? Before, you implied there was just one. Now you've switched to the plural.
Oh man, this is painful. You're looking for *any* excuse at this point to avoid the actual issues.
No, I'm waiting for you to make it possible for me to agree with you at all. I started out by asking you to take one image and draw the arcs you see on it, but you either ignored the request or simply refuse to do so. Until you can defined what a single arc looks like, I won't be able to follow along, Michael.quote:
It's a given in my model that "little" arcs occur all along the surface all the time, adn that large *groups* of coronal loop ocassionally shot high above the surface. Every image I've ever posted here shows *numberous* loops traversing the atmosphere Dave. Get real.
I'm asking you to show me the arcs and the reflections. In the Yohkoh/TRACE overlay, for example, I see brightish stuff shooting off to the right in blue, but not in yellow, and I've got no clue as to whether you think that's part of an "arc" or not. Until you help by defining an "arc," I won't be able to agree that your analysis is correct.quote:
I did the math for you Dave. I added up the photons for you and we can all see for ourselves that they line up pretty well.
Which means precisely nothing, as far as I can tell. The non-zero pixels in a 195A image are generally slightly hotter than the non-zero pixels in a 171A image, so adding them together shows nothing more than that the pixels in the dark areas are hot, too.quote:
If you think there is some hidden heat...
There is no hidden heat, Michael. A pixel value of 5 means that material existed in that pixel between 160,000 kelvin and 20 MK, just like a pixel value of 250. To be "hidden" heat, there'd have to be some zero-value pixels, but there are none.quote:
...or that the dark regions are hotter than the brightest regions, then you need to demonstrate this in some mathematically precise way that includes explanations of all the variables.
No, that's you once again shifting the burden of proof.quote:
By all the variables, I'm mostly refering to how you will explain which photons are represent which *exact* temperatures of *exact* ions...
Okay, I need you to do this for your model.quote:
...and how you account for absorbtion rates.
I need to know how your model accounts for the alleged absorbtion rates, Michael.quote:
If you can't do that, I'm not buying your invisible heat source story.
If you can't do those things, then I'm not buying your "brighter equals hotter" fairytale.
I'll answer your first post today later on. At least to explain how quantum theory impacts the images. |
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Michael Mozina
SFN Regular
1647 Posts |
Posted - 03/31/2006 : 16:22:47 [Permalink]
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quote:
Originally posted by Dave W.
You really don't get it, do you? In order to properly analyze the image, I need to know which pixels are reflections/re-emissions, and which are direct emissions.
No, it's you that don't get it actually. I don't really need to know what temperature each photon represents in my heat signature model, nor do I have to know all the relfection/re-emission atmospheric details in my model. All I need to do is to "correct" for some of the atmosopheric reflection in my model. I would do that simply by subtracting a specific amount from all the pixels of both images and then I would add the two images back together again. I'd still come up with pretty much the same results, only my "filter" might get rid of some of the atmospheric relections. I'm not even attempting to "measure" the temperature of the dark regions at all Dave since I don't believe we can measure these regions with these filters in the first place! You'd need to know what wavelengths of light these atoms actually do "peak" in to know the temperature of these areas.
You however are claiming that you can actually measure the temperature of the dark areas with these high energy wavelengths. Worse yet, you seem to be claiming that some of the dark regions are actually hotter than the brightest regions in these images, even though your own expert disagrees with you on this point and put the high temperature plasma *inside* the loops, just as I did.
I therefore want to see your math Dave. I want to see your work. I want some evidence that this method even works before I agree that it works. I see no evidence to suggest this method does work, or that your own expert put a lot of "faith" in the idea in the final analysis. In fact the only thing it could possibly reveal without a lot more information is the different absorbtion/reflection rates of the different wavelengths. The handwave of a method that you've supplied so far can't possibly reveal the actual temperature of the atmosphere outside of the arcs. |
| Edited by - Michael Mozina on 03/31/2006 16:30:37 |
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Michael Mozina
SFN Regular
1647 Posts |
Posted - 03/31/2006 : 17:29:37 [Permalink]
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quote:
Originally posted by Dave W.
No, I want you to demonstrate that the darker areas cannot be a higher temperature than the bright areas, because that's what you claimed.
I "claimed" this based on the visual evidence Dave.
There is no observational evidence to support an invisible heat source hiding in the shadows Dave. There's nothing there to be seen by either of these high energy satellite images. I can't prove that invisible things don't exist. This is your claim and now it's up to you to demonstrate that greater heat exists in the dark regions than in the bright areas of this image. I have no evidence to confirm this at all.
quote:
You need to show that Schryver is wrong, and do so without claiming that the corona can be modeled as a black body, which is a false premise.
Since you seem to be excluding the corona from the laws of math and physics in this statement, how would you suggest I go about such a thing?
quote:
That wasn't in defense of any claim of mine, it was for no other reason than to show that your "brighter equals hotter" claim is false.
But *his* statements don't confirm this Dave. *He* personally put the high temperature plasma in the same location I did. This is now your claim, not his. He only explained HOW the method *might* work. I can't disprove his final analysis, since his analysis came to the same conclusion as my analysis, and that expectation of finding the hot plasma in the coronal loops is verified by the composite image above.
I cannot however prove to you that invisible things do not exist. That is scientifically impossible. You then must show me some observation that confirm your claim.
quote:
Then why is it that you find it totally impossible to say, "I was wrong about brighter pixels always meaning hotter temperatures"?
Because you haven't proven anything of the sort Dave, certainly not as it relates to the two images in question. In these real world images, brighter is hotter and even Lockheed *assumes* this in their method which you can't even specify mathmatically in the first place! Like I've said many time, the math isn't the problem. I put my math on the table. You didn't. If math is the only criteria, your position is toast.
quote:
I don't care.
Well Dave, that pretty much says it all IMO. Your own expert agreed with me that we would likely find the plasma of the greatest temperature to be concentrated in the loops themselves. That analysis was confirmed by the Trace/Yohkoh composite image. You don't care what he thinks. You don't care what I think either. You don't even care what the images show us. You only seem to care about what you think, even though you can't even explain the math or know what math is being used in the first place! Talk about irrational and indefensible positions.....
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| Edited by - Michael Mozina on 03/31/2006 17:39:24 |
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Dave W.
Info Junkie
USA
26020 Posts |
Posted - 03/31/2006 : 19:14:01 [Permalink]
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quote:
Originally posted by Michael Mozina
You however are claiming that you can actually measure the temperature of the dark areas with these high energy wavelengths.
Well, this is just a repeat of what you've said before, and proves beyond any reasonable doubt that what I have to say on a subject is simply of no consequence to you, Michael. |
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Michael Mozina
SFN Regular
1647 Posts |
Posted - 04/01/2006 : 13:59:38 [Permalink]
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quote:
Originally posted by Dave W.
Well, this is just a repeat of what you've said before, and proves beyond any reasonable doubt that what I have to say on a subject is simply of no consequence to you, Michael.
Well Dave, we do seem to have reached an important impass. I'm not sure how it is possible to have a scientific discussion of this subject if you try to claim the corona is somehow immune from the laws of physics and math.
While I agree with the author's basic point that the light is a function of both temperature *and* density, I believe based on these observations that the coronal loops are considerably more dense and are operating at a considerably higher temperature than the lightly populated corona materials. While a less dense material like the corona will not emit as many photons as a more dense material like the photosphere, it isn't somehow immune from the laws of physics, just by virtue of being less dense than other materials.
My explanation of the coronal loops being the heat source of the corona works from a physics standpoint. It explains how the corona can be so much warmer than the photosphere quite elequently and it obeys all the laws of physics.
More importantly however, I've layed out the math I used before you and I've explained my position carefully and openly. I showed you the results in black and white, and in color as well. There is really little more that I can do to make my points clear on this subject.
Since you have not explained your position mathematically and you have not offered me any heat source alternatives to explain the heating of the corona, there is really no scientific way for me to judge the merit of your position. Moreover, I have no observational evidence to suggest there is anything wrong with my analysis, despite your footdragging and handwaving.
Since you didn't offer me any logical alternatives to explain these phenomenon, and the heating of the corona seems to be quite the enigma to gas model theoriests, I have few alternatives to work with Dave, and you haven't personally even offered me any. While you've "alledged" that there "could" be a problem with my analysis, you have yet to provide any observational support of any sort to demonstrate that your objection has merit. The one article you provided only suggested that density may play a role in the overall light emissions, but I already agree with that premise and I already agree that the coronal loops are more dense than the material outside of the loops. Since the author placed the highest temperature plasma to be *inside* the loops, and the Yohkoh/Trace system verifies this to be true, I can only conclude that you do not wish to embrace the choices that your own reference embraced. You seem to reject his position for 'personal' reasons that you cannot even adequately explain. I certainly cannot explain you reasoning either.
While I've enjoyed our discussions over the months, and I know you to be an intelligent and logical person in most circumstances, in this instance, I do not believe your position is logical, rational or defenseable. This is particularly true since you literally offered me no alternatives, and don't even know what math we might apply to these filters to demonstrate your position. You told me that math was important to you. I did the math for you in this case to make my point clear mathematically for you. You don't know any of the math forumulas that might support your case. If math was the key factor in your choice, then you would have no alternative but to concede this point, and yet you will not concede this point. I therefore now know that math isn't the key or the answer to getting through to you. Frankly I'm stumped on how I might proceed, particularly if you keep claiming that the corona is somehow immune from laws of math and physics.
I really have no idea how to proceed at this point. Do you have any ideas on how we might resolve the heat/light distribution question? |
| Edited by - Michael Mozina on 04/01/2006 14:11:00 |
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Dave W.
Info Junkie
USA
26020 Posts |
Posted - 04/01/2006 : 16:07:09 [Permalink]
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Apparently, you're incapable of comprehension. Since you continue to fabricate my end of the discussion (instead of listening to what I actually say), anything I might post would be utterly irrelevant. You are, of course, free to continue the discussion on your own (and continue to invent positions for other people to hold, as you have been), but there's no reason for me to post to these threads anymore. It doesn't matter what I say to your questions, you twist the answers into strawmen (such as that I've claimed "that the corona is somehow immune from laws of math and physics"). You don't need me in order to post garbage like that, and you display no interest whatsoever in the actual math behind temperature-dependent line ratios (there's quite a bit of literature on them, Lockheed didn't invent the process), the actual science of helioseismology, or discussing electrical arcs in general, so have fun by yourself.
Of course, everything I've written here will be mangled beyond recognition or dismissed, so this post represents nothing more than another ten minutes of my life which I won't get back. |
- Dave W. (Private Msg, EMail)
Evidently, I rock!
Why not question something for a change?
Visit Dave's Psoriasis Info, too. |
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Michael Mozina
SFN Regular
1647 Posts |
Posted - 04/01/2006 : 17:16:11 [Permalink]
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quote:
Originally posted by Dave W.
No, I want you to demonstrate that the darker areas cannot be a higher temperature than the bright areas, because that's what you claimed. You need to show that Schryver is wrong, and do so without claiming that the corona can be modeled as a black body, which is a false premise.
What exactly does this statement mean then Dave?
The corona is not going to act any differently from any other material as it relates to heat modeling and photon emission expectations. It will release a certain amount of energy in the form of light from each atom given a specific temperature. There may be fewer atoms in the corona and therefore fewer emissions in thin plasma, but each atom will still emit light. We should still be able to look at the corona and see which areas are likely to be more dense and more energetic based on which kinds of photons these different areas emit.
I don't see *all* the atoms in the atmosphere emitting yellow photons in great quantity in that composite image, nor do I see *any* evidence in the Trace/Yohkoh overlay to make me believe that the coronal loops are not more dense and more energetic than the rest of the plasma in the corona. If you won't accept that different wavelengths of light serve as a measure of heat, what will you accept as a measure of heat that *isn't* based on the method you are now trying to demonstrate? You can't "claim" the method works *without* demonstrating that it actually works! You can't apply math based on temperature, unless you agree that light can be equated with temperature! I feel like I'm stuck in a catch 22 and you're offering me no scientific way to demonstrate that your method works as advertized. How do *you* even know it works as advertized if you don't know what math forumulas they applied? I don't get it! |
| Edited by - Michael Mozina on 04/01/2006 17:20:04 |
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Dave W.
Info Junkie
USA
26020 Posts |
Posted - 04/01/2006 : 17:22:16 [Permalink]
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quote:
Originally posted by me
Of course, everything I've written here will be mangled beyond recognition or dismissed...
Another correct prediction on my part. Damn I'm good. |
- Dave W. (Private Msg, EMail)
Evidently, I rock!
Why not question something for a change?
Visit Dave's Psoriasis Info, too. |
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Michael Mozina
SFN Regular
1647 Posts |
Posted - 04/02/2006 : 15:59:14 [Permalink]
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http://www.skynightly.com/reports/Why_Is_The_Suns_Corona_Hotter_Than_The_Sun_Itself.html
Well Dave, while the heat source of the corona may seem like a mystery to you and other gas model theorists, you're literally looking at the cause of the coronal heating in this image. The coronal loops are the heat source of the corona, even based on *both* of the parameters that your own expert cited. The loops are in fact more dense *and* they are considerably hotter than the rest of the corona and they therefore release many more photons as all these images clearly show us. All the high energy satellite images put the high energy discharges *inside* the coronal loops (like your expert) or at the base of the coronal loops. The loops are the heat source of the corona, and it's all caused by *current flow*.
You have yet to offer any reasonable or scientific alternative to any of this. Instead you used a single source (Nitta) who ultimately put the high energy discharges inside the loop as I did, and pointed out theat light increases could be related to increase in density, which I also agree with. I don't see any evidence in the information you provided that suggests that the loops are not more dense and hotter than the rest of the corona, and even if the corona was relatively thinly populated, that does not prevent it from being examined using standard black body principles related to photon emissions. |
| Edited by - Michael Mozina on 04/02/2006 16:01:51 |
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